# SSC CGL 20th August Shift 2 - QA

**1. SSC CGL 20th August Shift 2 - QA | Number Theory - SSC**

What is the sum of the digits of the largest five digit number which is divisible by 5, 35, 39 and 65?

- A.
30

- B.
33

- C.
27

- D.
35

Answer: Option B

**Explanation** :

LCM of 5, 35, 39 and 65 = 1365

When the largest five digit number 99999 is divided by 1365, the remainder will be 354.

So, 99999 - 354 = 99645 is the largest five digit number divisible by 5, 35, 39 and 65.

Sum of the digits = 9 + 9 + 6 + 4 + 5 = 33

Hence, the correct answer is Option B

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Study the following table and answer the question:

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

The total marks obtained by Anuj in all the five subjects are ?

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

331

- A.
324

- B.
303

- C.
328

- D.

Answer: Option A

**Explanation** :

The total marks obtained by Anuj in all the five subjects = $\frac{80}{100}$ × 75 + $\frac{55}{100}$ × 80 + $\frac{68}{100}$ × 100 + $\frac{66}{100}$ × 50 + $\frac{84}{100}$ × 150

= 60 + 44 + 68 + 33 + 126

= 331

Hence, the correct answer is Option A

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**3. SSC CGL 20th August Shift 2 - QA | Triangles - SSC**

Points P and Q are on the sides AB and BC respectively of a triangle ABC, right angled at B. If AQ = 11 cm, PC = 8 cm, and AC = 13 cm, then find the length (in cm) of PQ.

- A.
$\sqrt{15}$

- B.
4.5

- C.
4

- D.
4$\sqrt{17}$

Answer: Option C

**Explanation** :

From right angled triangle ABC,

(x + y)^{2} + (w + z)^{2} = 13^{2}

(x + y)^{2} + (w + z)^{2} = 169 ...(1)

From right angled triangle ABQ,

(x + y)^{2} + w^{2} = 11^{2}

(x + y)^{2} + w^{2} = 121 ...(2)

From right angled triangle PBC,

x^{2} + (w + z)^{2} = 8^{2}

x^{2} + (w + z)^{2} = 64 ...(3)

Solving (2) + (3) - (1), we get

x^{2} + w^{2} = 121 + 64 - 169

x^{2} + w^{2} = 16 ...(4)

From right angled triangle PBQ,

PB^{2} + BQ^{2} = PQ^{2}

x^{2} + w^{2} = PQ^{2}

PQ^{2} = 16

PQ = 4 cm

Hence, the correct answer is Option C

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**4. SSC CGL 20th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

A shopkeeper sold two items. The selling price of the first item equals the cost price of the second item. He sold the first item at a profit of 20% and the second item at a loss of 10%. What is his overall profit’ loss percent?

- A.
Profit, 3$\frac{7}{11}$%

- B.
Loss, 4$\frac{6}{11}$%

- C.
Profit, 4$\frac{7}{11}$%

- D.
Loss, 8$\frac{1}{3}$%

Answer: Option A

**Explanation** :

Let the cost price of first item = 100C

Profit on first item = 20%

Selling price of first item = $\frac{120}{100}$ × 100C = 120C

The selling price of the first item equals the cost price of the second item.

Cost price of the second item = 120C

Loss on second item = 10%

Selling price of second item = $\frac{90}{100}$ = 120C = 108C

Total cost price = 100C + 120C = 220C

Total selling price = 120C + 108C = 228C

Overall profit percentage = $\frac{228C-220C}{220C}$ × 100

= $\frac{8C}{220C}$ × 100

= $\frac{40}{11}$

= 3$\frac{7}{11}$%

Hence, the correct answer is Option A

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**5. SSC CGL 20th August Shift 2 - QA | Percentage, Profit & Loss - SSC**

What price (in ₹) should Radha mark on a bag which costs ₹1680 so as to earn a profit of 25% after allowing a discount of 16% on the marked price?

