# SSC CHSL 10th June Shift 3 - QA

**1. SSC CHSL 10th June Shift 3 - QA | Tables & Graphs - SSC**

**In a school of 200 students, the following chart represents the percentage of students involved in different sports.**

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What is the number of students playing cricket?

- A.
17

- B.
83

- C.
24

- D.
34

Answer: Option D

**Explanation** :

The required number of students = 17% of 200 = 17/100 × 200 = 34.

Hence, option (d).

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

The mean of five numbers is 18. If one number is excluded, their mean is 16. The excluded number is:

- A.
36

- B.
32

- C.
26

- D.
30

Answer: Option C

**Explanation** :

Let the number excluded be 'x'

Sum of the 5 numbers = 5 × 18 = 90

Sum of remaining 4 numbers = 90 - x = 4 × 16

⇒ x = 90 - 64 = 26

**Alternately**,

Had the excluded number been 18, the average of the remaining numbers wouldn't have changed.

Since the average of remaining 4 numbers decreased by 2, it means the excluded number is 4 × 2 = 8 more than the original average.

∴ Excluded number = 18 + 8 = 26.

Hence, option (c).

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**3. SSC CHSL 10th June Shift 3 - QA | Tables & Graphs - SSC**

The table below shows the number of cakes sold by six different bakeries in a town on five different days of particular week.

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What is the total number of cakes sold by the bakery D on Monday, Thursday and Sunday, taken together?

- A.
779

- B.
668

- C.
530

- D.
686

Answer: Option D

**Explanation** :

Required sum = 221 + 185 + 280 = 686.

Hence, option (d).

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**4. SSC CHSL 10th June Shift 3 - QA | Tables & Graphs - SSC**

**Observe the graph and answer the question that follows.**

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How many students obtained marks more than 130?

- A.
19

- B.
14

- C.
17

- D.
20

Answer: Option D

**Explanation** :

We are asked how many students obtained more than 130 marks.

If we observe the graph, first bar represents those students who scored marks between 125 and 130, hence they definitely scored less than 130.

Second bar represents 3 students who scored marks between 130 and 135. Now these 3 students could have score exatly 130 or more than 130 as well, hence we are not sure if they scored 130 or more than 130.

∴ Number of students who definitely scored more than 130 = 2 + 8 + 5 + 4 + 1 = 20.

Hence, option (d).

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**5. SSC CHSL 10th June Shift 3 - QA | Ratio, Proportion & Variation - SSC**

A varies jointly with B and C. A = 6 when B = 3 and C = 2. Find A when B = 5 and C = 7.

- A.
105

- B.
70

- C.
17.5

- D.
35

Answer: Option D

**Explanation** :

A ∝ B × C

⇒ $\frac{\mathrm{A}}{\mathrm{B}\times \mathrm{C}}$ = constant

∴ $\frac{6}{3\times 2}$ = $\frac{\mathrm{A}}{5\times 7}$

⇒ A = 35

Hence, option (d).

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

A shopkeeper purchases 20000 units of a product at ₹1 each, 15000 units at ₹1.15 each and 5000 units at ₹2 each. What is the weighted average price of one unit?

(Correct to two decimal places)

- A.
₹1.20

- B.
₹1.36

- C.
₹1.38

- D.
₹1.18

Answer: Option D

**Explanation** :

Weighted average price = $\frac{\mathrm{Total}\mathrm{cost}}{\mathrm{Total}\mathrm{quantity}}$ = $\frac{20,000\times 1+15,000\times 1.15+5,000\times 2}{20,000+15,000+5,000}$ = $\frac{47,250}{40,000}$ = Rs. 1.18

Hence, option (d).

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**7. SSC CHSL 10th June Shift 3 - QA | Time, Speed & Distance - SSC**

After seeing a policeman from a distance of 500 m, a thief starts running at a speed of 10 km/h. After noticing, the policeman chases immediately with a speed of 12 km h, and the thief is caught. The distance run by the policeman is:

- A.
3.6 km

- B.
2.5 km

- C.
3 km

- D.
2.4 km

Answer: Option C

**Explanation** :

Initially the distance between policeman and thief = 500m = 0.5 km

Relative speed of policeman with respect to thief = 12 - 10 = 2 kmph

Time taken by the policeman to catch the thief = 0.5/2 = 0.25 hours.

