SNAP 2018 Question Paper

Probability of Sonal's selection = $$\frac{1}{7}$$

Probability of Meenal's selection = $$\frac{1}{5}$$

The probability that only one of them is selected = $$\frac{1}{7}$$*$$\frac{4}{5}$$ +  $$\frac{6}{7}$$*$$\frac{1}{5}$$

= $$\frac{10}{35}$$

= $$\frac{2}{7}$$

B is the correct answer.

$$\log_{10}{\left(\frac{1}{110}\right)}$$

$$\log_a\left(\ \frac{\ x}{y}\right)\ =\ \log_ax-\log_ay$$

$$\log_{10}{\left(\frac{1}{110}\right)}$$ = $$=\ \log_{10}1-\log_{10}110$$

= 0$$-\log_{10}110$$ 

=$$-\log_{10}11\times\ 10$$

=$$-\left(\log_{10}11+\log_{10}10\right)$$

= -(a+1)

D is the correct answer.

Let the cost of the fridge = 100x

The price of the TV = 150x (Since the cost of TV is 150% of the cost of the fridge)

The price of the fridge  $$\ \frac{\ 100x}{100x+150x}$$ times the price of TV and fridge

The price of the fridge =40% of the price of TV and fridge

B is the correct answer.

Given e = 8

Number of students shortlisted for A and B is 14

 e + b = 14

       b = 6

Number of students shortlisted for A and C is 12

e + d = 12

      d = 4

Number of students shortlisted for B and C is 15

e + f = 15

      f = 7

Given 

Number of students shortlisted for B is 26

b + f + e + c = 26

6 + 7 + 8 + c = 26

                  c = 5

Number of students shortlisted for C is 29

d + e + f + g = 29

                 g = 10

Number of students shortlisted for A is 21

b + e + d + a = 21

                  a = 3

Number of students shortlisted for only B = c = 5

Number of students shortlisted for only C = g = 10

Ratio = 5:10 = 1:2

Answer is option B

Let the fixed charge be Rs. x and variable charge/km is Rs. y.

As per the question,

x- 10y= 150 and x+ 15y = 220

On solving both the equations, we get 5y = 220-150= 70

y = Rs. 14/km and x = Rs. 10

Charge for 25 km = x+ 25y 

= 10+25*14 = Rs. 360

C is the correct answer.

we have :

28-7-7 =14 
So we have 4 semicircles with diameters 28,7,7 and 14 .
Now area of shaded region = Area of bigger semicircle (radius 14 ) - 2 (Area of smaller semicircles with radius 3.5 ) +Area of semicircle with radius 7
we get $$\pi\ \ \frac{14^2}{2}-\ 2\ \frac{\pi\left(3.5\right)^2}{2}+\pi\ \ \frac{\left(7\right)^2}{2}$$
= $$98\pi+24.5\pi\ -12.25\pi\ \ =110.25\pi\ $$

s = $$\ \frac{\ a+b+c}{2}$$ = $$\ \frac{\ 30+60+60}{2}$$ cm =75 cm
Using Heron's formula
Area=√[s(s-a)(s-b)(s-c)]
=√[75(75-60)(75-60)(75-30)]
=$$225\sqrt{\ 15}$$ cm
There are total 10 pieces 

First colour=5*($$225\sqrt{\ 15}$$) = $$1125\sqrt{15}$$ cm
Second colour=5*($$225\sqrt{\ 15}$$) = $$1125\sqrt{15}$$ cm

B is the correct answer.

$$\ \frac{\ R_c}{m}=\ln\left(1+\frac{R_m}{m}\right)$$

$$\ \frac{\ R_m}{m}+\ 1\ =\ e^{\ \frac{\ R_c}{m}}$$

$$\ \frac{\ R_m}{m}\ =\ e^{\ \frac{\ R_c}{m}}-1$$

$$R_m = m\left(e^{\frac{R_c}{m}} - 1 \right)$$

C is the correct answer.

Here $$a^0$$ and $$L^0$$ are both constants.

So, the output is independent of Labour.

C is the correct answer.

Sum of all (Petrol price per litre - taxes) in the last week = 35+34+35.5+37+37.5+38

=  217

Taxes on petrol = 1.25 times (the price of petrol - taxes)

2.25* taxes on petrol = 1.25 times the price of petrol

The price of petrol = 1.8 times the taxes on petrol.

1.8 times the taxes on petrol - taxes on petrol = 217

0.8 times the taxes on petrol = 217

Taxes on petrol = $$\ \frac{217}{0.8}$$ = 271.25

The average amount of tax collected = $$\ \frac{271.25}{6}$$ 

= 45.2

A is the correct answer.