Q. A company's profit increased by 50% this year. If last year's profit was $10,000, what is this year's profit?
A.
$15,000
B.
$12,000
C.
$14,000
D.
$16,000
Show solution
Solution
This Year's Profit = Last Year's Profit + (50% of Last Year's Profit) = 10000 + (0.5 * 10000) = 10000 + 5000 = $15,000.
Correct Answer:
A
— $15,000
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Q. A company's revenue increased from $1,000,000 to $1,200,000. What is the percentage increase in revenue?
A.
20%
B.
15%
C.
25%
D.
30%
Show solution
Solution
Percentage Increase = ((New Revenue - Original Revenue) / Original Revenue) * 100 = ((1200000 - 1000000) / 1000000) * 100 = (200000 / 1000000) * 100 = 20%.
Correct Answer:
A
— 20%
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Q. A cone has a radius of 2 cm and a height of 6 cm. What is its volume?
A.
12.57 cm³
B.
25.13 cm³
C.
8.42 cm³
D.
16.76 cm³
Show solution
Solution
Volume = (1/3)πr²h = (1/3)π(2)²(6) = (1/3)π(4)(6) = 8π ≈ 25.13 cm³.
Correct Answer:
B
— 25.13 cm³
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Q. A cone has a radius of 2 cm and a height of 9 cm. What is its volume?
A.
12π cm³
B.
18π cm³
C.
24π cm³
D.
36π cm³
Show solution
Solution
Volume = 1/3 × πr²h = 1/3 × π × (2 cm)² × 9 cm = 12π cm³.
Correct Answer:
B
— 18π cm³
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Q. A cone has a radius of 3 cm and a height of 4 cm. What is its volume?
A.
12π cm³
B.
36π cm³
C.
9π cm³
D.
18π cm³
Show solution
Solution
Volume = 1/3 × πr²h = 1/3 × π × (3 cm)² × 4 cm = 12π cm³.
Correct Answer:
A
— 12π cm³
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Q. A cone has a radius of 4 cm and a height of 9 cm. What is its volume?
A.
75.40 cm³
B.
113.10 cm³
C.
113.04 cm³
D.
150.80 cm³
Show solution
Solution
Volume = (1/3)πr²h = (1/3)π(4)²(9) = 48π ≈ 150.80 cm³.
Correct Answer:
B
— 113.10 cm³
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Q. A cone has a radius of 5 cm and a height of 12 cm. What is its volume?
A.
78.54 cm³
B.
94.25 cm³
C.
100.00 cm³
D.
120.00 cm³
Show solution
Solution
Volume = (1/3)πr²h = (1/3)π(5)²(12) = 100π/3 ≈ 104.72 cm³
Correct Answer:
B
— 94.25 cm³
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Q. A container has 80 liters of a solution that is 25% salt. How much salt is in the container?
A.
10 liters
B.
15 liters
C.
20 liters
D.
25 liters
Show solution
Solution
Salt in the solution = 25% of 80L = 0.25 * 80 = 20L.
Correct Answer:
A
— 10 liters
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Q. A container has a mixture of 60 liters of juice and water in the ratio 2:3. How much juice is in the container?
A.
24 liters
B.
36 liters
C.
30 liters
D.
18 liters
Show solution
Solution
In a 2:3 ratio, total parts = 2 + 3 = 5. Juice = (2/5) * 60 = 24 liters.
Correct Answer:
B
— 36 liters
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Q. A container has a mixture of two liquids in the ratio 7:3. If 20 liters of the mixture is removed, what is the new ratio if the remaining mixture is 50 liters?
A.
7:3
B.
6:4
C.
5:5
D.
8:2
Show solution
Solution
Initial volume = 50 + 20 = 70 liters. A = (7/10) * 70 = 49 liters, B = 21 liters. New ratio = 49:21 = 7:3.
Correct Answer:
A
— 7:3
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Q. A cube has a side length of 4 cm. What is its surface area?
A.
16 cm²
B.
32 cm²
C.
64 cm²
D.
48 cm²
Show solution
Solution
Surface Area = 6a² = 6(4)² = 96 cm²
Correct Answer:
C
— 64 cm²
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Q. A cyclist covers a distance of 30 km in 1.5 hours. What is the speed of the cyclist in km/h?
