SNAP 2013 Question Paper

Let the filling capacity of the pump be x $$m^3$$/min

Then the emptying capacity of the pump = (x+10) $$m^3$$/min

The time required for filling the tank = $$\ \frac{\ 2400}{x}$$

The time required for emptying the tank = $$\ \frac{\ 2400}{x+10}$$

 The pump needs 8 min less to empty the tank than to fill it.

$$\ \frac{\ 2400}{x}$$ - $$\ \frac{\ 2400}{x+10}$$ = 8

$$\ \frac{\ 300}{x}$$ - $$\ \frac{\ 300}{x+10}$$ = 1

300(x+10) - 300(x) = x(x+10)

$$x^2+10x-3000$$ = 0

(x+60)(x-50) = 0

x = 50 or -60

Since x cannot be negative, x = 50

Filling capacity of the pump = 50 $$m^3$$/min

C is the correct answer.

Let the usual speed of the plane be v kmph and the time taken by the plane to cover 1500m = t hrs

To reach the destination on time, the increased speed of the plane = v+250 kmph

Since he started 30 min later, the time taken should be t-$$\ \frac{\ 1}{2}$$ hr

v*t = (v+250)(t-$$\ \frac{\ 1}{2}$$) 

v = (250t-125)2

$$\ \frac{\ 1500}{v+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 1500}{500t-250+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 3}{t}=t-\ \frac{\ 1}{2}$$

On solving for t, we get

t= 2 hrs

His original speed = $$\ \frac{\ 1500}{2}$$ = 750 kmph

B is the correct answer.

Given,

Cost of each shirt = Rs. 43

Cost of each tie = Rs. 21

Let x, y be the number of shirts and ties purchased

x*43+ y*21 = 535

y = $$\ \frac{\ 535-43x}{21}$$

when x = 10, y is an integer i.e y =5

ratio of shirts to ties purchased = 10:5 =2:1

B is the correct answer.

The cost price of the mixture per kg = $$\ \frac{\ 12\times\ 16+4\times\ 2}{12+4}$$

= Rs. 12.5

Selling price of the mixture per kg = Rs 16

Profit made on selling 1 kg of mixture = 16-12.5=Rs. 3.5

Profit made o selling 40 kg of mixture = 40*3.5=Rs. 140

A is the correct answer.

Here we have two cases:

Case 1: If Y is a member of the committee

We have to select 3 other gentlemen and 3 ladies from 8 ladies and 5 gentlemen (Since If Y is a member of the committee, X will not be a member of the committee)

=$$^7C_3\times\ ^6C_3$$ =35*20=700.

Case 2: If Y is not a member of the committee

We have to select 4 gentlemen and 3 ladies from 8 ladies and 6 gentlemen (Since Y is not a member of the committee)

=$$^8C_3\times\ ^6C_4$$ = 56*15=840

Total cases = 700+840 = 1540

C is the correct answer.

Total no. of seats = 1 grandfather+ 5 sons and daughters + 8 grandchildren 

= 14.

The grandchildren can occupy the 4 seats on either side of the table in $$^8P_4$$*4! ways
The grandfather can occupy a seat in (5-1)= 4 ways (4 gaps between 5 sons and daughter).

The remaining seats can be occupied in 5!= 120 ways (5 seats for sons and daughter).

Total number of required ways = 8!*4*120

=8!*480

D is the correct answer.

Let the total number of chocolates be C

let chocolates received by each child = x

After distributing the sweets equally among 25 children, the number of chocolates left = 8
25x+8=C                - Eq 1

Had there been 28 children, the number of chocolates that would be required = 22

25x+8-22 should be divisible by 28            - Eq 2

25x-14 should be divisible by 28

28x-(3x+14) should be divisible by 28

(3x+14) should be divisible by 28.

==> x=14

The total number of chocolates = 25x+8 = 25*14+8=358

C is the correct answer.

Let the distance he has to travel to reach the place = x km

and the time given to reach the place = y minutes.

If he walks 4 km in an hour he is 15 minutes late ie if he walks with a speed 4 km/hour (4km/60min)he will take y+15 minutes

distance = time*speed

ie x=$$\ \frac{\ 4}{60}$$*(y+15)

60 x = 4* (y+15)

15 x = y+15    -- Eq (1)

if he walks 6km in an hour he is 10 minutes early ie if he walks with a speed of 6km/hour he will take y-10 minutes.

x=$$\ \frac{\ 6}{60}$$ * (y-10)

10x = (y-10)    -- Eq (2)

On solving Eq 1, 2, we get

y = 60 minutes , x = 5 km

B is the correct answer. 

We have MN || AB 
Now let area of CMN be A 
so area of ABMN will be 2A 
Now Area of triangle ABC will be 3A

Now since MN || AB 
so we can say CMN is similar to CAB
So we can say
(CM/CA)^2 = Area of CMN : Area of CAB
so we can say CM :CA =$$1:\sqrt{\ 3}$$
Now therefore CM :AM = $$1:\sqrt{\ 3}-1\ =\ \sqrt{\ 3}+1:2$$

$$\angle\ DCB\ =90^0$$ (Since each angle in a square is 90)

$$\angle\ BCE\ =60^0$$ (Since each angle in an equilateral triangle is 60)

$$\angle\ DCE\ =90^0+ 60^0$$  = $$150^0$$

In $$\triangle\ DCE$$, the sides DC and CE are equal, so the angles opposite to them are equal.

$$\angle\ DCE$$ = $$\angle\ CDE$$ = $$\ \frac{\ 180-150}{2}$$

=$$15^0$$

A is the correct answer.