CAT 1997 Question Paper

Anshuman travels at a minimum speed by car over A, so the time taken = $$\frac{2}{40}$$*60= 3 min
Anshuman travels stretch B at the fastest speed, so the time taken = $$\frac{2}{50}$$*60= 12/5 min
So in order to break the record, he has to cover the three stretches in less than 10 min
10-(3+$$\frac{12}{5}$$) = $$\frac{23}{5}$$
So the time taken to cover stretch C should be less than $$\frac{23}{5}$$ min.
Let x kmph be the speed at which he covers stretch C
$$\frac{2}{x}$$*60 < $$\frac{23}{5}$$
x > 26.08 kmph but the maximum speed at which he can cover stretch C is 20 kmph.
Hence it is not possible for him to break his previous record.
C is the correct answer.

Total time taken by Mr. Hare to complete the race = 50% more of (10 min.) = 15 min.= $$\frac{1}{4}$$ hr.
Or $$\frac{1}{4} = \frac{2}{40} + \frac{2}{40} + \frac{2}{v}$$ ( i.e. complete time for strech A,B and C Where v is velocity at strech C)
Or  v= 13.3 km./hr

Average speed to cover complete race = 20 km/hr = $$\frac{2+2+2}{t_1 + t_2 + t_3}$$ (Where $$t_1$$ is time taken to cover the distance A, $$t_2$$ is time taken to cover the distance B and $$t_3$$ is time taken to cover the distance C)

So total time =$$ t_1 + t_2 + t_3 = 18$$ min.

Avg. speed for first two streches = $$\frac{4}{t_1 + t_2}$$

Avg. speed for last strech = $$\frac{2}{t_3}$$

Given: $$\frac{4}{t_1 + t_2} = 4 \times \frac{2}{t_3}$$

Or $$2t_1 + 2t_2 = t_3$$

Or $$t_3 = 12$$ min.

So $$V_3 = \frac{2}{12} \times 60$$ = 10 km/hr

As data regarding weights of people is not given, hence we can't determine the avg. weight of people in group D 

As data regarding weights is not available, hence we can't say about the increment or decrement in group A an B
But avg. weight for overall class will remain same as no student went out or came inside to make any changes in total weight or total number of students.

Except option C, all are correct because if everyone has same weight than avg. weight of all groups remain same.
So avg. weight of D will be equal to avg. weight of A.

Let's say he scored marks as $$10x,9x,8x,7x,6x$$ or total of $$40x$$ which is 60% of total maximum marks(T).

$$\frac{T \times 60}{100}=40x$$

So T (total maximum marks)=$$\frac{400x}{6}$$

Or Individual max. marks = $$\frac{T}{5}=\frac{80x}{6}$$

Passing marks =50% of individual max. marks =$$\frac{40x}{6}=6.66x$$

Hence he scored more than passing marks in four subjects as $$10x,9x,8x$$ and $$7x$$ and failed in one subject as scoring $$6x$$ marks which is less than passing marks of $$6.66x$$

Chariman, Vice-Chairman and the director can be made as a group such that Chairman sits between the Vice-Chairman and the director. This group can be formed in 2 ways.

Each of the remaining 7 directors and the group can be arranged in 7! ways.

=> Total number of ways = 2 * 7!.

$$\log_{2}{\log_{7}{(x^2 - x+37)}}$$ = 1

$$\log_{7}{(x^2 - x+37)}$$ = $$2$$

$$(x^2 - x+37)$$ = $$7^{2}$$

Given eq. can be reduced to $$x^2 - x + 37 = 49$$

So x can be either -3 or 4.

Let's say cost price is 100
gain = 14.28
selling price = 114.28
Marked price = x(say)
So $$x- \frac{11.11x}{100} = \frac{8x}{9} = 114.28$$
Or $$x = 128.52$$
So marked price is 28.52% more than cost price.