# SSC CHSL 8th June Shift 3 - QA

**1. SSC CHSL 8th June Shift 3 - QA | Time & Work - SSC**

If Priya and Renu can do a job in 12 hours (working together at their respective constant speeds) and Priya can do the job alone in 18 hours, in how many hours can Renu do the job alone?

- A.
36

- B.
24

- C.
27

- D.
21

Answer: Option A

**Explanation** :

Let Renu alone takes R hours to complete the task.

∴ $\frac{1}{12}$ = $\frac{1}{18}$ + $\frac{1}{\mathrm{R}}$

⇒ $\frac{1}{\mathrm{R}}$ = $\frac{1}{18}$ - $\frac{1}{12}$

⇒ R = 36 hours.

Hence, option (a).

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**2. SSC CHSL 8th June Shift 3 - QA | Tables & Graphs - SSC**

The following table shows the sales of cars of four manufacturers for the period from 2015 to 2020 in a metropolitan city (all the figures are in thousands).

What are the total sales of cars of the four manufacturers under consideration in the city in the entire period of 6 years (from 2015 to 2020)?

- A.
1650000

- B.
1053000

- C.
1052000

- D.
1651000

Answer: Option B

**Explanation** :

Total sales of the car (in thousands) = 44 + 48 + 47 + 50 + 52 + 51 + 40 + 41 + 41 + 51 + 42 + 43 + 36 + 38 + 40 + 42 + 44 + 50 + 48 + 44 + 44 + 42 + 42 + 43 = 1053

∴ Total sales of the car = 1053000.

Hence, option (b).

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**3. SSC CHSL 8th June Shift 3 - QA | Trigonometry - SSC**

If $se{c}^{2}\theta =\frac{4}{3},$ then the value of cosec (θ + 30°) is __________. [θ is an acute angle.]

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

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

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

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

Answer: Option B

**Explanation** :

sec^{2 }θ = 3/4

⇒ cos^{2 }θ = 4/3

⇒ cos θ = √3/2 [since θ is an acute angle we take positive value]

⇒ θ = 30°

Now, cosec(θ + 30)° = cosec 60° = 2/√3

Hence, option (b).

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**4. SSC CHSL 8th June Shift 3 - QA | Simplification - SSC**

If $p{x}^{3}$ + ${x}^{2}$ + 3x + q is exactly divisible by (x + 2) and (x - 2), then the values of p and q are:

- A.
p = $-\frac{3}{4}$ q = 4

- B.
p =$\frac{3}{4}$ q = 4

- C.
p = $\frac{3}{4}$ q = -4

- D.
p = $-\frac{3}{4}$ q = -4

Answer: Option D

**Explanation** :

If f(x) is divisible by x - a, then f(a) = 0

f(x) = px^{3} + x^{2} + 3x + q is divisible by x - 2, hence f(2) = 0

∴ 8p + 4 + 6 + q = 0

⇒ 8p + q = -10 ...(1)

f(x) = px^{3} + x^{2} + 3x + q is divisible by x + 2, hence f(-2) = 0

∴ -8p + 4 - 6 + q = 0

⇒ -8p + q = 2 ...(2)

Solving (1) and (2), we get

q = -4 and p = -3/4

Hence, option (d).

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**5. SSC CHSL 8th June Shift 3 - QA | Tables & Graphs - SSC**

The following pie chart is the distribution of 36 marbles among three persons. The number of marbles that Madhuri has is ________.

- A.
18

- B.
20

- C.
16

- D.
12

Answer: Option A

**Explanation** :

Number of marbles with Madhuri = 180/360 × 36 = 18.

Hence, option (a).

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**6. SSC CHSL 8th June Shift 3 - QA | Tables & Graphs - SSC**

Study the given table and answer the question that follows.

Given that the total number of students is 100.

If at least 60% marks in B and C are required for pursuing Post-doc in B, how many students in total will be eligible to pursue higher studies in B and C?

- A.
38

- B.
36

- C.
46

- D.
48

Answer: Option C

**Explanation** :

Number of students getting 60% or more marks in B = 20

Number of students getting 60% or more marks in C = 26

Totla number of students getting 60% or more marks in B and C = 20 + 26 = 46

Hence, option (c).

