SSC CGL Previous Year Question Paper 23rd August Shift 2 | QA | Quantitative Ability questions | CGL Past Year Papers

1. SSC CGL 23rd August Shift 2 - QA | Triangles - SSC

Pie-Chart shows the degree wise breakup of expenditure of a family in a month. Total income of the family is ₹144000.

What is the expenditure (in ₹) on education?

A.
12000
B.
24000
C.
36000
D.
20000

Answer: Option B

Explanation :

The expenditure on education = $\frac{60}{300}$ × 144000

= ₹24000

Hence, the correct answer is Option B

2. SSC CGL 23rd August Shift 2 - QA | Lines & Angles - SSC

Tangent is drawn from a point P to a circle, which meets the circle at T such that PT = 8 cm. A secant PAB intersects the circle in points A and B. If PA= 5 cm, what is the length (in cm) of the chord AB?

A.
8.0
B.
8.4
C.
6.4
D.
7.8

3. SSC CGL 23rd August Shift 2 - QA | Tables & Graphs - SSC

Study the following table and answer the question:
Number of students enrolled for Vocational Courses (VC) in institutes A, B, C, D, E & F.

The difference between the average number of students enrolled for VC in institute F in 2015, 2017 and 2018 and the average number of students enrolled in all the six institutes in 2014, is:

A.
89
B.
88
C.
82
D.
85

Answer: Option D

Explanation :

The average number of students enrolled for VC in institute F in 2015, 2017 and 2018 = $\frac{230 + 220 + 210}{3}$ = 220

The average number of students enrolled for VC in all the six institutes in 2014 = $\frac{110 + 120 + 140 + 125 + 150 + 165 }{6}$ = $\frac{810}{6}$ = 135

Required difference = 220 - 135 = 85

Hence, the correct answer is Option D

4. SSC CGL 23rd August Shift 2 - QA | Trigonometry - SSC

If $\frac{\sin^{2} θ}{\tan^{2} θ - \sin^{2} θ}$ = 5, θ is an acute angle, then the value of $\frac{24 \sin^{2} θ - 15 s e c^{2} θ}{6 \cos e c^{2} θ - 7 c o t^{2} θ}$ IS:

A.
-2
B.
2
C.
-14
D.
14

Answer: Option C

Explanation :

$\frac{\sin^{2} θ}{\tan^{2} θ - \sin^{2} θ}$ = 5

$\frac{\sin^{2} θ}{\frac{\sin^{2} θ}{\cos^{2} θ} - \sin^{2} θ}$ = 5

$\frac{\sin^{2} θ}{\sin^{2} θ (\frac{1}{\cos^{2} θ} - 1)}$ = 5

$\frac{\cos^{2} θ}{1 - \cos^{2} θ}$ = 5

$\frac{\cos^{2} θ}{\sin^{2} θ}$ = 5

cot² θ = 5

cosec² θ = 1 + cot² θ = 1 + 5 = 6

sin² θ = $\frac{1}{\cos e c^{2} θ}$ = $\frac{1}{6}$

cos² θ = 1 - sin² θ = 1 - $\frac{1}{6}$ = $\frac{5}{6}$

sec² θ = $\frac{1}{\cos^{2} θ}$ = $\frac{6}{5}$

$\frac{24 \sin^{2} θ - 15 s e c^{2} θ}{6 \cos e c^{2} θ - 7 c o t^{2} θ}$ = $\frac{24 (\frac{1}{6}) - 15 (\frac{6}{5})}{6 (6) - 7 (5)}$

= $\frac{4 - 18}{36 - 35}$

= -14

Hence, the correct answer is Option C

5. SSC CGL 23rd August Shift 2 - QA | Circles - SSC

The area of a circular path enclosed by two concentric circles is 3080 m². If the difference between the radius of the outer edge and that of inner edge of the circular path is 10 m, what is the sum (in m) of the two radii?

(Take π = $\frac{22}{7}$ )

A.
112
B.
70
C.
84
D.
98

Answer: Option D

Explanation :

Let the radius of the outer circle and inner circle are r₀ or r_i respectively.

The difference between the radius of the outer edge and that of inner edge of the circular path is 10 m.

r₀ - r_i = 10 ....(1)

The area of a circular path enclosed by two concentric circles is 3080 m².

