Bank PO Questions

We already know that the calendar after a leap year repeats again after 28 years.

Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

speed = 12 km/h = 12×518=103m/s 

distance covered = 20×2×227×50=440007m 

time taken  = distance /speed =  440007×310sec=4400×37×160min=2207min 

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.

Using Trial and error method,

From the options  u = 1, v = 3/2 satisfies both the equations.

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.

Since, in one hour, two hands of a clock coincide only once, so, there will be value.
Required time T = 211H×30 + Ao minutes past H.
Here H - initial position of hour hand = 2 (since 2 O'clock)
A° = Required angle = 0° (Since it coincides)

T = 2112×30 + 0o minutes past 2 

=> 11011 minutes past 2

Let length = x meters, then breadth = 0.6x 

Given that  perimeter = 800 meters 

=> 2[ x + 0.6x] = 800 

=> x = 250 m 

Length = 250m and breadth = 0.6 x  250 = 150m 

Area = 250 x 150 = 37500 sq.m

distance covered in 1 revolution = 2πr = 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140