SNAP 2011 Question Paper

36 = 9*4 Since 9 and 4 are co-prime to each other, we can say that for a number to be divisible by 36 it must be divisible by both 9 and 4.

For G236G0, to be divisible by 4, last two digits should be divisible by 4.

Hence G0 should be a multiple of 4.

Possible values of G are 2, 4, 6, 8, 0

For the number to be divisible by 9, the sum of the digits should be a multiple of 9

G+2+3+6+G+0 = 11+2G should be multiple of 9

If G = 0, 11+2G is not a multiple of 9

If G = 2, 11+2G is not a multiple of 9

If G = 4, 11+2G is not a multiple of 9

If G = 6, 11+2G is not a multiple of 9

If G = 8, 11+2G is a multiple of 9.

Hence 8 is the correct answer.

Let the total work be LCM of (12, 15) = 60 units.

Amount of work done by Amit on one day = $$\ \frac{\ 60}{12}$$ = 5 units

Amount of work done by Sagar on one day = $$\ \frac{\ 60}{15}$$ = 4 units

Amount of work completed by them on 4 days = (5+4)*4 = 36 units

The amount of work left out = 60-36 = 24 units.

Fraction of work left out = $$\ \frac{\ 24}{60}$$

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

C is the correct answer.

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

Area of the crossroads = (2400 - 2109) m² = 291 m².

Let the width of the road be x metres.

Then, 60x + 40x - x² = 291

x² - 100x + 291 = 0

(x - 97)(x - 3) = 0

x = 3. 

B is the correct answer

 Let P(B₁) and P(B₂) denote the probabilities of the events of drawing balls from the bag I or bag II.

Since both the bags are equally likely to be selected, P(B₁) = P(B₂) = $$\ \frac{\ 1}{2}$$

Let E denote the event of drawing two balls of different colours from a bag.

The required probability

= P(B₁)P($$\ \frac{\ E}{B_1}$$)+ P(B₂)P($$\ \frac{\ E}{B_2}$$)

P($$\ \frac{\ E}{B_1}$$) denote the probability of drawing the two balls from bag I and as that from bag II.

= $$\ \frac{\ 1}{2}\left[\ \frac{\ ^5C_1\ \times\ \ ^3C_1}{^8C_2}\ +\ \ \frac{\ ^5C_1\ \times\ \ ^4C_1}{^9C_2}\right]$$

= $$\ \frac{\ 1}{2}\left[\ \frac{\ 15}{28}+\ \frac{\ 5}{9}\right]$$

= $$\ \frac{\ 275}{504}$$

A is the correct answer.

From the above table, it is clear that the percentage increase in the consumption of rice over the previous year was highest in 2008

B is the correct answer.

Population = $$\ \frac{\ Consumption}{Percapita\ consumption}$$

In 2008, the population of A = $$\ \frac{\ 230-138}{38.7}\ $$ = 2.79 million

C is the correct answer.

Exports to consumption ratio was highest in the year 2006

Answer is option A.

Population = $$\ \frac{\ Consumption}{Percapita\ consumption}$$

The population of country A was the highest in the year 2010.

The time spent in school corresponds to $105^0$$

Percentage of time spent in school = $$\ \frac{\ 105}{360}\times\ 100$$

= Approx 30%

B is the correct answer.

Time spent on games corresponds to $$30^0$$ and on sleeping to $$120^0$$

Required ratio = $$\ \frac{\ 30}{120}\times\ 100$$

=25%

C is the correct answer.