{"id":42965,"date":"2023-05-16T12:58:01","date_gmt":"2023-05-16T07:28:01","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=42965"},"modified":"2025-11-25T18:05:37","modified_gmt":"2025-11-25T12:35:37","slug":"ssc-chsl-quant-practice-questions-free-preparation","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/ssc-chsl-quant-practice-questions-free-preparation\/","title":{"rendered":"SSC CHSL Quant Practice Questions Free Preparation"},"content":{"rendered":"\n

SSC CHSL 2023 recruitment notification has been released featuring a total of 1600 vacancies. As per the notification the last date and time for receipt of online applications is 8th June 2023. So, we would like to suggest to the aspiring candidates not to wait for the last date and try filling the application forms as early as possible in order to avoid any last minute hassles. Those who  are done with the online application process can attempt the SSC CHSL free mock test<\/a> to know the kind of questions which are asked in the actual examination. Moreover, for those who have already started their exam preparation we have got for them SSC CHSL Quant Practice Questions<\/a> for Free Preparation<\/a> along with detailed solutions. Let\u2019s first take a look at the exam pattern.<\/p>\n\n\n\n

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

SSC CHSL Quant Practice Questions for Free Preparation<\/h2>\n\n\n\n

Topic: SSC_ Averages<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

1.) The mean of five numbers is 18. If one number is excluded, their mean is 16. The excluded number is:<\/strong><\/p>\n\n\n\n

a.) 30<\/strong><\/p>\n\n\n\n

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

c.) 36<\/strong><\/p>\n\n\n\n

d.) 32<\/strong><\/p>\n\n\n\n

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

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

Sum of five numbers = 18 \u00d7 5 = 90<\/strong><\/p>\n\n\n\n

Sum when one number is excluded = 16 \u00d7 4 = 64<\/strong><\/p>\n\n\n\n

So, excluded number = 90 \u2013 64 = 26<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Time, speed and distance<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

2.) After seeing a policeman from a distance of 500 metres, a thief starts running at a speed of 10 km\/h. After noticing, the policeman chases immediately with a speed of 12 km\/h, and the thief is caught. The distance run by the policeman is:<\/strong><\/p>\n\n\n\n

a.) 2.4 km<\/strong><\/p>\n\n\n\n

b.) 3 km<\/strong><\/p>\n\n\n\n

c.) 2.5 km<\/strong><\/p>\n\n\n\n

d.) 3.6 km<\/strong><\/p>\n\n\n\n

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

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

Relative speed of the policeman w.r.t the relative speed of the thief = 12 \u2013 10 = 2 km\/h<\/strong><\/p>\n\n\n\n

Time taken by the policeman to catch the thief = {(500\/1000) \u00f7 2} = (1\/4) hours<\/strong><\/p>\n\n\n\n

So, distance covered by the policeman = 12 \u00d7 (1\/4) = 3 km<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Ratio and proportions<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

3.) \u2018A\u2019 varies jointly with \u2018B\u2019 and \u2018C\u2019. A = 6 when B = 3 and C = 2. Find A when B = 5 and C = 7<\/strong><\/p>\n\n\n\n

a.) 17.5<\/strong><\/p>\n\n\n\n

b.) 35<\/strong><\/p>\n\n\n\n

c.) 105<\/strong><\/p>\n\n\n\n

d.) 70<\/strong><\/p>\n\n\n\n

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

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

ATQ:<\/strong><\/p>\n\n\n\n

A = k \u00d7 B \u00d7 C {Where \u2018k\u2019 is a constant}<\/strong><\/p>\n\n\n\n

So, 6 = k \u00d7 3 \u00d7 2<\/strong><\/p>\n\n\n\n

Or, k = 1<\/strong><\/p>\n\n\n\n

Required value = A = 1 \u00d7 5 \u00d7 7 = 35<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Simplification<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

