Ratio & Proportion — Multiple Choice Questions

Which of the following is the ratio between a number and the number obtained by adding one-fifth of that number to it?

1. 4 : 5

2. 5 : 4

3. 5 : 6

4. 6 : 5

Answer

Let the number be x.

Given,

Number obtained by adding number and one-fifth of that number.

$\Rightarrow x + \dfrac{x}{5} \\[1em] \Rightarrow x\Big(1 + \dfrac{1}{5}\Big) \\[1em] \Rightarrow x\Big(\dfrac{5 + 1}{5}\Big) \\[1em] \Rightarrow \Big(\dfrac{6x}{5}\Big).$

Ratio = x : $\dfrac{6}{5}$x

= $\dfrac{x}{\dfrac{6}{5}x}$

= $\dfrac{5x}{6x}$

= $\dfrac{5}{6}$

= 5 : 6.

Hence, option 3 is the correct option.

Govind spends ₹ 8,100 in buying some jeans at ₹ 1,200 each and some shirts at ₹ 300 each. The ratio of the number of shirts to that of jeans, when the maximum possible number of jeans is purchased is:

1. 1 : 2

2. 1 : 4

3. 2 : 1

4. 5 : 7

Answer

Let the number of jeans Gopal brought be x and shirts be y.

Given,

Cost of jeans = ₹ 1,200.

Cost of shirt = ₹ 300.

Total amount spent = ₹ 8,100.

⇒ 1200x + 300y = 8100

⇒ 1200x + 300y - 8100 = 0

⇒ 300(4x + y - 27) = 0

⇒ 4x + y - 27 = 0

To maximize x keep y ≥ 0

⇒ 4x = 27

Maximum value of x can be 6.

Substitute value of x to get y,

⇒ 4(6) + y - 27 = 0

⇒ 24 + y - 27 = 0

⇒ y - 3 = 0

⇒ y = 3.

Ratio of shirt(y) : jeans(x) = 3 : 6 = 1 : 2.

Hence, option 1 is the correct option.

which of the following represents xy = 64?

1. 8 : x = 8 : y

2. x : 16 = y : 4

3. x : 8 = y : 8

4. 32 : x = y : 2

Answer

If we consider,

$\Rightarrow \dfrac{32}{x} = \dfrac{y}{2}$

⇒ 32(2) = xy

⇒ xy = 64.

Hence, option 4 is the correct option.

If a carton containing a dozen mirrors is dropped, then which of the following cannot be the ratio of broken mirrors to unbroken mirrors?

1. 2 : 1

2. 7 : 5

3. 3 : 2

4. 3 : 1

Answer

Let ratio of broken mirrors to unbroken mirrors = p : q

Number of mirrors in dozen = 12

Number of broken mirrors = $\dfrac{p}{p + q} \times 12$

Number of broken mirrors = $\dfrac{q}{p + q} \times 12$

If we consider 3 : 2,

3 + 2 = 5

$\dfrac{3}{5} \times 12 = 7.2$

Since 7.2 is not integer. Thus, 3 : 2 is the ratio that cannot exist.

Hence, option 3 is the correct option.

A mixture of paint is prepared by mixing 2 parts of red pigments with 5 parts of the base. Using the given information in the following table, find the values of a, b & c to get the required mixture of paint.

1. a = 10, b = 10, c = 10

2. a = 5, b = 2, c = 5

3. a = 10, b = 5, c = 10

4. a = 10, b = 5, c = 15

Answer

Given,

2 parts of red pigments is mixed with 5 parts of the base.

$\therefore \dfrac{4}{a} = \dfrac{2}{5} \\[1em] \Rightarrow a = \dfrac{4 \times 5}{2} = \dfrac{20}{2} = 10. \\[1em] \therefore \dfrac{b}{12.5} = \dfrac{2}{5} \\[1em] \Rightarrow b = \dfrac{2}{5} \times 12.5 = 5. \\[1em] \therefore \dfrac{6}{c} = \dfrac{2}{5} \\[1em] \Rightarrow c = \dfrac{6 \times 5}{2} = 3 \times 5 = 15.$

Hence, option 4 is the correct option.

If (x + 1) : 8 = 3.75 : 7 then value of x is :

1. $1\dfrac{2}{7}$

2. $2\dfrac{2}{7}$

3. $3\dfrac{2}{7}$

4. $4\dfrac{2}{7}$

Answer

Given,

⇒ (x + 1) : 8 = 3.75 : 7

$\Rightarrow \dfrac{x + 1}{8} = \dfrac{3.75}{7} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{3.75}{7} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{375}{700} \\[1em] \Rightarrow \dfrac{x + 1}{8} = \dfrac{15}{28} \\[1em] \Rightarrow x + 1 = \dfrac{15}{28} \times 8 \\[1em] \Rightarrow x + 1 = \dfrac{30}{7} \\[1em] \Rightarrow x = \dfrac{30}{7} - 1 \\[1em] \Rightarrow x = \dfrac{30 - 7}{7} \\[1em] \Rightarrow x = \dfrac{23}{7} \\[1em] \Rightarrow x = 3\dfrac{2}{7}.$

Hence, option 3 is the correct option.

If $\sqrt2:(1 + \sqrt3)::\sqrt6:x$ then value of x is:

1. $\sqrt3 + 3$

2. $1 - \sqrt3$

3. $\sqrt3 - 3$

4. $1 + \sqrt3$

Answer

Given,

$\Rightarrow \sqrt2:(1 + \sqrt3)::\sqrt6:x \\[1em] \Rightarrow \dfrac{\sqrt{2}}{1 + \sqrt3} = \dfrac{\sqrt6}{x} \\[1em] \Rightarrow x = \dfrac{\sqrt6(1 + \sqrt3)}{\sqrt2} \\[1em] \Rightarrow x = \sqrt3(1 + \sqrt3) \\[1em] \Rightarrow x = \sqrt3 + 3.$

Hence, option 1 is the correct option.

0.5 of a number is equal to 0.07 of another. The ratio of the numbers is :

1. 1 : 14

2. 5 : 7

3. 7 : 50

4. 50 : 7

Answer

Let the numbers be x and y.

