# SSC CGL 23rd August Shift 3 - QA

**1. SSC CGL 23rd August Shift 3 - QA | Triangles - SSC**

△ABC is an equilateral triangle with side 18 cm. D is a point on BC such that BD = $\frac{1}{3}$BC. Then length(in cm) of AD is:

- A.
6$\sqrt{7}$

- B.
8$\sqrt{3}$

- C.
7$\sqrt{6}$

- D.
6$\sqrt{3}$

Answer: Option A

**Explanation** :

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

The given histogram represents the marks obtained by 128 students. Read the graph and answer the question that follows.

What percent of students got marks less than 60?

- A.
67.5%

- B.
72.5%

- C.
75%

- D.
62.5%

Answer: Option D

**Explanation** :

From the Histogram,

Number of students who got less than 60 marks = 8 + 14 + 28 + 30

= 80

Required percentage = $\frac{80}{128}$ × 100

= 62.5%

Hence, the correct answer is Option D

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**3. SSC CGL 23rd August Shift 3 - QA | Simplification - SSC**

If the 9-digit number 89x64287y is divisible by 72, then what is the value of (3x + 2y)?

- A.
28

- B.
30

- C.
25

- D.
31

Answer: Option A

**Explanation** :

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**4. SSC CGL 23rd August Shift 3 - QA | Average, Mixture & Alligation - SSC**

There are some children in a camp andtheir average weightis 40 kg. If 5 children with average weight 36 kg join the camp orif 5 children with average weight 43.2 kg leave the camp, the average weightof children in both cases is equal. How manychildren are there in the camp, initially?

- A.
45

- B.
50

- C.
40

- D.
35

Answer: Option A

**Explanation** :

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**5. SSC CGL 23rd August Shift 3 - QA | Percentage, Profit & Loss - SSC**

Lucky spends 85% of her income. If her expenditure increases by x %, savings increase by 60% and income increases by 26%, then what is the value of x ?

- A.
34

- B.
30

- C.
20

- D.
26

Answer: Option C

**Explanation** :

Let the income of Lucky = 100L

Lucky spends 85% of her income.

Expenditure of Lucky = $\frac{85}{100}$ × 100L = 85L

Savings of Lucky = 100L - 85L = 15L

According to the problem,

[100L + $\frac{26}{100}$ × 100L] = [15L + $\frac{60}{100}$ × 15L] + [85L + $\frac{\text{x}}{100}$ × 85L]

126L = 24L + 85L + $\frac{x}{100}$ × 85L

126L = 109L + $\frac{x}{100}$ × 85L

17L = $\frac{x}{100}$ × 85L

x = 20

Hence, the correct answer is Option C

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**6. SSC CGL 23rd August Shift 3 - QA | Simple & Compound Interest - SSC**

A person borrowed a sum of ₹30800 at 10% p.a. for 3 years, interest compounded annually. At the end of two years, he paid a sum of ₹13268. At the end of 3rd year, he paid ₹ x to clear of the debt. What is the value of x ?

- A.
26400

- B.
26510

- C.
26200

- D.
26620

Answer: Option A

**Explanation** :

Amount to be paid after 2 years = 30800${\left(1+\frac{10}{100}\right)}^{2}$

= 30800${\left(\frac{11}{10}\right)}^{2}$

= 30800$\left(\frac{121}{100}\right)$

= ₹37268

Amount paid by the person = ₹13268

Remaining amount = ₹37268 - ₹13268 = ₹24000

Amount to be paid at the end of 3rd year to clear debt(i.e, compound interest on ₹24000 for next 1 year) = 24000${\left(1+\frac{10}{100}\right)}^{1}$

⇒ x = 24000$\left(\frac{11}{10}\right)$

⇒ x = ₹26400

Hence, the correct answer is Option A

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**7. SSC CGL 23rd August Shift 3 - QA | Simplification - SSC**

