{"id":77358,"date":"2024-02-16T10:49:20","date_gmt":"2024-02-16T05:19:20","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=77358"},"modified":"2025-11-26T10:56:22","modified_gmt":"2025-11-26T05:26:22","slug":"ssc-gd-super-set-elementary-mathematics-important-question","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/ssc-gd-super-set-elementary-mathematics-important-question\/","title":{"rendered":"SSC GD 2024 Super Set: Elementary Mathematics Important Question\ud83d\udd22"},"content":{"rendered":"\n

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<\/a> 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!<\/p>\n\n\n\n

SSC GD Constable 2024 Free Practice Resource<\/strong><\/td><\/tr>
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SSC GD 2024 Important Questions Set<\/h2>\n\n\n\n

1.) \u2018A\u2019 can finish a piece of work in 10 days while \u2018B\u2019 can destroy the same piece of work in 16 days such that he destroys the work at the end of each day. If \u2018A\u2019 starts working on 1st January and \u2018B\u2019joins him 3 days later, the work will be finished on: (Note: The work once fully completed cannot be destroyed)<\/strong><\/p>\n\n\n\n

a.) 21stJanuary<\/p>\n\n\n\n

b.) 24th January<\/p>\n\n\n\n

c.) 20th January<\/p>\n\n\n\n

d.) 19th January<\/p>\n\n\n\n

Answer: c)<\/strong><\/p>\n\n\n\n

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Solution:<\/p>\n\n\n\n

Let the total work = 80 units {LCM (10, 16)}<\/p>\n\n\n\n

Efficiency of \u2018A\u2019 = (80\/10) = 8 units\/day<\/p>\n\n\n\n

Efficiency of \u2018B\u2019 = (80\/16) = 5 units\/day <\/p>\n\n\n\n

Efficiency of \u2018A\u2019 and \u2018B\u2019 while working together = 8 \u2013 5 = 3 units\/day <\/p>\n\n\n\n

Work done by A in 3 days = 3 \u00d7 8 = 24 units <\/p>\n\n\n\n

Remaining work = 80 \u2013 24 = 56 units <\/p>\n\n\n\n

\u2018A\u2019 and \u2018B\u2019 together will complete 48 units of work in (48\/3) = 16 days<\/p>\n\n\n\n

Next 8 unit of the work will be completed by \u2018A\u2019 in 1 day<\/p>\n\n\n\n

Total time taken = 3 + 16 + 1 = 20<\/p>\n\n\n\n

Hence, the work will be finished on 20th day i.e. 20th January<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

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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. <\/strong><\/p>\n\n\n\n

a.) 2.88 <\/p>\n\n\n\n

b.) 1.44<\/p>\n\n\n\n

c.) 1.36<\/p>\n\n\n\n

d.) 5.76<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Marked price of the lamp = 120 \u00d7 1.6 = Rs. 192<\/p>\n\n\n\n

Selling price of the lamp = 192 \u00d7 0.8 \u00d7 0.8 = Rs. 122.88<\/p>\n\n\n\n

Total profit earned = 122.88 \u2013 120 = Rs. 2.88<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

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3.) Find the area of the kite (in cm2) whose diagonals are 8cm and 6 cm long.<\/strong><\/p>\n\n\n\n

a.) 48<\/p>\n\n\n\n

b.) 96<\/p>\n\n\n\n

c.) 16<\/p>\n\n\n\n

d.) 24<\/p>\n\n\n\n

Answer: d)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Area of kite = (8 \u00d7 6) \u00d7 (1\/2) = 24 cm2<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

4.) 5 years ago from now, ages of Ram and Mohan were in ratio of 2:3, respectively. If15 years hence from now, Mohan\u2019s age will be 125% of Ram\u2019s age, then find the present age of Ram.<\/strong><\/p>\n\n\n\n

a.) 20 years<\/p>\n\n\n\n

b.) 30 years<\/p>\n\n\n\n

c.) 28 years<\/p>\n\n\n\n

d.) 25 years<\/p>\n\n\n\n

Answer: d)<\/strong><\/p>\n\n\n\n

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Solution:<\/p>\n\n\n\n

Let the ages of Ram and Mohan five years ago were \u20182x\u2019 years and \u20183x\u2019 years, respectively. <\/p>\n\n\n\n

