CAT Questions

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Assume Hebah has Rs. M

Since 25% more money at Poonam

=> Money at Poonam = M + (25% of M)

=> Money at Poonam = M + 0.25M = Rs. 1.25M

Money with Navaneet is thrice the money with Poonam,

=> Money at Navaneet = 3(1.25M) = Rs. 3.75M

Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M

But given the average of the money is Rs. 350

=> 6M/3 = 350

=> 2M = 350

=> M = 350/2 = Rs. 175

=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.

In computer network nodes are

* the computer that terminates the data

* the computer that originates the data

* the computer that routes the data

QUEUE is the only word with 5 letters but sounds like one letter 'Q'.

 Given A = log321875 and B = log2432187

B = log352187 = log321875

=> A

Therefore, A = B

Let no. of boys be 'B' and no. of girls be 'G'

Here from given data,

B + G = 1650 .....(1)

B - G = 400 .....(2)

From (1) & (2), we get

2B = 2050

=> B = 1025

=> G = 1650 - 1025 = 625

Hence, the % of girls = 625 x 100/1650

=> 38% (approx).

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

Computer memory - ROM is essentially empty. ROM is a Read Only Memory which cannot be retrieve.

Time taken by both Meghana and Ganesh to work together is given by =

t1 x t2

Where

$$\begin{gathered} t1 = 32 hrs \\ t2 = 12 12 hrs = 252 hrs \end{gathered}$$

Therefore, time took by both to work together = 

32 x 252 = 16 x 25 = 4 x 5 = 20 hrs