Practice Addition subtraction product on ratio and proportion - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   The ratio between two numbers is 3 : 4. If each number is increased by 6, the ratio becomes 4 : 5. The difference between the numbers is

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be 3x and 4x.

${3x + 6}/{4x + 6} = 4/5$

16x + 24 = 15x + 30

x = 30 - 24 = 6

Required difference = 6

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 = 3, b= 4, x = 6

c = 4, d = 5

The numbers are = ${xa(c-d)}/{ad-bc}$

= ${6.3(4 - 5)}/{3 ×5 - 4 × 4}$

= ${18 × -1}/{15 - 16}$ = 18

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

= ${6 × 4(4 - 5)}/{3 × 5 - 4 × 4}$

= ${24 × (-1)}/{15 - 16}$ = 24

Numbers are 24 and 18.

Their difference = 24 - 18 = 6

Q-2)   When a particular number is subtracted from each of 7, 9, 11 and 15, the resulting numbers are in proportion. The number to be subtracted is :

(a)

(b)

(c)

(d)

Explanation:

Let the number to be subtracted be x.

According to the question,

${7 - x}/{9 - x} ={11 - x}/{15 - x}$

Now, check through options

Clearly, putting x = 3,

Each ratio = $2/3$.

Note : Solve such questions orally by mental exercise.

Using Rule 32
Let 'x' be a number which is subtracted from a, b, c and d to make them proportional, then
x = ${ad - bc}/{(a+d) - (b+c)}$
Let 'x' be a number which is added to a, b, c and d to make them proportional, then
x = ${bc - ad}/{(a+d) - (b+c)}$
Here, a, b, c and d should always be in ascending order.

The number will be x

= ${ad - bc}/{(a+d) - (b+c)}$

= ${7 × 15 - 9 × 11}/{(7 + 15) - (9 + 11)}$

= ${105 - 99}/{22 - 20} = 6/2$ = 3

Q-3)   The number to be added to each of the numbers 7, 16, 43, 79 to make the numbers in proportion is

(a)

(b)

(c)

(d)

Explanation:

From the given options number = 5, because

${7 + 5}/{16 + 5} = {43 + 5}/{79 + 5}$

$12/21 = 48/84$

[check other options likewise]

Using Rule 32,

Here, a = 7, b=16, c = 43, d= 79

Required number

x= ${bc - ad}/{(a+d) - (b+c)}$

= ${16 × 43 - 7 × 79}/{(7 + 79) - (16 + 43)}$

= ${688 - 553}/{86 - 79}$

= $35/7$ = 5

Q-4)   Which number when added to each of the numbers 6, 7, 15, 17 will make the resulting numbers proportional ?

(a)

(b)

(c)

(d)

Explanation:

Let required number be x.

${6 +x}/{7 + x} = {15 + x}/{17 + x}$

$102 + 17x + 6x + x^2$

= $105 + 7x + 15x + x^2$

23x - 22x = 105 - 102

x = 3

Note : It is convenient to solve it orally using options

${6 + 3}/{7 + 3} = {15 + 3}/{17 + 3}$

= $9/10 = 18/20$

Using Rule 32
Let 'x' be a number which is subtracted from a, b, c and d to make them proportional, then
x = ${ad - bc}/{(a+d) - (b+c)}$
Let 'x' be a number which is added to a, b, c and d to make them proportional, then
x = ${bc - ad}/{(a+d) - (b+c)}$
Here, a, b, c and d should always be in ascending order.

Required Number

= ${bc - ad}/{(a+d) - (b+c)}$

Where a = 6, b = 7, c = 15, d = 17

= ${7 × 15 - 6 × 17}/{(6 + 17) - (7 + 15)}$

= ${105 - 102}/{23 - 22}$ = 3

Q-5)   What number should be subtracted from both terms of the ratio 15 : 19 in order to make it 3 : 4 ?

(a)

(b)

(c)

(d)

Explanation:

Let x be subtracted from each term of $15/19$.

${15 - x}/{19 - x} = 3/4$

57 - 3x = 60 - 4x

x = 3

Q-6)   What number should be added to each of 6, 14, 18 and 38 so that the resulting numbers make a proportion ?

(a)

(b)

(c)

(d)

Explanation:

${6 + x}/{14 + x} ={18 +x}/{38 + x}$

From the given alternatives

${6 + 2}/{14 + 2} = {18 + 2}/{38 + 2}$

$1/2 = 1/2$

Using Rule 32,

Here, a = 6, b = 14, c = 18, d = 38

Required number x

= ${bc - ad}/{(a+d) - (b+c)}$

= ${14 × 18 - 6 × 38}/{(6 + 38) - (14 + 18)}$

= ${252 - 228}/{44 - 32}$

= $24/12$ = 2

Q-7)   If two numbers are in the ratio 2 : 3 and the ratio becomes 3 : 4 when 8 is added to both the numbers, then the sum of the two numbers is

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be 2x and 3x respectively.

According to the question,

${2x + 8}/{3x + 8}= 3/4$

9x + 24 = 8x + 32

9x - 8x = 32 - 24 = 8

x = 8

Sum of numbers

= 2x + 3x = 5x

= 5 × 8 = 40

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 = 2, b = 3, x= 8,

c = 3, d = 4

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

= ${8 × 2(3 - 4)}/{2 × 4 - 3 × 3}$

= ${- 16}/{- 1}$ = 16

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

= ${8 × 3(3 - 4)}/{2 × 4 - 3 × 3}$

= ${- 24}/{- 1}$ = 24

Sum of numbers = 16 + 24 = 40

Q-8)   Two numbers are in the ratio 2 : 3. If 2 is subtracted from the first and 2 is added to the second, the ratio becomes 1 : 2. The sum of the numbers is :

(a)

(b)

(c)

(d)

Explanation:

Let the number be 2x and 3x.

Then. ${2x -2}/{3x + 2} = 1/2$

4x - 4 = 3x + 2

x = 6

Sum of numbers = 5x

= 5 × 6 = 30

Q-9)   Of the three numbers, the ratio of the first and the second is 8 : 9 and that of the second and third is 3 : 4. If the product of the first and third number is 2400, then the second number is :

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be a, b and c.

Now, a : b = 8 : 9

b : c = 3 : 4

a : b : c

= 8 × 3 : 9 × 3 : 9 × 4

= 24 : 27 : 36 = 8 : 9 : 12

$a/8 = b/9 = c/12= k$

a = 8k, b = 9k, c = 12k

According to the question,

8k × 12k = 2400

$k^2 = 2400/{8 × 12}$ = 25

k = 5

Second number

= 9k = 9 × 5 = 45

Q-10)   The product of two positive integers is 1575 and their ratio is 9 : 7. The smaller integer is

(a)

(b)

(c)

(d)

Explanation:

Let the integers be 9x and 7x respectively.

According to the question,

9x × 7x = 1575

$x^2 = 1575/63$

$x^2 = 25 ⇒ x = 5

[x being positive (+ve) integer]

Smaller integer

= 7x = 7 × 5 = 35