At what rate of interest per annum will ₹8,000 amount to ₹9,261 in 3 years, compounded annually?

The correct answer is B: 5%. Step 1: Use A = P(1 + r/100)^n. Step 2: 9261 = 8000(1 + r/100)^3. Step 3: (1 + r/100)^3 = 9261/8000 = 1.157625. Step 4: Taking cube root, 1 + r/100 = 1.05, so r = 5%. So option B is correct.

Compound interest on ₹10,000 at 10% per annum for 2 years is:

The correct answer is B: ₹2100. Amount = 10000(1.10)² = 10000 × 1.21 = ₹12100. CI = 12100 - 10000 = ₹2100

A shopkeeper marks an item at ₹1000 and offers a 20% discount. What is the selling price?

The correct answer is B: ₹800. SP = 1000 × (1 - 0.20) = 1000 × 0.80 = ₹800

A perfect square has how many odd divisors if the number is 144?

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

Two trains of lengths 200m and 150m travel towards each other at 60 km/h and 40 km/h respectively. Time to cross each other is?

The correct answer is B: 12.6 seconds. When two trains travel towards each other, their relative speed is the sum of their individual speeds, and they must cover a combined distance equal to the sum of their lengths. **Step 1: Convert Speeds to m/s** [Since speeds are given in km/h, we convert them to m/s by multiplying by 5/18] $$\text{

Corporate & IT Interview Preparation

Q.291 Easy HCF and LCM

Three bells ring at 8, 12, and 18-minute intervals. After how many minutes will they ring together if they start together?

A 36 minutes

B 48 minutes

C 60 minutes

D 72 minutes

Correct Answer: D. 72 minutes

Explanation:

To find when all three bells ring together, we need the Least Common Multiple (LCM) of their ringing intervals.

Step 1: Find prime factorization of each interval

\[8 = 2^3\]

12 = 2^2 \times 3

18 = 2 \times 3^2

Step 2: Identify highest powers of each prime factor

For LCM, take the highest power of each prime that appears:

- Highest power of 2: $2^3$ (from 8)

- Highest power of 3: $3^2$ (from 18)

Step 3: Calculate the LCM

\text{LCM}(8, 12, 18) = 2^3 \times 3^2 = 8 \times 9 = 72

Step 4: Verify the answer

- $72 \div 8 = 9$ ✓ (Bell 1 rings 9 times)

- $72 \div 12 = 6$ ✓ (Bell 2 rings 6 times)

- $72 \div 18 = 4$ ✓ (Bell 3 rings 4 times)

All three bells divide evenly into 72 minutes, confirming they ring together at this time.

Answer: The bells will ring together after $72$ minutes (Option D)

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Q.292 Easy HCF and LCM

The HCF of two numbers is 12 and their LCM is 180. If one number is 36, find the other number.

A 60

B 48

C 72

D 84

Correct Answer: A. 60

Explanation:

Using HCF × LCM = Product of two numbers. 12 × 180 = 36 × x. Therefore x = 2160/36 = 60

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Q.293 Easy HCF and LCM

Three numbers are in the ratio 2:3:4 and their LCM is 240. Find the HCF of these numbers.

A 20

B 30

C 15

D 10

Correct Answer: A. 20

Explanation:

Let numbers be 2k, 3k, 4k. LCM(2k, 3k, 4k) = 12k = 240, so k = 20. HCF = k = 20

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Q.294 Easy HCF and LCM

A product's cost price is ₹500. A trader marks it 60% above cost price and gives a discount of 25%. If HCF of profit and marked price is calculated, find the profit percentage.

A 20%

B 25%

C 30%

D 35%

Correct Answer: A. 20%

Explanation:

MP = 500 × 1.60 = ₹800. SP = 800 × 0.75 = ₹600. Profit = 100. Profit% = (100/500) × 100 = 20%

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Q.295 Easy HCF and LCM

HCF of two numbers is 11 and their sum is 99. If one number is 33, find the other number.

A 55

B 66

C 77

D 44

Correct Answer: B. 66

Explanation:

When two numbers share a common HCF (Highest Common Factor), both numbers must be multiples of that HCF. We can use this property along with the given sum to find the unknown number.

Step 1: Express both numbers as multiples of HCF

Since HCF = 11, both numbers can be written as:

\text{First number} = 11a, \quad \text{Second number} = 11b

where $a$ and $b$ are coprime integers (HCF of $a$ and $b$ is 1).

Step 2: Find the value of $a$ using the known number

One number is 33, so:

11a = 33 \Rightarrow a = 3

Step 3: Use the sum condition to find $b$

The sum of both numbers is 99:

\[11a + 11b = 99\]

\[11(3) + 11b = 99\]

\[33 + 11b = 99\]

\[11b = 66\]

\[b = 6\]

Step 4: Calculate the other number

\text{Other number} = 11b = 11 \times 6 = 66

Verification: HCF(33, 66) = 33... Wait, let me recalculate: 33 = 3 × 11 and 66 = 6 × 11 = 2 × 3 × 11. HCF = 11 ✓ and 33 + 66 = 99 ✓

Answer: The other number is 66 (Option B)

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Q.296 Easy HCF and LCM

The LCM of two coprime numbers is 143. If one number is 11, find the other.

A 13

B 15

C 17

D 19

Correct Answer: A. 13

Explanation:

For coprime numbers, HCF = 1. So LCM = Product. 143 = 11 × x. Therefore x = 13. Check: HCF(11,13) = 1 ✓

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Q.297 Easy HCF and LCM

A train 250 meters long passes a platform 150 meters long in 20 seconds. Find the speed of the train.

A 54 km/h

B 64.8 km/h

C 72 km/h

D 90 km/h

Correct Answer: C. 72 km/h

Explanation:

Total distance = 250 + 150 = 400 meters. Time = 20 seconds. Speed = 400/20 = 20 m/s = 20 × 18/5 = 72 km/h

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Q.298 Easy HCF and LCM

A person borrows ₹25,000 at 8% SI. After 3 years, how much total amount must he repay?

A ₹31,000

B ₹31,500

C ₹32,000

D ₹33,000

Correct Answer: A. ₹31,000

Explanation:

SI = (25000 × 8 × 3)/100 = ₹6000. Total = 25000 + 6000 = ₹31,000

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Q.299 Easy HCF and LCM

The HCF and LCM of two numbers are 15 and 360 respectively. If one number is 45, find the other number.

A 120

B 150

C 180

D 200

Correct Answer: A. 120

Explanation:

Using formula: HCF × LCM = Product of two numbers. 15 × 360 = 45 × x. Therefore x = 5400/45 = 120

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Q.300 Easy HCF and LCM

Find the HCF of 144, 180, and 216 using prime factorization method.

A 24

B 36

C 48

D 72

Correct Answer: B. 36

Explanation:

144 = 2⁴×3², 180 = 2²×3²×5, 216 = 2³×3³. HCF = 2²×3² = 4×9 = 36

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