# SSC CGL 24th August Shift 3 - QA

**1. SSC CGL 24th August Shift 3 - QA | Tables & Graphs - SSC**

Table shows District-wise data of number of primary school teachers posted in schools of a city.

What is the difference between the total number of male teachers in the districts East, North, West taken together and the total

number of female teachers in the districts East and South?

- A.
735

- B.
771

- C.
110

- D.
545

Answer: Option D

**Explanation** :

The total number of male teachers in the districts East, North, West taken together = 1650 + 1075 + 1280 = 4005

The total number of female teachers in the districts East and South = 2375 + 1085 = 3460

Required difference = 4005 - 3460 = 545

Hence, the correct answer is Option D

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**2. SSC CGL 24th August Shift 3 - QA | Simple Equation - SSC**

If x^{4} - 62x^{2} + 1 = 0, where x > 0, then the value of x^{3} + x^{-3} is:

- A.
500

- B.
512

- C.
488

- D.
364

Answer: Option C

**Explanation** :

x⁴ - 62x² + 1 = 0

x⁴ + 1 = 62x²

x² +$\frac{1}{{x}^{2}}$ = 62

x² + $\frac{1}{{x}^{2}}$ + 2 = 64

${\left(x+\frac{1}{x}\right)}^{2}$ = 64

x + $\frac{1}{x}$ = 8 ....(1)

${\left(x+\frac{1}{x}\right)}^{3}$ = 512

x³ + $\frac{1}{{x}^{3}}$ + 3.x. $\frac{1}{x}$$\left(x+\frac{1}{x}\right)$ = 512

x³ + $\frac{1}{{x}^{3}}$ + 3(8) = 512

x³ + $\frac{1}{{x}^{3}}$ + 24 = 512

x³ + $\frac{1}{{x}^{3}}$ = 488

Hence, the correct answer is Option C

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**3. SSC CGL 24th August Shift 3 - QA | Simplification - SSC**

Given that x^{8} - 34x^{4} + 1 = 0, x > 0. What is the value of (x^{3} - x^{-3})?

- A.
14

- B.
12

- C.
18

- D.
16

Answer: Option A

**Explanation** :

x⁸ - 34x⁴ + 1 = 0

x⁸ + 1 = 34x⁴

x⁴ + $\frac{1}{{x}^{4}}$ = 34

x⁴ + $\frac{1}{{x}^{4}}$ + 2 = 36

${\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}$ = 36

x² + $\frac{1}{{x}^{2}}$ = 6

x² + $\frac{1}{{x}^{2}}$ - 2 = 4

${\left(x-\frac{1}{x}\right)}^{2}$ = 4

x - $\frac{1}{x}$ = 2 ...(1)

${\left(x-\frac{1}{x}\right)}^{3}$ = 8

x³ - $\frac{1}{{x}^{3}}$ - 3.x.$\frac{1}{x}$$\left(x-\frac{1}{x}\right)$ = 8

x³ - $\frac{1}{{x}^{3}}$ - 3(2) = 8

x³ - $\frac{1}{{x}^{3}}$ - 6 = 8

x³ - $\frac{1}{{x}^{3}}$ = 14

Hence, the correct answer is Option A

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**4. SSC CGL 24th August Shift 3 - QA | Trigonometry - SSC**

If $\frac{\mathrm{sin}\theta +\mathrm{cos}\theta}{\mathrm{sin}\theta -\mathrm{cos}\theta}$ = 5, then the value of $\frac{4{\mathrm{sin}}^{2}\theta +3}{2{\mathrm{cos}}^{2}\theta +2}$ is:

