Bank Clerk Questions

At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand. 

 To be coincident, it must gain 15 min. spaces. 

 55 min. are gained in 60 min.

  Then 15 min spaces are gained in  = 6055×15 min. 

Therefore,  The hands are coincident at  16411 min. past 3 o'clock.

1. Rakesh

2. Ranga

3. Raju

4. Ravi

Given that, 

Number of windows = 50

Each window covering covers 15 windows

=> 50 windows requires 50/15 window coverings

= 50/15 = 3.333

Hence, more than 3 window coverings are required. In the options 4 is more than 3. 

Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.

Quantity 1:
Let the first pipe alone takes x hours to fill the
tank.
⇒The second and third pipes will take (x-5) and
(x-9) hours respectively.
According to the given information:
∴ 1x-9

⇒ (x-9)(2x-5) = x2 – 5x
⇒ 2x2 – 5x – 18x + 45 = x2 – 5x
⇒ x
2 -18x + 45 = 0
⇒ (x-15) (x-3) = 0
⇒ x = 15, 3

The first pipe can take 15 hours to fill the kund.
∵ 3 hours doesn’t satisfy the statement.
Quantity 2:
∴ Time taken by second pipe = x-5
⇒ Time taken by second pipe = 15-5 = 10hours
∴ Time taken by third pipe = x -9
⇒ Time taken by third pipe = 15- 9 = 6 hours
Now,

Net part filled in 1 hour = 115+110+16
⇒ Net part filled in 1 hour = 4+6+1060
⇒Net part filled in 1 hour = 2060=13
∴The Kund will be full in 3/1 hours if all the pipes are opened simultaneously
Now, comparing
15 > 3
Thus, Quantity 1 > quantity 2

Arakan yoma is the extension of the himalayas located in Myanmar. The Arakan Mountains are also known as the Rakhine Mountains.

These are a mountain range in western Myanmar, between the coast of Rakhine State and the Central Myanmar Basin, in which the Irrawaddy River flows.

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

From the given data,
how many goals scored = 5397....(1)
many more matches = 982.....(2)
he scored five = 163.....(3)
from (1)&(2), many = 9 ....(4)
from (1)&(3), scored = 3 ....(5)
Using (4)&(5)in(1), we get goals = 5 or 7.