SSC CHSL Mathematical Abilities Questions and Answers

2023-05-12

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SSC CHSL Mathematical Abilities Questions and Answers: Hey Guys!! here is a perfect way you can ace your SSC CHSL Exam. We know that many of you will be in search of the Maths Questions with Solution for SSC CHSL Exam, so here we have established the SSC CHSL Mathematical Ability Questions and Answers to help you in preparing for the exam.

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Maths Questions with Solution for SSC CHSL

Check the SSC CHSL Maths Questions and know what topics you should cover to prepare for the Mathematical Abilities section and then start your preparation from now.

1.How many of the following numbers are divisible by 132?

264, 396, 462, 792, 968, 2178, 5184, 6336

a. 4
b. 5
c. 6
d. 7

Answer: a

Explanation:

132 = 4 x 3 x 11
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264 11,3,4 (/)
396 11,3,4 (/)
462 11,3 (X)
792 11,3,4 (/)
968 11,4 (X)
2178 11,3 (X)
5184 3,4 (X)
6336 11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

2. If Q means ‘add’ to; J means ‘Multiply by’, T means ‘subtract from’ and K means ‘divide by’ then 30 K 2 Q 3 J 6 T 5 = ?

a.18
b.28
c.31
d.103

Answer: b

3. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

a. Rs. 1090
b. Rs. 1160
c. Rs. 1190
d. Rs. 1202

Answer: c

Explanation:

S.P. = 85% of Rs. 1400 = Rs.(85/100*1400) = Rs.1190

4. 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

Calculations:

According to the question,
Students obtained 135 marks = 2
Students obtained 140 marks = 8
Students obtained 145 marks = 5
Students obtained 150 marks = 4
Students obtained 155 marks = 1
Total students obtained marks more than 130 = 2 + 8 + 5 + 4 + 1 = 20
∴ 20 students obtained marks more than 130.

5. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:

a. 70 km/hr
b. 75 km/hr
c. 84 km/hr
d. 87.5 km/hr

Answer: d

Explanation:

Let the speed of two trains be 7x and 8x km/hr.

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

6. The efficiency of A, B, and C is 2 : 3 : 5. A alone can complete a work in 50 days. They all work together for 5 days and then C left the work, in how many days A and B together can complete the remaining work?

a. 50 days
b. 30 days
c. 20 days
d. 10 days

Answer: d

Explanation:

Given:
Efficiency of A, B and C = 2 : 3 : 5
A alone can complete the work in = 50 days
Formula:
Total work = Efficiency × Time
Calculation:
Let efficiency of A be 2 units/day
Efficiency of A, B and C = 2 : 3 : 5
Total work = 2 × 50 = 100 units
Work done by A, B and C in 5 days = (2 + 3 + 5) × 5 = 10 × 5 = 50 units
Remaining work = 100 – 50 = 50 units
∴ Time taken by A and B to complete the remaining work = 50/(2 + 3) = 50/5 = 10 days

In the figure given above, O is the centre of the circle and ∠AOD = 106°. What is ∠BCD equal to?

a. 53°
b. 43°
c. 40°
d. 37°

Answer: d

Explanation:

Calculation:
According to the question
⇒ ∠BOD = (180° – 106°)
⇒ ∠BOD = 74°

Since, BOD is an angle made by arc BD on centre.
Here, ∠BCD is an angle made by arc BD on circumference

∴ The required value is 37°

8. Rukmini has a farmland that is triangular in shape. What is the sum of all the exterior angles taken in the order of the farmland?

a. 90°
b. 180°
c. 360°
d. Can not be determined.

