APTITUDE
- TIME AND WORK
- SQUARE AND CUBE ROOTS
- DECIMAL FRACTIONS
- Boats and Streams
- CHAIN RULE
- Partnership
- RATIO AND PROPORTION
- Profit and Loss
- PERCENTAGES
- SURDS AND INDICES
- Problems on Ages
- Simplifications
- H.C.F AND L.C.M
- Numbers
SQUARE AND CUBE ROOTS
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
- 4
————–
9 2
————–
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
- 2
————-
9 6
————-
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
————
76
———–
12*12=144
76
+ 144
——————–
7744
——————–
Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
—————-
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
—————–
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
———————–
1 3 6 9
————————-
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
———————
1 1 2
——————–
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
Answer is 8.1
ii)86.491/2 + √(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.
3.√ 0.00004761 equals:
Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
Answer is 0.0069
4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.
Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.
8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35
10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.
18.3(4*12/125)1/2=?
Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
- 4
————–
9 2
————–
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
- 2
————-
9 6
————-
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
————
76
———–
12*12=144
76
+ 144
——————–
7744
——————–
Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
—————-
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
—————–
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
———————–
1 3 6 9
————————-
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
———————
1 1 2
——————–
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
Answer is 8.1
ii)86.491/2 + √(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.
3.√ 0.00004761 equals:
Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
Answer is 0.0069
4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.
Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.
8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35
10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.
18.3(4*12/125)1/2=?
Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
- 4
————–
9 2
————–
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
- 2
————-
9 6
————-
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
————
76
———–
12*12=144
76
+ 144
——————–
7744
——————–
Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
—————-
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
—————–
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
———————–
1 3 6 9
————————-
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
———————
1 1 2
——————–
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
Answer is 8.1
ii)86.491/2 + √(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.
3.√ 0.00004761 equals:
Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
Answer is 0.0069
4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.
Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.
8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35
10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.
18.3(4*12/125)1/2=?
Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
- 4
————–
9 2
————–
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
- 2
————-
9 6
————-
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
————
76
———–
12*12=144
76
+ 144
——————–
7744
——————–
Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
—————-
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
—————–
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
———————–
1 3 6 9
————————-
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
———————
1 1 2
——————–
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,√y=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
Answer is 8.1
ii)86.491/2 + √(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ √(5+x1/2 )=12.3
=> √(5+x1/2 ) =12.3-9.3
=> √(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.
3.√ 0.00004761 equals:
Solution: √ (4761/108)
√4761/√ 108
. 69/10000
0.0069.
Answer is 0.0069
4.If √18225=135,then the value of
√182.25 + √1.8225 + √ 0.018225 + √0.00018225.
Solution: √(18225/100) +√(18225/10000) +
√(18225/1000000) +√(18225/100000000)
=√(18225)/10 + (18225)1/2/100 +
√(18225)/1000 + √(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=√(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.
8.If √(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: √(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35
10.If 3a=4b=6c and a+b+c=27*√29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.
18.3(4*12/125)1/2=?
Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.
TIME AND WORK
TIME AND WORK
1.If A can do a piece of work in n days, then A’s 1 day work=1/n
2.If A’s 1 day’s work=1/n, then A can finish the work in n days.
Ex: If A can do a piece of work in 4 days,then A’s 1 day’s work=1/4.
If A’s 1 day’s work=1/5, then A can finish the work in 5 days
3.If A is thrice as good workman as B,then: Ratio of work done by
A and B =3:1. Ratio of time taken by A and B to finish a work=1:3
4.Definition of Variation: The change in two different variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation is called inverse variation.
5.Some Pairs of Variables:
i)Number of workers and their wages. If the number of workers
increases, their total wages increase. If the number of days
reduced, there will be less work. If the number of days is
increased, there will be more work. Therefore, here we have
direct proportion or direct variation.
ii)Number workers and days required to do a certain work is an
example of inverse variation. If more men are employed, they
will require fewer days and if there are less number of workers,
more days are required.
iii)There is an inverse proportion between the daily hours of a
work and the days required. If the number of hours is increased,
less number of days are required and if the number of hours is
reduced, more days are required.
6.Some Important Tips:
More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.
Number of days required to complete the given work=Total work/One
day’s work.
Since the total work is assumed to be one(unit), the number of days
required to complete the given work would be the reciprocal of one
day’s work.
Sometimes, the problems on time and work can be solved using the
proportional rule ((man*days*hours)/work) in another situation.
7.If men is fixed,work is proportional to time. If work is fixed,
then time is inversely proportional to men therefore,
(M1*T1/W1)=(M2*T2/W2)
Problems
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls —————?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A’s one hour work=1/8.
B’s one hour work=1/10
(A+B)’s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A’s one day work=1/18
B’s one day work=1/9
(A+B)’s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A’s one hour’s work=1/63 and
B’s one hour work=1/42
(A+B)’s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
Complex Problems
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C’s one day’s work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A’s share=Rs (600*4/8)=300
B’s share= Rs (600*3/8)=225
C’s share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A’s one day’s work=1/80
B’s one day’s work=1/48
(A+B)’s one day’s work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)P,Q and R are three typists who working simultaneously can type 216 pages
in 4 hours In one hour , R can type as many pages more than Q as Q can type more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 —————1
z-y=y-x => 2y=x+z ———–2
5z=7x => x=5x/7 —————3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to
type 32 pages on a computer, while Elan takes 5 hours to type 40 pages.
How much time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2×2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)’s one day’s work=1/12;
(B+C)’s one day’s work=1/15;
(A+C)’s one day’s work=1/20;
Adding we get 2(A+B+C)’s one day’s work=1/12+1/15+1/20=12/60=1/5
(A+B+C)’s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)’s one day’s work=1/6;
(A+B)’s one day’s work=1/8;
(B+C)’s one day’s work=1/12;
(A+C)’s one day’s work=2(A+B+C)’s one day’s work-((A+B)’s one day
work+(B+C)’s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)’s one day’s work=1/10;
C’s one day’s work=1/50
(A+B+C)’s one day’s work=(1/10+1/50)=6/50=3/25
Also, A’s one day’s work=(B+C)’s one day’s work
From i and ii ,we get :2*(A’s one day’s work)=3/25
=> A’s one day’s work=3/50
B’s one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A’s one day’s work=1/30;
B’s one day’s work=1/90;
(A+B)’s one day’s work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
Top
10) A can do a piece of work in 80 days. He works at it for 10 days and then
B alone finishes the remaining work in 42 days. In how much time will A&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A’s one day’s work =1/80 and
B’s one day’s work =1/48
(A+B)’s one day’s work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A’s one day’s work:B’s one days work=150:100 =3:2
Let A’s &B’s one day’s work be 3x and 2x days respectively.
Then C’s one day’s work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A’s one day’s work=3/100
B’s one day’s work=1/50
C’s one day’s work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B’s 10 day’s work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)’s one day’s work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)’s one day’s work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B’s one day’s work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) A and B can do a piece of work in 45 days and 40 days respectively. They
began to do the work together but A leaves after some days and then B completed
the remaining work n 23 days. The number of days after which A left the work was
Sol: (A+B)’s one day’s work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A’s share: B’s share: C’s share=1/2:1/3:1/6= 3:2:1
A’s share=(3/6)*1500=750
B’s share=(2/6)*1500=500
C’s share=(1/6)*1500=250
A’s daily wages=750/5=150/-
B’s daily wages=500/5=100/-
C’s daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)’s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A’s two day’s work=2/20=1/10
(A+B+C)’s one day’s work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man’s 1 day’s work=1/84
7 men’s 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men’s 1 day’s work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man’s 1 day work = 1/108
12 men’s 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men’s 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)’s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)12 men can complete a piece of work in 4 days, while 15 women can complete
the same work in 4 days. 6 men start working on the job and after working for
2 days, all of them stopped working. How many women should be put on the job
to complete the remaining work, if it is to be completed in 3 days.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man’s one day’s work= 1/48
one woman’s one day’s work=1/60
6 men’s 2 day’s work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child’s one day work= 1/192;
one adult’s one day’s work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)’s 1 day’s work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man’s one day’s work= 1/192
one child’s one day’s work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)’s 1 day’s work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man’s one day’s work= 1/384
one woman’s one day’s work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)’s two day’s work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man’s 1 day’s work =x days and
1 woman’s 1 day’s work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman’s 1 day’s work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man’s 1 hour’s work=x
1 woman’s 1 hour’s work=y
1 boy’s 1 hour’s work=z
Then, x+3y+4z=1/96 ———–(1)
2x+8z= 1/80 ———-(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ———(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)’s 1 hour’s work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C
1.Decimal fractions: Fractionin which denominations are powers
of 10 are decimal fractions.
