Quantitative Aptitude - MCQ Practice Questions

SI₁ = (1000 × 10 × 3)/100 = 300. SI₂ = (2000 × 12 × 2)/100 = 480. Total = 300 + 480 = ₹780 (closest: should verify - 300+480=780, option A)

A's rate = $\frac{1}{30}$, B's rate = $\frac{1}{40}$. Combined rate = $\frac{7}{120}$. Work in 10 days = $\frac{70}{120}$. Remaining = $\frac{50}{120}$. B alone = ($\frac{50}{120}$)/($\frac{1}{40}$) = 16.67 days. (Re-check: Remaining work = 1 - 10($\frac{1}{30}$ + $\frac{1}{40}$) = 1 - 10×$\frac{7}{120}$ = $\frac{50}{120}$. Time for B = ($\frac{50}{120}$)÷($\frac{1}{40}$) = $\frac{50}{120}$ × 40 = 16.67 ≈ 16-17 days, check options)

Amount = P(1.10)³ = P × 1.331 = 1331. P = ₹1000

Let CP = 100. MP = 140. For 12% profit, SP = 112. Discount = 140 - 112 = 28. Discount % = $\frac{28}{140}$ × 100 = 20%

New price = 1.25P. For constant expenditure: New consumption = $\frac{1}{1}$.25 = 0.8. Reduction = 20%.

SI = (50,000 × 10 × T)/100. 15,000 = 5000T. T = 3 years.

CP = 100, MP = 180, SP = 144 (for 44% profit). Discount = 180 - 144 = 36. Discount% = ($\frac{36}{180}$) × 100 = 20%.

Initial price × 1.16 = 465. Initial price = $\frac{465}{1}$.16 = 400.

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:

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

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

Step 3: Calculate Total Cost Price and Total Selling Price

Step 4: Find Overall Profit/Loss

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

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

For 20% gain at ₹1,200: CP = 1,$\frac{200}{1}$.20 = ₹1,000. For 20% loss at ₹1,200: CP = 1,$\frac{200}{0}$.80 = ₹1,500. Total CP = ₹2,500, Total SP = ₹2,400. Loss = ₹100. Loss% = ($\frac{100}{2}$,500) × 100 = 4%

# 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.

Step 3: Substitute values

Step 4: Calculate step-by-step

First, find $(0.85{)}^2$:

Then multiply by the initial value:

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

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

Let B = 100, A = 75. B is more than A by 25 on base of 75 = ($\frac{25}{75}$) × 100 = 33.33%

(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

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

SI paid by A = (5000 × 8 × 3)/100 = 1200. SI received by A = (5000 × 10 × 3)/100 = 1500. Profit = 1500 - 1200 = 300

Using compound growth: 100(1+r)³ = 160. (1+r)³ = 1.6. 1+r = 1.6^($\frac{1}{3}$) ≈ 1.1696. r ≈ 16.96% ≈ 16.5%.

Let n = previous innings. 40n + 65 = 45(n+1). 40n + 65 = 45n + 45. 20 = 5n. n = 4.

Total distance = 40 + 60 + 100 = 200 km. Time for first = $\frac{40}{20}$ = 2 hours. Time for second = $\frac{60}{30}$ = 2 hours. Time for third = $\frac{100}{50}$ = 2 hours. Total time = 6 hours. Average speed = $\frac{200}{6}$ = 33.33 km/h.