# SOLVING WORD PROBLEMS ON NUMBERS

Problem 1:

The sum of two numbers is 90  and they are in the  ratio is 4:5. Find those numbers.

(A) 20 and 70                          (B) 40 and 50

(C) 35 and 55                            (D)  28 and 62

Solution :

Let x and 90 - x be the second number.

x : (90- x)  =  4 : 5

x/(90-x)  =  4/5

5x  =  4(90 - x)

5x  =  360 - 4x

5x + 4x  =  360

9x  =  360

x  =  40

So, the first number x = 40

and the second number 90 - x => 90 - 40 ==> 50

Therefore the correct option is (B) 40 and 50.

Problem 2 :

If the same numbers is added to both the numerator and denominator of a fraction 3/5, then the result is 3/4. Find the number.

(A) 3        (B) 4      (C) 5    (D)  8

Solution :

Let x be the required number.

If it is added with numerator and denominator of the fraction 3/5. It will become (3+x)/(5+x).

(3+x)/(5+x)  =  3/4

4(3 + x)  =  3 (5 + x)

12 + 4x  =  15 + 3x

4 x - 3 x  =  15 - 12

x  =  3

Therefore the correct option is (A) 3

Problem 3 :

Two numbers are such that the ratio between them is 5 : 7. When 7 is added to each of them, the ratio becomes 2 : 3. Find the numbers.

(A) 40 and 56                            (B) 10 and 14

(C) -35 and -49                         (D)  30 and 42

Solution :

Let 5x and 7x be two numbers.

By adding 7, we get 5x + 7 and 7x + 7.

(5x + 7) : (7x + 7)  =  2 : 3

(5x + 7)/(7x + 7)  =  2/3

3 (5x + 7)  =  2 (7x + 7)

15x + 21  =  14x  + 14

15x - 14x  =  14 - 21

x = -7

5x  ==>  5(-7) ==>  -35

7x  ==>  7(-7) ==>  -49

Therefore the correct option is (C) -35 and -49

Problem 4 :

The sum of two numbers is 135 and they are in the ratio 7 : 8 . Find those numbers.

(A) 28 and 32                           (B)  35 and 40

(C) 63 and 72                           (D)  42 and 56

Solution :

Let 7x and 8x be two numbers.

Sum of two numbers  =  135

7x + 8x  =  135

15 x  =  135

x  =  135/15

x  =  9

7x ==> 7(9)  ==> 63

8x ==> 8(9)  ==> 72

Therefore the correct option is (C) 63 and 72

Problem 5 :

The difference of two numbers is 72 and the quotient obtained by dividing the one by other is 3. Find the numbers.

(A) 19 and 57                           (B)  36 and 108

(C) 163 and 72                          (D)  54 and 156

Solution :

Let x and y be the two numbers.

The difference between two numbers  =  72

So,  x - y  =  72

x  =  72 + y ------ (1)

If we divide first number by the second number, we will get 3 as quotient.

So,  x/y  =  3  --- (2)

Applying the value of x in the second equation

(72 + y)/y  =  3

72 + y  =  3y

3y - y  =  72

2y  =  72

y  =  36

Substituting the value of y in the first equation,we will get

x  =  72 + 36

x  =  108

Therefore the correct option is (B) 36 and 108.

Problem 6 :

In a certain fraction, the denominator is 4 less than the numerator. If the number 3 is added to both the numerator and denominator, the resulting fraction is equal to  9/7, find the original fraction.

(A)  7/9                                    (B)  8/19

(C) 2/21                                   (D)  15/11

Solution :

Let x be the numerator, then x - 4 be the denominator.

(x + 3)/(x - 4 + 3)  =  9/7

(x + 3)/(x - 1)  =  9/7

7(x + 3)  =  9 (x - 1)

7x + 21  =  9 x - 9

7x - 9x  =  -9 - 21

-2x  =  - 30

x  =  -30/(-2)

x  =  15

So, the required fraction is 15/11. Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

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