If 9(a² + b²) + c² + 20 = 12(a + 2b), then the value of $\sqrt{6a+9b+2c}$ is:

- A.
4

- B.
3

- C.
6

- D.
2

Answer: Option A

**Explanation** :

9(a^{2} + b^{2}) + c^{2} + 20 = 12(a + 2b)

9a^{2} + 9b^{2} + c^{2} + 20 = 12a + 24b

9a^{2} - 12a + 9b^{2} - 24b + c^{2} + 20 = 0

9a^{2} - 12a + 4 - 4 + 9b^{2} - 24b + 16 - 16 + c^{2} + 20 = 0

(3a - 2)^{2} - 4 + (3b - 4)^{2} - 16 + c^{2} + 20 = 0

(3a - 2)^{2} + (3b - 4)^{2} + c^{2} = 0

3a - 2 = 0, 3b - 4 = 0, c = 0

a = $\frac{2}{3}$, b = $\frac{4}{3}$, c = 0

$\sqrt{6a+9b+2c}$ = $\sqrt{6\left(\frac{2}{3}\right)+9\left(\frac{4}{3}\right)+2\left(0\right)}$

= $\sqrt{4+12}$

= $\sqrt{16}$

= 4

Hence, the correct answer is Option A

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**8. SSC CGL 23rd August Shift 3 - QA | Percentage, Profit & Loss - SSC**

The marked price of an article is ₹180. Renu sells it after 20% discount on its marked price and still gains 25%, The cost price (in ₹) of the article is:

- A.
120.80

- B.
110.80

- C.
115.20

- D.
125.50

Answer: Option C

**Explanation** :

Marked price of an article is ₹180.

Discount = 20%

Selling price of the article = $\frac{80}{100}$ × 180 = ₹144

Let the cost price of the article = C

Profit = 25%

Selling price of the article = $\frac{125}{100}$C

$\frac{125}{100}$C = 144

C = $\frac{576}{5}$

C = 115.2

Cost price of the article = ₹115.20

Hence, the correct answer is Option C

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**9. SSC CGL 23rd August Shift 3 - QA | Trigonometry - SSC**

If $\frac{1}{1-\mathrm{sin}\theta}$ + $\frac{1}{1+\mathrm{sin}\theta}$ = 4 sec θ, 0° < θ < 90°, then the value of cot θ + cosec θ is:

- A.
$\frac{4\sqrt{3}}{3}$

- B.
$\sqrt{3}$

- C.
$\frac{5\sqrt{3}}{3}$

- D.
3$\sqrt{3}$

Answer: Option B

**Explanation** :

$\frac{1}{1-\mathrm{sin}\theta}$ + $\frac{1}{1+\mathrm{sin}\theta}$ = 4 sec θ

$\frac{1+\mathrm{sin}\theta +1-\mathrm{sin}\theta}{1-{\mathrm{sin}}^{2}\theta}$ = $\frac{4}{\mathrm{cos}\theta}$

$\frac{2}{{\mathrm{cos}}^{2}\theta}$ = $\frac{4}{\mathrm{cos}\theta}$

0° < θ < 90°

⇒ θ = 60°

cot θ + cosec θ = cot 60° + cosec 60°

= $\frac{1}{\sqrt{3}}$ + $\frac{2}{\sqrt{3}}$

= $\frac{3}{\sqrt{3}}$

= $\sqrt{3}$

Hence, the correct answer is Option B

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**10. SSC CGL 23rd August Shift 3 - QA | Triangles - SSC**

The bisector of ∠A in △ABC meets side BC at D. If AB = 12 cm, AC = 15 cm and BC = 18 cm, then the length of DC is:

- A.
9 cm

- B.
6 cm

- C.
8 cm

- D.
10 cm

Answer: Option D

**Explanation** :

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**11. SSC CGL 23rd August Shift 3 - QA | Coordinate Geometry - SSC**

The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m²) of the path is:

(Take π = $\frac{22}{7}$)

- A.
8778

- B.
8558

- C.
8800

- D.
8756

Answer: Option A

**Explanation** :

Let the radius of the circular lawn = r

The perimeter of a circular lawn is 1232 m.

