# SSC CHSL 10th June Shift 1 - QA

**1. SSC CHSL 10th June Shift 1 - QA | Tables & Graphs - SSC**

Based on the data given in the following table find the arithematic mean of the marks obtained out of 10 in a class test by the students of a class.

- A.
5.1

- B.
6.4

- C.
5.2

- D.
5

Answer: Option C

**Explanation** :

Total number of students = 45

Total marks obtained = 3 × 0 + 2 × 1 + 2 × 2 + 4 × 3 + 6 × 4 + 7 × 5 + 7 × 6 + 5 × 7 + 3 × 8 + 4 × 9 + 2 × 10

= 0 + 2 + 4 + 12 + 24 + 35 + 42 + 35 + 24 + 36 + 20

= 234

∴ Average = 234/45 = 5.2

Hence, option (c).

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**2. SSC CHSL 10th June Shift 1 - QA | Simplification - SSC**

Simplify the folowing expression.

$\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)$ + $\left(\frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}\right)$ + (6 × 6 × 6) + 3$\left(\frac{1}{2}+\frac{3}{2}\right)$ $\left(\frac{1}{2}+6\right)$ $\left(6+\frac{3}{2}\right)$

- A.
521

- B.
64

- C.
512

- D.
256

Answer: Option C

**Explanation** :

Given, $\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)$ + $\left(\frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}\right)$ + (6 × 6 × 6) + 3$\left(\frac{1}{2}+\frac{3}{2}\right)$ $\left(\frac{1}{2}+6\right)$ $\left(6+\frac{3}{2}\right)$

= $\frac{1}{8}$ + $\frac{27}{8}$ + 216 + 6 × $\frac{13}{2}$ × $\frac{15}{2}$

= $\frac{28}{8}$ + 216 + $\frac{1170}{4}$

= $\frac{7}{2}$ + 216 + $\frac{585}{2}$

= 216 + $\frac{592}{2}$

= 216 + 296

= 512

Hence, option (c).

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**3. SSC CHSL 10th June Shift 1 - QA | Percentage, Profit & Loss - SSC**

A vendor bought oranges at 5 for ₹15 and sold them at 3 for ₹5. Find his loss percent.

- A.
$41\frac{4}{9}\%$

- B.
$42\frac{4}{9}\%$

- C.
$43\frac{4}{9}\%$

- D.
$44\frac{4}{9}\%$

Answer: Option D

**Explanation** :

Oranges were bought at 5 for ₹15, hence cost price of 15 oranges will be Rs. 45

Oranges were sold at 3 for ₹5, hence 15 oranges will be sold for Rs. 25

∴ Loss = 45 - 25 = 20

∴ Loss % = 20/45 × 100% = 44(4/9)%

Hence, option (d).

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**4. SSC CHSL 10th June Shift 1 - QA | Time, Speed & Distance - SSC**

For the first part of her journey,Sunita travelled at a speed of 450 m/min and,for the rest of the journey, at a speed 1.4 times of her initial speed. If Sunita travelled a total distance of 33.3 km in 1hour, what was the distance that Sunita travelled at a lower speed?

- A.
11.50 KM

- B.
11.15 KM

- C.
11.25 KM

- D.
11.20 KM

Answer: Option C

**Explanation** :

Let Sunita travels for t minutes at 450 m/min and remaining (60 - t) minutes at 1.4 × 450 = 630 m/min

∴ Total distance travelled = 33300 meters = 450 × t + 630 × (60 - t)

⇒ 3700 = 50t + 70(60 - t)

⇒ 3700 = 50t + 4200 - 70t

⇒ 20t = 500

⇒ t = 25 minutes

∴ Distance travelled by Sunita at slower speed = 450 × 25 = 11,250 meters = 11.25 kms.

Hence, option (c).

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**5. SSC CHSL 10th June Shift 1 - QA | Time & Work - SSC**

40 men can complete a piece of work in 18 days. 9 days after they start working together, 5 more men join them. How many days will they now take to complete the remaining work?

