HSBC Aptitude Question and Answers 2022 PDF

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

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

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

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

279936÷6=46656

46656÷6=7776

7776÷6=1296

1296÷6=216

216÷6=36

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

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

$\text{=Rs.125}$

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

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 $=\frac{\mathrm{1/(}}{9\times 12\times \mathrm{5)}}$.. (Equation 1)

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

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

Let x and  litres of milk be mixed from the first and second container respectively.

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

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

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

(.25x+6−.5x) / (.75x+6−.5x)= 3/5

⇒(6−.25x) / (.25x+6)=3/5

⇒30−1.25x  =.75x+18

⇒2x=12

⇒x=6
Since $x=6,12-x=12-\mathrm{6=6}$

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

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 $=10!\times 6!$

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

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.

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.