Answer: c

Explanation:

The sum of exterior angles of a triangle = 360°
Solution: The sum of exterior angle of the triangular farm land = (3 x 180°) – 180 = 360°

9. The perimeter of the base of a right prism is 22 cm and its height is 6 cm. It is also given that its whole surface area is 152 cm2. What is the area of its two plane ends?

a. 20 cm2
b. 10 cm2
c. 15 cm2
d. 25 cm2

Answer: a

Explanation:

Given:
The perimeter of the base of a right prism = 22 cm
Height = 6 cm
Total surface area = 152 cm2
Formula used:
Total surface area = (The perimeter of the base × h) + The area of its two plane ends
Calculation:
⇒ 152 = 22 × 6 + The area of its two plane ends
⇒ 152 – 132 = 2 × The area of its two plane ends
⇒ The area of its two plane ends = 20 cm2
∴ The area of its two plane ends is 20 cm2

10. Which of the following values suits for A to make the equation $$ {{ATan62°Sec 28°Cot 38°}\over {Cosec62°Tan11°}}=1$$ ?

a. $$ {{Tan38°Tan79°}\over {Tan 28°}}$$
b. $$ {{Tan28°Tan38°}\over {Tan 79°}}$$
c. $$ {{Tan38°}\over {Tan 79°Tan28°}}$$
d. $$ {{Tan28° Tan 79°}\over { Tan38°}}$$

Answer: b

11. The bar graph given below shows the sales of books (in thousand number) from six branches of a publishing company during two consecutive years 2000 and 2001.

Sales of Books (in thousand numbers) from Six Branches – B1, B2, B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.

What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

a. 2:3
b. 3:5
c. 4:5
d. 7:9

Answer: d

Explanation:

12. The marked price on an item was Rs 2000/- but the shopkeeper offered a double discount of 20% and 15%. How much did he finally sell the item for?

a. Rs. 640
b. Rs. 1300
c. Rs. 1360
d. Rs. 1600

Answer: c

Explanation:

First discount = 20%
∴ Price after First discount = 100-20% = 80% of M.P = 80% * 2000 = Rs. 1600
Second discount = 15%
∴ Price after Second discount = 100-15% = 85% of new price = 85% of 1600 = Rs. 1360/-
So, the item was finally sold for Rs. 1360/-

13. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

a. 9
b. 11
c. 13
d. 15

Answer: d

Explanation:

Let the three integers be x, x + 2 and x + 4.
Then, 3x = 2(x + 4) + 3 x = 11.
Third integer = x + 4 = 15.

14. From a point 375 meters away from the foot of a tower, the top of the tower is observed at an angle of elevation of 45°, then the height (in meters) of the tower is?

a. – 375
b. – 450
c. – 225
d. – 250

Answer: a

Explanation:

From the right angled triangle
Tan(45°)= X/375
=> X = 375 m

15. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

a. Rs. 500
b. Rs. 1500
c. Rs. 2000
d. None of these

Answer: c

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x – 3x = 1000
x = 1000.
B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

16. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs.854 in 4 years. The sum is:

a. Rs. 650
b. Rs. 690
c. Rs. 698
d. Rs. 700

Answer: c

Explanation:

S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.

= Rs. 5123.20
C.I. = Rs. (35123.20 – 25000) = Rs. 10123.20

17. The following pie-chart shows the percentage distribution of the expenditure incurred in publishing a book. Study the pie-chart and the answer the questions based on it.

Various Expenditures (in percentage) Incurred in Publishing a Book

If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, then what will be amount of royalty to be paid for these books?

a. Rs. 19,450
b. Rs. 21,200
c. Rs. 22,950
d. Rs. 26,150

Answer: c

Explanation:

Let the amount of Royalty to be paid for these books be Rs. r.

18. Let G be the centroid of the equilateral triangle ABC the length of AG is 8/√3 cm , then find the perimeter of ABC.

a. 24 cm
b. 14 cm
c. 28 cm
d. 26 cm

Answer: a

Explanation:

Given:
Length of the side AG = 8/√3 cm
Concept used:
Distance between the vertex of an equilateral triangle to the centroid = Side length/√3
Calculation:

According to the concept,
(AG) = AB/√3
⇒ 8/√3 = AB/√3
⇒ AB = 8
So, perimeter of the triangle = 3 × 8
⇒ 24
∴ The perimeter of ABC is 24 cm.

19. Which one of the following can’t be the square of natural number?

a. 30976
b. 75625
c. 28561
d. 143642

Answer: d

Explanation:

The square of a natural number nerver ends in 2.
143642 is not the square of natural number.

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