CAT 1991 Question Paper

We know that x + y + z = T and x + 2y + 3z = R, where
x = number of members belonging to exactly 1 set = 70
y = number of members belonging to exactly 2 sets
z = number of members belonging to exactly 3 sets = 10
T = Total number of members
R = Repeated total of all the members = (40+50+60) = 150
Thus we have two equations and two unknowns. Solving this we get y = 25

So, 25 people belong to exactly 2 clubs.

Let the side of the square which is cut be x.

Volume of the cuboid so formed =$$(10-2x)^2*x$$

Put x = 1, 2, 3 and so on till 5

Maximum volume would be at x = 2

Volume of the cuboid so formed =$$(10-2*2)^2*2 = 72$$

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xy and xyz+xz

The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.

Option a) is the correct answer.

The highest power of 5 in 80! = [80/5] + [$$80/5^2$$] = 16 + 3 = 19

So, the highest power of 5 which divides 80! exactly = 19

13*62 = 806

31*26 = 806

Hence the answer is option b

As 55 and 124 don't have any common factor, and n has to be a minimum integer, Hence, it should be 124 only. So that given equation won't have a remainder.

let's say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also  multiple of 7.
or 2n-4 should be a multiple of 7 
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )

A calculator has two memory buttons, A and B. Value 1 is initially stored in both memory locations. The following sequence of steps is carried out five times:
add 1 to B
multiply A to B
store the result in A
What is the value stored in memory location A after this procedure?

The value in A and B before the start of the process is 1 and 1 respectively.
In 1st iteration,
After step I, A = 1 and B = 2
After steps II and III, A = 1 * 2 = 2, B = 2

In 2nd iteration,
After step I, A = 2 and B = 3
After steps II and III, A = 2 * 3 = 6, B = 3

In 3rd iteration,
After step I, A = 6 and B = 4
After steps II and III, A = 6 * 4 = 24, B = 4

In 4th iteration,
After step I, A = 24 and B = 5
After steps II and III, A = 24 * 5 = 120, B = 5

In 5th iteration,
After step I, A = 120 and B = 6
After steps II and III, A = 120 * 6 = 720, B = 6

Hence, option D is the correct answer.

Considering the centre coin which is denoted by  C1.

The surrounding coins are of equal radius as the centre coin. Since each coin is placed next to each other and considering the total number of coins = n.

The pairs formed by the coins which touch the consecutive coins is given by :

(1, 2) , (2, 3) ........... (n, 1)

The triangles formed by joining the centres of all the consecutive circles at the centre of C1 = 360 degrees.

If we join centres of 2 outer circles with the centre circle, it will make an equilateral triangle.

A single pair forms 60 degrees as it forms an equilateral triangle. Hence $$\frac{360}{n}=\ 60$$

n = 6

Let's say the amount of money Gopal has is 100x rs. With this money he can buy either 50 oranges or 40 mangoes.

Hence, the cost of 1 orange = $$\dfrac{100x}{50}$$ = Rs. 2x

Similarly, the cost of 1 mango = $$\dfrac{100x}{40}$$ = Rs. 2.5x

Taxi fare = 0.1*100x = 10x

Remaining money with Gopal = 90x
Cost incurred in buying 20 mangoes = 2.5*20 = 50x

Hence, the amount of money left with Gopal = 100x - 10x - 50x = 40x

Therefore, we can say that with this much money he can but a maximum of $$\dfrac{40x}{2x}$$ = 20 oranges.