- A.
9 days

- B.
8 days

- C.
6 days

- D.
7 days

Answer: Option B

**Explanation** :

After 9 days, work left for 40 men is for 9 days.

Now, since 5 men join, let the work gets completed in another 'd' days.

∴ 40 × 9 = 45 × d

⇒ d = 8 days.

∴ The remaining work gets completed in 8 days.

Hence, option (b).

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**6. SSC CHSL 10th June Shift 1 - QA | Simple & Compound Interest - SSC**

A sum of money is invested at a rate of compounded interest that is paid out yearly. In the first two years, the interest was ₹400 and ₹420 respectively.The sum is:

- A.
₹7,500

- B.
₹9,530

- C.
₹8,000

- D.
₹8,765

Answer: Option C

**Explanation** :

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**7. SSC CHSL 10th June Shift 1 - QA | Percentage, Profit & Loss - SSC**

A man spends 20% of his monthly salary on food and one-fourth of the remaining salary on charity. If he saves ₹9,600 per month, which is equal to half of the balance after spending on food and charity, then his monthly salary is:

- A.
₹28,000

- B.
₹32,000

- C.
₹26,000

- D.
₹30,000

Answer: Option B

**Explanation** :

Let the income be Rs. I

He spends 20% i.e., 1/5^{th} on food.

∴ Amount remaining after food = 4I/5

He spends 1/4^{th} of the remaining on charity

∴ Amount remaining after charity = 4I/5 × 3/4 = 3I/5

He saves half of this remaining amount

∴ 3I/10 = 9600

⇒ I = 32,000

Hence, option (b).

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**8. SSC CHSL 10th June Shift 1 - QA | Coordinate Geometry - SSC**

What is the length of the arc of a circle whose radius is 35 cm and arc subtends an angle of 36° at the centre of the circle?

(Take π = 22/7)

- A.
44 cm

- B.
22 cm

- C.
10 cm

- D.
220 cm

Answer: Option B

**Explanation** :

Length of an arc = θ/360° × 2πr

= 36/360° × 2 × 22/7 × 35

= 1/10° × 2 × 22 × 5

= 22

Hence, option (b).

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**9. SSC CHSL 10th June Shift 1 - QA | Ratio, Proportion & Variation - SSC**

Calculate the third proportional to 4, 15 and 24.

- A.
$\frac{32}{5}$

- B.
$\frac{29}{5}$

- C.
$\frac{21}{5}$

- D.
$\frac{26}{5}$

Answer: Option A

**Explanation** :

Let x be the third proportion.

∴ $\frac{4}{15}$ = $\frac{x}{24}$

⇒ x = $\frac{4\times 24}{15}$ = $\frac{32}{5}$

Hence, option (a).

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**10. SSC CHSL 10th June Shift 1 - QA | Trigonometry - SSC**

Using the formula tan $\frac{x}{2}$ = $\frac{1-\mathrm{cos}x}{\mathrm{sin}x},$ find the value of 22.5°.

- A.
$\sqrt{2}$ - 1

- B.
$\sqrt{2}$ + 1

- C.
$\frac{\sqrt{3}+\sqrt{2}}{2}$

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

Answer: Option A

**Explanation** :

Given tan $\frac{x}{2}$ = $\frac{1-\mathrm{cos}x}{\mathrm{sin}x}$

Substitute x = 45°, we get

tan $\frac{45}{2}$ = $\frac{1-\mathrm{cos}45}{\mathrm{sin}45}$ = $\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{\frac{1}{\sqrt{2}}}$ = √2 - 1

Hence, option (a).

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**11. SSC CHSL 10th June Shift 1 - QA | Tables & Graphs - SSC**

What is the average amount invested in raw materials from year 1997 to 1999?

- A.
₹310 lakh

- B.
₹330 lakh

- C.
₹400 lakh

- D.
₹410 lakh

Answer: Option D

**Explanation** :

Average from 1997 to 1999 = (375 + 330 + 525)/3 = 1230/3 = 410

Hence, option (d).

