SSC MTS Quant Questions
20 questions on Quantitative Aptitude for SSC MTS, including questions from 2024 papers. Each comes with the correct answer and a worked explanation.
- Questions
- 20
- Easy
- 7
- Moderate
- 13
Quant topics
Quant questions with solutions
The circumference of a circle is 264 cm. What is its area? (Take π = 22/7)
- (A)1764 cm²
- (B)5676 cm²
- (C)5544 cm²
- (D)2772 cm²
Show answer and explanation
Correct answer: (C) 5544 cm²
Circumference = 2πr = 264, so r = 264 × 7 ÷ (2 × 22) = 42 cm. Area = πr² = 22/7 × 42 × 42 = 5544 cm².
Shortcut: With π = 22/7, a circumference of 44k gives radius 7k and area 154k².
The HCF and LCM of two numbers are 3 and 63 respectively. If one of the numbers is 9, what is the other number?
- (A)18
- (B)21
- (C)24
- (D)7
Show answer and explanation
Correct answer: (B) 21
For any two numbers, product of the numbers = HCF × LCM. So the other number = (3 × 63) ÷ 9 = 189 ÷ 9 = 21.
Shortcut: Other number = HCF × LCM ÷ given number.
Sneha can complete a piece of work in 10 days and Riya can complete the same work in 15 days. In how many days will they finish it working together?
- (A)8 days
- (B)5 days
- (C)12.5 days
- (D)6 days
Show answer and explanation
Correct answer: (D) 6 days
Sneha's one-day work = 1/10 and Riya's one-day work = 1/15. Together they do 1/10 + 1/15 = 25/150 = 1/6 of the work per day, so the work is finished in 6 days.
Shortcut: Days together = (a × b)/(a + b) = 150/25 = 6.
A train 600 m long is running at 90 km/h. How much time will it take to cross a pole?
- (A)48 seconds
- (B)24 seconds
- (C)7 seconds
- (D)28 seconds
Show answer and explanation
Correct answer: (B) 24 seconds
Convert the speed to metres per second: 90 × 5/18 = 25 m/s. To cross a pole the train covers its own length, so time = 600 ÷ 25 = 24 seconds.
Shortcut: km/h × 5/18 = m/s; time = length ÷ speed.
If x + 1/x = 8, what is the value of x² + 1/x²?
- (A)66
- (B)62
- (C)16
- (D)64
Show answer and explanation
Correct answer: (B) 62
Squaring both sides: (x + 1/x)² = x² + 1/x² + 2 = 64. Therefore x² + 1/x² = 64 − 2 = 62.
Shortcut: x² + 1/x² = k² − 2 when x + 1/x = k.
The perimeter of a square is 96 cm. What is its area?
- (A)625 cm²
- (B)576 cm²
- (C)600 cm²
- (D)192 cm²
Show answer and explanation
Correct answer: (B) 576 cm²
Each side of the square = perimeter ÷ 4 = 96 ÷ 4 = 24 cm. Area of a square = side² = 24 × 24 = 576 cm².
Shortcut: Area = (perimeter/4)².
If 50% of a number is 600, what is 60% of the same number?
- (A)660
- (B)600
- (C)360
- (D)720
Show answer and explanation
Correct answer: (D) 720
50% of the number is 600, so the number is 600 × 100 ÷ 50 = 1200. Now 60% of 1200 = 1200 × 60 ÷ 100 = 720. Hence the required value is 720.
Shortcut: Scale directly: 600 × 60/50 = 720.
The price of an item is first increased by 25% and then decreased by 20%. What is the net percentage change in the price?
- (A)10% increase
- (B)5% decrease
- (C)5% increase
- (D)No change
Show answer and explanation
Correct answer: (D) No change
Let the original price be 100. After a 25% increase it becomes 125. A 20% decrease on 125 gives 125 × 80 ÷ 100 = 100. The net change is 100 − 100 = 0, i.e. No change.