4.) Which of the following options has the greatest value?<\/strong><\/p>\n\n\n\n

a.) 20 + 4 + (\u22128) \u2013 2 + 3 + 6<\/strong><\/p>\n\n\n\n

b.) (\u221222) + (\u22124 \u2212 7)<\/strong><\/p>\n\n\n\n

c.) (\u221218) \u2013 45 + (\u22123 \u2212 2)<\/strong><\/p>\n\n\n\n

d.) (\u221299) + (\u221244) \u2212 12<\/strong><\/p>\n\n\n\n

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

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

Option A: 20 + 4 + (\u22128) \u2013 2 + 3 + 6<\/strong><\/p>\n\n\n\n

= 24 \u2013 8 \u2013 2 + 3 + 6 = 23<\/strong><\/p>\n\n\n\n

Option B: (\u221222) + (\u22124 \u2212 7)<\/strong><\/p>\n\n\n\n

= -22 \u2013 4 \u2013 7 = \u201333<\/strong><\/p>\n\n\n\n

Option C: (\u221218) \u2013 45 + (\u22123 \u2212 2)<\/strong><\/p>\n\n\n\n

= -18 \u2013 45 \u2013 3 \u2013 2 = \u201368<\/strong><\/p>\n\n\n\n

Option D: (\u221299) + (\u221244) \u2212 12<\/strong><\/p>\n\n\n\n

= \u201399 \u2013 44 \u2013 12 = \u2013155<\/strong><\/p>\n\n\n\n

So, option \u2018A\u2019 has the greatest value.<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Geometry and mensuration<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

5.) In \u0394ABC, if G is the centroid and AD is a median with length 9 cm, then the length of AG is:<\/strong><\/p>\n\n\n\n

a.) 1.8 cm<\/strong><\/p>\n\n\n\n

b.) 5 cm<\/strong><\/p>\n\n\n\n

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

d.) 6 cm<\/strong><\/p>\n\n\n\n

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

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

In a triangle with median AD and centroid \u2018G\u2019, the centroid divided GD and AG in ratio 1:2.<\/strong><\/p>\n\n\n\n

So, length of AG = 9 \u00d7 (2\/3) = 6 cm  <\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Compound interest and simple interest<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

6.) A sum of \u20b914,375, when invested at r% interest per year compounded annually, amounts to \u20b916,767 after two years. What is the value of r?<\/strong><\/p>\n\n\n\n

a.) 9<\/strong><\/p>\n\n\n\n

b.) 7<\/strong><\/p>\n\n\n\n

c.) 8<\/strong><\/p>\n\n\n\n

d.) 6<\/strong><\/p>\n\n\n\n

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

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

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

16767 = 14375 \u00d7 {1 + (r\/100)}<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

Or, {1 + (r\/100)}<\/strong>2<\/sup><\/strong> = (729\/625)<\/strong><\/p>\n\n\n\n

Or, 1 + (r\/100) = (27\/25)<\/strong><\/p>\n\n\n\n

Or, (r\/100) = (27\/25) \u2013 1<\/strong><\/p>\n\n\n\n

Or, (r\/100) = (2\/25)<\/strong><\/p>\n\n\n\n

Or, r = (2\/25) \u00d7 100 = 8<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ DI<\/strong><\/p>\n\n\n\n

Sub \u2013 Topic: Pie chart<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

7.) <\/strong>In a School of 200 students, the following charts represent the percentage of students involved in different sports.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

What is the number of students playing cricket?<\/strong><\/p>\n\n\n\n

a.) 34<\/strong><\/p>\n\n\n\n

b.) 24<\/strong><\/p>\n\n\n\n

c.) 83<\/strong><\/p>\n\n\n\n

d.) 17<\/strong><\/p>\n\n\n\n

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

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

Number of students playing cricket = 200 \u00d7 (17\/100) = 34<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Averages<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

8.) The average of 10 observations is 21. A new observation is included and the average of these 11 numbers is 1 less than the previous average. The 11th observation is ______.<\/strong><\/p>\n\n\n\n

