CAT 2006 Question Paper

$$2^{0.7x} * 3^{-1.25y} = 8\sqrt{6}/27$$ => $$2^{0.7x} * 3^{-1.25y}$$ = $$2^{3.5} * 3^{-2.5}$$

=> 0.7x = 3.5 => x = 5

=> -1.25y = -2.5 => y = 2

$$4^{0.3x} * 9^{0.2y} = 8*81^{1/5}$$ => $$2^{0.6x} * 3^{0.4y}$$ = $$2^3 * 3^{0.8}$$

=> 0.6x = 3 => x = 5

=> 0.4y = 0.8 => y = 2

=> (5,2) is the solution.

y = 38 => x = 1

y = 36 => x = 2

...

...

y = 14 => x = 13

y = 12 => x = 14 => Cases from here are not valid as x > y.

Hence, there are 13 solutions.

Let the areas be labelled as shown in the diagram above.

The number of people corresponding to "none of the three months" is 24. So, H is 24.

Only September is 18. So, G = 18

September but not August is 23. So, G + D = 23.

Hence, D = 23 - 18 = 5.

We know that September and July is 8. So, D + E = 8

This implies E = 3.

September = 28. So, D + E + F + G = 28.

So, F = 28 - 5 - 3 - 18 = 2.

July and August = 10.

So, B + E = 10.

E = 3. So, B = 7.

July = 48.

So, A + B + D + E = 48

A = 48 - 7 - 5 - 3 = 33.

There are 100 people in total. So, C = 100 - A - B - D - E - F - G - H = 100 - 33 - 7 - 5 - 3 - 2 - 18 - 24 = 8

So, number of people who read exactly two consecutive issues = (July & August) + (August & September) = B + F = 7 + 2 = 9

Maximum sum of the four numbers <= 384=99+97+95+93
384/10 = 38.4
So, the perfect square is a number less than 38.4
The possibilities are 36, 25, 16 and 9
For the sum to be 360, the numbers can be 87, 89, 91 and 93
The sum of four consecutive odd numbers cannot be 250
For the sum to be 160, the numbers can be 37,39,41 and 43
The sum of 4 consecutive odd numbers cannot be 90
So, from the options, the answer is 41.

For a normal graph with y and x-axis, the equation of the line passing through the origin is y =mx where m is the slope of the line.

m is +ve if the angle made by the line with the x-axis is < $$90^{\circ\ }$$

$$\therefore\ $$ The equation of the line in the given graph would be y-x = k( y+ x) since the axes are y-x and y+x and the line is passing through the origin.

k > 1 because the angle is greater than 45$$^{\circ\ }$$

$$y=\dfrac{x\left(k+1\right)}{1-k}$$

Since k>1

Therefore y<0 for x>-1 and y>0 for x<-1

Option d correctly satisfy this condition

The nth term is a + (n-1)d
1000 = 1 + (n-1)d
So, (n-1)d = 999
999 = 3^3 * 37
So, the number of factors is 4*2 = 8
Since there should be at least 3 terms in the series, d cannot be 999.
So, the number of possibilities is 7

Try to solve this type of questions using the options.

Subsitute 0 first => We ger -2 <=0, which is correct. Hence, 0 must be in the solution set.

Substitute 8 => 4 + 2 - 2 <=0 => 6 <= 0, which is false. Hence, 8 must not be in the solution set.

=> Option 1 is the answer.

The minimum value is obtained when 2x+1 = 3-4x => 6x = 2 => x = 1/3
So, f((x) = 2*1/3 + 1 = 5/3

Let the distance be D.
Time taken by Arun = D/30
Time taken by Barun = D/40
Now, D/40 = D/30 - 2
=> 3D = 4D - 240
=> D = 240
Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr
Time Kiranmala takes to cover 240 km = 240/60 = 4 hr
So, Kiranmala has to start 4 hours after Arun.

Let the number be xy
10y + x = 10x + y + 18
=> 9y - 9x = 18
=> y - x = 2
So, y can take values from 9 to 4 (since 3 is already counted in 13)
Number of possible values = 6