- A.
2800

- B.
2000

- C.
2100

- D.
2500

Answer: Option D

**Explanation** :

Cost price of the bag = ₹1680

Profit = 25%

Selling price of the bag = $\frac{125}{100}$ × 1680

Let the marked price of the bag = M

Discount = 16%

Selling price of the bag = $\frac{84}{100}$ × M

⇒ $\frac{84}{100}$ × M = $\frac{125}{100}$ × 1680

⇒ M = 2500

Marked price of the bag = M = ₹2500

Hence, the correct answer is Option D

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**6. SSC CGL 20th August Shift 2 - QA | Simple Equation - SSC**

If $\frac{x}{y}$ + $\frac{y}{x}$ = 2, (x, y ≠ 0), then the value of (x - y) is:

- A.
1

- B.
0

- C.
2

- D.
-2

Answer: Option B

**Explanation** :

$\frac{x}{y}$ + $\frac{y}{x}$ = 2

$\frac{{x}^{2}+{y}^{2}}{xy}$ = 2

x^{2} + y^{2} = 2xy

x^{2} + y^{2} - 2xy = 0

(x - y)^{2} = 0

x - y = 0

Hence, the correct answer is Option B

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

The data given in the table shows the number of students studying in 4 different disciplines in 5 institutes. Study the table and answer the question:

What is the ratio of number of students studying Science in institutes C and D taken together to the number of students studying Computer Science in institutes A and E taken together?

- A.
41 : 56

- B.
43 : 56

- C.
42 : 55

- D.
3 : 4

Answer: Option D

**Explanation** :

Number of students studying Science in institutes C and D taken together = 36 + 48 = 84

Number of students studying Computer Science in institutes A and E taken together = 57 + 55 = 112

Required ratio = 84 : 112

= 3 : 4

Hence, the correct answer is Option D

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**8. SSC CGL 20th August Shift 2 - QA | Lines & Angles - SSC**

Points A, D, C, B and E are concyclic. If ∠AEC = 50° and ∠ABD = 30°, then what is the measure(in degrees) of ∠CBD?

- A.
20

- B.
10

- C.
15

- D.
30

Answer: Option A

**Explanation** :

ADCE is a cyclic quadrilateral, so opposite angles are supplementary.

⇒ ∠ADC + ∠AEC = 180°

⇒ ∠ADC + 50° = 180°

⇒ ∠ADC = 130°

ABCD is a cyclic quadrilateral, so opposite angles are supplementary.

⇒ ∠ABC + ∠ADC = 180°

⇒ x + 30° + 130° = 180°

= x = 20°

⇒ ∠CBD = x = 20°

Hence, the correct answer is Option A

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**9. SSC CGL 20th August Shift 2 - QA | Trigonometry - SSC**

If 2 cos^{2} θ - 5 cos θ + 2 = 0, 0° < θ < 90°, then the value of (sec θ + tan θ) is:

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

- B.
1 - $\sqrt{3}$

- C.
1 + $\sqrt{3}$

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

Answer: Option A

**Explanation** :

2 cos^{2} θ - 5 cos θ + 2 = 0

2 cos^{2} θ - 4 cos θ - cos θ + 2 = 0

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

(cos θ - 2) (2 cos θ - 1) = 0

cos θ = 2 or cos θ = $\frac{1}{2}$

cos θ = 2 is not possible.

So, cos θ = $\frac{1}{2}$

sec θ = 2

tan θ = $\sqrt{se{c}^{2}\theta -1}$ = $\sqrt{4-1}$ = $\sqrt{3}$

∴ (sec θ + tan θ) = 2 + $\sqrt{3}$

Hence, the correct answer is Option A

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**10. SSC CGL 20th August Shift 2 - QA | Simplification - SSC**

If (56$\sqrt{7}$${x}^{3}$ - 2$\sqrt{2}$${y}^{3}$) ÷ (2$\sqrt{7}$x - $\sqrt{2}$y) = Ax^{2} + By^{2} - Cxy, then find the value of A + B - $\sqrt{14}$C.

- A.
38

- B.
10

- C.
19

- D.
58

Answer: Option D

**Explanation** :

(56$\sqrt{7}$${x}^{3}$ - 2$\sqrt{2}$${y}^{3}$) ÷ (2$\sqrt{7}$x - $\sqrt{2}$y) = Ax^{2} + By^{2} - Cxy

$\frac{\left(2\sqrt{7}x-\sqrt{2}y\right)\left(28{x}^{2}+2\sqrt{14}xy+2{y}^{2}\right)}{\left(2\sqrt{7}x-\sqrt{2}y\right)}$ = Ax^{2} + By^{2} - Cxy

28x^{2 }+ 2$\sqrt{14}$xy + 2y^{2 }= Ax^{2 }+ By^{2 }- Cxy

Comparing both sides,

A = 28, B = 2, C = -2$\sqrt{14}$

A + B - $\sqrt{14}$C = 28 + 2 - $\sqrt{14}$(-2 $\sqrt{14}$)

= 30 + 28

= 58

Hence, the correct answer is Option D

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**11. SSC CGL 20th August Shift 2 - QA | Coordinate Geometry - SSC**

The volume of a wall whose height is 10 times its width and whose length is 8 times its height is 51.2 m^{3}. What is the cost(in ₹) of painting the wall on one side at the rate of ₹100/m^{2}?

