SNAP 2006 Question Paper

The fraction of milk in A = 4/(4+3) = 4/7

The fraction of milk in B = 2/(2+3) = 2/5

Let them be mixed in the ratio x:y such that their fraction is 1/2
It can be written as 

$$\frac{4x}{7}$$ + $$\frac{2y}{5}$$ = $$\frac{x+y}{2}$$

Upon solving,

$$\frac{4x}{7}$$ - $$\frac{x}{2}$$ = $$\frac{y}{2}$$ - $$\frac{2y}{5}$$

$$\frac{x}{14}$$ = $$\frac{y}{10}$$

$$\frac{x}{y}$$ = $$\frac{7}{5}$$

Required ratio is 7:5

Let the cost of Picture be Rs X. Then the cost of frame is also Rs X.

As per question $$\frac{X+75}{2}$$ = $$2\cdot\left(X-100\right)$$

X + 75 = 2X - 200

X = 275

At 7 the hour hand is  $$\frac{7}{12}\cdot360$$ = 210 degrees ahead of the minute hand.

Let after x minutes they are at 180 degrees with each other (since they are exact opposite)

Total angle covered by hour hand = 210 + 0.5x ( as hour hand moves at 0.5 degrees/minute)

Total angle covered by minute hand = 6x ( speed of minute hand is 6 degrees/minute)

As per the question

    Angle by hour hand - Angle by minute hand = 180

(210 + 0.5x) - 6x = 180

210 - 5.5x = 180

5.5x = 30

x  = 5.45 minutes

The closest answer is 5.5 minutes

Out of all the 6 tickets, 4 are front row tickets and the rest 2 are not.

Out of 4 front tickets 2 can be selected in $$^{4\ }C\ _2$$

The remaining 1 ticket has to be selected in  $$^{2\ }C\ _1$$

Hence total ways of selecting 3 tickets such that 2 of them are front row are  $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$

3 out of 6 tickets can be selected in $$^{6\ }C\ _3$$ ways.

Required probability = $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$ / $$^{6\ }C\ _3$$

Which is equal to 0.6

Let the number be x

As per question 

 $$\ \frac{\ 75x}{100}$$ + 75 = x

$$\ \frac{\ 3x}{4}$$ + 75 = x

$$\ \frac{\ x}{4}$$ = 75

x = 4*75 = 300

Let x be the distance traveled by water

Distance by foot is $$\ \frac{\ 4x}{7}$$

Distance by horseback is $$\ \frac{\ 2x}{5}$$

Total distance = x + $$\ \frac{\ 4x}{7}$$ + $$\ \frac{\ 2x}{5}$$ = 3036

Hence x = 1540

$$\ \frac{\ 4x}{7}$$ = 880

$$\ \frac{\ 2x}{5}$$ = 616

Option A is correct

Let there be x boys and y girls in the family.

For each of the boys, number of brother = x-1 and number of sister = y

For each of the girls, number of brother = x and number of sisters = y-1

As per question,

x-1 = y ( equation of brother and sister for boys)

x = 2(y-1) (equation of brother and sister for girls

On solving we get x = 4 and y= 3, Hence option C

The probability of picking the white is 0.5*(Probability of white in Jar1) + 0.5*(Probability of white in Jar2)

To maximize the probability we can put 1 white marble in jar 1. Hence the probability of white in jar 1 = 1

The remaining 49 white and 50 black can be put in Jar 2. Thus, the Probability of white in Jar 2 = 49/99.

All other options give less probability

Since the bus has numbers from 1 to 500 and is a perfect square it has to be one of the numbers among

1 , 4 , 9 , ..... 484($$22^2$$)

When the plate is put upside down then also a number is obtained which is a perfect square.

It is possible when the digit on the plate contains only  digits 0,1,6,8 and 9

Among the options, 169 ($$13^2$$)is a perfect square which upon putting upside down gives 961($$31^2$$) which is also a perfect square

The rate at which spiders catch the flies remains the same. Hence multiplying the time frame and the number of spiders relative to the information provided gives the solution.

Distributing 5 spiders as a group. There are 20 groups and each group catches 5 spiders in 5 minutes ad hence 100 spiders are caught in 5 minutes and in 100 minutes.

100*(100)/5 = 2000 flies.

$$\frac{\left(M1\cdot D1\cdot\right)}{W1}\ =\frac{\left(M2\cdot D2\right)}{W2}$$

$$\frac{\left(5\cdot5\right)}{5}\ =\frac{\left(100\cdot100\right)}{W2}$$

=> W= 2000