So, present age of Ram = (2x + 5) years<\/p>\n\n\n\n

And, present age of Shyam = (3x + 5) years<\/p>\n\n\n\n

ATQ;<\/p>\n\n\n\n

1.25 \u00d7 (2x + 5 + 15) = (3x + 5 + 15) <\/p>\n\n\n\n

Or, 2.5x + 25 = 3x + 20<\/p>\n\n\n\n

Or, 5 = 0.5x<\/p>\n\n\n\n

Or, x = 10<\/p>\n\n\n\n

So, present age of Ram = (10 \u00d7 2) + 5 = 25 years <\/p>\n\n\n\n

Hence, option d. <\/p>\n\n\n\n

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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 \u03c0 = 3)<\/strong><\/p>\n\n\n\n

a.) 256 cm3<\/p>\n\n\n\n

b.) 244 cm3<\/p>\n\n\n\n

c.) 288 cm3<\/p>\n\n\n\n

d.) 225 cm3<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution: <\/p>\n\n\n\n

Radius of the sphere inscribed in a cube with side \u2018a\u2019 cm = (a\/2) cm<\/p>\n\n\n\n

So, radius of the sphere = (8\/2) = 4 cm<\/p>\n\n\n\n

Volume of a sphere with radius \u2018r\u2019 = (4\/3)\u00d7\u03c0 \u00d7radius3<\/p>\n\n\n\n

= (4\/3) \u00d7 3 \u00d7 4 \u00d7 4 \u00d7 4 = 256 cm3<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

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]<\/strong><\/p>\n\n\n\n

a.) 17.1<\/p>\n\n\n\n

b.) 10.3<\/p>\n\n\n\n

c.) 13.3<\/p>\n\n\n\n

d.) 9.6<\/p>\n\n\n\n

Answer: c)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Speed of train = 144 \u00d7 (5\/18) = 40 m\/s<\/p>\n\n\n\n

Length of train = 15 \u00d7 40 = 600 metres <\/p>\n\n\n\n

Speed of cyclist = 18 \u00d7 (5\/18) = 5 m\/s<\/p>\n\n\n\n

Relative speed of train w.r.t. cyclist = 40 + 5 = 45 m\/s<\/p>\n\n\n\n

Time taken to cross the cyclist = {600\/45} ~13.3 seconds <\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

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7.) Simplify: 32 \u00f7 8 \u00d7 92 + [{99 \u00f7 33 \u00d7 24 + 32 of (16\/8)} \u00f7 {\u221a169 + 5 of 11}] \u00d7 12<\/strong>   <\/p>\n\n\n\n

a.) 392<\/p>\n\n\n\n

b.) 440<\/p>\n\n\n\n

c.) 364<\/p>\n\n\n\n

d.) 378<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

32 \u00f7 8 \u00d7 92 + [{99 \u00f7 33 \u00d7 24 + 32 of (16\/8)} \u00f7 {\u221a169 + 5 of 11}] \u00d7 12   <\/p>\n\n\n\n

= 4 \u00d7 92 + [{3 \u00d7 24 + 32 \u00d7 2} \u00f7 {13 + 5 \u00d7 11}] \u00d7 12<\/p>\n\n\n\n

