CAT 2001 Question Paper

Since the product is constant, 

(a+b+c+d)/4 >= $$(abcd)^{1/4}$$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$$(a+1)(b+1)(c+1)(d+1)$$ 
= $$1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$$
We know that $$abcd = 1$$
Therefore, $$a = 1/bcd, b = 1/acd, c = 1/bda$$ and $$d = 1/abc$$
Also, $$cd = 1/ab, bd = 1/ac, bc = 1/ad$$
The expression can be clubbed together as $$1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$$
For any positive real number $$x$$, $$x + 1/x \geq 2$$
Therefore, the least value that $$(a+1/a), (b+1/b)....(ad+1/ad)$$ can take is 2.
$$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$$
=> $$(a+1)(b+1)(c+1)(d+1) \geq 16$$
The least value that the given expression can take is 16. Therefore, option C is the right answer.

Let the time taken working together be t.
Time taken by Arnold = t+1
Time taken by Asit = t+6
Time taken by Afzal = 2t
Work done by each person in one day = $$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}$$
Total portion of workdone in one day $$=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{2-1}{2t}$$
$$2t+7=\frac{(t+1)\cdot(t+6)}{2t}$$
$$3t^2-7t+6=0 \longrightarrow\ t=\frac{2}{3} $$or $$t=-3$$
Therefore total time = $$\frac{2}{3}$$hours = 40mins

Alternatively,
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
From the options, if time $$= 40$$ min, that is, $$t = \frac{2}{3}$$
LHS = $$\frac{3}{5} + \frac{3}{20} + \frac{3}{4} = \frac{(12+3+15)}{20} = \frac{30}{20} = \frac{3}{2}$$
RHS = $$\frac{1}{t}=\frac{3}{2}$$
The equation is satisfied only in case of option C
Hence, C is correct

We know that area = 0.5*h*10 ; we get h = 16=oa.

Using pythagoras we find

$$ob^2+oa^2 = ab^2$$

$$ob^2+16^2 = 20^2$$

ob = 12

Using pythagoras in triangle oam we get

am = $$\sqrt {2^2 + 16^2 } $$ = $$\sqrt {260}$$. 

It is given that in a Fibonacci sequence, from the third term on wards, each term in the sequence is the sum of the previous two terms in that sequence.

Let x and y be the 1st and 2nd term respectively.

3rd term = x+y

4th term = x+2y

5th term = 2x+3y

6th term = 3x+5y

7th term = 5x+8y

We know that difference of the squares of 6th and 7th terms is 517 = 47*11 .

And $$a^2-b^2=(a+b)(a-b)$$.

Applying above formula we get (8x+13y)(2x+3y) = 47*11.

So only possible way is (8x+13y)=47 and

2x+3y=11 .

Solving we get x=1 and y=3 .

Using the concept that every term is the sum of the previous two terms, as used in the beginning of the solution, we get 10th term as 21x+34y, which gives 10th term as 123.

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg. 

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

Distance between A-B , A-C, C-B is 180, 120 and 60 km respectively.

Let x be the distance from A where the 2 trains meet.

According to given condition we have

$$\frac{x}{70}=\frac{60}{50} + \frac{1}{4} + \frac{120-x}{50}$$.

Solving the equation we get x around 112 km.

Since the number starts from 1 if there are n numbers then initial average = $$\dfrac{n+1}{2}$$.

Average of N natural number can be either an integer {ab} or {ab.50} type. For example average of first 10 number = 5.5 whereas the average of first 11 natural numbers is 6. 
Even if we erased the largest number change in average will be always less than 0.5. 
Here we are given the average is 602/17 or 35$$\frac{7}{17}$$ Hence we can say that average must have been 35.5 or 35 before. 
Case 1: If the average was 35.5 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35.5 = $$\dfrac{N+1}{2}$$
N = 70. 
Sum of these 70 numbers = 70*71/2 = 35*71 = 2485.
Sum of the 69 numbers which we are left with after removing a number = (602/17)*69 = 2443.41. Which is not possible as the sum of natural numbers will always be an integer. Hence, we can say that case is not possible. 

Case 1: If the average was 35 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35 = $$\dfrac{N+1}{2}$$
N = 69. 
Sum of these 69 numbers = 69*70/2 = 35*69 = 2415.
Sum of the 68 numbers which we are left with after removing a number = (602/17)*68 = 2408. 
Hence, we can say that the erased number = 2415 - 2408 = 7. 

Let angle EAC = x, so angle AEC = x and angle ACE = 180-2x
Let angle FBC = y, so angle BFC = y and angle BCF = 180-2y
So, angle ACB = 180-(180-2x+180-2y) = 2(x+y) - 180
x+y = 180 - 40 = 140
So, angle ACB = 280 - 180 = 100

Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $$x^2/100 * P = 441$$
Let the new price be N => $$P - x^2/100 * P = N$$
Price after the second cycle = $$N - x^2/100 * N$$ = 1944.81
=> $$(P - x^2/100 * P)(1 - x^2/100) = 1944.81$$
=> $$(P - 441)(1 - 441/P) = 1944.81$$
=> $$P - 441 - 441 + 441^2/P = 1944.81$$
=> $$P^2 - (882 + 1944.81)P + 441^2 = 0$$
=> $$P^2 - 2826.81P + 441^2 = 0$$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25

Let x be the required distance.

Let a,b,c be speed of the A,B and C respectively.

From the given conditions we have, 

$$\frac{a}{b}=\frac{x}{x-12}$$ and $$\frac{a}{c}=\frac{x}{x-18}$$ and $$\frac{b}{c}=\frac{x}{x-8}$$ . From first 2 equations we can deduce $$\frac{b}{c}=\frac{x-12}{x-18}$$. 

$$\frac{b}{c}=\frac{x-12}{x-18} = \frac{x}{x-8}$$

x = 48 satisfy the equation.