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.
Divide 30*60 seconds by LCM of Numbers =15
15km/hr
2cm b)
Slice the cake
Spade 13/52 = 1/4
King 4/52 = 1/13
Spade OR King 13/52 + 4/52 – 1/52 (because the King of Spades is in both groups) = 16/52 = 4/13
consider the tank capacity as 90 litres.
to fill the tank in 3 hrs first pipe must flow at a rate of 30 litres/hr
2nd pipe has to flow at a rate of 45 litres/hr
if two pipes are opened at a time the flow rate will become 75 litres/hr
90/75= 6/5 = 1.2= (i.e) 1hr and 12 minutes 1 1/5 hr
No loss no gain.
first fill 3quart pail and pour in 5quart pail.fill again 3
quart pail and fill remaining 2 quarts in 5 quart
pail.hence 1 quart remains in 3quart pail.empty 5quart pail
and pour 1 quart from 3 quart pail. also add another 3
quart from 3 quart pail. hence 1 + 3 quarts = 4 quarts.
3000
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
Correct Deven,
democrate=1008
each pencil 100/25 = 4
each book 100/15 = 6.66
100 – 15 = 85
85-5*4 = 65
65/6.66 = 9
he can buy 9 books that is not in given options currently so
answer is NONE
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