CAT 1990 Question Paper

If the number of mynahs is 7 or more and if 7 birds escaped, there is a possibility that no pigeon escaped. But the bird-keeper was able to conclude that atleast one pigeon escaped. So, the number of mynahs is less than 7.

From the options, we can see that only option a) violates this condition.

$$2^{60}$$ or $$4^{30}$$ when divided by 5
So according to remainder theorem 
remainder will be $$(-1)^{30}$$ = 1.

Take 64 for example.

$$\sqrt{64}$$ = 8

$$\sqrt{8}$$ = 2.828

$$\sqrt{2.828}$$ is close to 1.7

So, we can see that the result is tending towards 1.

Given series can be written as:

$$\sum_{n=1}^{100} (\frac{1}{n\times (n+1)})$$

or $$\sum_{n=1}^{100} (\frac{(n+1)-n}{n\times (n+1)})$$

or $$\sum_{n=1}^{100} (\frac{1}{n} - \frac{1}{n+1})$$

After putting values of n from 1 to 100, all terms will cancel out, only first and last terms will be there

i.e. $$1-\frac{1}{101}$$

or $$\frac{100}{101}$$

$$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{4}{1-x^4}+\frac{4}{1+x^4}$$
or $$\frac{8}{1-x^8}$$

Putting value of x=0 or 1 and solving all four options,
We will find that only option A satisfies the equation with both values, hence answer will be A.

Total ways when all 6 boxes have only black balls = 1
Total ways when 5 boxes have black balls = 2
Total ways when 4 boxes have black balls = 3
Total ways when 3 boxes have black balls = 4
Total ways when 2 boxes have black balls = 5
Total ways when only 1 box has black ball = 6
So total ways of putting a black ball such that all of them come consecutively = (1+2+3+4+5+6) = 21

1. x=1 ; y=2

2. x=2 ; y=3

3. x=6 ; y=4

4. x=24 ; y=5

Hence when y=5 , x will be 24

Let's say total products maufactured by M1, M2 and M3 are 100.

So M1 produced 40, M2 produced 30 and M3 produced 30

Defective pieces for M1 = $$\frac{120}{100}$$

Defective pieces for M2 = $$\frac{30}{100}$$

Defective pieces for M3 = $$\frac{150}{100}$$

So total defective pieces are $$\frac{150+30+120}{100}$$ = $$\frac{300}{100}$$ = 3% of total products.

$$x*x < y*y$$
or $$x + 0.5x - x^2 < y + 0.5y - y^2$$
$$y^2 - x^2 + 1.5x - 1.5y < 0$$
$$(y - x)(y + x) - 1.5 (y - x) < 0$$
$$(y - x)(y + x -1.5) < 0$$
$$(x - y)(1.5 - (x + y)) < 0$$
Now there will be two possibilities
$$x < y$$ and $$(x + y) < 1.5$$ ...........(i)
or $$x > y$$ and $$(x + y) > 1.5$$ ............(ii)
Among all options only option B satisfies (ii).
Hence, option B is the correct answer.