GATE Questions

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Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Given total sum = Rs. 427
And given that 3 times A’s share, 4 times B’s share and 7 times C’s share are all equal.
=> 3A = 4B = 7C
But given
=> A + B + C = 427
Now express A and B in terms of C i.e
=> (7C/3) + (7C/4) + C = 427
=> C = 84

(385/1001) = 5/13

First Number is 5

=> Fourth number is 13 

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Frugal and parsimonious are synonyms of each other and bear the meaning 'being stingy' or 'sparing'. Similarly, joy and ecstasy are related to happiness.

The relation is father and daughter. As we go with male gender, will be son of A. But it is given in the question that B is not the son of A. Hence, te only possile is B is female and B is daughter of A.

Quantity of alohol in the mixture = 40 x 5/8 = 25 lit

Quantity of water = 40 - 25 = 15 lit

According to question,

Required ratio = 20 - 40 x 20100 x 5815 - 40 x 20100 x 38 + 40 x 20100 = 2015 - 3 + 8 = 1 : 1