GATE Questions

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Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264

Arithmatic mean = sum/members 

=> (1x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7) / (1 + 2 + 3 + 4 + 5 + 6 + 7)

=> 140/28 = 5

Let first part is 'K' and second part is 'L'
then K + L = 50000---------eq1
Total profit = profit on x + profit on y
7000 = (K x 10 x 1)/100 + (L x 20 x 1)/100
70000 = K + 2L-----------------------------------eq2
70000 = 50000 + L
so L = 20000 then K = 50000 - 20000 = 30000
first part on 10% is = Rs.30000

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

Second is a bony protective covering for the first. Here pericardium protects Heart and Cranium protect Brain.

Hence, Heart : Pericardium :: Brain : Cranium.

Let his monthly salary be 'P'

20% on children's education = 20P100 = P5

50% on food and transport = 50P100 = P2

1000 on rent

=> $$\begin{gathered} P - P5 - P2 - 1000 = 2000 \\ => 3P = 30000 \\ => P = 10000 \end{gathered}$$

Therefore, his monthly salary = Rs. 10,000