CAT Questions

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Given the salaries are in the ratio of 4 : 5 : 6

Now the salaries are increased by 50%, 60% and 50% respectively

The New ratio of salaries will be

4 x (150/100) : 5 x (160/100) : 6 x (150/100)

= 6 : 8 : 9

The new ratio of salaries of Akhil, Arun and Karthik is 9 : 8 : 6

Required number of signals = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

= 5 + 20 + 60 + 120 + 120 = 325

Let a, b and c be the amounts invested in schemes X, Y and Z respectively. Then,

As we know:

Simple interest (S.I.) = PTR/100

(a × 10 × 1/100) + (b × 12 × 1/100) + (c × 15 × 1/100) = 3200

= 10a + 12b + 15c = 320000 .........(1)

Now, c = 240% of b = 12b/5 .........(2)

And, c = 150% of a = 3a/2 => a = 2/3 c = (2 × 12)b/(3 × 5) = 8b/5 .......(3)

From (1), (2) and (3), we have

16b + 12b + 36b = 320000 => 64b = 320000 => b = 5000

∴ Sum invested in Scheme Y = Rs.5000.

Let filling capacity of the pump be 'F' m3/min

Then, the filling capacity of the pump will be (F + 10) m3/min

From the given data,

$$\begin{gathered} 2400F - 2400F+10= 8 \\ ⇒ 2400 F+10-FF(F+10) =8 \\ ⇒ F2 + 10F - 3000 = 0 \\ ⇒ F = 50 m3/min \end{gathered}$$

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

Here total distance = 154 + 101 = 255 mts
Given speed = 120 kmph = 120x5/18 m/sec
Hence, Time = d/s = 255x18/120x5 = 7.65 sec

There are two kingdoms of bacteria. They are

1. Eubacteria  and 

2. Archaea

The members of these two kingdoms appear similar in shape and appearance, even under the extreme magnification of the electron microscope . However, they are very different from each other in a number of molecular and biochemical aspects. It is these differences that have resulted in the microorganisms being grouped into separate kingdoms.

For example, eubacteria contain the rigid, stress-bearing network known as the peptidoglycan . The only exceptions are the bacteria from the genera Mycoplasma and Chlamydia. Archaebacteria do not contain peptidoglycan. Instead, they contain a different structure that is called pseudomurein.

The common bacteria belongs to Eubacteria kingdom.

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

5 letters which can be arranged in  5P5=5!

Vowels A,I = 3!2!

No.of arrangements = 5! x 3!2!=360