CAT 1997 Question Paper

$$7^{(3^2)} = 7^9$$

$$(7^3)^2 = 7^6$$

So $$7^{(3^2)} > (7^3)^2$$

Let a be the number who watch only one channel, b be the number who watch only 2 channels and c be the number who watch all channels.

a+b+c = 100

a+2b+3c = 80+22+15  =117

Subtracting both equations,

b+2c = 117-100 = 17

Maximum c occurs when b = 0

2c = 17

c = 8.5

Applying AUBUC formula

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.

100 = 80+22+15-(10+5+x)+y

x-y = 2

Hence only 2% people watch BBC and CNN only.

Applying AUBUC formula

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.

100 = 80+22+15-(10+5+x)+y

x-y = 2

We cannot find the exact value of y.

Hence, the answer is "cannot be determined".

He earns x% on first 2000 and y% on rest of his income.

So on 4000 rs. , he will earn as follows:

$$2000 \frac{x}{100} + 2000 \frac{y}{100}$$ = 700

Or $$x+y = 35$$

Similarly on 5000 rs. ,he will earn 900 as follows:

$$2000 \frac{x}{100} + 3000 \frac{y}{100} = 900$$

Or $$20x + 30y = 900$$

On solving above equations, we will get value of x = 15

Since ACBD is cyclic quadrilateral with diagonals as AB = 15 ad CD.
So area = $$\frac{1}{2} \times (AB) \times (CD)$$                  eq.(1)
Triangle ACB and ABD is right angled triangle
Let's say angle CBA = $$\theta$$
so $$CD = 2 \times (9 sin\theta)$$  (Where  $$sin\theta$$ = $$\frac{12}{15}$$)
CD = $$\frac{72}{5}$$
Putting values in eq. (1) , we will get area = 108 sq.cm.
Hence answer will be D)

Let's say three numbers are (a-2),a,(a+2)

So 3(a-2) = 2(a+2) - 3

a=7 and a+2 = R = 9

Best approach to these type question remain assuming values and checking

Case - 1.x=8 ; y=7 ; z = 5

la (x,y,z) = 12

le (x,y,z) = 2

ma (x,y,z) = 7

Case -2: Let us try to find values for which la(x,y,z) and le(x,y,z) would be equal. In such a case, ma(x,y,z) would also be the same.

So max(x-y,y-z)= min(x+y, y+z)

As x>y>z>0, min(x+y, y+z) = y+z

So max(x-y, y-z) =y+z

Either x-y=y+z or y-z = y+z

So x=2y+z or z=0

But z cannot be 0 according to given condition.

So, x=2y+z

Let us assume y=2 and z=1

So x=5

la (x,y,z)= 3

le (x,y,z) = 3

ma (x,y,z)= 3

based on these two cases we can deduce that non of the given options holds true.

So the correct option to choose is D - None of these.

$$Ma(10, 4, le((la10, 5, 3), 5, 3))$$
Or $$Ma(10, 4, le(8, 5, 3))$$
Or $$Ma (10,4,3)$$
Or $$\frac{1}{2} (6+7) = 6.5$$

Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9