# SSC CHSL 9th June Shift 2 - QA

**1. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

The difference between a discount of 35% and two successive discounts of 20% and 20% on a certain bill, was ₹22. Find the amount of the bill.

- A.
₹2,800

- B.
₹2,200

- C.
₹3,200

- D.
₹2,000

Answer: Option B

**Explanation** :

Two successive discount of 20% is equivalent to = - 20 - 20 + (-20 × -20)/100 = -36% i.e., one single discount of 36%

∴ The difference in two discounts is 1% (= 36% - 35%) of total bill.

⇒ 1% of total bill = 22

⇒ Total bill = Rs. 2200

Hence, option (b).

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**2. SSC CHSL 9th June Shift 2 - QA | Time, Speed & Distance - SSC**

Riya runs 3/2 times as fast as Prerna. In a race, if Riya gives a lead of 100 m to Prerna, find the distance Riya has to run before both of them meet.

- A.
315 m

- B.
300 m

- C.
265 m

- D.
240 m

Answer: Option B

**Explanation** :

Ratio of speeds of Riya and Prerna = 3 : 2

Let Prerna run P meters before Riya catches up with her.

⇒ Distance travelled by Riya = P + 100

∴ $\frac{\mathrm{Distance}\mathrm{covered}\mathrm{by}\mathrm{Riya}}{\mathrm{Distance}\mathrm{covered}\mathrm{by}\mathrm{Prerna}}$ = $\frac{\mathrm{P}+100}{\mathrm{P}}$ = $\frac{3}{2}$

⇒ 2P + 200 = 3P

⇒ P = 200

∴ Distance travelled by Riya = P + 100 = 300 meters.

Hence, option (b).

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

Simplify the following: 50% of [6 - {15 - [6 + 8 ÷ (5 - 3)] + 2}].

- A.
12.5

- B.
1.5

- C.
4

- D.
200

Answer: Option B

**Explanation** :

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**4. SSC CHSL 9th June Shift 2 - QA | Tables & Graphs - SSC**

**The given pie chart presents the monthly expenses on various heads and the savings of Mr.X's family. Study the pie chart and answer the question that follows.**

In the total income of Mr.X is ₹1,20,000, then how much does he pay for transport.

- A.
₹15,000

- B.
₹12,000

- C.
₹20,000

- D.
₹10,000

Answer: Option D

**Explanation** :

Amount paid for transport = $\frac{30\xb0}{360\xb0}$ × 1,20,000 = 10,000

Hence, option (d).

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**5. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

While selling to the retailer, a company allows 30% discount on the marked price of its products. If the retailer sells those products at marked price, his profit per cent will be ________.

- A.
$42\frac{8}{7}$

- B.
$42\frac{6}{7}$

- C.
$42\frac{4}{7}$

- D.
$42\frac{2}{7}$

Answer: Option B

**Explanation** :

Let the marked price of the product = Rs. 100

Cost price of product for retainler = 70% of 100 = Rs. 70

Retailer sells it for Rs. 100.

∴ Retailer's profit = 100 - 70 = Rs. 30

⇒ Retailer's profit % = 30/70 × 100% = 42(6/7)%

Hence, option (b).

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**6. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

A shopkeeper sold an article for ₹1,326 after allowing a discount of 15% on its marked price. Find the marked price of the article.

- A.
₹1,153

- B.
₹ 1,560

- C.
₹1,525

- D.
₹1,650

Answer: Option B

**Explanation** :

Let marked price be M

∴ M × 0.85 = 1,326

⇒ M = 132600/85 = 26520/17 = 1560

Hence, option (b).

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

If (x + 2) and (x - 3) are the factors of x^{2} + k_{1}x + k_{2}, then:

- A.
k

_{1}= 1 and k_{2}= -6 - B.
k

_{1}= -1 and k_{2}= -6 - C.
k

_{1}= -1 and k_{2}= 6 - D.
k

_{1}= 1 and k_{2}= 6

Answer: Option B

**Explanation** :

If (x + 2) and (x - 3) are the factors of x^{2} + k_{1}x + k_{2}, then for x = -2 and 3, the given expression should be zero.

