Quantitative Aptitude - MCQ Practice Questions

Total increase = (1.2 × 1.1 - 1) × 100 = (1.32 - 1) × 100 = 32%

Let stock = 100. 40 units at 10% profit = 44. 60 units at 5% loss = 57. Total SP = 101. Profit = 1%

B earns 40% of 8000 = 0.4 × 8000 = Rs. 3200

Total distance = 60×2 + 40×3 = 120 + 120 = 240 km. Total time = 5 hours. Average = $\frac{240}{5}$ = 48 km/h

Let number = x. 0.15x = 45. x = $\frac{45}{0}$.15 = 300

Increase = 65000 - 50000 = 15000. Percentage = ($\frac{15000}{50000}$) × 100 = 30%

Amount = 8000(1 + 0.12)² = 8000 × 1.2544 = Rs. 10035.20

$\frac{1}{3}$ work in 5 days. Full work = 5 × 3 = 15 days

Total distance = 240 + 160 = 400m. Speed = $\frac{400}{20}$ = 20 m/s = 72 km/h

To find profit percentage, we track the price through markup and discount stages, then compare final selling price to cost price.

Step 1: Assume a Cost Price

Let Cost Price = $\text{₹}100$ (assume convenient value)

Step 2: Calculate Marked Price (25% markup)

The shopkeeper marks goods 25% above cost price:

Step 3: Calculate Selling Price (after 10% discount)

A discount of 10% is given on the marked price:

Step 4: Calculate Profit Percentage

Profit = Selling Price − Cost Price:

Profit percentage:

**Answer: The shopkeeper makes a profit of $12.5\%$ (Option A)**

A's rate = $\frac{1}{12}$, B's rate = $\frac{1}{18}$. Combined rate = $\frac{1}{12}$ + $\frac{1}{18}$ = $\frac{5}{36}$. Time = $\frac{36}{5}$ = 7.2 hours

To find the speed of the boat in still water, we use the relationship between upstream/downstream speeds and the boat's speed relative to water.

Step 1: Find upstream and downstream speeds

Upstream speed is distance divided by time:

Downstream speed:

Step 2: Set up equations using boat and current speeds

Let $b$ = speed of boat in still water and $c$ = speed of current.

When going upstream, the current opposes motion:

When going downstream, the current aids motion:

Step 3: Solve for boat speed

Add equations (1) and (2) to eliminate $c$:

**Answer: The speed of the boat in still water is $5.5\text{km/h}$ (Option A)**

To find the final amount with compound interest, use the formula $A=P(1+r{)}^n$ where principal is compounded annually.

Step 1: Identify the given values

Step 2: Apply the compound interest formula

The amount after compound interest is:

Step 3: Substitute the values

Step 4: Calculate the final amount

Answer: The amount is Rs. 5832 (Option A)

Speed = Distance/Time = $\frac{160}{8}$ = 20 m/s = 20 × 3.6 = 72 km/h

Combined rate = $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{7}{60}$. Time = $\frac{60}{7}$ ≈ 8.57 days

SP = CP × (1 - loss%) = 8000 × 0.85 = Rs. 6800

Relative speed = 80 - 60 = 20 km/h. Distance = 20 × 3 = 60 km

If boys:girls = 3:2 and boys = 15, then girls = ($\frac{2}{3}$) × 15 = 10

Let original number = x. x × 1.20 = 480. x = $\frac{480}{1}$.20 = 400

Net rate = $\frac{1}{8}$ - $\frac{1}{12}$ = (3-2)/24 = $\frac{1}{24}$. Time = 24 hours