Quantitative Aptitude - MCQ Practice Questions

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)

HCF(60, 90) = 30. For three numbers, LCM = 1800. If third number is x, then HCF(60, 90, x) = 12 and LCM(60, 90, x) = 1800. LCM(60,90) = 180. We need LCM(180, x) = 1800. So x = 180 or factor to make 1800. Testing: 180 fits

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

Total distance = 250 + 150 = 400 meters. Time = 20 seconds. Speed = $\frac{400}{20}$ = 20 m/s = 20 × $\frac{18}{5}$ = 72 km/h

When a number divides three quantities leaving the same remainder, the greatest such divisor is the GCD of the differences between pairs of those numbers.

If a number leaves the same remainder when dividing 2070, 2415, and 2760, then it must exactly divide the differences of these numbers.

Differences:

2415−2070=345

2760−2415=345

2760−2070=690

Now find the HCF of 345,345, and 690:

gcd(345,690)=345

Therefore, the greatest number is 345.

Step 1: Find differences between pairs

If divisor $d$ leaves the same remainder $r$ when dividing 2070, 2415, and 2760, then $d$ must divide their pairwise differences:

Step 2: Find GCD of the differences

The greatest divisor must divide all differences. We need:

Since $690=345\times 2$:

Step 3: Verify the answer

Check that 345 divides each number with the same remainder:

All leave remainder $r=0$. ✓

**Answer: The greatest number is $345$ (Option D)**

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