SSC CHSL Quant Practice Questions Free Preparation

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 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 for Free Preparation along with detailed solutions. Let’s first take a look at the exam pattern.

SSC CHSL Tier I Exam Pattern

SSC CHSL Quant Practice Questions for Free Preparation

Topic: SSC_ Averages

Level: 0

1.) The mean of five numbers is 18. If one number is excluded, their mean is 16. The excluded number is:

a.) 30

b.) 26

c.) 36

d.) 32

Answer: b)

Solution:

Sum of five numbers = 18 × 5 = 90

Sum when one number is excluded = 16 × 4 = 64

So, excluded number = 90 – 64 = 26

Hence, option b.

Topic: SSC_ Time, speed and distance

Level: 0

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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:

a.) 2.4 km

b.) 3 km

c.) 2.5 km

d.) 3.6 km

Answer: b)

Solution:

Relative speed of the policeman w.r.t the relative speed of the thief = 12 – 10 = 2 km/h

Time taken by the policeman to catch the thief = {(500/1000) ÷ 2} = (1/4) hours

So, distance covered by the policeman = 12 × (1/4) = 3 km

Hence, option b.

Topic: SSC_ Ratio and proportions

Level: 0

3.) ‘A’ varies jointly with ‘B’ and ‘C’. A = 6 when B = 3 and C = 2. Find A when B = 5 and C = 7

a.) 17.5

b.) 35

c.) 105

d.) 70

Answer: b)

Solution:

ATQ:

A = k × B × C {Where ‘k’ is a constant}

So, 6 = k × 3 × 2

Or, k = 1

Required value = A = 1 × 5 × 7 = 35

Hence, option b.

Topic: SSC_ Simplification

Level: 0

4.) Which of the following options has the greatest value?

a.) 20 + 4 + (−8) – 2 + 3 + 6

b.) (−22) + (−4 − 7)

c.) (−18) – 45 + (−3 − 2)

d.) (−99) + (−44) − 12

Answer: a)

Solution:

Option A: 20 + 4 + (−8) – 2 + 3 + 6

= 24 – 8 – 2 + 3 + 6 = 23

Option B: (−22) + (−4 − 7)

= -22 – 4 – 7 = –33

Option C: (−18) – 45 + (−3 − 2)

= -18 – 45 – 3 – 2 = –68

Option D: (−99) + (−44) − 12

= –99 – 44 – 12 = –155

So, option ‘A’ has the greatest value.

Hence, option a.

Topic: SSC_ Geometry and mensuration

Level: 0

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5.) In ΔABC, if G is the centroid and AD is a median with length 9 cm, then the length of AG is:

a.) 1.8 cm

b.) 5 cm

c.) 7 cm

d.) 6 cm

Answer: d)

Solution:

In a triangle with median AD and centroid ‘G’, the centroid divided GD and AG in ratio 1:2.

So, length of AG = 9 × (2/3) = 6 cm  

Hence, option d.

Topic: SSC_ Compound interest and simple interest

Level: 0

6.) A sum of ₹14,375, when invested at r% interest per year compounded annually, amounts to ₹16,767 after two years. What is the value of r?

a.) 9

b.) 7

c.) 8

d.) 6

Answer: c)

Solution:

ATQ;

16767 = 14375 × {1 + (r/100)}²

Or, {1 + (r/100)}² = (729/625)

Or, 1 + (r/100) = (27/25)

Or, (r/100) = (27/25) – 1

Or, (r/100) = (2/25)

Or, r = (2/25) × 100 = 8

Hence, option c.

Topic: SSC_ DI

Sub – Topic: Pie chart

Level: 0

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7.) In a School of 200 students, the following charts represent the percentage of students involved in different sports.

What is the number of students playing cricket?

a.) 34

b.) 24

c.) 83

d.) 17

Answer: a)

Solution:

Number of students playing cricket = 200 × (17/100) = 34

Hence, option a.

Topic: SSC_ Averages

Level: 0

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 ______.

a.) 21

b.) 11

c.) 10

d.) 12

Answer: c)

Solution:

Sum of first 10 observations = 10 × 21 = 210

Sum of 11 observations = 11 × 20 = 220

So, 11^th observation = 220 – 210 = 10

Hence, option c.

