ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: A person crosses a 750 m long street in 5 minutes, What is his speed in km per hour?

Solution:
Speed = 750/(5 × 60) m/sec
= 2.5 m/sec
= 2.5 × (18/5) km/h
= 9 km/h

Question: If a person walks at 18 km/h instead of 12 km/h, he would have walked 24 km more. The actual distance travelled by him is-

Solution:
Let
the actual distance travelled be x km.

Then,
x/12 = (x + 24)/18
⇒ 18x = 12x + 288
⇒ 6x = 288
∴ x = 48 km

Question: In a race of 1200 m, X can beat Y by 120 m. In a 600 m, Y beats Z by 60 m. In a race of 600 m. X will beat Z by-

Solution:
While X covers 1200 Y covers 1080
while X covers 600 Y covers 540m

While Y covers 600, Z covers 540m
While Y covers 540, Z covers (540 × 540)/600
= 486 m

∴ in a 600 m race X will beat Z by = (600 - 486) m
= 114 m

Question: A monkey climbs a 36 m high pole. In first minute he climbs 6 m and slips down 3 m in the next minute. How much time is required by it to reach the top?

Solution:
According to the question,
For 1^st min, he climbs 6 m and comes down 3 m in 2^nd
Effectively in 2 min, he climbs 3 m.

In 1 min, he climbs (3/2) m

So, in 20 mins he climbs = (3/2) × 20 = 30 m

So, in the 21^th minute, he climbs the next 6 m.

∴ The required value is 21 minutes.

Question: Two trains 115 meters and 95 meters long, run at the speeds of 50 kmph and 76 kmph respectively, in opposite directions on parallel tracks. The time which they take to cross each other, is-

Solution:
Length of the 1st train = 115 m
Length of the 2nd train = 95 m
Relative speed of the trains = (50 + 76) = 126 kmph
= (126 × 5)/18 m/sec
= 35 m/sec

Time taken to cross each other = (Length of 1st train + length of 2nd train)/relative speed of the trains
= (115 + 95)/35 sec
= 210/35 sec
= 6 sec

Question: A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

Solution:
A's 2 day's work = (1/20) × 2 = 1/10
(A + B + C)'s 1 day's work = 1/20 + 1/30 + 1/60
= 6/60
= 1/10

Work done in 3 days = 1/10 + 1/10
= 2/10 part
= 1/ 5 part

Now,
1/5 work is done in 3 days.
∴ Whole work will be done in (3 × 5) = 15 days

Question: An old man is walking on a foggy road at a speed of x km/h. Due to low visibility, the old man see only up to 600 meters. If a car overtakes the man from behind with the speed of 15 km/hr then the man can see the car for 216 seconds. Find the speed of the man?

Solution:
Distance up to old man see = 600/1000 = 0.6km
Time for which the man can see the car = 216/(60 × 60) = 0.06 hour

ATQ,
0.6/(15 - x) = 0.06
⇒ 15 - x = 10
∴ x = 5

So the old man is walking at a speed of 5 km/h

Question: In a race, the speeds of A and B are in the ratio of 5 : 6. A takes 20 minutes more than B takes to reach the destination. What is the time taken by A to reach the destination in hours?

Solution:
Ratio of speeds = 5/6
⇒ Ratio of times taken = 6/5

Let,
A and B take 6x and 5x hours to reach the destination.

ATQ,
6x - 5x = 20/60
∴ x = 1/3

∴ Time taken by A = 6x = (6 × 1/3) hours
= 2 hours.

Question: 10 women can do a work in 8 days. 8 men can complete the same work in 7 days. What is the ratio between the capacity of a man and a woman?

Question: 
(10 × 8) = 80 women can complete the work in 1 day.
∴ 1 woman's 1 day's work = 1/80 part

(8 × 7) = 56 men can complete the work in 1 day.
1 man's 1 day's work = 1/56 part

So, required ratio = 1/56 : 1/80
= (1/56)/(1/80)
= 80/56
= 10/7
= 10 : 7

Question: An individual is cycling at a speed of 30 km per hour. He catches his predecessor who had started earlier in three hours. What is the speed of his predecessor who had started 3 hours earlier?

Solution:
The distance covered in three hour = 3 × 30
= 90 km

Time taken by first individual = (3h + 3h) = 6h

Then, the speed of predecessor, = 90/6 kmph
= 15 kmph.

Question: A job can done by 3 skilled men in 15 days or by 5 boys in 30 days. How many days will they take if they work together?

Solution:
3 men's 1 day work = 1/15 part
and 5 boys 1 day's work = 1/30 part

∴ (3 men's + 5 boy's) 1 day work = (1/15 + 1/30) = (2 +1)/30 = 3/30 = 1/10 part

∴ 3 men and 5 boys will complete the work in 10 days.

Question: The speed of three cars is in the ratio of 2 : 3 : 4. The ratio of the times taken by these cars to travel the same distance is-

Solution:
Speed is always inversely proportional to time,
⇒ s ∝ (1/t)

∴ Ratio of times takes = 1/2 : 1/3 : 1/4
= (1/2) × 12 : (1/3) × 12 : (1/4) × 12
= 6 : 4 : 3

Question: A man completes a certain journey by a car. If he covered 40% of the distance at the speed of 20kmph. 30% of the distance at 15 kmph and the remaining of the distance at 30 kmph, his average speed is-

Solution:
Suppose, total distance = 100 km

He covered 40% of the distance at the speed of 20 kmph
His time taken = 40/20 = 2 hours

He covered 30% of the distance at 15 kmph
His time taken = 30/15 = 2 hours

He covered the remaining of the distance at 30 kmph
The remaining of the distance = 100% - (40 + 30)% = 30%
His time taken = 30/30 = 1 hour

∴ Average speed = Total distance/time
= 100/(2 + 2 + 1)
= 20 kmph

Question: A, B and C start together from the same place to walk round a circular path of length 12km. A walks at the rate of 6 km/h, B 3 km/h and C 3/2 km/h. They will meet together at the starting place at the end of-

Solution:
We know,
Time = Distance/speed

Time taken to complete the revolution
A = 12/6 = 2 hours
B = 12/3 = 4 hours
C = 12 × (2/3) = 8 hours

Now, required time = LCM of 2, 4, and 8
= 8 hours

Question: The speed of a car increases by 2 kms after every one hours. If the distance travelled in the first one hour was 35 kms, what was the total distance travelled in 12 hours?

Solution:
Total distance travelled in 12 hours = (35 + 37 + 39 +..... upto 12 terms)
This is an A.P with first term, a = 35 ,
number of terms, n = 12
d = 2

Required distance = (12/2){(2 × 35) + (12 - 1) × 2}
= 6 × (70 + 22)
= 552 kms.