- A.
$\frac{75}{17}$

- B.
$\frac{75}{34}$

- C.
$\frac{1}{2}$

- D.
$\frac{3}{2}$

Answer: Option B

**Explanation** :

$\frac{\mathrm{sin}\theta +\mathrm{cos}\theta}{\mathrm{sin}\theta -\mathrm{cos}\theta}$ = 5

sin θ + cos θ = 5 sin θ - 5 cos θ

4 sin θ = 6 cos θ

tan θ = $\frac{3}{2}$

sec θ = $\sqrt{{\left(\frac{3}{2}\right)}^{2}+1}$ = $\frac{\sqrt{13}}{2}$

cos θ = $\frac{2}{\sqrt{13}}$

sin θ = $\sqrt{1-{\left(\frac{2}{\sqrt{13}}\right)}^{2}}$ = $\frac{3}{\sqrt{13}}$

$\frac{4{\mathrm{sin}}^{2}\theta +3}{2{\mathrm{cos}}^{2}\theta +2}$ = $\frac{4{\left({\displaystyle \frac{3}{\sqrt{13}}}\right)}^{2}+3}{2{\left({\displaystyle \frac{2}{\sqrt{13}}}\right)}^{2}+2}$

$=\frac{{\displaystyle \frac{36}{13}}+3}{{\displaystyle \frac{8}{13}}+2}$

$=\frac{{\displaystyle \frac{36+39}{13}}}{{\displaystyle \frac{8+26}{13}}}$

= $\frac{75}{34}$

Hence, the correct answer is Option B

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**5. SSC CGL 24th August Shift 3 - QA | Circles - SSC**

In a circle with centre O, points A, B, C and D in this order are concyclic such that BD is a diameter of the circle. If ∠BAC = 22 , then find the measure (in degrees) of ∠COD.

- A.
79

- B.
136

- C.
158

- D.
68

Answer: Option B

**Explanation** :

Angle in a semicircle is right angle.

⇒ ∠BAD = 90°

⇒ ∠BAC + ∠CAD = 90°

⇒ 22° + ∠CAD = 90°

⇒ ∠CAD = 68°

Angle subtended by a chord at the center of the circle is twice the angle subtended by the chord on the point of a circle in the same segment.

⇒ ∠COD = 2∠CAD

⇒ ∠COD = 136°

Hence, the correct answer is Option B

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**6. SSC CGL 24th August Shift 3 - QA | Simplification - SSC**

What is the coefficient of x² in the expansion of ${\left(5-\frac{{x}^{2}}{3}\right)}^{3}$?

- A.
-25

- B.
$-\frac{25}{3}$

- C.
25

- D.
$-\frac{5}{3}$

Answer: Option A

**Explanation** :

${\left(5-\frac{{x}^{2}}{3}\right)}^{3}$ = $\left(5-\frac{{x}^{2}}{3}\right)$${\left(5-\frac{{x}^{2}}{3}\right)}^{2}$

= $\left(5-\frac{{x}^{2}}{3}\right)$$\left(25+\frac{{x}^{4}}{9}-\frac{10{x}^{2}}{3}\right)$

= 125 + $\frac{5{x}^{4}}{9}$ - $\frac{50{x}^{2}}{3}$ - $\frac{25{x}^{2}}{3}$ - $\frac{{x}^{6}}{27}$ + $\frac{10{x}^{4}}{9}$

= - $\frac{{x}^{6}}{27}$ + $\frac{15{x}^{4}}{9}$ - $\frac{75{x}^{2}}{3}$ + 125

= -$\frac{{x}^{6}}{27}$ + $\frac{5{x}^{4}}{3}$ - 25x² + 125

The coefficient of x² in the expansion = -25

Hence, the correct answer is Option A

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**7. SSC CGL 24th August Shift 3 - QA | Percentage, Profit & Loss - SSC**

A shopkeeper marks an article at a price such that after giving a discount of x%, he gains 20%. If the cost price and the marked price of the article are ₹920 and ₹1472 respectively, then what is the value of x?

