SSC GD Exam: Practice Questions For Quantitative Aptitude

The Quantitative Aptitude section of the SSC GD exam is a fundamental part of the assessment that evaluates candidates’ Mathematical and problem-solving skills. This section plays a crucial role in determining a candidate’s overall score and performance. To excel in the SSC GD Quantitative Aptitude section, it is essential to have a solid grasp of Mathematical concepts, formulas and problem-solving techniques. More information about the SSC GD notification can be checked in the linked article.

By familiarising themselves with the expected questions and practising them regularly, aspirants can enhance their speed, accuracy and confidence in tackling quantitative problems. These questions cover a wide range of topics, including simplification, percentage, profit and loss, algebra, time and work, geometry, time, speed, distance and more. 

In this article, we will explore a comprehensive set of expected questions in the SSC GD Quantitative Aptitude section. 

To know more about Govt Exams, check out the linked article.

SSC GD Maths Exam: Important Topics

The Quantitative Aptitude section is often considered challenging in competitive government exams, including the SSC GD exam. While many candidates may find it daunting, with thorough preparation and effective time management, this section can become the scoring area.

The SSC GD Math exam covers a wide range of topics from various mathematical concepts. While it is important to have a comprehensive understanding of all the topics, some areas tend to carry more weightage and are considered more important.

Candidates can check the SSC GD Syllabus in the linked article.

This section primarily assesses the candidate’s knowledge of high-school-level mathematical problems. According to the SSC GD syllabus, the topics are given below:

• Time and Work
• Time and Distance
• Relationship between Numbers
• Ratio and Time
• Ratio and Proportion
• Profit and Loss
• Problems related to the number
• Percentages
• Number Systems
• Mensuration
• Interest
• Fundamental Arithmetical Operations
• Discount
• Decimals and Fractions
• Data Interpretation
• Computation of Whole Numbers
• Averages

It is crucial to cover these topics thoroughly and practice a variety of questions from each area to build a strong foundation and enhance problem-solving skills. Understanding the concepts and practising with the previous year’s question papers and mock tests will help in familiarising oneself with the exam pattern and gaining confidence in taking the SSC GD Math exam effectively.

Check the 10 Simple Maths Tricks and Shortcuts to save time in solving the Quantitative Aptitude section for the upcoming Government exams.

SSC GD Maths Practice Questions

The types of questions asked in the SSC GD exam from the Maths section are given below:

1.Two numbers are respectively 25% and 50% more than a third number. The ratio of the first number to the second number is

A. 1 : 2

B. 2 : 1

C. 6 : 5

D. 5 : 6

2. The captain of a cricket team of 11 members is 28 years old and the wicketkeeper is 4 years older than him. If the ages of these two are excluded, the average age of the remaining players is two years less than the average age of the whole team. What is the average age of the team?

A. 15 years

B. 22 years

C. 21 years

D. 29 years

3. A trader allows two successive discounts of 5% each on the marked price of a sofa set for Rs.24,500. The selling price of the sofa set is:

A. Rs.23,274.75

B. Rs.23,275

C. Rs.1,163.75

D. Rs.22,111.25

4. The distance between the two stations is 500 km. A train starts from station ‘A’ at 9 am and travels towards station ‘B’ at 60 km/hrs. Another train starts from station ‘B’ at 10 am and travels towards ‘A’ at 40 km/hrs. At what time do both the train meet to each other.

A. 12:45 PM

B. 1:36 PM

C. 3:36 PM

D. 2:24 PM

5. Rama and Hari can together finish a piece of work in 15 days. Rama works twice as fast as Hari, then Hari alone can finish work in

A. 45 days

B. 30 days

C. 25 days

D. 20 days

6. In a university, the number of students studying science, mathematics and language are in the ratio of 2:4:9. If the number of students in Science, Mathematics and Language be increased by 10%, 20% and 40% respectively. What will be the new ratio?