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

b.) 11<\/strong><\/p>\n\n\n\n

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

d.) 12<\/strong><\/p>\n\n\n\n

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

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

Sum of first 10 observations = 10 \u00d7 21 = 210<\/strong><\/p>\n\n\n\n

Sum of 11 observations = 11 \u00d7 20 = 220<\/strong><\/p>\n\n\n\n

So, 11<\/strong>th<\/sup><\/strong> observation = 220 \u2013 210 = 10<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ DI<\/strong><\/p>\n\n\n\n

Sub \u2013 Topic: Bar<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

9.) Observe the graph and answer the question that follows:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

How many students obtained marks more than 130?<\/strong><\/p>\n\n\n\n

a.) 14<\/strong><\/p>\n\n\n\n

b.) 19<\/strong><\/p>\n\n\n\n

c.) 20<\/strong><\/p>\n\n\n\n

d.) 17<\/strong><\/p>\n\n\n\n

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

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

Required sum = 2 + 8 + 5 + 4 + 1 = 20<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Number system<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

10.) Which of the following is the least 6-digit number that is divisible by 93?<\/strong><\/p>\n\n\n\n

a.) 100068<\/strong><\/p>\n\n\n\n

b.) 100070<\/strong><\/p>\n\n\n\n

c.) 100075<\/strong><\/p>\n\n\n\n

d.) 100065<\/strong><\/p>\n\n\n\n

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

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

We know that 100000 \u00f7 93 ~ 1075.26<\/strong><\/p>\n\n\n\n

So, required number must be 93 \u00d7 1076<\/strong><\/p>\n\n\n\n

So, least 6-digit number that is divisible by 93 = 93 \u00d7 1076 = 100068<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Ratio and proportions<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

11.) A number \u2018x\u2019 is three times another number \u2018y\u2019. If the sum of both the numbers is 20, then x and y, respectively, are:<\/strong><\/p>\n\n\n\n

a.)  8 and 12<\/strong><\/p>\n\n\n\n

b.) 15 and 5<\/strong><\/p>\n\n\n\n

c.) 5 and 15<\/strong><\/p>\n\n\n\n

d.) 2 and 18<\/strong><\/p>\n\n\n\n

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

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

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

x = 3y<\/strong><\/p>\n\n\n\n

So, x + y = 3y + y = 20<\/strong><\/p>\n\n\n\n

Or, 4y = 20<\/strong><\/p>\n\n\n\n

So, y = 5<\/strong><\/p>\n\n\n\n

So, x = 20 \u2013 5 = 15<\/strong><\/p>\n\n\n\n

Therefore, \u2018x\u2019 and \u2018y\u2019 are 15 and 5, respectively<\/strong><\/p>\n\n\n\n

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

Topic: SSC_ Percentage, profit and loss<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

12.) A person could save 20% of his income. A year later, his income increased by 25% but he could save the same amount only as before. By what percentage has his expenditure increased?<\/strong><\/p>\n\n\n\n

a.) 27.5%<\/strong><\/p>\n\n\n\n

b.) 32.5%<\/strong><\/p>\n\n\n\n

c.) 31.25%<\/strong><\/p>\n\n\n\n

d.) 29.75%<\/strong><\/p>\n\n\n\n

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

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

Let the initial income of the person be Rs. 100x<\/strong><\/p>\n\n\n\n

So, amount saved = 100x \u00d7 0.2 = Rs. 20x<\/strong><\/p>\n\n\n\n

So, original expenditure = 100x \u2013 20x = Rs. 80x<\/strong><\/p>\n\n\n\n

New income = 100x \u00d7 1.25 = Rs. 125x<\/strong><\/p>\n\n\n\n

So, new expenditure = 125x \u2013 20x = Rs. 105x<\/strong><\/p>\n\n\n\n

Increase in expenditure = 105x \u2013 80x = Rs. 25x<\/strong><\/p>\n\n\n\n

So, required percentage = (25x\/80x) \u00d7 100 = 31.25%<\/strong><\/p>\n\n\n\n

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

Topic: SSC_Percentages, profit and loss<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