Given,

⇒ 0.5x = 0.07y

⇒ $\dfrac{x}{y} = \dfrac{0.07}{0.5} = \dfrac{7}{50}$

⇒ x : y = 7 : 50

Hence, option 3 is the correct option.

25% of x is equal to 35% of y, then x : y equals to :

1. 5 : 7

2. 7 : 5

3. 13 : 15

4. 15 : 13

Answer

Given,

25% of x = 35% of y

$\Rightarrow \dfrac{25}{100}x = \dfrac{35}{100}y \\[1em] \Rightarrow 25x = 35y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{35}{25} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{7}{5}.$

Thus,

x : y = 7 : 5

Hence, option 2 is the correct option.

If A = $\dfrac{1}{3}$ B and B = $\dfrac{1}{2}$ C, Then A : B : C is equal to :

1. 1 : 2 : 6

2. 1 : 3 : 6

3. 1 : 2 : 3

4. 3 : 2 : 1

Answer

Given,

A = $\dfrac{1}{3}$B

B = 3A

B = $\dfrac{1}{2}$ C

C = 2B

C = 2(3A)

C = 6A

A : B : C = A : 3A : 6A

= 1 : 3 : 6.

Hence, option 2 is the correct option.

If 2p = 3q = 4r then p : q : r is equal to:

1. 2 : 3 : 4

2. 3 : 4 : 6

3. 4 : 3 : 2

4. 6 : 4 : 3

Answer

Let, 2p = 3q = 4r = k

p = $\dfrac{k}{2}$

q = $\dfrac{k}{3}$

r = $\dfrac{k}{4}$

$\Rightarrow p : q : r = \dfrac{k}{2}:\dfrac{k}{3}:\dfrac{k}{4} \\[1em] = \dfrac{k}{2} \times 12 : \dfrac{k}{3} \times 12 : \dfrac{k}{4} \times 12 \\[1em] = 6k : 4k : 3k \\[1em] = 6 : 4 : 3.$

Hence, option 4 is the correct option.

If $\dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{7}$, then $\dfrac{a + b + c}{c}$ = ?

1. $\dfrac{1}{2}$

2. $\dfrac{1}{7}$

3. 2

4. 7

Answer

Let, $\dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{7} = k$

We can express,

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

Substituting values in $\dfrac{a + b + c}{c}$, we get :

$\Rightarrow \dfrac{3k + 4k + 7k}{7k} \\[1em] \Rightarrow \dfrac{14k}{7k} \\[1em] \Rightarrow 2.$

Hence, option 3 is the correct option.

If x, 5.4, 5, 9 are in proportion, then x is equal to:

1. 3

2. 9.72

3. 25

4. $\dfrac{25}{3}$

Answer

Given,

x, 5.4, 5, 9 are in proportion.

⇒ $\dfrac{x}{5.4} = \dfrac{5}{9}$

⇒ $x = \dfrac{5}{9} \times 5.4$

⇒ x = 0.6 × 5

⇒ x = 3.

Hence, option 1 is the correct option.

The fourth proportional to 5, 8, 15 is :

1. 18

2. 20

3. 21

4. 24

Answer

Let forth proportional be x.

⇒ 5 : 8 = 15 : x

⇒ $\dfrac{5}{8} = \dfrac{15}{x}$

⇒ 5x = 8 × 15

⇒ x = $\dfrac{120}{5}$

⇒ x = 24.

Hence, option 4 is the correct option.

The fourth proportional to 0.12, 0.21 and 8 is :

1. 8.9

2. 14

3. 17

4. 56

Answer

Let fourth proportional be x.

⇒ 0.12 : 0.21 = 8 : x

⇒ $\dfrac{0.12}{0.21} = \dfrac{8}{x}$

⇒ 0.12x = 0.21 × 8

⇒ x = $\dfrac{1.68}{0.12}$

⇒ x = 14.

Hence, option 2 is the correct option.

The third proportional to 38 and 15 is :

1. $\dfrac{15}{38 \times 38}$

2. $\dfrac{38 \times 38}{15}$

3. $\dfrac{15 \times 15}{38}$

4. $\dfrac{38 \times 15}{2}$

Answer

Let third proportional be x.

⇒ 38 : 15 = 15 : x

⇒ $\dfrac{38}{15} = \dfrac{15}{x}$

⇒ 38x = 15 × 15

⇒ x = $\dfrac{15 \times 15}{38}$

Hence, option 3 is the correct option.

The third proportional to (x² − y²) and (x − y) is :

1. $\dfrac{x - y}{x + y}$

2. $\dfrac{x + y}{x - y}$

3. (x + y)

4. (x − y)

Answer

Let third proportional be p.

⇒ (x² − y²) : (x − y) = (x − y) : p

$\Rightarrow \dfrac{x^2 - y^2}{x - y} = \dfrac{x - y}{p} \\[1em] \Rightarrow p = \dfrac{(x - y)^2}{x^2 - y^2} \\[1em] \Rightarrow p = \dfrac{(x - y)^2}{(x + y)(x - y)} \\[1em] \Rightarrow p = \dfrac{x - y}{x + y}.$

Hence, option 1 is the correct option.

The mean proportion between 9 and 16 is:

1. 7

2. 12

3. 25

4. 144

Answer

Let mean proportional be x.

⇒ 9 : x = x : 16

⇒ $\dfrac{9}{x} = \dfrac{x}{16}$

⇒ x² = 9(16)

⇒ x = $\sqrt{144}$

⇒ x = 12.

Hence, option 2 is the correct option.

The mean proportion between 0.02 and 0.32 is :

1. 0.08

2. 0.16

3. 0.3

4. 0.34

Answer

Let mean proportion be x.

⇒ 0.02 : x = x : 0.32

⇒ $\dfrac{0.02}{x} = \dfrac{x}{0.32}$

⇒ x² = 0.02 × 0.32

⇒ x² = 0.0064

⇒ x = $\sqrt{0.0064}$

⇒ x = 0.08

Hence, option 1 is the correct option.

The mean proportion between (3 + $\sqrt{2}$) and (12 − $\sqrt{32}$) is:

1. 6

2. $2\sqrt{7}$

3. $\sqrt{7}$

4. $\dfrac{15 - 3\sqrt{2}}{2}$

Answer

Let mean proportion be x.