Example:1 /10 = 0.1, 1 / 100 = 0.01
2.Convertion of Decimal into fraction:-
Example: 0.25 = 25/100 = 1/4
3.i) If numerator and denominator contain same number of decimal
places, then we remove decimal sign. Thus, 1.84/2.99 =184/299
Problems
1.0.75 =75/100 =3/4
2.Find porducts= 6.3204*100
= 632.04
3.2.61*1.3=261*13=3393 some of decimal places 2 +1 =3
sol: 3.393
4.If 1/3.718 =0.2689,then find value of 1/0.0003718 ?
Sol: 10000/3.718 =10000*1/3.718
=10000*0.2689
= 2689
5.Find fractions :
i) 0.37 = 37/99
ii)3.142857 =3+0.142857
=3 +142857/999999
= 3 142857/ 999999
iii) 0.17=17-1/90 =16/90=8/45
iv)0.1254 =1254 -12/9900 =1242/9900=69/550
6.Fraction 101 27/100000
Sol: 101+27/100000
=101+0.00027
=101.00027
7.If 47.2506 =4A + 7/B +2C + 5/D + 6E then 40+7+0.2+0.05+0.0006
Sol: compairing terms
4A= 40 => A=10
7/B = 7 => B=1
2C= 0.2=> C=0.1
5/D= 0.05=>D=5/0.05 =>5*100/5 =100
6E= 0.0006=> E= 0.0001
5A + 3B+6C+ D+ 3E = 5*10+ 3*1+ 6*0.1 + 100+ 3*0.0001
=50+3+0.6+100+0.0003
=153.6003
8.4.036 divided by 0.04
Sol: 4.036/0.04 =4036/4 =100.9
9.[ 0.05/0.25 + 0.25/ 0.05]3
Sol: =>[5/25 + 25/5]
= [1/5+ 5]3
=26/53
=5.23
= 140.603
10.The least among the following :-
a. 0.2 b.1/0.2 c. 0.2 d. 0.22
sol:10/2 =5 0.2222 0.04 0.04 <>
11.Let F= 0.84181
Sol: when F is written as a fraction in lowest terms, denominator
exceeds numerator by
84181 -841 /99000 = 83340/99000 =463/550
Required distence = (550 – 463) = 87
12.2 .75 + 3.78
Sol: [-2+0.75]+[-3+0.78]
=-5+[0.75+0.78]
= -5+1.53
=-5+1+0.53
= -4+0.53
= 4.53
13.the sum of first 20 terms of series is 1/5*6 +1/6*7+1/7*8—–
Sol: [1/5 -1/6]+[1/6-1/7]+[1/7-1/8]+————————
= [1/5-1/25]
=4/25=0.16
14.13 +23+ ————+93 =2025
Sol: value of (0.11) 3+ (0.22) 3+———(0.99)3 =>
(0.11) [1+2+--------+9]
=0.001331*2025
=2.695275
15.(0.96)3 – (0.1)3/ (0.96)2 +0.096 +(0.1)2
Sol: formula => a3 -b3/a2 +ab +b2 =a -b
(0.96-0.1)=0.86
16.3.6*0.48*2.50 / 0.12*0.09*0.5
Sol: 36*48*250/12*9*5=800
17.find x/y = 0.04/1.5
= 4/150 =2/75
find y-x/y+x
(1- x/y) / (1+ x/y)
1 - 2/75 /1 +2/75 =73/77
18.0.3467+0.1333
Sol: 3467 -34/9900 + 1333-13/9900
= 3433 +1320/9900
= 4753/9900
= 4801 -48/9900 =0.4301
PROBLEMS ON BOATS AND STREAMS
Important facts:
1)In water, the direction along the stream is called down stream.
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of
stream is V km/hr then
speed down stream =U + V km/hr
speed up stream = U – V km/hr
Formulae:
If the speed down stream is A km/hr and the speed up stream is
B km/hr then speed in still water = ½(A+B) km/hr
rate of stream =1/2(A-B) km/hr
Problems:
1. In one hour a boat goes 11 km long the stream and 5 km
against the stream. The speed of the boat in still water is?
Sol:
Speed in still water = ½ ( 11+5) km/hr= 8 kmph
2.A man can row 18 kmph in still water. It takes him thrice
as long as row up as to row down the river. find the rate
of stream.
Sol:
Let man’s rate up stream be xkmph
then, in still water =1/2[3x+x]=2x kmph
so, 2x= 18, x=9
rate upstream =9kmph
rate downstream =27 kmph
rate of stream = ½ [27-9]
= 9kmph
3.A man can row 71/2kmph in still water . if in a river running at 1.5 km an hour, if takes him 50 min to row to place and back. how far off is the place?
Sol: speed down stream =7.5+1.5=9kmph
speed upstream =7.5-1.5=6kmph
let the required distence x km. then ,
x/9+x/6=50/60 = 2x+3x= 5/6*18
5x=15, x=3
Hence, the required distence is 3 km
4.A man can row 3 quarters of a km aganist the stream is
111/4 min. the speed of the man in still water is ?
Sol: rate upstream = 750/625 m/sec =10/9 m/sec
rate downstream =750/450 m/sec = 5/3 m/sec
rate in still water =1/2[10/9+5/3] = 25/18 m/sec
= 25/18*18/5=5 kmph
5.A boat can travel with a speed of 13 kmph in still water.
if the speed of stream is 4 kmph,find the time taken by the boat to go 68 km downstream?
Sol: Speed down stream = 13+4= 17 kmph
time taken to travel 68km downstream =68/17 hrs
= 4 hrs
6.A boat takes 90 min less to travel 36 miles downstream then
to travel the same distence upstream. if the speed of the boat in still water is 10 mph . The speed of the stream is :
Sol: Let the speed of the stream be x mph .
then, speed downstream = [10+x]mph
speed upstream =[10-x] mph
36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2]
(x+50)(x-2) =0
x=2 kmph
7.At his usual rowing rate, Rahul 12 miles down stream in a certain river in 6 hrs less than it takes him to travel the same distence upstream. but if he could double his usual rowing rate for his 24 miles roundthe down stream 12 miles would then take only one hour less than the up stream 12 miles.what is the speed of the current in miles per hours?
Sol: Let the speed in still water be x mph and the speed of
the curren be y mph.
then, speed upstream = (x-y)
speed downstream =(x+y)
12/(x-y) - 12/(x+y) = 6
6(x2 – y2) m= 2xy => x2 – y2 =4y -(1)
and 12/(2x-y) - 12/(2x+y) =1 => 4×2 – y2 = 24y
x2= ( 24y + y2)/4 –>(2)
from 1 and 2 we have
4y+ y2 =( 24y+y2)/4
y=8/3 mph
y= 22/3 mph
8.There is a road beside a river. two friends started from
a place A, moved to a temple situated at another place B and then returned to A again. one of them moves on a cycle at a speed of 12 kmph, while the other sails on a boat at a speed of 10 kmph . if the river flows at the speedof 4 kmph,which of the two friends will return to place A first ?
Sol: Clearly, The cyclist moves both ways at a speed of 12 kmph
so, average speed of the cyclist = 12 kmph
the boat sailor moves downstream = (10+4) = 14 kmph
upstream =(10-4) = 6 kmph
So, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph
=42/5 kmph =8.4 kmph
Since, the average speed of the cyclist is greater, he will
return to A first.
9.A boat takes 19 hrs for travelling downstream from point A to
point B. and coming back to a point C midway between A and B.
if the velocity of the sream is 4 kmph. and the speed of the
boat in still water is 14 kmph. what is the distence between
A and B?
Sol:
speed downstream =14+4 =18 kmph
speed upstream = 14 -4 = 10 kmph
let the distence between A and B be x km. then,
x/18 + (x/2)/10 = 19
x/18 + x/20 =19
19x/180 =19 =>x = 180km
Hence, the distence between A and B bw 180 km
PROBLEMS ON CHAIN RULE
CHAIN RULE
1.DIRECT PROPORTION: Two Quantities are said to be directly proportional,if on the increase (or decrease) of th one, the other increases(or decreases) to the same extent.
Ex:(i) Cost is directly proportional to the number of articles.(More articles,More cost).
(ii)Work done is directly proportional to the number of men working on it. (More men, more work).
2.INDIRECT PROPORTION: Two Quantities are said to be indirectly proportional,if on the increase of the one , the other decreases to the same extent and vice-versa.
Ex:(i) The time taken by a car covering a certain distance is inversely
proportional to th speed of the car.(More speed, less is the time taken to cover the distance).
(ii)Time taken to finish a work is inversely proportional to the number
of persons working at it.(More persons, less is the time taken to finish a job).
NOTE: In solving Questions by chain rule, we compare every item with the term to be found out.
SIMPLE PROBLEMS
1)If 15 toys cost Rs.234, what do 35 toys cost ?
Sol: Let the required cost be Rs. x then
more toys more cost(direct proportion)
15:35:: 234:x
(15*x)=(234*35)
x=(234*35) /(15)= 546 Rs
2)If 36 men can do a piece of work in 25hours, in how many hours will 15men do it?
Sol: Let the required number of hours be x.
less men more hours(Indirect proportion).
