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Aptitude Made Easy | Problems on Numbers – codewindow.in

Various competitive examinations ask questions regularly based on Problems on
Numbers.
Many students face challenges in understanding and solving problems as they are not
able to covert problem statements from words into equation of numbers.
Let us take real life example and will try to understand concept of Problem of numbers.

Real life example for Problems on Numbers:

Assume you have 1000 Rs with you and your friend has some amount of Rs. Total
money you and your friend has is 2000 Rs. What is amount your friend have?
You will feel this thing very simple 1000 + 1000 = 2000
So your friend has 2000 Rs.
We just need to solve similar problems but before solving we just need to convert
problem statement given in words to equation.
Here we can say equation as:
Money you have + Money friend have = 2000
Assume your friend has x Rs with him/her.
1000 + x = 2000
x = 1000

So basically, if we are able to convert problem statement into equation we have solved
80% of the problem. We will practice initially for this before solving problem. Assume original number as x and we will convert problem statement into equation.

Practice these similar statements and most of the time in numbers problem, you will
either see 1 or more number of statement combination.
First step in problem would be convert problem statement into equation and then solve
it.
Some places it might be easy to substitute answers given in option and check whether
our equation satisfies it or not.
We will solve problems now.

Problem 1:

The sum of rational number and its reciprocal is 13/6. Find the number.

Solution :

Let us assume number as x

Reciprocal of number = 1/x

x + 1/x = 13/6

(x²+1)/x = 13/6

6x² + 6 = 13x

6x² – 13x + 6 = 0

Note: I will create 1 separate document on solving quadratic equations

6x² – 9x – 4x + 6 = 0

3x(2x – 3) – 2(2x – 3) = 0

(2x – 3)(3x – 2) = 0

So either (2x – 3) = 0 or (3x – 2) = 0
x = 2/3 or x = 3/2
Answer is number is 2/3 or 3/2

Problem 2:

The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers?

Solution :

Let us assume number as x and y
Difference of 2 numbers = 11
x – y = 11 (Equation 1)

1/5(x + y) = 9

x + y = 45 (Equation 2)
Adding 2 equations:
x – y = 11 (Equation 1)
x + y = 45 (Equation 2)
————————-
2x = 56
x = 28

Put value of x in equation 1
x – y = 11
28 – y = 11
y = 17

Answer is the numbers are 28 and 17

Problem 3:

The sum of two numbers is 15 and sum of their squares is 113. Find the numbers?

Solution :

Let us assume number as x and y
Sum of number is 15
x + y = 15 (Equation 1)
Sum of squares is 113

x² + y² = 113 (Equation 2)

We need to represent y in form of x so that we can get equation to solve.
As x +y = 15 means y = 15 – x
Put this value of y in Equation 2 

x² + (15 – x)² = 113

x² + 225 – 30x + x² = 113

2x² – 30x + 112 = 0

2x² – 16x – 14x + 112 = 0

2x(x – 8) – 14(x – 8) = 0

(2x – 14) (x – 8) = 0

So 2x – 14 = 0 or x – 8 = 0
2x =14 or x = 8
x=7

When x =7 then y = 15 -7 =8 and vice versa [you can cross check your answer by putting these values in Equation1 and Equation 2)

Answer is numbers are 7 and 8 or 8 and 7

Problem 4:

The average of 4 consecutive even numbers is 27. Find the largest of these numbers.

Solution :

As you can see numbers are consecutive even.
Let us assume 1st smallest number x so next numbers would be x+2 , x+4 and x+6
Average of 4 number is 27
So total of number would be 4 * 27 =108
x + (x + 2) + (x + 4) + (x + 6) = 108
4x + 12 = 108
4x = 96
x = 24
So smallest number is 24
Largest number = x+ 6 = 24 + 6 = 30
Answer is largest number is 30

Problem 5:

Sum of squares of three consecutive odd number is 2531. Find the numbers.

Solution :

As you can see numbers are consecutive odd.
Let us assume 1st smallest number x so next numbers would be x+2 , x+4
Sum of squares of these three numbers = 2531

x² + (x + 2)² + (x + 4)² = 2531

x² + x² + 4x +4 + x² + 8x + 16 = 2531

3x² + 12x + 20 = 2531

3x² + 12x + 2511 = 0

Divide by 3 throughout

x² + 4x − 837 = 0

x² + 31x − 27x − 837 = 0
x(x + 31) – 27(x + 31) = 0
(x + 31)(x – 27) = 0
x = 27 or x = -31

Take positive number
Smallest number = x = 27
Next number = x + 2 = 27 + 2 =29
Last number = x + 4 = 27 + 4 = 31

Answer is numbers are 27, 29 and 31.

Problem 6:
The difference between a number and its three-fifth is 50. What is the number?

