Quantitative Aptitude - MCQ Practice Questions

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%

When a boat travels upstream, it moves against the current, so its effective speed decreases. Use the relationship: $\text{Upstream Speed}=\text{Speed in still water}-\text{Stream speed}$

Step 1: Find the upstream speed from distance and time

Given: Distance = 500 km, Time = 20 hours

Step 2: Set up the upstream speed equation

Let $s$ = speed of the stream (in km/hr)

The boat's speed in still water is 50 km/hr, so:

Step 3: Solve for stream speed

From Steps 1 and 2:

Verification: Upstream speed = $50-25=25$ km/hr ✓

Distance covered in 20 hours = $25\times 20=500$ km ✓

Answer: The speed of the stream is 25 km/hr (Option B)

Relative speed = 60 + 40 = 100 km/h = 100×$\frac{5}{18}$ = $\frac{250}{9}$ m/s; Total distance = 150+100 = 250m; Time = 250/($\frac{250}{9}$) = 9 seconds

When three numbers share a common HCF and LCM, we can use the fundamental property that for any number, both the HCF and LCM must divide it appropriately, and the product relationship: HCF × LCM = product of pairs.

Step 1: Verify the given numbers satisfy HCF and LCM conditions

For numbers 45 and 60:

Both must divide the LCM of all three numbers (360) and be divisible by the HCF (15). ✓

Step 2: Use the property relating HCF, LCM, and numbers

For any three numbers $a$, $b$, $c$ with HCF $h$ and LCM $l$:

For 45 and 60: $45=15\times 3$ and $60=15\times 4$

Step 3: Find prime factorizations in terms of HCF multiples

The third number $c$ must satisfy:

Step 4: Determine the third number

From LCM, the third number must contribute the $2^3$ factor (since 45 and 60 together only have $2^2$).

Let $c=2^a\times 3^b\times 5^d\times \dots$

For HCF to be exactly 15: we need $b=1$ and $d=1$

For LCM to be 360: we need $a=3$ (the maximum power of 2)

Verification: $\text{HCF}(45,60,120)=15$ ✓ and $\text{LCM}(45,60,120)=360$ ✓

**Answer: The third number is $120$ (Option A)**

After 15% discount: 85%; After 10% on that: 85×$\frac{90}{100}$ = 76.5%; Single discount = 100-76.5 = 23.5%

Let CP = x; At 20% profit: SP = 1.2x; At 30% profit: SP = 1.3x; 1.3x - 1.2x = 60; 0.1x = 60; x = 600