TNPSC Group 4 Maths Questions with Solutions – 15 Important Quant Problems

 

TNPSC Group 4 Quantitative Aptitude

15 Important Maths Questions with Detailed Solutions

TNPSC Group 4 | Maths Practice | Competitive Exams

Prepare for the TNPSC Group 4 examination with these 15 important quantitative aptitude questions and solved problems. Practise average speed, time and work, percentages, discounts, LCM, averages, algebra, number systems and other important mathematics topics.

Try to solve each question yourself before checking the answer and explanation.

Note: Questions 2, 11 and 12 contain mathematical notation that may have been lost in the supplied text. Verify the original expressions and answer choices before publishing those questions.

Important TNPSC Group 4 Maths Questions

Question 1: Average Speed

A train travels from station A to B at 120 kmph, from B to C at 100 kmph and from C to D at 140 kmph. If the time taken between each successive pair of stations is equal, find the average speed.

1. 120 kmph
2. 100 kmph
3. 140 kmph
4. 360 kmph

Correct Answer: Option 1 — 120 kmph

Detailed Solution

Let the time taken for each journey be t hours.

Total distance = 120t + 100t + 140t = 360t km.

Total time = 3t hours.

Average speed = Total distance ÷ Total time
= 360t ÷ 3t = 120 kmph

Therefore, the average speed is 120 kmph.

Question 3: Time and Work

A and B can build a wall in 20 and 35 days, respectively. If they work on alternate days, beginning with A, in how many days will the work be completed?

1. 25¼ days
2. 24 days
3. 25 days
4. 24½ days

Correct Answer: Option 1 — 25¼ days

Detailed Solution

Total work = LCM of 20 and 35 = 140 units.

A's one-day work = 140 ÷ 20 = 7 units.

B's one-day work = 140 ÷ 35 = 4 units.

In two days, they complete 7 + 4 = 11 units.

In 24 days, there are 12 such cycles.

Work completed = 12 × 11 = 132 units.

Remaining work = 140 − 132 = 8 units.

On day 25, A completes 7 units. The remaining work is 1 unit.

B completes 4 units per day, so the time required for 1 unit is 1/4 day.

Total time = 24 + 1 + 1/4 = 25¼ days

Question 4: Percentage

a is 10% of b, and b is 30% of c. If c = 2100, find a.

1. 630
2. 490
3. 63
4. 49

Correct Answer: Option 3 — 63

Detailed Solution

b = 30% of 2100 = (30/100) × 2100 = 630.

a = 10% of 630 = (10/100) × 630 = 63.

Therefore, a = 63.

Question 5: Successive Discounts

A laptop is sold for ₹28,512 after successive discounts of 10%, 12% and 25%. What would be its selling price if a single discount of 57% were offered on the same marked price?

1. ₹20,640
2. ₹25,860
3. ₹25,680
4. ₹22,460

Correct Answer: Option 1 — ₹20,640

Detailed Solution

Let the marked price be M.

M × 90/100 × 88/100 × 75/100 = 28,512

Therefore, M = ₹48,000.

After a single discount of 57%, the customer pays 43% of the marked price.

Selling price = 48,000 × 43/100 = ₹20,640

Question 6: LCM

The sum of two numbers is 50 and their product is 525. Find their LCM.

1. 85
2. 125
3. 115
4. 105

Correct Answer: Option 4 — 105

Detailed Solution

The two numbers are 35 and 15 because 35 + 15 = 50 and 35 × 15 = 525.

Prime factorisation:

35 = 5 × 7

15 = 3 × 5

LCM = 3 × 5 × 7 = 105

Question 7: Percentage Change in Savings

Anuj spends 90% of his income. His income increases by 20%, while his expenditure increases by 10%. Find the percentage increase or decrease in his savings.

1. 100%
2. 110%
3. 120%
4. 90%

Correct Answer: Option 2 — 110% increase

Detailed Solution

Assume the initial income is ₹100.

Initial expenditure = ₹90.

Initial savings = 100 − 90 = ₹10.

New income = 100 × 1.20 = ₹120.

