Quantitative Aptitude - MCQ Practice Questions

Step 1: SI for first 2 years = (15000 × 10 × 2) / 100 = ₹3000.

Step 2: After 2 years, remaining principal = 15000 - 3000 = ₹12000.

Step 3: SI on ₹12000 for 1 year = (12000 × 10 × 1) / 100 = ₹1200.

Step 4: Total SI = 3000 + 1200 + ₹1500 (interest on ₹3000 for 1 year) = ₹5700.

We use the simple interest formula $I=\frac{P\times R\times T}{100}$ to find interest on equal principal amounts invested at different rates and periods, then set up an equation using the given difference.

Step 1: Write the interest formula for each investment

Let the principal be $P$ (same for both).

For Investment 1 (12% p.a. for 4 years):

For Investment 2 (15% p.a. for 3 years):

Step 2: Find the difference in interests

Since $I_1>I_2$ (higher rate × longer time):

Step 3: Use the given difference to find P

We're told the difference is ₹540:

Step 4: Verify the answer

$I_1=0.48\times 18000=8640$

$I_2=0.45\times 18000=8100$

Difference: $8640-8100=540$ ✓

Let each principal amount be P.

Using Simple Interest formula:

SI=

100

P×R×T

​

First investment

SI

1

​

=

100

P×12×4

​

=

100

48P

​

Second investment

SI

2

​

=

100

P×15×3

​

=

100

45P

​

Difference in interests:

100

48P

​

−

100

45P

​

=540

100

3P

​

=540

3P=54000

P=18000

Therefore, each principal amount is ₹18,000.

Answer: Each principal amount is ₹18,000 (Option B)

This question asks us to find the smallest positive number that is divisible by both 12 and 18.

Step 1: Find Prime Factorization of Both Numbers

Break down each number into its prime factors.

Step 2: Identify Highest Powers of All Prime Factors

The LCM uses the highest power of each prime factor that appears.

Step 3: Calculate the LCM

Multiply the highest powers together.

The LCM of 12 and 18 is 36, which is the smallest number divisible by both numbers.

Numbers of form 7k+3: when k=5, number=7(5)+3=38.

Check: 38÷7=5 remainder 3 ✓.

Check others: 24÷7=3 rem 3 (close but let's verify 38 first), 38÷7 gives remainder 3 ✓

Let smaller odd number = x.

Then x+(x+2)=56.

So 2x+2=56, 2x=54, x=27.

Check: 27+29=56 ✓

36 = 2² × 3², 48 = 2⁴ × 3. HCF = 2² × 3 = 12, LCM = 2⁴ × 3² = 144.

Product = 36 × 48 = HCF × LCM = 12 × 144 = 1728.

Using formula: HCF × LCM = Product of two numbers. 8 × 96 = 24 × x. 768 = 24x. x = 32.

100 = 2² × 5², 150 = 2 × 3 × 5², 200 = 2³ × 5².

Common factors: 2¹ × 5² = 2 × 25 = 50.

This question asks us to find the least common multiple (LCM) of two numbers using prime factorization.

Step 1: Find prime factorization of 84

Break 84 into its prime factors by dividing by smallest primes.

Step 2: Find prime factorization of 140

Break 140 into its prime factors by dividing by smallest primes.

Step 3: Calculate LCM using highest powers of all prime factors

LCM is found by taking the highest power of each prime factor that appears in either number.

The LCM of 84 and 140 is 420, which is option (A).

A's rate = $\frac{1}{12}$, B's rate = $\frac{1}{18}$.

Combined rate = $\frac{1}{12}$ + $\frac{1}{18}$ = $\frac{3}{36}$ + $\frac{2}{36}$ = $\frac{5}{36}$.

Time = $\frac{36}{5}$ = 7.2 days

Combined rate = $\frac{1}{10}$ + $\frac{1}{15}$ + $\frac{1}{30}$ = $\frac{3}{30}$ + $\frac{2}{30}$ + $\frac{1}{30}$ = $\frac{6}{30}$ = $\frac{1}{5}$.

Time = 5 days

A completes $\frac{1}{3}$ work in 5 days, so rate = $\frac{1}{15}$ per day.

Remaining work = $\frac{2}{3}$.

Days needed = ($\frac{2}{3}$)/($\frac{1}{15}$) = ($\frac{2}{3}$) × 15 = 10 days

Combined rate = $\frac{1}{8}$, A's rate = $\frac{1}{12}$. B's rate = $\frac{1}{8}$ - $\frac{1}{12}$ = $\frac{3}{24}$ - $\frac{2}{24}$ = $\frac{1}{24}$. B takes 24 days

Let numbers be 5x and 7x.

Then 5x + 7x = 120, so 12x = 120, x = 10.

Larger number = 7 × 10 = 70

Odd numbers between 10 and 30: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

Count = 10

Even numbers from 2 to 50: 2, 4, 6, ..., 50.

This is an AP with first term = 2, last term = 50, common difference = 2.

Number of terms = 25.

Sum = 25(2+50)/2 = 25 × 26 = 650

Let original number = x.

Then x + 25% of x = 500.

So x + 0.25x = 500, 1.25x = 500, x = 400

Let smaller number = x, larger = 2x.

Then 2x - x = 45, so x = 45.

Numbers are 45 and 90.

Let integers be n-1, n, n+1.

Sum = 3n = 81, so n = 27.

The three integers are 26, 27, 28.

Largest = 28

Sum of squares = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55.

Formula: n(n+1)(2n+1)/6 = 5×6×$\frac{11}{6}$ = 55