ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

B is 60% more efficient than A, which means that the rate of B is 1.6 times as greater as that of A.
So, if A can complete the job in 12 days,
then B completes the same job in 12/1.6 = 12/(8/5)
= 15/2
= 7.5 days.

Alternative method:
Ratio of times taken by
A and B = 160 : 100 = 8 : 5
Suppose B alone takes x days to do the job.
Then, 8 : 5 :: 12 : x
or, 8x = 5 × 12
or, x = 7.5 days

• 2 men and 3 boys can do a piece of work in 10 days.
Thus, 20 men and 30 boys can do a piece of work in 1 day......(i)

• 3 men and 2 boys can do the same work in 8 days.
Thus, 24 men and 16 boys can do the same work in 1 day....(ii)

• Equating (i) and (ii) we get -
o 20 men + 30 boys = 24 men + 16 boys
o 4 men = 14 boys
o 2 men = 7 boys

• Substituting this in equation (i) we get

10 boys can do a piece of work in 10 days.
But we need to find out in how many days 2 men and 1 boy can do the work, which is equivalent to 8 boys.

8 boys can do the same work in (10 × 10/8) = 12.5 days.

We Know
Total work = rate × Time

therefore
30 = Rate × 6
or Rate = 30/6 = 5

Now we need to find the number of computers
total work given = 80 and total time = 3 hrs

therefore
80 = 5 × 3 × No. of computers
or No. of computers = 80/15 = 5.333 , but no. of computers cannot be fraction i.e , so we have to consider as 1.

Total no. of computers = 5 + 1 = 6.

Time taken by Javed =100/{18 × (5/18)} = 20 seconds
Time taken by Naveed = 20 + 5 = 25 seconds
Speed of Naveed = 100/25 × 18/5 = 14.4 km/h

The ratio of speeds of A, B, C = 6 : 3 : 1
The ratio of time taken by A, B, C = 1/6 : 1/3 : 1
To simplify it, we will multiply it by the LCM of ratio of speeds given.
Hence, the ratio of time taken by A, B, C = 1 : 2 : 6

[Speed is inversely proportional to time, meaning if speed increases, time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ]

Time taken by C to covered given distance = 78 = 6 × 13
The ratio of time of A and C = 1 : 6
Thus, time taken by A = 13 min.

Alternative method:
Let C's speed be x metres/min
Let time taken by A be y min
Then B's speed = 3x metres/min
And, A's speed = 6x metres/min
Ratio of speed of A and C = Ratio of time by C and A
6x : x = 78 : y
6x/x = 78/y
y = 13 min

d = v × t
= (4 - 3) × 4
= 4 km

V = (2 × 40 × 60)/(40 + 60) = 48kmph

Alternative method:
Average speed is total distance travelled divided by total time taken.

Let the distance travelled is 2x or half of distance is x.

The first half of distance x is travelled with speed of 40km/hr, so time taken is x/40 hr.

Similarly time taken to travel the remaining x km is at speed of 60km/hr, so time taken for this 2nd half of distance is x/60.

Now, the total time taken to travel 2x distance is = x/40 + x/60 hr
= 3x/120 + 2x/120hr
= 5x/120 hr
= x/24 hr.

The average speed of car is 2x/(x/24) or 2 × 24 km/hr or 48km/hr.

In the first minute the monkey climbs 1 meter.
In the second minute it slips 1/2 meter.
For every two minute it climbs 1/2 meter.

So Average speed = 1 meter/4 minutes
For 11 meters, time taken = 44 minutes.

For the last 1 meter jump add 1 minute.
So time taken = 45 minutes.

(d/80 + d/40) - (d/60 + d/60) = 2
(3d + 6d - 4d - 4d)/240 = 2
d = 480 km

Length of bridge = Distance travelled by the person in 4 min
= speed X time
Speed = 54 Km/hr = 54×5/18 = 3 × 5 = 15 m/s.
Time =4 min = 4 × 60 = 240 s
Required length = 15 × 240 = 3600 m.

Let the speed of first car = 7x
2nd car = 8x.

According to the question
8x = 200/5 = 40 [using speed = distance time]
x = 40/8 = 5

Speed of first car = 7x = 7 X 5 = 35 Km/hr

We know that, speed = Distance/Time
Therefore, Required distance = Speed × Time
=37 × 8
= 296km

Part filled by (A + B + C) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40

Part filled by (A + B) in 1 min
= 1/30 + 1/40
= (4 + 3)/120
= 7/120

Therefore, Part filled by C in 1 min
= 1/10 - 7/120
= (12 - 7)/120
= 5/120
= 1/24
Therefore, Tap C will fill the cistern in 24 min.

Tap A can fill the tank in 30 minutes. Therefore, it fills 1/30th of the tank every minute.
Tap B can fill the tank in 60 minutes. Therefore, it fills 1/60th of the tank every minute.
Together, the two taps will fill
1/30 + 1/60
= (2 + 1)/60
= 3/60
= 1/20th of the tank every minute.

Therefore, when both the taps are opened simultaneously, they will fill the tank in 20 minutes. As the tank is already half full, they need to fill only half the tank.
Therefore, the tank will overflow 10 minutes after both the taps are opened.

We know that, when a pipe empties a cistern in 'n' min, then the part emptied by the pipe in 1 min = 1/n
Here, n = 30
Therefore, Required part of the tank emptied in 1 min = 1/30part

Let, the total capacity of the tank be 96 units (LCM of 24 & 32).

A fills = 96/24 = 4 units / minute.
B fills = 96/32 = 3 units / minute

Let the time for which B is opened is x minutes.
According to the question, 
4 × 9 + 3 × x = 96
3x = 96 - 36
3x = 60
x = 20 minutes.
The pipe B should be closed after 20 minutes

Let the total distance covered = x
= Length of journey
According to the question,
x/2 × 1/40 + x/2 × 1/60 = 16
x/80 + x/120 = 160
(3x + 2x)/240 = 16
5x = 16 × 240
x = (16 × 240)/5
x = 16 × 48
x = 768 km

Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20

Net part filled in 1h when both (tap and leakage) work.
= 1/12 - 1/20
= (5 - 3)/60
= 2/60
= 1/30
Therefore, Required time to fill the tank = 30 h