- A.
12800

- B.
12750

- C.
12250

- D.
12500

Answer: Option A

**Explanation** :

Let the width of the wall = b

Height of the wall = 10b

Length of the wall = 8 x Height = 8 x 10b = 80b

Volume of the wall = 51.2 m^{3}

length x width x height = 51.2

80b x b x 10b = 51.2

8000b^{3} = 512

b = $\frac{8}{20}$

b = $\frac{2}{5}$m

Area of one side of the wall = length x height = 80b x 10b = 800b^{3} = 800 × $\frac{4}{25}$ = 128 m^{2}

The cost of painting the wall on one side at the rate of ₹100/m^{2} = 128 × 100 = ₹12800

Hence, the correct answer is Option A

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**12. SSC CGL 20th August Shift 2 - QA | Ratio, Proportion & Variation - SSC**

The ratio of monthly incomes of A and B is 4 : 5 and that of their monthly expenditures is 3 : 8. If the income of A is equal to the expenditure of B, then what is the ratio of savings of A and B?

- A.
3 : 8

- B.
2 : 5

- C.
8 : 3

- D.
5 : 2

Answer: Option D

**Explanation** :

The ratio of monthly incomes of A and B is 4 : 5.

Let the monthly incomes of A and B are 4p and 5p respectively.

The income of A is equal to the expenditure of B.

Monthly expenditure of B = 4p

The ratio of monthly expenditures of A and B is 3 : 8.

Monthly expenditure of A = $\frac{3}{8}$ × 4p = $\frac{3}{2}$p

The ratio of savings of A and B = (4p - $\frac{3}{2}$p) : (5p - 4p)

= $\frac{5}{2}$p : p

= 5 : 2

Hence, the correct answer is Option D

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**13. SSC CGL 20th August Shift 2 - QA | Time & Work - SSC**

Two pipes A and can fill an empty tank in 10 hours and 16 hours respectively. They are opened alternately for 1 hour each, opening pipe first. In how many hours, will the empty tank be filled?

- A.
14$\frac{2}{5}$

- B.
16$\frac{2}{5}$

- C.
10$\frac{2}{5}$

- D.
12$\frac{2}{5}$

Answer: Option D

**Explanation** :

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**14. SSC CGL 20th August Shift 2 - QA | Trigonometry - SSC**

If cot θ = $\frac{15}{8}$, θ is an acute angle, then find the value of $\frac{(1-\mathrm{cos}\theta )(2+2\mathrm{cos}\theta )}{(2-2\mathrm{sin}\theta )(1+\mathrm{sin}\theta )}$.

- A.
$\frac{225}{64}$

- B.
$\frac{64}{225}$

- C.
$\frac{16}{15}$

- D.
$\frac{8}{15}$

Answer: Option B

**Explanation** :

cot θ = $\frac{15}{8}$

tan θ = $\frac{8}{15}$

sec θ = $\sqrt{1+{\mathrm{tan}}^{2}\theta}$ = $\sqrt{1+{\left(\frac{8}{15}\right)}^{2}}$ = $\sqrt{\frac{289}{225}}$ = $\frac{17}{15}$

cos θ = $\frac{15}{17}$

sin θ = $\sqrt{1-{\mathrm{cos}}^{2}\theta}$ = $\sqrt{1-{\left(\frac{15}{17}\right)}^{2}}$ = $\sqrt{1-\frac{225}{289}}$ = $\frac{64}{289}$ = $\frac{8}{17}$

$\frac{(1-\mathrm{cos}\theta )(2+2\mathrm{cos}\theta )}{(2-2\mathrm{sin}\theta )(1+\mathrm{sin}\theta )}$ = $\frac{\left(1-{\displaystyle \frac{15}{17}}\right)\left[2+2\left({\displaystyle \frac{15}{17}}\right)\right]}{\left[2-2\left({\displaystyle \frac{8}{17}}\right)\right]\left(1+{\displaystyle \frac{8}{17}}\right)}$

= $\frac{\left({\displaystyle \frac{2}{17}}\right)\left[{\displaystyle \frac{64}{17}}\right]}{\left[{\displaystyle \frac{18}{17}}\right]\left({\displaystyle \frac{25}{17}}\right)}$

= $\frac{2\times 64}{18\times 25}$

= $\frac{64}{225}$

Hence, the correct answer is Option B

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**15. SSC CGL 20th August Shift 2 - QA | Average, Mixture & Alligation - SSC**

The present population of a village is 15280. If the number of males increases by 25% and the number of females increases by 15%, then the population will become 18428. The difference between present population of males and females in the village is:

- A.
1840

- B.
920

- C.
2760

- D.
1380

Answer: Option A

**Explanation** :

The present population of a village is 15280.