(3 + $\sqrt{2}$) : x = x : (12 − $\sqrt{32}$)

$\Rightarrow \dfrac{(3 + \sqrt{2})}{x} = \dfrac{x}{(12 − \sqrt{32})} \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times (12 − \sqrt{32}) \\[1em] \Rightarrow x^2 = (3 + \sqrt{2}) \times 4(3 − \sqrt{2}) \\[1em] \Rightarrow x^2 = 4[(3)^2 − (\sqrt{2})]^2 \\[1em] \Rightarrow x^2 = 4(9 - 2) \\[1em] \Rightarrow x^2 = 4 \times 7 \\[1em] \Rightarrow x^2 = 28 \\[1em] \Rightarrow x = \sqrt{28} \\[1em] \Rightarrow x = 2\sqrt7.$

Hence, option 2 is the correct option.

The mean proportion between x and y is 6. The third proportional to x and y is 48. Then x : y equals :

1. 1 : 3

2. 1 : 4

3. 1 : 6

4. 1 : 9

Answer

Given,

Mean proportion between x and y is 6.

⇒ x : 6 = 6 : y

⇒ $\dfrac{x}{6} = \dfrac{6}{y}$

⇒ y = $\dfrac{36}{x}$ ......(1)

Given,

The third proportional to x and y is 48.

⇒ x : y = y : 48

⇒ $\dfrac{x}{y} = \dfrac{y}{48}$

⇒ y² = 48x .......(2)

Substitute value of y from equation 1 in equation 2

$\Rightarrow 48x = \Big(\dfrac{36}{x}\Big)^2 \\[1em] \Rightarrow 48x = \Big(\dfrac{1296}{x^2}\Big) \\[1em] \Rightarrow 48x \times x^2 = 1296 \\[1em] \Rightarrow 48x^3 = 1296 \\[1em] \Rightarrow x^3 = \dfrac{1296}{48} \\[1em] \Rightarrow x^3 = 27 \\[1em] \Rightarrow x = \sqrt[3]{27} \\[1em] \Rightarrow x = 3.$

Substituting value of x in equation 1, we get :

⇒ y = $\dfrac{36}{3}$

⇒ y = 12

⇒ x : y = 3 : 12

⇒ x : y = 1 : 4.

Hence, option 2 is the correct option.

The ratio between the third proportional to 12 and 30 and mean proportion of 9 and 25, is :

1. 2 : 1

2. 5 : 1

3. 7 : 15

4. 9 : 14

Answer

Let, third proportional to 12 and 30 be t.

If 12 : 30 :: 30 : t, then,

$\Rightarrow \dfrac{12}{30} = \dfrac{30}{t} \\[1em] \Rightarrow t = \dfrac{30 \times 30}{12} = \dfrac{30^2}{12} \\[1em] \Rightarrow t = \dfrac{900}{12} \\[1em] \Rightarrow t = 75.$

Let Mean proportion between 9 and 25 be m.

$\Rightarrow \dfrac{9}{m} = \dfrac{m}{25} \\[1em] \Rightarrow m^2 = 9 \times 25 \\[1em] \Rightarrow m = \sqrt{9 \times 25} \\[1em] \Rightarrow m = \sqrt{225} \\[1em] \Rightarrow m = 15.$

Required ratio = t : m = 75 : 15 = 5 : 1.

Hence, option 2 is the correct option.

When 30% of one number is subtracted from another number, the second number reduces to its four-fifths. What is the ratio of the first to the second number?

1. 2 : 5

2. 2 : 3

3. 4 : 7

4. cannot be determined

Answer

Let the first number be x and the second number be y.

According to question,

⇒ y - 30% of (x) = $\dfrac{4}{5}$y

$\Rightarrow y - \dfrac{30}{100}x = \dfrac{4}{5}y \\[1em] \Rightarrow y - \dfrac{3}{10}x = \dfrac{4}{5}y \\[1em] \Rightarrow y - \dfrac{4}{5}y = \dfrac{3}{10}x \\[1em] \Rightarrow \dfrac{y}{5} = \dfrac{3}{10}x \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{5} \times \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{10}{15} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3}.$

⇒ x : y = 2 : 3.

Hence, option 2 is the correct option.

Five mangoes and four guavas cost as much as three mangoes and seven guavas. What is the ratio of the cost of one mango to the cost of one guava?

1. 1 : 3

2. 3 : 2

3. 4 : 3

4. 5 : 2

Answer

Let the cost of a mangoe be M and the cost of a guava be G.

Given,

Five mangoes and four guavas cost as much as three mangoes and seven guavas.

$\Rightarrow 5M + 4G = 3M + 7G \\[1em] \Rightarrow 5M - 3M = 7G - 4G \\[1em] \Rightarrow 2M = 3G \\[1em] \Rightarrow \dfrac{M}{G} = \dfrac{3}{2}.$

Hence, option 2 is the correct option.

Of 132 examinees of a certain class, the ratio of passed to failed students is 9 : 2. If 4 more students passed, what would have been the ratio of passed to failed students?

1. 3 : 28

2. 28 : 5

3. 4 : 25

4. 25 : 4

Answer

Given,

Total examinees of a class = 132 and ratio of passed to failed = 9 : 2.

Let number of passed students be 9x and failed students be 2x.

Then,

⇒ 9x + 2x = 132

⇒ 11x = 132

⇒ x = $\dfrac{132}{11}$ = 12.

The number of passed students = 9 × 12 = 108

The number of failed students = 2 × 12 = 24

If 4 more students passed, then

New number of passed students = 108 + 4 = 112

New number of failed students = 24 - 4 = 20

Then the new ratio of passed to failed = 112 : 20 = 28 : 5.

Hence, option 2 is the correct option.

There are three boxes – P, Q and R, containing marbles in the ratio 1 : 2 : 3. Total number of marbles is 60. The above ratio can be changed to 3 : 4 : 5 by transferring :

1. 2 marbles from P to Q and 1 from R to Q

2. 3 marbles from Q to R

3. 4 marbles from R to Q

4. 5 marbles from R to P

Answer

Given,

Initial ratio,

P : Q : R = 1 : 2 : 3

Let initial no. of marbles in P, Q and R be x, 2x and 3x respectively.

Total number of marbles = 60.

