type 7 finding sum difference product based ratio & proportion problems Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11 [SSC CGL Tier-I 2010]

Two numbers are in the ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. The smaller number is

a) 66

b) 77

c) 49

d) 39

Answer: (c)

Let the numbers be 7x and 11x respectively.

${7x + 7}/{11x + 7} = 2/3$

22x + 14 = 21x + 21

x = 7

Smaller number

= 7x = 7 × 7 = 49

Using Rule 34
Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d. Then the two numbers will be
${xa(c-d)}/{ad-bc}$ and ${xb(c-d)}/{ad-bc}$

Here, a = 7, b = 11, x = 7, c= 2, d = 3

1st Number = ${xa(c-d)}/{ad-bc}$

= ${7 × 7(2 - 3)}/{7 × 3 - 11 × 2}$

= ${49 × -1}/{21 - 22}$ = 49

2nd Number = ${xb(c-d)}/{ad-bc}$

= ${7 × 11(2 - 3)}/{7 × 3 - 11 × 2}$

= ${77 × -1}/{21 - 22}$ = 77

Smallest number = 49

Question : 12 [SSC Constable 2012]

Two numbers are such that the ratio between them is 4 : 7. If each is increased by 4, the ratio becomes 3 : 5. The larger number is

a) 56

b) 64

c) 48

d) 36

Answer: (a)

Let the numbers be 4x and 7x.

${4x + 4}/{7x + 4} = 3/5$

21x + 12 = 20x + 20

21x - 20x = 20 - 12

x = 8

Larger number

= 7x = 7 × 8 = 56

Using Rule 34,

a = 4, b = 7, c = 3,

d = 5, x = 4

Larger number = ${xb(c-d)}/{ad-bc}$

= ${4 ×7(3 -5)}/{4 × 5 - 3 × 7}$

= ${4 × 7 × (-2)}/{20 - 21}$= 56

Question : 13 [SSC CGL Prelim 2007]

In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the school, the ratio of boys and girls becomes

a) 8 : 7

b) 4 : 3

c) 10 : 7

d) 12 : 7

Answer: (b)

Using Rule 21,

Initially number of boys

= $8/{8 + 5} × 286 = 8/13 × 286$ = 176

Number of girls

= $5/13 × 286$ = 110

22 more girls get admitted.

∴ Required ratio

= $176/{110 + 22} = 176/132 = 4/3$ = 4 : 3

Question : 14 [SSC CPO S.I.2003]

Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is

a) 15

b) 18

c) 12

d) 6

Answer: (c)

Let the numbers be 5x and 7x.

Now, ${5x - 9}/{7x - 9} = 7/11$

11 (5x - 9) = 7 (7x - 9)

55x - 99 = 49x - 63

55x - 49x = 99 - 63

6x = 36 ⇒ x = 6

Required difference

= 7x - 5x = 2x = 2 × 6 = 12

Using Rule 35
Two numbers are in the ratio a:b and if x is subtracted from each number the ratio becomes c:d. The two numbers will be
= ${xa(d-c)}/{ad-bc}$ and ${xb(d-c)}/{ad-bc}$

Here, a = 5, b = 7, x = 9, c = 7, d = 11

1st Number = ${xa(d-c)}/{ad-bc}$

= ${9 ×5(11 - 7)}/{5 ×11 - 7 × 7}$

= ${45 × 4}/{55 - 49}$

= ${45 × 4}/6$= 30

2nd Number = ${xa(d-c)}/{ad-bc}$

= ${9 ×7(11 - 7)}/{5 ×11 - 7 × 7}$

= ${63 × 4}/{55 - 49}$

= ${63 × 4}/6$= 42

Their difference = 42 - 30 = 12

Question : 15 [SSC CGL Tier-II 2015]

The average of 11 numbers is 36, whereas average of 9 of them is 34. If the remaining two numbers are in the ratio of 2 : 3, find the value of the smaller number (between remaining two numbers).

a) 54

b) 36

c) 48

d) 45

Answer: (b)

According to the question,

Sum of remaining two numbers

= 11 × 36 - 9 × 34

= 396 - 306 = 90

Ratio of the remaining two numbers = 2 : 3

Smaller number

= $2/5$ × 90 = 36

IMPORTANT quantitative aptitude EXERCISES

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