# SSC CHSL 10th June Shift 2 - QA

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

What is the smallest natural number that should be added to 54321 so that the sum is divisible by 6?

- A.
7

- B.
3

- C.
1

- D.
5

Answer: Option B

**Explanation** :

For a number to be divisible by 6, it should be divisible by both 2 and 3.

For any number to be divisible by 2, its unit's digit should be even and for any number to be divisible by 3, sum of its digits should be divisible by 3.

For 54321, units digit is odd. To make unit's digit even the least number that can be added is 1 but then the sum of the digits will not be divisible by 3.

∴ Least number to be added is 3. This will make the unit's digit even while sum of the digits will be divisible by 3.

Hence, option (b).

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**2. SSC CHSL 10th June Shift 2 - QA | Ratio, Proportion & Variation - SSC**

Rice worth ₹96 per kg and ₹104 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth ₹113 per kg, the price of the third variety of rice per kg will be:

- A.
₹109

- B.
₹128

- C.
₹126

- D.
₹117

Answer: Option C

**Explanation** :

Let the quantity mixed of each variety is 1 kg, 1kg and 2 kgs respectively.

Total quantity = 4 kgs

Total cost = 4 × 113 = 1 × 96 + 1 × 104 + 2 × P

⇒ 452 - 96 - 104 = 2P

⇒ P = 252/2 = Rs. 126

Hence, option (c).

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**3. SSC CHSL 10th June Shift 2 - QA | Mensuration - SSC**

How many spherical balls of radius 5 cm can be made by melting a spherical clay ball having a radius of 15 cm?

- A.
3

- B.
27

- C.
9

- D.
18

Answer: Option B

**Explanation** :

Let 'n' spherical balls can be made.

∴ Volume of n spherical balls = Volume of the original clay ball

⇒ 4/3 × π × 5^{3} × n = 4/3 × π × 15^{3}

⇒ 5^{3} × n = 15^{3}

⇒ n = 3^{3}

⇒ n = 27

Hence, option (b).

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**4. SSC CHSL 10th June Shift 2 - QA | Simple & Compound Interest - SSC**

Find the compound interest on ₹5,50,000 at 7% per annum for 2 years, compounded annually.

- A.
₹79,695

- B.
₹80,605

- C.
₹62,695

- D.
₹79,690

Answer: Option A

**Explanation** :

Interest for 1^{st} year = 7% of 5,50,000 = 38,500

Interest for 2^{nd} year = 7% more than 38,500 = 41,195

∴ Total interest for 2 years = 41,195 + 38,500 = 79,695

Hence, option (a).

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**5. SSC CHSL 10th June Shift 2 - QA | Time & Work - SSC**

A and B together can finish a job in 40 days. A can do the same job on her own in 60 days. How long will B take to do the three-fourth of the same work all alone?

- A.
120 days

- B.
80 days

- C.
90 days

- D.
100 days

Answer: Option C

**Explanation** :

Let the total work to be done = LCM(40, 60) = 120 units.

Efficiency of A alone = 120/60 = 2 units/day

Efficiency of A and B together = 120/40 = 3 units/day

∴ Efficiency of B alone = 3 - 2 = 1 unit/day

⇒ Time taken by B alone to complete the work = 120/1 = 120 days.

⇒ Time taken by B alone to complete 3/4th work = 120 × 3/4 = 90 days.

**Alternately**,

Let B taken B days to complete the work alone.

⇒ $\frac{1}{40}$ = $\frac{1}{60}$ + $\frac{1}{\mathrm{B}}$

⇒ $\frac{1}{40}$ - $\frac{1}{60}$ = $\frac{1}{\mathrm{B}}$

⇒ $\frac{1}{\mathrm{B}}$ = $\frac{3-2}{120}$

⇒ B = 120

⇒ Time taken by B alone to complete 3/4th work = 120 × 3/4 = 90 days.

Hence, option (c).

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

The average monthly expenditure of a family for the first four months is ₹13,750, for the next three months is ₹11,750 and for the last five months is ₹31,750. If the family saves ₹15,550 during the whole year, find the average monthly income of the family during the year.

