SNAP 2011 Question Paper

The intercept form of the line = $$\ \frac{\ x}{a}+\ \frac{\ y}{b}\ =\ 1$$ where a is x-intercept and b is y-intercept.

3x + 4y - 12 = 0

3x + 4y = 12 

$$\ \frac{3\ x}{12}+\ \frac{\ 4y}{12}\ =\ 1$$

$$\ \frac{\ x}{4}+\ \frac{\ y}{3}\ =\ 1$$

So the intercepts are 4 and 3.

B is the correct answer.

Given, 

The average of the four prime numbers = 35.

a + b + c + d = 35 * 4 = 140.

Since a and d are equidistant from 36.

a + d = 72 --- Eq (1)

b + c = 68 --- Eq (2)

a + b = 60 --- Eq (3) and c + d = 80 --- Eq (4)

Using the equation (3) let us look for the prime values of a and b and the corresponding values of c and d using Eq 2 and 1.

Also given that a < b < c < d.

 (a, b, c, d) = 29, 31, 37, 43

d - a = 43 - 29 = 14

B is the correct answer.

SP of 1kg rasgulla= Rs.15
Profit= $$\frac{200}{3}$$%
CP of 1 kg rasgulla = ($$\ \frac{\ 15\times100}{\ \frac{\ 500}{3}}$$) = Rs. 9
CP of 1 kg of sugar= 7x
CP of 1 kg of floor= 3x
3x            7x
         9
7x-9       9-3x
$$\ \frac{\ 7x-9}{9-3x}\ =\ \ \frac{\ 5}{3}$$
x=2
CP of sugar= 7x= 14 per kg

D is the correct answer.

Let the initial price of 100 kg Sugar was Rs. 100.
Now, a 20% decrease in rate then rate of 100 kg sugar would be Rs. 80.
So, Rs. 80 = 100 kg sugar.
Rs. 100 = 100× $$\ \ \frac{100}{80}$$ = 125 Kg sugar.
Increment in sugar = 25 kg.
% increment = $$\ \ \frac{125-100}{100}\times\ 100$$ = 25%.
25% increment = 6 kg of sugar

total sugar initially = 6*4 =24 kg

Original price of the sugar = $$\ \ \frac{240}{24}$$

= Rs. 10 per kg

A is the correct answer.

Let the radius of solid sphere be r.

The volume of sphere $$V_s$$= $$\ \frac{\ 4}{3}$$ πr³

It is melted and recast into a right circular cone of radius r and height h.

The volume of cone $$V_c$$= $$\ \frac{\ 1}{3}$$ πr²h

The same volume of material is used for recasting. Therefore,

$$V_s=V_c$$

$$\ \frac{\ 4}{3}$$πr³ =  $$\ \frac{\ 1}{3}$$ πr²h

h = 4r

$$\ \frac{\ h}{r}$$  = 4
D is the correct answer.

Let the distance covered be 16 km, and the velocities be 1kmph, 4kmph and 16kmph.

Then the time taken: 16h, 4h, and 1h respectively

Ratio of time and velocity for each vehicle: $$\frac{16}{1}:\frac{4}{4}:\frac{1}{16}$$

=$$16:1:\frac{1}{16}$$

=256:16:1.

A can do a piece of work in 15 days.
B is twice as efficient as A.
So B can do the same piece of work in 7.5 days.
Assume the total work to be 15 units
A does 1 unit per day and B does 2 units per day.
For the first x days, A worked alone and hence work done per day is
1 * x = x units.
For the next (11 - x) days, A and B worked together. In 1 day, they will do (1 + 2) = 3 units.
So in 11 - x days, they will do 33 - 3x units.
33 - 3x + x = 15
2x = 18 and x = 9.
Thus A worked alone for 9 days the A and B worked together for 2 days.

A+B+C+D = 56  -- Eq 1

B+C+D = 4.6A   -- Eq 2
A+C+D = $$\ \ \frac{11B}{3}$$   --- Eq 3

A+B+D = 2.5C  --Eq 4

Using Eq 1 and 2, we get A = 10

Using Eq 1 and 4, we get C = 16

Using Eq 1 and 3, we get B = 12

On substituting the values of A, B, C in Eq 1, we get D = 18

D is the correct answer.

Let the salary of Saroj be 100x

The salary of Raju = 80x

The salary of Ram =  70x

the percent by which the salary of Raju more than the salary of Ram = $$\ \frac{\ 80x-70x}{70x}\times\ 100$$

=$$\ \frac{\ 1}{7}\times\ 100$$

=14.28%

D is the correct answer.

We know that wire is in the shape of a cylinder.
So, the volume of wire = $$\pi\ r^2h$$

$$\pi\ r^2h_1$$ = $$\pi\ (\frac{r}{3})^2h_2$$

Cancelling π on both sides,

$$\ r^2h_1$$ = $$\frac{\ r}{9}^{^2}h_2$$
$$9h_1$$ = $$h_2$$
length is increased by 9 times

D is the correct answer.