# NUMBER WORD PROBLEMS WITH SOLUTIONS

Problem 1 :

The sum of 5 times a number and 8 is 48. Find the number.

Solution :

Let x be the number.

5x + 8 = 48

5x = 40

x = 8

Therefore, the number is 8.

Problem 2 :

A number consists of three digits of which the middle one is zero  and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297. Find the number.

Solution :

Let x0y be the required three digit number.

Given : The sum of the other digits is 9.

x + y = 9

y = 9 - x ----(1)

Given : The number formed by interchanging the first and third digits is more than the original number by 297.

y0x - x0y = 297

[100(y) + 10(0) + 1(x)] - [100(x) + 10(0) + 1(y)] = 297

(100y + 0 + x) - (100x + 0 + y) = 297

(100y + x) - (100x + y) = 297

100y + x - 100x - y = 297

-99x + 99y = 297

Divide both sides by 99.

-x + y = 3 ----(2)

Substitute y = 9 - x.

-x + 9 - x = 3

-2x + 9 = 3

-2x = -6

x = 3

Substitute x = 3 into (1).

y = 9 - 3

y = 6

x0y = 306

Therefore, the three-digit number is 306.

Problem 3 :

Of two numbers, th of a the greater equal to rd of the smaller and their sum is 16. Find the numbers.

Solution :

Let x and y be the two numbers such that x > y.

Given : th of a the greater equal to rd of the smaller.

()x  =  (y

3x = 5y

3x - 5y = 0 -----(1)

Given : The sum of the two numbers is 16.

x + y = 16

y = 16 - x ----(2)

Substitute x = 16 - y into (1)

3x - 5(16 - x) = 0

3x - 80 + 5x = 0

8x - 80 = 0

8x = 80

x = 10

Substitute x = 10 into (2).

y = 16 - 10

y = 6

So, the two numbers are 10 and 6.

Problem 4 :

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number.

Solution :

Let xy be the number between 10 and 100.

Given : The number between 10 and 100 is five times the sum of its digits.

xy = 5(x + y)

10(x) + 1(y) = 5x + 5y

10x + y = 5x + 5y

5x - 4y = 0 ----(1)

Given : When 9 is added to the number, the digits are reversed.

xy + 9 = yx

10(x) + 1(y) + 9 = 10(y) + 1(x)

10x + y + 9 = 10y + x

9x - 9y = -9

x - y = -1

x = y - 1 ----(2)

Substitute x = y - 1 into (1).

5(y - 1) - 4y = 0

5y - 5 - 4y = 0

y - 5 = 0

y = 5

Substitute y = 5 into (2).

x = 5 - 1

x = 4

xy = 45

Therefore, the required number is 45.

Problem 5 :

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

Solution :

Let x and y be the two numbers such that x > y.

Given : One number is greater than thrice the other number by 2.

x = 3y + 2 ----(1)

Given : 4 times the smaller number exceeds the greater by 5.

4y - x = 5

Substitute x = 3y + 2.

4y - (3y + 2) = 5

4y - 3y - 2 = 5

y = 7

Substitute y = 7 into (1).

x = 3(7) + 2

x = 23

Therefore the two numbers are 23 and 7.

Problem 6 :

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the two digit number.

Solution :

Let xy be the required two digit number.

Given : The two digit number is 7 times the sum of its digits.

xy = 7(x + y)

10(x) + 1(y) = 7x + 7y

10x + y = 7x + 7y

3x = 6y

x = 2y ----(1)

Given : The number formed by reversing the digits is 18 less than the given number.

xy - yx = 18

[10(x) + 1(y)] - [10(y) + 1(x)] = 18

(10x + y) - (10y + x) = 18

10x + y - 10y - x = 18

9x - 9y = 18

x - y = 2

Substitute x = 2y into (2).

2y - y = 2

y = 2

Substitute y = 2 into (1).

x = 2(2)

x = 4

xy = 42

Therefore, the two digit number is 42.

Problem 7 :

If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

Solution :

Let x be the number.

x = 296k + 75

where k is quotient, when x is divided by 296.

In the above sentence we have 296 is multiplied by the constant k, 75 is added to that. In this form , we consider the number 75 as remainder, when the number x is divided by 296.

We want to find the remainder, when we divide the number x by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.

So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.

x = 37×8k + 37×2 + 1

x = 37(8k + 2) + 1

Therefore, the remainder is 1 when the number x is divided by 37.

Problem 8 :

Find the least number which when divided by 35, leaves a remainder 25, when divided by 45, leaves a remainder 35 and when divided by 55, leaves a remainder 45.

Solution :

For each divisor and corresponding remainder, we have to find the difference.

35 - 25 = 10

45 - 35 = 10

55 - 45 = 10

We get the difference 10 (for all divisors and corresponding remainders)

Find the least common multiple of (35, 45, 55) and subtract the difference from the least common multiple.

Least common multiple of (35, 45, 55) = 3465

Therefore, the required least number is

= 3465 - 10

= 3455

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