A.
20 km/h
B.
25 km/h
C.
30 km/h
D.
35 km/h
Show solution
Solution
Speed = Distance / Time = 30 km / 1.5 h = 20 km/h.
Correct Answer:
B
— 25 km/h
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Q. A cyclist covers a distance of 30 km in 2 hours. What is his speed in km/h?
A.
10 km/h
B.
12 km/h
C.
15 km/h
D.
18 km/h
Show solution
Solution
Speed = Distance / Time = 30 km / 2 h = 15 km/h.
Correct Answer:
B
— 12 km/h
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Q. A cyclist covers a distance of 45 km in 1.5 hours. What is the cyclist's speed in km/h?
A.
25 km/h
B.
30 km/h
C.
35 km/h
D.
40 km/h
Show solution
Solution
Speed = Distance / Time = 45 km / 1.5 hours = 30 km/h.
Correct Answer:
B
— 30 km/h
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Q. A cyclist covers a distance of 45 km in 1.5 hours. What is the speed of the cyclist?
A.
25 km/h
B.
30 km/h
C.
35 km/h
D.
40 km/h
Show solution
Solution
Speed = Distance / Time = 45 km / 1.5 hours = 30 km/h.
Correct Answer:
B
— 30 km/h
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Q. A cyclist covers a distance of 90 km at a speed of 15 km/h. How long does the journey take?
A.
4 hours
B.
5 hours
C.
6 hours
D.
7 hours
Show solution
Solution
Time = Distance / Speed = 90 km / 15 km/h = 6 hours.
Correct Answer:
B
— 5 hours
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Q. A cyclist covers a distance of 90 km in 4.5 hours. What is the average speed of the cyclist? (2020)
A.
15 km/h
B.
20 km/h
C.
25 km/h
D.
30 km/h
Show solution
Solution
Average Speed = Total Distance / Total Time = 90 km / 4.5 h = 20 km/h
Correct Answer:
C
— 25 km/h
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Q. A cyclist travels at a speed of 15 km/h. How long will it take to cover a distance of 45 km?
A.
2 hours
B.
2.5 hours
C.
3 hours
D.
3.5 hours
Show solution
Solution
Time = Distance / Speed = 45 km / 15 km/h = 3 hours.
Correct Answer:
C
— 3 hours
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Q. A cyclist travels at a speed of 15 km/h. If he travels for 4 hours, how far does he go?
A.
50 km
B.
60 km
C.
70 km
D.
80 km
Show solution
Solution
Distance = Speed × Time = 15 km/h × 4 h = 60 km.
Correct Answer:
B
— 60 km
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Q. A cylinder has a height of 10 cm and a base radius of 2 cm. What is its surface area?
A.
75.40 cm²
B.
62.83 cm²
C.
40.00 cm²
D.
50.27 cm²
Show solution
Solution
Surface Area = 2πr(h + r) = 2π(2)(10 + 2) = 2π(2)(12) = 48π ≈ 150.80 cm².
Correct Answer:
A
— 75.40 cm²
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Q. A discount of 10% is offered on a product priced at $150. What is the sale price after the discount?
A.
$135
B.
$140
C.
$145
D.
$130
Show solution
Solution
Sale Price = Original Price - (10% of Original Price) = 150 - (0.1 * 150) = 150 - 15 = $135.
Correct Answer:
A
— $135
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Q. A discount of 10% on a product priced at $250 results in a selling price of what?
A.
$220
B.
$225
C.
$230
D.
$240
Show solution
Solution
Discount = 10% of 250 = 0.1 * 250 = 25. Selling Price = 250 - 25 = 225.
Correct Answer:
D
— $240
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Q. A discount of 15% is given on a product priced at $200. What is the selling price after the discount?
A.
$170
B.
$180
C.
$185
D.
$190
Show solution
Solution
Discount = 15% of 200 = 0.15 * 200 = 30. Selling Price = Cost Price - Discount = 200 - 30 = 170.
Correct Answer:
A
— $170
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Q. A discount of 25% on a product brings its price down to $75. What was the original price?
A.
$100
B.
$90
C.
$80
D.
