HSBC Aptitude Question and Answers 2022 PDF

1. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water?

4 hr

10.76%

2 hr 30 min

61.39%

2 hr

10.13%

1 hr 15 min

17.72%

Speed upstream = $\frac{2}{2} = 1 km/hr$

Speed downstream = $\frac{1}{(\frac{20}{60})} = 3 km/hr$

Speed in still water = $\frac{1}{2} (3 + 1) = 2 km/hr$

Time taken to travel 5 km in still water $= \frac{5}{2} = 2 \frac{1}{2} hours$ 2 hour 30 minutes

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2. Find the missing number $279936, 46656, 7776, 1296, 216, ?$

279936, 46656, 7776, 1296

66

20.25%

36

64.42%

60

8.59%

46

6.75%

279936÷6=46656

46656÷6=7776

7776÷6=1296

1296÷6=216

216÷6=36

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3. The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 9000, brokerage being ¹⁄₄ %, is:

Rs. 125.25

28.89%

Rs. 124.75

57.78%

Rs. 112.20

6.67%

Rs. 108.25

6.67%

Assume that face value = Rs.100

Dividend per share = Rs.10.5 (as it is 10.5% stock)

By investing Rs.9000, earnings = Rs.756

$To get an earning of Rs.10.5, investment required = \frac{9000 \times 10.5}{756} = Rs.125$

ie,market value of Rs. 100 stock = Rs.(125- 1/4) = Rs.124.75

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4. 9 examiners can examine a certain number of answer books in 12 days by working 5 hours a day. How many hours in a day should 4 examiners work to examine twice the number of answer books in 30 days?

9

62.38%

11

18.81%

12

12.87%

10

5.94%

Given that work done by 9 examiners in 12 days by working 5 hours a day = 1
=> Work done by 1 examiner in 1 day in 1 hour $= \frac{1}{9 \times 12 \times 5}$ .. (Equation 1)

Work needs to be done by 4 examiners in 30 days working $x$ hours a day = 2 (∵ twice work to be completed)
=> Work needs to be done by 1 examiner in 1 day working $x$ hours a day $= \frac{2}{4 \times 30}$ ... (Equation 2)

From (Equation 1) and (Equation 2),
$x = \frac{(\frac{2}{4 \times 30})}{(\frac{1}{9 \times 12 \times 5})} = \frac{2 \times 9 \times 12 \times 5}{4 \times 30} = \frac{9 \times 12 \times 5}{2 \times 30} = \frac{9 \times 6 \times 5}{30} = 9$

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5. if log₁₀2 = 0.3010, what is the value of log₁₀1600 ?

None of these

15.04%

5.204

15.04%

3.204

61.95%

1.204

7.96%

log₁₀1600
= log₁₀(16 × 100)
= log₁₀(16) + log₁₀(100)
= log₁₀(2⁴) + log₁₀(10²)
= 4 log₁₀(2) + 2
= (4 × 0.3010) + 2
= 1.204 + 2
= 3.204

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6. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

5,7

17.11%

4,8

27.63%

6,6

48.68%

7,4

6.58%

Let x and $(12 - x)$ litres of milk be mixed from the first and second container respectively.

Amount of milk in $x$ litres of the the first container $= .75 x$
Amount of water in $x$ litres of the the first container $= .25 x$

Amount of milk in $(12 - x)$ litres of the the second container $= .5 (12 - x)$
Amount of water in ( $(12 - x)$ litres of the the second container $= .5 (12 - x)$

Ratio of water to milk
$= [.25 x + .5 (12 - x)] : [.75 x + .5 (12 - x)]$ $= 3 : 5$ $= 3 : 5$

$\Rightarrow \frac{(.25 x + 6 - .5 x)}{(.75 x + 6 - .5 x)} = \frac{3}{5} \Rightarrow \frac{(6 - .25 x)}{(.25 x + 6)} = \frac{3}{5} \Rightarrow 30 - 1.25 x = .75 x + 18 \Rightarrow 2 x = 12 \Rightarrow x = 6$ (.25x+6−.5x) / (.75x+6−.5x)= 3/5

⇒(6−.25x) / (.25x+6)=3/5

⇒30−1.25x =.75x+18

⇒2x=12

⇒x=6
Since $x = 6, 12 - x = 12 - 6 = 6$

Hence 6 and 6 litres of milk should mixed from the first and second container respectively.

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7. A company has 10 software engineers and 6 civil engineers. In how many ways can they be seated around a round table so that all the civil engineers are together?

(10!)²

4.17%

11! × 6!

25%

10! × 6!

57.29%

9! × 6!

13.54%

Group all the 6 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 11. These 11 engineers can be arranged around a round table in $(11 - 1)! = 10!$ ways ...(A)

We had grouped 6 civil engineers. These 6 civil engineers can be arranged among themselves in $6!$ ways ...(B)

From (A) and (B),
required number of ways $= 10! \times 6!$

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8. How many minimum terms in the G.P.: 1, 1.2, 1.44, 1.728, . . . are needed so that the sum of the terms is greater than 25? (Take log 2 = 0.301, log 3 = 0.4771

8

14.29%

9

22.22%

10

55.56%

13

7.94%

The given GP is 1, 1.2, 1.44, 1.728, -----
Here a = 1, r = 1.2 Now Sn = > 25
⇒ > 25 ⇒ 1.2ⁿ - 1 > 5
⇒ 1.2ⁿ> 6 ⇒ n log (1.2) > log 6
⇒ n[log12 – log10] > log 6
⇒ n [2 log 2 + log 3 – log 10) > log 2 + log 3
⇒ n [2 × 0.301 + 0.4771 – 1) > 0.301 + 0.4771
⇒ 0.07491 n > 0.7781
⇒ = 9.83
⇒ n > 9.83
So the least value of n is 10

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9. A merchant has three different types of milk- 324 litres, 351 litres and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.

17

22.67%

61

14.67%

42

48%

51

14.67%

Since a minimum number of casks are required, the size of the cask is greatest. Also the cask in three cases is of equal size. The size of the cask is the H.C.F. of 324 litres, 351 litres and 459 litres which is 27. Now, the number of casks required for storing the milk = (324 + 351+ 459)÷ 27 = 42.

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10. Train A leaves Mumbai for Pune at 5:30 a.m. and reaches Pune at 9:30 a.m. Train B leaves Pune for Mumbai at 7:30 a.m. and reaches Mumbai at 13:30 hrs. At what time will these two trains meet?

7:42 a.m.

10.81%

8:42 a.m.

71.62%

9:12 a.m.

9.46%

8:48 a.m.

8.11%

Train A takes 4 hrs to complete its journey and train B takes 6 hrs to complete its journey. Let the total journey = D.
Speed of train A = D/4 and speed of train B = D/6.

Distance covered by train A in 2 hrs = D/4 × 2 = D/2.
∴ Remaining distance = D/2

∴ time taken to meet = D/2 / ((D/4)+(D/6))= 6/5 hours= 1 hr 12 min.
∴ Trains will meet at 7.30 + 1 hr 12 min. = 8.42 A.M.

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