CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31

Practice MCQs for Class 10 Mathematics Commercial Mathematics

Access targeted multiple-choice questions for Commercial Mathematics designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Commercial Mathematics Questions and Solutions

Access the complete set of multiple-choice questions for Commercial Mathematics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: A leak in the bottom of a tank can empty it in 6 hours. A pipe fills the tank at 4 litres per minute. When the tank is full, the inlet is opened, but because of the leak the tank is emptied in 8 hours. What is the capacity of the tank?
(a) 5260 litres
(b) 5760 litres
(c) 5846 litres
(d) 6970 litres
Show Answer & Explanation

Answer: (b) 5760 litres

Explanation:
1. Inlet = 240 litres per hour. Let the capacity be C: C/6 − 240 = C/8.
2. C/24 = 240, so C = 5760 litres.

Teacher's Note:
1. Net emptying rate = leak rate minus inlet rate.

Question 2: Because of a leak, a tank which is usually filled in 3 hours now takes 3.5 hours to be filled. The leak alone will empty the full tank in
(a) 18 hours
(b) 21 hours
(c) 15 hours
(d) Cannot be determined
Show Answer & Explanation

Answer: (b) 21 hours

Explanation:
1. Leak rate = 1/3 − 1/3.5 = 1/3 − 2/7 = 1/21.
2. The leak empties the tank in 21 hours.

Teacher's Note:
1. The leak is the difference between the two filling rates.

Question 3: Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is
(a) 50 m
(b) 72 m
(c) 80 m
(d) 82 m
Show Answer & Explanation

Answer: (a) 50 m

Explanation:
1. Relative speed = 10 km/hr = 25/9 m/s, so distance covered in 36 s = 100 m.
2. This is the sum of the lengths, so each train is 50 m.

Teacher's Note:
1. Overtaking covers the sum of the two lengths.

Question 4: A train X speeding at 120 km/hr crosses another train Y, running in the same direction, in 2 minutes. If the lengths of X and Y are 100 m and 200 m respectively, what is the speed of train Y?
(a) 111 km/hr
(b) 123 km/hr
(c) 127 km/hr
(d) 129 km/hr
Show Answer & Explanation

Answer: (a) 111 km/hr

Explanation:
1. Relative speed = 300 m / 120 s = 2.5 m/s = 9 km/hr.
2. Speed of Y = 120 − 9 = 111 km/hr.

Teacher's Note:
1. Same direction means subtract the speeds.

Question 5: A train overtakes two persons walking along a railway track in the same direction as the train. The first walks at 4.5 km/hr and the other at 5.4 km/hr. The train needs 8.4 seconds and 8.5 seconds respectively to overtake them. What is the speed of the train?
(a) 66 km/hr
(b) 72 km/hr
(c) 78 km/hr
(d) 81 km/hr
Show Answer & Explanation

Answer: (d) 81 km/hr

Explanation:
1. Let the speed be v: (v − 4.5) × 8.4 = (v − 5.4) × 8.5, both equal to the train length.
2. 8.4v − 37.8 = 8.5v − 45.9, so v = 81 km/hr.

Teacher's Note:
1. The train length is the same in both cases.

Question 6: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively, and they cross each other in 23 seconds. The ratio of their speeds is
(a) 1 : 3
(b) 3 : 2
(c) 3 : 4
(d) None of these
Show Answer & Explanation

Answer: (b) 3 : 2

Explanation:
1. Lengths are 27a and 17b for speeds a and b. Crossing each other: (27a + 17b)/(a + b) = 23.
2. 4a = 6b, so a : b = 3 : 2.

Teacher's Note:
1. Opposite directions add the speeds.

Question 7: A man is travelling by car at 40 km/hr. After every 80 km he rests for 20 minutes. How long will he take to cover 240 km?
(a) 6 hr 40 min
(b) 6 hr
(c) 6 hr 20 min
(d) 7 hr
Show Answer & Explanation

Answer: (a) 6 hr 40 min

Explanation:
1. Driving time = 240/40 = 6 hours.
2. He rests after 80 km and 160 km (not at the end), so 2 × 20 = 40 minutes: 6 hr 40 min.

Teacher's Note:
1. Count the number of rests carefully.

Question 8: At what percent per annum will simple interest double a sum of money in 12 years?
(a) 8 1/3 %
(b) 24 %
(c) 25/4 %
(d) 8.25 %
Show Answer & Explanation

Answer: (a) 8 1/3 %

Explanation:
1. Interest = principal, so P × r × 12/100 = P.
2. r = 100/12 = 8 1/3 %.

Teacher's Note:
1. Doubling means the interest equals the principal.