$70
Show solution
Solution
Let the original price be x. After a 25% discount, Selling Price = x - 0.25x = 0.75x. 0.75x = 75, x = 75/0.75 = 100.
Correct Answer:
A
— $100
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Q. A dress is marked at $120. If a discount of 15% is applied, what is the selling price?
A.
$102
B.
$100
C.
$105
D.
$110
Show solution
Solution
Selling Price = Marked Price - (Discount % of Marked Price) = 120 - (15/100 * 120) = 120 - 18 = $102.
Correct Answer:
A
— $102
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Q. A dress is marked at $200. If it has a 30% discount followed by an additional 10% discount, what is the final price?
A.
$140
B.
$130
C.
$150
D.
$120
Show solution
Solution
First discount: 200 - (30/100 * 200) = 200 - 60 = $140. Second discount: 140 - (10/100 * 140) = 140 - 14 = $126.
Correct Answer:
A
— $140
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Q. A dress is marked at $250 and has a discount of 15% followed by another discount of 10%. What is the final price?
A.
$212.50
B.
$215.00
C.
$220.00
D.
$225.00
Show solution
Solution
First discount: 250 - (15/100 * 250) = 250 - 37.5 = $212.5. Second discount: 212.5 - (10/100 * 212.5) = 212.5 - 21.25 = $191.25.
Correct Answer:
A
— $212.50
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Q. A dress is marked at $250 and has a discount of 15%. What is the selling price?
A.
$212.50
B.
$225
C.
$237.50
D.
$240
Show solution
Solution
Selling Price = Marked Price - (Discount % of Marked Price) = 250 - (15/100 * 250) = 250 - 37.5 = $212.50.
Correct Answer:
A
— $212.50
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Q. A dress is marked at $250 and has a discount of 20% followed by an additional discount of 10%. What is the final selling price?
A.
$200
B.
$225
C.
$210
D.
$215
Show solution
Solution
First discount = 250 * 20/100 = 50. New price = 250 - 50 = 200. Second discount = 200 * 10/100 = 20. Final price = 200 - 20 = $180.
Correct Answer:
C
— $210
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Q. A dress is marked at $250 and has two successive discounts of 10% and 20%. What is the final selling price?
A.
$200
B.
$180
C.
$190
D.
$210
Show solution
Solution
First discount: 250 - (10/100 * 250) = 250 - 25 = $225. Second discount: 225 - (20/100 * 225) = 225 - 45 = $180.
Correct Answer:
C
— $190
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Showing 151 to 180 of 1468 (49 Pages)
Quantitative Aptitude (SSC) MCQ & Objective Questions
Quantitative Aptitude is a crucial component of various exams, especially for students preparing for the SSC (Staff Selection Commission) exams. Mastering this subject not only enhances problem-solving skills but also boosts confidence in tackling objective questions. Regular practice with MCQs and practice questions is essential for scoring better and understanding important concepts effectively.
What You Will Practise Here
Number Systems and their properties
Percentage, Ratio, and Proportion calculations
Time, Speed, and Distance problems
Simple and Compound Interest concepts
Algebraic expressions and equations
Data Interpretation and analysis
Mensuration and Geometry basics
Exam Relevance
Quantitative Aptitude is a significant part of the syllabus for CBSE, State Boards, and competitive exams like NEET and JEE. In these exams, students can expect questions that assess their ability to apply mathematical concepts to real-world scenarios. Common question patterns include direct problem-solving, data interpretation, and application of formulas, making it essential for students to be well-prepared.
Common Mistakes Students Make
Misunderstanding the problem statement leading to incorrect assumptions
Neglecting to apply the correct formulas in calculations
Overlooking units of measurement in word problems
Rushing through questions without double-checking calculations
FAQs
Question: What are the best ways to prepare for Quantitative Aptitude in SSC exams?Answer: Regular practice with MCQs, understanding key concepts, and solving previous years' question papers are effective strategies.
Question: How can I improve my speed in solving Quantitative Aptitude questions?Answer: Practicing timed quizzes and focusing on shortcut methods can significantly enhance your speed and accuracy.
Start your journey towards mastering Quantitative Aptitude today! Solve practice MCQs and test your understanding to achieve your exam goals. Remember, consistent practice is the key to success!