∴ The distance covered by the policeman = 0.25 × 12 = 3 kms

Hence, option (c).

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

The average of 10 observations is 21. A new observation is included and the average of these 11 numbers is 1 less than the prevoius average. The observation is ______.

- A.
11

- B.
10

- C.
21

- D.
12

Answer: Option B

**Explanation** :

Original average = 21, new average = 21 - 1 = 20

Sum of the 10 observations = 10 × 21 = 210

Let the number added be 'x'.

New sum = 210 + x = 11 × 20

⇒ x = 220 - 210 = 10

Hence, option (b).

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**9. SSC CHSL 10th June Shift 3 - QA | Mensuration - SSC**

The diameter of the base of a right-circular cylinder is 12 cm and the height of the cylinder is 2.45 times the radius of its base. Find the curved surface area of the cylinder.

[Use $\pi =\frac{22}{7}$]

- A.
554.4 cm

^{2} - B.
552.4 cm

^{2} - C.
556.4 cm

^{2} - D.
544.4 cm

^{2}

Answer: Option A

**Explanation** :

Diameter of the base = 12 cm, hence radius of the base = 12/2 = 6 cm.

Height of the cylinder 2.45 × 6 = 14.7 cm

⇒ Curved surface area = 2πrh

= 2 × 22/7 × 6 × 14.7

= 2 × 22 × 6 × 2.1

= 554.4 cm^{2}.

Hence, option (a).

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**10. SSC CHSL 10th June Shift 3 - QA | Simple Equation - SSC**

One number x is thrice the other number y. If the sum of both the numbers is 20, then what is the values of x and y are respectively ?

- A.
8 and 12

- B.
5 and 15

- C.
15 and 5

- D.
2 and 18

Answer: Option C

**Explanation** :

Given, x = 3y ...(1)

Also, x + y = 20 ...(2)

Substituting y = 3x from (1) in (2)

⇒ 3y + x = 20

⇒ 4y = 20

⇒ y = 5

∴ x = 3y = 15

Hence, option (c).

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

A pen passing through two hands is finally sold at a profit of 44% of the original cost price. If the first dealer makes a profit of 20%, then the profit made by the second dealer is:

- A.
24%

- B.
36%

- C.
20%

- D.
27%

Answer: Option C

**Explanation** :

Let the original cost prime of the pen = Rs. 100

First person sells the pen for 20% profit, hence selling price for first person = Rs. 120 [This is the cost price for 2^{nd} person.]

Now, overall the price increases by 44% after 2 hands, it means the second person sold the pen for Rs. 144.

∴ Profit of 2^{nd} person = $\frac{144-120}{120}$ × 100 = 20%

Hence, option (c).

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**12. SSC CHSL 10th June Shift 3 - QA | Number Theory - SSC**

Which of the following is the least 6-digit number that is divisible by 93?

- A.
100065

- B.
100070

- C.
100075

- D.
100068

Answer: Option D

**Explanation** :

Option (b): 100070 is not divisible by 3, hence it will also not be divisible by 93.

Option (c): 100075 is not divisible by 3, hence it will also not be divisible by 93.

Option (a): 100065 is not divisible by 31, hence it will also not be divisible by 93.

Option (d): 100068 is divisible by both 3 as well as 31, hence it will also not divisible by 93.

Hence, option (d).

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**13. SSC CHSL 10th June Shift 3 - QA | Simple & Compound Interest - SSC**

A sum of ₹14,375, when invested at r% interest per year compounded annually, amounts to ₹16,767 after two years. What is the value of r?

