# SSC CHSL 9th June Shift 3 - QA

**1. SSC CHSL 9th June Shift 3 - QA | Time, Speed & Distance - SSC**

The average of a car is $1\frac{3}{5}$ times the average of a bus. A tracor covers 1150 km in 23 hours. how much distance will the car cover in 4 hours if the speed of the bus is twice the speed of the tractor?

- A.
540 km

- B.
740 km

- C.
440 km

- D.
640 km

Answer: Option D

**Explanation** :

Speed of tractor = 1150/23 = 50 kmph

Speed of bus is twice the speed of tractor = 2 × 50 = 100 kmph

Speed of car is 8/5 times the speed of bus = 8/5 × 100 = 160 kmph

∴ Distance travelled by car in 4 hours = 4 × 160 = 640 kms.

Hence, option (d).

Workspace:

**2. SSC CHSL 9th June Shift 3 - QA | Number Theory - SSC**

If x^{2} - 9x + 1 = 0, what is the value of x^{8} - 6239x^{4} + 1?

- A.
1

- B.
0

- C.
-1

- D.
2

Answer: Option B

**Explanation** :

Given, x^{2} - 9x + 1 = 0

⇒ x^{2} + 1 = 9x

Now, squaring both sides, we get

⇒ x^{4} + 1 + 2x^{2} = 81x^{2}

⇒ x^{4} + 1 = 79x^{2}

Again, squaring both sides

⇒ x^{8} + 1 + 2x^{4} = 6241x^{2}

⇒ x^{8} - 6239x + 1 = 0

Hence, option (b).

Workspace:

**3. SSC CHSL 9th June Shift 3 - QA | Trigonometry - SSC**

(cosec θ - cot θ)^{2} = ?. 0° < θ < 90°

- A.
$\frac{1-\mathrm{sin}\theta}{1+\mathrm{cos}\theta}$

- B.
$\frac{1-\mathrm{cos}\theta}{1+\mathrm{cos}\theta}$

- C.
$\frac{1+\mathrm{cos}\theta}{1-\mathrm{cos}\theta}$

- D.
$\frac{1+\mathrm{cos}\theta}{1-\mathrm{sin}\theta}$

Answer: Option B

**Explanation** :

(cosec θ - cot θ)2

= ${\left(\frac{1}{\mathrm{sin\theta}}-\frac{\mathrm{cos\theta}}{\mathrm{sin\theta}}\right)}^{2}$

= ${\left(\frac{1-\mathrm{cos}\theta}{\mathrm{sin\theta}}\right)}^{2}$

= $\frac{{(1-\mathrm{cos}\theta )}^{2}}{{\mathrm{sin}}^{2}\theta}$

= $\frac{{(1-\mathrm{cos}\theta )}^{2}}{1-{\mathrm{cos}}^{2}\theta}$

= $\frac{(1-\mathrm{cos}\theta )(1-\mathrm{cos}\theta )}{(1-\mathrm{cos}\theta )(1+\mathrm{cos}\theta )}$

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

Hence, option (b).

Workspace:

**4. SSC CHSL 9th June Shift 3 - QA | Average, Mixture & Alligation - SSC**

The weekly earnings of a shopkeeper in February 2021 are ₹21,980, ₹20,580, ₹22,380, ₹22,616, respectively. What is his average daily income in February 2021?

- A.
₹3,227

- B.
₹3,120

- C.
₹3,127

- D.
₹3,027

Answer: Option C

**Explanation** :

Total earning in February = 21,980 + 20,580 + 22,380 + 22,616 = 87556

Febraury of 2021 will have 28 days as 2021 is not a leap year.

∴ Average daily earnings = 87556/28 = 3,127

Hence, option (c).

Workspace:

**5. SSC CHSL 9th June Shift 3 - QA | Time & Work - SSC**

3 men and 7 women can complete a task in 10 days. On the other hand, 4 men and 6 women, take 8 days to complete the same task. How long will it take 10 women to complete the same task?

- A.
36 days

- B.
43 days

- C.
48 days

- D.
40 days

Answer: Option D

**Explanation** :

Let the efficiency of each man and woman is m and w respectivley.

∴ Work done by 3 men and 7 women in 10 days = work done by 4 men and 6 women in 8 days = work done by 10 women in say 'd' days.

