SSC GD 2024 Super Set: Elementary Mathematics Important Question🔢

Are you preparing for the SSC GD 2024 exam and feeling overloaded by elementary mathematics? Don’t worry! The SSC GD 2024 Super Set: Elementary Mathematics Important Questions is here to help. This set contains essential questions that cover basic math concepts in a straightforward manner. Whether you’re struggling with addition, subtraction, multiplication, or division, this resource will guide you through the fundamentals step by step. With clear explanations and plenty of practice problems, you’ll gain the confidence you need to tackle the math section of the exam with ease. So, let’s dive in and conquer those math questions together!

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SSC GD 2024 Important Questions Set

1.) ‘A’ can finish a piece of work in 10 days while ‘B’ can destroy the same piece of work in 16 days such that he destroys the work at the end of each day. If ‘A’ starts working on 1st January and ‘B’joins him 3 days later, the work will be finished on: (Note: The work once fully completed cannot be destroyed)

a.) 21stJanuary

b.) 24th January

c.) 20th January

d.) 19th January

Answer: c)

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Solution:

Let the total work = 80 units {LCM (10, 16)}

Efficiency of ‘A’ = (80/10) = 8 units/day

Efficiency of ‘B’ = (80/16) = 5 units/day 

Efficiency of ‘A’ and ‘B’ while working together = 8 – 5 = 3 units/day 

Work done by A in 3 days = 3 × 8 = 24 units 

Remaining work = 80 – 24 = 56 units 

‘A’ and ‘B’ together will complete 48 units of work in (48/3) = 16 days

Next 8 unit of the work will be completed by ‘A’ in 1 day

Total time taken = 3 + 16 + 1 = 20

Hence, the work will be finished on 20th day i.e. 20th January

Hence, option c.

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2.) A shopkeeper bought a lamp for Rs. 120 and marked it 60% above its cost price. If he sold the lamp to a customer after allowing two successive discounts of 20% each, then find the profit (in Rs.) earned by the shopkeeper after selling the lamp. 

a.) 2.88 

b.) 1.44

c.) 1.36

d.) 5.76

Answer: a)

Solution:

Marked price of the lamp = 120 × 1.6 = Rs. 192

Selling price of the lamp = 192 × 0.8 × 0.8 = Rs. 122.88

Total profit earned = 122.88 – 120 = Rs. 2.88

Hence, option a.

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3.) Find the area of the kite (in cm2) whose diagonals are 8cm and 6 cm long.

a.) 48

b.) 96

c.) 16

d.) 24

Answer: d)

Solution:

Area of kite = (8 × 6) × (1/2) = 24 cm2

Hence, option d.

4.) 5 years ago from now, ages of Ram and Mohan were in ratio of 2:3, respectively. If15 years hence from now, Mohan’s age will be 125% of Ram’s age, then find the present age of Ram.

a.) 20 years

b.) 30 years

c.) 28 years

d.) 25 years

Answer: d)

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Solution:

Let the ages of Ram and Mohan five years ago were ‘2x’ years and ‘3x’ years, respectively. 

So, present age of Ram = (2x + 5) years

And, present age of Shyam = (3x + 5) years

ATQ;

1.25 × (2x + 5 + 15) = (3x + 5 + 15) 

Or, 2.5x + 25 = 3x + 20

Or, 5 = 0.5x

Or, x = 10

So, present age of Ram = (10 × 2) + 5 = 25 years 

Hence, option d. 

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5.)The length of the edge of a cube is 8 cm. Find the volume of a sphere that is circumscribed by the cube. (Take π = 3)

a.) 256 cm3

b.) 244 cm3

c.) 288 cm3

d.) 225 cm3

Answer: a)

Solution: 

Radius of the sphere inscribed in a cube with side ‘a’ cm = (a/2) cm

So, radius of the sphere = (8/2) = 4 cm

Volume of a sphere with radius ‘r’ = (4/3)×π ×radius3

= (4/3) × 3 × 4 × 4 × 4 = 256 cm3

Hence, option a.