15:36::25:x
(15*x)=(36*25)
x=(36*25) /15
x=60
For 15 men it takes 60 hours.
3)If 9 engines consume 24metric tonnes of coal, when each is working 8 hours a day, how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of former type
consume as much as 4 engines of latter type?
Sol: Let 3 engines of former type consume 1 unit in 1 hour.
4 engines of latter type consume 1 unit in 1 hour.
1 engine of former type consumes 1/3 unit in 1 hour.
1 engine of latter type consumes ¼ unit in 1 hour.
Let required consumption of coal be x units.
Less engines, less coal consumed.(direct)
More working hours, more coal consumed(direct)
Less rate of consumption, less coal consumed (direct)
9:8
8:13 :: 24:x
1/3:1/4
(9*8*(1/3)*x)=(8*13*(1/4)*24)
24x=624
x=26 metric tonnes.
COMPLEX PROBLEMS
1)A contract is to be completed in 46 days and 117 men were set to work,each working 8 hours a day. After 33 days, 4/7 of the work is completed.How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
Sol: 4/7 of work is completed .
Remaining work=1- 4/7
=3/7
Remaining period= 46-33
=13 days
Less work, less men(direct proportion)
less days, more men(Indirect proportion)
More hours/day, less men(Indirect proportion)
work 4/7:3/7
Days 13:33 :: 117:x
hrs/day 9:8
(4/7)*13*9*x=(3/7)*33*8*117
x=(3*33*8*117) / (4*13*9)
x=198 men
So, additional men to be employed=198 -117=81
2)A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that?
Sol: Let initially there be x men having food for y days.
After, 10 days x men had food for ( y-10)days
Also, (x -x/5) men had food for y days.
x(y-10)=(4x/5)*y
=> (x*y) -50x=(4(x*y)/5)
5(x*y)-4(x*y)=50x
x*y=50x
y=50
3)A contractor undertook to do a certain piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?
Sol: 40 days- 35 days=5 days
=>(100*35)+(200*5) men can finish the work in 1 day.
4500 men can finish it in 4500/100= 45 days
This s 5 days behind the schedule.
4)12 men and 18 boys,working 7 ½ hors a day, can do a piece if work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working
9 hours a day?
Sol: 1man =2 boys
12men+18boys=>(12*2+18)boys=42 boys
let the required number of boys=x
21 men+x boys
=>((21*2)+x) boys
=>(42+x) boys
less days, more boys(Indirect proportion)
more hours per day, less boys(Indirect proportion)
days 50:60
hrs/day 9:15/2 :: 42:(42+x)
work 1:2
(50*9*1*(42+x))=60*(15/2)*2*42
(42+x)= (60*15*42)/(50*9)= 84
x=84-42= 42
=42
42 days behind the schedule it will be finished.
Important Facts:
Partnership:When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership.
Ratio of Division of Gains:
1.When the investments of all the partners are do the same time, the
gain or loss is distributed among the partners in the ratio of their
investments.
Suppose A and B invest Rs x and Rs y respectively for a year in a
business, then at the end of the year:
(A’s share of profit):(B’s share of profit)=x:y
2.When investments are for different time periods, then equivalent
capitals are calculated for a unit of time by taking (capital*number
of units of time). Now gain or loss is divided in the ratio of these
capitals.
Suppose A invests Rs x for p months and B invests Rs y for q months,
then (A’s share of profit):(B’s share of profit)=xp:yq
3.Working and sleeping partners:A partner who manages the business is
known as working partner and the one who simply invests the money is
a sleeping partner.
Formulae
1.When investments of A and B are Rs x and Rs y for a year in a
business ,then at the end of the year
(A’s share of profit):(B’s share of profit)=x:y
2.When A invests Rs x for p months and B invests Rs y for q months,
then A’s share profit:B’s share of profit=xp:yq
Short cuts:
1.In case of 3 A,B,C investments then individual share is to be found
then A=16000 , B=32,000 , C=40,000
Sol: A:B:C=16:32:40
=2:4:5`
then individual share can be easily known.
2.If business mans A contributes for 5 months and B contributes for 9 months then share of B in the total profit of Rs 26,8000 ,A = Rs 15000,B =Rs 12000
Sol: 15000*5 : 12000*9
25 : 36
for 36 parts=268000*(36/61)
=Rs 158.16
Difficult problems:
1.P and Q started a business investing Rs 85,000 and Rs 15,000
respectively. In what ratio the profit earned after 2 years be divided
between P and Q respectively?
Sol: 85,000*2 : 15,000*2
17*2 : 3*2=34:6
2.A,B and C started a business by investing Rs 1,20,000, Rs 1,35,000 and Rs 1,50,000.Find the share of each ,out of an annual profit of Rs 56,700?
Sol: Ratio of shares of A,B and C=Ratio of their investments
120,000:135,000:150,000
=8:9:10
A’s share=Rs 56,700*(8/27)=Rs 16,800
B’s share =Rs 56,700*(9/27)=Rs 18,900
C’s share =Rs 56,700*(10/27)=Rs 21,000
3.3 milkman A,B,C rented a pasture A grazed his 45 cows for 12 days B grazed his 36 cows for 15 days and c 60 cows for 10 days.If b’s
share of rent was Rs 540 What is the total rent?
Sol: 45*12:36*15:60*10
=9:9:10
9 parts is equal to Rs 540
then one part is equal to Rs 60
total rent=60*28=Rs 1680
4.Ramu and Krishna entered into a partnership with Rs 50,000 and Rs 60,000, after 4 months Ramu invested Rs 25,000 more while Krishna withdraw Rs 20,000 . Find the share of Ramu in the annual profit of Rs 289,000.
Sol: Ramu : Krishna=50,000*4+75,000*8:60,000*4+40,000*8
=10:7
Ramu’s annual profit=289000*(10/17)=Rs 170,000
5.A,B,C enter into partnership .A invests 3 times as much as B invests and B invests two third of what C invests. At the end of the year ,the profit earned is Rs 6600. what is the share of B?
Sol: let C’s capital =Rs x
B’s capital=Rs (2/3)*x
A’s capital =3*(2/3)*x=Rs 2x
ratio of their capitals=2x:(2/3)*x:x
=6x:2x;3x
B’s share =Rs 6600(2/11)=Rs 1200
6.A,B and C enter into a partnership by investing in the ratio of 3:2:4.After one year ,B invests another Rs 2,70,000 and C,at the end of 2 years, also invests Rs 2,70,000.At the end of 3 years ,profit are shared
in the ratio of 3:4:5.Find the initial investment of each?
Sol: Initial investments of A,B,c be Rs 3x, Rs 2x, Rs 4x then for 3 years
(3x*36):[(2x*12)+(2x+270000)*24]:[(4x*24)+(4x+270000)*12]=3:4:5
108x:(72x+640,000):(144x+3240000)=3:4:5
108x:72x+6480000:144x+3240000=3:4:5
(108x)/(72x+6480000)=3/4
432x=216x+19440000
216x=19440000
x=Rs 90000
A’s initial investment=3x=3*90,000=Rs 2,70,000
B’s initial investment=2x=2*90,000=Rs 1,80,000
C’s initial investment=4x=4*90,000=Rs 3,60,000
PROBLEMS ON RATIO AND PROPORTION
1.Ratio : The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio, a:b, we call a as the first term of antecedent and b, the second term consequent.
Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9
3Rule: The multiplication or division of each term of 9 ratio by the same non-zero number does not affect the ratio.
4.Proportion: The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms.
Product of means=product of extremes
Thus, a:b::c:d => (b*c)=(a*d)
5.Fourth proportional: If a:b::c:d, then d is called the fourth proportional to a,b and c.
6.Third proportional: If a:b::b:c, then c is called third proportional to a and b.
7.Mean proportional: Mean proportional between a and b is SQRT(a*b).
COMPARISION OF RATIOS:
We say that (a:b)>(c:d) => (a/b)>(c/d)
8.Compounded ratio: The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf).
9.Duplicate Ratio: If (a:b) is (a2: b2 )
10.Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))
11.Triplicate ratio of (a:b) is (a3: b3 )
12.Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).
13.If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (component and dividend o)
VARIATION:
14.we say that x is directly proportional to y, if x=ky for some constant k and we write.
15.We say that x is inversely proportional to y, if xy=k for some constant and we write.
16. X is inversely proportional to y.
If a/b=c/d=e/f=g/h=k then
k=(a+c+e+g)/(b+d+f+h)
If a1/b1,a2/b2, a3/b3…………..an/bn are unequal
fractions then the ratio.
SIMPLE PROBLEMS
1.If a:b =5:9 and b:c=4:7 Find a:b:c?