A. 12:23:63

B. 11:24:63

C. 24:11:63

D. 63:11:24

7. In how many years, a sum will be thrice of it at the rate of interest 5% per annum?

A. 25 years

B. 40 years

C. 30 years

D. 20 years

8. The difference between the upper limit and the lower limit of the class interval is called as:

A. Range

B. Mode

C. Frequency distribution

D. Class interval

9. The areas of two squares are 16:9 . The ratio of their perimeter is

A. 9:16

B. 9:12

C. 12:16

D. 16:12

10. A shopkeeper selling an article for Rs.46 loses 8%. In order to gain 6%, what should be the selling price of the article?

A. Rs.65

B. Rs.56

C. Rs.53

D. Rs.85

11. A wheel makes 4000 revolutions covering a distance of 60 km. The radius of the wheel is

A. 8 m

B. 8.25 m

C. 4.68 m

D. 2.39 m

12. If 15 apples cost as much as 6 strawberries, 2 strawberries cost as much as 16 bananas, and 6 bananas cost as much as 15 potatoes, then what is the cost of 1 potato if an apple costs ₹20?

A. ₹25

B. ₹22.50

C. ₹2.50

D. ₹2.25

13. If the selling price is tripled, the profit becomes 5 times. What is the profit per cent?

A. 100%

B. 200%

C. 50%

D. 150%

14. If A is equal to 20% of C and B is equal to 40% of C, then which one of the following is 150% of B?

A. 50% of C

B. 250% of A

C. 65% of C

D. 300% of A

15. Find the value of

[(7×6 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

A. 12

B. 14

C. 16

D. 18

16. The volume of a solid sphere is 4851 m3. What is the surface area of the sphere? (Take Π = 22/7)

A. 1386 m2

B. 1364 m2

C. 1260 m2

D. 1408 m2

17. The ratio between the perimeter and the breadth of a rectangle is 3: 1. If the area of the rectangle is 310 sq. cm, the length of the rectangle is near

A. 11.45 cm

B. 10.45 cm

C. 12.45 cm

D. 13.45 cm

18. A and together have Rs 1800. If 1/3 of A’s amount is equal to 2/3 of B’s amount, how much does B have?

A. Rs. 1200

B. Rs. 900

C. Rs. 750

D. Rs. 600

19. The value of [4 ÷ 2 + 3 x 6 – 7] is

A. 13

B. 14

C. 11

D. 12

20. A batsman scored 120 runs which included 15 fours and 2 sixes. What per cent of runs were scored by him running between the wickets?

A. 40%

B. 35%

C. 37.50%

D. 25%

21. A cloth was 50cm broad and 8 cm long. When washed, it was found to have lost 25% of its length and 14% of its breadth, then the percentage decreased in its area is ……….

A. 34.5%

B. 35.5%

C. 36%

D. 35%

22. Study the bar chart and answer the question based on it.

The average production of 1999 and 2000 was less than the average production of which of the following pairs of years?

A. 1998 and 2000

B. 1996 and 1997

C. 2000 and 2001

D. 1995 and 2001

23. If a shopkeeper gives two successive discounts of 20% and 15% on a book whose marked price is Rs. 850, then what is the selling price of the book?

A. Rs. 758

B. Rs. 587

C. Rs. 785

D. Rs. 578

24. A is twice as good as a workman as B. And together, they finish a piece of work in 20 days. In how many days, will A alone finish the work?

A. 30 days

B. 25 days

C. 26 days

D. 28 days

25. Simplify: [7.8 − 0.4 of (5.1 − 3.8) + 9.3 × 1.5]

A. 23.12

B. 12.23

C. 21.23

D. 23.21

26. A car’s price increased from $10,000 to $12,500. What is the percentage increase in price?

A. 25%

B. 20%

C. 30%

D. 40%

27. If 5 workers can complete a task in 8 days, how many workers are needed to complete the same task in 4 days?

A. 6

B. 10

C. 8

D. 4

Refer to the following table and answer the questions based on the data provided.

Number of Books Sold by a Bookstore in Different Months

28. What was the total number of books sold in the first three months?

A. 450

B. 500

C. 300

D. 400

29. In which month were the most books sold?

A. January

B. April

C. May

D. February

30. What was the average number of books sold per month?

A. 150

B. 200

C. 250

D. 300

For more information on questions on Quantitative Aptitude, visit the linked article.