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**12. SSC CHSL 10th June Shift 1 - QA | Simplification - SSC**

Find the value of ‘a’ using the concept of BODMAS.

42 ÷ 2 + a × 3 - 22 = 8

- A.
4

- B.
6

- C.
5

- D.
3

Answer: Option D

**Explanation** :

42 ÷ 2 + a × 3 - 22 = 8

⇒ 21 + 3a - 22 = 8

⇒ 3a - 1 = 8

⇒ 3a = 9

⇒ a = 3

Hence, option (d).

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**13. SSC CHSL 10th June Shift 1 - QA | Average, Mixture & Alligation - SSC**

In an examination, the average marks of 7 children was 20, that of 3 children was 45, that of another 4 children was 15 and that of 2 children was 0. What is the average marks of all the 16 children?

(Correct to one decimal place)

- A.
20

- B.
20.9

- C.
4

- D.
4.9

Answer: Option B

**Explanation** :

Average of all 16 children = $\frac{7\times 20+3\times 45+4\times 15+2\times 0}{7+3+4+2}$ = $\frac{335}{16}$ = 20.9

Hence, option (b).

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**14. SSC CHSL 10th June Shift 1 - QA | Percentage, Profit & Loss - SSC**

A washing machine is sold at a discount of 10% on its marked price. A further discount of 5% on the selling price is offered if a person makes the payment in cash. If the marked price of the machine is ₹54,000, how much will a customer paying in cash have to pay for purchasing it?

- A.
₹46,200

- B.
₹45,800

- C.
₹46,170

- D.
₹46,000

Answer: Option C

**Explanation** :

Price after 10% discount = 54,000 - 5,400 = 48,600

Price after further 5% discount = 48,600 - 2,430 = 46,170

Hence, option (c).

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**15. SSC CHSL 10th June Shift 1 - QA | Mensuration - SSC**

The length, breadth and height of a room are 7 m, 4 m and 9 m, respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of ₹7.50 per m^{2}.

- A.
₹1,965

- B.
₹1,695

- C.
₹1,968

- D.
₹1,698

Answer: Option B

**Explanation** :

Here we need to calculate the total surface area except the floor

∴ Total area painted = Total surface area - Area of floor

= 2(7 × 4 + 4 × 9 + 7 × 9) - 7 × 4

= 226 m^{2}

Total Cost of painting = 226 × 7.5 = Rs. 1,695

Hence, option (b).

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**16. SSC CHSL 10th June Shift 1 - QA | Percentage, Profit & Loss - SSC**

Find the selling price of an article if a shopkeeper allows two successive discounts of 5% each on the marked price of ₹80.

- A.
₹76.60

- B.
₹72

- C.
₹72.20

- D.
₹74.40

Answer: Option C

**Explanation** :

Selling price after first 5% discount = 80 - 4 = 76

Selling price after second 5% discount = 76 - 3.8 = 72.2

Hence, option (c).

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**17. SSC CHSL 10th June Shift 1 - QA | Number Theory - SSC**

Which of the following numbers is NOT divisible by 15?

- A.
2595

- B.
3195

- C.
3465

- D.
2995

Answer: Option D

**Explanation** :

Since last digit of all options is 5, all options are divisible by 5.

If the number is also divisible by 3, it will also be divisible by 15.

Option (d): 2995 is not divisible by 3, hence it will not be divisible by 15 as well.

Hence, option (d).

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**18. SSC CHSL 10th June Shift 1 - QA | Trigonometry - SSC**

If sin θ = $\frac{\sqrt{3}}{2},$ and θ is an acute angle, find the value of cos 3θ.

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

- B.
1

- C.
-1

- D.
0

Answer: Option C

**Explanation** :

Given, sin θ = $\frac{\sqrt{3}}{2}$

∴ θ = 60°

⇒ cos 3θ = cos 180° = -1.

Hence, option (c).