Sol: a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9
a:b:c=5:9:63/9=20:36:63
2.Find the fourth proportion to 4,9,12
Sol: d is the fourth proportion to a,b,c
a:b=c:d
4:9=12:x
4x=9*12=>x=27
3.Find third proportion to 16,36
Sol: if a:b=b:c then c is the third proportion to a,b
16:36=36:x
16x=36*36
x=81
4.Find mean proportion between 0.08 and 0.18
Sol: mean proportion between a and b=square root of ab
mean proportion =square root of 0.08*0.18=0.12
5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is
Sol: a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4
c:d=6:7=6*15/24:7*15/24=15/4:35/8
a:b:c:d=2:3:15/4:35/8=16:24:30:35
6.2A=3B=4C then A:B:C?
Sol: let 2A=3B=4C=k then
A=k/2, B=k/3, C=k/4
A:B:C=k/2:k/3:k/4=6:4:3
7.15% of x=20% of y then x:y is
Sol: (15/100)*x=(20/100)*y
3x=4y
x:y=4:3
8.a/3=b/4=c/7 then (a+b+c)/c=
Sol: let a/3=b/4=c/7=k
(a+b+c)/c=(3k+4k+7k)/7k=2
9.Rs 3650 is divided among 4 engineers, 3 MBA’s and 5 CA’s such that 3 CA’s get as much as 2 MBA’s and 3 Eng’s as much as 2 CA’s .Find the share of an MBA.
Sol: 4E+3M+5C=3650
3C=2M, that is M=1.5C
3E=2C that is E=.66 C
Then, (4*0.66C)+(3*1.5C)+5C=3650
C=3650/12.166
C=300
M=1.5 and C=450.
DIFFICULT PROBLEMS
1.Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together.
Sol: Assume that there are 500,400 and 500 liters respectively in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.
Thus, the total milk is 441.66 liters. Hence, the amount of water in the mixture is 1400-441.66=958.33liters.
Hence, the ratio of milk to water is 441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.
2.A certain number of one rupee,fifty parse and twenty five paise coins are in the ratio of 2:5:3:4, add up to Rs 210. How many 50 paise coins were there?
Sol: the ratio of 2.5:3:4 can be written as 5:6:8
let us assume that there are 5 one rupee coins,6 fifty paise coins and 8 twenty-five paise coins in all.
their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10
If the total is Rs 10,number of 50 paise coins are 6.
if the total is Rs 210, number of 50 paise coins would be 210*6/10=126.
3.The incomes of A and B are in the ratio of 4:3 and their expenditure are in the ratio of 2:1 . if each one saves Rs 1000,what are their incomes?
Sol: Ratio of incomes of A and B=4:3
Ratio of expenditures of A and B=2:1
Amount of money saved by A=Amount of money saved by B
=Rs 1000
let the incomes of A and B be 4x and 3x respectively
let the expense of A and B be 2y and 1yrespectively
Amount of money saved by A=(income-expenditure)=4x-2y=
Rs 1000
Amount of money saved by B=3x-y=Rs 1000
this can be even written as 6x-2y=Rs 2000
now solve 1 and 3 to get
x=Rs 500
therefore income of A=4x=4*500=Rs 2000
income of B=3x=3*500=Rs 1500
4.A sum of Rs 1162 is divided among A,B and C. Such that 4 times A’s share share is equal to 5 times B’s share and 7 times C’s share . What is the share of C?
Sol: 4 times of A’s share =5 times of B’s share=7 times of C’s share=1
therefore , the ratio of their share =1/4:1/5:1/7
LCM of 4,5,7=140
therefore, ¼:1/5:1/7=35:28:20
the ratio now can be written as 35:28:20
therefore C’s share=(20/83)*1162=20*14=Rs 280.
5.The ratio of the present ages of saritha and her mother is 2:9, mother’s age at the time of saritha’s birth was 28 years , what is saritha’s present age?
Sol: ratio of ages of saritha and her mother =2:9
let the present age of saritha be 2x years. then the mother’s present age would be 9x years
Difference in their ages =28 years
9x-2x=28 years
7x=28=>x=4
therefore saritha’s age =2*4=8 years
Important Facts:
Cost Price: The price at which an article is purchased,
is called its cost price,abbreviated as C.P.
Selling Price: The price at which an article is sold,
is called its selling price,abbreviated as S.P.
Profit or Gain: If S.P. Is greater than C.P. The seller
is said to have a profit or gain.
Loss:if S.P. Is less than C.P., the seller is said to
have incurred a loss.
Formulae
1.Gain=(S.P-C.P)
2.Loss=(C.P-S.P)
3.Loss or Gain is always reckoned on C.P.
4.Gain%=(gain*100)/C.P
5.Loss%=(loss*100)/C.P
6.S.P=[(100+gain%)/100]*C.P
7.S.P=[(100-loss%)/100]*C.P
8.C.P=(100*S.P)/(100+gain%)
9.C.P=(100*S.P)/(100-loss%)
10.If an article is sold at a gain of say,35%,then S.P=135% of C.P.
11.If an article is sold at a loss of say,35%,then S.P=65% of C.P.
12.When a person sells two similar items, one at a gain of say,x%,and the other at a loss of x%,then the seller always incurs a loss given by Loss%=[common loss and gain %/10]2=(x/10)2.
13.If a trader professes to sell his goods at cost price,but uses false weight,then Gain%=[(error/(true value-error))*100]%.
14.Net selling price=Marked price-Discount.
Simple Problems
1.A man buys an article for Rs.27.50 and sells it for Rs.28.60
Find the gain percent.
Sol: C.P=Rs 27.50 S.P=Rs 28.60
then Gain=S.P-C.P=28.60-27.50=Rs 1.10
Gain%=(gain*100)/C.P% =(1.10*100)/27.50%=4%
2.If a radio is purchased for Rs 490 and sold for Rs 465.50 Find the loss%?
Sol: C.P=Rs 490 S.P=Rs 465.50
Loss=C.P-S.P=490-465.50=Rs 24.50
Loss%=(loss*100)/C.P% =(24.50*100)/490%=5%
3.Find S.P when C.P=Rs 56.25 and Gain=20%
Sol: S.P=[(100+gain%)/100]*C.P
S.P=[(100+20)/100]56.25=Rs 67.50
4.Find S.P when C.P=Rs 80.40,loss=5%
Sol: S.P=[(100-loss%)/100]*C.P
S.P=[(100-5)/100]*80.40=Rs 68.34
5.Find C.P when S.P=Rs 40.60,gain=16%?
Sol: C.P=(100*S.P)/(100+gain%)
C.P=(100*40.60)/(100+16)=Rs 35
6.Find C.P when S.P=Rs 51.70 ,loss=12%?
Sol: C.P=(100*S.P)/(100-loss%)
C.P=(100*51.70)/(100-12)=Rs 58.75
7.A person incurs 5% loss by selling a watch for Rs 1140 . At what price should the watch be sold to earn 5% profit?
Sol: Let the new S.P be Rs x then,
(100-loss%):(1st S.P)=(100+gain%):(2nd S.P)
(100-5)/1140=(100+5)/x
x=(105*1140)/95=Rs 1260
8.If the cost price is 96% of the selling price,then what is the profit percent?
Sol: let S.P=Rs 100 then C.P=Rs 96
profit=S.P-C.P=100-96=Rs 4
profit%=(profit*C.P)/100%
=(4*96)/100=4.17%
9.A discount dealer professes to sell his goods at cost price but uses a weight of 960 gms for a Kg weight .Find his gain %?
Sol: Gain%=[(error*100)/(true value-error)]%
=[(40*100)/1000-40)]%=25/6%
10.A man sold two flats for Rs 675,958 each .On one he gains
16% while on the other he losses 16%.How much does he gain or
lose in the whole transaction?
Sol: loss%=[common loss or gain%/10]2=(16/10)2=2.56%
11.A man sold two cows at Rs 1995 each. On one he lost 10% and
on the other he gained 10%.what his gain or loss percent?
Sol: If loss% and gain% is equal to 10 then there is no loss or no gain.
12.The price of an article is reduced by 25% in order to restore the must be increased by ?
Sol: [x/(100-x)]*100 =[25/(100-25)]*100
=(25/75)*100=100/3%
13.Two discounts of 40% and 20% equal to a single discount of?
Sol: {[(100-40)/100]*[(100-20)/100]}%=(60*80)/(100*100)%
=48%
single discount is equal to (100-48)%=52%
Difficult Problems
1.The cost of an article including the sales tax is Rs 616.The rate of sales tax is 10%,if the shopkeeper has made a profit of 12%,then the cost price of the article is?
Sol: 110% of S.P=616
S.P=(616*100)/110=Rs 560
C.P=(100*S.P)/(100+gain%)
C.P. =(100*560)/(100+12)=Rs 500
2.Sam purchased 20 dozens of toys at the rate of 375 Rs per dozen.He sold each one of then at the rate of Rs 33.What was his
percentage profit?
Sol: C.P of one toy=Rs 375/12=Rs 31.25
S.P of one toy=Rs 33
profit=S.P-C.P=33-31.25=Rs 1.75
profit %=(profit/C.P)*100
=(1.75/31.25)*100
profit% =5.6%
3.Two third of consignment was sold at a profit of 5% and the
remainder at a loss of 2%.If the total was Rs 400,the value of the
consignment was?