= 368 + [(72 + 64) \u00f7 (68)] \u00d7 12<\/p>\n\n\n\n

= 368 + 2 \u00d7 12<\/p>\n\n\n\n

= 368 + 24<\/p>\n\n\n\n

= 392 <\/p>\n\n\n\n

Hence, option a. <\/p>\n\n\n\n

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8.) If \u2018a\u2019, \u2018b\u2019, \u2018c\u2019 and \u2018d\u2019 are prime factors of 1020 such that a < b < c < d, then find the value of {a\u00d7(b + d) \u00f7 c}<\/strong><\/p>\n\n\n\n

a.) 16<\/p>\n\n\n\n

b.) 12<\/p>\n\n\n\n

c.) 10<\/p>\n\n\n\n

d.) 8<\/p>\n\n\n\n

Answer: d)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

1020 = 22 \u00d7 3 \u00d7 5 \u00d7 17<\/p>\n\n\n\n

So, a = 2, b = 3, c = 5 and d = 17<\/p>\n\n\n\n

So, {a \u00d7 (b + d) \u00f7 c}<\/p>\n\n\n\n

 = 2 \u00d7 (3 + 17) \u00f7 5 = 40 \u00f7 5 = 8<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

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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?<\/strong><\/p>\n\n\n\n

a.) Rs. 264<\/p>\n\n\n\n

b.) Rs. 300<\/p>\n\n\n\n

c.) Rs. 240<\/p>\n\n\n\n

d.) Rs. 288<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the sum invested be Rs. \u2018P\u2019<\/p>\n\n\n\n

Using formula of simple interest, we have;<\/p>\n\n\n\n

240 = {(P \u00d7 20 \u00d7 2)\/100}<\/p>\n\n\n\n

Or, P = 240\/0.4<\/p>\n\n\n\n

Or, P = 600<\/p>\n\n\n\n

Amount received after 2 years, when Rs. 600 is invested at compound interest of 20% p.a. = 600 \u00d7 (1.2)2 = Rs. 864<\/p>\n\n\n\n

Desired interest earned = 864 \u2013 600 = Rs. 264<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

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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 \u03c0 = 22\/7]<\/strong><\/p>\n\n\n\n

a.) 3:7<\/p>\n\n\n\n

b.) 2:3<\/p>\n\n\n\n

c.) 3:2<\/p>\n\n\n\n

d.) 7:3<\/p>\n\n\n\n

Answer: d)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let radius and height of the cylinder is \u2018r\u2019 cm and \u2018h\u2019 cm, respectively. <\/p>\n\n\n\n

Curved surface area of cylinder = 2\u03c0 \u00d7 r \u00d7 h<\/p>\n\n\n\n

Volume of the cylinder = \u03c0 \u00d7r2\u00d7 h <\/p>\n\n\n\n

ATQ;<\/p>\n\n\n\n

132 = 2 \u00d7 (22\/7) \u00d7 r \u00d7 h<\/p>\n\n\n\n

Or, 21 = r \u00d7 h \u2026\u2026\u2026… (1)<\/p>\n\n\n\n

198 = (22\/7) \u00d7 r2 \u00d7 h<\/p>\n\n\n\n

Or, 63 = r2\u00d7h \u2026\u2026\u2026\u2026\u2026. (2)<\/p>\n\n\n\n

Dividing equation (2) by equation (1), we get;<\/p>\n\n\n\n

r = 3 <\/p>\n\n\n\n

On putting r = 3 in equation (1), we have<\/p>\n\n\n\n

h = 21\/3 = 7<\/p>\n\n\n\n

Required ratio = 7:3<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

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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.<\/strong><\/p>\n\n\n\n

a.)77.5%<\/p>\n\n\n\n

b.) 9%<\/p>\n\n\n\n

c.) 22.5%<\/p>\n\n\n\n

d.) 27%<\/p>\n\n\n\n

Answer: c)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the number of people in the town be \u2018100x\u2019<\/p>\n\n\n\n

Number of literate people in the town = 0.70 \u00d7 100x = 70x<\/p>\n\n\n\n

Number of males in the town = 0.60 \u00d7 100x = 60x<\/p>\n\n\n\n

Number of literate males in the town = 60x \u00d7 0.65 = 39x<\/p>\n\n\n\n

Number of literate females in the town =70x \u2013 39x = 31x<\/p>\n\n\n\n

Total number of females in the town = 100x \u2013 60x = 40x<\/p>\n\n\n\n

Number of illiterate females in the town = 40x \u2013 31x = 9x<\/p>\n\n\n\n

Required percentage = (9x\/40x) \u00d7 100 = 22.5%<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

