CAT 1990 Question Paper

Let the number of gold bars be 7a and the number of silver bars be 27a.
The total value of the gold and silver bars is 500000 - 225000 = 275000
Therefore, 4000*7a + 1000*27a = 275000
Or, 55000a = 275000
Or, a = 5

Therefore the number of silver bars with Ghosh Babu is 27a = 135

After my first move of retaining right board will be like

Now second person will force me to retain with the lowest value possible i.e. 2 or 3
So if he retains lower half and i retain with left one (as mentioned) then he can only force me to pick 3.
Or if he retains upper half and i retain with left one(as mentioned) then he can retain upper half so i will be having minimized profit of 2.
Hence second person will have moves as "Retain (upper half) and retain (upper half)"

After second move i will be having from a quarter cell of board to make a move.

Since left quarters have cells with values 1, Hence i will choose to retain right in my first move so that second person can force my profit to be either 2 or 3 not 1.

So now after my first move of retaining right, we have two quarters

Now intelligently he will make a move to retain with lower half so that at fourth move i can retain with choices of 2 and 3 not 2 and 4.

After retaining right in first move, second person will have 2 quarters to choose in which minimum value is 2 and 3.
If he retains upper half, i will retain right as to maximize my profit of either 4 or 7.
If he retains lower half, i will retain left as to maximize my profit of either 3 or 8.
Hence whatever he retains, i won't have a profit less than 3.

If roots of given equation are reciprocal to each other than product of roots should be equal to 1.
i.e. $$\frac{6}{a} = 1$$
hence a=6

Time difference, when second time car's engine failed at a distance of 30 km., is of 9 min.
Hence putting this in equation:
$$\frac{12}{\frac{4v}{5}} - \frac{12}{v} = \frac{9}{60}$$ hr. (Because difference of time is considered with extra travelling of 12 km. in second case)
We will get $$v (velocity) = 20$$ km/hr.
Now for distance $$ \frac{d-18}{16} + \frac{18}{20} - \frac{d}{20} = \frac{45}{60}$$ hr. (As car is 45 min. late after engine's faliure in first case)
So $$d$$ = 78 km. 
Hence none of these will be our answer.

Money will be distributed in the ratio of work done in an hour.
i.e. $$\frac{1}{6} : \frac{1}{8} : \frac{1}{15}$$
or 20:15:8
Hence part of A will be = $$\frac{20}{43} \times 94.6 = 44$$
part of B will be = $$\frac{15}{43} \times 94.6 = 33$$
part of C will be = $$\frac{8}{43} \times 94.6 = 17.60$$

let's $$l_s$$ is length of slower train and $$l_f$$ is length of faster train.
So according to second condition when two trains are moving in same direction
$$l_s = v_{fs} \times t$$  (where $$v_{fs}$$ is relative velocity of faster train w.r.t. slower train and t is time taken to cross it)
or $$l_s = \frac{(60-50) \times 5}{18} \times 18$$ = 50 meter
Only option which has length of slower train as 50 is C.
 

According to given question $$S_{50}$$ will have 50 terms 
And its first term will be 50th number in the series 1,2,4,7,.........$$T_{50}$$
$$T_1 = 1$$
$$T_2 = 1+1$$
$$T_3 = 1+1+2$$
$$T_4 = 1+1+2+3$$
$$T_n = 1+(1+2+3+4+5....(n-1))$$
       = $$1+\frac{n(n-1)}{2}$$
So $$T_{50} = 1+1225 = 1226$$
Hence $$S_{50} = (1226,1227,1228,1229........)$$
And summation will be = $$\frac{50}{2} (2\times 1226 + 49 \times 1 ) = 62525$$

Side of first square = 8cm. 

Side of second square made by joining mid-points of first square = $$\frac{8}{\sqrt2}$$

Similarly side of third square =  $$\frac{8}{\sqrt2\times\sqrt2}$$ and so on.

Now summation of areas will be $$8^2+(\frac{8}{\sqrt2})^2+(\frac{8}{\sqrt2\times\sqrt2})^2$$ .........

or $$8^2 ( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}.....)$$

or $$64 \times (\frac{1}{1-\frac{1}{2}})$$ (As we know sum of an infinite G.P. is $$\frac{a}{1-r}$$ where a is first term and r is common ratio)

or 128 sq. cm.