Sol: let the total value be Rs x
value of 2/3=2x/3, value of 1/3=x/3
total S.P value be Rs x
total S.P=[(105% of 2x/3)+(98% of x/3)]
=(105*2x)/(100*3)+(98/100)+x/3
=308x/300
(308x/300)-x=400
8x/300=400
x=(300*400)/8=Rs 15000
4.Kunal bought a suitcase with 15% discount on the labelled price. He sold the suitcase for Rs 2880 with 20% profit on the labelled
price .At what price did he buy the suitcase?
Sol: let the labelled price be Rs x
then 120% of x=2880
x=(2880*100)/120=Rs 2400
C.P=85% of the 2400
(85*2400)/100=Rs 2040
5.A tradesman gives 4% discount on the marked price and gives
article free for buying every 15 articles and thus gains 35%. The
marked price is above the cost price by
Sol: let the C.P of each article be Rs 100
then C.P of 16 articles=Rs (100*16)=Rs 1600
S.P of 15 articles =1600*(135/100)=Rs 2160
S.P of each article =2160/15=Rs 144
If S.P is Rs 96, marked price =Rs 100
If S.P is Rs 144,marked price =(100/96)*144=Rs 15000
therefore marked price=50% above C.P
6.By selling 33m of cloth ,one gains the selling price of 11m.Find
the gain percent?
Sol: gain=S.P of 33m-C.P of 33m
=11m of S.P
S.P of 22m=C.P of 33m
let C.P of each meter be Rs 1,then C.P of 22m=Rs 22
S.P of 22m=Rs 33
gain=S.P-C.P=33-22=Rs 11
gain%=(gain/C.P)*100
=(11/22)*100=50%
7.The price of a jewel, passing through three hands, rises on the whole by 65%.if the first and second sellers earned 20% and 25%
profit respectively,find the percentage profit earned by the
third seller?
Sol: let the original price of the jewel be Rs P and
let the profit earned by the third seller be x%
then (100+x)% of 125% of P=165% of P
[(100+x)/100]*(125/100)*(120/100)*P=(165/100)*P
100+x=(165*100*100)/(125*120)
100+x=110
x=10%
8.When a producer allows 36% commission on the retail price of his product ,he earns a profit of 8.8%.what would be his profit
percent if the commission is reduced by 24%
Sol: let retail price =Rs 100
commission=Rs 36
S.P=retail price-commission=100-36=Rs 64
But profit=8.8%
C.P=(100*C.P)/(gain+100)=(100*64)/(100+8.8)=Rs 1000/17
new commission=Rs 12
new S.P=100-12=Rs 88
gain=88-(1000/17)=Rs 496/17
gain%=gain*100/C.P
=(496*17*100)(17*1000)
gain%=49.6%
9.Vikas bought paper sheets for Rs 7200 and spent Rs 200 on
transport. Paying Rs 600,he had 330 boxes made,which he sold
at Rs 28 each. His profit percentage is
Sol: total
- investments=7200+200+600=Rs 8000
total receipt=330*28=Rs 9240
gain=S.P-C.P
=total receipt-total investments
gain=9240-8000=Rs 1240
gain% =gain*100/C.P=1240*100/8000=15.5%
10.A person earns 15% on investment but loses 10% on another investment .If the ratio of the two investments be 3:5 ,what is the gain or loss on the two investments taken together?
Sol: let the investments be 3x and 5x
then total investment=8x
total receipt=115% of 3x+90% of 5x
=115*3x/100+90*5x/100=7.95x
loss=C.P-S.P=8x-7.95x=0.05x
loss%=.05x*100/8x=0.625%
11.The profit earned by selling an article for Rs 900 is double the loss incurred when the same article is sold for Rs 490 .At what
price should the article be sold to make 25% profit?
Sol: let C.P be Rs x
900-x=2(x-450)
3x=1800
x=Rs 600
C.P=Rs 600 , gain required=25%
S.P=(100+gain%)*C.P/100
S.P=(100+25)*600/100=Rs 750
12.If an article is sold at 5% gain instead of 5% loss,the seller
gets Rs 6.72 more. The C.P of the article is?
Sol: let C.P be Rs x
105% of x-95% of x=6.72
(105/100)*x-(95/100)*x=6.72
x/10=6.72
x=Rs 67.21
EXAMPLE PROBLEMS:
1 . Express the following as a fraction.
a) 56%
SOLUTION:
56/100=14/25
b) 4%
SOLUTION:
4/100=1/25
c) 0.6%
SOLUTION:
0.6/100=6/1000=3/500
d) 0.08%
SOLUTION:
0.08/100=8/10000=1/1250
2.Express the following as decimals
a) 6%
SOLUTION:
6% = 6/100=0.06
b) 0.04%
SOLUTION:
0.04% = 0.04/100=0.0004
3 . Express the following as rate percent.
i).23/36
SOLUTION:
= (23/36*100) %
= 63 8/9%
ii).6 ¾
SOLUTION:
6 ¾ =27/4
(27/4 *100) % =675 %
4.Evaluate the following:
28% of 450 + 45% of 280 ?
SOLUTION:
=(28/100) *450 + (45/100) *280
= 28 * 45 / 5
= 252
5.2 is what percent of 50?
SOLUTION:
Formula : (IS / OF ) *100 %
= 2/50 *100
= 4%
6.½ is what percent of 1/3?
SOLUTION:
=( ½) / (1/3) *100 %
= 3/2 *100 %
= 150 %
7.What percent of 2 Metric tonnes is 40 Quintals?
SOLUTION:
1 metric tonne =10 Quintals
So required percentage = (40/(2*10)) *100 %
= 200%
8.Find the missing figure .
i) ? % of 25 = 2.125
SOLUTION :
Let x% of 25 = 2.125. then
(x/100) *25 =2.125
x = 2.125 * 4
= 8.5
ii) 9% of ? =6.3
SOLUTION:
Let 9 % of x = 6.3.
Then 9/100 of x= 6.3
so x = 6.3 *100/7
= 70.
9.Which is the greatest in 16 2/3 %, 2/15,0.17?
SOLUTION:
16 2/3 % = 50/3 %
=50/3 * 1/100
=1/6
= 0.166
2 / 15 =0.133
So 0.17 is greatest number in the given series.
10.If the sales tax be reduced from 3 ½ % to 3 1/3 % ,then what difference does it make to a person who purchases an article with marked price of RS 8400?
SOLUTION:
Required difference = 3 ½ % of 8400 – 3 1/3 % of 8400
=(7/2-10/3)% of 8400
=1/6 % of 8400
= 1/6* 1/100* 8400
= Rs 14.
11. A rejects 0.08% of the meters as defective .How many will he examine to reject 2?
SOLUTION:
Let the number of meters to be examined be x.
Then 0.08% of x=2.
0.08/100*x= 2
x= 2 * 100/0.08
=2 * 100 * 100/8
= 2500
12.65 % of a number is 21 less than 4/5 of that number. What is the number?
SOLUTION: Let the number be x.
4/5 x- (65% of x) = 21
4/5x – 65/100 x=21
15x=2100
x=140
13. Difference of two numbers is 1660.If 7.5 % of one number is 12.5% of the other number. Find two numbers?
SOLUTION:
Let the two numbers be x and y.
7.5% of x=12.5% of y
So 75x=125 y
3x=5y
x=5/3y.
Now x-y=1660
5/3y-y=1660
2/3y=1660
y=2490
So x= 2490+1660
=4150.
So the numbers are 4150 , 1660.
14. In expressing a length 81.472 KM as nearly as possible with 3 significant digits ,Find the % error?
SOLUTION:
Error= 81.5-81.472=0.028
So the required percentage = 0.028/81.472*100%
= 0.034%
15. In an election between two persons ,75% of the voters cast their votes out of which 2% are invalid. A got 9261 which 75% of the total valid votes. Find total number of votes?
SOLUTION:
Let x be the total votes.
valid votes are 98% of 75% of x.
So 75%(98%(75% of x))) = 9261
==> 75/100 *98 /100 * 75 100 *x = 9261
x= 1029 * 4 *100 *4 / 9
= 16800
So total no of votes = 16800
16 . A’s maths test had 75 problems i.e 10 arithmetic, 30 algebra and 35 geometry problems. Although he answered 70% of arithmetic , 40% of algebra and 60 % of geometry problems correctly he didn’t pass the test because he got less than 60% of the problems right. How many more questions he would have needed to answer correctly to get a 60% passing grade.
SOLUTION:
70% of 10 =70/100 * 10
=7
40% of 30 = 40 / 100 * 30
= 12
60 % of 35 = 60 / 100 *35
= 21
So correctly attempted questions = 7 + 12 + 21
=40.
Questions to be answered correctly for 60% grade
=60% of 75
= 60/100 *75
=45.
So required questions=45-40 = 5
17 . If 50% of (x – y) = 30% of (x + y) then what percent of x is y ?