6.) Train ‘While traveling at a speed of 144 km/h crosses a pole in 15 seconds. How long (in seconds) will it take to cross a cyclist (whose length is negligible) traveling towards it at a speed of 18 km/h? [Correct up to one decimal place]

a.) 17.1

b.) 10.3

c.) 13.3

d.) 9.6

Answer: c)

Solution:

Speed of train = 144 × (5/18) = 40 m/s

Length of train = 15 × 40 = 600 metres 

Speed of cyclist = 18 × (5/18) = 5 m/s

Relative speed of train w.r.t. cyclist = 40 + 5 = 45 m/s

Time taken to cross the cyclist = {600/45} ~13.3 seconds 

Hence, option c.

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7.) Simplify: 32 ÷ 8 × 92 + [{99 ÷ 33 × 24 + 32 of (16/8)} ÷ {√169 + 5 of 11}] × 12   

a.) 392

b.) 440

c.) 364

d.) 378

Answer: a)

Solution:

32 ÷ 8 × 92 + [{99 ÷ 33 × 24 + 32 of (16/8)} ÷ {√169 + 5 of 11}] × 12   

= 4 × 92 + [{3 × 24 + 32 × 2} ÷ {13 + 5 × 11}] × 12

= 368 + [(72 + 64) ÷ (68)] × 12

= 368 + 2 × 12

= 368 + 24

= 392 

Hence, option a. 

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8.) If ‘a’, ‘b’, ‘c’ and ‘d’ are prime factors of 1020 such that a < b < c < d, then find the value of {a×(b + d) ÷ c}

a.) 16

b.) 12

c.) 10

d.) 8

Answer: d)

Solution:

1020 = 22 × 3 × 5 × 17

So, a = 2, b = 3, c = 5 and d = 17

So, {a × (b + d) ÷ c}

 = 2 × (3 + 17) ÷ 5 = 40 ÷ 5 = 8

Hence, option d.

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9.) A certain sum invested at 20% p.a. Simple interest for 2 years yields an interest of Rs. 240. If the same sum is invested at compound interest (compounded annually)for the same duration at 20% p.a., then the interest earned will be?

a.) Rs. 264

b.) Rs. 300

c.) Rs. 240

d.) Rs. 288

Answer: a)

Solution:

Let the sum invested be Rs. ‘P’

Using formula of simple interest, we have;

240 = {(P × 20 × 2)/100}

Or, P = 240/0.4

Or, P = 600

Amount received after 2 years, when Rs. 600 is invested at compound interest of 20% p.a. = 600 × (1.2)2 = Rs. 864

Desired interest earned = 864 – 600 = Rs. 264

Hence, option a.

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10.)The volume of a cylinder is 198 cm3. If the curved surface area of the cylinder is 132 cm2, then find the ratio between height of the cylinder and radius of the cylinder, respectively. [Take π = 22/7]

a.) 3:7

b.) 2:3

c.) 3:2

d.) 7:3

Answer: d)

Solution:

Let radius and height of the cylinder is ‘r’ cm and ‘h’ cm, respectively. 

Curved surface area of cylinder = 2π × r × h

Volume of the cylinder = π ×r2× h 

ATQ;

132 = 2 × (22/7) × r × h

Or, 21 = r × h ………… (1)

198 = (22/7) × r2 × h

Or, 63 = r2×h ……………. (2)

Dividing equation (2) by equation (1), we get;

r = 3 

On putting r = 3 in equation (1), we have

h = 21/3 = 7

Required ratio = 7:3

Hence, option d.

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11.) 60% of the total number of people living in a town are males and 65% of the males are literate. If 70 out of every 100 people in the town are literate, then find the percentage of females in the town who are illiterate.

a.)77.5%

b.) 9%

c.) 22.5%

d.) 27%

Answer: c)

Solution:

Let the number of people in the town be ‘100x’

Number of literate people in the town = 0.70 × 100x = 70x

Number of males in the town = 0.60 × 100x = 60x

Number of literate males in the town = 60x × 0.65 = 39x

Number of literate females in the town =70x – 39x = 31x

Total number of females in the town = 100x – 60x = 40x

Number of illiterate females in the town = 40x – 31x = 9x

Required percentage = (9x/40x) × 100 = 22.5%

Hence, option c.