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**7. SSC CHSL 8th June Shift 3 - QA | Coordinate Geometry - SSC**

The surface area of a cube is 726 cm^{2}. Find the volume of the cube

- A.
2744 cm

^{3} - B.
1729 cm

^{3} - C.
1331 cm

^{3} - D.
2197 cm

^{3}

Answer: Option C

**Explanation** :

Surface area of a cube of side 'a' = 6a^{2} = 726

⇒ a^{2} = 121

⇒ a = 11

Volume of the cube = a^{3} = 11^{3} = 1331 cm^{3}

Hence, option (c).

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**8. SSC CHSL 8th June Shift 3 - QA | Simplification - SSC**

If a number 7P323Q is completely divisible by 88, then the values of P and Q, respectively, are:

- A.
3; 2

- B.
2; 3

- C.
4; 2

- D.
9; 2

Answer: Option Q

**Explanation** :

Since 7P323Q is divisible by 88, it will be divisible by 8 as well as 11.

For 7P323Q to be divisible by 8, 23Q should be divisible by 8.

Hence, Q has be 2

Now, the number is 7P3232,

u = 2 + 2 + P = 4 + P

t = 3 + 3 + 7 = 13

∴ u - t = 4 + P - 13 = P - 9

⇒ P = 9

Hence, option (d).

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**9. SSC CHSL 8th June Shift 3 - QA | Ratio, Proportion & Variation - SSC**

If a : b = 2 : 5 and b : c = 3 : 4, find the value of a : b : c.

- A.
6 : 16 : 21

- B.
2 : 5 : 7

- C.
6 : 15 : 20

- D.
3 : 8 : 10

Answer: Option C

**Explanation** :

a : b = 2 : 5 = 6 : 15

b : c = 3 : 4 = 15 : 20

∴ a : b : c = 6 : 15 : 20

Hence, option (c).

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**10. SSC CHSL 8th June Shift 3 - QA | Coordinate Geometry - SSC**

Find the length of diagonal of a cuboid 13 m long, 7 m broad and 3 m high.

- A.
$\sqrt{207}$ m

- B.
$\sqrt{232}$ m

- C.
$\sqrt{327}$ m

- D.
$\sqrt{227}$ m

Answer: Option D

**Explanation** :

Length of body diagonal = $\sqrt{{13}^{2}+{7}^{2}+{3}^{2}}$

= $\sqrt{169+49+9}$

= $\sqrt{227}$

Hence, option (d).

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**11. SSC CHSL 8th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

If half of the quantity of sugar in a bag was sold at a profit of 20%, a quarter of the quantity of sugar in the bag was sold at a profit of 16%, and the remaining sugar in the bag was sold at a profit of 24%, and the total profit obtained by selling all the sugar in the bag was ₹90, what was the cost price of the entire quantity of sugar in the bag?

- A.
₹450

- B.
₹750

- C.
₹500

- D.
₹400

Answer: Option A

**Explanation** :

Let the total quantity of the sugar in bag = 100 kg while cost price of sugar is Rs. x/kg.

50 kg is sold at 20% profit, hence profit = 50 × 0.2x = 10x

25 kg is sold at 16% profit, hence profit = 25 × 0.16x = 4x

25 kg is sold at 24% profit, hence profit = 25 × 0.24x = 6x

∴ Total profit = 10x + 4x + 6x = 20x

Now, 20x = 90

⇒ x = 4.5

∴ Total cost price of the bag of sugar = 100x = Rs. 450

Hence, option (a).

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**12. SSC CHSL 8th June Shift 3 - QA | Circles - SSC**

If two circles of radii 28 cm and 18 cm touch each other externally, then the length of a common tangent is _________. [Give your answer correct to 2 decimal places.]

- A.
40.90 cm

- B.
42.00 cm

- C.
44.90 cm

- D.
44.12 cm

Answer: Option C

**Explanation** :

Length of Direct Common Tangent = $\sqrt{{d}^{2}-{\left({r}_{1}-{r}_{2}\right)}^{2}}$

∴ $\sqrt{{(28+18)}^{2}-{\left(28-18\right)}^{2}}$ = $\sqrt{{46}^{2}-{10}^{2}}$ = 44.90

Hence, option (c).

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**13. SSC CHSL 8th June Shift 3 - QA | Ratio, Proportion & Variation - SSC**

How much water (in ml) should be added to 300 ml of a 75% milk and water mixture so that it becomes a 45% milk and water mixture?