2 × $\frac{22}{7}$ × r = 1232

r = 196 m

Area of the path = π(r + 7)² - π r²

= $\frac{22}{7}$[(r + 7)² - r²]

= $\frac{22}{7}$[(196 + 7)² - 196²]

= $\frac{22}{7}$[203² - 196²]

= $\frac{22}{7}$[(203 + 196)(203 - 196)]

= $\frac{22}{7}$[(399)(7)]

= 8778 m³

Hence, the correct answer is Option A

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**12. SSC CGL 23rd August Shift 3 - QA | Triangles - SSC**

Triangles ABC and DBCareright angled triangles with common hypotenuse BC. BD and ACintersect at P when produced. If PA = 8 cm, PC = 4 cm and PD = 3.2 cm, then the length of BD,in cm, is:

- A.
5.6

- B.
6.8

- C.
7.2

- D.
6.4

Answer: Option B

**Explanation** :

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**13. SSC CGL 23rd August Shift 3 - QA | Trigonometry - SSC**

Find the value of cosec (60° + A) - sec(30° - A) + $\frac{\mathrm{cos}ec49\xb0}{sec41\xb0}$.

- A.
1

- B.
0

- C.
-1

- D.
2

Answer: Option A

**Explanation** :

cosec(60° + A) - sec(90° - 60° - A) + $\frac{\mathrm{cos}ec49\xb0}{sec(90-49)\xb0}$

[sec(90 − θ) = cosec θ]

= cosec (60° + A) - sec (90° - (60° + A)) + $\frac{\mathrm{cos}ec49\xb0}{\mathrm{cos}ec49\xb0}$

= cosec (60° + A) - cosec (60° + A) + 1

= 1

Hence, the correct answer is Option A

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**14. SSC CGL 23rd August Shift 3 - QA | Simplification - SSC**

The value of

54 ÷ 16 of 3 × [12 ÷ 4 of {6 × 3 ÷ (11 − 2)}] ÷ (12 ÷ 8 × 2) is:

- A.
$\frac{3}{4}$

- B.
$\frac{9}{8}$

- C.
$\frac{3}{8}$

- D.
$\frac{9}{16}$

Answer: Option D

**Explanation** :

54 ÷ 16 of 3 × [12 ÷ 4 of {6 × 3 ÷ (11 − 2)}] ÷ (12 ÷ 8 × 2)

= 54 ÷ 16 of 3 × [12 ÷ 4 of {6 × 3 ÷ 9}] ÷ $\left(\frac{12}{8}\times 2\right)$

= 54 ÷ 16 of 3 × [12 ÷ 4 of $\left\{6\times \frac{3}{9}\right\}$] ÷ 3

= 54 ÷ 16 of 3 × [12 ÷ 4 of 2] ÷ 3

= 54 ÷ 16 of 3 × [12 ÷ 8] ÷ 3

= 54 ÷ 16 of 3 × $\frac{12}{8}$ ÷ 3

= 54 ÷ 48 × $\frac{12}{8}$ ÷ 3

= $\frac{54}{48}$ × $\frac{12}{8\times 3}$

= $\frac{9}{16}$

Hence, the correct answer is Option D

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Study the following table and answer the question:

Number of students enrolled for Vocational Courses (VC) in five institutes - A, B, C, D & E.

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

What is the sum of the average number of students enrolled for VC in institute B in 2014, 2015 and 2017 and the average number of students enrolled in institute E in 2013 and 2018?