⇒ $\frac{22}{7}$ ${r_{0}}^{2}$ - $\frac{22}{7}$ ${r_{i}}^{2}$ = 3080

⇒ (r₀ + r_i) (r₀ - r_i) = 140 × 7

⇒ (r₀ + r_i) (10) = 140 × 7

⇒ (r₀ + r_i) = 98

Sum of the two radii = 98 m

Hence, the correct answer is Option D

6. SSC CGL 23rd August Shift 2 - QA | Percentage, Profit & Loss - SSC

Three persons A, B and C donate 10%, 7% and 9% respectively of their monthly salaries to a charitable trust. Monthly salaries of A and B are equal and the difference between the donations of A and B is ₹900. If the total donation by A and B is ₹600 more than that of C, then what is the monthly salary (in ₹) of C?

A.
45000
B.
55000
C.
50000
D.
60000

Answer: Option C

Explanation :

Monthly salaries of A and B are equal.

Let the monthly salaries of A and B are 'p'.

The difference between the donations of A and B is ₹900.

$\frac{10}{100}$ p - $\frac{7}{100}$ p = 900

$\frac{3}{100}$ p = 900

p = 30000

The monthly salaries of A and B are ₹30000.

The total donation by A and B is ₹600 more than that of C.

$\frac{10}{100}$ × 30000 + $\frac{7}{100}$ × 30000 = Donation by C + 600

$\frac{17}{100}$ × 30000 =

Donation by C + 600

5100 = Donation by C + 600

Donation by C = ₹4500

Let the monthly salary of C = t

$\frac{9}{100}$ × t = 4500

t = 50000

Monthly salary of C = ₹50000

Hence, the correct answer is Option C

7. SSC CGL 23rd August Shift 2 - QA | Ratio, Proportion & Variation - SSC

If x is subtracted from each of 24, 40, 33 and 57, the numbers, so obtained are in proportion. The ratio of (5x + 12) to (4x + 15) is:

A.
7 : 4
B.
4 : 3
C.
14 : 13
D.
7 : 5

8. SSC CGL 23rd August Shift 2 - QA | Time & Work - SSC

To do a certain work, efficiencies of A and B are in the ratio 7:5. Working together, they can complete the work in 17 $\frac{1}{2}$ days. In how many days, will B alone complete 50% of the same work?

A.
42
B.
21
C.
15
D.
30

Answer: Option B

Explanation :

Let the total work = 700 units

Efficiencies of A and B are in the ratio 7:5.

Let the efficiency of A and B are 7p and 5p respectively.

Efficiencies of A and B together = 7p + 5p = 12p units/day

Working together, they can complete the work in 17 $\frac{1}{2}$ days.

$\frac{700}{12 p}$ = 17 $\frac{1}{2}$

$\frac{700}{12 p}$ = $\frac{35}{2}$

p = $\frac{10}{3}$

Efficiency of B = 5p = $\frac{50}{3}$

units/day

Number of days required for B alone complete 50% of the same work = $\frac{350}{\frac{50}{3}}$

= $\frac{350 \times 3}{50}$

= 21 days

Hence, the correct answer is Option B

9. SSC CGL 23rd August Shift 2 - QA | Simple & Compound Interest - SSC

At what rate percent per annum will ₹7200 amountto ₹7938 in one year, if interest is compounded half yearly?

A.
5
B.
8
C.
12
D.
10

Answer: Option D

Explanation :

Let the rate of interest per annum = R

According to the problem,

7938 = 7200 ${(1 + \frac{\frac{R}{2}}{100})}^{2}$

3969 = 3600 ${(1 + \frac{\frac{R}{2}}{100})}^{2}$

441 = 400 ${(1 + \frac{\frac{R}{2}}{100})}^{2}$

${(1 + \frac{\frac{R}{2}}{100})}^{2}$ = $\frac{441}{400}$

1 + $\frac{\frac{R}{2}}{100}$ = $\frac{21}{20}$

$\frac{R}{200}$ = $\frac{1}{20}$

R = 10%

Hence, the correct answer is Option D

10. SSC CGL 23rd August Shift 2 - QA | Tables & Graphs - SSC

Study the following table and answer the question:
Number of students enrolled for Vocational Courses (VC) in institutes A, B, C, D, E & F.