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12.)If A:B = 1:4 and B:C = 2:3, then find the value of {(9B – A)/(A + C)}

a.) 11

b.) 9

c.) 7

d.) 5

Answer: d)

Solution:

Let A = x

Then, B =(4/1) × x = 4x

So, C = (3/2) × 4x = 6x

So, {(9B – A)/(A + C)} = {(9 × 4x – x)/(x + 6x)}= 35x/7x = 5

Hence, option d.

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13.) If the area of a regular pentagon is 110.08 cm2, then find the perimeter of the same pentagon.

a.) 20 cm

b.) 40 cm

c.) 30 cm

d.) 48 cm

Answer: b)

Solution:

Since, area of a regular pentagon = [(1/4) ×√{5 × (5 + 2√5)}] × a2, where ‘a’ is the length of each side of the regular pentagon

Since, [(1/4) ×√{5 × (5 + 2√5)}] ~ 1.72

So, area of the regular pentagon with side ‘a’ = 1.72 ×a2

Let length of each side of the given pentagon is ‘a’ cm

So, 110.08 = 1.72 × a2

Or, a2 = 64 

Or, a = ± 8 (length of side cannot be negative, so we may discard a = -8)

So, a = 8

Perimeter of the pentagon = 5 × a = 5 × 8 = 40 cm

Hence, option b.

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14.)Mr. ‘Y’ spent 12.5% of his income on groceries and 40% of the remaining income on rent. Out of the remaining amount he saved 20% and invested the rest of the money. If he invested Rs. 8,400, then how much amount (in Rs.) did he spend on rent?

a.) 5,000

b.) 6,000

c.) 7,000

d.) 7,500

Answer: c)

Solution:

Let the income of Mr. ‘Y’ be Rs. ‘100x’

Amount spent on groceries = .0125 × 100x = Rs. 12.5x

Amount spent on rent = 87.5x × 0.4 = Rs. 35x

Amount saved = (100x – 12.5x – 35x) × 0.2 = Rs. 10.5x

Amount invested = 100x – 12.5x – 35x – 10.5x = Rs. 42x

ATQ;

42x = 8400

Or, x = 200

Therefore, amount spent on rent = 200 × 35 = Rs. 7,000 

Hence, option c.

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15.)Which of the following numbers is completely divisible by all 7, 9 and 13?

a.) 1764

b.) 1638

c.) 1456

d.)1872

Answer: b)

Solution:

LCM of(7 × 9 × 13) = 819

So, the number must be divisible by 819.

From the options we can see that 1638 = 819 × 2 

Therefore, 1638 is divisible by all 7, 9 and 13.

Hence, option b.

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16.) A certainsum when invested at compound interest of 15% p.a., compounded every 4 months, becomes Rs. 9,261 after one year. Find the sum(in Rs.) invested.

a.) 8,000

b.) 8,400

c.) 7,800

d.) 9,600

Answer:a)

Solution:

Let the sum invested be Rs. ‘P’.

Effective rate = 15/3 = 5%

Effective time = 3 term, where each term consists of 4 months.

ATQ;

9261 = P× {1+ (5/100)}3

Or, 9261 = P×(1.05)3

Or, P = {9261/(1.05)3}

Or, P = 8000

Hence, option a.

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17.)The Lateral surface area of a right-angled prism whose base is an equilateral triangle is 144 cm2. If the height of the prism is 8 cm, then find the volume of the prism (in cm3).

a.) 72√3

b.) 36√6

c.) 18√6

d.) 36√3

Answer: a)

Solution:

Lateral surface area of right prism = perimeter of base (P) × height 

Let perimeter of base of the given prism is ‘P’ cm

ATQ;

144 = P × 8

Or, P = 18

Since, the base is an equilateral triangle. So, the length of each side of the triangle will be = 18 ÷ 3 = 6 cm 

Volume of prism (V) = Area of base × height 

Desired volume = (√3/4) × 6 × 6 × 8= 72√3 cm3

Hence, option a.