13.) The marked price of a table fan is Rs. 3750 and is available to the retailer at a discount of 20%. At what price should the retailer sell it to earn a profit of 15%?<\/strong><\/p>\n\n\n\n

a.) Rs. 3250<\/strong><\/p>\n\n\n\n

b.) Rs. 3500<\/strong><\/p>\n\n\n\n

c.) Rs. 3700<\/strong><\/p>\n\n\n\n

d.) Rs. 3450<\/strong><\/p>\n\n\n\n

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

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

Price at which the retailer purchases the table fan = 3750 \u00d7 0.8 = Rs. 3000<\/strong><\/p>\n\n\n\n

So, price at which the retailer must sell the fan to earn a profit of 15% = 3000 \u00d7 1.15 = Rs. 3450<\/strong><\/p>\n\n\n\n

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

Topic: Time, speed and distance<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

14.) Ravi drove for 4 hours at a speed of 70 miles per hour and for 2 hours at a speed of 40 miles per hour. What was his average speed (in miles per hour) for the whole journey?<\/strong>
<\/strong>a.) 60 miles per hour<\/strong><\/p>\n\n\n\n

b.) 45 miles per hour<\/strong><\/p>\n\n\n\n

c.) 55 miles per hour<\/strong><\/p>\n\n\n\n

d.) 65 miles per hour<\/strong><\/p>\n\n\n\n

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

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

Total distance travelled by Ravi = 4 \u00d7 70 + 2 \u00d7 40 = 360 miles<\/strong><\/p>\n\n\n\n

Total time taken for travel = 4 + 2 = 6 hours<\/strong><\/p>\n\n\n\n

So, average speed = total distance covered \u00f7 total time taken = (360\/6) = 60 miles per hour <\/strong><\/strong><\/strong><\/p>\n\n\n\n

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

Topic: Algebra<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

15.) What is the product of (x + a) and (x + b)?<\/strong><\/p>\n\n\n\n

a.) x<\/strong>2<\/sup><\/strong> + (a + b)x + ab<\/strong><\/p>\n\n\n\n

b.) x<\/strong>2<\/sup><\/strong> + (a \u2013 b)x + ab<\/strong><\/p>\n\n\n\n

c.) x<\/strong>2<\/sup><\/strong> + (a + b)x \u2013 ab<\/strong><\/p>\n\n\n\n

d.) x<\/strong>2<\/sup><\/strong> + (a \u2013 b)x \u2013 ab<\/strong><\/p>\n\n\n\n

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

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

(x + a) \u00d7 (x + b) = x<\/strong>2<\/sup><\/strong> + ax + bx + ab<\/strong><\/p>\n\n\n\n

= x<\/strong>2<\/sup><\/strong> + (a + b)x + ab<\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Mensuration<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

16. What is the height of a solid right circular cylinder whose radius is 3 cm and total surface area is 60\u03c0 cm<\/strong>2<\/sup><\/strong>.<\/strong><\/p>\n\n\n\n

a.) 4.2 cm<\/strong><\/p>\n\n\n\n

b.) 5 cm<\/strong><\/p>\n\n\n\n

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

d.) 9 cm<\/strong><\/p>\n\n\n\n

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

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

Total surface area of a cylinder = 2 \u00d7 \u03c0 \u00d7 radius \u00d7 (height + radius)<\/strong><\/p>\n\n\n\n

= 2 \u00d7 \u03c0 \u00d7 3 \u00d7 (height + 3)<\/strong><\/p>\n\n\n\n

Or, 6\u03c0(height + 3) = 60\u03c0<\/strong><\/p>\n\n\n\n

So, height + 3 = 60\u03c0 \u00f7 6\u03c0 = 10<\/strong><\/p>\n\n\n\n

So, height of the cylinder = 10 \u2013 3 = 7 cm <\/strong><\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Time and work<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