The total number of students enrolled for VC in institutes D and F in 2014 is what percent of the total number of students enrolled in institutes A, B and C in 2018? (correct to one decimal point)

A.
43.2
B.
42.8
C.
43.8
D.
44.6

Answer: Option D

Explanation :

The total number of students enrolled for VC in institutes D and F in 2014 = 125 +165 = 290

The total number of students enrolled for VC in institutes A, B and C in 2018 = 205 + 220 + 225 = 650

Required percentage = $\frac{290}{650}$ × 100

= 44.61(approximately)

Hence, the correct answer is Option D

11. SSC CGL 23rd August Shift 2 - QA | Simplification - SSC

If (x + 6)³ + (2x + 3)³ + (3x + 5)³ = (3x + 18)(2x + 3)(3x + 5), then what is the value of x?

A.
- $\frac{5}{3}$
B.
$\frac{5}{3}$
C.
- $\frac{7}{3}$
D.
$\frac{7}{3}$

Answer: Option C

Explanation :

(x + 6)³ + (2x + 3)³ + (3x + 5)³ = (3x + 18)(2x + 3)(3x + 5)

(x + 6)³ + (2x + 3)³ + (3x + 5)³ = [3(x + 6)] (2x + 3)(3x + 5)

(x + 6)³ + (2x + 3)³ + (3x + 5)³ - 3(x + 6) (2x + 3) (3x + 5) = 0

This is in the form of a³ + b³ + c³ - 3abc = 0, a ≠ b ≠ c then a + b + c = 0

⇒ (x + 6) + (2x + 3) + (3x + 5) = 0

⇒ 6x + 14 = 0

⇒ x = - $\frac{7}{3}$

Hence, the correct answer is Option C

12. SSC CGL 23rd August Shift 2 - QA | Triangles - SSC

In triangle ABC, D and E are the mid points ofAB and BCrespectively. If area (△CED) = 8 cm², then what is the area(ADEC) in cm²?

A.
24
B.
21
C.
32
D.
16

13. SSC CGL 23rd August Shift 2 - QA | Percentage, Profit & Loss - SSC

Radha bought a fridge and a washing machine together for ₹57300. She sold the fridge at a profit of 15% and washing machine at a loss of 24% and both are sold at the same price. The cost price of washing machine(in ₹) is:

A.
28650
B.
34500
C.
24500
D.
22800

Answer: Option B

Explanation :

Let the cost price of fridge = x

Profit = 15%

Selling price of fridge = $\frac{115}{100} x$

Cost price of washing machine = 57300 - x

Loss = 24%

Selling price of washing machine = $\frac{76}{100}$ (57300 - x)

According to the problem, both are sold at same price.

$\frac{115}{100}$ x = $\frac{76}{100}$ (57300 - x)

115x + 76x = 76\times×57300

191x = 76\times×57300

x = 76\times×300

x = 22800

Cost price of fridge = ₹22800

Cost price of washing machine = 57300 - x

= 57300 - 22800

= ₹34500

Hence, the correct answer is Option B

14. SSC CGL 23rd August Shift 2 - QA | Simple Equation - SSC

If x⁴ + x²y² + y⁴ = 21 and x² + xy + y² = 3, then what is the value of (-xy)?

A.
-1
B.
2
C.
1
D.
-2

15. SSC CGL 23rd August Shift 2 - QA | Average, Mixture & Alligation - SSC

The average weight of students of section A and B having 40 students each is 45.5 kg and 44.2 kg respectively. Two students ofsection A having average weight 48.75 kg were shifted to section B and 2 students of section B were shifted to section A, making the average weight of both the sections equal, What is the average weight(in kg) of the students who were shifted from section B to section A?

A.
35.75
B.
34.25
C.
34.5
D.
35

16. SSC CGL 23rd August Shift 2 - QA | Time, Speed & Distance - SSC

A car can cover a distance of 144 km in 1.8 hours. In what time(in hours) will it cover double the distance when its speed is increased by 20% ?

A.
3
B.
2.5
C.
2
D.
3.2

Answer: Option A

Explanation :

Speed of the car = $\frac{144}{1.8}$ = 80 km/hr

Speed of the car when increased by 20% = $\frac{120}{100}$ × 80 = 96 km/hr

Required time = $\frac{288}{96}$

= 3 hours

Hence, the correct answer is Option A

17. SSC CGL 23rd August Shift 2 - QA | Triangles - SSC

In △ABC, D is a point on BC such that ∠BAD = $\frac{1}{2}$ ∠ADC and ∠BAC = 77° and ∠C = 45°. What is the measure of ∠ADB?

