CAT Questions

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The device required for the internet connection is a MODEM.

A modem (modulator–demodulator) is a network hardware device that modulates one or more carrier wave signals to encode digital information for transmission.

Relative speed = (72 - 36) x 5/18 = 2 x 5 = 10 mps.
Distance covered in 32 sec = 32 x 10 = 320 m.

The length of the faster train = 320 m.

8 S on SS means 8 Sides on a Stop Sign.

This is one type of Letter Equations which are well known phrases or facts where the key words have been replaced with the first letter of that word. These are often in the form of an equation, which contain a number, an = sign and the rest of the obscured phrase or fact.

Some more of them are ::

90 D in a RA

26 L of the A 

12 S of the Z.

Given that SP = Rs. 12000 - 10%  = Rs. 10,800

Loss% = 4

We know that, C.P = 100/(100 - Loss%) x 100

=> 100/100-4 x 10800

=> 1080000/96

C.P = Rs. 11,250

There were all parrots but six means that six birds were not parrots but only pigeons and koels.

Similarly, number of parrots + number of koels = 6 and number of parrots + number of pigeons = 6.

This is possible when there are 3 parrots, 3 pigeons and 3 koels i.e. 9 birds in all.

Let the runs scored by P, Q, R, S and T be  a, b, c, d and e

From given data,

a + b + c + d + e = 36 x 5 = 180
b + c = 107
Let a = x
e = x-8
d = x-3
now 3x - 11 + 107 = 180
3x = 84
x = 28
e's score = 28-8 = 20

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

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.