∴ (-2)^{2} + (-2)k_{1} + k_{2} = 0, and

3^{2} + 3k_{1} + k_{2} = 0

Solving the two equations above we get,

k_{1} = -1 and k_{2} = -6

Hence, option (b).

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**8. SSC CHSL 9th June Shift 2 - QA | Mensuration - SSC**

In a right circular cone, the radius of the base is 14 cm and the height is 48 cm. A cross-section is made through the midpoint of the height parallel to the base. The volume of the upper portion is:

(Take π = 22/7).

- A.
1102 cm

^{2} - B.
1232 cm

^{2} - C.
1120 cm

^{2} - D.
1442 cm

^{2}

Answer: Option B

**Explanation** :

We know, for frustum of a cone ratio of volume of smaller cone to bigger cone is cube of the ratio of heights.

∴ $\frac{\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{larger}\mathrm{cone}}$ = ${\left(\frac{\mathrm{h}/2}{\mathrm{h}}\right)}^{3}$ = $\frac{1}{8}$

⇒ Volume of smaller cone = $\frac{1}{8}$ × $\frac{1}{3}$πr2h

⇒ Volume of smaller cone = $\frac{1}{8}$ × $\frac{1}{3}$ × $\frac{22}{7}$ × 142 × 48 = 1232

Hence, option (b).

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**9. SSC CHSL 9th June Shift 2 - QA | Simple & Compound Interest - SSC**

On a sum of money, when invested for 2 years, compound interest and simple interest are ₹300 and ₹250, respectively. For both simple and compound interests the rate of interest per annum is the same, and for compound interest, interest is compounded annually. Find the rate of interest per annum.

- A.
10%

- B.
20%

- C.
40%

- D.
30%

Answer: Option C

**Explanation** :

Let the rate of interest be r% p.a.

Simple interest for 2 years = Rs. 250, hence simple interest every year = Rs. 125

Compound interset for 1st year will be same as Simple intrest for 1st year = Rs. 125

Compound interset for 2nd year = 300 - 125 = Rs. 175.

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

∴ r% = (175 - 125)/125 × 100% = 40% p.a.

Hence, option (c).

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

Find the greatest possible value of (a + b) for which the 8-digit number 143b203a is divisible by 15.

- A.
15

- B.
17

- C.
16

- D.
14

Answer: Option D

**Explanation** :

143b203a should be divisible by 15.

Hence, it should be divisible by both 5 and 3.

∴ For 143b203a to be divisible by 15, a should be either 0 or 5.

**Case 1**: a = 0

Now, 143b2030 should be divisible by 3.

∴ Sum of digits = 1 + 4 + 3 + b + 2 + 0 + 3 + 0 = 13 + b should be divisble by 3.

Hence, highest value of b can be 8

∴ Highest value of a + b = 0 + 8 = 8

**Case 2**: a = 5

Now, 143b2035 should be divisible by 3.

∴ Sum of digits = 1 + 4 + 3 + b + 2 + 0 + 3 + 5 = 18 + b should be divisble by 3.

Hence, highest value of b can be 9

∴ Highest value of a + b = 5 + 9 = 14

Hence, option (d).

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**11. SSC CHSL 9th June Shift 2 - QA | Time, Speed & Distance - SSC**

Gopal travels from A to B at the speed of 5km/h, from B to C at 10 km/h, and from C to D at 15 km/h. If AB = BC = CD, then find Gopal’s average speed.

- A.
$8\frac{2}{11}$

- B.
$70\frac{2}{11}$

- C.
$60\frac{2}{11}$

- D.
$9\frac{2}{11}$

Answer: Option A

**Explanation** :

Let AB = BC = CD = 30 kms

Time taken from A to B = 30/5 = 6 hours.

Time taken from B to C = 30/10 = 3 hours.

Time taken from C to D = 30/15 = 2 hours.

∴ Total time taken = 6 + 3 + 2 = 11 hours.

Total distance travelled = 90 kms

Average speed = 90/11 = 8(2/11) kmph.

Hence, option (a).