SOLUTION:
50/100(x-y) =30/100(x+y)
½ (x-y)= 3/10(x+y)
5x-5y=3x+3y
x=4y
So Required percentage =y/x*100 %
=y/4y *100 %
= 25%.
18 . If the price of tea is increased by 20% ,find how much percent must a householder reduce her consumption of tea so as not to increase the expenditure?
SOLUTION:
Reduction in consumption= R/(100+R) *100%
=20/120 *100
= 16 2/3 %
19.The population of a town is 176400 . If it increases at the rate of 5% per annum ,what will be the population 2 years hence? What was it 2 years ago?
SOLUTION:
Population After 2 years = 176400[1+5/100]2
=176400 * 21/20 *21/20
=194481
Population 2 years ago = 176400/(1+5/100)2
= 176400 * 20/21 *20/ 21
=160000
20.1 liter of water is add to 5 liters of a 20 % solution of alcohol in water . Find the strength of alcohol in new solution?
SOLUTION:
Alcohol in 5 liters = 20% of 5
=1 liter
Alcohol in 6 liters of new mixture = 1liter
So % of alcohol is =1/6 *100=16 2/3%
21. If A earns 33 1/3 more than B .Then B earns less than A by what percent?
SOLUTION:
33 1/3 =100 / 3
Required Percentage = (100/3)/(100 + (100/3)) *100 %
= 100/400 *100 = 25 %
22. A school has only three classes which contain 40,50,60 students respectively . The pass percent of these classes are 10, 20 and 10 respectively . Then find the pass percent in the school.
SOLUTION:
Number of passed candidates =
10/100*40+20/100 *50+10/100 * 60
=4+10+6
=20
Total students in school = 40+50+60 =150
So required percentage = 20/150 *100
= 40 /3
=13 1/3 %
23. There are 600 boys in a hostel . Each plays either hockey or football or both .If 75% play hockey and 45 % play football ,Find how many play both?
SOLUTION:
n(A)=75/100 *600
=450
n(B) = 45/100 *600
= 270
n(A^B)=n(A) + n(B) – n(AUB)
=450 + 270 -600
=120
So 120 boys play both the games.
24.A bag contains 600 coins of 25p denomination and 1200 coins of 50p denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the percentage of money removed from the bag ?
SOLUTION:
Total money = (600 * 25/100 +1200 *50/100)
=Rs 750
25p coins removed = 12/100 *600
=72
50p coins removed = 24/100 *1200
=288
So money removed =72 *1/4 +288 *1/2
= Rs 162
So required percentage=162/750 *100
=21 .6%
25. P is six times as large as Q.Find the percent that Q is less than P?
SOLUTION:
Given that P= 6Q
So Q is less than P by 5Q.
Required percentage= 5Q/P*100 %
=5/6 * 100 %
=83 1/3%
26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume?
SOLUTION:
Surface area = 4 *22/7 *r2
= 3/r(4/3 * 22/7 * r3)
=3/r * VOLUME
Where r = 10 cm
So we have S= 3/10 V
=3/10 *100 % of V
= 30 % of V
So surface area is 30 % of Volume.
27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5 kg more for Rs 100.What is the reduced price per kg.
SOLUTION:
Let the original price = Rs x/kg
Reduced price =79/100x /kg
==> 100/(79x/100)-100/x =10.5
==> 10000/79x-100/x=10.5
==> 10000-7900=10.5 * 79 x
==> x= 2100/10.5 *79
So required price = Rs (79/100 *2100/10.5 *79) /kg
= Rs 2 per kg.
28.The length of a rectangle is increased by 60 % .By what percent would the width have to be decreased to maintain the same area?
SOLUTION:
Let the length =l,Breadth= b.
Let the required decrease in breadth be x %
then 160/100 l *(100-x)/100 b=lb
160(100-x)=100 *100
or 100-x =10000/160
=125/2
so x = 100-125/2
Simple problems:
1. Laws of Indices:
(i) am * an = a(m+n)
(ii) am / an = a(m-n)
(iii) (am)n = a(m*n)
(iv) (ab)n = an * bn
(v) (a/b)n = an / bn
(vi) a0 = 1
2.Surds :Let ‘a’ be a rational number & ‘n’ be a positive integer such that a1/n = nth root a is irrational.Then nth root a is called ‘a’ surd of ‘n’.
Problems:-
(1)
(i) (27)2/3 = (33)2/3 = 32 = 9.
(ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256.
(iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16
(2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .
Solution: 2x/2+2x.2 = 1280
2x(1+22) = 2*1280
2x = 2560/5
2x = 512 => 2x = 29
x = 9
(3) Find the value of [5[81/3+271/3]3]1/4
Solution: [5[(23)1/3+(33)1/3]3]1/4
[5[2+3]3]1/4
[54]1/4 => 5.
(4) If (1/5)3y= 0.008 then find the value of (0.25)y
Solution: (1/5)3y = 0.008
(1/5)3y =[0.2]3
(1/5)3y =(1/5)3
3y= 3 => y=1.
(0.25)y = (0.25)1 => 0.25 = 25/100 = 1/4
(5) Find the value of (243)n/5 * 32n+1 / 9n * 3 n-1
Solution: (35)n/5 * 32n +1 / (32)n * 3n-1
33n+1 / 33n-1 3
33n+1 * 3-3n+1 => 32 =>9.
(6) Find the value of (21/4-1)( 23/4 +21/2+21/4+1)
Solution: Let us say 21/4 = x
(x-1)(x3+x2+x+1)
(x-1)(x2(x+1)+(x+1))
(x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)]
(x2+1) (x2-1) => (x4-1)
((21/4))4 - 1) = > (2-1) = > 1.
(7) If x= ya , y = zb , z = xc then find the value of abc.
Solution: z= xc
z= (ya)c [ x= ya ]
z= (y)ac
z= (zb)ac [y= zb]
z= zabc
abc = 1
(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2
Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2
[ (a-b)(a2+ab+b2) = a3-b3]
from the above formula
=> xa3-b3 xb3-c3 xc3-a3
=> xa3-b3+b3-c3+c3-a3
=> x0 = 1
(9) (1000)7 /1018 = ?
(a) 10 (b) 100 (c ) 1000 (d) 10000
Solution: (1000)7 / 1018
(103)7 / (10)18 = > (10)21 / (10)18
=> (10)21-18 => (10)3 => 1000
Ans :( c )
(10) The value of (8-25-8-26) is
(a) 7* 8-25 (b) 7*8-26 (c ) 8* 8-26 (d) None
Solution: ( 8-25 - 8-26 )
=> 8-26 (8-1 )
=> 7* 8-26
Ans: (b)
(11) 1 / (1+ an-m ) +1/ (1+am-n) = ?
(a) 0 (b) 1/2 (c ) 1 (d) an+m
Solution: 1/ (1+ an/am) + 1/ ( 1+ am/an)
=> am / (am+ an ) + an /(am +an )
=> (am +an ) /(am + an)
=> 1
Ans: ( c)
(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=?
(a) 0 (b) 1 ( c ) xa-b-c (d) None of the above
Solution: 1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) +
1/(1+xb/xc +xa/xc)
=> xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc)
=>(xa +xb+xc) /(xa +xb+xc)
=>1
Ans: (b)
(13) If x=3+2 √2 then the value of (√x – 1/ √x)
is [ √=root]
(a) 1 (b) 2 (c ) 2√2 ( d) 3√3
Solution: (√x-1/√x)2 = x+ 1/x-2
=> 3+2√2 + (1/3+2√2 )-2
=> 3+2√2 + 3-2√2 -2
=> 6-2 = 4
(√x-1/√x)2 = 4
=>(√x-1/√x)2 = 22
(√x-1/√x) = 2.
Ans : (b)
(14) (xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c = ? (a) xabc (b) 1 ( c) xab+bc+ca (d) xa+b+c
Solution: [xb-c]b+c-a [xc-a]c+a-b [xa-b]a+b-c
=>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c)
=>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc)
=>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc)
=> x0
=>1
Ans: (b)
(15) If 3x-y = 27 and 3x+y = 243 then x is equal to
(a) 0 (b) 2 (c ) 4 (d) 6
Solution: 3x-y = 27 => 3x-y = 33
x-y= 3
3x+y = 243 => 3x+y = 35
x+y = 5
From above two equations x = 4 , y=1
Ans: (c )
(16) If ax = by = cz and b2 = ac then ‘y’equals
(a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z
Solution: Let us say ax = by = cz = k
ax =k => [ax]1/x = k1/x
=> a = k1/x
Simillarly b = k1/y
c = k1/z
b2 = ac
[k1/y]2=k1/xk1/z
=>k2/y = k1/x+1/z
=> 2/y = 1/x+1/y
=>y= 2xz/x+z
Ans: (d)
(17) ax = b,by = c ,cz = a then the value of xyz is is
(a) 0 (b) 1 (c ) 1/abc (d) abc
Solution: ax = b
(cz)x = b [cz = a]
by)xz = b [by = c]
=>xyz =1
Ans: (b)
(18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then the value of ‘z’ is
(a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64
Solution: 2x = 4y=8z
2x = 22y = 23z
x= 2y = 3z
Multiply above equation with ‘ 2’
2x = 4y= 6z
(1/2x+1/4y+1/6z) = 24/7
=>(1/6z+1/6z+1/6z) = 24/7
=> 3 / 6z = 24/7
=> z= 7/48
Ans: ( c)
PROBLEMS ON AGES
Simple problems:
1.The present age of a father is 3 years more than three times
the age of his son.Three years hence,father’s age will be 10
years more than twice the age of the son.Find the present age
of the father.