- A.
18

- B.
30

- C.
20

- D.
25

Answer: Option D

**Explanation** :

Cost price of the article = ₹920

Gain = 20%

Selling price of the article = $\frac{120}{100}$ × 920

= ₹1104

Marked price of the article = ₹1472

Discount = x%

Selling price of the article = $\frac{100-x}{100}$ × 1472

1104 = $\frac{100-x}{100}$ × 1472

69 = $\frac{100-x}{100}$ × 92

6900 = 9200 - 92x

92x = 2300

x = 25

Hence, the correct answer is Option D

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**8. SSC CGL 24th August Shift 3 - QA | Tables & Graphs - SSC**

In the table, the production and the sale (in 1000 tonnes) of a certain product of a company over5 years is given.

In which year(s) the production increases by more than 10% of that in the previous year?

- A.
2018, 2019

- B.
2017, 2018

- C.
2019

- D.
2016

Answer: Option D

**Explanation** :

Percentage increase in production of 2016 compared to 2015 = $\frac{150}{1250}$ ×100 = 12%

Percentage increase in production of 2017 compared to 2016 = $\frac{50}{1400}$ ×100 = 3.57%

Percentage increase in production of 2018 compared to 2017 = $\frac{50}{1450}$ ×100 = 3.44%

Percentage increase in production of 2019 compared to 2018 = $\frac{100}{1500}$ ×100 = 6.67%

∴ Only in 2016, the production increases by more than 10% of that in the previous year.

Hence, the correct answer is Option D

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**9. SSC CGL 24th August Shift 3 - QA | Trigonometry - SSC**

Simplify:

$\frac{{(\mathrm{sin}\theta +sec\theta )}^{2}+{(\mathrm{cos}\theta +\mathrm{cos}ec\theta )}^{2}}{{(1+sec\theta \mathrm{cos}ec\theta )}^{2}},$ 0° < θ < 90°

- A.
0

- B.
1

- C.
-1

- D.
2

Answer: Option B

**Explanation** :

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**10. SSC CGL 24th August Shift 3 - QA | Triangles - SSC**

Triangle ABC is right angled at B and D is a point of BC such that BD = 5 cm, AD = 13 cm and AC = 37 cm. then find the length of DC in cm.

- A.
25

- B.
30

- C.
5

- D.
35

Answer: Option B

**Explanation** :

From right angled triangle ABD,

AD^{2} AB^{2} + BD^{2}

13^{2} = AB^{2} + 5^{2}

169 = AB^{2} + 25

AB^{2} = 144

AB = 12 cm

From right angled triangle ABC,

AC^{2} = AB^{2} + BC^{2}

37^{2} = 12^{2} + BC^{2}

1369 = 144 + BC^{2}

BC^{2} = 1225

BC = 35

BD + DC = 35

5 + DC = 35

DC = 30 cm

Hence, the correct answer is Option B

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**11. SSC CGL 24th August Shift 3 - QA | Tables & Graphs - SSC**

Study the following table and answer the question:

Percentage of marks obtained by six students A, B, C, D, E and in five subjects.

Total marks obtained by student in all the five subjects are ?

- A.
306

- B.
330

- C.
316

- D.
340

Answer: Option A

**Explanation** :

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**12. SSC CGL 24th August Shift 3 - QA | Trigonometry - SSC**

If sec 31° = x, then sin² 59° + $\frac{1}{se{c}^{2}31\xb0}$ - $\frac{1}{se{c}^{2}59\xb0\mathrm{cos}e{c}^{2}59\xb0}$ is equal to:

- A.
$\frac{{x}^{2}-2}{{x}^{2}}$

- B.
$\frac{2-{x}^{2}}{x}$

- C.
$\frac{2-{x}^{2}}{{x}^{2}}$

- D.
$\frac{{x}^{2}-2}{x}$

Answer: Option C

**Explanation** :

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**13. SSC CGL 24th August Shift 3 - QA | Ratio, Proportion & Variation - SSC**

A certain sum is divided among A, B, C and D such that the ratio of the shares is A : B : C : D = 4 : 12 : 30 : 45. If the difference between the shares of A and D is ₹5,535, then the total sum (in ₹) is:

- A.
12785

- B.
13550

- C.
12285

- D.
11000

Answer: Option C

**Explanation** :

Ratio of shares of A, B, C and D is 4 : 12 : 30 : 45 respectively.

