CAT 2003 Question Paper (Leaked)

If there is only 1 green ball, it can be done in 6 ways
If there are 2 green balls, it can be done in 5 ways.
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If there are 6 green balls, it can be done in 1 way.
So, the total number of possibilities is 6*7/2 = 21

Equate the 2 equations we get value of x = 1 and -1 . Also we notice that there is intersection at x=0 . hence D

Let there only be 2 questions.

Thus there are $$2^{2-1}$$ = 2 students who have done 1 or more questions wrongly, 2$$^{2-2}$$ = 1 students who have done all 2 questions wrongly .

Thus total number of wrong answers = 2 + 1 = 3= $$2^n - 1$$.

Now let there be 3 questions. Then j = 1,2,3

Number of students answering 1 or more questions incorrectly = 4

Number of students answering 2 or more questions incorrectly = 2

Number of students answering 3 or more questions incorrectly = 1

Total number of incorrect answers = 1(3)+(2-1)*2+(4-2)*1 = 7 = $$2^3-1$$

According to the question , the total number of wrong answers = 4095 = $$2^{12} - 1$$.

Hence Option A.

For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .

$$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$$ = x/z + x/y + y/z + y/x + z/y + z/x

Applying AM greater than or equal to GM, we get minimum sum = 6

Take any 12 points.

The maximum number of edges which can be drawn through these 12 points are $$^{12}C_2$$ = 66

The minimum number of edges which can be drawn through these 12 points are 12-1 = 11 as the resulting figure need not be closed. It might be open.

positive integers n in the range $$12 \leq n \leq 40$$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30. 

Let a,d,j,g be the shots put by Ashish, Dhanraj, Ganesh and Jugraj respectively.

According to given conditions we have ,
g = a-8; d + r = 37; j = d + 8; a = 5 + d; a + g = 40

Solving these equations, we have a = 24, d = 19 and j = 27 and r = 18. Hence option A is the correct answer.

We know that the total time taken is 1.5 hrs. Calculating the individual time taken and the adding and then equating to 1.5.

Let R be the radius of the outer-ring road.

$$\frac{\pi*R}{2*30*\pi} + \frac{\sqrt{5}*R}{2*15*\sqrt{5}}$$ = 1.5 solving we get R=30.

Let the radii of 2 circles be R and r respectively such that R=2*r. Triangle $$ON_2E_1$$ and all the other 3 similar triangles form a right angle at the centre. So, using Pythagoras theorem, the value of chords comes out to be $$\sqrt{\ 5}\times\ \dfrac{R}{2}$$ . Hence, the total distance travelled is $$\sqrt{\ 5}\times\ \dfrac{R}{2}$$ + $$0.5\ \times\ R\ \times\ \pi\ $$. The total time required can be calculated by distance/speed, which comes out to be 3.5 * R. Among options, only 105 is an integral multiple of 3.5.