A.
77°
B.
64°
C.
58°
D.
45°

Answer: Option B

Explanation :

∠ADC = x

Given, ∠BAD = $\frac{1}{2}$ ∠ADC

∠BAD = $\frac{x}{2}$

From triangle ADC,

∠ADC + ∠ACD + ∠DAC = 180°

x + 45° + 77 - $\frac{x}{2}$ = 180°

$\frac{x}{2}$ = 58°

x = 116°

∠ADB = 180° - x

= 180° - 116°

= 64°

Hence, the correct answer is Option B

18. SSC CGL 23rd August Shift 2 - QA | Coordinate Geometry - SSC

A square has the perimeter equal to the circumference of a circle having radius 7 cm. What is the ratio of the area of the circle to area of the square?

A.
121 : 44
B.
7 : 2
C.
14 : 11
D.
7 : 11

Answer: Option C

Explanation :

(Use π = $\frac{22}{7}$ )

Radius of the circle = 7 cm

Circumference of the circle = 2 × $\frac{22}{7}$ × 7 = 44 cm

Perimeter of square is equal to the circumference of the circle.

Perimeter of the square = 44 cm

Let the side of the square = a

4a = 44

a = 11 cm

Ratio of the area of the circle to area of the square = $\frac{22}{7}$ × 7² : 11²

= 14 : 11

Hence, the correct answer is Option C

19. SSC CGL 23rd August Shift 2 - QA | Simplification - SSC

If x + $\frac{1}{x}$ = 2 $\sqrt{5}$ , then what is the value of $\frac{(x^{4} + \frac{1}{x^{2}})}{x^{2} + 1}$ ?

A.
14
B.
17
C.
20
D.
23

Answer: Option B

Explanation :

x + $\frac{1}{x}$ = 2 $\sqrt{5}$ ....(1)

${(x + \frac{1}{x})}^{3}$ = 40 $\sqrt{5}$

x³ + $\frac{1}{x^{3}}$ + 3.x. $\frac{1}{x}$ $(x + \frac{1}{x})$ = 40 $\sqrt{5}$

x³ + $\frac{1}{x^{3}}$ + 3(2 $\sqrt{5}$ ) = 40 $\sqrt{5}$ [From (1)]

x³ + $\frac{1}{x^{3}}$ + 6 $\sqrt{5}$ = 40 $\sqrt{5}$

x³ + $\frac{1}{x^{3}}$ + 34 $\sqrt{5}$ ....(2)

$\frac{(x^{4} + \frac{1}{x^{2}})}{x^{2} + 1}$ = $\frac{x (x^{3} + \frac{1}{x^{3}})}{x (x + \frac{1}{x})}$

= $\frac{x^{3} + \frac{1}{x^{3}}}{x + \frac{1}{x}}$

= $\frac{34 \sqrt{5}}{2 \sqrt{5}}$

= 17

Hence, the correct answer is Option B

20. SSC CGL 23rd August Shift 2 - QA | Simplification - SSC

The value of 423 ÷ [270 ÷ $\frac{3}{7}$ × 35 + (17 ÷ $\frac{1}{3}$ ) - (8 $\frac{1}{2}$ - $\frac{5}{2}$ )] is:

A.
$\frac{51}{2455}$
B.
$\frac{47}{2455}$
C.
$\frac{43}{2455}$
D.
$\frac{41}{2455}$

Answer: Option B

Explanation :

423 ÷ [270 ÷ $\frac{3}{7}$ × 35 + (17 ÷ $\frac{1}{3}$ ) - (8 $\frac{1}{2}$ - $\frac{5}{2}$ )]

= 423 ÷ [270 ÷ $\frac{3}{7}$ × 35 + 51 - 6]

= 423 ÷ [270 × $\frac{7}{3}$ × 35 + 51 - 6]

= 423 ÷ [22050 + 51 - 6]

= 423 ÷ 22095

= $\frac{423}{22095}$

= $\frac{47}{2455}$

Hence, the correct answer is Option B

21. SSC CGL 23rd August Shift 2 - QA | Percentage, Profit & Loss - SSC

The cost price of an article is ₹250. A shopkeeper gains 20% by selling it at a discount of 36% on its marked price. What is the marked price (in ₹) of the article?