Let the shares of A, B, C and D are 4p, 12p, 30p and 45p respectively.

The difference between the shares of A and D is ₹5,535.

45p - 4p = 5535

41p = 5535

p = 135

Total sum = 4p + 12p + 30p + 45p

= 91p

= 91 x 135

= ₹12285

Hence, the correct answer is Option C

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**14. SSC CGL 24th August Shift 3 - QA | Time, Speed & Distance - SSC**

The speed of a motor boat in still water is 20 km/h.It travels 150 km downstream and then returns to the starting point. If the round trip takes a total of 16 hours, what is the speed (in km/h) of the flow of river?

- A.
8

- B.
6

- C.
5

- D.
4

Answer: Option C

**Explanation** :

Let the speed of the flow of the river = s

The speed of motor boat in still water(m) = 20 km/h

Downstream speed = 20 + s

Upstream speed = 20 - s

Time taken for boat to travel 150 km downstream = $\frac{150}{20+s}$

Time taken for boat to travel 150 km upstream = $\frac{150}{20-s}$

Total time taken = 16 hours

$\frac{150}{20+s}$ + $\frac{150}{20-s}$ = 16

150$\left[\frac{20-s+20+s}{(20+s)(20-s)}\right]$ = 16

$\frac{150\times 40}{16}$ = (20 + s)(20 - s)

375 = 400 - s²

s² = 25

s = 5 km/h

The speed of the flow of the river = s = 5 km/h

Hence, the correct answer is Option C

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**15. SSC CGL 24th August Shift 3 - QA | Simplification - SSC**

If a nine-digit number 7698x138y is divisible by 72, then the value of $\sqrt{4x+y}$ is:

- A.
8

- B.
6

- C.
9

- D.
5

Answer: Option B

**Explanation** :

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**16. SSC CGL 24th August Shift 3 - QA | Time & Work - SSC**

A tank is filled in 4 hours by three pipes A, B and C. The pipe C is 1$\frac{1}{2}$times as fast as B and is 3 times as fast as A. How many hours will pipe A alone taketo fill the tank?

- A.
17

- B.
30

- C.
34

- D.
15

Answer: Option C

**Explanation** :

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**17. SSC CGL 24th August Shift 3 - QA | Triangles - SSC**

Triangle ABC is right angled at B. BDis an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm, then find the measure of AB (in cm).

- A.
3$\sqrt{6}$

- B.
3

- C.
6$\sqrt{3}$

- D.
6

Answer: Option A

**Explanation** :

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**18. SSC CGL 24th August Shift 3 - QA | Percentage, Profit & Loss - SSC**

A shopkeeper sold an article for ₹455 at a loss (in ₹). If he sells it for ₹490, then he would gain an amount four times the loss. At what price (in ₹) should he sell the article to gain 25%?

- A.
575

- B.
577.50

- C.
570.50

- D.
115.50

Answer: Option B

**Explanation** :

Let the loss when the shopkeeper sold the article for ₹455 = L

and the Cost price of the article = C

⇒ C - 455 = L

⇒ C = L + 455 ......(1)

According to the problem, when the shopkeeper sells it for ₹490, then he would gain an amount four times the loss.

⇒ 490 - C = 4L

⇒ 490 - (L + 455) = 4L

⇒ 35 = 5L

⇒ L = 7

From (1),

C = L + 455 = 7 + 455 = ₹462

Cost price of the article = ₹462

Selling price of the article when the shopkeeper sells at 25% gain = $\frac{125}{100}$ × C

= $\frac{125}{100}$ × 462

= ₹577.50

Hence, the correct answer is Option B

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**19. SSC CGL 24th August Shift 3 - QA | Average, Mixture & Alligation - SSC**

What is the average of numbers from 1 to 50 which are multiples of 2 or 5?