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**12. SSC CHSL 9th June Shift 2 - QA | Ratio, Proportion & Variation - SSC**

P, Q and R are batsmen. The ratio of the runs scored by them in a certain match was P ∶ Q = 16 : 17 and Q : R = 15 ∶ 16. At the end of the match, they scored a total of 956 runs. The number of runs scored by R is (nearest to an integer):

- A.
440

- B.
335

- C.
339

- D.
430

Answer: Option C

**Explanation** :

Given, P ∶ Q = 16 : 17 and Q : R = 15 ∶ 16

∴ P : Q : R = 16 × 15 : 17 × 15 : 16 × 17 = 240 : 255 : 272

⇒ R = 272/767 × 956 = 339

Hence, option (c).

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**13. SSC CHSL 9th June Shift 2 - QA | Tables & Graphs - SSC**

**Study the following table and answer the question that follows:**

The following table gives the month-wise number of different types of scooters produced by a company during the first six months of 1992.

In which month, did the company produce an equal number of scooters of each type?

- A.
May

- B.
January

- C.
March

- D.
June

Answer: Option B

**Explanation** :

In January the company produced equal number of scooters of each type.

Hence, option (b).

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**14. SSC CHSL 9th June Shift 2 - QA | Average, Mixture & Alligation - SSC**

The average weight of 8 men is increased by 1.5 kg when one of the men who weighs 65 kg is replaced by a new man. The weight of the new man is:

- A.
71 kg

- B.
87 kg

- C.
81 kg

- D.
77 kg

Answer: Option D

**Explanation** :

Let the weight of the person coming in is X kgs.

The average of 8 men increases by 1.5 kg, hence the total weight increases by 8 × 1.5 = 12 kgs.

This is becase the person coming is 12 kgs heavier than the person going out.

∴ X - 65 = 12

⇒ X = 77

Hence, option (d).

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

If k + $\left(\frac{1}{k}\right)$ = 2, find the value of 8k × k × k.

- A.
8

- B.
1

- C.
4

- D.
2

Answer: Option A

**Explanation** :

For any positive number k, k + $\left(\frac{1}{k}\right)$ = 2 only when k = 1/k = 1

∴ 8k × k × k = 8 × 1 × 1 × 1 = 8

Hence, option (a).

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**16. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

There is a continuous growth in the population of a village at the rate of 5% per annum. If its present population is 18522, what was the population of the village 3 years ago?

- A.
17500

- B.
16400

- C.
16000

- D.
17200

Answer: Option C

**Explanation** :

Initial Population = $\frac{18522}{1.05\times 1.05\times 1.05}$ = 16000

Hence, option (c).

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**17. SSC CHSL 9th June Shift 2 - QA | Time & Work - SSC**

If 15 boys earn ₹900 in 5 days, then how much will 20 boys earn in 7 days?

- A.
₹1,580

- B.
₹1,680

- C.
₹1,540

- D.
₹1,650

Answer: Option B

**Explanation** :

15 boys earn ₹900 in 5 days

∴ Earning per boy per day = 900/(5 × 15) = Rs. 12

∴ Earnings of 20 boys in 7 days = 12 × 20 × 7 = Rs. 1680

Hence, option (b).

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**18. SSC CHSL 9th June Shift 2 - QA | Circles - SSC**

In a circle, AB and CD are two diameters which are perpendicular to each other. Find the length of chord AC.

- A.
$\sqrt{2}$ CD

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

- C.
$\frac{CD}{2}$

- D.
2 AB

Answer: Option B

**Explanation** :

Let the radius of the circle be 'r'.

∴ AB = CD = 2r

Let O be the center of the circle

AOC is a right triangle, where AO = OC = r

⇒ AC^{2} = AO^{2} + OC^{2}

⇒ AC^{2} = r^{2} + r^{2} = 2r^{2}

⇒ AC = √2 × r = √2 × AB/2

⇒ AC = AB/√2

Hence, option (b).

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**19. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

The expenditure of a company increases by 25%, then increases by 30%, then further decreases by 20%.The overall percentage change in expenditure is:

- A.
10

- B.
30

- C.
34

- D.
22

Answer: Option B

**Explanation** :

Let the initial expenditure = Rs. 400

Expenditure after 25% increase = Rs. 500

Expenditure after another 30% increase = Rs. 650

Expenditure after 20% decrease = Rs. 520

∴ Overall % change = 120/400 × 100% = 30%

Hence, option (b).

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**20. SSC CHSL 9th June Shift 2 - QA | Tables & Graphs - SSC**

The following table shows the marks (in percentages) obtained by six students in four different subjects in an examination.

