# SSC CGL 24th August Shift 2 - QA

**1. SSC CGL 24th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

Avinash has 20% less coins of different countries than Gaurav has. Gaurav has 40% more such coins than Chetan has. By what percent the number of coins which Chetan hasis less than the number of coins which Avinash has? (correct to one decimalplace)

- A.
10.7

- B.
10.6

- C.
10.5

- D.
12

Answer: Option A

**Explanation** :

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**2. SSC CGL 24th August Shift 2 - QA | Simplification - SSC**

If x - $\frac{2}{x}$ = 15, then what is the value of $\left({x}^{2}+\frac{4}{{x}^{2}}\right)$?

- A.
229

- B.
227

- C.
221

- D.
223

Answer: Option A

**Explanation** :

x - $\frac{2}{x}$ = 15

Squaring on both sides,

x² + $\frac{4}{{x}^{2}}$ - 2.x.$\frac{2}{x}$ = 225

x² + $\frac{4}{{x}^{2}}$ - 4 = 225

x² + $\frac{4}{{x}^{2}}$ = 229

Hence, the correct answer is Option A

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

Simplify the following expression:

$\frac{7}{12}$ ÷ $\frac{1}{10}$ of $\frac{2}{3}$ - $\frac{5}{3}$ × $\frac{9}{10}$ + $\frac{5}{8}$ ÷ $\frac{3}{4}$ of $\frac{2}{3}$

- A.
8$\frac{1}{2}$

- B.
-4

- C.
3$\frac{23}{36}$

- D.
7$\frac{29}{36}$

Answer: Option A

**Explanation** :

$\frac{7}{12}$ ÷ $\frac{1}{10}$ of $\frac{2}{3}$ - $\frac{5}{3}$ × $\frac{9}{10}$ + $\frac{5}{8}$ ÷ $\frac{3}{4}$ of $\frac{2}{3}$

= $\frac{7}{12}$ ÷ $\frac{1}{15}$ - $\frac{5}{3}$ × $\frac{9}{10}$ + $\frac{5}{8}$ ÷ $\frac{1}{2}$

= $\frac{7}{12}$ ÷ $\frac{15}{1}$ - $\frac{5}{3}$ × $\frac{9}{10}$ + $\frac{5}{8}$ × $\frac{1}{2}$

= $\frac{35}{4}$ - $\frac{3}{2}$ + $\frac{5}{4}$

= $\frac{35-6+5}{4}$

= $\frac{34}{4}$

= $\frac{17}{2}$

= 8$\frac{1}{2}$

Hence, the correct answer is Option A

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**4. SSC CGL 24th August Shift 2 - QA | Tables & Graphs - SSC**

The data given in the table shows the number of boys and girls enrolled in three different streams in a school over 5 years.

What is the ratio of the total number of boys in the year 2014 to the total number of girls in the year 2020?

- A.
58 : 57

- B.
55 : 57

- C.
58 : 53

- D.
1 : 1

Answer: Option A

**Explanation** :

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**5. SSC CGL 24th August Shift 2 - QA | Trigonometry - SSC**

If sin² θ = 2 sin θ - 1, 0° ≤ θ ≤ 90°, then find the value of:

$\frac{1+\mathrm{cos}ec\theta}{1-\mathrm{cos}\theta}$

- A.
-2

- B.
1

- C.
2

- D.
-1

Answer: Option C

**Explanation** :

sin² θ = 2 sin θ - 1

sin² θ - 2 sin θ + 1 = 0

(sin θ - 1)² = 0

sin θ - 1 = 0

sin θ = 1

0° ≤ θ ≤ 90°

⇒ θ = 90°

$\frac{1+\mathrm{cos}ec\theta}{1-\mathrm{cos}\theta}$ = $\frac{1+\mathrm{cos}ec90\xb0}{1-\mathrm{cos}90\xb0}$

= $\frac{1+1}{1-0}$

= 2

Hence, the correct answer is Option C

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**6. SSC CGL 24th August Shift 2 - QA | Simplification - SSC**

In △ABC and △DEF we have $\frac{AB}{DF}$ = $\frac{BC}{DE}$ = $\frac{AC}{EF},$ then which of the following is true?

