CAT 1999 Question Paper

Let the distribution of votes for each of the proposal be as given below.
From the information given, we know that 

a+b+c+d+e+f+g = 78 --- (1)
a+b+e+f = 50  ---- (2)
b+c+f+g = 30 ----  (3)
e+f+g+d = 20 ---- (4) and 
f = 5 --- (5)

We need to find b+e+g+f = ?
In the above equations, (2)+(3)+(4) - (1) implies

(a+b+e+f)+(b+c+f+g)+(e+f+g+d) - (a+b+c+d+e+f+g) = 50+30+20-78 = 22
Or, b+e+g+2f=22.
As, f = 5, it implies that b+e+g+f=17

Let p and q be any two elements of the set A.
For the computation of the GCF of elements of the set A, we can replace both p and q by just the GCF(p,q) and the result is unchanged.

So, for every application of the function h, we are reducing the number of elements of the set A by 1. (In this case two numbers p and q are replaced by one number GCF(p,q)).
Expanding this concept further, the minimum number of times the function h should be called is n-1

By symmetry, it is safe to assume that the polygon ABCD is a square. So, AB = PO. The perimeter of the inner square = 4 AB. The perimeter of the outer circle = $$ 2 \pi \times AB$$

So, ratio = $$ \frac{2 \pi \times AB}{4AB}$$ = $$ \frac{\pi}{2}$$

All of them can be mislabeled in 2 ways: 

Red box  - white label

White box - Red and white label

Red and white box - Red label

Red box - red and white label

White box - Red label

Red and white box - White label

So, we would try the box with the red and white label and if it has a white ball, labelling to the boxes is done as per case 1. If it has a red ball labelling is done as per case 2. 

Note: It's not a good idea to try the white label as if we get a red ball, we can't make out if we are picking from red box or red and white box. Similarly if we try from the box with red label and we get a white ball, again we can't make out if it is coming from the white box or the red and white box. 

Observe the pattern given below.
$$11^2 = 121$$
$$111^2 = 12321$$
$$1111^2 = 1234321$$ and so on.

So, $$11111111^2 = 123456787654321$$

Mrs. Abraham can't be sitting next to him as per the seating arrangement. Wives sit three places away from their husbands.

Mrs. Charlie is sitting to the left of Mr. Abraham. So, she can't be sitting to his right.

Mrs. Elenor is sitting two places to the right of Mrs. Border (and not Mrs. Charlie). So, she can't be sitting right next to Mr. Abraham.

Mrs. Border and Mrs. Dennis are the remaining two wives and each is equally likely to to be sitting to the right of Mr. Abraham.

The distance between Ahmedabad and Baroda is 100 Km
Navjivan express starts at 6:30 am at 50 Km/hr and Howrah expresses starts at 7:00 am at 40 Km/hr.
Distance covered by Navjivan express in 30 minutes (by 7 am) is 25 Km/hr.

So, at 7 am, the distance between the two trains is 75 Kms and they are travelling towards each other a relative speed of 50+40=90 Km/hr.

So, time taken them to meet is 75/90*60 = 50 minutes.

As, Mr. Shah realizes the problem after thirty minutes, time left to avoid collision is 50-30 = 20 minutes

The perimeter of the circle is equal to 2$$\pi $$.

The perimeter of the polygon inscribing the circle is always greater than the perimeter of the circle => L1(13) > 2$$\pi $$

The perimeter of the polygon inscribed in the circle is always less than the perimeter of the circle => L2(13) < 2$$\pi $$

=> $$\frac{L1(13)+2\pi }{L2(17)}$$ > 2

Let EF be the fence.

As the field is in the shape of a square, the straight line face that is put up will cordon a field of the form right angled triangle.

The area of a right angled triangle is maximum when the sides that contain the right angle are equal.

Let the side be a.

$$a^2 + a^2$$ = $$100^2$$ => $$a = 50\sqrt{2}$$

Area = $$\frac{1}{2}*50\sqrt{2}*50\sqrt{2}$$ = 2500 sq m.

From the given passage, it is given that Q>P, R>S.

According to the information Q = 2 when P = 1 or Q = 4 when P = 2.

In first case R = 4 and S = 3 and in the second case R = 3 and S = 1.

In both the instances S is odd.