# Indian Bank Po Mains Quant 11TH Oct 2018

**Directions (1-5): Study the information given below carefully and answer the questions related to it.**

**In sector 40 of Cyber park, Gurugram, there are 2400 people who play three different type of sports Tennis, Volleyball and Badminton. 56% of total population play Badminton. 44% of total population play Volleyball. 240 people play both Badminton and Volleyball. 8% of total population play Tennis and Volleyball both. The total population who play Tennis is 24%. The people who play all the three types of games viz Tennis, Volleyball and Badminton are 4% of total population.**

**Q1. The population who play Tennis and Badminton both is approximately what percent of people who play Tennis?**

40%

42%

48%

49%

45%

**Q2. What is the average number of people who play Badminton only, Volleyball only and Tennis only?**

648

662

640

650

658

**Q3. What is the total number of people who play only Tennis and Volleyball both and only Tennis and Badminton both ?**

240

246

236

256

242

**Q4. If the ratio of male to female is 13 : 9 in that population who play Volleyball then the total number of males who play Volleyball are what percent of total population of given sector?**

24%

26%

28%

30%

32%

**Q5. What is the difference between people who play Badminton only and Volleyball only?**

244

238

235

240

245

**Q6. A and B started a business in partnership by investing Rs 10,000 and Rs 4000 respectively. Condition of partnership is that B got Rs 100 per month for management of the business. Remaining profit has to be distributed in the ratio of their investments. Find the share of their profit, if the annual profit is Rs 4000.**

Rs 3000 each

Rs 2500 each

Rs 1500 each

Rs 2000 each

Rs 1800 each

Solution:

Ratio of investments of A and B =10,000: 4000

= 5: 2

Profit of A = 5x

Profit of B = 2x + 12 × 100

A/q,

7x + 1200 = 4000

x = 400

Profit of A = 2000

Profit of B = 4000 – 2000 = 2000

= 5: 2

Profit of A = 5x

Profit of B = 2x + 12 × 100

A/q,

7x + 1200 = 4000

x = 400

Profit of A = 2000

Profit of B = 4000 – 2000 = 2000

**Q7. A cylindrical flask, whose circular part has a diameter of 35 cm is filled with water up to a height of 24 cm. A solid iron spherical ball of radius 12 cm is dropped in the flask to submerge completely in water. Find the increase in the level of water (in cm) (rounded off to two decimal places).**

8.56

8.24

7.88

7.52

6.52

**Q8. Marked price of two articles is Rs.2800 each. One is sold at a discount of 24% and the other for Rs. 1200 more than the marked price. If the net profit is 25%. Find the total cost price of articles.**

Rs. 4,500.5

Rs. 4,902.4

Rs. 4,800

Rs. 4,600.25

Rs. 5,400.6

**Q9. Several litres of alcohol drawn off a 72 litre vessel full of alcohol and equal amount of water added. Again, the same volume of the mixture was drawn off and replaced by water. As a result, the vessel contained 18 litres of pure alcohol. How much the alcohol was drawn off initially?**

30 litres

37 litres

39 litres

36 litres

35 litres

**Q10. The simple interest on a sum of money will be Rs. 450 after five years. What will be the total interest at the end of 12th year?**

Rs. 1200

Rs. 1080

Rs. 1100

Rs. 1000

Rs. 800

Solution:

Simple interest on sum after 5 years = 450

Simple Interest for 1 year = 90

Simple Interest for 12 year = 90 × 12 = 1080

Simple Interest for 1 year = 90

Simple Interest for 12 year = 90 × 12 = 1080

**Directions (11-15): Find the wrong number in the following number series.**

**Q11. 3, 8, 27, 112, 560, 3396**

8

560

27

3396

112

Solution:

Pattern is ×2+2, ×3+3, ×4+4, ×5+5, ×6+6

∴ Wrong term = 560 ≠ 112 × 5 + 5 = 565

∴ Wrong term = 560 ≠ 112 × 5 + 5 = 565

**Q12. 1162, 1167, 1182, 1207, 1242, 1285**

1167

1182

1285

1242

1207

Solution:

Pattern is +5×1, +5×3, ×5+5, +5×7, ×5×9

∴ Wrong term = 1285 ≠ 1242 + 45 = 1287

∴ Wrong term = 1285 ≠ 1242 + 45 = 1287

**Q13. 7, 29, 5, 31, 3, 33, 2**

2

31

33

5

29

Solution:

Pattern is +22, –24, +26, –28, +30, -32

**Q14. 223, 287, 359, 439, 529, 623**

439

287

623

529

359

Solution:

Series is

15² – 2 = 223

17² - 2 = 287

19² – 2 = 359

21² – 2 = 439

23² - 2 = 527

25² – 2 = 623

∴ wrong term = 529

15² – 2 = 223

17² - 2 = 287

19² – 2 = 359

21² – 2 = 439

23² - 2 = 527

25² – 2 = 623

∴ wrong term = 529

**Q15. 2, 3, 11, 65 , 519, 5190**

3

5190

11

65

519

Solution:

Pattern is ×2- 1, ×4–1, ×6 – 1, ×8–1, ×10–1

∴ Wrong term = 5190 ≠ 519 × 10 – 1 = 5189

∴ Wrong term = 5190 ≠ 519 × 10 – 1 = 5189

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