# SSC CGL 23rd August Shift 1 - QA

**1. SSC CGL 23rd August Shift 1 - QA | Simple & Compound Interest - SSC**

A sum of ₹25600 is invested on simple interest partly at 7% per annum and the remaining at 9% per annum. The total interest at the end of 3 years is ₹5832. How much money(in ₹) was invested at 9% per annum?

- A.
18000

- B.
7600

- C.
9600

- D.
16000

Answer: Option B

**Explanation** :

Let the amount invested in 9% per annum = P

Amount invested in 7% per annum = 25600 - P

The total interest at the end of 3 years is ₹5832.

P × 3 × $\frac{9}{100}$ + (25600 - P) × 3 × $\frac{7}{100}$ = 5832

$\frac{9}{100}$P + 1792 - $\frac{7}{100}$P = 1944

$\frac{2}{100}$P = 152

P = Rs. 7600

∴ Amount invested at 9% per annum = P = ₹7600

Hence, the correct answer is Option B

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**2. SSC CGL 23rd August Shift 1 - QA | Tables & Graphs - SSC**

Study the following table and answer the question:

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

The average marks of Manju, Rekha and Abhi in subject B are?

- A.
54

- B.
60

- C.
56

- D.
62

Answer: Option B

**Explanation** :

Marks obtained by Majnu in subject B = $\frac{85}{100}$× 80 = 68

Marks obtained by Rekha in subject B = $\frac{75}{100}$ × 80 = 60

Marks obtained by Abhi in subject B = $\frac{65}{100}$ × 80 = 52

The average marks of Manju, Rekha and Abhi in subject B = $\frac{68+60+52}{3}$

= $\frac{180}{3}$

= 60

Hence, the correct answer is Option B

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**3. SSC CGL 23rd August Shift 1 - QA | Percentage, Profit & Loss - SSC**

Radha saves X % of her income. If her income increases by 28%and the expenditure increases by 20%, then her savings increase by 40%. What is the value of x ?

- A.
35

- B.
40

- C.
50

- D.
25

Answer: Option B

**Explanation** :

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**4. SSC CGL 23rd August Shift 1 - QA | Tables & Graphs - SSC**

Study the following table and answer the question:

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

Total marks obtained by Amit, Abhi and Anuj in subject E is what percent more than the total marks obtained by all the six students in subject B? (correct to one decimal! place)

- A.
8.4

- B.
7.2

- C.
7.5

- D.
8.5

Answer: Option C

**Explanation** :

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**5. SSC CGL 23rd August Shift 1 - QA | Simple Equation - SSC**

If 8 + 2px^{2} - 36x - 27^{3} = (2 - 3x)^{3}, then what is the value of p?

- A.
27

- B.
54

- C.
9

- D.
-27

Answer: Option A

**Explanation** :

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**6. SSC CGL 23rd August Shift 1 - QA | Time & Work - SSC**

Pipes A and B can fill a tank in 12 hours and 16 hours respectively and pipe C can empty the full tank in 24 hours. All three pipes are opened together, but after 4 hours pipe B is closed. In how manyhours, the empty tank will be completely filled?

- A.
18

- B.
32

- C.
28

- D.
14

Answer: Option A

**Explanation** :

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**7. SSC CGL 23rd August Shift 1 - QA | Ratio, Proportion & Variation - SSC**

Fourth proportion to 12, 18, 6 is equal to the third proportion to 4, k. What is the value of k?

- A.
6

- B.
4$\sqrt{3}$

- C.
6.5

- D.
4

Answer: Option A

**Explanation** :

Let the fourth proportion to 12, 18, 6 is 't'.

12 x t = 18 x 6

t = 9

According to the problem 't' is the third proportion to 4, k.

$\frac{4}{k}$ = $\frac{k}{t}$

$\frac{4}{k}$ = $\frac{k}{9}$

k² = 36

k = 6

Hence, the correct answer is Option A

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**8. SSC CGL 23rd August Shift 1 - QA | Trigonometry - SSC**

If sin^{2} θ - cos^{2} θ - 3 sin θ + 2 = 0, 0° < θ < 90°, then what is the value of $\frac{1}{\sqrt{sec\theta -\mathrm{tan}\theta}}$ is:

- A.
$\sqrt[4]{3}$

- B.
$\sqrt[2]{2}$

- C.
$\sqrt[2]{3}$

- D.
$\sqrt[4]{2}$

Answer: Option A

**Explanation** :

sin^{2} θ - cos^{2} θ - 3 sin θ + 2 = 0

sin^{2} θ - (1 - sin^{2} θ) - 3 sin θ + 2 = 0

2 sin^{2} θ - 3 sin θ + 1 = 0

2 sin^{2} θ - 2 sin θ - sin θ + 1 = 0

2 sin θ (sin θ - 1) - 1(sin θ - 1) = 0

(sin θ - 1) (2 sin θ - 1) = 0

sin θ = 1 or sin θ = $\frac{1}{2}$

θ = 90° or θ = 30°

Given, 0° < θ < 90°

⇒ θ = 30°

$\frac{1}{\sqrt{sec\theta -\mathrm{tan}\theta}}$ = $\frac{1}{\sqrt{sec30\xb0-\mathrm{tan}30\xb0}}$

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

= $\frac{1}{\sqrt{{\displaystyle \frac{1}{\sqrt{3}}}}}$

= $\sqrt[4]{3}$

Hence, the correct answer is Option A

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**9. SSC CGL 23rd August Shift 1 - QA | Tables & Graphs - SSC**

Table shows the number of trees planted in 4 cities from 2016 to 2020.

In which city were maximum trees planted in 2016 and 2019 taken together?

- A.
Chandigarh

- B.
Ahmedabad

- C.
Pune

- D.
Kolkata

Answer: Option B

**Explanation** :

Trees planted in Chandigarh in 2016 and 2019 together = 1800 + 2440 = 4240

Trees planted in Ahmedabad in 2016 and 2019 together = 2500 + 1950 = 4450

Trees planted in Pune in 2016 and 2019 together = 1800 + 1900 = 3700

Trees planted in Kolkata in 2016 and 2019 together = 2000 + 1600 = 3600

∴ In Ahmedabad, maximum trees planted in 2016 and 2019 taken together

Hence, the correct answer is Option B

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**10. SSC CGL 23rd August Shift 1 - QA | Percentage, Profit & Loss - SSC**

In festival season, a shopkeeper allows a discount of 10% on every item. Even after giving the discount, he makes a profit of 20%. If he does not give any discount, then what will be his profit percent?

(correct to 2 decimal places)

- A.
33

- B.
25

- C.
33.33

- D.
33.43

Answer: Option C

**Explanation** :

Let the cost price of the article = 100C

Profit = 20%

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

Discount = 10%

$\frac{90}{100}$ × Marked price = 120C

Marked price of the article = $\frac{400}{3}$C

When no discount is provided,

Selling price = $\frac{400}{3}$C

Profit% = $\frac{{\displaystyle \frac{400}{3}}C-100C}{100C}$ × 100

= 33.33%

Hence, the correct answer is Option C

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**11. SSC CGL 23rd August Shift 1 - QA | Number Theory - SSC**

If the 5 - digit number 593ab is divisible by 3, 7 and 11, then whatis the value of (a^{2} - b^{2} + ab)?

- A.
35

- B.
31

- C.
25

- D.
29

Answer: Option D

**Explanation** :

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

In △ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:

- A.
130°

- B.
65°

- C.
50°

- D.
115°

Answer: Option D

**Explanation** :

BO is the bisector of angle B.

Let ∠OBC = ∠OBA = y

CO is the bisector of angle C.

Let ∠OCA = ∠OCB = x

From △ABC,

∠A + ∠B + ∠C = 180°

50° + 2y + 2x = 180°

2(x + y) = 130°

x + y = 65° ......(1)

From △OBC,

∠OBC + ∠OCB + ∠BOC = 180°

y + x + ∠BOC = 180°

65° + ∠BOC = 180° [From (1)]

∠BOC = 115°

Hence, the correct answer is Option D

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**13. SSC CGL 23rd August Shift 1 - QA | Trigonometry - SSC**

Find the value of $\frac{{\mathrm{tan}}^{2}30\xb0}{se{c}^{2}30\xb0}$ + $\frac{\mathrm{cos}e{c}^{2}45\xb0}{co{t}^{2}45\xb0}$ - $\frac{se{c}^{2}60\xb0}{\mathrm{cos}e{c}^{2}60\xb0}$

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

- B.
$\frac{5}{4}$

- C.
$\frac{13}{4}$

- D.
$\frac{23}{12}$

Answer: Option A

**Explanation** :

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

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

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

= $\frac{-3}{4}$

Hence, the correct answer is Option A

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**14. SSC CGL 23rd August Shift 1 - QA | Average, Mixture & Alligation - SSC**

In an examination, the average score of a student was 67.6. If he would have got 27 more marks in Mathematics, 10 more marks in Computer Science, 18 more marks in History and retained the same marks in other subjects, then his average score would have been 72.6. How may papers were there in the examination?

- A.
11

- B.
10

- C.
12

- D.
9

Answer: Option A

**Explanation** :

Let the number of papers = n

The average score of a student was 67.6.

Sum of the scores of the student = 67.6n

Average after the increase in marks = 72.6

Sum of the scores of the student after increase in marks = 72.6n

⇒ 67.6n + 27 + 10 + 18 = 72.6n

⇒ 5n = 55

⇒ n = 11

Number of papers in the examination = n = 11

Hence, the correct answer is Option A

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**15. SSC CGL 23rd August Shift 1 - QA | Simple Equation - SSC**

If 2x^{2} - 7x + 5 = 0, then what is the value of x^{3} + $\frac{125}{8{x}^{3}}$?

- A.
12$\frac{5}{8}$

- B.
16$\frac{5}{8}$

- C.
10$\frac{5}{8}$

- D.
18$\frac{5}{8}$

Answer: Option B

**Explanation** :

2x^{2} - 7x + 5 = 0

2x^{2} - 2x - 5x + 5 = 0

2x(x - 1) - 5(x - 1) = 0

(x - 1) (2x - 5) = 0

x - 1 = 0 or 2x - 5 = 0

x = 1 or x = $\frac{5}{2}$

When x = 1,

x^{3} + $\frac{125}{8{x}^{3}}$ = (1)^{3} + $\frac{125}{8{\left(1\right)}^{3}}$ = 1 + $\frac{125}{8}$ = $\frac{133}{8}$ = 16$\frac{5}{8}$

Hence, the correct answer is Option B

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**16. SSC CGL 23rd August Shift 1 - QA | Simplification - SSC**

Simplify the following expression:

7 × 4 ÷ 21 of 4 − 5 ÷ 4 × (9 − 13) + 2 − 2 ÷ 8

- A.
7$\frac{1}{12}$

- B.
5$\frac{1}{3}$

- C.
12$\frac{1}{2}$

- D.
5$\frac{1}{16}$

Answer: Option A

**Explanation** :

7 × 4 ÷ 21 of 4 − 5 ÷ 4 × (9 − 13) + 2 − 2 ÷ 8

= 7 × 4 ÷ 84 − 5 ÷ 4 × (−4) + 2 − 2 ÷ 8

= 7 × $\frac{4}{84}$ - $\frac{5}{4}$ × (-4) + 2 - $\frac{2}{8}$

= $\frac{1}{3}$ + 5 + 2 - $\frac{1}{4}$

= 7 + $\frac{4-3}{12}$

= 7$\frac{1}{12}$

Hence, the correct answer is Option A

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

In triangle ABC, AD is the bisector of ∠A. If AB = 5 cm, AC = 7.5 cm and BC = 10 cm, then what is the distance of D from the mid-point of BC (in cm)?

- A.
2

- B.
1.5

- C.
2.2

- D.
1

Answer: Option D

**Explanation** :

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**18. SSC CGL 23rd August Shift 1 - QA | Percentage, Profit & Loss - SSC**

A fruit merchant bought some bananas. Onefifth of them got rotten and were thrown away. Hesold two fifth of the bananas with him at 15% profit and the remaining bananasat 10% profit. Find his overall loss or profit percent?

- A.
Profit, 9.6%

- B.
Loss, 10.4%

- C.
Loss, 9.6%

- D.
Profit, 10.4%

Answer: Option B

**Explanation** :

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**19. SSC CGL 23rd August Shift 1 - QA | Coordinate Geometry - SSC**

Vertices A, B, C and D of a quadrilateral ABCD lie on a circle. ∠A is three times ∠C and ∠D is two times ∠B. What is the difference between the measures of ∠D and ∠C?

- A.
55°

- B.
65°

- C.
75°

- D.
45°

Answer: Option C

**Explanation** :

Given,

∠A is three times ∠C and ∠D is two times ∠B.

∠A = 3∠C .......(1)

∠D = 2∠B .......(2)

In a cyclic quadrilateral, opposite angles are supplementary.

∠A + ∠C = 180° and ∠B + ∠D = 180°

∠A + ∠C = 180°

3∠C + ∠C = 180° [From (1)]

4∠C = 180°

∠C = 45°

∠A = 3∠C = 135°

∠B + ∠D = 180°

∠B + 2∠B = 180° [From (2)]

3∠B = 180°

∠B = 60°

∠D = 2∠B = 120°

Difference between the measures of ∠D and ∠C = 120° - 45°

= 75°

Hence, the correct answer is Option C

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**20. SSC CGL 23rd August Shift 1 - QA | Tables & Graphs - SSC**

Table shows income(in Rs) received by 4 employees of a company during the month of December 2020 and the income sources.

Whose income from all sources except salary is more than 25% of his salary?

- A.
Amit and Nitin

- B.
Varun

- C.
Amit

- D.
None

Answer: Option A

**Explanation** :

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**21. SSC CGL 23rd August Shift 1 - QA | Coordinate Geometry - SSC**

The area of a square shaped field is 1764 m^{2}. The breadth of a rectangular park is $\frac{1}{6}$th of the side of the square field and the length is four times its breadth.

What is the cost (in ₹) of levelling the park at ₹30 per m^{2}.

- A.
5880

- B.
4768

- C.
2940

- D.
6342

Answer: Option A

**Explanation** :

Let the side of the square shaped field = a

The area of square shaped field is 1764 m^{2}

a^{2} = 1764

a = 42 m

ide of the square shaped field = a = 42 m

The breadth of a rectangular park is $\frac{1}{6}$th of the side of the square field.

Breadth of the rectangular park = $\frac{1}{6}$ × 42 = 7 m

The length is four times its breadth.

Length of the rectangular park = 4 x 7 = 28 m

Area of the rectangular park = length x breadth

= 28 x 7

= 196 m^{2}

The cost (in ₹) of levelling the park at ₹30 per m^{2 }= 196 x 30

= ₹5880

Hence, the correct answer is Option A

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**22. SSC CGL 23rd August Shift 1 - QA | Circles - SSC**

A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC CD and DA at P, Q, R and S,respectively. If AS = 6cm, BC = 12 cm, and CR = 5 cm,then the length ofAB (in cm) is:

- A.
13

- B.
11

- C.
15

- D.
12

Answer: Option A

**Explanation** :

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

If x is a real quantity, whatis the minimum value of (25 cos^{2} x + 9 sec^{2} x)

- A.
30

- B.
20

- C.
15

- D.
40

Answer: Option A

**Explanation** :

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**24. SSC CGL 23rd August Shift 1 - QA | Simple Equation - SSC**

If x + $\frac{1}{x}$ = $\frac{17}{4}$, x > 1, then what is the value of x - $\frac{1}{x}$?

- A.
$\frac{9}{4}$

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

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

- D.
$\frac{15}{4}$

Answer: Option D

**Explanation** :

x + $\frac{1}{x}$ = $\frac{17}{4}$

${\left(x+\frac{1}{x}\right)}^{2}$ = $\frac{289}{16}$

x^{2} + $\frac{1}{{x}^{2}}$ + 2 = $\frac{289}{16}$

x^{2} + $\frac{1}{{x}^{2}}$ = $\frac{289}{16}$ - 2

x^{2} + $\frac{1}{{x}^{2}}$ = $\frac{257}{16}$

x^{2} + $\frac{1}{{x}^{2}}$ - 2 = $\frac{257}{16}$ - 2

${\left(x-\frac{1}{x}\right)}^{2}$ = $\frac{257-32}{16}$

${\left(x-\frac{1}{x}\right)}^{2}$ = $\frac{225}{16}$

x - $\frac{1}{x}$ = $\frac{15}{4}$

Hence, the correct answer is Option D

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**25. SSC CGL 23rd August Shift 1 - QA | Time, Speed & Distance - SSC**

A boat covers a roundtrip journey between twopoints A and B ina river in T hours.If its speed in still water becomes 2 times, it would take $\frac{80}{161}$T hours for the same journey. Find the ratio of its speed in still water to the speed of the river.

- A.
11 : 1

- B.
161 : 40

- C.
1 : 11

- D.
2 : 1

Answer: Option A

**Explanation** :

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