SNAP 2011 Question Paper

Speed of the Train = 36 km/hr.
= 36 × 5/18 [Changing into m/s]
= 2 × 5
= 10 m/s.

Time taken to cross the Platform = 20 seconds.
Time taken to cross the man = 10 seconds.
∵ Speed of the Train = Length of the Train/Time to cross the Man

∴ Length of the Train = Speed of the Train × Time to cross the Man.
= 10 × 10
= 100 m.
Sum of the Length of Platform and train = Speed of the Train × Time taken by the Train to cross the Platform.
= 10 × 20
= 200 m.
∵ Length of the Train + Length of the Platform = 200
∴ 100 + Length of the Platform = 200
∴ Length of the Platform = 200 - 100
= 100 m.
∴ Length of the Platform is 100 m.

Let the usual speed and the time taken by the man = 5v, 4t respectively.

If the man walks at $$\frac{4}{5}th$$ of his usual speed, a man reaches office 10 minutes later than usual

i.e the time taken by the man = $$\ \frac{\ 5}{4}$$ of the usual time (Speed and time are inversely proportional to each other)

=5t

5t-4t = 10 minutes.

t = 10 minutes.

Usual time taken by the man = 4t = 40 minutes

B is the correct answer.

Time taken to meet = $$\ \frac{\ Dis\tan ce\ between\ them\ }{Relative\ speed}$$

Relative Speed = 5 +4 = 9 miles per hour.
Time take to cover 81 miles = $$\ \frac{\ 81\ }{9}$$ = 9 hours.
Woman travelled in 9 hours = 9*4 = 36 miles.

1 person :
$$6^2 + 8^2 = OB^2$$
√(36+64) = OB
√(100)= OB
10 = OB...
The distance of the first person from his starting point is 10 miles.
Similarly, the distance of the second person from his starting point is 10 miles.
B is the correct answer

(4M+ 3W)6 = (5M+6W)4 = W

4M = 6W

2M = 3W

let x be the number of days taken by 2 women and 3 men take to do the job

(4M+3W)6=(2W+3M)x

9W*6=6.5W*x

x=8.3 days

C is the correct answer.

Let the total work be 25x units

Ram completes 60% of the task in 15 days.

Ram's efficiency = x units per day

Rahim is 50% as efficient as Ram is and Rachel is 50% as efficient as Rahim is

Efficiency of Rahim = 0.5x units per day

Efficiency of Rahim = 0.25x units per day

Ram will come 15x units in 15 days.

The remaining 10x units of work will be done by Ram, Rahim, Rachel.

=$$\ \frac{10x}{x+0.5x+0.25x}$$

= $$\ \frac{10}{1.75}$$

=$$\ \frac{1000}{175}$$

=$$\frac{40}{7}$$

C is the correct answer.

Let the total work be LCM of (21, 24) = 168 units.

Amount of work done by A on 1 day = $$\ \frac{\ 168}{21}$$ = 8 units

Amount of work done by B on 1 day = $$\ \frac{\ 168}{24}$$ = 7 units

Let p be the number of days for which they worked together.

In P days, the amount of work done by A and B = 15p

Remaining work = 168x-15p, which is completed by B in 9 days.

168x-15p=9*7

p=7

So A left after 7 days.

B is the correct answer.

It is given that profit on 100 articles = SP of 75 articles

100(SP-CP) = 75*SP

4(SP-CP) = 3SP

SP = 4CP

Profit = $$\ \frac{\ SP-CP}{CP}\times\ 100$$

=$$\ \frac{\ 4CP-CP}{CP}\times\ 100$$

=300%

C is the correct answer.

Let us consider A covers 100 m in t sec.

Speed of A = $$\ \frac{100}{t}$$

Since A gave B a headstart of 20m, B covers 80m in (t+5) sec (Given)

Speed of B =  $$\ \frac{80}{t+5}$$

If A gives head start of 40m to B, then the time taken by A to cover 100m is the same as the time taken by B to cover 60m

$$\ \frac{100}{\ \frac{\ 100}{t}}$$ = $$\ \frac{60}{\ \frac{\ 80}{t+5}}$$

n solving for t, we get 

t =  15 sec

Speed of A =  $$\ \frac{100}{15}$$ m/sec

Time taken by A to cover 200m(x) = 

200=time * $$\ \frac{100}{15}$$

x = 30 sec

C is the correct answer.

Initial surface area of the cube = 6$$(side)^2$$ 

= 6$$(4)^2$$ 

= 96$$cm^2$$

No. of new cubes = $$\ \ \frac{\ Volume\ of\ older\ cube}{Volume\ of\ 1\ new\ cube}$$

=$$\ \ \frac{\ 4\times\ 4\times\ 4}{1\times\ 1\times\ 1}$$

=64 cubes 

Newer surface area of 64 cubes = 64*6$$(1)^2$$ = 384$$cm^2$$

Percentage increase in surface area = $$\ \ \frac{384-96}{96}$$ *100

=$$\ \ \frac{28800}{96}$$

 = 300%

B is the correct answer