17.) 45 people can repair a road in 10 days working 6 hours a day. In how many days can 30 people working 6 hours a day complete the same work?<\/strong><\/p>\n\n\n\n

a.) 12<\/strong><\/p>\n\n\n\n

b.) 10<\/strong><\/p>\n\n\n\n

c.) 18  <\/strong><\/p>\n\n\n\n

d.) 15<\/strong><\/p>\n\n\n\n

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

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

Let the hourly efficiency of 1 person be \u2018x\u2019 units\/hour<\/strong><\/p>\n\n\n\n

Then, total work = 45x \u00d7 10 \u00d7 6 = 2700x<\/strong><\/p>\n\n\n\n

So, number of days taken by 30 people working 6 hours a day to complete the same work<\/strong><\/p>\n\n\n\n

= 2700x \u00f7 (30x \u00d7 6) = 15 days<\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Mensuration<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

18.) What is the area (in cm<\/strong>2<\/sup><\/strong>) of an equilateral triangle of side 20 cm.<\/strong><\/p>\n\n\n\n

a.) 100\u221a3 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

b.) 200 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

c.) 100 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

d.) 100\u221a2 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

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

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

For an equilateral triangle with side length of \u2018a\u2019 units<\/strong><\/p>\n\n\n\n

Area of the triangle = (\u221a3\/4)a<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

= (\u221a3\/4) \u00d7 20<\/strong>2<\/sup><\/strong> = 100\u221a3 cm<\/strong>2<\/sup><\/strong> <\/strong><\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Percentages profit and loss<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

19.) Under a discount scheme, a dozen pair of gloves quoted at Rs. 200 is available at a discount of 20%. How many pairs of gloves can be bought for Rs. 320?<\/strong><\/p>\n\n\n\n

a.) 15<\/strong><\/p>\n\n\n\n

b.) 20<\/strong><\/p>\n\n\n\n

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

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

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

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

Selling price of 1 dozen pair of gloves = 200 \u00d7 0.8 = Rs. 160<\/strong><\/p>\n\n\n\n

So, number of dozens of pairs of gloves that can be bought for Rs. 320 = 320 \u00f7 160 = 2<\/strong><\/p>\n\n\n\n

So, number of pairs of gloves that can be bought for Rs. 320 = 2 \u00d7 12 = 24 <\/strong><\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Trignometry<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

20.)  The value of sin 73<\/strong>o<\/sup><\/strong> + cos 137<\/strong>o<\/sup><\/strong> is:<\/strong><\/p>\n\n\n\n

a.) sin 13<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

b.) cos 13<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

c.) cos 18<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

d.) sin 18<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

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

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

sin 73<\/strong>o<\/sup><\/strong> + cos 137<\/strong>o  <\/sup><\/strong>= sin 73<\/strong>o<\/sup><\/strong> + cos (90 + 47)<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

= sin 73<\/strong>o<\/sup><\/strong> \u2013 sin 47<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

We know that, sin A \u2013 sin B = 2cos{(A+ B)\/2} sin{(A \u2013 B)\/2}<\/strong><\/p>\n\n\n\n

So, sin 73<\/strong>o<\/sup><\/strong> \u2013 sin 47<\/strong>o <\/sup><\/strong>= 2cos{(73 + 47)\/2} sin{(73 \u2013 47)\/2}<\/strong><\/p>\n\n\n\n

= 2cos 60<\/strong>o<\/sup><\/strong> sin 13<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

= 2 \u00d7 (1\/2) \u00d7 sin 13<\/strong>o<\/sup><\/strong><\/p>\n\n\n\n

= sin 13<\/strong>o<\/sup><\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Mensuration<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