⇒ x + 2x + 3x = 60

⇒ 6x = 60

⇒ x = $\dfrac{60}{6}$

⇒ x = 10

P = 1 × 10 = 10, Q = 2 × 10 = 20, R = 3 × 10 = 30.

Given,

New ratio of marbles = 3 : 4 : 5.

Let now the no. of marbles in box P, Q and R be 3x, 4x and 5x respectively.

⇒ 3x + 4x + 5x = 60

⇒ 12x = 60

⇒ x = $\dfrac{60}{12}$ = 5.

P = 3 × 5 = 15, Q = 4 × 5 = 20, R = 5 × 5 = 25.

Thus, if initially 5 marbles are transferred from R to P then the ratio changes from 1 : 2 : 3 to 3 : 4 : 5.

Hence, option 4 is the correct option.

If x² + 4y² = 4xy, then x : y is :

1. 1 : 1

2. 1 : 2

3. 1 : 4

4. 2 : 1

Answer

Given,

x² + 4y² = 4xy

Dividing the equation by y²,

$\therefore \Big(\dfrac{x}{y}\Big)^2 + 4 = 4\Big(\dfrac{x}{y}\Big)$

Let $\dfrac{x}{y} = k$. Then

⇒ k² + 4 = 4k

⇒ k² - 4k + 4 = 0

⇒ k² - 2k - 2k + 4 = 0

⇒ k(k - 2) - 2(k - 2) = 0

⇒ (k - 2)(k - 2) = 0

⇒ (k - 2)= 0      [Using Zero-product rule]

⇒ k = 2

Therefore, $\dfrac{x}{y} = k = \dfrac{2}{1}$

Hence, option 4 is the correct option.

If 3A = 5B and 4B = 6C, then A : C is equal to :

1. 2 : 5

2. 3 : 5

3. 4 : 5

4. 5 : 2

Answer

Given,

3A = 5B

$\Rightarrow \dfrac{A}{B} = \dfrac{5}{3}$ ....(1)

Also, 4B = 6C

$\Rightarrow \dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2}$ ....(2)

From (1) and (2):

$\Rightarrow \dfrac{A}{C} = \dfrac{A}{B} \times \dfrac{B}{C} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{5}{3} \times \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{5}{2}.$

Therefore, A : C = 5 : 2.

Hence, option 4 is the correct option.

If, A : B = 7 : 9 and B : C = 5 : 4, then A : B : C is:

1. 7 : 45 : 36

2. 28 : 36 : 35

3. 35 : 45 : 36

4. none of these

Answer

Given,

A : B = 7 : 9 and B : C = 5 : 4

L.C.M of two values of B, that are 9 and 5 is 45.

$\Rightarrow \dfrac{A}{B} = \dfrac{7 \times 5}{9\times 5} = \dfrac{35}{45} \\[1em] \Rightarrow \dfrac{B}{C} = \dfrac{5 \times 9}{4 \times 9} = \dfrac{45}{36}.$

A : B : C = 35 : 45 : 36.

Hence, option 3 is the correct option.

If 8a = 9b, then the ratio of $\dfrac{a}{9}$ to $\dfrac{b}{8}$ is:

1. 1 : 1

2. 1 : 2

3. 2 : 1

4. 64 : 81

Answer

Given,

8a = 9b

$\therefore \dfrac{a}{b} = \dfrac{9}{8}$

Solving,

$\Rightarrow \dfrac{\dfrac{a}{9}}{\dfrac{b}{8}} \\[1em] \Rightarrow \dfrac{a}{9} \times \dfrac{8}{b} \\[1em] \Rightarrow \dfrac{8a}{9b} \\[1em] \Rightarrow \dfrac{8}{9} \times \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{9}{9} \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1 : 1.$

Hence, option 1 is the correct option.

If a, b, c and d are proportional, then $\dfrac{a + b}{a - b}$ is equal to:

1. $\dfrac{c}{d}$

2. $\dfrac{c-d}{c+d}$

3. $\dfrac{d}{c}$

4. $\dfrac{c+d}{c-d}$

Answer

Given,

a, b, c and d are proportional.

∴ a : b = c : d

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

Hence, option 4 is the correct option.

If x : y = 3 : 2, then the ratio (2x² + 3y²) : (3x² − 2y²) is equal to:

1. 5 : 3

2. 6 : 5

3. 12 : 5

4. 30 : 19

Answer

Given,

x : y = 3 : 2

Let x = 3k and y = 2k for some k.

For ratio (2x² + 3y²) : (3x² − 2y²),

Substituting value of x and y in the antecedent,

⇒ 2x² + 3y²

⇒ 2(3k)² + 3(2k)²

⇒ 2 × 9k² + 3 × 4k²

⇒ 18k² + 12k²

⇒ 30k².

Substituting value of x and y in the consequent,

⇒ 3x² - 2y²

⇒ 3(3k)² - 2(2k)²

⇒ 3 × 9k² - 2 × 4k²

⇒ 27k² - 8k²

⇒ 19k².

Therefore the required ratio is:

$\Rightarrow \dfrac{2x^2 + 3y^2}{3x^2 - 2y^2} \\[1em] \Rightarrow \dfrac{30k^2}{19k^2} \\[1em] \Rightarrow \dfrac{30}{19} \\[1em] \Rightarrow 30 : 19.$

Hence, option 4 is the correct option.

If (5a + 3b) : (2a − 3b) = 23 : 5, then the value of a : b is:

1. 1 : 2

2. 1 : 4

3. 2 : 1

4. 4 : 1

Answer

Given,

$\therefore \dfrac{5a + 3b}{2a - 3b} = \dfrac{23}{5}$

Cross multiplying:

$\Rightarrow 5(5a + 3b) = 23(2a - 3b) \\[1em] \Rightarrow 25a + 15b = 46a - 69b \\[1em] \Rightarrow 25a - 46a = -69b - 15b \\[1em] \Rightarrow -21a = -84b \\[1em] \Rightarrow 21a = 84b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{84}{21} = 4.$

Thus,

a : b = 4 : 1.

Hence, option 4 is the correct option.