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12.)If A:B = 1:4 and B:C = 2:3, then find the value of {(9B \u2013 A)\/(A + C)}<\/strong><\/p>\n\n\n\n

a.) 11<\/p>\n\n\n\n

b.) 9<\/p>\n\n\n\n

c.) 7<\/p>\n\n\n\n

d.) 5<\/p>\n\n\n\n

Answer: d)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let A = x<\/p>\n\n\n\n

Then, B =(4\/1) \u00d7 x = 4x<\/p>\n\n\n\n

So, C = (3\/2) \u00d7 4x = 6x<\/p>\n\n\n\n

So, {(9B \u2013 A)\/(A + C)} = {(9 \u00d7 4x \u2013 x)\/(x + 6x)}= 35x\/7x = 5<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

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13.) If the area of a regular pentagon is 110.08 cm2, then find the perimeter of the same pentagon.<\/strong><\/p>\n\n\n\n

a.) 20 cm<\/p>\n\n\n\n

b.) 40 cm<\/p>\n\n\n\n

c.) 30 cm<\/p>\n\n\n\n

d.) 48 cm<\/p>\n\n\n\n

Answer: b)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Since, area of a regular pentagon = [(1\/4) \u00d7\u221a{5 \u00d7 (5 + 2\u221a5)}] \u00d7 a2, where \u2018a\u2019 is the length of each side of the regular pentagon<\/p>\n\n\n\n

Since, [(1\/4) \u00d7\u221a{5 \u00d7 (5 + 2\u221a5)}] ~ 1.72<\/p>\n\n\n\n

So, area of the regular pentagon with side \u2018a\u2019 = 1.72 \u00d7a2<\/p>\n\n\n\n

Let length of each side of the given pentagon is \u2018a\u2019 cm<\/p>\n\n\n\n

So, 110.08 = 1.72 \u00d7 a2<\/p>\n\n\n\n

Or, a2 = 64 <\/p>\n\n\n\n

Or, a = \u00b1 8 (length of side cannot be negative, so we may discard a = -8)<\/p>\n\n\n\n

So, a = 8<\/p>\n\n\n\n

Perimeter of the pentagon = 5 \u00d7 a = 5 \u00d7 8 = 40 cm<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

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14.)Mr. \u2018Y\u2019 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?<\/strong><\/p>\n\n\n\n

a.) 5,000<\/p>\n\n\n\n

b.) 6,000<\/p>\n\n\n\n

c.) 7,000<\/p>\n\n\n\n

d.) 7,500<\/p>\n\n\n\n

Answer: c)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the income of Mr. \u2018Y\u2019 be Rs. \u2018100x\u2019<\/p>\n\n\n\n

Amount spent on groceries = .0125 \u00d7 100x = Rs. 12.5x<\/p>\n\n\n\n

Amount spent on rent = 87.5x \u00d7 0.4 = Rs. 35x<\/p>\n\n\n\n

Amount saved = (100x \u2013 12.5x \u2013 35x) \u00d7 0.2 = Rs. 10.5x<\/p>\n\n\n\n

Amount invested = 100x \u2013 12.5x \u2013 35x \u2013 10.5x = Rs. 42x<\/p>\n\n\n\n

ATQ;<\/p>\n\n\n\n

42x = 8400<\/p>\n\n\n\n

Or, x = 200<\/p>\n\n\n\n

Therefore, amount spent on rent = 200 \u00d7 35 = Rs. 7,000 <\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

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15.)Which of the following numbers is completely divisible by all 7, 9 and 13?<\/strong><\/p>\n\n\n\n

a.) 1764<\/p>\n\n\n\n

b.) 1638<\/p>\n\n\n\n

c.) 1456<\/p>\n\n\n\n

d.)1872<\/p>\n\n\n\n

Answer: b)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

LCM of(7 \u00d7 9 \u00d7 13) = 819<\/p>\n\n\n\n

So, the number must be divisible by 819.<\/p>\n\n\n\n

From the options we can see that 1638 = 819 \u00d7 2 <\/p>\n\n\n\n

Therefore, 1638 is divisible by all 7, 9 and 13.<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