21. The diameter of the base of a right circular cylinder in 12 cm and the height of the cylinder is 2.45 times the radius of its base. Find the curved surface area of the cylinder.<\/strong><\/p>\n\n\n\n

a.) 550.4 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

b.) 548.8 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

c.) 578.2 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

d.) 554.4 cm<\/strong>2<\/sup><\/strong><\/p>\n\n\n\n

Answer: 4<\/strong><\/p>\n\n\n\n

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

Radius of the given cylinder = diameter \u00f7 2 = 12 \u00f7 2 = 6 cm<\/strong><\/p>\n\n\n\n

So, height of the cylinder = 6 \u00d7 2.45 = 14.7 cm<\/strong><\/p>\n\n\n\n

Curved surface area of the cylinder = 2 \u00d7 \u03c0 \u00d7 radius \u00d7 height<\/strong><\/p>\n\n\n\n

= 2 \u00d7 (22\/7) \u00d7 6 \u00d7 14.7<\/strong><\/p>\n\n\n\n

= 554.4 cm<\/strong>2<\/sup><\/strong> <\/strong><\/strong> <\/strong><\/p>\n\n\n\n

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

Topic: SSC_Algebra<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

22.) If x + y + z = 10, x<\/strong>2<\/sup><\/strong> + y<\/strong>2<\/sup><\/strong> + z<\/strong>2<\/sup><\/strong> = 30, then the value of x<\/strong>3<\/sup><\/strong> + y<\/strong>3<\/sup><\/strong> + z<\/strong>3<\/sup><\/strong> \u2013 3xyz is?<\/strong><\/p>\n\n\n\n

a.)  -25<\/strong><\/p>\n\n\n\n

b.) -50<\/strong><\/p>\n\n\n\n

c.) -40<\/strong><\/p>\n\n\n\n

d.) -10<\/strong><\/p>\n\n\n\n

Answer: 2<\/strong><\/p>\n\n\n\n

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

(x + y + z)<\/strong>2<\/sup><\/strong> = 10<\/strong>2<\/sup><\/strong> = x<\/strong>2<\/sup><\/strong> + y<\/strong>2<\/sup><\/strong> + z<\/strong>2<\/sup><\/strong> + 2(xy + yz + zx)<\/strong><\/p>\n\n\n\n

So, 2(xy + yz + zx) = 100 \u2013 30 = 70<\/strong><\/p>\n\n\n\n

So, (xy + yz + zx) = (70\/2) = 35<\/strong><\/p>\n\n\n\n

x<\/strong>3<\/sup><\/strong> + y<\/strong>3<\/sup><\/strong> + z<\/strong>3<\/sup><\/strong> \u2013 3xyz = (x + y + z)(x<\/strong>2<\/sup><\/strong> + y<\/strong>2<\/sup><\/strong> + z<\/strong>2<\/sup><\/strong> \u2013 xy \u2013 yz \u2013 zx)<\/strong><\/p>\n\n\n\n

= 10 \u00d7 (30 \u2013 35)<\/strong><\/p>\n\n\n\n

= 10 \u00d7 – 5<\/strong><\/p>\n\n\n\n

= – 50<\/strong><\/p>\n\n\n\n

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

Topic: SSC_Percentages, profit and loss<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

23.) A pen passing through 2 hands is finally sold at a profit of 44% of the original cost price. If the first dealer makes a profit of 20%, then the profit made by the second dealer is?<\/strong><\/p>\n\n\n\n

a.) 36%<\/strong><\/p>\n\n\n\n

b.) 24%<\/strong><\/p>\n\n\n\n

c.) 20%<\/strong><\/p>\n\n\n\n

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

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

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

Let the original cost price of the article be Rs. 100x<\/strong><\/p>\n\n\n\n

Then, price at which first dealer sold the article = 100x \u00d7 1.2 = Rs. 120x<\/strong><\/p>\n\n\n\n