If (4x² − 3y²) : (2x² + 5y²) = 12 : 19 then x : y is:

1. 2 : 3

2. 1 : 2

3. 2 : 1

4. 3 : 2

Answer

Given,

$\Rightarrow \dfrac{4x^2 - 3y^2}{2x^2 + 5y^2} = \dfrac{12}{19} \\[1em] \Rightarrow 19(4x^2 - 3y^2) = 12(2x^2 + 5y^2) \\[1em] \Rightarrow 76x^2 - 57y^2 = 24x^2 + 60y^2 \\[1em] \Rightarrow 76x^2 - 24x^2 = 60y^2 + 57y^2 \\[1em] \Rightarrow 52x^2 = 117y^2 \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{117}{52} \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{9}{4} \\[1em] \Rightarrow \dfrac{x}{y} = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{3}{2}.$

Thus,

x : y = 3 : 2.

Hence, option 4 is the correct option.

If, a : b = b : c then a⁴ : b⁴ would be equal to:

1. ac : b²

2. a² : c²

3. b² : ac

4. c² : a²

Answer

Given,

a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$

⇒ b² = ac

⇒ b⁴ = (ac)²

⇒ b⁴ = a²c².

$\Rightarrow \dfrac{a^4}{b^4} = \dfrac{a^4}{a^2c^2} \\[1em] = \dfrac{a^4}{a^2c^2} \\[1em] = \dfrac{a^{4 - 2}}{c^2} \\[1em] = \dfrac{a^{2}}{c^2}.$

⇒ a⁴ : b⁴ = a² : c²

Hence, option 2 is the correct option.

If a : b : c = 2 : 3 : 4, then $\dfrac{1}{a} : \dfrac{1}{b} : \dfrac{1}{c}$ is equal to :

1. $\dfrac{1}{4} : \dfrac{1}{3} : \dfrac{1}{2}$

2. 4 : 3 : 2

3. 6 : 4 : 3

4. none of these

Answer

Given,

a : b : c = 2 : 3 : 4

Let a = 2k, b = 3k, c = 4k.

Now,

$\Rightarrow \dfrac{1}{a} : \dfrac{1}{b} : \dfrac{1}{c} = \dfrac{1}{2k} : \dfrac{1}{3k} : \dfrac{1}{4k} \\[1em] = \dfrac{1}{2k} \times 12k : \dfrac{1}{3k} \times 12k : \dfrac{1}{4k} \times 12k \\[1em] = 6 : 4 : 3.$

Hence, option 3 is the correct option.

If $\dfrac{1}{x} : \dfrac{1}{y} : \dfrac{1}{z} = 2 : 3 : 5$, then x : y : z is equal to:

1. 2 : 3 : 5

2. 5 : 3 : 2

3. 15 : 10 : 6

4. 6 : 10 : 15

Answer

Given,

$\dfrac{1}{x} : \dfrac{1}{y} : \dfrac{1}{z} = 2 : 3 : 5$

Thus,

x : y : z = $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{5}$

Since, L.C.M of 2, 3, 5 is 30.

$x : y : z = \dfrac{1}{2} \times 30 : \dfrac{1}{3} \times 30 : \dfrac{1}{5} \times 30$

= 15 : 10 : 6.

Hence, option 3 is the correct option.

If (x + y) : (x − y) = 4 : 1, then (x² + y²) : (x² − y²) is :

1. 8 : 17

2. 17 : 8

3. 16 : 1

4. 25 : 9

Answer

Given,

⇒ (x + y) : (x − y) = 4 : 1

$\Rightarrow \dfrac{(x + y)}{(x - y)} = \dfrac{4}{1}$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(x + y) + (x - y)}{(x + y) - (x - y)} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{x + y + x - y}{x + y - x + y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{2y} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{5}{3}.$

Let x = 5k and y = 3k for some constant k.

Substituting value of x and y in $\dfrac{x^2 + y^2}{x^2 - y^2}$, we get:

$\Rightarrow \dfrac{(5k)^2 + (3k)^2}{(5k)^2 - (3k)^2} \\[1em] \Rightarrow \dfrac{25k^2 + 9k^2}{25k^2 - 9k^2} \\[1em] \Rightarrow \dfrac{34k^2}{16k^2} \\[1em] \Rightarrow \dfrac{17}{8}.$

Thus, (x² + y²) : (x² − y²) = 17 : 8.

Hence, option 2 is the correct option.

If a : b : c = 2 : 3 : 4 and 2a − 3b + 4c = 33, then the value of c is :

1. 6

2. 9

3. $\dfrac{66}{7}$

4. 12

Answer

Given,

a : b : c = 2 : 3 : 4

Let a = 2k, b = 3k and c = 4k for some constant k.

Substituting value of a, b and c in 2a − 3b + 4c = 33, we get :

⇒ 2(2k) − 3(3k) + 4(4k) = 33

⇒ 4k - 9k + 16k = 33

⇒ 11k = 33

⇒ k = $\dfrac{33}{11}$

⇒ k = 3.

⇒ 4k = 4(3) = 12.

Hence, option 4 is the correct option.

If x, y, z are in continued proportion, then (y² + z²) : (x² + y²) is equal to :

1. z : x

2. x : z

3. x : y

4. (y + z) : (x + y)

Answer

Given,

x, y, z are in continued proportion

Let $\dfrac{x}{y} = \dfrac{y}{z} = k$ for some constant ratio k.

⇒ $\dfrac{x}{y} = k, \dfrac{y}{z} = k$

⇒ y = zk and x = yk = (zk)k = zk²

Substitute value of x and y in $\dfrac{y^2 + z^2}{x^2 + y^2}$ we get:

$\Rightarrow \dfrac{(zk)^2 + z^2}{(zk^2)^2 + (zk)^2} \\[1em] \Rightarrow \dfrac{z^2k^2 + z^2}{z^2k^4 + z^2k^2} \\[1em] \Rightarrow \dfrac{z^2(k^2 + 1)}{z^2k^2(k^2 + 1)} \\[1em] \Rightarrow \dfrac{1}{k^2} \\[1em] \Rightarrow \dfrac{1}{\dfrac{x}{y} \times \dfrac{y}{z}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{x}{z}} \\[1em] \Rightarrow \dfrac{z}{x} \\[1em] \Rightarrow z : x.$

Hence, option 1 is the correct option.

If $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d}$, then $\dfrac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3}$ is equal to:

1. $\dfrac{a}{b}$

2. $\dfrac{b}{c}$

3. $\dfrac{c}{d}$

4. $\dfrac{d}{a}$

Answer

Let $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d}$ = k where k is the constant ratio.