- A.
△BCA∼△DEF

- B.
△DEF∼△ABC

- C.
△DEF∼△BAC

- D.
△CAB∼△DEF

Answer: Option A

**Explanation** :

Hence, the correct answer is Option A

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**7. SSC CGL 24th August Shift 2 - QA | Circles - SSC**

In a circle with centre O, AB is a chord of length 10 cm. Tangents at points A and B intersect outside the circle at P. If OP = 2 OA, then find the length (in cm) of AP.

- A.
12.5

- B.
10

- C.
12

- D.
15

Answer: Option B

**Explanation** :

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**8. SSC CGL 24th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

Monthly salaries of Anil and Kumud are in the ratio 19 : 17. If Anil and Kumud get salary hike of ₹2000 and ₹1000 respectively, then the ratio in their salaries become 8 : 7. What is the present salary of Kumud(in ₹)?

- A.
38000

- B.
18000

- C.
34000

- D.
35000

Answer: Option C

**Explanation** :

Monthly salaries of Anil and Kumud are in the ratio 19 : 17.

Let the monthly salaries of Anil and Kumud are 19p and 17p respectively.

Anil and Kumud get salary hike of ₹2000 and ₹1000 respectively, then the ratio in their salaries become 8 : 7.

$\frac{19p+2000}{17p+1000}$ = $\frac{8}{7}$

133p + 14000 = 136p + 8000

3p = 6000

p = 2000

Present salary of Kumud = 17p = ₹34000

Hence, the correct answer is Option C

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**9. SSC CGL 24th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

Trader A gives a single discount of 25% and Trader B gives two successive discounts of 20% and 5% on an identical item. If the discount given by A is ₹320 more than the discount given by B, then what is the marked price (in ₹) of the item?

- A.
3200

- B.
32000

- C.
30000

- D.
25000

Answer: Option B

**Explanation** :

Let the marked price of the item = M

i) Trader A gives a single discount of 25%.

Discount = $\frac{25}{100}$M = $\frac{1}{4}$M

ii) Trader B gives two successive discounts of 20% and 5%.

Price of the item after 20% discount = $\frac{80}{100}$ × M

Price of the item after 5% discount = $\frac{95}{100}$ × $\frac{80}{100}$ × M = $\frac{19}{25}$M

Total discount given trader B = M - $\frac{19}{25}$M = $\frac{6}{25}$M

According to the problem, discount given by A is ₹320 more than the discount given by B.

$\frac{1}{4}$M = $\frac{6}{25}$M + 320

$\frac{25M-24M}{100}$ = 320

M = ₹32000

Hence, the correct answer is Option B

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**10. SSC CGL 24th August Shift 2 - QA | Time & Work - SSC**

Eighteen men can complete a work in 14 days. Three women do as much work as two men. Five men and six women started the work and continued for 4 days. Subsequently 3 more men joined the group. In how manytotal days was the work completed?

- A.
21$\frac{1}{3}$

- B.
17$\frac{1}{3}$

- C.
18

- D.
22

Answer: Option D

**Explanation** :

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**11. SSC CGL 24th August Shift 2 - QA | Trigonometry - SSC**

The value of

$\frac{se{c}^{2}60\xb0{\mathrm{cos}}^{2}45\xb0+\mathrm{cos}e{c}^{2}30\xb0}{cot30\xb0se{c}^{2}45\xb0-\mathrm{cos}e{c}^{2}30\xb0\mathrm{tan}45\xb0}$ is:

- A.
-3(2 + $\sqrt{3}$)

- B.
3(2 - $\sqrt{3}$)

- C.
-3(2 - $\sqrt{3}$)