- A.
250

- B.
265

- C.
260

- D.
255

Answer: Option C

**Explanation** :

Average number of students enrolled for VC in institute B in 2014, 2015 and 2017 = $\frac{132+138+135}{3}$

= $\frac{405}{3}$

= 135

Average number of students enrolled in institute E in 2013 and 2018 = $\frac{105+145}{2}$

= $\frac{250}{2}$

= 125

Required sum = 135 + 125

= 260

Hence, the correct answer is Option C

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

The total number of students enrolled for VC in institutes A, B and D in 2015 is what percent more than the total number of students enrolled in institutes C and E in 2018? (correct to one decimal point)

- A.
28.2

- B.
36.8

- C.
35.7

- D.
39.3

Answer: Option D

**Explanation** :

Total number of students enrolled for VC in institutes A, B and D in 2015 = 130 + 138 + 122

= 390

Total number of students enrolled for VC in institutes C and E in 2018 = 135 + 145

= 280

Required percentage = $\frac{390-280}{280}$ × 100

= $\frac{110}{280}$ × 100

= 39.3% (approximately)

Hence, the correct answer is Option D

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**17. SSC CGL 23rd August Shift 3 - QA | Time, Speed & Distance - SSC**

A train running at 48 km/h crosses a man going with the speed of 12 km/h, in the same direction, in 18 seconds and passes a woman coming from the opposite direction in 12 seconds. The speed (in km/h) of the woman is :

- A.
6

- B.
8

- C.
9

- D.
10

Answer: Option A

**Explanation** :

Let the length of the train = L

Relative speed between train and man = 48 - 12 = 36 km/h

= 36 x $\frac{5}{18}$ m/sec

= 10 m/sec

Time taken by train to cross the man = 18 seconds

$\frac{L}{10}$ = 18

L = 180 m

Length of the train = 180 m

Let the speed of the woman = s km/h

Relative speed between train and woman = (48 + s) km/h

= (48 + s) x $\frac{5}{18}$ m/sec

Time taken by train to cross the woman = 12 seconds

$\frac{L}{(48+s)\times {\displaystyle \frac{5}{18}}}$ = 12

$\frac{180}{(48+s)\times {\displaystyle \frac{5}{18}}}$ = 12

$\frac{180\times 18}{(48+s)5}$ = 12

48 + s = 54

s = 6

Speed of the woman = 6 km/h

Hence, the correct answer is Option A

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**18. SSC CGL 23rd August Shift 3 - QA | Coordinate Geometry - SSC**

ABCD is cyclic quadrilateral in which ∠A = x°, ∠B = 5y°, ∠C = 2x° and ∠D = y°. What is the value of (3x - y)?

- A.
120

- B.
90

- C.
150

- D.
60

Answer: Option C

**Explanation** :

ABCD is cyclic quadrilateral.

Opposite angles in a cyclic quadrilateral are supplementary.

∠A + ∠C = 180°+

x + 2x = 180°

3x = 180°

x = 60°

∠B + ∠D = 180°

5y + y = 180°

6y = 180°

y = 30°

3x - y = 3(60) - 30 = 150

Hence, the correct answer is Option C

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**19. SSC CGL 23rd August Shift 3 - QA | Time & Work - SSC**

To do a certain work, the ratio of efficiencies of X and Y is 5 : 7. Working together, X and Y can complete the same work in 70 days. X alonestarted the work andleft after 42 days. Y alone will complete the remaining work in:

- A.
90 days

- B.
72 days

- C.
96 days

- D.
80 days

Answer: Option A

**Explanation** :

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**20. SSC CGL 23rd August Shift 3 - QA | Simplification - SSC**

If x + y + z = 2, x^{3} + y^{3} + z^{3} - 3xyz = 74, then (x^{2} + y^{2} + z^{2} ) is equal to:

- A.
24

- B.
26

- C.
29

- D.
22

Answer: Option B

**Explanation** :

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**21. SSC CGL 23rd August Shift 3 - QA | Percentage, Profit & Loss - SSC**

A dealer bought some toys for ₹1800. He sold 40% of these at a loss of 15% and 33$\frac{1}{3}$% of the remaining toys at 20% profit. At what percent profit should he sell the remaining toys to earn an overall profit of 10%?