Therefore,

c = dk

b = ck = (dk)k = dk²

a = bk = (dk²)k = dk³

k³ = $\dfrac{a}{d}$

Substitute value of a,b and c in $\dfrac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3}$, we get:

$\Rightarrow \dfrac{(dk^2)^3 + (dk)^3 + d^3}{(dk^3)^3 + (dk^2)^3 + (dk)^3} \\[1em] \Rightarrow \dfrac{d^3k^6 + d^3k^3 + d^3}{d^3k^9 + d^3k^6 + d^3k^3} \\[1em] \Rightarrow \dfrac{d^3(k^6 + k^3 + 1)}{d^3k^3(k^6 + k^3 + 1)} \\[1em] \Rightarrow \dfrac{1}{k^3} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a}{d}} \\[1em] \Rightarrow \dfrac{d}{a}.$

Hence, option 4 is the correct option.

If a : b = c : d, then $\dfrac{ma + nc}{mb + nd}$ is equal to :

1. m : n

2. dm : cn

3. an : mb

4. a : b

Answer

Given,

⇒ a : b = c : d

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$

Let, $\dfrac{a}{b} = \dfrac{c}{d}$ = k where k is the constant ratio.

Therefore, a = bk and c = dk

Substituting values of a and c in $\dfrac{ma + nc}{mb + nd}$, we get :

$\Rightarrow \dfrac{m(bk) + n(dk)}{mb + nd} \\[1em] \Rightarrow \dfrac{k(mb + nd)}{mb + nd} \\[1em] \Rightarrow k \\[1em] \Rightarrow \dfrac{a}{b} \\[1em] \Rightarrow a : b.$

Hence, option 4 is the correct option.

In an alloy, the ratio of copper and zinc is 5 : 2. If 1.250 kg of zinc is mixed in 17 kg 500 g of alloy, then the ratio of copper and zinc in the alloy will be :

1. 1 : 2

2. 2 : 1

3. 2 : 3

4. 3 : 2

Answer

Given,

Total weight of alloy = 17 kg 500 g = 17.5 kg

Given,

Ratio of copper and zinc is 5 : 2.

Let initial quantity of copper be 5x and zinc be 2x.

$\text{Initial quantity of copper} = \dfrac{5x}{5x + 2x} \times 17.5 \\[1em] = \dfrac{5x}{7x} \times 17.5 \\[1em] = 5 \times 2.5 \\[1em] = 12.5\text{ kg} \\[1em] \text{Initial quantity of zinc} = \dfrac{2x}{5x + 2x} \times 17.5 \\[1em] = \dfrac{2x}{7x} \times 17.5 \\[1em] = 2 \times 2.5 \\[1em] = 5\text{ kg}.$

1.250 kg of zinc is mixed in 17 kg 500 g of alloy, so new quantity of zinc = 5 + 1.250 = 6.25 kg.

Ratio of copper to zinc = 12.5 : 6.25 = 2 : 1

Hence, option 2 is the correct option.

A sum of ₹ 6,400 is divided among three workers in the ratio $\dfrac{3}{5} : 2 : \dfrac{5}{3}$. The share of the second worker is :

1. ₹ 2,560

2. ₹ 3,000

3. ₹ 3,200

4. ₹ 3,840

Answer

Wages ₹ 6,400 divided into 3 workers in ratio = $\dfrac{3}{5} : 2 : \dfrac{5}{3}$

L.C.M of 5 and 3 is 15

$\Rightarrow \dfrac{3}{5} \times 15:15 \times 2: \dfrac{5}{3} \times 15$

⇒ 9 : 30 : 25

Sum of the ratio = 9 + 30 + 25 = 64

The share of the second worker = $\dfrac{30}{64} \times 6400$

= 30 × 100

= ₹ 3,000.

Hence, option 2 is the correct option.

If a sum of ₹ x is divided between A and B in the ratio $\dfrac{a}{b} : \dfrac{c}{d}$, then A gets:

1. ₹$\Big(\dfrac{abx}{ac + bd}\Big)$

2. ₹$\Big(\dfrac{abx}{ad + bc}\Big)$

3. ₹$\Big(\dfrac{adx}{ad + bc}\Big)$

4. ₹$\Big(\dfrac{adx}{ab + cd}\Big)$

Answer

The sum ₹ x is divided between A and B in the ratio :

A : B = $\dfrac{a}{b}:\dfrac{c}{d}$

L.C.M of denominators bd:

A : B = $\dfrac{a}{b} \times bd :\dfrac{c}{d} \times bd$

A : B = ad : bc

Sum of the ratio = ad + bc.

The sum of money A get = $\dfrac{ad}{ad + bc} \times x = \dfrac{adx}{ad + bc}$.

Hence, option 3 is the correct option.

The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

1. 90

2. 120

3. 220

4. 240

Answer

Given,

The total number of students is 720.

The initial ratio of boys (B) to girls (G) is 7 : 5.

The total number of ratio parts is 7 + 5 = 12.

Initial number of boys = $\dfrac{7}{12} \times 720$ = 420.

Initial number of girls = $\dfrac{5}{12} \times 720$ = 300.

In order to make the ratio 1 : 1, the number of girls and boys must be equal.

Thus, 420 - 300 = 120, more girls should be admitted.

Hence, option 2 is the correct option.

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:

1. 39

2. 49

3. 66

4. 77

Answer

Let the two numbers be 7x and 11x.

Given,

When 7 is added to each number, the new ratio is 2 : 3.

$\therefore \dfrac{7x + 7}{11x + 7} = \dfrac{2}{3}$

⇒ 3(7x + 7) = 2(11x + 7)

⇒ 21x + 21 = 22x + 14

⇒ 22x - 21x = 21 - 14

⇒ x = 7.

The smaller number is 7x = 7(7) = 49.

Hence, option 2 is the correct option.

What must be subtracted from each of 7, 9, 11 and 15, so that the resulting numbers are in proportion?

1. 1

2. 2

3. 3

4. 5

Answer

Let the number to be subtracted be x.

Thus, the numbers 7 - x, 9 - x, 11 - x and 15 - x are in proportion.