Question 9: Find the amount of Rs. 12000 after 2 years compounded annually, the rate being 5% p.a. in the first year and 6% p.a. in the second year, and the compound interest.
(a) Rs. 1356
(b) Rs. 1200
(c) Rs. 1256
(d) None of these
Show Answer & Explanation

Answer: (a) Rs. 1356

Explanation:
1. Amount = 12000 × 1.05 × 1.06 = Rs. 13356.
2. Compound interest = 13356 − 12000 = Rs. 1356.

Teacher's Note:
1. Successive rates multiply.

Question 10: Find the compound interest on Rs. 31250 at 8% per annum for 2 3/4 years.
(a) Rs. 8637
(b) Rs. 7387
(c) Rs. 1250
(d) None of these
Show Answer & Explanation

Answer: (b) Rs. 7387

Explanation:
1. After 2 years: 31250 × 1.08² = Rs. 36450.
2. Simple interest for 3/4 year: 36450 × 0.08 × 0.75 = Rs. 2187, so the amount is Rs. 38637.
3. Compound interest = 38637 − 31250 = Rs. 7387.

Teacher's Note:
1. Use compound interest for whole years and simple interest for the fraction.

Question 11: A certain sum amounts to Rs. 72900 in 2 years at 8% per annum compound interest. Find the sum.
(a) Rs. 72000
(b) Rs. 62500
(c) Rs. 135000
(d) None of these
Show Answer & Explanation

Answer: (b) Rs. 62500

Explanation:
1. P × 1.08² = 72900, so P = 72900/1.1664 = Rs. 62500.

Teacher's Note:
1. Divide the amount by the growth factor.

Question 12: In what time will Rs. 64000 amount to Rs. 68921 at 5% per annum, interest being compounded half-yearly?
(a) 1 1/2 years
(b) 2/3 years
(c) 2 years
(d) None of these
Show Answer & Explanation

Answer: (a) 1 1/2 years

Explanation:
1. 68921/64000 = 1.076890625 = (1.025)³.
2. So there are 3 half-years, i.e. 1 1/2 years.

Teacher's Note:
1. Find the number of half-yearly periods first.

Question 13: Reena borrowed a certain sum from Kamila for two years at simple interest. Reena lent it to Hamid at the same rate for two years at compound interest. At the end of two years she received Rs. 110 as compound interest but paid Rs. 100 as simple interest. Find the sum and the rate of interest.
(a) Rs. 250 and 20%
(b) Rs. 250 and 25%
(c) Rs. 200 and 20%
(d) Rs. 200 and 25%
Show Answer & Explanation

Answer: (a) Rs. 250 and 20%

Explanation:
1. Simple interest for 2 years: 2Pr = 100, so Pr = 50.
2. CI − SI = Pr² = 10, so r = 10/50 = 0.2 = 20% and P = 250.

Teacher's Note:
1. For two years, CI − SI = P(r/100)².

Question 14: The value of a refrigerator purchased 2 years ago depreciates at 12% p.a. If its present value is Rs. 9680, how much was it purchased for?
(a) Rs. 11500
(b) Rs. 10500
(c) Rs. 12000
(d) Rs. 12500
Show Answer & Explanation

Answer: (d) Rs. 12500

Explanation:
1. P × 0.88² = 9680, so P = 9680/0.7744 = Rs. 12500.

Teacher's Note:
1. Depreciation is compound with a multiplier below 1.

Question 15: The compound interest on Rs. 10,000 in 2 years at 4% per annum, being compounded half-yearly, is
(a) Rs. 832.24
(b) Rs. 828.82
(c) Rs. 824.32
(d) Rs. 912.86
Show Answer & Explanation

Answer: (c) Rs. 824.32

Explanation:
1. Rate per half-year = 2%, with 4 periods: 10000 × 1.02⁴ = 10824.32.
2. CI = Rs. 824.32.

Teacher's Note:
1. Four half-year periods at 2%.

Question 16: A sum of Rs. 500 was lent for two years at 2% compound interest. The interest for the two years will be
(a) Rs. 20.00
(b) Rs. 25.00
(c) Rs. 50.20
(d) Rs. 20.20
Show Answer & Explanation

Answer: (d) Rs. 20.20

Explanation:
1. 500 × 1.02² = 520.20, so the interest = Rs. 20.20.

Teacher's Note:
1. Subtract the principal from the amount.

Question 17: Find the compound interest on Rs. 24000 at 15% per annum for 2 1/3 years.
(a) Rs. 9327
(b) Rs. 9723
(c) Rs. 9372
(d) Rs. 9237
Show Answer & Explanation

Answer: (a) Rs. 9327

Explanation:
1. After 2 years: 24000 × 1.15² = Rs. 31740.
2. Simple interest for 1/3 year: 31740 × 0.15/3 = Rs. 1587, so the amount is Rs. 33327 and CI = Rs. 9327.