Solution: Let the present age be ‘x’ years.
Then father’s present age is 3x+3 years.
Three years hence
(3x+3)+3=2(x+3)+10
x=10
Hence father’s present age = 3x+3 = 33 years.
2. One year ago the ratio of Ramu & Somu age was 6:7respectively. Four years hence their ratio would become 7:8. How old is Somu.
Solution: Let us assume Ramu &Somu ages are x &y respectively.
One year ago their ratio was 6:7
i.e x-1 / y-1 = 7x-6y=1
Four years hence their ratios,would become 7:8
i.e x-4 / y-4 = 7 / 8
8x-7y=-4
From the above two equations we get y= 36 years.
i.e Somu present age is 36 years.
3. The total age of A &B is 12 years more than the total age of
B&C. C is how many year younger than A.
Solution: From the given data
A+B = 12+(B+C)
A+B-(B+C) = 12
A-C=12 years.
C is 12 years younger than A
4. The ratio of the present age of P & Q is 6:7. If Q is 4 years
old than P. what will be the ratio of the ages of P & Q after
4 years.
Solution: The present age of P & Q is 6:7 i.e
P / Q = 6 / 7
Q is 4 years old than P i.e Q = P+4.
P/ P+4 = 6/7
7P-6P = 24,
P = 24 , Q = P+4 =24+4 = 28
After 4 years the ratio of P &Q is
P+4:Q+4
24+4 : 28+4 = 28:32 = 7:8
5. The ratio of the age of a man & his wife is 4:3.After 4 years this ratio will be 9:7. If the time of marriage the ratio was 5:3,
then how many years ago were they married.
Solution: The age of a man is 4x .
The age of his wife is 3x.
After 4 years their ratio’s will be 9:7 i.e
4x+4 / 3x+4 = 9 / 7
28x-27x=36-28
x = 8.
Age of a man is 4x = 4*8 = 32 years.
Age of his wife is 3x = 3*8 = 24 years.
Let us assume ‘y’ years ago they were married ,
the ratio was 5:3 ,i.e
32-y / 24-y = 5/ 3
y=12 years
i.e 12 years ago they were married
6. Sneh’s age is 1/6th of her father’s age.Sneh’s father’s age will be twice the age of Vimal’s age after 10 years. If Vimal’s eight
birthday was celebrated two years before,then what is Sneh’s
present age.
a) 6 2/3 years b) 24 years c) 30 years d) None of the above
Solution: Assume Sneh’s age is ‘x’ years.
Assume her fathers age is ‘y’ years.
Sneh’s age is 1/6 of her fathers age i.e x = y /6.
Father’s age will be twice of Vimal’s age after 10
years.
i.e y+10 = 2( V+10)( where ‘V’ is the Vimal’s age)
Vimal’s eight birthday was celebrated two years before,
Then the Vimal’s present age is 10 years.
Y+10 = 2(10+10)
Y=30 years.
Sneh’s present age x = y/6
x = 30/6 = 5 years.
Sneh’s present age is 5 years.
7.The sum of the ages of the 5 children’s born at the intervals of 3 years each is 50 years what is the age of the youngest child.
a) 4 years b) 8 years c) 10 years d)None of the above
Solution: Let the age of the children’s be
x ,x+3, x+6, x+9, x+12.
x+(x+3)+(x+6)+(x+9)+(x+12) = 50
5x+30 = 50
5x = 20
x=4.
Age of the youngest child is x = 4 years.
8. If 6 years are subtracted from the present age of Gagan and
the remainder is divided by 18,then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years,then what is Gagan’s present age.
a) 48 years b)60 years c)84 years d)65 years
Solution: Let us assume Gagan present age is ‘x’ years.
Anup age = 5-2 = 3 years.
(x-6) / 18 = 3
x-6 = 54
x=60 years
9.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my father was 26 years of age when i was born. If my sister was 4 years of age when my brotherwas born,then what was the age my father and mother respectively when my brother was born.
a) 32 yrs, 23yrs b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs
Solution: My brother was born 3 years before I was born & 4
years after my sister was born.
Father’s age when brother was born = 28+4 = 32 years.
Mother’s age when brother was born = 26-3 = 23 years.
Introduction:
‘BODMAS’ rule: This rule depicts the correct sequence in which
the operations are to be executed, so as to find out the value of
a given expression.
Here B stands for Bracket, O for Of, D for Division, M for
Multiplication, A for Addition and S for Subtraction.
First of all the brackets must be removed, strictly in the
order () , {} , [].
After removing the brackets, we want use the following operations:
1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction
Modulus of a real number:
Modulus of a real number is a defined as
|a| = a, if a>0 or -a, if a <>
Problems:
1.(5004 /139) – 6= ?
Sol: Expression = 5004/ 139 – 6 = 36 – 6 = 30;
2.What mathematical operations should come at the place of ? in the equation : (2 ? 6 – 12 / 4 + 2 = 11) ?
Sol: 2 ? 6 = 11 + 12 / 4 – 2
= 11 + 3 – 2
= 12
2 * 6 = 12
3.( 8 / 88) * 8888088 = ?
Sol : (1/11) * 8888088 = 808008
4.How many 1/8’s are there in 371/2 ?
Sol: (371/2) /(1/8)= (75/2) /(1/8) = 300
5.Find the values of 1/2*3 +1/3*4 +1/4*5+ ……………..+1/9*10 ?
Sol: 1/2*3 +1/3*4+1/4*5+ ………………+1/9*10
= [½ -1/3] +[ 1/3 – ¼] + [¼- 1/5] +……………+[1/9-1/10]
= [ ½ – 1/10]
= 4/15 = 2/5
6.The value of 999 of 995/999* 999 is:
Sol: [1000- 4/1000]*999 = 999000-4
= 998996
7.Along a yard 225m long, 26 trees are planted at equal distance, one
tree being at each end of the yard. what is the distance between two
consecutive trees ?
Sol: 26 trees have 25 gaps between them.
Hence , required distance = 225/ 25 m= 9m
8.In a garden , there are 10 rows and 12 columns of mango trees. the
distance between the two trees is 2 m and a distance of one meter is
left from all sides of the boundary of the length of the garden is :
Sol: Each row contains 12 plants.
leaving 2 corner plants, 10 plants in between have 10 * 2 meters and
1 meter on each side is left.
length = (20 + 2) m = 22m
9.Eight people are planning to share equally the cost of a rental car,
if one person with draws from the arrangement and the others share
equally the entire cost of the car, then the share of each of the
remaining persons increased by?
Sol: Original share of one person = 1/8
new share of one person = 1/7
increase = 1/7 – 1/8 = 1/56
required fractions = (1/56)/(1/8) = 1/7
10.A piece of cloth cost Rs 35. if the length of the piece would
have been 4m longer and each meter cost Re 1 less , the cost
would have remained unchanged. how long is the piece?
Sol: Left the length of the piece be x m.
then, cost of 1m of piece = Rs [35 / x]
35/ x – 35 /x+4 = 1
x + 4 – x = x(x+ 4)/35
x2 + 4x – 140 = 0
x= 10
11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews.
If each daughter receives four times as much as each nephew, and
each son receives five as much as each nephew. how much does each
daughter receive ?
Sol:
Let the share of each nephew be Rs x.
then, share of each daughter Rs 4x.
share of each son = 5x Rs
so, 5 *5x+ 4 * 4x + 2x =8600
2x + 16x + 25x= 8600
43x = 8600
x = 200
share of each daughter = 4 * 200 = Rs 800
12.A man spends 2/5 of his salary on house rent, 3/10 of his salary on food, and 1/8 of his salary on conveyance. if he has Rs 1400 left with him, find his expenditure on food and conveyance?
Sol: Part of the salary left = 1-[2/5 +3/10+1/9]
= 1- 33/40
=7/40
Let the monthly salary be rs x
then, 7/40 of x = 1400
x= [1400*40]/7
x= 8000
Expenditure on food = 3/10*8000 =Rs 2400
Expenditure on conveyance= 1/8*8000 =Rs 1000
Facts And Formulae:
1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :
The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly.
There are two methods :
1.Factorization method: Express each one of the given numbers as the product of prime factors.
The product of least powers of common prime factors gives HCF.
Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72
Sol: The prime numbers given common numbers are 2,5,7
Therefore HCF is 22 * 5 *72 .