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18.) If the cost price of 10 articles is equal to selling price of 7 articles, then find the profit/loss percentage in the transaction. [Correct up to two decimal places]

a.)48.22%

b.) 42.85%

c.) 38.67%

d.) 44.56%

Answer: b)

Solution:

Let the cost price of each article be Rs. ‘7x’

Cost price of 10 articles = 7x × 10 = Rs. 70x

ATQ’

70x = selling price of 7 articles 

So, selling price of each article = (70x/7) = Rs. 10x

Desired profit percentage = {(10x – 7x)/7x} × 100 ~42.85%

Hence, option b.

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19.) Pipe ‘A’ can fill a tank in 4 hours while pipe ‘B’ and ‘C’ together can fill the same tank in 6 hours. If pipe ‘A’ is opened at 12:00 p.m. when the tank is empty and pipe ‘B’ and ‘C’ are opened simultaneously when the tank is half filled, then at what time will the tank be full?

a.) 3:12 p.m.

b.) 3:00 p.m.

c.) 3:30 p.m.

d.) 2:20 p.m.

Answer: a)

Solution:

Let the capacity of the tank = 12 litres {LCM (4, 6)}

Efficiency of pipe ‘A’ = (12/4) = 3 litres/hour

Efficiency of pipe ‘B’ and ‘C’ together = (12/6) = 2 litres/hour

Time taken by pipe ‘A’ to fill 50% of the tank = (6/3) = 2 hours

Time required to fill rest of the tank = {6/(3+2)} = 1.2 hours 

Total time taken to fill the tank = 3.2 hours = 3 hours and 12 minutes 

Therefore, the tank will be filled at 3:12 p.m.

Hence, option a.

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20.) An article is listed at Rs. 250, if the seller sold it at 45% gain making a profit of Rs. 67.5, then find the discount offered by the shopkeeper.

a.) 20%

b.) 26%

c.) 13%

d.) 10%

Answer: c)

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Solution:

Cost price of the article = 67.5 ÷ 0.45 = Rs. 150

Selling price of the article = 150 + 67.5 = Rs. 217.5

Discount offered = 250 – 217.5 = Rs. 32.5

Desired discount percentage = {32.5/250} × 100 = 13%

Hence, option c.

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SSC GD 2024 Tips for Elementary Mathematics

Preparing for the Elementary Mathematics section of the SSC GD 2024 exam? Here are some helpful tips to ace this crucial part of the test:

Master the Basics: Ensure you have a solid understanding of basic mathematical concepts like addition, subtraction, multiplication, and division. Practice solving problems involving these operations to build a strong foundation.

Focus on Problem-solving Skills: Elementary Mathematics questions in SSC GD exams often require logical reasoning and problem-solving skills rather than complex calculations. Practice solving different types of problems to improve your analytical skills.

Memorize Important Formulas: Memorize key formulas and concepts for topics such as percentages, ratios, averages, and basic geometry. Being familiar with these formulas will help you solve questions more efficiently during the exam.

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Use Shortcut Techniques: Learn shortcut techniques for solving mathematical problems quickly. For example, learn tricks for mental calculation, simplification, and finding patterns in numerical sequences.

Review Mistakes: After practicing, review your mistakes carefully to understand where you went wrong. Identify any weak areas and focus on improving them through targeted practice.

Practice Regularly: Practice is key to success in mathematics. Dedicate regular study sessions to solving math problems from previous years’ question papers and practice sets. This will help you become faster and more accurate in solving problems.

Time Management: Practice solving math problems within the time constraints of the exam. Work on improving your speed while maintaining accuracy with PracticeMock Mock Test. Time management is crucial during the exam to ensure you can attempt all the questions.

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By following these tips and practicing the important questions consistently, you can improve your performance in the Elementary Mathematics section of the SSC GD 2024 exam and boost your overall score. Good luck with your preparation!

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