Let the number of males and females of the village are M and F respectively.

M + F = 15280............(1)

If the number of males increases by 25% and the number of females increases by 15%, then the population will become 18428.

$\frac{125}{100}$M + $\frac{115}{100}$F = 18428

25M + 115F = 18428 x 100

25M + 23F = 368560.........(2)

Solving 25 x (1) - (2), we get

25F - 23F = 382000 - 368560

2F = 13440

F = 6720

Substituting F = 6720 in equation (1),

M + 6720 = 15280

M = 8560

The difference between present population of males and females in the village = 8560 - 6720 = 1840

Hence, the correct answer is Option A

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**16. SSC CGL 20th August Shift 2 - 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 total marks obtained by Amit in subjects A, B and C is what percent less than the total marks obtained by Vikram in subjects B, C, D and E?

- A.
42

- B.
35

- C.
40

- D.
38

Answer: Option C

**Explanation** :

Total marks obtained by Amit in subjects A, B and C = $\frac{64}{100}$ × 75 + $100$ × 80 + $\frac{80}{100}$ × 100 = 48 + 52 + 80 = 180

Total marks obtained by Vikram in subjects B, C, D and E = $\frac{70}{100}$ × 80 + $\frac{73}{100}$ × 100 + $\frac{84}{100}$ × 50 + $\frac{86}{100}$ × 150 = 56 + 73 + 42 + 129 = 300

Required percentage = $\frac{300-180}{300}$ × 100 = 40%

Hence, the correct answer is Option C

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**17. SSC CGL 20th August Shift 2 - QA | Simplification - SSC**

Simplify the following expression:

15 ÷ 3 of 2 × 4 + 9 ÷ 18 of 2 × 3 − 4 ÷ 8 × 2

- A.
12$\frac{3}{4}$

- B.
39$\frac{3}{4}$

- C.
42$\frac{3}{4}$

- D.
9$\frac{3}{4}$

Answer: Option D

**Explanation** :

15 ÷ 3 of 2 × 4 + 9 ÷ 18 of 2 × 3 − 4 ÷ 8 × 2

= 15 ÷ 6 × 4 + 9 ÷ 36 × 3 − 4 ÷ 8 × 2

= $\frac{15}{6}$ × 4 + $\frac{9}{36}$ × 3 - $\frac{4}{8}$ × 2

= 10 + $\frac{3}{4}$ - 1

= 9$\frac{3}{4}$

Hence, the correct answer is Option D

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

The table shows the daily income (in ₹) of 50 persons.

Study the table and answer the question:

What is the percentage of persons earning ₹250 or more?

- A.
52

- B.
68

- C.
32

- D.
48

Answer: Option D

**Explanation** :

Number of persons earning less than ₹200 = 12

Number of persons earning between ₹200 and ₹250 = 26 - 12 = 14

Number of persons earning between ₹250 and ₹300 = 34 - 26 = 8

Number of persons earning between ₹300 and ₹350 = 40 - 34 = 6

Number of persons earning between ₹350 and ₹400 = 50 - 40 = 10

Percentage of persons earning ₹250 or more = $\frac{8+6+10}{50}$ × 100 = 48%

Hence, the correct answer is Option D

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**19. SSC CGL 20th August Shift 2 - QA | Simple Equation - SSC**

If $\sqrt{x}$ - $\frac{1}{\sqrt{x}}$ = $\sqrt{7}$, then the value of x^{2} + $\frac{1}{{x}^{2}}$ is:

- A.
81

- B.
60

- C.
79

- D.
75

Answer: Option C

**Explanation** :

$\sqrt{x}$ - $\frac{1}{\sqrt{x}}$ = $\sqrt{7}$

${\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)}^{2}$ = ${\left(\sqrt{7}\right)}^{2}$

x + $\frac{1}{x}$ - 2 = 7

x + $\frac{1}{x}$ = 9

${\left(x+\frac{1}{x}\right)}^{2}$ = 9^{2}

x^{2} + $\frac{1}{{x}^{2}}$ + 2 = 81

x^{2} + $\frac{1}{{x}^{2}}$ = 79

Hence, the correct answer is Option C

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**20. SSC CGL 20th August Shift 2 - QA | Trigonometry - SSC**

If cos(2θ + 54°) = sin θ, 0° < (2θ + 54°) < 90°, then what is the value of $\frac{1}{cot5\theta +sec{\displaystyle \frac{5\theta}{2}}}$?