2.Division Method : Divide the larger number by
smaller one. Now divide the divisor by remainder. Repeat the process
of dividing preceding number last obtained till zero is obtained as
number. The last divisor is HCF.
Example: Find HCF of 513, 1134, 1215
Sol:
1134) 1215(1
1134
———-
81)1134(14
81
———–
324
324
———–
0
———–
HCF of this two numbers is 81.
81)513(6
486
——–
27)81(3
81
—–
0
—
HCF of 81 and 513 is 27.
3.Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM.
There are two methods for this:
1.Factorization method : Resolve each one into product of prime factors.
Then LCM is product of highest powers of all factors.
2.Common division method.
Problems:
1.The HCF of 2 numbers is 11 and LCM is 693.If one of numbers is 77.find other.
Sol: Other number = 11 * 693/77=99.
2.Find largest number of 4 digits divisible by 12,15,18,27
Sol: The largest number is 9999.
LCM of 12,15,18,27 is 540.
on dividing 9999 by 540 we get 279 as remainder.
Therefore number =9999 - 279 =9720.
3.Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.
Sol:
20–14=6
25-19=6
35-29=6
40-34=6
Therefore number =LCM of (20,25,35,40) - 6=1394
4.252 can be expressed as prime as :
2 252
2 126
3 63
3 21
7
prime factor is 2 *2 * 3 * 3 *7
5.1095/1168 when expressed in simple form is
1095)1168(1
1095
——
73)1095(15
73
———
365
365
———
0
———-
So, HCF is 73
Therefore
1095/1168 = 1095/73/1168/73= 15/16
6.GCD of 1.08,0.36,0.9 is
Sol:
HCF of 108,36,90
36)90(2
72
—-
18)36(2
36
—-
0
—-
HCF is 18.
HCF of 18 and 108 is 18
18)108(6
108
——-
0
——–
Therefore HCF =0.18
7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.
Sol:
Let the numbers be x.
Three numbers are x,2x,3x
Therefore HCF is
2x)3x(1
2x
—–
x)2x(2
2x
——–
0
————-
HCF is x so, x is 12
Therefore numbers are 12,24,36.
8.The sum of two numbers is 216 and HCF is 27.
Sol:
Let numbers are
27a + 27 b =216
a + b =216/27=8
Co-primes of 8 are (1,7) and (3,5) numbers=(27 * 1 ), (27 * 7)
=27,89
9.LCM of two numbers is 48..The numbers are in ratio 2:3.The sum of numbers is
Sol:
Let the number be x.
Numbers are 2x,3x
LCM of 2x,3x is 6x
Therefore
6x=48
x=8.
Numbers are 16 and 24
Sum=16 +24=40.
10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is
Sol:
Let the numbers be x,4x
Then x * 4x = 84 * 21
x2 =84 * 21 /4
x = 21
Largest number is 4 * 21.
11.HCF of two numbers is 23,and other factors of LCM are 13,14.Largest number is
Sol:
23 * 14 is Largest number.
12.The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and pencils is ?
Sol:
HCF of 1001 and 910
910)1001(1
910
————
91)910(10
910
——–
0
———
Therefore HCF=91
13.The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ?
Sol: LCM of 5,6,4,3 is 60.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to added is 60 - 37 =23.
Answer is 23.
14.The least number which is a perfect square and is divisible by each of numbers 16,20,24 is ?
Sol: LCM of 16,20,24 is 240.
2 * 2*2*2*3*5=240
To make it a perfect square multiply by 3 * 5
Therefore 240 * 3 * 5=3600
Answer is 3600.
Natural Numbers:
All positive integers are natural numbers.
Ex 1,2,3,4,8,……
There are infinite natural numbers and number 1 is the least natural number.Based on divisibility there would be two types of natural numbers. They are Prime and composite.
Prime Numbers:
A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.
Properties Of Prime Numbers:
->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime.
->The remainder of division of the square of a prime number p>=5 divide by 24 is 1.
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1.
Process to Check A Number s Prime or not:
Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number will be prime.
Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.
Composite Numbers:
The numbers which are not prime are known as composite numbers.
Co-Primes:
Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11)…..
Place value or Local value of a digit in a Number:
place value:
Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.
Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.
Tests of Divisibility:
Divisibility by 2:-A number is divisible by 2,if its unit’s digit is
any of 0,2,4,6,8.
Example 84932 is divisible by 2,while 65935 is not.
Divisibility by 3:-A number is divisible by 3,if the sum of its digits is divisible by 3.
Example 1.592482 is divisible by 3,since sum of its digits
5+9+2+4+8+2=30 which is divisible by 3.
Example 2.864329 is not divisible by 3,since sum of its digits
8+6+4+3+2+9=32 which is not divisible by 3.
Divisibility by 4:-A number is divisible by 4,if the number formed by last two digits is divisible by 4.
Example 1.892648 is divisible by 4,since the number formed by the last
two digits is 48 divisible by 4.
Example 2.But 749282 is not divisible by 4,since the number formed by
the last two digits is 82 is not divisible by 4.
Divisibility by 5:-A number divisible by 5,if its unit’s digit is either 0 or 5.
Example 20820,50345
Divisibility by 6:-If the number is divisible by both 2 and 3.
example 35256 is clearly divisible by 2
sum of digits =3+5+2+5+21,which is divisible by 3
Thus the given number is divisible by 6.
Divisibility by 8:-A number is divisible by 8 if the last 3 digits of the number are divisible by 8.
Divisibility by 11:-If the difference of the sum of the digits in the odd places and the sum of the digitsin the even places is zero or divisible
by 11.
Example 4832718
(8+7+3+4) - (1+2+8)=11 which is divisible by 11.
Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.
Divisibility by 7,11,13:-The difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7,11,13.
BASIC FORMULAE:
->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c
DIVISION ALGORITHM
If we divide a number by another number ,then
Dividend = (Divisor * quotient) + Remainder
MULTIPLICATION BY SHORT CUT METHODS
1.Multiplication by distributive law:
a)a*(b+c)=a*b+a*c
b)a*(b-c)=a*b-a*c
Example
a)567958*99999=567958*(100000-1)
567958*100000-567958*1
56795800000-567958
56795232042
b)978*184+978*816=978*(184+816)
978*1000=978000
2.Multiplication of a number by 5n:-Put n zeros to the right of the
multiplicand and divide the number so formed by 2n
Example 975436*625=975436*54=9754360000/16=609647500.
PROGRESSION:
A succession of numbers formed and arranged in a definite order according to certain definite rule is called a progression.
1.Arithmetic Progression:-If each term of a progression differs from its
preceding term by a constant.
This constant difference is called the common difference of the A.P.
The n th term of this A.P is Tn=a(n-1)+d.
The sum of n terms of A.P Sn=n/2[2a+(n-1)d].
Important Results:
a.1+2+3+4+5………………….=n(n+1)/2.
b.12+22+32+42+52………………….=n(n+1)(2n+1)/6.
c.13+23+33+43+53………………….=n2(n+1)2/4
2.Geometric Progression:-A progression of numbers in which every
term bears a constant ratio with ts preceding term.
i.e a,a r,a r2,a r3……………
In G.P Tn=a r n-1
Sum of n terms Sn=a(1-r n)/1-r
Problems
1.Simplify
a.8888+888+88+8
b.11992-7823-456
Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713
2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114
Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)
3.Simplify: a.5793405*9999 b.839478*625
Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.
4.Evaluate 313*313+287*287
Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338
5.Which of the following is a prime number?
a.241 b.337 c.391
Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.
6.Find the unit’s digit n the product 2467 153 * 34172?
Solution: Unit’s digit in the given product=Unit’s digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit’s digit in the product =7*1=7.
7.Find the total number of prime factors in 411 *7 5 *112 ?
Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29
8.Which of the following numbers s divisible by 3?
a.541326
b.5967013
Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.
9.What least value must be assigned to * so that th number 197*5462 is divisible by 9?
Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
10.What least number must be added to 3000 to obtain a number exactly divisible by 19?
Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.
11.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
12.On dividing 15968 by a certain number the quotient is 89 and the remainder is 37.Find the divisor?
Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.
13.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder?
Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.
14.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed?
Solution:Let the number be x.
3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.
15.Find the remainder when 231 is divided by 5?
Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
16.How many numbers between 11 and 90 are divisible by 7?
Solution:The required numbers are 14,21,28,………..,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
17.Find the sum of all odd numbers up to 100?
Solution:The given numbers are 1,3,5………99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
18.How many terms are there in 2,4,6,8……….,1024?
Solution:Clearly 2,4,6……..1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.
19.2+22+23+24+25……….+28=?
Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.
20.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.The positive integer is?
a.1 b.3 c.5 d.7
Solution:1000+N>1000N
clearly N=1.
21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?
a.18 b.22 c.36 d. none
Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.
22.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d67.
Solution:x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15.
The smaller number is?
a.240 b.270 c.295 d.360
Solution:Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
25.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20
Solution:Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
No comments:
Post a Comment