- A.
20%

- B.
25%

- C.
30%

- D.
24%

Answer: Option C

**Explanation** :

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**22. SSC CGL 23rd August Shift 3 - QA | Trigonometry - SSC**

If sin α + sin β = cos α + cos β = 1, then sin α + cos α = ?

- A.
-1

- B.
0

- C.
1

- D.
2

Answer: Option C

**Explanation** :

sin α + sin β = 1

sin² α + sin² β + 2 sin α sin β = 1 ...(1)

cos α + cos β = 1

cos² α + cos² β + 2 cos α cos β = 1 ...(2)

Adding (1) and (2),

(sin² α + cos² α) + (sin² β + cos² β) + 2 sin α sin β + 2 cos α cos β = 1 + 1

1 + 1 + 2 sin α sin β + 2 cos α cos β = 2

2 [cos α cos β + sin α sin β] = 0

cos (β - α) = 0

β - α = 90°

β = 90° + α

sin α + sin β = 1

sin α + sin (90° - α) = 1

sin α + cos α = 1

Hence, the correct answer is Option C

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**23. SSC CGL 23rd August Shift 3 - QA | Simplification - SSC**

If (2x + y)^{3} - (x - 2y)^{3} = (x + 3y)[Ax^{2} + By^{2} + Cxy], then what is the value of (A + 2B + C)?

- A.
13

- B.
14

- C.
7

- D.
10

Answer: Option D

**Explanation** :

(2x + y)^{3} - (x - 2y)^{3} = (x + 3y)[Ax^{2} + By^{2} + Cxy]

[2x + y - (x - 2y)] [(2x + y)^{2} + (2x + y)(x - 2y) + (x - 2y)^{2}] = (x + 3y)[Ax^{2} + By^{2} + Cxy]

[x + 3y] [4x^{2} + y^{2} + 4xy + 2x^{2} - 3xy - 2y^{2} + x^{2} + 4y^{2} - 4xy] = (x + 3y)[Ax^{2} + By^{2} + Cxy]

(x + 3y) [7x^{2} + 3y^{2} - 3xy] = (x + 3y) [Ax^{2} + By^{2} + Cxy]

Comparing both sides,

A = 7, B = 3 and C = -3

A + 2B + C = 7 + 2(3) - 3 = 10

Hence, the correct answer is Option D

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

The following bar graph shows the amount(in Lakh Rs.) invested by a Company in purchasing raw material over the years and the values (in Lakh Rs.) of finished goods sold by the Company over the years.

The ratio of total amount invested for purchasing raw material from 2013 to 2015 to the total sales of finished goods in 2014, 2016 and 2017 is:

- A.
27 : 56

- B.
56 : 27

- C.
64 : 37

- D.
37 : 64

Answer: Option D

**Explanation** :

Total amount invested for purchasing raw material from 2013 to 2015 = 250 + 350 + 325 = 925

The total sales of finished goods in 2014, 2016 and 2017 = 475 + 600 + 525 = 1600

Required ratio = 925 : 1600

= 37 : 64

Hence, the correct answer is Option D

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**25. SSC CGL 23rd August Shift 3 - QA | Ratio, Proportion & Variation - SSC**

When x is subtracted from each of the numbers 54, 49, 22 and 21, the numbers so obtained are in proportion. The ratio of (8x - 25) to (7x - 26) is:

- A.
5 : 4

- B.
27 : 26

- C.
29 : 24

- D.
15 : 13

Answer: Option C

**Explanation** :

According to the problem,

(54 − x) (21 − x) = (49 − x) (22 − x)

1134 − 75x + x^{2} = 1078 − 71x + x^{2}

4x = 56

x = 14

(8x − 25) : (7x − 26) = 8(14) − 25 : 7(14) − 26

= 87 : 72

= 29 : 24

Hence, the correct answer is Option C

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