Topic: SSC_ DI

Sub – Topic: Bar

Level: 0

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9.) Observe the graph and answer the question that follows:

How many students obtained marks more than 130?

a.) 14

b.) 19

c.) 20

d.) 17

Answer: c)

Solution:

Required sum = 2 + 8 + 5 + 4 + 1 = 20

Hence, option c.

Topic: SSC_ Number system

Level: 0

10.) Which of the following is the least 6-digit number that is divisible by 93?

a.) 100068

b.) 100070

c.) 100075

d.) 100065

Answer: a)

Solution:

We know that 100000 ÷ 93 ~ 1075.26

So, required number must be 93 × 1076

So, least 6-digit number that is divisible by 93 = 93 × 1076 = 100068

Hence, option a.

Topic: SSC_ Ratio and proportions

Level: 0

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11.) A number ‘x’ is three times another number ‘y’. If the sum of both the numbers is 20, then x and y, respectively, are:

a.)  8 and 12

b.) 15 and 5

c.) 5 and 15

d.) 2 and 18

Answer: b)

Solution:

ATQ;

x = 3y

So, x + y = 3y + y = 20

Or, 4y = 20

So, y = 5

So, x = 20 – 5 = 15

Therefore, ‘x’ and ‘y’ are 15 and 5, respectively

Hence, option b.

Topic: SSC_ Percentage, profit and loss

Level: 0

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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?

a.) 27.5%

b.) 32.5%

c.) 31.25%

d.) 29.75%

Answer: c)

Solution:

Let the initial income of the person be Rs. 100x

So, amount saved = 100x × 0.2 = Rs. 20x

So, original expenditure = 100x – 20x = Rs. 80x

New income = 100x × 1.25 = Rs. 125x

So, new expenditure = 125x – 20x = Rs. 105x

Increase in expenditure = 105x – 80x = Rs. 25x

So, required percentage = (25x/80x) × 100 = 31.25%

Hence, option c.

Topic: SSC_Percentages, profit and loss

Level: 0

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%?

a.) Rs. 3250

b.) Rs. 3500

c.) Rs. 3700

d.) Rs. 3450

Answer: d

Solution:

Price at which the retailer purchases the table fan = 3750 × 0.8 = Rs. 3000

So, price at which the retailer must sell the fan to earn a profit of 15% = 3000 × 1.15 = Rs. 3450

Hence, option d.

Topic: Time, speed and distance

Level: 0

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?
a.) 60 miles per hour

b.) 45 miles per hour

c.) 55 miles per hour

d.) 65 miles per hour

Answer: a

Solution:

Total distance travelled by Ravi = 4 × 70 + 2 × 40 = 360 miles

Total time taken for travel = 4 + 2 = 6 hours

So, average speed = total distance covered ÷ total time taken = (360/6) = 60 miles per hour

Hence, option a.

Topic: Algebra

Level: 0

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15.) What is the product of (x + a) and (x + b)?

a.) x² + (a + b)x + ab

b.) x² + (a – b)x + ab

c.) x² + (a + b)x – ab

d.) x² + (a – b)x – ab

Answer: a

Solution:

(x + a) × (x + b) = x² + ax + bx + ab

= x² + (a + b)x + ab

Hence, option a.

Topic: SSC_Mensuration

Level: 0

16. What is the height of a solid right circular cylinder whose radius is 3 cm and total surface area is 60π cm².

a.) 4.2 cm

b.) 5 cm

c.) 7 cm

d.) 9 cm

Answer: c

Solution:

Total surface area of a cylinder = 2 × π × radius × (height + radius)

= 2 × π × 3 × (height + 3)

Or, 6π(height + 3) = 60π

So, height + 3 = 60π ÷ 6π = 10

So, height of the cylinder = 10 – 3 = 7 cm

Hence, option c.

Topic: SSC_Time and work

Level: 0

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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?

a.) 12

b.) 10

c.) 18  

d.) 15

Answer: d

Solution:

Let the hourly efficiency of 1 person be ‘x’ units/hour

Then, total work = 45x × 10 × 6 = 2700x

So, number of days taken by 30 people working 6 hours a day to complete the same work

= 2700x ÷ (30x × 6) = 15 days

Hence, option d.