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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.<\/strong><\/p>\n\n\n\n

a.) 8,000<\/p>\n\n\n\n

b.) 8,400<\/p>\n\n\n\n

c.) 7,800<\/p>\n\n\n\n

d.) 9,600<\/p>\n\n\n\n

Answer:a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the sum invested be Rs. \u2018P\u2019.<\/p>\n\n\n\n

Effective rate = 15\/3 = 5%<\/p>\n\n\n\n

Effective time = 3 term, where each term consists of 4 months.<\/p>\n\n\n\n

ATQ;<\/p>\n\n\n\n

9261 = P\u00d7 {1+ (5\/100)}3<\/p>\n\n\n\n

Or, 9261 = P\u00d7(1.05)3<\/p>\n\n\n\n

Or, P = {9261\/(1.05)3}<\/p>\n\n\n\n

Or, P = 8000<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

\ud83c\udfc3 Mock Test \u092d\u093e\u0917 \u0932\u0947\u0902 \u0914\u0930 \u0905\u092a\u0928\u0940 \u092a\u094d\u0930\u0924\u093f\u092d\u093e \u0915\u093e \u092e\u0942\u0932\u094d\u092f\u093e\u0902\u0915\u0928 \u0915\u0930\u0947\u0902 \ud83c\udfaf\ud83d\udcaf\ud83e\udd47<\/a><\/em><\/h2>\n\n\n\n

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).<\/strong><\/p>\n\n\n\n

a.) 72\u221a3<\/p>\n\n\n\n

b.) 36\u221a6<\/p>\n\n\n\n

c.) 18\u221a6<\/p>\n\n\n\n

d.) 36\u221a3<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Lateral surface area of right prism = perimeter of base (P) \u00d7 height <\/p>\n\n\n\n

Let perimeter of base of the given prism is \u2018P\u2019 cm<\/p>\n\n\n\n

ATQ;<\/p>\n\n\n\n

144 = P \u00d7 8<\/p>\n\n\n\n

Or, P = 18<\/p>\n\n\n\n

Since, the base is an equilateral triangle. So, the length of each side of the triangle will be = 18 \u00f7 3 = 6 cm <\/p>\n\n\n\n

Volume of prism (V) = Area of base \u00d7 height <\/p>\n\n\n\n

Desired volume = (\u221a3\/4) \u00d7 6 \u00d7 6 \u00d7 8= 72\u221a3 cm3<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

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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]<\/strong><\/p>\n\n\n\n

a.)48.22%<\/p>\n\n\n\n

b.) 42.85%<\/p>\n\n\n\n

c.) 38.67%<\/p>\n\n\n\n

d.) 44.56%<\/p>\n\n\n\n

Answer: b)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the cost price of each article be Rs. \u20187x\u2019<\/p>\n\n\n\n

Cost price of 10 articles = 7x \u00d7 10 = Rs. 70x<\/p>\n\n\n\n

ATQ\u2019<\/p>\n\n\n\n

70x = selling price of 7 articles <\/p>\n\n\n\n

So, selling price of each article = (70x\/7) = Rs. 10x<\/p>\n\n\n\n

Desired profit percentage = {(10x \u2013 7x)\/7x} \u00d7 100 ~42.85%<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Sign Up to Grab SSC GD 2024 FREE Mock Test\ud83c\udd93\ud83c\udd93<\/a><\/em><\/h2>\n\n\n\n

19.) Pipe \u2018A\u2019 can fill a tank in 4 hours while pipe \u2018B\u2019 and \u2018C\u2019 together can fill the same tank in 6 hours. If pipe \u2018A\u2019 is opened at 12:00 p.m. when the tank is empty and pipe \u2018B\u2019 and \u2018C\u2019 are opened simultaneously when the tank is half filled, then at what time will the tank be full?<\/strong><\/p>\n\n\n\n

a.) 3:12 p.m.<\/p>\n\n\n\n

b.) 3:00 p.m.<\/p>\n\n\n\n

c.) 3:30 p.m.<\/p>\n\n\n\n

d.) 2:20 p.m.<\/p>\n\n\n\n

Answer: a)<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Let the capacity of the tank = 12 litres {LCM (4, 6)}<\/p>\n\n\n\n

