GATE Questions

---

Given that, upstream distance = 7 kms

Upstream speed = 7/42 x 60 = 10 kms

Speed of the stream = 3 kmph

Let speed in still water = M kmph, then

Upstream speed = M - 3 = 10

=> M = 13 kmph.

From the assumption and trial and error method,

last three digits sum must leave the same digit as the answer. It is possible only if the answer must be units digit number leaving tens digit as carry.

And the possible 3rd digit is 5. Since 5 + 5 + 5 = 15 

Now, 2nd step is  1 + x + x + x = 5

=> 3x = 4

It is possible with x = 8. 8 + 8 + 8 + 1 = 25 leaving 2 as carry. 

Now, 3rd step is 2 + x + x + x = 5 => x = 1

Hence, the three dgits are 1, 8, 5.

185 + 185 + 185 = 555. 

One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20

Aru efficiency = 2/3 of (Raghu + Sam)

Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days. 

Given share of L is 2/7 th of profit. Then, their profits are divided in to 5:2 ratio.

Ratio of K and L is

16000×8x×4=52

x = Rs. 12,800.

The intersection of a column and a row in a spread sheet is called as a cell or a bon.

Let his present earnings be Rs. p

ATQ,

(125/100) x (96/100) x (125/100) x (96/100) x (125/100) x p = 72000

p = Rs. 40,000.

Let distance between A & B = d km
Let speed in still water = x kmph
Let speed of current = y kmph

from the given data,
d/x = 2

From A) we get d
From B) we get d/x+y
From C) we get y

So, Any one pair of A and B, B and C or C and A is sufficient to give the answer i.e, the speed of upstream.

Here the number of cogs is inveresly proportional to number of revolutions

=> More cogs less revolution
=> 6 : 14 :: x : 21
=> (21x6)/14 = 9