Quantitative Aptitude - MCQ Practice Questions

Using the property: GCD(a,b) × LCM(a,b) = a × b. Therefore: 12 × 144 = 36 × b. So 1728 = 36b, which gives b = 48.

Let number be 10a + b. Reversed number is 10b + a. Given: (10b + a) - (10a + b) = 27, so 9b - 9a = 27, thus b - a = 3. Also a + b = 9. Solving: b = 6, a = 3. Original number = 36.

Let three consecutive odd numbers be x, x+2, x+4. Their sum: x + (x+2) + (x+4) = 51. So 3x + 6 = 51, thus 3x = 45, and x = 15.

Prime factorizations: 24 = 2³ × 3, 36 = 2² × 3², 60 = 2² × 3 × 5. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.

Factoring: x² - 5x + 6 = (x - 2)(x - 3) = 0. Therefore x = 2 or x = 3.

We need to find a number that satisfies two remainder conditions simultaneously. This is a Chinese Remainder Theorem problem.

Step 1: Express the first condition

When the number is divided by 7, remainder is 4:

where $k$ is a non-negative integer.

This means $n\in \{4,11,18,25,32,39,46,53,60,67,74,81,88,95,...\}$

Step 2: Apply the second condition

The same number divided by 11 gives remainder 6:

where $m$ is a non-negative integer.

This means $n\in \{6,17,28,39,50,61,72,83,94,...\}$

Step 3: Find the common value

We need a number that appears in both lists and lies between 1 and 100.

From Step 1: $\{4,11,18,25,32,39,46,53,60,67,74,81,88,95\}$

From Step 2: $\{6,17,28,39,50,61,72,83,94\}$

The common element is $n=39$.

Step 4: Verify the answer

✓

✓

Answer: The number is 39 (Option B)

A number is divisible by 9 if sum of its digits is divisible by 9. Here sum is 12, which is not divisible by 9. However, any number divisible by 9 is also divisible by 3. But the given condition states sum of digits is 12, and divisible by 9, which is contradictory. Re-reading: if divisible by 9, then sum must be divisible by 9. Since sum is 12 and divisible by 3, the number is divisible by 3.

Perfect squares from 1 to 1000 are 1², 2², 3², ..., n² where n² ≤ 1000. So n ≤ √1000 ≈ 31.62. Therefore n can be 1, 2, 3, ..., 31. Total = 31 perfect squares.

Let the two consecutive even numbers be n and n+2. Then n(n+2) = 528. So n² + 2n - 528 = 0. Using quadratic formula or factoring: (n+24)(n-22) = 0. So n = 22 (taking positive value). The two numbers are 22 and 24. Larger = 24.

Let number = 10a + b. Reversed = 10b + a. Given: (10b + a) - (10a + b) = 45, so 9b - 9a = 45, thus b - a = 5. Also |a - b| = 5 or a - b = 5. From b - a = 5 and a - b could be -5 or 5. Testing: if b - a = 5 and digits sum conditions... Let a = 2, b = 7: number = 27. Reversed = 72. 72 - 27 = 45. ✓

Using Euclidean algorithm: 180 = 144×1 + 36; 144 = 36×4 + 0. Therefore HCF = 36.

[When a number is divisible by two coprime numbers, it must be divisible by their product.]

Step 1: Identify the Given Conditions

We are told that a number is divisible by both 9 and 11, meaning both divide evenly into this number with no remainder.

Step 2: Check if 9 and 11 are Coprime

Two numbers are coprime if their greatest common divisor (GCD) is 1. Since 9 = 3² and 11 is prime, they share no common factors.

Step 3: Apply the Divisibility Rule for Coprime Numbers

When a number is divisible by two coprime numbers, it must be divisible by their product (by the Fundamental Theorem of Arithmetic).

The answer is (B) 99.

Digit sum = 9 + 8 + 7 + 6 + 5 + 4 = 39.

360 = 2³×3²×5¹. Number of divisors = (3+1)(2+1)(1+1) = 4×3×2 = 24.

Let numbers be x and y. x + y = 50 and x - y = 10. Adding: 2x = 60, so x = 30.

A number is divisible by 8 if its last 3 digits form a number divisible by 8. 456 ÷ 8 = 57. So 2456 is divisible by 8.

1071 = 462×2 + 147; 462 = 147×3 + 21; 147 = 21×7 + 0. Therefore HCF = 21.

When a number leaves remainder 3 when divided by 5, we can express it as $n=5k+3$ for some integer $k$. To find the remainder when divided by 15, we need to examine what values this number can take.

Step 1: Express the general form

Any number satisfying the given condition can be written as:

where $k$ is a non-negative integer.

Step 2: Test specific values

Let's substitute different values of $k$ and find remainders when divided by 15:

Step 3: Analyze the pattern

The possible remainders are $\{3,8,13,3,8,13,\dots \}$

We get three different remainders: 3, 8, and 13 (cycling as $k$ varies).

Step 4: Conclusion

Since the same condition (remainder 3 when divided by 5) can produce different remainders when divided by 15 depending on which specific number we choose, the remainder cannot be uniquely determined from the given information alone.

Answer: The remainder cannot be determined uniquely. (Option D)

To find the smallest number to add to 5800 to make it a perfect square, we need to find the smallest perfect square greater than 5800.

Step 1: Find the approximate square root of 5800

Calculate $\sqrt{5800}$ to determine which perfect square is nearest:

Since $\sqrt{5800}$ lies between 76 and 77, the next perfect square will be $77^2$.

Step 2: Calculate the next perfect square

Step 3: Find the difference

The number we must add is:

Step 4: Verify the answer

Check that $5800+129=5929=77^2$ ✓

Answer: The smallest number to be added is 129 (Option D)

Even numbers: 2, 4, 6, ..., 100. This is AP with a=2, l=100, d=2. n = 50 terms. Sum = (n/2)(a+l) = ($\frac{50}{2}$)(2+100) = 25×102 = 2550.