Solution :
Let us assume number as x
Difference between number and its 3/5th is 50.

x – 3x/5 = 50

x(1 – 3/5) = 50

x (2/5) = 50

2x = 250

x = 125

Answer is number is 125

Problem 7:
If one-third of one fourth of a number is 15. Then three-tenth of number is?
Solution :
Let us assume number as x
one-third of one fourth of a number = 15

1/3 * 1/4 * x = 15

x/12 = 15

x = 180
Three tenth of number:

= 3/10 * 180 = 3 * 18 = 54

Answer is three-tenth of number is 54

Problem 8:

A number is doubled and 9 is added. If the resultant is trebled, it becomes 75. What is
that number?

Solution :

Let us assume number as x
Number is doubled and 9 added = 2x + 9
Resultant is trebled = 3 * (2x + 9) = 6x + 27 = 75
6x + 27 =75
6x = 48
x = 8

Answer is number is 8

Problem 9:

When 24 is subtracted from a number, it reduces to its four-seventh. What is the sum of
digits of that number?

Solution :

Let us assume number as x.
24 subtracted from number : x – 24

4/7th  of number : 4x/7

x – 24 = 4x/7

x – 4x/7 = 24

3x/7 = 24

3x = 24 *  7

x = 8 * 7

x = 56

The number is 56
Sum of digits of 56= 5 + 6 = 11

Answer is sum of digits of number is 11

Problem 10:

Find number which when multiplied by 15 is increased by 196.

Solution :

Let us assume number as x.
Number is multiplied by 15 : 15x
It is increased by 196.
This means :
Number multiplied by 15 – number = 196
15x – x = 196
14x = 196
x = 14
Answer is number is 14

Problem 11:

A number whose 5th part increased by 4 is equal to its fourth part diminished by 10, is?
number which when multiplied by 15 is increased by 196.

Solution :

Let us assume number as x.

5th part of the number: x/5

5th part Increased by 4: (x/5) + 4 —————- Equation 1

4th part of number : x/4

4th part of number diminished by 10 : (x/4) – 10 —————- Equation 2

As per problem statement both equation 1 and equation 2 are equal

(x/5) + 4 = (x/4) -10

(x/4) – (x/5) = 4 + 10

x((1/4) – (1/5)) = 14

x(1/20) = 14

x = 14*20

x = 280

Answer is number is 280

Problem 12:

The sum of two numbers is 25 and their difference is 13. Find their product.

Solution :

Let us assume numbers are x and y.
Sum of numbers is 25: x + y = 25  ——————– Eqaution1
Difference of numbers is 13: x – y = 13 ———————– Equation2

Add Equation1 and Equation2
x + y = 25
+ x – y = 13
—————
2x = 38
x = 19

Put value of x in Equation1

x + y = 25
19 + y = 25
y = 6
Numbers are 19 and 6
Product of numbers: 19 * 6 = 114
Answer is product of numbers is 114

Problem 13:

Three numbers are in ratio 4 : 5 : 6 and their average is 25. Largest number is?

Solution :

Numbers are in proportion 4 : 5 : 6 so largest number would be 3rd (with 6 in ratio)
Let us assume common multiple of ratio as x.
So our numbers are 4x : 5x : 6x
Average of number is 25 means their sum is 25 * 3 = 75
4x + 5x + 6x = 75
15x = 75
x = 5 [Common multiple is 5]
Largest number = 6x = 6 * 5 = 30

Answer is largest number is 30

Problem 14: 

Find positive integer which when increased by 17 is equal to 60 times the reciprocal of
the number.

Solution :

Let us assume number as x.

Reciprocal of number : 1/x

Integer increased by 17 : x + 17   ——————- Equation 1
60 times of reciprocal: 60 * 1/x = 60/x  ——————- Equation 2

As per problem statement Equation 1 and Equation 2 are equal

x + 17 = 60/x

x(x + 17) = 60

x² + 17x – 60 = 0

x² + 20x – 3x – 60 = 0

x(x + 20) – 3(x + 20) = 0

(x + 20)(x – 3) = 0

x – 3 = 0  or  x + 20 = 0
x = 3  or  x = -20

As per problem statement, we need positive integer.
So x = 3

Answer is number is 3

Problem 15:

The sum of four consecutive even integers is 1284. The greatest of them is ?

Solution :

Let us assume smallest number as x.
As next consecutive even would be x + 2, x + 4, x + 6.
Sum of 4 consecutive digit is 1284
x + (x + 2) + (x + 4) + (x + 6) = 1284
4x + 12 = 1284
4x = 1272
x = 318
Smallest number is 318.
Largest number : x + 6 = 318 + 6 = 324

Answer is largest number is 324

Problem 16:

What is the sum of consecutive even numbers, the difference of whose square is 84?

Solution :

Let us assume smallest number as x.
Next even number would be x + 2
Difference of square is 84

(x + 2)² – x² = 84

x² + 4x + 4 – x² = 84

4x + 4 = 84

4(x + 1) = 84

x + 1 = 21

x = 20

Smaller number = x = 20
Next number = x + 2 = 20 + 2 = 22
Numbers are 20 and 22.
Sum of numbers = 20 + 22 = 42

Answer is sum of number is 42

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