The SSC GD Preparation online in a proper way is very crucial as there are a lot of concepts across core subjects like General Intelligence & Reasoning, Quantitative Aptitude, General Knowledge and the English language. Check out important SSC GD Books for reference in the linked article.  

Answers And Solutions

Answer 1 : D

Solution:

Let the third number be 100x.

∴ First number = 25% of 100x + 100x

First number = (25/100)×100x + 100x = 25x + 100x = 125x

Second number = 50% of 100x

Second number = (50/100)×100x + 100x = 50x + 100x = 150x

So, the required ratio will be –

= 125x : 150x

= 5 : 6

Answer 2: C

Solution:

Let the average age of 11 team members is x.

So, the total age of the team = 11x.

Also, the age of the captain = 28 years

Age of the wicket-keeper = 28 + 4 = 32 years

According to the question,

Age of the team – Age of captain and Wicket-keeper = Age of other 9 members

⇒ 11x – (28 + 32) = 9×(x – 2)

⇒ 11x – 60 = 9x – 18

⇒ 11x – 9x = 60 – 18

⇒ 2x = 42

⇒ x = 42/2 = 21

∴ Average age of the team is 21 years.

Answer 3 : D

Solution:

Marked price of a sofa set = Rs.24,500.

Trader allows two successive discounts of 5% each on the marked price.

The selling price of sofa set = 24500 x (95/100) x (95/100) 

= Rs. 22111.25

Answer 4 : D

Solution: 

Distance between two stations = 500 km.

A train starts from station ‘A’ at 9 am and travels towards station ‘B’ at 60 km/hrs. Another train starts from station ‘B’ at 10 am and travels towards ‘A’ at 40 km/hrs.

Let train P starts from station ‘A’ and travels towards station ‘B’ and train Q starts from station ‘B’ and travels towards station ‘A’

Distance covered by train P from 9 am to 10 am = 60 km /hours × 1 hour = 60 km

Remaining distance between station A and station B = 500 km – 60 km = 440 km

After 10 am both trains are moving towards each other.

So, Relative speed = 60 + 40 = 100 km/hrs

Time taken to cover 440 km at the relative speed of 100 km/hr = 440/100 = 4 hour 24 minutes

Both trains will meet after 4 hours and 24 minutes post 10 am.

Hence, Both trains will meet at 2.24 PM.

Answer 5: A

Solution: 

Let the Number of days taken by Hari alone to finish work = 2x

Hari’s one-day work = 1/2x unit

Number of days taken by Rama alone to finish work = x

Rama’s one-day work = 1/x unit

It is given that Rama and Hari can together finish a piece of work in 15 days.

According to the question,

One day Work of Hari + One day work of Rama = One day work of Hari and Rama Together

⇒ (1/ 2x) + (1/ x) = (1/15)

⇒ 15 + 30 = 2x

⇒ 2x = 45

Number of days taken by Hari alone to finish work = 45 days

Answer 6: B

Solution: 

Let the number of students studying science = 2x

The Number of students studying mathematics = 4x

Number of students studying language = 9x

The number of students in Science, Mathematics and Language be increased 10%, 20% and 40% respectively. 

The new number of students studying science = 2x + 10% of 2x

= 2x + (10/100)×2x = 2x + x/5 = 11x/5

The new number of students studying mathematics = 4x + 20% of 4x

= 4x + (20/100)×4x = 4x + 4x/5 = 24x/5

The new number of students studying language = 9x + 40% of 9x

= 9x + (40/100)×9x = 9x + 18x/5 = 63x/5

New ratio = 11x/5 : 24x/5 : 63x/5 = 11 : 24 : 63

Answer 7: B

Solution: 

Given, Rate of interest = 5%

Let the number of years required = t
Let Principle = P
Amount = 3P
Interest = Amount – principle =3P – P = 2P
Now using the formula, 

Simple Interest (Principal x Rate x Time)/ 100

⇒ 2P = (Px5xt)/ 100

⇒ t = 200/5 = 40 years

Answer 8: D

Solution: 

We know that difference between the upper limit and lower limit of the class interval is called class size.

Answer 9: D

Solution: 

Given, the areas of two squares are 16:9

Let area of first square = 16x2

Area of the second square = 9x2

Hence,

Side of the first square = √16x2 = 4x

Side of the second square = √9x2 = 3x

Perimeter of first square = 16x

Perimeter of second square = 12x

Ratio of their perimeter = 16 : 12

Answer 10: C

Solution: 

Given, If Shopkeeper sells an article for Rs.46, he loses 8%.

Let cost price of article = 100 unit

Loss% = 8%

So, 8% of 100 unit = 8 unit

Selling price = 100 unit – 8 unit = 92 unit

According to the question,

92 unit → Rs. 46

1 unit → Rs. 0.5

100 unit → Rs. 50

Hence, the Cost price of the article = Rs. 50

If Profit% = 6%

Then Selling price of article = 50 + 6% of 50 = 50 + 3 = 53

Answer 11: D

Solution:

A wheel makes 4000 revolution is covering a distance of 60 km.

60 km = 60000 m

Distance covered in 1 revolution = 60000/4000 = 15 meter

We know, Distance covered in one revolution = Perimeter of Wheel

⇒ 2Πr = 15

⇒ r = 15/2Π

⇒ r = (15×7)/(2×22)

⇒ r = 2.39 m

Hence, radius of the wheel = 2.39 m

Answer 12: C

Solution:

ATQ, an apple costs ₹20.

So, Cost price of 15 apples = 15 × 20 = ₹300

Also, 15 apples cost as much as 6 strawberries.

Cost price of 6 strawberries = ₹300

Cost price of 2 strawberries = (300/6)×2 = ₹100

Now, 2 strawberries cost as much as 16 bananas.

Cost price of 16 bananas = ₹100

Cost price of 6 bananas = (100/16)×6 = ₹37.5

And 6 bananas cost as much as 15 potatoes.

Cost price of 15 potatoes = ₹37.5

Cost price of a potato = 37.5/15 = ₹2.5

Answer 13: A

Solution:

Let cost price = x unit

And Selling price = y unit

Let Profit = (y−x) unit

If the selling price is tripled, the profit becomes 5 times.

New Selling price = 3y unit

New profit = 5(y – x) unit ………(i)

But

if S.P. = 3y, then Profit = (3y – x) unit ………….(ii)

From (i) and (ii):

⇒ 3y − x = 5y – 5x

⇒ 2y = 4x

⇒ y = 2x

Hence Original Profit = y – x = 2x – x = x

Hence, Profit percentage = (x/x)×100 = 100%

Answer 14: D

Solution:

Given,

A = 20% of C

A = 20C/100 = C/5

C = 5A ……………….(i)

And B = 40% of C

B = 40C/100 = 2C/5 …………..(2)

Also, 150% of B = 150B/100

Using (1) and (2):

= (150/100) x (2C/5)

=(150/100) x (2/5) x 5A

=(300/100) x A

= 300% of A

Answer 15: A

Solution: 

Consider: [(7×6 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

Using BODMAS:

⇒ [(42 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

⇒ [32÷4 + 6÷3 – 4 + 18÷3]

⇒ 8 + 2 – 4 + 6 = 12

Answer 16: A

Solution:

Volume of a solid sphere = 4851 m3

⇒4∏r3/3 = 4851

⇒r3 = (4851 × 3)/4∏

⇒r3 = (4851 × 3)/4∏

⇒ r3 = 4851 x (3/4)x (7/22)

⇒ r3 = 1157.625

⇒ r = 10.5 m

Surface area of the sphere = 4∏r2 

= 4x (22/ 7) x 10.5 x 10.5

= 1386 m2

Answer 17: C

Solution: 

The ratio between the perimeter and the breadth of a rectangle is 3 : 1.

Let perimeter of a rectangle = 3x cm

Breadth of a rectangle = x cm

We know that, Perimeter of a rectangle = 2(L + B)

Hence, 3x = 2(Length + x)

⇒2 × Length = x

⇒Length = x/2 cm

Now, Area of the rectangle is 310 sq. cm.

So,

Length × Breadth = 310

⇒(x/2) ×(x) = 310

⇒x2/2 = 310

⇒x2= 620

⇒x = 24.9 cm

Hence, Length of rectangle = x/2 = 24.9 = 12.45 cm

Answer 18: D

Solution: 

Let A and B have a and b amount respectively.

So,

(a + b) = 1800 —- (i)

Also,

a/3 = 2b/3

⇒ a = 2b —-(ii)

Putting the value of (ii) in (i)-

⇒ 2b + b = 1800

⇒ 3b = 1800

⇒ b = 1800/3 = Rs. 600

Answer 19: A

Solution:

= [4÷2+3 x 6–7]

= [(4/2) + (3 × 6) – 7]

= (2 + 18) – 7

= 20 – 7 = 13

Answer 20: A

Solution:

Let the runs taken between the wickets by the batsman is x.

According to the question,

Total Runs = (15 × 4) + (2 × 6) + x = 120

⇒ 60 + 12 + x = 120

⇒ 72 + x = 120

⇒ x = 120 – 72 = 48

∴ Required percentage will be = (48/120)×100 = 40%

Answer 21. B

Solution: 

Area of the cloth = length × breadth = 50 × 8 = 400 cm2

After getting washed,

New length = 50 – 25% of 50 = 50 – 25/2 = 75/2 cm

New breadth = 8 – 14% of 8

= 8- (14/100) x 8

= 172/25 cm

New area = (75/2) x (172/ 25)

= 3 × 86 = 258 cm2

Net change in area = (400 – 258) = 142 cm2

∴ Percentage change in its area =

= (142/ 400) x 100

= 142/4 = 35.5%

Answer 22: C

Solution: 

Average production of 1999 and 2000 = (65+50)/2 = 57.5

Average production of 1998 and 2000 = (45+50)/2 = 47.5

Average production of 1996 and 1997 = (40+60)/2 = 50

Average production of 2000 and 2001 = (50+75)/2 = 62.5

Average production of 1995 and 2001 = (25+75)/2 = 50

Thus, it is clear that in average production of 2000 and 2001 is more than that of 1999 and 2000.

Answer 23: D

Solution:

Since SP = MP x (100-d)/ 100

∴ Required selling price of the book is

= 850 x {(100-20)/ 100} x {(100-15)/ 100}

= 5780000/10000

= Rs. 578

Answer 24: A

Solution:

Let the efficiency of B is x.

∴ Efficiency of A is 2x.

Efficiency of A and B together = x + 2x = 3x

We know that,

Work = Time × Efficiency

So, work = 20 × 3x = 60x

Thus, A will alone finish the work in = 60x/2x = 30 days

Answer 25: C

Solution:

[7.8 − 0.4 of (5.1 − 3.8) + 9.3 × 1.5]

= [7.8 – (0.4 × 1.3) + (9.3 × 1.5)]

= [7.8 – 0.52 + 13.95]

= (21.75 – 0.52)

= 21.23

Answer 26: A

Solution: 

Percentage Increase = [(New Price – Old Price) / Old Price] × 100 

= [(12,500 – 10,000) / 10,000] × 100 = 2,500 / 10,000 × 100 = 25%

Answer 27: B
Solution: 

Workers × Days = Constant (Work done) 5 workers × 8 days 

= x workers × 4 days 40 

= 4x x = 40 / 4 x = 10

Therefore, 10 workers are needed to complete the task in 4 days.

Answer 28: A

Solution:

Total number of books sold in the first three months: January + February + March = 150 + 200 + 100 = 450 books

Answer 29: C

Solution:

Month with the most books sold: The month with the highest number of books sold is May (300 books).

Answer 30: B

Solution: 

Average number of books sold per month: Total number of books sold / Number of months = (150 + 200 + 100 + 250 + 300) / 5 = 1000 / 5 = 200 books

These practice questions will help you get started with your SSC GD Quantitative Aptitude preparation. Make sure to practice a variety of questions and solve previous year’s papers to strengthen your skills and improve your performance in the actual exam.

For comprehensive preparation of the vast syllabus of the SSC GD Constable Exam, candidates are advised to take free SSC GD Mock Tests online.

SSC GD Practice Questions For Maths

After solving the above questions candidates can also practise from the following set of questions for increasing speed and understanding of the concepts.

Candidates who are willing to prepare for the SSC GD 2023 exam, can check the Tips to prepare for SSC exams without Coaching in the linked article.

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