(Consider integral part only)

- A.
₹21,055

- B.
₹22,045

- C.
₹22,000

- D.
₹23,040

Answer: Option B

**Explanation** :

Total annual expenditure of the family = 4 × 13750 + 3 × 11750 + 5 × 31750 = 2,49,000

Annual savings of the family = Rs. 15,550

∴ Total annual income of the family = 2,49,000 + 15,550 = Rs. 2,64,550

⇒ Average monthly income = 2,64,550/12 = Rs. 22,045

Hence, option (b).

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**7. SSC CHSL 10th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

Successive discounts of 10% and 20% are given on the purchase of a purse . If the price of the purse is ₹2,250, find the selling price.

- A.
₹1,620

- B.
₹1,320

- C.
₹1,290

- D.
₹1,520

Answer: Option A

**Explanation** :

Price of the purse = Rs. 2,250

Price after 10% discount = 2250 - 225 = 2025

Price after 20% discount = 2025 - 405 = 1620

Hence, option (a).

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**8. SSC CHSL 10th June Shift 2 - QA | Quadrilaterals & Polygons - SSC**

The base of a parallelogram is twice as long as its corresponding height. If the area of the parallelogram is 144 , find the mentioned height.

- A.
6√2 cm

- B.
3√2 cm

- C.
2√2 cm

- D.
8√2 cm

Answer: Option A

**Explanation** :

Let the height of the parallelogram = h, and its base will be 2h

Area of a parallelogram = base × height

⇒ 144 = 2h × h

⇒ 72 = h^{2}

⇒ h = 6√2

Hence, option (a).

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**9. SSC CHSL 10th June Shift 2 - QA | Tables & Graphs - SSC**

A householder spent his monthly salary of ₹7,200 on different items. If he spent ₹4,000 on food and ₹400 on education, then the central angles respectively are:

- A.
30°, 60°

- B.
50°, 200°

- C.
200°, 20°

- D.
200°, 60°

Answer: Option C

**Explanation** :

% fraction of amount spent on food = 4000/7200 = 5/9

∴ central angle = 5/9 × 360° = 200°

% fraction of amount spent on education = 400/7200 = 1/18

∴ central angle = 1/18 × 360° = 200°

Hence, option (c).

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

A shopkeeper mixes three types of rice costing ₹50 per kg, ₹60 per kg and ₹75 per kg in the ratio of 3 : 1 : 2. The average cost of the mixture per kg is:

- A.
₹ 50

- B.
₹ 75

- C.
₹ 60

- D.
₹ 65

Answer: Option C

**Explanation** :

Let the quantity mixed is 3 kgs, 1 kg and 2 kgs respectively.

∴ Average cost per kg = $\frac{3\times 50+1\times 60+2\times 75}{3+1+2}$ = $\frac{360}{6}$ = Rs. 60/kg

Hence, option (c).

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**11. SSC CHSL 10th June Shift 2 - QA | Tables & Graphs - SSC**

Study the given bar-chart and answer the question that follows. The bar chart shows the production and sale of cars (in thousands) over the years 2011 to 2015

The percentage by which sales exceeds production in 2014 is:

- A.
35

- B.
25

- C.
20

- D.
30

Answer: Option D

**Explanation** :

Sales in 2014 = 13

Production in 2014 = 10

Required % = (13 - 10)/10 × 100% = 30%

Hence, option (d).

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**12. SSC CHSL 10th June Shift 2 - QA | Triangles - SSC**

One of the angles of a triangle is 108°, and the other two angles are equal. What is the measure of each of these equal angles?

- A.
36°

- B.
72°

- C.
78°

- D.
39°

Answer: Option A

**Explanation** :

Let the measure of other two angle be x° each.

∴ Sum of the three angles of a triangle is 180° = 108° + x + x

⇒ 2x = 180° - 108° = 72°

⇒ x = 36°

Hence, option (a).

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

What is the value of 12 - 8 ÷ 2 - {16 of - 2 + 3 × 5 - 4}?

- A.
0

- B.
1

- C.
45

- D.
29

Answer: Option D

**Explanation** :

Applying BODMAS Rule

12 - 8 ÷ 2 - {16 of - 2 + 3 × 5 - 4}

= 12 - 8 ÷ 2 - {-32 + 3 × 5 - 4}

= 12 - 8 ÷ 2 - {-32 + 15 - 4}

= 12 - 8 ÷ 2 - {-21}

= 12 - 8 ÷ 2 + 21

= 12 - 4 + 21

= 29

Hence, option (d).