- A.
9

- B.
8

- C.
7

- D.
6

Answer: Option B

**Explanation** :

Given, 16767 = 14375${\left(1+\frac{\mathrm{r}}{100}\right)}^{2}$

⇒ ${\left(1+\frac{\mathrm{r}}{100}\right)}^{2}$ = $\frac{16767}{14375}$

⇒ ${\left(1+\frac{\mathrm{r}}{100}\right)}^{2}$ = 1.1664

⇒ $\left(1+\frac{\mathrm{r}}{100}\right)$ = 1.08

⇒ r = 8%

Hence, option (b).

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

Under a discount scheme, a dozen pair of gloves quoted at ₹200 is available at a discount of 20%. How many pairs of gloves can be bought for ₹320?

- A.
20

- B.
15

- C.
18

- D.
24

Answer: Option D

**Explanation** :

Discounted price of dozen i.e., 12 gloves = 200 × 0.8 = Rs. 160

For Rs. 160, 12 gloves can be bought, hence for Rs. 320, 24 gloves can be purchased.

Hence, option (d).

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**15. SSC CHSL 10th June Shift 3 - QA | Simplification - SSC**

Which of the following options has the greatest value?

- A.
(−99) + (−44) − 12

- B.
20 + 4 + (−8) − 2 + 3 + 6

- C.
(−18) − 45 + (− 3 − 2)

- D.
(−22) + (− 4 − 7)

Answer: Option B

**Explanation** :

Option (a): (-99) + (-44) - 12 = -ve

Option (b): 20 + 4 + (-8) - 2 + 3 + 6 = 33 - 10 = 23 (the only +ve value)

Option (c): (-18) - 45 + (- 3 - 2) = -ve

Option (d): (-22) + (- 4 - 7) = -ve

Hence, option (b).

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**16. SSC CHSL 10th June Shift 3 - QA | Time, Speed & Distance - SSC**

Ravi drove for 4 hours at a speed of 70 miles per hour and for 2 hours at 40 miles per hour. What was his average speed (in miles per hour) for the whole journey?

- A.
50

- B.
60

- C.
55

- D.
45

Answer: Option B

**Explanation** :

Total distance travelled by Ravi = 4 × 70 + 2 × 40 = 360 miles

Total time taken by Ravi = 4 + 2 = 6 hours.

∴ Average speed of Ravi = 360/6 = 60 miles/hour.

Hence, option (b).

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**17. SSC CHSL 10th June Shift 3 - QA | Triangles - SSC**

In ΔABC, if G is the centroid and AD is a median with length 9 cm, then the length of AG is:

- A.
5 cm

- B.
6 cm

- C.
8 cm

- D.
7 cm

Answer: Option B

**Explanation** :

We know, centroid divides the median in the ratio of 2 : 1 [The part closer to vetex will be longer.]

∴ AG : GD = 2 : 1

Also, AD = AG + GD = 9

⇒ AG = 2/3 × 9 = 6 cm.

Hence, option (b).

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

A person could save 20% of his income. A year later, his income increased by 25% but he could save the same amount only as before. By what percentage has his expenditure increased?

- A.
29.75 %

- B.
31.25 %

- C.
27.5 %

- D.
32.5 %

Answer: Option B

**Explanation** :

Let the initial income of the person be Rs. 100.

∴ His initial savings = 20% of income = Rs. 20

∴ His initial expenditure = 100 - 20 = Rs. 80

New income of the preson is 25% more than initial income = Rs. 125

∴ His new savings is same = Rs. 20

∴ His new expenditure = 125 - 20 = Rs. 105

⇒ % Increase in his expenditure = $\frac{105-80}{80}$ × 100 = 31.25%

Hence, option (b).

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**19. SSC CHSL 10th June Shift 3 - QA | Number Theory - SSC**

What is the product of (x + a) and (x + b)?

- A.
x

^{2}+ (a - b) x + ab - B.
x

^{2}+ (a - b) x - ab - C.
x

^{2}+ (a + b) x + ab - D.
x

^{2}+ (a - b) x - ab

Answer: Option C

**Explanation** :

(x + a)(x + b)

= x(x + b) + a(x + b)

= x^{2} + bx + ax + ab

= x^{2} + (a + b)x + ab

Hence, option (c).