Efficiency of pipe \u2018A\u2019 = (12\/4) = 3 litres\/hour<\/p>\n\n\n\n

Efficiency of pipe \u2018B\u2019 and \u2018C\u2019 together = (12\/6) = 2 litres\/hour<\/p>\n\n\n\n

Time taken by pipe \u2018A\u2019 to fill 50% of the tank = (6\/3) = 2 hours<\/p>\n\n\n\n

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

Total time taken to fill the tank = 3.2 hours = 3 hours and 12 minutes <\/p>\n\n\n\n

Therefore, the tank will be filled at 3:12 p.m.<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

\ud83c\udfc3 Mock Test \u092d\u093e\u0917 \u0932\u0947\u0902 \u0914\u0930 \u0905\u092a\u0928\u0940 \u092a\u094d\u0930\u0924\u093f\u092d\u093e \u0915\u093e \u092e\u0942\u0932\u094d\u092f\u093e\u0902\u0915\u0928 \u0915\u0930\u0947\u0902 \ud83c\udfaf\ud83d\udcaf\ud83e\udd47<\/a><\/em><\/h2>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

a.) 20%<\/p>\n\n\n\n

b.) 26%<\/p>\n\n\n\n

c.) 13%<\/p>\n\n\n\n

d.) 10%<\/p>\n\n\n\n

Answer: c)<\/strong><\/p>\n\n\n\n

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Solution:<\/p>\n\n\n\n

Cost price of the article = 67.5 \u00f7 0.45 = Rs. 150<\/p>\n\n\n\n

Selling price of the article = 150 + 67.5 = Rs. 217.5<\/p>\n\n\n\n

Discount offered = 250 \u2013 217.5 = Rs. 32.5<\/p>\n\n\n\n

Desired discount percentage = {32.5\/250} \u00d7 100 = 13%<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

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SSC GD 2024 Tips for Elementary Mathematics<\/h2>\n\n\n\n

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:<\/p>\n\n\n\n

Master the Basics:<\/strong> 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.<\/p>\n\n\n\n

Focus on Problem-solving Skills:<\/strong> Elementary Mathematics questions<\/a> 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.<\/p>\n\n\n\n

Memorize Important Formulas:<\/strong> 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.<\/p>\n\n\n\n

\ud83c\udfc3 Mock Test \u092d\u093e\u0917 \u0932\u0947\u0902 \u0914\u0930 \u0905\u092a\u0928\u0940 \u092a\u094d\u0930\u0924\u093f\u092d\u093e \u0915\u093e \u092e\u0942\u0932\u094d\u092f\u093e\u0902\u0915\u0928 \u0915\u0930\u0947\u0902 \ud83c\udfaf\ud83d\udcaf\ud83e\udd47<\/a><\/em><\/h2>\n\n\n\n

Use Shortcut Techniques: <\/strong>Learn shortcut techniques for solving mathematical problems quickly. For example, learn tricks for mental calculation, simplification, and finding patterns in numerical sequences.<\/p>\n\n\n\n

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

Practice Regularly: <\/strong>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.<\/p>\n\n\n\n

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

Want to Track Your Preparation? Sign up! Take Free Mock Test & Analyze Your Performance\ud83d\udc68\u200d\ud83d\udcbb<\/a><\/em><\/h4>\n\n\n\n

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!<\/p>\n\n\n\n

<\/p>\n\n\n\n

Disclaimer:<\/strong> The mathematics questions and solutions provided are for practice and guidance only. They are illustrative, not official SSC GD exam content. Actual exam questions may differ in wording, format, or difficulty. Candidates should always consult official SSC notices<\/a> for authentic and updated information before preparation.<\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

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