GATE Questions

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In the given puzzle the logic is

2 x 1 + 1 = 2 + 1 = 3

3 x 2 + 2 = 6 + 2 = 8

4 x 3 + 3 = 12 + 3 = 15

5 x 4 + 4 = 20 + 4 = 24

6 x 5 + 5 = 30 + 5 = 35

Similarly,

7 x 6 + 6 = 42 + 6 = 48

7 = 48.

We need to find the total cost to send r messsages, r > 1000.

The first 1000 messsages will cost Rs.p each (Or) 

The total cost of first 1000 messsages = Rs.1000p

The remaining (r - 1000) messsages will cost Rs.q each (Or)

The cost of the (r - 1000) = Rs.(r - 1000)y

Therefore, total cost = 1000p + rq - 1000q

= 1000(p - q) + qr

The given digits are 1, 2, 3, 5, 7, 9  

Number of even numbers = ⁵P₃ = 60.

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

 0.48z = 1.40

 z = 140/18 = 35/12 = 2.9 (approx.)

We know that,

The equivalent K =  L1 + L2L1K1+L2K2

where L1 & L2 are plate1 & Plate2 thicknesses, K1 & K2 are coefficient thermal conductivity of plate1 & Plate2 respectively.

But here L1 = L2 = L (say)

Then K = L+L/(L/K1 + L/K2)  =  2K1K2K1+K2

Let the total Profit be Z.
Ramu's profit share is 2/3 of profit (i.e 2Z/3)
Ravi's profit is (Z-2Z/3) =Z/3
Hence the profit ratio is , Ravi : Ramu = Z/3 : 2Z/3 = 1:2

Let total Capital be Rs.X and Ramu has contributed for Y months.Since Ramu's profit share is 2/3, his invest share will be 2/3 in capital.

Ravi's invest for 16 months / Ramu's invest for Y months = Ravi's profit share / Ramu's profit share
i.e. (X/6 x 16) / (2X/3 x Y) = 1/2

Solving the above equation, we get Y = 8.
So Ramu's money has been used for 8 months.

Ozone molecule has one single bond and one double bond.