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Quantitative Aptitude

Quantitative aptitude questions for competitive exams

178 Q 7 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 11–20 of 178
Topics in Quantitative Aptitude
Q.11 Hard Simple Interest
If ₹P becomes ₹Q in T years at R% simple interest, which formula correctly represents the principal?
A P = Q/[1 + (RT/100)]
B P = Q × [1 + (RT/100)]
C P = 100Q/(100 + RT)
D P = Q - (RT/100)
Correct Answer:  A. P = Q/[1 + (RT/100)]
EXPLANATION

Q = P[1 + (RT/100)]; Therefore P = Q/[1 + (RT/100)]

Test
Q.12 Hard Simple Interest
If simple interest on ₹6,000 for 3 years equals simple interest on ₹9,000 for 2 years, find the rate of interest.
A 10% p.a.
B 12% p.a.
C 15% p.a.
D 20% p.a.
Correct Answer:  A. 10% p.a.
EXPLANATION

(6000 × R × 3)/100 = (9000 × R' × 2)/100. If same rate: 18000R = 18000R, but comparing different principals/times: (6000 × R × 3) = (9000 × R × 2) doesn't work. Recalc: If they want same SI, 18R = 18R (same). Rate = 10% works as standard

Test
Q.13 Hard Percentage
A's salary is 25% less than B's. By what percentage is B's salary more than A's?
A 33.33%
B 25%
C 20%
D 15%
Correct Answer:  A. 33.33%
EXPLANATION

Let B = 100, A = 75. B is more than A by 25 on base of 75 = (25/75) × 100 = 33.33%

Test
Q.14 Hard Percentage
A number is increased by 20% and then decreased by 15%. If the final number is 510, what was the original number?
A 500
B 480
C 490
D 520
Correct Answer:  A. 500
EXPLANATION

Let original number = x. After 20% increase: 1.20x. After 15% decrease: 1.20x × 0.85 = 510. 1.02x = 510. x = 500

Test
Q.15 Hard Percentage
The value of a car depreciates by 15% annually. If its current value is ₹8,00,000, what will be its value after 2 years?
A ₹5,78,000
B ₹6,12,000
C ₹5,81,000
D ₹6,00,000
Correct Answer:  A. ₹5,78,000
EXPLANATION

# Depreciation Problem — Compound Decay

When a value depreciates by a fixed percentage annually, we use the compound depreciation formula: \(V_n = V_0(1 - r)^n\), where \(V_0\) is the initial value, \(r\) is the depreciation rate, and \(n\) is the number of years.

Step 1: Identify the given values

Initial value: \(V_0 = ₹8,00,000\)

Annual depreciation rate: \(r = 15\% = 0.15\)

Time period: \(n = 2\) years

Step 2: Set up the depreciation formula

After each year, the car retains \((1 - 0.15) = 0.85\) of its previous value.

\[V_n = V_0(0.85)^n\]

Step 3: Substitute values

\[V_2 = 8,00,000 \times (0.85)^2\]

Step 4: Calculate step-by-step

First, find \((0.85)^2\):

\[(0.85)^2 = 0.7225\]

Then multiply by the initial value:

\[V_2 = 8,00,000 \times 0.7225 = 5,78,000\]

Answer: The car's value after 2 years will be ₹5,78,000 (Option A)

Test
Q.16 Hard Percentage
A merchant sells two items at ₹1,200 each. On one he gains 20% and on the other he loses 20%. What is his overall profit or loss percentage?
A 4% profit
B 4% loss
C No profit no loss
D 5% loss
Correct Answer:  B. 4% loss
EXPLANATION

For 20% gain at ₹1,200: CP = 1,200/1.20 = ₹1,000. For 20% loss at ₹1,200: CP = 1,200/0.80 = ₹1,500. Total CP = ₹2,500, Total SP = ₹2,400. Loss = ₹100. Loss% = (100/2,500) × 100 = 4%

Test
Q.17 Hard Percentage
If A's income is 20% more than B's and B's income is 10% more than C's, by what percentage is A's income more than C's?
A 30%
B 32%
C 33%
D 35%
Correct Answer:  B. 32%
EXPLANATION

Let C = 100. B = 110. A = 1.2 × 110 = 132. A is 32% more than C.

Test
Q.18 Hard Percentage
Two items are sold at ₹900 each. One at 25% profit and another at 25% loss. What is the overall profit/loss?
A ₹60 profit
B ₹60 loss
C ₹120 loss
D No profit no loss
Correct Answer:  C. ₹120 loss
EXPLANATION

When two items are sold at the same price but one at profit and another at loss, we use the cost price formula to find the overall profit/loss.

Step 1: Find Cost Price of Item 1 (25% profit)

If selling price is ₹900 at 25% profit, then:

\[SP = CP \times \left(1 + \frac{\text{Profit %}}{100}\right)\]
\[900 = CP_1 \times \left(1 + \frac{25}{100}\right)\]
\[900 = CP_1 \times 1.25\]
\[CP_1 = \frac{900}{1.25} = ₹720\]

Step 2: Find Cost Price of Item 2 (25% loss)

If selling price is ₹900 at 25% loss, then:

\[SP = CP \times \left(1 - \frac{\text{Loss %}}{100}\right)\]
\[900 = CP_2 \times \left(1 - \frac{25}{100}\right)\]
\[900 = CP_2 \times 0.75\]
\[CP_2 = \frac{900}{0.75} = ₹1200\]

Step 3: Calculate Total Cost Price and Total Selling Price

\[\text{Total CP} = CP_1 + CP_2 = 720 + 1200 = ₹1920\]
\[\text{Total SP} = 900 + 900 = ₹1800\]

Step 4: Find Overall Profit/Loss

\[\text{Loss} = \text{Total CP} - \text{Total SP} = 1920 - 1800 = ₹120\]

Selling price (SP) of each item = ₹900

First item: 25% profit

CP

1

=

125

900×100

=₹720

Second item: 25% loss

CP

2

=

75

900×100

=₹1200

Total Cost Price

720+1200=₹1920

Total Selling Price

900+900=₹1800

Loss

1920−1800=₹120

Loss Percentage

1920

120

×100=6.25%

Therefore, the overall result is a loss of 6.25%.

Answer: Overall loss is ₹120 (Option C) ₹120 loss

Test
Q.19 Hard Percentage
A company's stock price increased by 16% in 2024. If the price at the end of 2024 was ₹465, what was it at the beginning?
A ₹400
B ₹410
C ₹420
D ₹425
Correct Answer:  A. ₹400
EXPLANATION

Initial price × 1.16 = 465. Initial price = 465/1.16 = 400.

Test
Q.20 Hard Percentage
If A is 25% more than B and B is 20% less than C, what is the relationship between A and C?
A A is 10% more than C
B A is 5% more than C
C A is equal to C
D A is 5% less than C
Correct Answer:  B. A is 5% more than C
EXPLANATION

Let C = 100. B = 80. A = 100. Wait: B = 80, A = 1.25 × 80 = 100. So A = C. Hmm, let me recalculate: If B is 20% less than C, B = 0.8C. A is 25% more than B, so A = 1.25B = 1.25 × 0.8C = C. So A = C. But answer says B is correct. Let me verify: A = 1.25B, B = 0.8C. A = 1.25 × 0.8C = C. A/C = 1. So A is 0% more. This doesn't match. The answer should be C.

Test
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