The maximum marks in each subject is 100.

Answer the following questions based on the table:

In which of the following subjects is the average of the percentage marks obtained by the six students the highest?

- A.
Maths

- B.
Geography

- C.
Chemistry

- D.
Physics

Answer: Option B

**Explanation** :

Sum of % for physics = (85 + 75 + 80 + 90 + 95 + 90) = 515

Sum of % for geography = (80 + 85 + 85 + 90 + 90 + 90) = 520

Since sum of % is highest for geography, average will also be highest for geography.

Hence, option (b).

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**21. SSC CHSL 9th June Shift 2 - QA | Simplification - SSC**

Simplify:

$\frac{{(r-s)}^{3}+{(s-t)}^{3}+{(t-r)}^{3}}{6(r-s)(s-t)(t-r)}$

- A.
$\frac{1}{6}$

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

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

- D.
$\frac{1}{4}$

Answer: Option B

**Explanation** :

Given, $\frac{{(r-s)}^{3}+{(s-t)}^{3}+{(t-r)}^{3}}{6(r-s)(s-t)(t-r)}$

We know, if a + b + c = 0, then a^{3} + b^{3 }+ c^{3} = 3abc

Here, (r - s) + (s - t) + (t - r) = 0

∴ (r - s)^{3} + (s - t)^{3} + (t - r)^{3} = 3(r - s)(s - t)(t - r)

∴ $\frac{{(r-s)}^{3}+{(s-t)}^{3}+{(t-r)}^{3}}{6(r-s)(s-t)(t-r)}$ = $\frac{3(r-s)(s-t)(t-r)}{6(r-s)(s-t)(t-r)}$ = $\frac{1}{2}$

Hence, option (b).

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**22. SSC CHSL 9th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

By using faulty weight, a shopkeeper cheats to the extent of 6% while buying and selling rice. Find his gain percentage (rounded to two decimal places).

- A.
14.66%

- B.
13.65%

- C.
12.77%

- D.
11.25%

Answer: Option C

**Explanation** :

While buying 100 gms, the shopkeeper actually gets 106 gms.

Whlie selling 100 gms, the shopeeker actually gives 94 gms.

∴ The shopeeker makes of profit of 12 gms by selling only 94 gms.

⇒ Required profit % = 12/94 × 100% = 12.77%

Hence, option (c).

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**23. SSC CHSL 9th June Shift 2 - QA | Mensuration - SSC**

The radius of a roller is 14 cm and its length 20 cm. It takes 235 complete revolutions to move once over to level a playground. Find the area of the playground.

(Use π = 22/7)

- A.
4136 cm

^{2} - B.
4136 × 10

^{3}cm^{2} - C.
41360 cm

^{2} - D.
4136 × 10

^{2}cm^{2}

Answer: Option D

**Explanation** :

Area of the playground will be 235 times the curved surface area of the roller

= 235 × 2πrh

= 235 × 2 × 22/7 × 14 × 20

= 413600

Hence, option (d).

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**24. SSC CHSL 9th June Shift 2 - QA | Trigonometry - SSC**

If α is an acute angle, which of the following options will NOT necessarily be equal to the value of cosec α?

- A.
$\frac{1}{\mathrm{cos}\alpha}$

- B.
$\sqrt{1+co{t}^{2}\alpha}$

- C.
$\frac{1}{\mathrm{sin}\alpha}$

- D.
$\frac{cot\alpha}{\mathrm{cos}\alpha}$

Answer: Option A

**Explanation** :

We know, $\frac{1}{\mathrm{cos}\alpha}$ = sec α

Hence, option (a).

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**25. SSC CHSL 9th June Shift 2 - QA | Mensuration - SSC**

Find the total surface area of a cylinder with diameter of base 28 cm and height 70 cm.

- A.
7300 cm

^{2} - B.
7932 cm

^{2} - C.
8000 cm

^{2} - D.
7392 cm

^{2}

Answer: Option D

**Explanation** :

Total surface area = 2πrh + 2πr^{2}

= 2 × 22/7 × 14 × 70 + 2 × 22/7 × 14^{2}

= 6160 + 1232

= 7392

Hence, option (d).

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