State Exam — Quantitative Aptitude
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Showing 311–320 of 499 questions
Q.311 Medium HCF and LCM
Three bells ring at intervals of 8, 12, and 18 minutes. If they ring together at 9:00 AM, when will they ring together again?
A9:72 AM
B10:12 AM
C11:00 AM
D11:30 AM
Correct Answer:  B. 10:12 AM
Explanation:

LCM(8,12,18) = 72 minutes. 9:00 AM + 72 minutes = 10:12 AM.

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Q.312 Medium HCF and LCM
HCF of three numbers is 8. If the numbers are 56, 72, and x, find the largest possible value of x less than 100.
A80
B88
C96
D92
Correct Answer:  C. 96
Explanation:

HCF(56,72) = 8. x must be multiple of 8 and have HCF with 56,72 as 8. Largest multiple of 8 < 100 is 96.

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Q.313 Medium HCF and LCM
A trader sells an item at 15% profit. If he had sold it for ₹100 more, his profit would be 25%. Find cost price.
A₹1000
B₹1200
C₹1500
D₹2000
Correct Answer:  A. ₹1000
Explanation:

Let CP = x. SP₁ = 1.15x. SP₂ = 1.25x = 1.15x + 100. 0.1x = 100. x = 1000.

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Q.314 Medium HCF and LCM
Two trains of length 100m each run towards each other at 50 km/h and 70 km/h. Time to cross each other completely?
A3 seconds
B4 seconds
C5 seconds
D6 seconds
Correct Answer:  D. 6 seconds
Explanation:

Relative speed = 50+70 = 120 km/h = 100/3 m/s. Distance = 100+100 = 200m. Time = 200÷(100/3) = 6 seconds.

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Q.315 Medium HCF and LCM
A tank can be filled by pipes A, B, C in 10, 15, 20 hours respectively. Pipe D empties in 30 hours. All open together, how long to fill?
A4 hours
B5 hours
C6 hours
D7 hours
Correct Answer:  C. 6 hours
Explanation:

Net rate = 1/10 + 1/15 + 1/20 - 1/30 = 6/60 + 4/60 + 3/60 - 2/60 = 11/60. Time = 60/11 ≈ 5.45, closest is 6.

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Q.316 Medium HCF and LCM
A boat takes 4 hours to travel 60 km downstream and 6 hours to travel 60 km upstream. Speed in still water?
A12.5 km/h
B13.5 km/h
C14 km/h
D15 km/h
Correct Answer:  A. 12.5 km/h
Explanation:

Downstream = 60/4 = 15 km/h. Upstream = 60/6 = 10 km/h. Speed in still water = (15+10)/2 = 12.5 km/h.

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Q.317 Medium HCF and LCM
HCF and LCM of two numbers are 12 and 288 respectively. If one number is 48, find the other.
A60
B72
C84
D96
Correct Answer:  B. 72
Explanation:

HCF×LCM = number1×number2. 12×288 = 48×other. Other = 3456/48 = 72.

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Q.318 Medium HCF and LCM
A principal of ₹10,000 becomes ₹12,100 in 2 years at CI. Find the rate of interest per annum.
A8%
B10%
C12%
D15%
Correct Answer:  B. 10%
Explanation:

12100 = 10000(1+r)². (1+r)² = 1.21 = (1.1)². r = 10% p.a.

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Q.319 Medium HCF and LCM
A merchant offers successive discounts of 20% and 15%. Find equivalent single discount.
A32%
B33%
C34%
D35%
Correct Answer:  A. 32%
Explanation:

When successive discounts are applied, each discount is calculated on the remaining amount after the previous discount. We can find the equivalent single discount using the formula for compound discounts.

Step 1: Apply the first discount of 20%

If the marked price is \(MP\), after a 20% discount, the remaining price is:

\[P_1 = MP \times (1 - 0.20) = MP \times 0.80\]

Step 2: Apply the second discount of 15% on the reduced price

The second discount of 15% is applied to \(P_1\):

\[P_{\text{final}} = P_1 \times (1 - 0.15) = MP \times 0.80 \times 0.85\]

Step 3: Calculate the combined multiplier

\[P_{\text{final}} = MP \times 0.80 \times 0.85 = MP \times 0.68\]

This means the customer pays 68% of the marked price.

Step 4: Find the equivalent single discount

The equivalent single discount is the percentage reduction from the original price:

\[\text{Equivalent Discount} = (1 - 0.68) \times 100\% = 0.32 \times 100\% = 32\%\]

For successive discounts, use:

Equivalent Discount=a+b−

100

ab

where

a=20% and b=15%

=20+15−

100

20×15

=35−3

=32%

Therefore, the equivalent single discount is 32%.

Answer: The equivalent single discount is \(32\%\) (Option A)

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Q.320 Medium HCF and LCM
Two pipes can fill a tank in 10 hours and 15 hours respectively. A drain empties it in 20 hours. If all three are opened, how long to fill the tank?
A7.5 hours
B8.57 hours
C9 hours
D10 hours
Correct Answer:  B. 8.57 hours
Explanation:

Net rate = 1/10 + 1/15 - 1/20 = 6/60 + 4/60 - 3/60 = 7/60; Time = 60/7 ≈ 8.57 hours

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