⇒ (3m + 7w) × 10 = (4m + 6w) × 8 = 10w × d

Solving first two expressions, we get

⇒ (3m + 7w) × 10 = (4m + 6w) × 8

⇒ 11w = m

Now, equating the last two expressions, we get

⇒ (4m + 6w) × 8 = 10w × d

⇒ (44w + 6w) × 8 = 10w × d

⇒ 400w = 10w × d

⇒ d = 40 days

Hence, option (d).

Workspace:

**6. SSC CHSL 9th June Shift 3 - QA | Mensuration - SSC**

What is the volume of a cylinder with radius 15 cm and height 119 cm?

(Take π = 22/7)

- A.
84150 cm

^{3} - B.
85140 cm

^{3} - C.
84510 cm

^{3} - D.
85500 cm

^{3}

Answer: Option A

**Explanation** :

Volume of a cylinder = πr^{2}h

= 22/7 × 15^{2} × 119

= 22 × 225 × 17

= 84150

Hence, option (a).

Workspace:

**7. SSC CHSL 9th June Shift 3 - QA | Tables & Graphs - SSC**

The pie-chart below gives the spending of a country on various sports during a particular year as percentages of total spending of all sports by the country in that year.

Based on the information given in the pie-chart, find the total amount spent (in crores of rupees) by the country on volleyball and gymnastic taken together in that year, if the total spending by the country in that year on all sports taken together was ₹286 crores.

- A.
₹89.2 crore

- B.
₹78.56 crore

- C.
₹69.84 crore

- D.
₹74.36 crore

Answer: Option D

**Explanation** :

Volleyball and Gymnastics together is (14 + 12) = 26% of total spending

= 26/100 × 286 crores = 74.26 crores

Hence, option (d).

Workspace:

**8. SSC CHSL 9th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

A table fan, passing through two hands, is sold finally at a profit of 30% of the original cost price. If the first dealer makes a profit of 25%, then the profit made by the second dealer is:

- A.
5%

- B.
10%

- C.
15%

- D.
4%

Answer: Option D

**Explanation** :

Let the cost price of first dealer is Rs. 100.

He sold it to second dealer at 25% profit i.e., at Rs. 125.

Finally the secon dealer sold it for Rs. 130 [overall profit is 30%, given]

∴ Profit % for second dealer = 5/125 × 100 = 4%.

Hence, option (d).

Workspace:

**9. SSC CHSL 9th June Shift 3 - QA | Mensuration - SSC**

A rectangular piece of paper is 52 cm long and 18 cm wide. A cylinder is formed by rolling the paper along its breadth. Find the volume of the cylinder.

[Taken π = 22/7]

- A.
1870 cm

^{3} - B.
2002 cm

^{3} - C.
1030 cm

^{3} - D.
1290 cm

^{3}

Answer: Option B

**Explanation** :

Workspace:

**10. SSC CHSL 9th June Shift 3 - QA | Average, Mixture & Alligation - SSC**

A few boys and a few girls appeared for a test. The average score of the boys was 52, the average score of the girls was 72, while the combined average was 60. What was the ratio of the number of boys to the number of girls who took the test?

- A.
3 : 2

- B.
13 : 18

- C.
18 : 13

- D.
2 : 3

Answer: Option A

**Explanation** :

Using alligation,

$\frac{\mathrm{Boys}}{\mathrm{Girls}}$ = $\frac{72-60}{60-52}$ = $\frac{12}{8}$ = $\frac{3}{2}$

Hence, option (a).

Workspace:

**11. SSC CHSL 9th June Shift 3 - QA | Number Theory - SSC**

The number 123456 is divisible by:

- A.
4

- B.
14

- C.
56

- D.
7

Answer: Option A

**Explanation** :

123456 is divisible by 4 as well as 8.

123456 is not divisible by 7, hence it will not be divisible by either 14 or 56.

Hence, option (a).

Workspace:

**12. SSC CHSL 9th June Shift 3 - QA | Tables & Graphs - SSC**

The given pie charts show the proportion of the population and the proportion of population below poverty line of six towns P, Q, R, S, T and U in 2018. Study the pie charts and answer the question that follows.

If the population of the town 'U' is 12,00,000 then the population below poverty line of town 'P' is:

- A.
74000

- B.
68,000

- C.
56,000

- D.
72,000

Answer: Option D

**Explanation** :

Workspace:

**13. SSC CHSL 9th June Shift 3 - QA | Time, Speed & Distance - SSC**

Anjali and Babita are running on a circular track in opposite directions from same time at same point with speeds of 8 m/sec and 6 m/sec, respectively. If the length of the circular track is 960 m, how many times distinct points they will meet?