Shortcut: Net % change = a − b − ab/100 = 25 − 20 − 5 = 0%.
A shopkeeper buys an article for ₹1,900 and sells it for ₹2,660. What is the profit percentage?
- (A)35%
- (B)29%
- (C)50%
- (D)40%
Show answer and explanation
Correct answer: (D) 40%
Profit = selling price − cost price = 2660 − 1900 = 760. Profit percentage is always calculated on the cost price: 760 ÷ 1900 × 100 = 40%.
Shortcut: Profit % = (SP − CP)/CP × 100.
The marked price of a watch is ₹1,600. If a discount of 10% is allowed, what is its selling price?
- (A)₹1,760
- (B)₹1,360
- (C)₹1,440
- (D)₹1,590
Show answer and explanation
Correct answer: (C) ₹1,440
Discount = 10% of 1600 = 160. Selling price = marked price − discount = 1600 − 160 = 1440. So the watch is sold for ₹1,440.
Shortcut: SP = MP × (100 − d)/100 = 1600 × 90/100.
What is the simple interest on ₹7,500 at 4% per annum for 5 years?
- (A)₹9,000
- (B)₹1,500
- (C)₹1,875
- (D)₹1,800
Show answer and explanation
Correct answer: (B) ₹1,500
Simple interest = P × R × T ÷ 100 = 7500 × 4 × 5 ÷ 100 = 1500. Therefore the interest earned over 5 years is ₹1,500.
Shortcut: 1% of 7500 is 75; multiply by 4 × 5 = 20.
Find the compound interest on ₹18,000 at 20% per annum for 2 years, compounded annually.
- (A)₹7,380
- (B)₹7,920
- (C)₹7,200
- (D)₹7,830
Show answer and explanation
Correct answer: (B) ₹7,920
Amount after 2 years = 18000 × (1 + 20/100)² = 25920. Compound interest = amount − principal = 25920 − 18000 = 7920. Note that the simple interest for the same period would be only 7200.
Shortcut: For 2 years, effective rate = 2r + r²/100 = 44% of 18000.
The average of 11 numbers is 59. If one number is removed, the average of the remaining numbers becomes 60. What is the removed number?
- (A)38
- (B)60
- (C)59
- (D)49
Show answer and explanation
Correct answer: (D) 49
Sum of all 11 numbers = 11 × 59 = 649. Sum of the remaining 10 numbers = 10 × 60 = 600. The removed number is the difference: 649 − 600 = 49.
Shortcut: Removed number = old average + (n − 1) × (old average − new average) = 59 + 10 × (-1).
Two numbers are in the ratio 3 : 5 and their sum is 56. What is the larger number?
- (A)42
- (B)35
- (C)53
- (D)14
Show answer and explanation
Correct answer: (B) 35
Let the numbers be 3x and 5x. Then 3x + 5x = 56, so 8x = 56 and x = 7. The larger number is 5x = 5 × 7 = 35.
Shortcut: Larger number = 5/8 of 56.
Divya can complete a piece of work in 15 days and Sneha can complete the same work in 60 days. In how many days will they finish it working together?
- (A)11 days
- (B)45 days
- (C)12 days
- (D)37.5 days
Show answer and explanation
Correct answer: (C) 12 days
Divya's one-day work = 1/15 and Sneha's one-day work = 1/60. Together they do 1/15 + 1/60 = 75/900 = 1/12 of the work per day, so the work is finished in 12 days.
Shortcut: Days together = (a × b)/(a + b) = 900/75 = 12.
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Frequently asked questions
How many SSC MTS Quant questions can I practise here?
20 questions — 7 easy and 13 moderate. 15 are shown on this page with answers and explanations, and the practice set draws from all of them.
How should I practise Quant for SSC MTS?
Read through a few solved questions first to see the method, then take a timed practice set. Afterwards, check which questions you got wrong or spent longest on and redo those before moving on.