(correct to one decimal place)

- A.
25.9

- B.
25.8

- C.
25.4

- D.
26.4

Answer: Option B

**Explanation** :

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**20. SSC CGL 24th August Shift 3 - QA | Simplification - SSC**

Simplify the following expression:

$\left(\frac{3}{4}-\frac{1}{4}\xf7\frac{1}{4}\mathrm{of}\frac{2}{5}\right)$ ÷ $\left(\frac{3}{4}\xf7\frac{2}{3}\mathrm{of}\frac{3}{5}\right)$

- A.
$\frac{14}{75}$

- B.
$\frac{32}{75}$

- C.
$-\frac{70}{27}$

- D.
$-\frac{14}{15}$

Answer: Option D

**Explanation** :

$\left(\frac{3}{4}-\frac{1}{4}\xf7\frac{1}{4}\mathrm{of}\frac{2}{5}\right)$ ÷ $\left(\frac{3}{4}\xf7\frac{2}{3}\mathrm{of}\frac{3}{5}\right)$

= $\left(\frac{3}{4}-\frac{1}{4}\xf7\frac{2}{20}\right)$ ÷ $\left(\frac{3}{4}\xf7\frac{6}{15}\right)$

= $\left(\frac{3}{4}-\frac{1}{4}\times \frac{20}{2}\right)$ ÷ $\left(\frac{3}{4}\times \frac{15}{6}\right)$

= $\left(\frac{3}{4}-\frac{5}{2}\right)$ ÷ $\left(\frac{15}{8}\right)$

= $\left(\frac{3-10}{4}\right)$ ÷ $\left(\frac{15}{8}\right)$

= $\left(\frac{-7}{4}\right)$ ÷ $\left(\frac{15}{8}\right)$

= $\left(\frac{-7}{4}\right)$ ÷ $\left(\frac{8}{15}\right)$

= -$\frac{14}{15}$

Hence, the correct answer is Option D

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**21. SSC CGL 24th August Shift 3 - QA | Coordinate Geometry - SSC**

The surface area of a cube is 13.5 m^{2}. What is the length (in m) of its diagonal?

- A.
2√3

- B.
1.5√3

- C.
1.5

- D.
2

Answer: Option B

**Explanation** :

Let the side of the cube = a

The surface area of a cube is 13.5 m²

6a² = 13.5

12a² = 27

a² = $\frac{9}{4}$

a = $\frac{3}{2}$

a = 1.5 m

Length of the diagonal of the cube = $\sqrt{3}$a

= 1.5$\sqrt{3}$ m

Hence, the correct answer is Option B

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**22. SSC CGL 24th August Shift 3 - QA | Number Theory - SSC**

By mistake, the reciprocal of a positive fraction got typed in place of itself, and thereby its value got reduced by $\frac{175}{4}$%. What was the value of the fraction?

- A.
$\frac{1}{4}$

- B.
$\frac{1}{2}$

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

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

Answer: Option C

**Explanation** :

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**23. SSC CGL 24th August Shift 3 - QA | Lines & Angles - SSC**

In a circle with centre O, a diameter AB is produced to a point P lying outside the circle and PT is a tangent to the circle at a point C on it. If ∠BPT = 28°, then what is the measure of ∠BCP?

- A.
62°

- B.
45°

- C.
28°

- D.
31°

Answer: Option D

**Explanation** :

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

Study the following table and answer the question:

Percentage of marks obtained by six students A, B, C, D, E and F in five subjects.

The total marks obtained by students C, D and F in Science is what percent more than the total marks obtained by B in Science, Hindi and Social Studies?

- A.
12.2

- B.
10.5

- C.
12.5

- D.
11.1

Answer: Option C

**Explanation** :

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**25. SSC CGL 24th August Shift 3 - QA | Simple & Compound Interest - SSC**

What is the difference between the compound interest (in ₹) compounded yearly and compounded half yearly for 18 months at 20% per annum on a sum of ₹12,000?

- A.
121

- B.
132

- C.
145

- D.
165

Answer: Option B

**Explanation** :

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