- D.
3(2 + $\sqrt{3}$)

Answer: Option A

**Explanation** :

$\frac{se{c}^{2}60\xb0{\mathrm{cos}}^{2}45\xb0+\mathrm{cos}e{c}^{2}30\xb0}{cot30\xb0se{c}^{2}45\xb0-\mathrm{cos}e{c}^{2}30\xb0\mathrm{tan}45\xb0}$ = $\frac{{\left(2\right)}^{2}.{\left(\frac{1}{\sqrt{2}}\right)}^{2}+{\left(2\right)}^{2}}{\left(\sqrt{3}\right){\left(\sqrt{2}\right)}^{2}-{\left(2\right)}^{2}.\left(1\right)}$

= $\frac{4\times {\displaystyle \frac{1}{2}}+4}{2\sqrt{3}-4}$

= $\frac{6}{2\sqrt{3}-4}$

= $\frac{3}{\sqrt{3}-2}$ × $\frac{\sqrt{3}+2}{\sqrt{3}+2}$

= $\frac{3\left(\sqrt{3}+2\right)}{3-4}$

= -3(2 + $\sqrt{3}$)

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**12. SSC CGL 24th August Shift 2 - QA | Simplification - SSC**

Simplify the following expression.

$\frac{5{({a}^{6}-{b}^{6})}^{3}+5{({b}^{6}-{c}^{6})}^{3}+5{({c}^{6}-{a}^{6})}^{3}}{2{({a}^{3}-{b}^{3})}^{3}+2{({b}^{3}-{c}^{3})}^{3}+2{({c}^{3}-{a}^{3})}^{3}}$

- A.
$\frac{5}{2}$(a³ - b³)(b³ - c³)(c³ - a³)

- B.
$\frac{5}{2}$(a³ - b³)(b³ + c³)(c³ + a³)

- C.
$\frac{5}{2}$(a³ - b³)(b³ - c³)(c³ + a³)

- D.
$\frac{5}{2}$(a³ + b³)(b³ + c³)(c³ + a³)

Answer: Option D

**Explanation** :

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**13. SSC CGL 24th August Shift 2 - QA | Simple & Compound Interest - SSC**

Atul borrowed a sum of ₹12000 and agreed to repay it by paying ₹4800 at the end of first year and ₹9240 at the end of second year. Whatis the rate of compound interest compounded annually?

- A.
10%

- B.
8%

- C.
12%

- D.
$\frac{8}{5}$%

Answer: Option A

**Explanation** :

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**14. SSC CGL 24th August Shift 2 - QA | Average, Mixture & Alligation - SSC**

What is the product of the average of first ten positive odd numbers and the average of first fifteen positive even numbers?

- A.
44

- B.
150

- C.
160

- D.
85.25

Answer: Option C

**Explanation** :

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**15. SSC CGL 24th August Shift 2 - QA | Tables & Graphs - SSC**

Bar graph shows the number of males and females in five organizations A, B, C, D and E.

What is the ratio of number of males working in organizations C, D and E taken together to that of females working in organizations A, B and C taken together?

- A.
10 : 11

- B.
49 : 46

- C.
11 : 10

- D.
46 : 49

Answer: Option A

**Explanation** :

Number of males working in organizations C, D and E taken together = 325 + 275 + 150 = 750

Number of females working in organizations A, B and C taken together = 300 + 275 + 250 = 825

Required ratio = 750 : 825

= 10 : 11

Hence, the correct answer is Option A

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**16. SSC CGL 24th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

A shop keeper sold an article at four-fifth of the marked price and suffered a loss of 3$\frac{1}{3}$ %. Find the profit percent, if he sold the article at the marked price.

(correct to nearest integer)

- A.
22

- B.
18

- C.
21

- D.
20

Answer: Option C

**Explanation** :

Let the cost price of the article = 100C

Loss = $3\frac{1}{3}\%$ = $\frac{10}{3}\%$

Selling price of the article = 100C - $\frac{\frac{10}{3}}{100}$ × 100C

= $\frac{290}{3}$ C

Shop keeper sold the article at four-fifth of the marked price.

$\frac{290}{3}$C = $\frac{4}{5}$ × Marked price of the article

Marked price of the article = $\frac{725}{6}$C

Profit percentage when article is sold at marked price = $\frac{\frac{725}{6}C-100C}{100C}$ × 100

= $\frac{125C}{6\times 100C}$ × 100

= 20.833%

= 21% (approximately)

Hence, the correct answer is Option C

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**17. SSC CGL 24th August Shift 2 - QA | Triangles - SSC**

Points A, B and C are on circle with centre O such that ∠BOC = 84°. If AC is produced to a point D such that ∠BDC = 40°, then find the measure of ∠ABC (in degrees).

- A.
98

- B.
92

- C.
56

- D.
102

Answer: Option A

**Explanation** :

Angle subtended by chord BC at the centre is twice the angle subtended by chord BC on the point A of the circle.

∠BOC = 2∠BAC

84° = 2∠BAC

∠BAC = 42°

From triangle BAD,

∠BAD + ∠ABD + ∠BDA = 180°

42° + ∠ABD + 40° = 180°

∠ABD = 98°

Hence, the correct answer is Option A

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**18. SSC CGL 24th August Shift 2 - QA | Time, Speed & Distance - SSC**

A man walking at a speed of 3 km/h crosses a square field diagonally in 5 minutes. What is the area of the field (in m²)?

- A.
31250

- B.
3125

- C.
312.5

- D.
3.125

Answer: Option A

**Explanation** :

Let the side of the square field = a

Speed of the man = 3 km/h = 3 x $\frac{1000}{60}$m/min = 50 m/min

Time taken to cross the field diagonally = 5 minutes

Length of the diagonal of the square field = 50 x 5 = 250 m

$\sqrt{2}$a = 250

a = $\frac{250}{\sqrt{2}}$ m

Area of the square field = ${a}^{2}$

= ${\left(\frac{250}{\sqrt{2}}\right)}^{2}$

= $\frac{62500}{2}$

= 31250 m²

Hence, the correct answer is Option A

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

If y = 2x + 1, then what is the value of (8x^{3} - y^{3} + 6xy)?

- A.
1

- B.
-1

- C.
15

- D.
-15

Answer: Option B

**Explanation** :

y = 2x + 1

2x - y = -1 ...(1)

Cubing on both sides, we get

8x^{3} - y^{3} -3.2x.y(2x - y) = -1

8x^{3} - y^{3} - 6xy(-1) = -1 [From (1)]

8x^{3} - y^{3} + 6xy = -1

Hence, the correct answer is Option B

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**20. SSC CGL 24th August Shift 2 - QA | Tables & Graphs - SSC**

The pie graph shows the distribution of employees working in five departments A, B, C, D and E of a company.

Total number of employees = 9000

If the number of employees working in department A is x and the total number of employees working in departments C and E is y, then the value of y - 2x is:

- A.
725

- B.
850

- C.
1000

- D.
915

Answer: Option D

**Explanation** :

Number of employees working in department A = x = $\frac{64.2\xb0}{360\xb0}$ × 9000 = 1605

x = 1605

Number of employees working in department C = $\frac{{72}^{\circ}}{{360}^{\circ}}$ × 9000 = 1800

Number of employees working in department E = $\frac{{93}^{\circ}}{{360}^{\circ}}$ × 9000 = 2325

The total number of employees working in departments C and E = y = 1800 + 2325 = 4125

y = 4125

y - 2x = 4125 - 2(1605) = 4125 - 3210 = 915

Hence, the correct answer is Option D

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**21. SSC CGL 24th August Shift 2 - QA | Time, Speed & Distance - SSC**

Places A and are 45 km apart from each other. A car starts from place A and another car starts from place at the same time. If they move in the same direction, they meet in 4 and a half hour and if they move towardseach other, they meet in 27 minutes. Whatis the speed (in km/h) of the car which moves faster?

