CAT 1996 Question Paper

Akshay can complete 1600 - 128 = 1472 m and Bhairav can completer 1600 m in the same time.

Bhairav can complete 100 m and Chinmay can complete 96 m in the same time.

=> Bhairav can complete 1600 m and Chinmay can complete 1536 m in the same time.

=> Akshay can complete 1472 m and Chinmay can complete 1536 m in the same time.

1.5 miles => 2400 m

Distance travelled by Akshay by the time Chinmay completes 1.5 miles = $$\frac{1472}{1536}*2400$$ = 2300 m

=> Akshay lost by 100 m, which is $$\frac{1}{16}th$$ of a mile.

Fraction of A in contained 1 = $$\frac{5}{6}$$

Fraction of A in contained 2 = $$\frac{1}{4}$$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($$\frac{5}{6}$$) + (1-x)($$\frac{1}{4}$$) = $$\frac{1}{2}$$

$$\frac{7x}{12}$$ = $$\frac{1}{4}$$

=> x = $$\frac{3}{7}$$

=> Ratio = 3:4

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

(Where x is distance between A and B; $$\frac{\frac{3x}{5}}{3a}$$ = time taken to cover the distance with speed 3a ; $$\frac{\frac{2x}{5}}{2b}$$ = time taken to cover the distance with speed 2b; $$\frac{2x}{5c}$$ = time taken to cover the distance x from B to A then return.)

$$\frac{\frac{3x}{5}}{3a} + \frac{\frac{2x}{5}}{2b} = \frac{2x}{5c}$$

Or $$\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$$

Total time taken to reach at D:

$$\frac{12}{x} + x + \frac{12}{2x} + 2x + \frac{12}{4x} = 16$$

Or $$3x^2 - 16x + 21 = 0$$

From the options we can see that only, $$x$$ = 3hr satisfies the equation. Thus, A is the right choice.

Total matches played = 17+3 = 20

Total matches = 20*$$\frac{3}{2}$$ = 30

Number of wins required = 75% of 30 = 22.5 = 23 wins

23-17 = 6 more wins are required out of 10 matches to maintain 75% win recor which means there would be 4 losses.

let's say price of maruti car is x rs.
Sales = y
revenue = xy
Changed price = 1.3x
changed value of sales = 0.8y
new revenue = 1.04 xy
Percentage change in revenue = 4%

For points to be coincident, value of determinant should not be equal zero, so that they have a unique value of system.

Here value of determinant is not equal to zero, simultaneously not any two lines are parallel or perpendicular.

So system has a unique value

Hence points are coincident.

Selecting 3 girls from 5 girls can be done in $$^5C_3$$ ways => 10 ways

Each of the boys may or may not be selected => 2 * 2 * 2 * 2 = 16 ways

=> 16 * 10 = 160 ways

In a normal watch, the minute hand crosses the hour's hand after every 1 hour 5 minutes and 27 seconds.
So, the third time the hour's hand crosses the minute's hand is after 3 hours 16 minutes and 21 seconds.

In this watch, the time taken for this to happen is 3 hours 18 minutes and 15 seconds.
Hence, the watch loses 1 minute and 54 seconds after every 3 hours 18 minutes and 15 seconds.
18 minutes and 15 seconds = 1095 seconds = 1095/3600 $$\approx$$ .304 hours.
=> 3 hours 18 minutes and 15 seconds = 3.304 hours
So, time lost in a day is $$1\frac{54}{60}*\frac{24}{3.304} = \frac{114}{60}*\frac{24}{3.304} \approx 13.8$$
So, the time lost by the watch in a day is approximately equal to 13 minutes and 48 seconds.

Selling price of first watch = 300
Profit = 10%
cost price = $$\frac{300}{1.1}$$
Selling price of second watch = 300
Loss = 10%
cost price = $$\frac{300}{0.9}$$

Total selling price of transaction= 600
Total cost price of transaction = $$300(\frac{10}{11} + \frac{10}{9}) = 600 (\frac{100}{99})$$
Loss = $$600 (\frac{100}{99} - 1)$$
%loss = $$(600 (\frac{100}{99} - 1)) \div (600(\frac{100}{99})) \times 100 = 1$$