- A.
300

- B.
250

- C.
150

- D.
200

Answer: Option D

**Explanation** :

Amount of milk initially = 300 × 75%

Let x liters of water is added.

Amount of milk finallly = (300 + x) × 45%

Amount of milk finallly = Amount of milk initially

∴ (300 + x) × 45% = 300 × 75%

⇒ 45x = 300 × 30

⇒ x = 200 liters

Hence, option (d).

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**14. SSC CHSL 8th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

The marked price of a shirt is ₹1,200. The shopkeeper gives successive discounts of 25% and 20%. Find the selling price of the shirt.

- A.
₹660

- B.
₹640

- C.
₹720

- D.
₹680

Answer: Option C

**Explanation** :

Price after 25% discount = 3/4 × 1200

Price after another 20% discount = 4/5 × 3/4 × 1200 = Rs. 720

Hence, option (c).

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**15. SSC CHSL 8th June Shift 3 - QA | Average, Mixture & Alligation - SSC**

A batch of 30 students took an intelligence test in an institution. 20 of them scored an average of 75 marks and the remaining students scored an average of 87 marks. What is the average score of the entire batch of students?

- A.
79

- B.
75

- C.
78

- D.
77

Answer: Option A

**Explanation** :

Total marks of all 30 students = 20 × 75 + 10 × 87 = 2370

Average marks of all 30 students = 2370/30 = 79.

Hence, option (a).

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**16. SSC CHSL 8th June Shift 3 - QA | Simplification - SSC**

Find the value of p from the following.

1$\frac{2}{3}$ ÷ $\frac{4}{9}$ ÷ $\frac{p}{9}$ = 1$\frac{1}{4}$ × $\frac{2}{3}$ ÷ $\frac{1}{6}$

- A.
12

- B.
18

- C.
20

- D.
6

Answer: Option A

**Explanation** :

1$\frac{2}{3}$ ÷ $\frac{4}{9}$ ÷ $\frac{p}{9}$ = 1$\frac{1}{4}$ × $\frac{2}{3}$ ÷ $\frac{1}{6}$

⇒ $\frac{5}{3}$ ÷ $\frac{4}{\mathrm{p}}$ = $\frac{5}{4}$ × 4

⇒ $\frac{5\mathrm{p}}{12}$ = 5

⇒ p = 12

Hence, option (a).

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**17. SSC CHSL 8th June Shift 3 - QA | Simple & Compound Interest - SSC**

Sundar lends a sum of ₹6,000 to Mahesh at an interest rate of 10% p.a., compounded annually. What will be the amount at the end of 2 years?

- A.
₹7,160

- B.
₹7,250

- C.
₹7,200

- D.
₹7,260

Answer: Option D

**Explanation** :

Amount after two years = 6000 × 1.1^{2} = 6000 × 1.21 = Rs. 7260

Hence, option (d).

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**18. SSC CHSL 8th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

At 19% discount, the selling price of a book is ₹453.60. What is its marked price (in ₹)?

- A.
690

- B.
490

- C.
520

- D.
560

Answer: Option D

**Explanation** :

Selling price = Marked price × (1 - 19%)

⇒ 453.60 = MP × 0.81

⇒ MP = 453.60/0.81 = 560

Hence, option (d).

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**19. SSC CHSL 8th June Shift 3 - QA | Average, Mixture & Alligation - SSC**

Kumar’s salary is ₹ 30,000 per month. He spends 25% of this on food and rent and 20% of the remaining on education. But he pays 5% of his total salary on paying the income tax and saves ₹ 8,000 after all these three deductions. What is the amount (in ₹.) available with him after his savings?

- A.
8500

- B.
7500

- C.
7250

- D.
8550

Answer: Option A

**Explanation** :

Amount spend on Food = 25% of 30,000 = 7,500

Amount remaining = Rs. 22,500

Amount spent on education = 20% of 22,500 = Rs. 4,500

Amount spent on income tax = 5% of 30,000 = Rs. 1,500

Savings = Rs. 8,000

Amount left after all deductions = 30,000 - 7,500 - 4,500 - 1,500 - 8,000 = Rs. 8,500

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**20. SSC CHSL 8th June Shift 3 - QA | Simplification - SSC**

If x^{3} + y^{3} = 416 and x + y = 8, then find x^{4} + y^{4}.