Topic: SSC_Mensuration

Level: 0

18.) What is the area (in cm²) of an equilateral triangle of side 20 cm.

a.) 100√3 cm²

b.) 200 cm²

c.) 100 cm²

d.) 100√2 cm²

Answer: a

Solution:

For an equilateral triangle with side length of ‘a’ units

Area of the triangle = (√3/4)a²

= (√3/4) × 20² = 100√3 cm²

Hence, option a.

Topic: SSC_Percentages profit and loss

Level: 0

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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?

a.) 15

b.) 20

c.) 18

d.) 24

Answer: d

Solution:

Selling price of 1 dozen pair of gloves = 200 × 0.8 = Rs. 160

So, number of dozens of pairs of gloves that can be bought for Rs. 320 = 320 ÷ 160 = 2

So, number of pairs of gloves that can be bought for Rs. 320 = 2 × 12 = 24

Hence, option d.

Topic: SSC_Trignometry

Level: 0

20.)  The value of sin 73^o + cos 137^o is:

a.) sin 13^o

b.) cos 13^o

c.) cos 18^o

d.) sin 18^o

Answer: a

Solution:

sin 73^o + cos 137^o  = sin 73^o + cos (90 + 47)^o

= sin 73^o – sin 47^o

We know that, sin A – sin B = 2cos{(A+ B)/2} sin{(A – B)/2}

So, sin 73^o – sin 47^o = 2cos{(73 + 47)/2} sin{(73 – 47)/2}

= 2cos 60^o sin 13^o

= 2 × (1/2) × sin 13^o

= sin 13^o

Hence, option a.

Topic: SSC_Mensuration

Level: 0

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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.

a.) 550.4 cm²

b.) 548.8 cm²

c.) 578.2 cm²

d.) 554.4 cm²

Answer: 4

Solution:

Radius of the given cylinder = diameter ÷ 2 = 12 ÷ 2 = 6 cm

So, height of the cylinder = 6 × 2.45 = 14.7 cm

Curved surface area of the cylinder = 2 × π × radius × height

= 2 × (22/7) × 6 × 14.7

= 554.4 cm²  

Hence, option d.

Topic: SSC_Algebra

Level: 0

22.) If x + y + z = 10, x² + y² + z² = 30, then the value of x³ + y³ + z³ – 3xyz is?

a.)  -25

b.) -50

c.) -40

d.) -10

Answer: 2

Solution:

(x + y + z)² = 10² = x² + y² + z² + 2(xy + yz + zx)

So, 2(xy + yz + zx) = 100 – 30 = 70

So, (xy + yz + zx) = (70/2) = 35

x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

= 10 × (30 – 35)

= 10 × – 5

= – 50

Hence, option b.

Topic: SSC_Percentages, profit and loss

Level: 0

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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?

a.) 36%

b.) 24%

c.) 20%

d.) 27%

Answer: c

Solution:

Let the original cost price of the article be Rs. 100x

Then, price at which first dealer sold the article = 100x × 1.2 = Rs. 120x

Final selling price = 100x × 1.44 = Rs. 144x

So, profit made by the second dealer = (144x – 120x) ÷ 120x × 100 = 20%

Hence, option c.

Topic: SSC_Averages

Level: 0

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) 

a.) Rs. 1.36

b.) Rs. 1.28

c.) Rs. 1.18

d.) Rs. 1.22

Answer: c

Solution:

Total price of all products purchased = 20000 × 1 + 15000 × 1.15 + 5000 × 2

= Rs. 20000 + 17250 + 10000 = Rs. 47250

Total number of products purchased = 20000 + 15000 + 5000 = 40000

So, required average = 47250 ÷ 40000 ~ 1.18

Hence, option c.

Topic: SSC_DI

Sub-Topic: Table

Level: 0

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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.

What is the total number of cakes sold by bakery ‘D’ on Monday, Thursday and Sunday taken together?

a.) 686

b.) 712

c.) 672

d.) 654

Answer: a

Solution:

Required sum = 221 + 185 + 280 = 686

Hence, option a.

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

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