New expenditure = 90 × 1.10 = ₹99.

New savings = 120 − 99 = ₹21.

Increase in savings = 21 − 10 = ₹11.

Percentage increase = (11/10) × 100 = 110%

Question 8: Average

What is the average of the first 100 positive integers?

1. 49
2. 50
3. 50.5
4. 51.5

Correct Answer: Option 3 — 50.5

Detailed Solution

For consecutive integers from 1 to n:

Average = (First number + Last number) ÷ 2

Average = (1 + 100) ÷ 2 = 101 ÷ 2 = 50.5.

Question 9: LCM and Remainders

Find the smallest positive number that leaves a remainder of 3 when divided by 4, 6, 10 and 15.

1. 58
2. 126
3. 37
4. 63

Correct Answer: Option 4 — 63

Detailed Solution

The required number minus 3 must be divisible by all four divisors.

LCM(4, 6, 10, 15) = 60.

Required number = 60 + 3 = 63

Check: 63 leaves a remainder of 3 when divided by each of 4, 6, 10 and 15.

Question 10: Modulus

When x = 2, find the value of |3x − 1| + |x − 2|.

1. 4
2. 5
3. 1
4. 3

Correct Answer: Option 2 — 5

Detailed Solution

Substitute x = 2.

|3(2) − 1| + |2 − 2|
= |6 − 1| + |0|
= 5 + 0 = 5

Question 11: Simplification of Mixed Fractions

Simplify the following expression. The expression below is reconstructed from the supplied solution and should be checked against the original question.

5¼ + 4¾ + 7⅝ + 6⅞ ÷ 11¾

1. 6847/376
2. 10/849
3. 98/45
4. 96/47

Answer from the supplied calculation: Option 1 — 6847/376

Detailed Solution

Apply BODMAS: perform division before addition.

= 21/4 + 19/4 + 61/8 + (55/8 ÷ 47/4)
= 21/4 + 19/4 + 61/8 + 55/94
= 6847/376

Note: Verify that the reconstructed mixed-fraction expression and options match the original paper.

Question 13: Geometric Progression (G.P.)

If x, 2x + 2 and 3x + 3 are in G.P., then 11x, 22x + 22 and 33x + 33 form:

1. An A.P.
2. A G.P.
3. A constant sequence
4. Neither A.P. nor G.P.

Expected Answer: Option 2 — A G.P.

Detailed Solution

For three terms a, b and c to be in G.P., the middle term squared equals the product of the other two terms.

(2x + 2)² = x(3x + 3)

Expanding and simplifying:

4x² + 8x + 4 = 3x² + 3x

x² + 5x + 4 = 0

(x + 1)(x + 4) = 0

Thus, x = −1 or x = −4.

For x = −4, the new terms are −44, −66 and −99. Their consecutive ratios are both 3/2, so they form a G.P.

Qualification: The original condition also permits x = −1, giving the sequence −11, 0, 0, which is degenerate under the usual nonzero-ratio definition of a G.P. The question should specify its intended domain or convention.

Question 14: Profit and Loss

Gain or loss percentage is always calculated on:

1. Cost price
2. Selling price
3. Gain
4. Loss

Correct Answer: Option 1 — Cost price

Detailed Solution

Profit % = (Profit ÷ Cost Price) × 100

Loss % = (Loss ÷ Cost Price) × 100

Both profit percentage and loss percentage are calculated using cost price as the base.

Question 15: Sum of Squares

Find the sum of the squares of the natural numbers from 1 to 20.

1. 1120
2. 1978
3. 2870
4. 579

Correct Answer: Option 3 — 2870

Detailed Solution

Formula for the sum of squares of the first n natural numbers:

Sum = n(n + 1)(2n + 1) ÷ 6

Here, n = 20.

= 20 × 21 × 41 ÷ 6
= 2870

Quick Answer Key

1. 120 kmph
2. 
3. 25¼ days
4. 63
5. ₹20,640
6. 105
7. 110% increase
8. 50.5
9. 63
10. 5
11. 6847/376
12. 
13. G.P. (with qualification noted above)
14. Cost price
15. 2870

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