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**19. SSC CHSL 10th June Shift 1 - QA | Simplification - SSC**

Simplify (5 × 5 × 5 × 5 × 5)^{5} × (5 × 5 × 5)^{5 }÷ 5 = (125)^{?}.

- A.
15

- B.
21

- C.
13

- D.
14

Answer: Option C

**Explanation** :

(5 × 5 × 5 × 5 × 5)^{5} × (5 × 5 × 5)^{5 }÷ 5

= (5^{5})^{5} × (5^{3})^{5 }÷ 5

= 5^{5×5} × 5^{3×}^{5 }÷ 5

= 5^{25} × 5^{1}^{5 }÷ 5

= 5^{25+15} ÷ 5 = 5^{40} ÷ 5

= 5^{40-1 }= 5^{39}

= 5^{3×13 } = (5^{3})^{13}

= (125)^{13}

Hence, option (c).

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**20. SSC CHSL 10th June Shift 1 - QA | Tables & Graphs - SSC**

Study the given graph and answer the question that follows.

For which of the following years was the percentage rise/fall in production from the previous year maximum for country R?

- A.
2013

- B.
2011

- C.
2012

- D.
2015

Answer: Option D

**Explanation** :

This question can be solved by simple observation.

The change is highest for 2014-15 while the base is smallest in 2014. Hence, % change will highest for 2015.

Hence, option (d).

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**21. SSC CHSL 10th June Shift 1 - QA | Mensuration - SSC**

If the length, breadth and height of a cuboid are 7.5 m, 22 m and 13 m, respectively, then find the volume of the cuboid.

- A.
2145 m

^{3} - B.
1245 m

^{3} - C.
4215 m

^{3} - D.
2154 m

^{3}

Answer: Option A

**Explanation** :

Volume of a cuboid = l × b × h

= 7.5 × 22 × 13

= 2145 m^{3}

Hence, option (a).

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**22. SSC CHSL 10th June Shift 1 - QA | Average, Mixture & Alligation - SSC**

What is the average of first 10 multiples of 6?

- A.
55

- B.
66

- C.
60

- D.
33

Answer: Option D

**Explanation** :

The first 10 multiples of 6 are:

6, 12, 18, ..., 54 and 60

Since these numbers are in Arithmetic Progression, their average is same as average of first and last terms

∴ Required Average = (6 + 60)/2 = 33

Hence, option (d).

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**23. SSC CHSL 10th June Shift 1 - QA | Percentage, Profit & Loss - SSC**

The price of a bike is ₹75,500. What will be the price of the bike after a reduction of 17%?

- A.
₹68,650

- B.
₹68,680

- C.
₹62,665

- D.
₹62,650

Answer: Option C

**Explanation** :

17% of the price of bike = 17/100 × 75,500 = 12,835

Final price of bike = 75,500 - 12,835 = 62,665

Hence, option (c).

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**24. SSC CHSL 10th June Shift 1 - QA | Simple & Compound Interest - SSC**

If the compound interest on a principal for one year is ₹350 and the compound interest for the second year is ₹420, find the rate of interest.

- A.
30%

- B.
20%

- C.
25%

- D.
15%

Answer: Option B

**Explanation** :

We know, compound interest for each year increases by r%.

∴ 420 is r% more than 350

⇒ r = (420 - 350)/350 × 100% = 70/350 × 100% = 20%

Hence, option (b).

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**25. SSC CHSL 10th June Shift 1 - QA | Triangles - SSC**

The area of quadrilateral is 336 and the perpendiculars drawn to one diagonal from the opposite vertices are 16 m and 12 m long. Find the length of this diagonal.

- A.
28 cm

- B.
26 cm

- C.
21 cm

- D.
24 cm

Answer: Option D

**Explanation** :

Area of a quadrilateral = ½ × diagonal × (sum of perpendiculars on this diagonal)

⇒ 336 = ½ × d × (16 + 12)

⇒ 336 = ½ × d × 28

⇒ 336 = 14 × d

⇒ d = 336/14 = 24

Hence, option (d).

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