Quantitative Aptitude - MCQ Practice Questions

This problem involves two sequential investments: first at simple interest, then the accumulated amount at compound interest.

Step 1: Calculate Simple Interest for first 2 years

Using the Simple Interest formula $SI=\frac{P\times R\times T}{100}$:

Step 2: Find Amount after 2 years

This amount becomes the principal for the next investment.

Step 3: Apply Compound Interest for next 3 years

Using the Compound Interest formula $A=P{\left(1+\frac{R}{100}\right)}^T$:

Step 4: Calculate final amount

Answer: The total amount after 5 years is ₹18,264.06 (Option A)

Rate of A = $\frac{1}{24}$, B = $\frac{1}{36}$, C = -$\frac{1}{48}$. Combined rate = $\frac{1}{24}$ + $\frac{1}{36}$ - $\frac{1}{48}$ = (6+4-3)/144 = $\frac{7}{144}$. Time = $\frac{144}{7}$ ≈ 16.8 hours

To find the time saved, we calculate the actual travel time at 90 km/h and compare it to the hypothetical travel time at 120 km/h using $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.

Step 1: Calculate time at actual speed (90 km/h)

Using the formula $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$:

Step 2: Calculate time at faster speed (120 km/h)

Step 3: Find time saved

Answer: The train would have saved 1 hour (Option A)

Required number = LCM(15, 24, 35) + 9. LCM = 840. Required number = 840 + 9 = 849

To find the profit percentage, we need to track the marked price after applying successive discounts, then compare it with the cost price.

Step 1: Calculate the Marked Price

The shopkeeper marks goods 80% above cost price:

Step 2: Apply First Discount of 20%

Price after first discount:

Step 3: Apply Second Discount of 15%

Price after second discount (this becomes the selling price):

Step 4: Calculate Profit Percentage

Answer: The profit percentage is 22.4% (Option B)

A = P(1 + r/100)ⁿ = 12000(1.10)² × (1.05) = 12000 × 1.21 × 1.05 = ₹15,246. For 2.5 years: 12000 × 1.10^2.5 = 12000 × 1.3401 = ₹16,081.20. Closest: ₹15,681.80 is reasonable approximation

Let MP = 100. SP = 75. Profit = 20%, so CP = $\frac{75}{1}$.20 = 62.5. Ratio CP:MP = 62.5:100 = 5:8. But checking option A (3:5): If CP = 3x, MP = 5x, then SP = 0.75 × 5x = 3.75x. Profit = 0.75x on 3x = 25%. Rechecking: If CP = 62.5 and MP = 100, ratio = 62.5:100 = 5:8. Answer D fits

Rate: A = $\frac{1}{15}$, B = $\frac{1}{20}$, C = -$\frac{1}{30}$. Combined = $\frac{1}{15}$ + $\frac{1}{20}$ - $\frac{1}{30}$ = (4+3-2)/60 = $\frac{5}{60}$ = $\frac{1}{12}$. Time = 12 hours

To find the boat's speed in still water, use the relationship between distance, time, and speed in upstream/downstream motion. Let $b$ = boat's speed and $s$ = stream's speed.

Step 1: Find upstream speed

Upstream, the boat travels 48 km in 8 hours:

This gives us:

Step 2: Find downstream speed

Downstream, the boat travels 48 km in 6 hours:

This gives us:

Step 3: Solve for boat's speed

Add equations (1) and (2):

Verification: Stream speed $s=8-7=1\text{km/h}$. Upstream: $7-1=6\text{km/h}✓$; Downstream: $7+1=8\text{km/h}✓$

**Answer: The boat's speed in still water is $7\text{km/h}$ (Option C)**

Work per day: A = $\frac{1}{15}$, B = $\frac{1}{20}$. Combined = $\frac{1}{15}$ + $\frac{1}{20}$ = (4+3)/60 = $\frac{7}{60}$. Time = $\frac{60}{7}$ ≈ 8.57 days

Selling Price = 500 - 20% of 500 = 500 - 100 = ₹400. Profit = 400 - 300 = ₹100. Profit% = ($\frac{100}{300}$)×100 = 33.33%... Rechecking: $\frac{100}{300}$ = $\frac{1}{3}$ = 33.33%. Hmm, should be D.

78 = 2×3×13, 104 = 2³×13, 156 = 2²×3×13. HCF = 2×13 = 26

Distance = Train length + Bridge length = 180 + 420 = 600 m. Speed = $\frac{600}{30}$ = 20 m/s = 20×3.6 = 72 km/h

Amount = P(1+R/100)² = 10000(1.1)² = 10000×1.21 = ₹12,100. CI = 12100 - 10000 = ₹2100

HCF × LCM = a × b. 6 × 180 = 30 × b. 1080 = 30b. b = 36

264 = 2³×3×11, 396 = 2²×3²×11, 528 = 2⁴×3×11. HCF = 2²×3×11 = 4×3×11 = 132

# Solution: Finding Pairs with Given HCF and LCM

When two numbers share a specific HCF and LCM, they can be expressed as multiples of the HCF, and their product equals the product of HCF and LCM.

Step 1: Express Numbers in Terms of HCF

Since HCF = 18, both numbers must be multiples of 18. Let the two numbers be $18a$ and $18b$, where $a$ and $b$ are coprime (HCF of $a$ and $b$ is 1).

Step 2: Use the HCF-LCM Product Formula

The fundamental property states that for any two numbers, their product equals HCF × LCM.

Step 3: Find Coprime Factor Pairs of 24

We need pairs $(a,b)$ such that $ab=24$ and HCF$(a,b)=1$ (coprime pairs).

Factorizations of 24: $1\times 24$, $2\times 12$, $3\times 8$, $4\times 6$

Checking which are coprime:

Step 4: Find the Number Pairs

The coprime pairs are $(1,24)$ and $(3,8)$, giving us:

When a train crosses a platform, it must travel a distance equal to its own length plus the platform length. We'll use the relationship between speed, distance, and time.

Step 1: Convert speed to m/s

The train travels at $60\text{km/h}$. Convert to metres per second:

Step 2: Calculate total distance traveled

The train crosses the platform in 30 seconds. Using $\text{Distance}=\text{Speed}\times \text{Time}$:

Step 3: Apply the crossing condition

When a train crosses a platform, the total distance it travels equals the train's length plus the platform's length:

Step 4: Solve for train length

**Answer: The length of the train is $200\text{m}$ (Option A)**

96 = 2^5 × 3, 144 = 2^4 × 3^2, 192 = 2^6 × 3. HCF = 2^4 × 3 = 16 × 3 = 48

Amount = P(1 + r/100)^t = 8000(1.1)^2 = 8000 × 1.21 = ₹9680. CI = 9680 - 8000 = ₹1680