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**20. SSC CHSL 10th June Shift 3 - QA | Number Theory - SSC**

If x + y + z = 10, x^{2} + y^{2} + z^{2} = 30, then the value of x^{3} + y^{3} + z^{3} - 3xyz is ________.

- A.
-10

- B.
-70

- C.
-50

- D.
-30

Answer: Option C

**Explanation** :

Given, x + y + z = 10

Squaring both sides, we get

⇒ x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx = 100

⇒ 30 + 2xy + 2yz + 2zx = 100

⇒ 2xy + 2yz + 2zx = 70

⇒ xy + yz + zx = 35

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

⇒ x^{3} + y^{3} + z^{3} - 3xyz = (10)(30 - 35)

⇒ x^{3} + y^{3} + z^{3} - 3xyz = (10)(-5)

⇒ x^{3} + y^{3} + z^{3} - 3xyz = -50

Hence, option (c).

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**21. SSC CHSL 10th June Shift 3 - QA | Mensuration - SSC**

What is the height of a solid right circular cylinder whose radius is 3 cm and total surface area is 60π cm^{2}?

- A.
3 cm

- B.
9 cm

- C.
7 cm

- D.
5 cm

Answer: Option C

**Explanation** :

Total surface area of a cylinder = 2πrh + 2πr^{2} = 60π

⇒ 2rh + 2r^{2} = 60

⇒ 2 × 3 × h + 2 × 3^{2} = 60

⇒ 2 × 3 × h + 18 = 60

⇒ 6 × h = 42

⇒ h = 7

Hence, option (c).

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**22. SSC CHSL 10th June Shift 3 - QA | Trigonometry - SSC**

The value of sin 73° + cos 137° is:

- A.
cos 13°

- B.
sin 13°

- C.
cos 18°

- D.
sin 18°

Answer: Option B

**Explanation** :

We know, sin A - sin B = 2 × $\mathrm{cos}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)$ × $\mathrm{sin}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$

Given, sin 73° + cos 137° = sin 73° + cos (90 + 47)° = sin 73° - sin 47°

⇒ sin 73° - sin 47° = 2 × $\mathrm{cos}\left(\frac{73+47}{2}\right)$ × $\mathrm{sin}\left(\frac{73-47}{2}\right)$

⇒ sin 73° - sin 47° = 2 × cos 60° × sin 13°

⇒ sin 73° - sin 47° = 2 × 1/2 × sin 13°

⇒ sin 73° - sin 47° = sin 13°

Hence, option (b).

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**23. SSC CHSL 10th June Shift 3 - QA | Triangles - SSC**

What is the area (in cm^{2}) of an equilateral triangle of side 20 cm?

- A.
100√3

- B.
200

- C.
100√2

- D.
100

Answer: Option A

**Explanation** :

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{a}^{2}$ [where a is the side of the triangle]

∴ Required area = $\frac{\sqrt{3}}{4}\times {20}^{2}$ = 100√3

Hence, option (a).

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**24. SSC CHSL 10th June Shift 3 - QA | Time & Work - SSC**

45 people can repair a road in 10 days, working 6 hours a day. In how many days can 30 people, working 6 hours a day, complete the same work?

- A.
15

- B.
12

- C.
10

- D.
18

Answer: Option A

**Explanation** :

We know when work to be done is same, M_{1}H_{1}D_{1} = M_{2}H_{2}D_{2}

⇒ 45 × 6 × 10 = 30 × 6 × D_{2}

⇒ 450 = 30 × D_{2}

⇒ D_{2} = 450/30 = 15 days

Hence, option (a).

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

The marked price of a table fan is ₹3,750 and is available to the retailer at a discount of 20%. At what price should the retailer sell it to earn a profit of 15%?

- A.
₹3,450

- B.
₹3,350

- C.
₹3,300

- D.
₹3,400

Answer: Option A

**Explanation** :

Cost price of retailer = 3750 × 4/5 = Rs. 3000

To earn a profit of 15%, retailer should sell it for 3000 × 1.15 = Rs. 3,450

Hence, option (a).

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