- A.
$\frac{\sqrt{3}}{2}$

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

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

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

Answer: Option C

**Explanation** :

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**21. SSC CGL 20th August Shift 2 - QA | Triangles - SSC**

The angles of a triangle are in AP (arithmetic progression). If measure of the smallest angle is 50° less than that of the largest angle, then find the largest angle (in degrees).

- A.
90

- B.
85

- C.
80

- D.
75

Answer: Option B

**Explanation** :

The angles of triangle are in AP (arithmetic progression).

Let the angles are a, a + r, a + 2r.

Measure of the smallest angle is 50° less than that of the largest angle.

a = a + 2r - 50°

2r = 50°

r = 25°

Sum of the angles of triangle = 180°

a + a + r + a + 2r = 180°

3a + 3r = 180°

3a + 75° = 180°

3a = 105°

a = 35°

Largest angle of triangle = a + 2r = 35° + 50° = 85°

Hence, the correct answer is Option B

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**22. SSC CGL 20th August Shift 2 - QA | Average, Mixture & Alligation - SSC**

The average monthly salary of 60 employees of a factory is ₹29900. If two officers are getting ₹90000 each and the average salary of 8 supervisors is ₹65000, then what is the average salary (in ₹) of the remaining employees?

- A.
22680

- B.
29080

- C.
21080

- D.
21880

Answer: Option D

**Explanation** :

The average monthly salary of 60 employees of the factory is ₹29900.

Total monthly salary of 60 employees of the factory = 29900 x 60 = ₹1794000

Two officers are getting ₹90000 each.

Sum of the salary of two officers = 2 x 90000 = ₹180000

The average salary of 8 supervisors is ₹65000.

Total salary of 8 supervisors = 65000 x 8 = ₹520000

Total salary of remaining 50 employees of the factory = 1794000 - 180000 - 520000 = ₹1094000

Average of remaining 50 employees of the factory = $\frac{1094000}{50}$ = ₹21880

Hence, the correct answer is Option D

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**23. SSC CGL 20th August Shift 2 - QA | Circles - SSC**

Two circles of radii 18 cm and 16 cm intersect each other and the length of their common chord is 20 cm. What is the distance (in cm) between their centres?

- A.
4$\sqrt{2}$ - 2$\sqrt{39}$

- B.
4$\sqrt{10}$ - 2$\sqrt{39}$

- C.
4$\sqrt{10}$ + 2$\sqrt{39}$

- D.
4$\sqrt{14}$ + 2$\sqrt{39}$

Answer: Option D

**Explanation** :

From triangle AGH,

AH^{2} + GH^{2} = AG^{2}

AH^{2} + 10^{2} = 16^{2}

AH^{2} + 100 = 256

AH^{2} = 156

AH = 2$\sqrt{39}$

From triangle CGH,

CH^{2} + GH^{2} = CG^{2}

CH^{2} + 10^{2} = 18^{2}

CH^{2} + 100 - 324

CH^{2} = 224

CH = 4$\sqrt{14}$

Distance between centres of circles = AC = AH + CH = 2$\sqrt{39}$ + 4$\sqrt{14}$

Hence, the correct answer is Option D

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**24. SSC CGL 20th August Shift 2 - QA | Simple & Compound Interest - SSC**

The interest (in ₹) to be paid on a sum of ₹30000 at 15% p.a. after 2$\frac{2}{3}$ years, if interest compounded yearly, is:

- A.
14362.50

- B.
12364.50

- C.
16342.50

- D.
13642.50

Answer: Option D

**Explanation** :

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**25. SSC CGL 20th August Shift 2 - QA | Time, Speed & Distance - SSC**

A boat can go 5 km upstream and 7$\frac{1}{2}$ km downstream in 45 minutes.It can also go 5 km downstream and 2.5 km upstream in 25 minutes. How muchtime(in minutes) will it take to go 6 km downstream?

- A.
10

- B.
12

- C.
15

- D.
20

Answer: Option B

**Explanation** :

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