- A.
3002

- B.
3204

- C.
3004

- D.
3104

Answer: Option D

**Explanation** :

We know, (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)

⇒ 8^{3} = 416 + 3 × xy × 8

⇒ 512 - 416 = 24xy

⇒ xy = 96/24 = 4

Now, x^{2} + y^{2} = (x + y)^{2} - 2xy = 64 - 8 = 56

Squaring both sides we get,

(x^{2} + y^{2})^{2} = 56^{2}

⇒ x^{4} + y^{4} + 2x^{2}y^{2} = 3136

⇒ x^{4} + y^{4} + 32 = 3136

⇒ x^{4} + y^{4} = 3104

Hence, option (d).

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**21. SSC CHSL 8th June Shift 3 - QA | Simplification - SSC**

Simplify the following expression.

32^{3} - 3(32^{2}) (12) + 3(32) (12^{3}) - 12^{3}

- A.
8000

- B.
4000

- C.
16000

- D.
800

Answer: Option A

**Explanation** :

We know, (x - y)^{3} = x^{3} - 3x^{2}y + 3xy^{2} - y^{3}

∴ 32^{3} - 3(32^{2}) (12) + 3(32) (12^{3}) - 12^{3} = (32 - 12)^{3} = 20^{3} = 8000

Hence, optin (a).

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**22. SSC CHSL 8th June Shift 3 - QA | Time, Speed & Distance - SSC**

A ship sails out to a mark at the speed of 15 km/h and sails back to the starting point at the speed of 10 km/h. The average speed of sailing of the ship is:

- A.
12 km/h

- B.
15 km/h

- C.
13 km/h

- D.
14 km/h

Answer: Option A

**Explanation** :

Let the distance between starting point and the mark be 150 kms.

Time taken to go to the mark = 150/15 = 10 hours

Time taken to come back = 150/10 = 15 hours

∴ Average speed = (150 + 150)/(10 + 15) = 300/25 = 12 kmph

Hence,, option (a).

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**23. SSC CHSL 8th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

A woman spends 25% of her income. If her expenditure is ₹1,250, then find her income.

- A.
₹2,500

- B.
₹5,000

- C.
₹1,250

- D.
₹1,000

Answer: Option B

**Explanation** :

Expenditure = 25% of Income = Rs. 1,250

⇒ 1/4 × Income = 1250

∴ Income 1250 × 4 = Rs. 5000

Hence, option (b).

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**24. SSC CHSL 8th June Shift 3 - QA | Time, Speed & Distance - SSC**

Three cities, A, B, and C are located such that they form the vertices of an equilateral triangle if joined by straight lines. Rashid travels from A to B at the speed of 40 km/h, from B to C at the speed of 60 km/h and from C to A at the speed of 72 k /h. Find the average speed of Rashid for the entire journey.

- A.
54 km/h

- B.
56$\frac{2}{3}$

- C.
55 km/h

- D.
57$\frac{1}{3}$

Answer: Option A

**Explanation** :

Let the sides of the equilaterl triangle be 360 kms each.

Time taken to travel from A to B = 360/40 = 9 hours

Time taken to travel from B to C = 360/60 = 6 hours

Time taken to travel from C to A = 360/72 = 5 hours

∴ Total time taken = 9 + 6 + 5 = 20 hours to travel a total of 360 + 360 + 360 = 1080 kms.

∴ Average speed = 1080/20 = 54 kmph.

Hence option (a).

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**25. SSC CHSL 8th June Shift 3 - QA | Triangles - SSC**

The perimeter of an equilateral triangle is 75 cm. Find its area.

- A.
$\frac{625\sqrt{2}}{4}$ cm²

- B.
$\frac{625\sqrt{3}}{4}$ cm²

- C.
$\frac{625\sqrt{3}}{3}$ cm²

- D.
$\frac{125\sqrt{3}}{4}$ cm²

Answer: Option B

**Explanation** :

Perimeter = 3 × Side = 75 cm.

∴ Side = 25 cm.

Area of an equlateral triangle = $\frac{\sqrt{3}}{4}$(side)^{2}

= $\frac{\sqrt{3}}{4}\times {25}^{2}$

= $\frac{625\sqrt{3}}{4}$

Hence, option (b).

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