- A.
7 times

- B.
6 times

- C.
12 times

- D.
14 times

Answer: Option A

**Explanation** :

Number of disting meetings points for two people whose simplified ratio of speeds is a : b is

⇒ (a + b), when going in opposite directions

⇒ |a - b|, when going in same direction

Ratio of speed of Anajli and Babita = 8 : 6 = 4 : 3.

∴ Number of distinct points on the circle where they will meet = 4 + 3 = 7.

Hence, option (a).

Workspace:

**14. SSC CHSL 9th June Shift 3 - QA | Simplification - SSC**

The value of 272 ÷ 16 + [119 ÷ {1491 ÷ 3(21 - 13 × 8 + 4 × 3)}] is.

- A.
-9

- B.
0

- C.
8

- D.
11

Answer: Option B

**Explanation** :

Given, 272 ÷ 16 + [119 ÷ {1491 ÷ 3(21 - 13 × 8 + 4 × 3)}]

= 272 ÷ 16 + [119 ÷ {1491 ÷ 3(21 - 104 + 12)}]

= 272 ÷ 16 + [119 ÷ {1491 ÷ 3(-71)}]

= 272 ÷ 16 + [119 ÷ {1491 ÷ (-213)}]

= 272 ÷ 16 + [119 ÷ {-7}]

= 272 ÷ 16 + [-17]

= 272 ÷ 16 - 17

= 17 - 17

= 0

Hence, option (b).

Workspace:

**15. SSC CHSL 9th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

Khushi went to purchase a mixer. The marked price of the mixer was ₹6,000 and 20% flat discount was offered on it by the shopkeeper. How much does Khushi have to pay for the mixer?

- A.
₹4,200

- B.
₹4,500

- C.
₹4,800

- D.
₹5,000

Answer: Option C

**Explanation** :

20% of 6000 = 1200.

∴ Final price of mixer = 6000 - 1200 = Rs. 4800

Hence, option (c).

Workspace:

**16. SSC CHSL 9th June Shift 3 - QA | Circles - SSC**

The ratio in which a transverse common tangent drawn to two circles with radii 4 cm and 6 cm, respectively, divides the line joining their centres is:

- A.
2 : 3

- B.
1 : 1

- C.
1: 2

- D.
3 : 4

Answer: Option A

**Explanation** :

The transverse common tangent divides the line joining the two centers in the same ratio as radii of the two circles.

Hence, option (a).

Workspace:

**17. SSC CHSL 9th June Shift 3 - QA | Tables & Graphs - SSC**

The table below gives the numbers of units of four different products manufactured by a company during four different months of a particular year. The total number of units manufactured in March of all four products taken together is what percentage of the total number of spoons manufactured by the company in these four given months taken together?

- A.
108.4%

- B.
109.6%

- C.
95.78%

- D.
10.96%

Answer: Option B

**Explanation** :

Total units manufactures in March = 300 + 268 + 236 + 292 = 1096

Total spoons manufactured during the given 4 months = 284 + 236 + 220 + 260 = 1000

∴ Required % = 1096/1000 × 100% = 109.6%

Hence, option (b).

Workspace:

**18. SSC CHSL 9th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

The ratio of income and expenditure of a person is 8 : 5. If her income increases by 25% and her expenditure increases by 28%, then what is the percentage increase in her savings?

- A.
18%

- B.
20%

- C.
$16\frac{2}{3}\%$

- D.
$22\frac{1}{2}\%$

Answer: Option B

**Explanation** :

Ratio of income and expenditure is 8 : 5.

Let the income be Rs. 800 and Expenditure be Rs. 500. Therefore, his savings = Rs. 300.

New income = 800 × 5/4 = Rs. 1000

New expenditure = 500 × 1.28 = Rs. 640

∴ New savings = 1000 - 640 = 360

∴ % increase in savings = (360 - 300)/300 × 100 = 20%.

Hence, option (b).

Workspace:

**19. SSC CHSL 9th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

The annual salary of Suhas has increased from ₹18,00,000 to ₹22,00,000. What is the percentage increase?

(Correct to two decimal places)

- A.
33:33%

- B.
18.18%

- C.
81.81%

- D.
22:22%

Answer: Option D

**Explanation** :

Required % = 4,00,000/18,00,000 × 100% = 400/18% = 200/9% = 22.22%

Hence, option (d).

