Quantitative Aptitude - MCQ Practice Questions

SI = (25000 × 8 × 3)/100 = ₹6000. Total = 25000 + 6000 = ₹31,000

When two numbers have a given HCF and LCM, we can express them using the relationship: if HCF = $h$ and LCM = $l$, then the numbers are $h\cdot a$ and $h\cdot b$ where $\gcd (a,b)=1$ and $a\cdot b=\frac{l}{h}$.

Step 1: Apply the fundamental relationship

For any two numbers with $\text{HCF}=18$ and $\text{LCM}=540$, we can write:

where $\gcd (a,b)=1$ (a and b are coprime) and $a\cdot b=\frac{\text{LCM}}{\text{HCF}}$

Step 2: Find the product of coprime factors

Step 3: Prime factorize 30 to find coprime pairs

All pairs $(a,b)$ where $a\cdot b=30$ and $\gcd (a,b)=1$:

Step 4: Count distinct unordered pairs

Since we need pairs of numbers (unordered), we count:

This gives us the number pairs:

**Answer: There are $4$ pairs of such numbers (Option C**)

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

Using formula: HCF × LCM = Product of two numbers. 15 × 360 = 45 × x. Therefore x = $\frac{5400}{45}$ = 120

144 = 2⁴×3², 180 = 2²×3²×5, 216 = 2³×3³. HCF = 2²×3² = 4×9 = 36

Speed = $\frac{360}{6}$ = 60 km/h. Time = Distance/Speed = $\frac{600}{60}$ = 10 hours

24 = 2³×3, 36 = 2²×3². LCM = 2³×3² = 8×9 = 72

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.

SI = (P×R×T)/100 = (8000×6×2.5)/100 = $\frac{120000}{100}$ = ₹1200

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

12 = 2²×3, 18 = 2×3², 24 = 2³×3. LCM = 2³×3² = 8×9 = 72

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

When two trains travel towards each other, their relative speed is the sum of their individual speeds, and they must cover a combined distance equal to the sum of their lengths.

Step 1: Convert Speeds to m/s

[Since speeds are given in km/h, we convert them to m/s by multiplying by $\frac{5}{18}$]

Step 2: Find Relative Speed and Time

[When trains move towards each other, relative speed = sum of individual speeds, and time = total distance ÷ relative speed]

The time to cross each other is 12.6 seconds.

The answer is (B) 12.6 seconds

Profit = 520 - 400 = ₹120. Profit% = ($\frac{120}{400}$)×100 = 30%

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