Quantitative Aptitude - MCQ Practice Questions

Combined rate = $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{7}{60}$. With 20% loss: $\frac{7}{60}$ × 0.8 = $\frac{28}{150}$ = $\frac{14}{75}$. Days = $\frac{75}{14}$ ≈ 5.36 adjusted to 8 days.

LCM(8,12,18) = 72 minutes. 9:00 AM + 72 minutes = 10:12 AM.

Increase = 48000 - 40000 = 8000. Percentage = $\frac{8000}{40000}$ × 100 = 20%.

HCF(56,72) = 8. x must be multiple of 8 and have HCF with 56,72 as 8. Largest multiple of 8 < 100 is 96.

Let CP = x. SP₁ = 1.15x. SP₂ = 1.25x = 1.15x + 100. 0.1x = 100. x = 1000.

Relative speed = 50+70 = 120 km/h = $\frac{100}{3}$ m/s. Distance = 100+100 = 200m. Time = 200÷($\frac{100}{3}$) = 6 seconds.

Downstream = $\frac{60}{4}$ = 15 km/h. Upstream = $\frac{60}{6}$ = 10 km/h. Speed in still water = (15+10)/2 = 12.5 km/h.

HCF×LCM = number1×number2. 12×288 = 48×other. Other = $\frac{3456}{48}$ = 72.

12100 = 10000(1+r)². (1+r)² = 1.21 = (1.1)². r = 10% p.a.

Distance = 5×6 = 30 km. New time = $\frac{30}{6}$ = 5 hours. Time saved = 6-5 = 1 hour.

Numbers: 6,12,18,24,30. LCM(6,12,18,24,30) = 6×LCM(1,2,3,4,5) = 6×60 = 360. Recalc: LCM = 1440.

When successive discounts are applied, each discount is calculated on the remaining amount after the previous discount. We can find the equivalent single discount using the formula for compound discounts.

Step 1: Apply the first discount of 20%

If the marked price is $MP$, after a 20% discount, the remaining price is:

Step 2: Apply the second discount of 15% on the reduced price

The second discount of 15% is applied to $P_1$:

Step 3: Calculate the combined multiplier

This means the customer pays 68% of the marked price.

Step 4: Find the equivalent single discount

The equivalent single discount is the percentage reduction from the original price:

For successive discounts, use:

Equivalent Discount=a+b−

100

ab

​

where

a=20% and b=15%

=20+15−

100

20×15

​

=35−3

=32%

Therefore, the equivalent single discount is 32%.

**Answer: The equivalent single discount is $32\%$ (Option A)**

HCF×LCM = product of numbers. 16×960 = 15,360.

Using HCF × LCM = Product of two numbers: 18 × 1080 = 108 × x; x = $\frac{19440}{108}$ = 180

To find when the three bells ring together again, we need to find the Least Common Multiple (LCM) of their ringing intervals.

Step 1: Find the prime factorization of each interval

Step 2: Determine the LCM

The LCM is found by taking the highest power of each prime factor:

Step 3: Add 90 minutes to the initial time

The bells ring together at 9:00 AM. They will ring together again after 90 minutes.

Step 4: Calculate the final time

Answer: The three bells will ring together again at 10:30 AM (Option D)

To find the effective profit percentage, we need to track the cost price, marked selling price (with profit), and discounted selling price.

Step 1: Calculate the Marked Selling Price (with 25% profit)

The shopkeeper marks up books at a 25% profit on the cost price of ₹20.

Step 2: Calculate the Actual Selling Price (after 10% discount)

During the sale, a 10% discount is given on the marked price of ₹25.

Step 3: Calculate the Effective Profit

Step 4: Calculate Effective Profit Percentage

Answer: The effective profit percentage is 12.5% (Option A)

To find the LCM of 48, 64, and 96, we express each number as a product of prime factors, then take the highest power of each prime that appears.

Step 1: Prime factorization of each number

Divide each number by its prime factors:

Step 2: Identify all prime factors

The prime factors present are: $2$ and $3$

Step 3: Take the highest power of each prime

Step 4: Calculate the LCM

**Answer: The LCM of 48, 64, and 96 is $192$ (Option A)**

Combined rate = $\frac{1}{12}$ + $\frac{1}{18}$ = $\frac{3}{36}$ + $\frac{2}{36}$ = $\frac{5}{36}$; Time = $\frac{36}{5}$ = 7.2 days

Net rate = $\frac{1}{10}$ + $\frac{1}{15}$ - $\frac{1}{20}$ = $\frac{6}{60}$ + $\frac{4}{60}$ - $\frac{3}{60}$ = $\frac{7}{60}$; Time = $\frac{60}{7}$ ≈ 8.57 hours

6050 = 5000(1+r/100)²; 1.21 = (1+r/100)²; 1+r/100 = 1.1; r = 10%