SNAP 2011 Question Paper

The time spent in playing games and in doing homework corresponds to 30° and 45° respectively.

If the time spent in playing games is equal to that spent on doing homework, then there should be an increase of 15° in the time spent in games.

If this 15 ° is obtained by decreasing the time spent in sleeping (with the other activities remaining constant), 

The percentage decrease in the time spent in sleeping = $$\ \frac{\ 15^0}{120^0}\times\ 100$$

= $$12.5^0$$

B is the correct answer.

School corresponds to  $$105^0$$ and homework corresponds to $$45^0$$

Difference =  $$105^0$$ - $$45^0$$

=$$60^0$$

= $$\ \frac{\ 60}{360}\times\ 24$$

= 4 hrs

C is the correct answer.

The total number of hours = 24

Time spent on homework = $$45^0$$

=$$\ \frac{\ 45}{360}\times\ 24$$ = 3 hrs

It is given that she spent $$\frac{1}{3}rd$$ time of homework in Mathematics.

The time she spent on Mathematics = $$\ \frac{\ 1}{3}\times\ 3$$

=1 hr

The time she spent on the rest of the subjects = 3-1 = 2 hr

B is the correct answer.

AD is parallel to BE  is parallel to CF 

$$\ \frac{\ AB}{BC}\ =\ \ \frac{\ DE}{EF}$$   [Basic proportionality theorem]

$$\ \frac{\ 2}{4}\ =\ \ \frac{\ 1.5}{EF}$$

EF = 3 cm

B is the correct answer.

$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$

Since $$\log_aa\ $$ = 1

$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$ = 1+2+....n

=$$\ \frac{\ n\left(n+1\right)}{2}$$ 

=$$\frac{(n^{2} + n)}{2}$$

D is the correct answer.

Let the number be a

It is given that sum of a number and its reciprocal is thrice the difference of the number and its reciprocal.

a+$$\ \frac{\ 1}{a}$$ = 3(a-$$\ \frac{\ 1}{a}$$ )

2a = $$\ \frac{\ 4}{a}$$

a = $$\ \frac{\ 2}{a}$$

$$\ \ a^2=2$$

a = $$\pm \surd{2}$$

A is the correct answer.

The first natural number between 10 and 300 which is divisible by 9 is 18

The last natural number which is divisible by 9 and is less than 300 is 297

This is an AP with first term 18, the last term 297 and common difference = 9

297=18+(n-1)9

279=(n-1)9

n-1 = 31

n=32

A is the correct answer.

$$^nP_r$$ = $$^nC_r$$*r!

$$^nP_x = 336$$

$$^nC_x *x!$$ = 336

x!= $$\ \frac{336}{56}$$

x!=6

x=3

$$^nC_x = 56$$

$$\ \ \frac{\ n!}{\left(n-x\right)!\times\ x!}\ =\ 56$$

$$\ \ \frac{\ n!}{\left(n-3\right)!\times\ 3!}\ =\ 56$$

n=8

$$\therefore$$ n= 8, x=3

C is the correct answer.

Side of an equilateral triangle= 24 cm

The mid-points of its sides are joined to form another triangle whose mid-points are joined to form another triangle. This process continues indefinitely.

Side of 2nd equilateral triangle=12 cm [ Line segment joining midpoint of two sides of a triangle is parallel to the third side and half of it.]

Similarly, Side of third equilateral triangle=6 cm

Sides are 3, 1.5,3/4,....

Perimeter of 1st triangle=24+24+24=72 cm

Perimeter of 2nd triangle=12+12+12=36 cm

Perimeter of 3rd triangle=6+6+6=18 cm

..........

The sum of the perimeter of all the triangles=72+36+18+9+9/2+9/4+........

This is a geometric progression having a common ratio =36/72=1/2

Sum of an infinite G.P=

= 72 × 2

=144 which is the sum of the perimeter of all triangles.

A is the correct answer.

Total number of days in a leap year = 366

It will contain 52 weeks and 2days

These two days can be (Sunday, Monday); (Monday, Tuesday); (Tuesday, Wednesday); (Wednesday, Thursday); (Thursday, Friday); (Friday, Saturday); (Saturday, Sunday)

For 53 Sundays, probability = $$\ \ \frac{2}{7}$$

Similarly for 53 Mondays, probability =$$\ \ \frac{2}{7}$$

This includes one way where Sunday and Monday occur simultaneously (i.e) Sunday, Monday 

Probability for this =$$\ \ \frac{1}{7}$$

Hence required probability = $$\ \ \frac{2}{7}$$ +$$\ \ \frac{2}{7}$$-$$\ \ \frac{1}{7}$$

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

C is the correct answer.