Quantitative Aptitude - MCQ Practice Questions

To find how many numbers between 50 and 150 are divisible by 7, we identify the smallest and largest multiples of 7 in this range, then count them using the arithmetic sequence formula.

Step 1: Find the smallest multiple of 7 greater than 50

Divide 50 by 7:

The next whole number is 8, so the smallest multiple is:

Step 2: Find the largest multiple of 7 less than 150

Divide 150 by 7:

The largest whole number is 21, so the largest multiple is:

Step 3: Identify the arithmetic sequence

The multiples of 7 from 56 to 147 form an A.P. with:

Step 4: Use the A.P. formula to find the count

For an A.P., $a_n=a_1+(n-1)d$

Answer: There are 14 numbers between 50 and 150 divisible by 7 (Option A)

Let numbers be 3k and 5k. Since gcd(3,5)=1, LCM = 3k×5k/1 = 15k. Given LCM = 150, so 15k = 150, k = 10. Numbers are 30 and 50.

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.

[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.

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

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.

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.

To find the value of this expression, we'll factor out the smallest power of 2 and simplify using algebraic factoring.

**Step 1: Factor out $2^7$ from all terms**

Each term contains at least $2^7$ as a factor, so we can write:

Step 2: Simplify the expression in parentheses

Calculate each power of 2:

Step 3: Multiply by the factored term

**Answer: $2^{10}-2^9-2^8-2^7=128$ (Option B)**

The digital root is found by repeatedly summing the digits of a number until a single digit remains.

Step 1: Sum all digits of 9875

Add each digit:

Step 2: Sum the digits of the result

Since 29 is not a single digit, sum its digits:

Step 3: Sum again until single digit

Since 11 is still two digits, sum once more:

Step 4: Verify the digital root

We have reached a single digit. The digital root of 9875 is $2$.

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**Answer: The digital root of 9875 is $2$ (Option A)**

Let consecutive odd numbers be (2n-1) and (2n+1). (2n-1)(2n+1) = 323. 4n² - 1 = 323, so 4n² = 324, n² = 81, n = 9. Numbers are 17 and 19.

Let number = 5a + 2 = 7b + 3. From 5a + 2 = 7b + 3, we get 5a = 7b + 1. Testing b = 2: 7(2) + 1 = 15, a = 3. Number = 5(3) + 2 = 17. Check: 17 ÷ 5 = remainder 2, 17 ÷ 7 = remainder 3. ✓

Let number = 10a + b. (10a + b) - (10b + a) = 45. So 9a - 9b = 45, a - b = 5. Also a + b = 9. Solving: a = 7, b = 2. Number = 72.

144 = 2⁴ × 3², 108 = 2² × 3³. HCF = 2² × 3² = 4 × 9 = 36.

For any two numbers a and b: a × b = HCF(a,b) × LCM(a,b). So 2160 = 12 × LCM. LCM = $\frac{2160}{12}$ = 180.

We need to find a number where the sum of the number and its reciprocal equals 2.1.

Step 1: Set up the equation

Let the number be $x$. Then:

Step 2: Convert to standard form

Multiply both sides by $x$:

Rearrange to standard quadratic form:

Step 3: Convert decimal to fraction and solve

Rewrite $2.1=\frac{21}{10}$:

Multiply by 10 to clear fractions:

Using the quadratic formula:

Step 4: Factor and find exact solutions

Actually, factor directly: $10x^2-21x+10=(5x-2)(2x-5)=0$

This gives:

Step 5: Verify

For $x=\frac{2}{5}$: $\frac{2}{5}+\frac{5}{2}=\frac{4+25}{10}=\frac{29}{10}=2.9$ ✗

For $x=\frac{5}{2}$: $\frac{5}{2}+\frac{2}{5}=\frac{25+4}{10}=\frac{29}{10}=2.9$ ✗

Both values work identically (they are reciprocals of each other).

**Answer: The number could be $\frac{2}{5}$ or $\frac{5}{2}$ (Option C)**