A.
450
B.
380.50
C.
475
D.
468.75

Answer: Option D

Explanation :

Cost price of the article is ₹250.

Profit = 20%

Selling price of the article = $\frac{120}{100}$ × 250 = ₹300

Let the marked price of the article = M

Discount = 36%

$\frac{64}{100}$ × M = 300

M = 468.75

Marked price of the article = ₹468.75

Hence, the correct answer is Option D

22. SSC CGL 23rd August Shift 2 - QA | Trigonometry - SSC

If 0° < θ < 90°, $\sqrt{\frac{s e c^{2} θ + \cos e c^{2} θ}{\tan^{2} θ - \sin^{2} θ}}$ is equal to:

A.
sec³ θ
B.
cosec³ θ
C.
sin² θ
D.
sec² θ

Answer: Option B

Explanation :

= $\sqrt{\frac{s e c^{2} θ + \cos e c^{2} θ}{\tan^{2} θ - \sin^{2} θ}}$ = $\sqrt{\frac{\frac{1}{\cos^{2} θ} + \frac{1}{\sin^{2} θ}}{\frac{\sin^{2} θ}{\cos^{2} θ} - \sin^{2} θ}}$

= $\sqrt{\frac{\frac{\sin^{2} θ + \cos^{2} θ}{\cos^{2} θ \sin^{2} θ}}{\frac{\sin^{2} θ - \sin^{2} θ \cos^{2} θ}{\cos^{2} θ}}}$

= $\sqrt{\frac{1}{\cos^{2} θ \sin^{2} θ} \times \frac{\cos^{2} θ}{\sin^{2} θ (1 - \cos^{2} θ)}}$

= $\sqrt{\frac{1}{\sin^{2} θ} \times \frac{1}{\sin^{2} θ (\sin^{2} θ)}}$

= $\sqrt{\frac{1}{\sin^{6} θ}}$

= $\sqrt{\cos e c^{6} θ}$

= cosec³ θ

Hence, the correct answer is Option B

23. SSC CGL 23rd August Shift 2 - QA | Tables & Graphs - SSC

The following Pie chart represents the percentage-wise distribution of 800 students of class XII in a school in six different sections A, B, C, D, E and F.

The table given below shows the number of girls of class XII in six different sections A, B, C, D, E and F.

The total number of girls in sections B, C and D together is what percent more than the total number of boys in sections A, B and D together?

A.
50
B.
76.25
C.
65.75
D.
80

Answer: Option B

Explanation :

Number of students in section A = $\frac{20}{100}$ × 800 = 160

Number of boys in section A = 160 - 102 = 58

Number of students in section B = $\frac{18}{100}$ × 800 = 144

Number of boys in section B = 144 - 80 = 64

Number of students in section D = $\frac{17}{100}$ × 800 = 136

Number of boys in section D = 136 - 98 = 38

Total number of boys in sections A, B and D together = 58 + 64 + 38 = 160

Total number of girls in sections B, C and D together = 80 + 104 + 98 = 282

Required percentage = $\frac{282 - 160}{160}$ × 100

= $\frac{122}{160}$ × 100

= 76.25%

Hence, the correct answer is Option B

24. SSC CGL 23rd August Shift 2 - QA | Trigonometry - SSC

(cosec A − cot A) (1 + cos A) = ?

A.
cos A
B.
sin A
C.
cot A
D.
cosec A

Answer: Option B

Explanation :

(cosec A − cot A )(1 + cos A) = $(\frac{1}{\sin A} - \frac{\cos A}{\sin A})$ (1 + cos A)

= $(\frac{1 - \cos^{2} A}{\sin A})$

= $\frac{\sin^{2} A}{\sin A}$

= sin A

Hence, the correct answer is Option B

25. SSC CGL 23rd August Shift 2 - QA | Number Theory - SSC

If the six-digit number 5z3x4y is divisible by 7, 11 and 13, then whatis the value of (x + y - z)?

A.
3
B.
6
C.
5
D.
4

SSC CGL 23rd August Shift 2 - QA

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