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**14. SSC CHSL 10th June Shift 2 - QA | Simple & Compound Interest - SSC**

A sum of ₹5,000 was deposited for 3 years at 10% per annum, compounded annually. The difference between the interest for 2 years and that for 3 years is:

- A.
₹506

- B.
₹605

- C.
₹560

- D.
₹650

Answer: Option B

**Explanation** :

Interest for 2 years = 5,000${\left(1+\frac{10}{100}\right)}^{2}$ - 5000 = 6050 - 5000 = Rs. 1,050

Interest for 3 years = 5,000${\left(1+\frac{10}{100}\right)}^{3}$ - 5000 = 6655 - 5000 = Rs. 1,655

Difference in interest for 3 years and 2 years = 1655 - 1050 = Rs. 605

Hence, option (b).

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**15. SSC CHSL 10th June Shift 2 - QA | Time, Speed & Distance - SSC**

‘A’ and ‘B’ are two stations 494 km apart. A train starts from station ‘A’ at 3 p.m. and travels towards station ‘B’ at 83 km/h. Another train starts from station ‘B’ at 4 p.m. and travels towards station ‘A’ at 54 km/h. At what time do they meet?

- A.
7 p.m.

- B.
9 a.m.

- C.
7 a.m.

- D.
8 p.m

Answer: Option A

**Explanation** :

Distance travelled by A till 4 pm = 83 × 1 = 83 kms.

Distance between A and B at 4 pm = 494 - 83 = 411 kms

Relative speed of A and B after 4 pm = 83 + 54 = 137 km/hr

∴ Time taken for A and B to meet = $\frac{\mathrm{Relative}\mathrm{Distance}\mathrm{between}\mathrm{them}}{\mathrm{Their}\mathrm{relative}\mathrm{speed}}$ = $\frac{411}{137}$ = 3 hours.

∴ They meet 3 hours after 4 pm i.e., at 7 pm.

Hence, option (a).

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**16. SSC CHSL 10th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

One single discount which is equivalent to 20%, 10% and 5% is given by:

- A.
32.60%

- B.
31.60%

- C.
30.60%

- D.
32.80%

Answer: Option B

**Explanation** :

Let's say the initial vaue is 100.

Value after 20% discount = 100 - 20 = 80

Value after 10% discount = 80 - 8 = 72

Value after 5% discount = 72 - 3.6 = 68.4

∴ Total discount = 100 - 68.4 = 31.6

∴ Discount % = 31.6%

Hence, option (b).

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

Simplify the fo llowing expression.

$\frac{(62\times 62\times 62)-3(62\times 62\times 22)+3(62\times 22\times 22)-(22\times 22\times 22)}{8\times 8\times 8}$

- A.
1250

- B.
25

- C.
125

- D.
225

Answer: Option C

**Explanation** :

$\frac{(62\times 62\times 62)-3(62\times 62\times 22)+3(62\times 22\times 22)-(22\times 22\times 22)}{8\times 8\times 8}$

$\frac{{62}^{3}-3\times {62}^{2}\times 22+3\times 62\times {22}^{2}-{22}^{3}}{{8}^{3}}$

We know, (a - b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}

∴ 62^{3} - 3 × 62^{2 }× 22 + 3 × 62 × 22^{2} - 22^{3} = (62 - 22)^{3}

∴ $\frac{{(62-22)}^{3}}{{8}^{3}}$ = $\frac{{40}^{3}}{{8}^{3}}$ = 53 = 125

Hence, option (c).

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**18. SSC CHSL 10th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

P sold an item to Q at a 20% gain, and Q sold it to R at a loss of 10%. If R bought the item for ₹1,080, then at what price did P purchase it?

- A.
₹1000

- B.
₹5000

- C.
₹3000

- D.
₹800

Answer: Option A

**Explanation** :

Q sold it for 10% loss for Rs, 1,080

∴ His cost price = 1080/0.9 = Rs. 1,200

⇒ P sold it for 20% loss for Rs, 1,200

∴ His cost price = 1200/1.2 = Rs. 1,000

Hence, option (a).

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**19. SSC CHSL 10th June Shift 2 - QA | Trigonometry - SSC**

If cot θ = cot 30° cot 60° and θ is an acute angle, then 2θ is equal to:

- A.
45°

- B.
60°

- C.
30°

- D.
90°

Answer: Option D

**Explanation** :

cot θ = cot 30° cot 60°

We konw, cot 30° = √3 and cot 60° = 1/√3

∴ cot θ = √3 × 1/√3 = 1

⇒ θ = 45°

⇒ 2θ = 90°

Hence, option (d).