Teacher's Note:
1. Fractional year: simple interest on the amount of the whole years.

Question 18: If a, b, c are in continued proportion, the expression (a² + ab + b²)/(b² + bc + c²) can be simplified to
(a) (a + c)/(a + b)
(b) a/c
(c) c/a
(d) (a − c)/b
Show Answer & Explanation

Answer: (b) a/c

Explanation:
1. Continued proportion means b = ar and c = ar².
2. The numerator is a²(1 + r + r²) and the denominator is a²r²(1 + r + r²), giving 1/r² = a/c.

Teacher's Note:
1. Put the terms in the form a, ar, ar².

Question 19: In a school, the average score of all 1400 students in an examination was 69.5. The average score of boys was 68 and that of girls was 72. The number of boys and the number of girls respectively are
(a) 680 and 720
(b) 720 and 680
(c) 875 and 525
(d) 800 and 600
Show Answer & Explanation

Answer: (c) 875 and 525

Explanation:
1. Let boys = b: 68b + 72(1400 − b) = 69.5 × 1400 = 97300.
2. 100800 − 4b = 97300, so b = 875 and girls = 525.

Teacher's Note:
1. Use the weighted average.

Question 20: The distance between two spots A and B on the same bank of a river is 75 km. The speed of the boat in still water is twice the speed of the river current. The boat travels from A to B and returns to the starting spot in 16 hours. The speed of the boat in still water is
(a) 12.5 km/h
(b) 15 km/h
(c) 16 km/h
(d) 18 km/h
Show Answer & Explanation

Answer: (a) 12.5 km/h

Explanation:
1. Let the current be c and the boat 2c: time = 75/(3c) + 75/c = 100/c = 16.
2. c = 6.25 km/h, so the boat speed = 12.5 km/h.

Teacher's Note:
1. Upstream speed = boat − current, downstream = boat + current.

Question 21: Y varies inversely as x. If x is increased by 25%, then Y becomes what percentage of its original value?
(a) 80%
(b) 75%
(c) 60%
(d) 62.5%
Show Answer & Explanation

Answer: (a) 80%

Explanation:
1. xy is constant, so y becomes 1/1.25 = 0.8 of the original, which is 80% (a 20% decrease).

Teacher's Note:
1. Inverse variation: the product stays constant.

Question 22: Scientists in an R and D company made three design improvements to a car: the first saves 50% of fuel, the second saves 30% and the third saves 20%. If all three changes are implemented at once, the new car will consume fuel that is ___% of the fuel consumption of a normal car.
(a) 50%
(b) 100%
(c) 28%
(d) 20%
Show Answer & Explanation

Answer: (c) 28%

Explanation:
1. Successive savings multiply the consumption: 0.5 × 0.7 × 0.8 = 0.28, i.e. 28%.

Teacher's Note:
1. Savings combine multiplicatively, not additively.

Question 23: A water filter advertisement claims to provide 8 litres of water per hour. How much time does it take to fill four bottles of 1.5 litres each?
(a) 2 hr
(b) 1 hr
(c) 30 min
(d) 45 min
Show Answer & Explanation

Answer: (d) 45 min

Explanation:
1. Volume = 4 × 1.5 = 6 litres, time = 6/8 h = 45 minutes.

Teacher's Note:
1. Time = volume divided by rate.

Question 24: If a : b = c : d, how many of the following statements are true? (i) c(a + b) = a(c + d); (ii) d(a − b) = b(c − d); (iii) (a² + b²)(ac − bd) = (a² − b²)(ac + bd); (iv) (a²/b²) + (c²/d²) = 2ac/(bd)
(a) 1
(b) 2
(c) 3
(d) All
Show Answer & Explanation

Answer: (d) All

Explanation:
1. Let a = kb and c = kd. (i) and (ii) reduce to ad = bc, which is true.
2. (iii) both sides equal b³d(k⁴ − 1), and (iv) both sides equal 2k², so all four are true.

Teacher's Note:
1. Substitute a = kb, c = kd to check each identity.

Free study material for Mathematics

Download Chapter MCQs: Class 10 Mathematics Commercial Mathematics

Chapter MCQs with Answers for Class 10 Mathematics

Explore reliable practice questions for Commercial Mathematics tailored for Class 10 Mathematics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Expert Practice Material for Class 10 Mathematics

Cross-reference your completed choices with comprehensive NCERT solutions for Class 10 Mathematics to ensure absolute clarity across all sub-topics in this chapter.

Complete Your Chapter Revision

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in CBSE assessments.

FAQs

Where can I access latest CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31?

You can get most exhaustive CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31 for free on StudiesToday.com. These MCQs for Class 10 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31?

Yes, Mathematics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

Can I practice these Mathematics Class 10 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 10 Mathematics Commercial Mathematics MCQs Set 31 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.