# SOLVING WORD PROBLEMS ON NUMBERS

Problem 1 :

Twice a number is 500 more than six times the number. What is the number?

Solution :

Let x be the number.

2x = 6x + 500

2x - 6x = 500

-4x = 500

x = -500/4

x = -125

So, the required number is -125.

Problem 2 :

Three-sevenths of a number is 24. Find the number.

Solution :

Let x be the number.

3/7 of x = 24

(3/7)  x= 24

x = 24 (7/3)

x = 8 ⋅ 7

x = 56

So, the required number is 56.

Problem 3 :

What number increased by ¼ of itself is equal to 30?

Solution :

Let x be the required number.

x + (1/4) of x = 30

x + x/4 = 30

(4x + x)/4 = 30

5x / 4 = 30

5x = 120

x = 120/5

x = 24

So, the required number is 24.

Problem 4 :

Find a number such that ¼ of the number is 50 less than 2/3 of the number.

Solution :

Let x be the number,

1/4 of x = (2/3) of x - 50

x/4 = (2x/3) - 50

x/4 = (2x - 150)/3

Doing cross multiplication, we get

3x = 4(2x - 150)

3x = 8x - 600

3x - 8x = -600

-5x = -600

x = 600/5

x = 120

So, the required number is 120.

Problem 5 :

The denominator of a fraction exceeds the numerator of a fraction by 25. The value of the fraction is 3/8 . Find the fraction.

Solution :

Let x be the numerator.

denominator = x + 25

x/(x + 25) = 3/8

8x = 3(x + 25)

8x = 3x + 75

8x - 3x = 75

5x = 75

x = 75/5

x = 15

So, the required fraction is 15/40

Problem 6 :

If 6 times a number is decreased by 6, the result is the same as when 3 times the number is increased by 12. Find the number.

Solution :

Let x be the number,

6x - 6 = 3(x + 12)

6x - 6 = 3x + 36

6x - 3x = 36 + 6

3x = 42

x = 42/3

x = 14

So, the required number is 14.

Problem 7 :

Separate 84 into two parts such that one part will be 12 less than twice the other.

Solution :

Let x be the number. Then the other number 2x - 12

Sum of those two numbers = 84

x + 2x - 12 = 84

3x - 12 = 84

3x = 84 + 12

3x = 96

x = 96/3

x = 32

2x - 12 ==> 2(32) - 12 ==> 52

So, the required numbers are 32 and 52.

Problem 8 :

The difference between two numbers is 24. Find the numbers if their sum is 88.

Solution :

Let x and x + 24 be two numbers.

Sum of the numbers = 88

x + x + 24 = 88

2x = 88 - 24

2x = 64

x = 64/2

x = 32

x + 24 ==> 32 + 24 ==> 76

So, the required numbers are 32 and 76.

Problem 9 :

One number is 3 times another number. If 17 is added to each, the first resulting number is twice the second resulting number. Find the two numbers.

Solution :

Let x and 3x be the numbers.

x + 17 and 3x + 17

3x + 17 = 2(x + 17)

3x + 17 = 2x + 34

3x - 2x = 34 - 17

x = 17

The other number = 3x ==> 3(17) = 51

So, the two numbers are 17 and 51.

Problem 10 :

The larger of two numbers is 1 less than 3 times the smaller. If 3 times the larger is 5 more than 8 times the smaller, find the numbers.

Solution :

Let x be the smaller number

Larger number = 3x - 1

3(3x - 1) = 8x + 5

9x - 3 = 8x + 5

9x - 8x = 5 + 3

x = 8

the other number = 3x - 1==> 3(8) - 1 ==> 23

So, the required numbers are 8 and 23.

Problem 11 :

The second of three numbers is one less than the first. The third number is 5 less than twice the second. If the third number exceeds the first number by 12, find the three numbers.

Solution :

Let x be the first number.

Second number = x - 1

Third number = 2(x - 1) - 5

2x - 7 = x + 12

2x - x = 12 + 7

x = 19

x - 1 = 18

2(x - 1) - 5 = 31

So, the required numbers are 18, 19 and 31.

Problem 12 :

One number is 4 more than 5 times another number. If 6 is added to each, the first resulting number is three times the second resulting number. Find the two numbers.

Solution :

Let x be the other number

one number = 5x + 4

6 is added to each, then x + 6 and 5x + 4 + 6

5x + 10 = 3(x + 6)

5x + 10 = 3x + 18

5x - 3x = 18 - 10

2x = 8

x = 4

5x + 4 ==> 5(4) + 4

= 24

So, the required numbers are 4 and 24.

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