AIEEE Questions

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One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

Given that,

Members in football team = n(f) = 10

Members in science club = n(s) = 14

Members in both football team and science club = n(f ^ s) = 5

Then, Members in only football team = n(F) = 10 - 5 = 5

Members in only science team = n(S) = 14 - 5 = 9

Hence, members in only football or science team = n(FUS) = n(F) + n(S) - n(f^s) 

= 9 + 5 - 5

= 9.

Hence, many students belong to only science club team or football team = 9.

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

Therefore, the number of boys in the class = 8.

Let A worked for x days.

x/21 + 4/28 = 1  => x = 18

A worked for 18 days. So, A can complete the remaining work in 18 - 4 = 14 days.

We know that Time is inversely proportional to Efficiency

Here given time ratio of P & P+Q as

P/P+Q = 150/100 = 3:2

Efficiency ratio of P & Q as

P:Q = 2:1 ....(1)

Given Efficiency ratio of Q & R as

Q:R = 3:1 ....(2)

From (1) & (2), we get

P:Q:R = 6:3:1

P alone can finish the work in

22.5(3+1)/6 = 15 days.

Better is compared with worst and slower must be compared with quickest since fast is the extreme of it.

Hence, Better is to worst as slower is to quickest.

From the given diagram,

Tank 10 will be first filled because from tank 1 it will go through tank 4 and tank 10.