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

For what value(s) of k will the expression p + $\frac{1}{9}\sqrt{p}$ + ${k}^{2}$ have equal roots.

- A.
$k=\pm \frac{1}{21}$

- B.
$k=\pm \frac{1}{18}$

- C.
$k=\pm \frac{1}{9}$

- D.
$k=\pm \frac{1}{8}$

Answer: Option B

**Explanation** :

For roots to be equal Discriminant should be 0

⇒ (1/9)^{2} - 4 × 1 × k^{2} = 0

⇒ 4k^{2} = (1/9)^{2}

⇒ k^{2} = (1/18)^{2}

⇒ k = ± (1/18)

Hence, option (b).

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**21. SSC CHSL 10th June Shift 2 - QA | Tables & Graphs - SSC**

The following bar graph shows the sales (in thousands) of books from six different branches of a publishing company in 2000 and 2001.

Answer the following question based on the bar graph:

What is the ratio of the total sales of C3 for both years to the total sales of C4 for both years?

- A.
14 : 15

- B.
15 : 14

- C.
13 : 14

- D.
12 : 13

Answer: Option A

**Explanation** :

Total sales of C3 = 110 + 100 = 210

Total sales of C4 = 115 + 110 = 225

Required ratio = 210 : 225 = 42 : 45 = 14 : 15

Hence, option (a).

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**22. SSC CHSL 10th June Shift 2 - QA | Percentage, Profit & Loss - SSC**

In a panchayat election Candidate A secured 30% of the total votes and did NOT qualify by 6 votes. Candidate B secured 40% of the total votes and got 6 votes more than the bare minimum to qualify. The total number of votes were _________.

- A.
100

- B.
160

- C.
140

- D.
120

Answer: Option D

**Explanation** :

Let the minimum votes required to qualify = Q and Total votes = T

Candidate A secured 30% of the total votes and did NOT qualify by 6 votes.

∴ 0.3T = Q - 6 ...(1)

Candidate B secured 40% of the total votes and got 6 votes more than the bare minimum to qualify

∴ 0.4T = Q + 6 ...(2)

(2) - (1), we get

0.1T = 12

⇒ T = 120

**Alternately**,

The difference between number of votes recieved by A and B is 12 which is 10% (40% - 30%) of total votes

∴ 10% of Total votes = 12

⇒ Total votes = 120

Hence, option (d).

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

If x + 2y = 10 and 2xy = 9, then one of the value of x − 2y is:

- A.
10

- B.
12

- C.
6

- D.
8

Answer: Option D

**Explanation** :

Given x + 2y = 10 ⇒ x = 10 - 2y

Also, 2xy = 9, [substituting x = 10 - 2y]

⇒ 2(10 - 2y)y = 9

⇒ 4y^{2} - 20y + 9 = 0

⇒ 4y^{2} - 18y - 2y + 9 = 0

⇒ (2y - 1)(2y + 9) = 0

⇒ y = -9/2 or 1/2

⇒ x = 28 or 9

∴ x - 2y = 28 - (-9) or 9 - (1)

⇒ x - 2y = 37 or 8

Hence, option (d).

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**24. SSC CHSL 10th June Shift 2 - QA | Mensuration - SSC**

What is the surface area of a sphere whose diameter is 30 cm?

[Use π = 3.14.]

- A.
1413

- B.
1130

- C.
2826

- D.
1134

Answer: Option C

**Explanation** :

Surface area of a sphere = 4πr^{2}

= 4 × 3.14 × 15^{2}

= 4 × 3.14 × 225

= 900 × 3.14

= 2826

Hence, option (c).

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**25. SSC CHSL 10th June Shift 2 - QA | Ratio, Proportion & Variation - SSC**

What is the mean proportional between 64 and 4096?

- A.
8

- B.
512

- C.
128

- D.
192

Answer: Option B

**Explanation** :

Let mean proportion be x.

∴ $\frac{64}{\mathrm{x}}$ = $\frac{\mathrm{x}}{4096}$

⇒ x^{2} = 64 × 4096

⇒ x = 8 × 64 = 512

Hence, option (b).

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