Final selling price = 100x \u00d7 1.44 = Rs. 144x<\/strong><\/p>\n\n\n\n

So, profit made by the second dealer = (144x \u2013 120x) \u00f7 120x \u00d7 100 = 20%<\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_Averages<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

24. A shopkeeper purchases 20000 units of a product at Rs. 1 per unit, 15000 units at Rs. 1.15 per unit and 5000 units at Rs. 2 per unit. What is the weight average price of one unit of product bought? (correct to two decimal places)<\/strong> <\/strong><\/p>\n\n\n\n

a.) Rs. 1.36<\/strong><\/p>\n\n\n\n

b.) Rs. 1.28<\/strong><\/p>\n\n\n\n

c.) Rs. 1.18<\/strong><\/p>\n\n\n\n

d.) Rs. 1.22<\/strong><\/p>\n\n\n\n

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

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

Total price of all products purchased = 20000 \u00d7 1 + 15000 \u00d7 1.15 + 5000 \u00d7 2<\/strong><\/p>\n\n\n\n

= Rs. 20000 + 17250 + 10000 = Rs. 47250<\/strong><\/p>\n\n\n\n

Total number of products purchased = 20000 + 15000 + 5000 = 40000<\/strong><\/p>\n\n\n\n

So, required average = 47250 \u00f7 40000 ~ 1.18<\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Topic: SSC_DI<\/strong><\/p>\n\n\n\n

Sub-Topic: Table<\/strong><\/p>\n\n\n\n

Level: 0<\/strong><\/p>\n\n\n\n

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

25. The table below shows the number of cakes sold by six different bakeries in a town on five different days of a particular week.<\/strong><\/p>\n\n\n\n

Days\/Bakery<\/strong><\/td>Monday<\/strong><\/td>Tuesday<\/strong><\/td>Thursday<\/strong><\/td>Saturday<\/strong><\/td>Sunday<\/strong><\/td><\/tr>
A<\/strong><\/td>222<\/strong><\/td>255<\/strong><\/td>215<\/strong><\/td>250<\/strong><\/td>266<\/strong><\/td><\/tr>
B<\/strong><\/td>205<\/strong><\/td>275<\/strong><\/td>314<\/strong><\/td>295<\/strong><\/td>260<\/strong><\/td><\/tr>
C<\/strong><\/td>245<\/strong><\/td>266<\/strong><\/td>305<\/strong><\/td>195<\/strong><\/td>235<\/strong><\/td><\/tr>
D<\/strong><\/td>221<\/strong><\/td>230<\/strong><\/td>185<\/strong><\/td>300<\/strong><\/td>280<\/strong><\/td><\/tr>
E<\/strong><\/td>312<\/strong><\/td>325<\/strong><\/td>298<\/strong><\/td>272<\/strong><\/td>254<\/strong><\/td><\/tr>
F<\/strong><\/td>175<\/strong><\/td>205<\/strong><\/td>255<\/strong><\/td>240<\/strong><\/td>308<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

What is the total number of cakes sold by bakery \u2018D\u2019 on Monday, Thursday and Sunday taken together?<\/strong><\/p>\n\n\n\n

a.) 686<\/strong><\/p>\n\n\n\n

b.) 712<\/strong><\/p>\n\n\n\n

c.) 672<\/strong><\/p>\n\n\n\n

d.) 654<\/strong><\/p>\n\n\n\n

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

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

Required sum = 221 + 185 + 280 = 686<\/strong> <\/strong><\/strong><\/p>\n\n\n\n

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

Attempt SSC CHSL Tier I 2023 Free Mock Test<\/a><\/p>\n\n\n\n

This brings us to the end of the article. So, here are few SSC CHSL Quant Practice Questions Free Preparation<\/a> based on latest exam patern. Keep preparing for the upcoming SSC CHSL 2023 Tier I exam. <\/p>\n\n\n\n

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

Disclaimer:<\/strong> The Quantitative Aptitude practice questions and solutions provided are for guidance only. They are illustrative, not official SSC CHSL 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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