- A.
56

- B.
55

- C.
45

- D.
50

Answer: Option B

**Explanation** :

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**22. SSC CGL 24th August Shift 2 - QA | Triangles - SSC**

Points M and N are on the sides PQ and QR respectively of a triangle PQR. right angled at Q. If PN = 9 cm, MR = 7 cm, and MN = 3 cm, then find the length of PR (in cm).

- A.
11

- B.
13

- C.
12

- D.
$\sqrt{41}$

Answer: Option A

**Explanation** :

From right angled triangle QMN,

b^{2} + c^{2} = 3^{2}

b^{2} + c^{2} = 9..........(1)

From right angled triangle PQN,

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

a^{2} + b^{2} + 2ab + c^{2} = 81

a^{2} + 2ab + 9 = 81 [From (1)]

a^{2} + 2ab = 72..........(2)

From right angled triangle MQR,

b^{2} + (c + d)^{2} = 7^{2}

b^{2} + c^{2} + d^{2} + 2cd = 49

9 + d^{2} + 2cd = 49 [From (1)]

d^{2} + 2cd = 40..........(3)

From right angled triangle PQR,

(a + b)^{2} + (c + d)^{2} = PR^{2}

a^{2} + 2ab + b^{2} + c^{2} + d^{2} + 2cd = PR^{2}

72 + 9 + 40 = PR^{2}

PR^{2} = 121

PR = 11 cm

Hence, the correct answer is Option A

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**23. SSC CGL 24th August Shift 2 - QA | Number Theory - SSC**

Find the sum of all the possible values of (a + b), so that the number 4a067b is divisible by 11.

- A.
21

- B.
11

- C.
16

- D.
5

Answer: Option A

**Explanation** :

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**24. SSC CGL 24th August Shift 2 - QA | Trigonometry - SSC**

Find the value of sin^{2} 60° + cos^{2} 30° - sin^{2} 45° - 3 sin^{2} 90°.

- A.
$\frac{1}{3}$

- B.
-1$\frac{3}{4}$

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

- D.
-2

Answer: Option D

**Explanation** :

sin^{2} 60° + cos^{2} 30° - sin^{2} 45° - 3 sin^{2} 90° = ${\left(\frac{\sqrt{3}}{2}\right)}^{2}$ + ${\left(\frac{\sqrt{3}}{2}\right)}^{2}$ - ${\left(\frac{1}{\sqrt{2}}\right)}^{2}$ - 3(1)²

= $\frac{3}{4}$ + $\frac{3}{4}$ - $\frac{1}{2}$ - 3

= $\frac{3+3-2-12}{4}$

= $\frac{-8}{4}$

= -2

Hence, the correct answer is Option D

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**25. SSC CGL 24th August Shift 2 - QA | Tables & Graphs - SSC**

The bar graph given below shows the sales of Newspapers (in lakh number) from six branches of a Media Publication Company during two consecutive years 2017 and 2018.

(Note: The data shown below is only for mathematical exercise. They do not represent the actual figures).

Total Sales of U for both the years is what percent (correct to one place of decimal) of the combined Sales of the branches Q and R for 2017 and 2018?

- A.
48.6%

- B.
67.1%

- C.
44.4%

- D.
41.0%

Answer: Option C

**Explanation** :

Total Sales of U for both the years = 80 + 100

= 180

Total sales of Q for 2017 and 2018 = 85 + 85

= 170

Total sales of R for 2017 and 2018 = 105 + 130

= 235

The combined Sales of the branches Q and R for 2017 and 2018 = 170 + 235

= 405

Required percentage = $\frac{180}{405}$ × 100

= 44.4%

Hence, the correct answer is Option C

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