$\Rightarrow \dfrac{7 - x}{9 - x} = \dfrac{11 - x}{15 - x}$

⇒ (7 - x)(15 - x) = (11 - x)(9 - x)

⇒ 105 - 7x - 15x + x² = 99 - 11x - 9x + x²

⇒ 105 - 22x + x² = 99 - 20x + x²

⇒ 105 - 99 = 22x - 20x

⇒ 6 = 2x

⇒ x = $\dfrac{6}{2}$

⇒ x = 3.

Hence, option 3 is the correct option.

If p, q, and r are in continued proportion, then :

1. p:q = q:r

2. q:r = p²:q²

3. p:q² = r:p²

4. p:r = p²:q²

Answer

Since, p, q, and r are in continued proportion.

$\therefore \dfrac{p}{q} = \dfrac{q}{r}$

⇒ q² = pr.....(1)

Solving,

p:r = p²:q²

$\Rightarrow \dfrac{p}{r} = \dfrac{p^2}{q^2} \\[1em] \Rightarrow q^2 = \dfrac{p^2 \times r}{p} \\[1em] \Rightarrow q^2 = pr...(2)$

Since, equation (1) and (2) are equal.

Hence, Option 4 is the correct option.

The ratio of milk to water in 80 litres of a mixture is 7 : 3. The quantity of water (in litres) to be added to it to make the ratio 2 : 1 is :

1. 4

2. 5

3. 6

4. 8

Answer

Let the initial quantities of milk and water be 7x and 3x respectively.

Given,

The total volume of the mixture is 80 litres.

⇒ 7x + 3x = 80

⇒ 10x = 80

⇒ x = $\dfrac{80}{10}$

⇒ x = 8.

Initial Quantity of Milk : 7x = 7 × 8 = 56 litres

Initial Quantity of Water : 3x = 3 × 8 = 24 litres

Let W be the quantity of water added to the mixture.

The quantity of milk remains unchanged.

Water = 24 + W

The new ratio of milk to water is required to be 2 : 1.

⇒ $\dfrac{56}{24 + W} = \dfrac{2}{1}$

⇒ 56 = 2(24 + W)

⇒ 56 = 48 + 2W

⇒ 2W = 56 - 48

⇒ 2W = 8

⇒ W = $\dfrac{8}{2}$

⇒ W = 4.

The quantity of water to be added is 4 litres.

Hence, option 1 is the correct option.

What number must be added to each of the numbers 7, 11 and 19 so that the resulting numbers may be in continued proportion?

1. −4

2. −3

3. 3

4. 4

Answer

Let the number to be added be x.

Thus, numbers 7 + x, 11 + x and 19 + x will be in continued proportion.

$\therefore \dfrac{7 + x}{11 + x} = \dfrac{11 + x}{19 + x} \\[1em] \Rightarrow (7 + x)(19 + x) = (11 + x)(11 + x) \\[1em] \Rightarrow 133 + 7x + 19x + x^2 = 121 + 11x + 11x + x^2 \\[1em] \Rightarrow x^2 + 26x + 133 = x^2 + 22x + 121 \\[1em] \Rightarrow x^2 - x^2 + 26x - 22x = 121 - 133 \\[1em] \Rightarrow 4x = -12 \\[1em] \Rightarrow x = \dfrac{-12}{4} \\[1em] \Rightarrow x = -3.$

Hence, option 2 is the correct option.

Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each, they are in the ratio of 12 : 23. The larger number is :

1. 40

2. 45

3. 55

4. 60

Answer

Given,

Let the two numbers be 3k and 5k.

After subtracting 9 from each number, the ratio changes to 12 : 23.

$\therefore \dfrac{3k - 9}{5k - 9} = \dfrac{12}{23} \\[1em] \Rightarrow 23(3k - 9) = 12(5k - 9) \\[1em] \Rightarrow 69k - 207 = 60k - 108 \\[1em] \Rightarrow 69k - 60k = 207 - 108 \\[1em] \Rightarrow 9k = 99 \\[1em] \Rightarrow k = \dfrac{99}{9} \\[1em] \Rightarrow k = 11.$

The larger number = 5k = 5 × 11 = 55.

Hence, option 3 is the correct option.

What number must be added to the terms of 3 : 5 to make the ratio 5 : 6 ?

1. 6

2. 7

3. 12

4. 13

Answer

Let the number to be added be x.

$\therefore \dfrac{3 + x}{5 + x} = \dfrac{5}{6} \\[1em] \Rightarrow 6(3 + x) = 5(5 + x) \\[1em] \Rightarrow 18 + 6x = 25 + 5x \\[1em] \Rightarrow 6x - 5x = 25 - 18 \\[1em] \Rightarrow x = 7.$

Hence, option 2 is the correct option.

Six numbers a, b, c, d, e, f are such that ab = 1, bc = $\dfrac{1}{2}$, cd = 6, de = 2 and ef = $\dfrac{1}{2}$. What is the value of (ad : be : cf)?

1. 4 : 3 : 27

2. 6 : 1 : 9

3. 8 : 9 : 9

4. 72 : 1 : 9

Answer

Given,

ab = 1, bc = $\dfrac{1}{2}$, cd = 6, de = 2, ef = $\dfrac{1}{2}$.

$\Rightarrow ef = \dfrac{1}{2} \\[1em] \Rightarrow e = \dfrac{1}{2f} \\[1em] \Rightarrow de = 2 \\[1em] \Rightarrow d = \dfrac{2}{e} = \dfrac{2}{\dfrac{1}{2f}} = 4f \\[1em] \Rightarrow cd = 6 \\[1em] \Rightarrow c = \dfrac{6}{d} = \dfrac{6}{4f} = \dfrac{3}{2f} \\[1em] \Rightarrow bc = \dfrac{1}{2} \\[1em] \Rightarrow b = \dfrac{1}{2c} = \dfrac{1}{2 \times \dfrac{3}{2f}} = \dfrac{f}{3} \\[1em] \Rightarrow ab = 1 \\[1em] \Rightarrow a = \dfrac{1}{b} = \dfrac{1}{\dfrac{f}{3}} = \dfrac{3}{f}$

ad : be : cf = $\dfrac{3}{f} \times 4f : \dfrac{f}{3} \times \dfrac{1}{2f}: \dfrac{3}{2f} \times f$

= 12 : $\dfrac{1}{6} : \dfrac{3}{2}$

Since, L.C.M. of 2 and 6 is 6.

= 12 × 6 : $\dfrac{1}{6} \times 6 : \dfrac{3}{2} \times 6$

⇒ 72 : 1 : 9.

Hence, option 4 is the correct option.

If a = $\dfrac{4xy}{x+y}$, then $\Big(\dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y}\Big)$ equals :

1. 1

2. 2

3. $\dfrac{1}{2}$

4. none of these

Answer

Given,

a = $\dfrac{4xy}{x+y}$

Solving,

$\Rightarrow \dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y} = \dfrac{\dfrac{4xy}{x+y} + 2x}{\dfrac{4xy}{x+y} - 2x} + \dfrac{\dfrac{4xy}{x+y} + 2y}{\dfrac{4xy}{x+y} - 2y} \\[1em] \Rightarrow \text{Multiply numerator and denominator by (x + y)} \\[1em] \Rightarrow \dfrac{\Big(\dfrac{4xy}{x+y} + 2x\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2x\Big) \times (x + y)} + \dfrac{\Big(\dfrac{4xy}{x+y} + 2y\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2y\Big) \times (x + y)}\\[1em] \Rightarrow \dfrac{4xy + 2x(x+y)}{4xy - 2x(x+y)} + \dfrac{4xy + 2y(x+y)}{4xy - 2y(x+y)} \\[1em] \Rightarrow \dfrac{4xy + 2x^2 + 2xy}{4xy - 2x^2 - 2xy} + \dfrac{4xy + 2xy + 2y^2}{4xy - 2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x^2 + 6xy}{-2x^2 + 2xy} + \dfrac{6xy + 2y^2}{2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x(x + 3y)}{2x(y - x)} + \dfrac{2y(3x+y)}{2y(x-y)} \\[1em] \Rightarrow \dfrac{x + 3y}{y - x} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{x + 3y}{-(x - y)} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-(x + 3y)}{x - y} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x - 3y + 3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x + 3x + y - 3y}{x - y} \\[1em] \Rightarrow \dfrac{2x - 2y}{x - y} \\[1em] \Rightarrow \dfrac{2(x - y)}{x - y} \\[1em] \Rightarrow 2.$

Hence, option 2 is the correct option.

If $\dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b}$, then x equals:

1. $\dfrac{a^2}{a^2 + b^2}$

2. $\dfrac{b^2}{a+b}$

3. $\dfrac{ab}{a^2 + b^2}$

4. $\dfrac{2ab}{a^2 + b^2}$

Answer

Given,

$\Rightarrow \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \dfrac{a}{b} \\[1em] \Rightarrow b\Big(\sqrt{1+x} + \sqrt{1-x}\Big) = a\Big(\sqrt{1+x} - \sqrt{1-x}\Big) \\[1em] \Rightarrow b\sqrt{1+x} + b\sqrt{1-x} = a\sqrt{1+x} - a\sqrt{1-x} \\[1em] \Rightarrow b\sqrt{1+x} - a\sqrt{1+x} = -a\sqrt{1-x} - b\sqrt{1-x} \\[1em] \Rightarrow (b-a)\sqrt{1+x} = -(a+b)\sqrt{1-x} \\[1em] \text{Squaring on Both Sides,} \\[1em] \Rightarrow (b-a)^2(1+x) = (a+b)^2(1-x) \\[1em] \Rightarrow (b^2 - 2ab + a^2)(1+x) = (a^2 + 2ab + b^2)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab)(1+x) = (a^2 + b^2 + 2ab)(1-x) \\[1em] \Rightarrow (a^2 + b^2 - 2ab) + (a^2 + b^2 - 2ab)x = (a^2 + b^2 + 2ab) - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow (a^2 + b^2 - 2ab) - (a^2 + b^2 + 2ab) = -(a^2 + b^2 - 2ab)x - (a^2 + b^2 + 2ab)x \\[1em] \Rightarrow a^2 + b^2 - 2ab - a^2 - b^2 - 2ab = - a^2x - b^2x + 2abx - a^2x - b^2x - 2abx \\[1em] \Rightarrow a^2 - a^2 - b^2 + b^2 - 2ab - 2ab = - a^2x - a^2x - b^2x - b^2x + 2abx - 2abx \\[1em] \Rightarrow -4ab = - 2a^2x - 2b^2x \\[1em] \Rightarrow -4ab = -2(a^2 + b^2)x \\[1em] \Rightarrow x = \dfrac{-4ab}{-2(a^2 + b^2)} \\[1em] \Rightarrow x = \dfrac{2ab}{a^2 + b^2}.$

Hence, option 4 is the correct option.

If x = $\dfrac{\sqrt{2a+1} + \sqrt{2a-1}}{\sqrt{2a+1} - \sqrt{2a-1}}$, then which of the following is true?

1. x² + 2ax + 1 = 0

2. x² − 2ax + 1 = 0

3. x² + 4ax − 1 = 0

4. x² − 4ax + 1 = 0

Answer

Given,

$x = \dfrac{\sqrt{2a+1} + \sqrt{2a-1}}{\sqrt{2a+1} - \sqrt{2a-1}} \\[1em] x\Big(\sqrt{2a+1} - \sqrt{2a-1}\Big) = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] x\sqrt{2a+1} - x\sqrt{2a-1} = \sqrt{2a+1} + \sqrt{2a-1} \\[1em] (x-1)\sqrt{2a+1} = (x+1)\sqrt{2a-1} \\[1em] \text{Squaring on Both Sides,} \\[1em] (x-1)^2(2a+1) = (x+1)^2(2a-1) \\[1em] (2a+1)(x^2 - 2x + 1) = (2a-1)(x^2 + 2x + 1) \\[1em] 2ax^2 - 4ax + 2a + x^2 - 2x + 1 = 2ax^2 + 4ax + 2a - x^2 - 2x - 1 \\[1em] 2ax^2 - 2ax^2 + x^2 + x^2 - 4ax - 4ax + 2a - 2a - 2x + 2x + 1 + 1 = 0\\[1em] 2x^2 - 8ax + 2 = 0 \\[1em] 2(x^2 - 4ax + 1) = 0 \\[1em] x^2 - 4ax + 1 = 0$