time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
79
20
Prob(choosing two people who are a couple from 6 couples) = 1/6
We have 5 couple left (10 people)
Prob(choosing a person A from 10 people) = 1/10
Prob(choosing 1 person who is not a couple of A) = 1/8
Prob(choosing 4 people that has exactly one couple) = 1/6*1/8*1/10 = 0.0021
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
360 times