Workspace:

**20. SSC CHSL 9th June Shift 3 - QA | Simple & Compound Interest - SSC**

A man borrows a sum of ₹1,025 and pays back in two equal yearly instalments. If the rate of interest is 5% p.a, compounded yearly, then how much is each instalments?

- A.
₹425.25

- B.
₹525.32

- C.
₹451.32

- D.
₹551.25

Answer: Option D

**Explanation** :

Let the equal installment be Rs. x.

∴ 1025 × 1.05^{2} = x × 1.05 + x

⇒ 1025 × 1.1025 = 2.05x

⇒ x = 551.25

Hence, option (d).

Workspace:

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

Find the mean proportional between 0.04 and 0.0036.

- A.
0.012

- B.
0.12

- C.
0.0012

- D.
0.004

Answer: Option A

**Explanation** :

Let the mean proportional be x.

⇒ $\frac{0.04}{\mathrm{x}}$ = $\frac{\mathrm{x}}{0.0036}$

⇒ x^{2} = 0.04 × 0.0036

⇒ x = 0.2 × 0.06

⇒ x = 0.012

Hence, option (a).

Workspace:

**22. SSC CHSL 9th June Shift 3 - QA | Simplification - SSC**

Evaluate 107 × 93.

- A.
10047

- B.
9951

- C.
10049

- D.
9953

Answer: Option B

**Explanation** :

107 × 93 = (100 + 7) × (100 - 7)

= 100^{2} - 7^{2}

= 10000 - 49

= 9951

Hence, option (b).

Workspace:

**23. SSC CHSL 9th June Shift 3 - QA | Number Theory - SSC**

**Simplify:**

(a^{-1} + b^{-1}) ÷ (a^{-3} + b^{-3})

- A.
$\frac{ab}{({a}^{2}-ab+{b}^{2})}$

- B.
$\frac{{a}^{3}{b}^{3}}{({a}^{2}-ab+{b}^{2})}$

- C.
$\frac{{a}^{2}{b}^{2}}{({a}^{2}+ab+{b}^{2})}$

- D.
$\frac{{a}^{2}{b}^{2}}{({a}^{2}-ab+{b}^{2})}$

Answer: Option D

**Explanation** :

Given, (a^{-1} + b^{-1}) ÷ (a^{-3} + b^{-3})

= $\frac{{\displaystyle \frac{1}{\mathrm{a}}}+{\displaystyle \frac{1}{\mathrm{b}}}}{\frac{1}{{\mathrm{a}}^{3}}+\frac{1}{{\mathrm{b}}^{3}}}$

= $\frac{{\displaystyle \frac{a+b}{ab}}}{{\displaystyle \frac{{a}^{3}+{b}^{3}}{{\left(ab\right)}^{3}}}}$

= $\frac{{\displaystyle (a+b){\left(ab\right)}^{2}}}{{\displaystyle (a+b)({a}^{2}-ab+{b}^{2})}}$

= $\frac{{\displaystyle {\left(ab\right)}^{2}}}{{\displaystyle ({a}^{2}-ab+{b}^{2})}}$

Hence, option (d).

Workspace:

**24. SSC CHSL 9th June Shift 3 - QA | Triangles - SSC**

The length of the hypotenuse of a right-angled triangle is 37 cm and the length of one of its other two sides is 12 cm. What is the area of the triangle?

- A.
204 cm

^{2} - B.
216 cm

^{2} - C.
210 cm

^{2} - D.
222 cm

^{2}

Answer: Option D

**Explanation** :

Using pythagorus theorem,

37^{2} = 12^{2} + (3^{rd} side)^{2}

⇒ (3^{rd} side)^{2} = 1369 - 144 = 1225

⇒ 3^{rd} side = 35

∴ Area of triangle = 1/2 × 12 × 35 = 210 cm^{2}

Hence, option (c).

Workspace:

**25. SSC CHSL 9th June Shift 3 - QA | Percentage, Profit & Loss - SSC**

A washing machine marked at ₹28,000 is sold at two successive discounts of 15% and 10%. An additional discount of 5% is offered for cash payment. The selling price of the washing machine at cash payment is:

- A.
₹21,458

- B.
₹19,568

- C.
₹20,349

- D.
₹18,851

Answer: Option C

**Explanation** :

Price after 15% discount = 28,000 - 4,200 = 23,800

Price after another 10% discount = 23,800 - 2,380 = 21,420

Price after another 5% discount = 21,420 - 1,071 = 20,349.

Hence, option (c).

Workspace: