Direct or Indirect Proportion MCQ Quiz - Objective Question with Answer for Direct or Indirect Proportion - Download Free PDF

Given:

Salaries of Rahim and Karim are in the ratio 2:3.

Salary of each is increased by ₹2000, and the new ratio becomes 40:57.

Formula Used:

New Ratio = (Old Salary + Increment) / (Old Salary + Increment)

Calculation:

Let Rahim's present salary be 2x and Karim's present salary be 3x.

After increasing ₹2000, their salaries become:

Rahim: 2x + 2000

Karim: 3x + 2000

New ratio = 40:57

So, (2x + 2000) / (3x + 2000) = 40 / 57

Cross multiply:

57 × (2x + 2000) = 40 × (3x + 2000)

⇒ 114x + 114000 = 120x + 80000

⇒ 120x - 114x = 114000 - 80000

⇒ 6x = 34000

⇒ x = 34000 / 6 = 5666.67

Karim's present salary:

3x = 3 × 5666.67 = ₹17000

∴ Karim's present salary is ₹17000

Given:

Let the total number of apples = x

A gets one-fourth of the total, so apples A gets = \(\dfrac{x}{4}\)

Let there be n other people who get an equal number of apples, say y apples each.

A gets thrice as many apples as each of the others, so: \(\dfrac{x}{4} = 3y\)

Formula used:

Total apples = Apples with A + Apples with others

Total apples = \(\dfrac{x}{4} + n \times y\)

Calculation:

From \(\dfrac{x}{4} = 3y\), we get: \(\dfrac{x}{4} = 3y \Rightarrow x = 12y\)

Substitute this value of x in the total apples equation:

x = \(\dfrac{x}{4} + n \times y\)

⇒ 12y = \(\dfrac{12y}{4} + n \times y\)

⇒ 12y = 3y + n × y

⇒ 12y - 3y = n × y

⇒ 9y = n × y

⇒ n = 9

So, there are 9 other people, and the total number of apples is 12y.

The total number of apples must be a multiple of 12. Therefore, any number that is not a multiple of 12 could not be the total number of apples.

∴ The number 16 could NOT be a possible number of apples.

Given:

E is 25% greater than B.

A : B = 5 : 7

D is 18 greater than C.

C = 60% of A

C : D = 4 : 7

Formula Used:

Percentage increase: New Value = Original Value × (1 + Percentage/100)

Calculations:

Sequential order of steps: C, E, A, B, F, D

(C) Let C and D be 4k and 7k respectively

(E) D - C = 18 ⇒ 7k - 4k = 18 ⇒ 3k = 18 ⇒ k = 6

(A) C = 4 × 6 = 24

(F) A = (100/60) × 24 = 40

(B) B = (7/5) × 40 = 56

(D) E = (125/100) × 56 = 1.25 × 56 = 70

The correct sequence is: CEAFBD.

Given:

64 : x :: x : 169

Formula used:

a : b :: c : d means a \times d = b \times c

Calculation:

64 : x :: x : 169

⇒ \(64 \times 169 = x \times x\)

⇒ \(64 \times 169 = x^2\)

⇒ \(x^2 = 64 \times 13^2\)

⇒ \(x^2 = (8 \times 13)^2\)

⇒ x = 104

∴ The correct answer is option 1.

Given:

9 : 12 :: 12 : k

Formula used:

a : b :: c : d ⇒ b/a = d/c

Calculation:

12/9 = k/12

⇒ k = (12 × 12)/9

⇒ k = 16

⇒ So, k + 1 = 16 + 1

⇒ k + 1 = 17

∴ The correct answer is option (4).

Given:

Gold is 12 times as heavy as aluminum and copper  is 5 times as heavy as aluminum.

Calculation:

Let the ratio of gold and copper that taken be x : y

According to the question,

⇒ \(\frac{{12x + 5y}}{{x + y}} = \frac{8}{1}\)

⇒ 12x + 5y = 8x + 8y

⇒ 4x = 3y

⇒ x/y = 3/4

∴ Gold and copper are taken in the ratio 3 : 4.  Alternate Method

Allegation method use in this question,

∴ Gold and copper are taken in the ratio 3 : 4. 

Given:

\(\frac{1}{a}:\frac{1}{7} = \frac{1}{{3.43}}:\frac{1}{a}\)

Calculation:

\(\frac{1}{a}:\frac{1}{7} = \frac{1}{{3.43}}:\frac{1}{a}\)

⇒ 7/a = a/3.43

⇒ a² = 3.43 × 7

⇒ a = \(\sqrt {3.43 \times 7}\)

⇒ a = 4.9 (Considering the positive value only)

∴ The value of a is 4.9.

Given

ratio of the prices of tea to coffee is 3 ∶ 5

ratio of the quantities consumed by a family is 5 ∶ 7

Formula used

expenditure =price \(\times\)quantities

Calculation

\(\frac{E1}{E2}=\frac{P1}{P2}\times \frac{Q1}{Q2}\)

\(\frac{E1}{E2}=\frac{3}{5}\times \frac{5}{7}=\frac{3}{7}\)

The ratio of the expenditure is 3:7

Formula used:

a3 + b3 = (a + b)(a2 - ab + b2)

a² - b² = (a + b)(a - b)

Calculation:

According to the question

(a3 + b3) ∝ (a2 – b2)

Using the above identities

⇒ (a + b)(a 2 - ab + b2) ∝ (a + b)(a - b)

⇒ (a 2 - ab + b2) ∝ (a - b)

∴ (a 2 - ab + b2) will be proportional to the  a - b   

Given:

The price of two articles are in the ratio of 5 : 6 

The price of first article is increased by 30%

The price of second article is decreased by X%

So the Ratio becomes 13 : 11

Calculation:

Let the price of the articles initially be 50 and 60 respectively

The price of first article is increased by 30%

⇒ 50 + (30/100) × 50 = 65

The price of second article is decreased by X%

⇒ 60 - {(X/100) × 60} = 60 × (100 - X)/100

So the Ratio becomes 13 : 11

⇒ 65/{60× (100 - X)/100} = 13/11

⇒ 715 = 780 × (100 - X)/100

⇒ 71500 = 78000 - 780X

⇒ 780X = 6500

∴ X = 8.33

Given :

A = k(B + 18), A = 108, B = 36

Calculation : 

⇒ A = k(B + 18)

⇒ 108 = k(36 + 18)

⇒ k = 108/54 = 2

A = 2 × (68 + 18)

⇒ 2 × 86 = 172

∴ The correct answer is 172.

Given:

Ratio of amount invested in share market and gold is 6 : 7

Profit earned on the investment after one year is 30%

Ratio of amount reinvested in share market and gold is 4 : 5

Amount reinvested in gold is Rs. 94,500

Calculation:

Since the amount reinvested is in the ratio = 4 : 5

Then total amount reinvested,

⇒ (Amount of gold reinvested × total of ratio)/part of gold in ratio

⇒ (94,500 × 9)/5 = Rs. 1,70,100

Since there was a profit in this amount = 30%

So the Original amount,

⇒ (Total amount reinvested × 100)/130 = Rs. 130846

Original amount invested in gold,

⇒ (130846 × 7)/13

⇒ Rs. 70,455.53 ≈ Rs. 70,456

∴ Original amount invested in gold is Rs. 70,456

Given

Bag contains Rs. 1, 50 paise, coins

The ratio of value is 30 : 11

The total no. of coins are 520

Concept used

Ratio concept used

Calculation

Value of Rs. 1 and 50 paise coins = 30 : 11

Number of coins = 30 : 22

⇒ 52 part = 520

⇒ 1p = 10

⇒ 22p = 10 × 22

⇒ 220

Calculation:

Rs. 1 = 100 paise 

Let the total number of 100 paise coins be x

Then 50 paise coins are 520 – x

According to the question,

⇒ 100x/50(520 – x) = 30/11

⇒ 11 × 100x = 30 × 50(520 – x)

⇒ 1,100x = 7,80,000 – 1,500x

⇒ 2,600x = 7,80,000

⇒ x = 7,80,000/2,600

⇒ x = 300

50 paise coins,

⇒ 520 – 300 = 220

∴ 50 paise coins are 220.

Calculation:

Let the first part be "a", second part be "b" and third part be "c".

According to the question-

⇒ \(1\over2\) × a = \(1\over5\) × b = \(3\over8\) × c 

Let  \(1\over2\) × a = \(1\over5\) × b = \(3\over8\) × c = k

⇒ a = 2k

⇒ b = 5k

⇒ c = \(8\over3\)k

⇒ 2k + 5k + \(8\over3\)k = Rs 5800

⇒ 29k = Rs 5800 × 3

⇒ k = Rs 600

Substituting the value of k in the equation of "c"-

⇒  \(3\over8\) × c = Rs 600

⇒ c = Rs 1600

∴ The value of third part is Rs 1600.

Given,

The ratio of the present salaries of Vinod and Manoj is 6 : 7

If both of them get their salaries increased by Rs. 16,000, then the ratio becomes 8 : 9.

Calculation:

Let the present salaries of Vinod and Manoj be 6x and 7x respectively

According to the question

(6x + 16000) / (7x + 16000) = 8/9

⇒ 9 (6x + 16000) = 8 (7x + 16000)

⇒ 54x + 144000 = 56x + 128000

⇒ 56x – 54x = 144000 – 128000

⇒ 2x = 16000

⇒ x = 16000/2

⇒ x = 8000

∴ Present salary of Manoj = 7 × 8000 = Rs. 56,000.