Quantitative Aptitude - MCQ Practice Questions

Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024

Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28

2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9

Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.

144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.

We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = $\frac{120}{4}$ = 30.

Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.

For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: $\frac{58}{7}$ = 8 rem 2, not 3. Try 38: $\frac{38}{7}$ = 5 rem 3 ✓, $\frac{38}{11}$ = 3 rem 5 ✓. Answer is A=38.

To find the remainder when $2^{100}$ is divided by 7, we use Fermat's Little Theorem, which states that if $p$ is prime and $\gcd (a,p)=1$, then $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Step 1: Apply Fermat's Little Theorem

Since 7 is prime and $\gcd (2,7)=1$:

Step 2: Express the exponent in terms of 6

Divide 100 by 6:

Therefore:

Step 3: Simplify using the congruence

**Step 4: Calculate $2^4\mathrm{mod}7$

Therefore:

**Answer: The remainder is $2$ (Option B)**

# Unit Digit of 7^2019

To find the unit digit of any power, we identify the cyclical pattern of unit digits for that base number.

Step 1: Find the Cyclical Pattern of Unit Digits for Powers of 7

The unit digit of powers of 7 repeats in a cycle. Let's calculate the first few powers:

The pattern repeats: 7, 9, 3, 1, 7, 9, 3, 1, ...

The cycle length is 4.

Step 2: Find the Position of 7^2019 in the Cycle

Divide the exponent 2019 by the cycle length 4 to find which position in the pattern it corresponds to:

Since the remainder is 3, we need the 3rd unit digit in our cycle pattern (7, 9, 3, 1).

The 3rd position corresponds to unit digit 3.

The unit digit of 7^2019 is 3.

Answer: (A) 3

Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.

To find $3^5\mathrm{mod}11$, we calculate successive powers of 3 and reduce modulo 11 at each step to keep numbers manageable.

**Step 1: Calculate $3^2\mathrm{mod}11$

Since $9<11$, we have $3^2\equiv 9(\mathrm{mod}11)$

**Step 2: Calculate $3^4\mathrm{mod}11$

Dividing: $81=7\times 11+4$, so $3^4\equiv 4(\mathrm{mod}11)$

**Step 3: Calculate $3^5\mathrm{mod}11$

Step 4: Final reduction

**Answer: $3^5\equiv 1(\mathrm{mod}11)$ (Option A)**

HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.

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.