SNAP 2018 Question Paper

Number of letters in ASTRONAUT = 9

Here A and T are repeated twice. 

So the number of ways of arrangement = $$\ \frac{\ 9!}{2!\times\ 2!}$$

= 90720

D is the correct answer.

Total fund collected by 20 students = 20*10000 = 2,00,000 

Investment in riskless government securities = 2,00,000*$$\ \frac{\ 50}{100}$$ = 100000

Investment in equities = 2,00,000 -  1,00,000 = 1,00,000

Gain by equities = 1,00,000*$$\ \frac{\ 1}{100}$$ = 1000

Gain by riskless government securities = 1,00,000*$$\ \frac{\ 7.8}{100}$$ = 7800

Total gain = 1000+7800 = 8800

gain/student = $$\ \frac{\ 8800}{20}$$ = 440

Gain% = $$\ \frac{\ 440}{10000}\times\ 100$$

= 4.4%

B is the correct answer.

Given AD = 30 cm

In triangle ACD,

$$\tan\angle ACD\ =\ \ \frac{\ AD}{DC}$$
$$\sqrt{\ 3}\ =\ \ \frac{\ 30}{DC}$$

DC = $$10\sqrt{\ 3}$$

radius of cone = DC = $$10\sqrt{\ 3}$$

Volume of cone = $$\frac{1}{3}\pi\ r^2h$$ = $$\frac{1}{3}\pi\ \left(10\sqrt{\ 3}\right)^2\left(30\right)$$ = $$3000\pi\ cm^3$$

Answer is option A.

Let x, y, z be the number of items purchased by Nisha of price Rs. 5, 2, 1 respectively.

5x+2y+z = 20

Since the shopkeeper had no change with him/her; he/she gave Nisha three more products of price Rs. 1 each i.e Nisha purchased items of Rs. 17

5x+2y+z =17

She purchased all three types of products in more than one quantity, 

If she purchases three items of Rs. 5, she cannot purchase 2 items of Rs 2 and 1

So Nisha purchases 2 items of Rs. 5.

If Nisha purchases 3 items of Rs 2., then she cannot purchase more than 1 item of Rs. 1.

Hence she must have purchased 2 items of Rs. 2 and 3 items of Rs. 1.

The number of items Nisha will have at the end of the transaction = 2+2+3+3(due to no change)

= 10

D is the correct answer.

Let the sides of the rectangle be 2x, 7x

It is given that the perimeter = 360m

2(2x+7x) = 360

x= 20m

Length = 7*20 = 140m, breadth = 2*20 = 40m

A is the correct answer.

Let the initial price and expenditure of sugar be 100p, 100e respectively.

Consumption = $$\ \frac{\ Expenditure\ }{\Pr ice}$$

=$$\ \frac{\ e\ }{p}$$

The increased price of sugar = 120p

Increased expenditure on sugar = 105e

Changed consumption = $$\ \frac{105\ e\ }{120p}$$

=0.875 $$\ \frac{\ e\ }{p}$$

Percentage decrease in consumption = $$\ \frac{(0.875-1)}{1}$$*100

=12.5

B is the correct answer.

If P is the principal and r is the rate of interest, then the difference between simple interest and compound interest for 2  years = $$\ \frac{\ \Pr^2}{100^2}$$

It is given that the difference is Rs 50 and the rate of interest is 5%  per annum.

Principle P = Rs. 20000

Amount at the end of 3 years = $$P\left(1+\ \frac{\ 5}{100}\right)^3$$

= $$20000\left(1+\ \frac{\ 5}{100}\right)^3$$

= Rs.23152.5

D is the correct answer.

$$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$.

=$$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ ........ + \log \tan 45^\circ$$.

=$$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ...........................\log\ \left(\tan\ 45^0\right)$$

tan $$45^0$$ = 1

$$\log\ \left(\tan\ 45^0\right)\ =\ 0$$

$$\therefore$$ $$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$ = 0

Let the  other multiplier be p

Instead of calculating 25*p, he calculated 34*p

It is given that because of the mistake, the answer was 405 more than the correct answer.

34p-25p=405

9p = 405

p=$$\ \frac{\ 405\ }{9}$$

Now we have to find the value of 34p

=34*$$\ \frac{\ 405\ }{9}$$

=1530

A is the correct answer.

Number of different words that can be formed using the letters of the word CUSTOM and starting with M are 

M,_, _, _, _, _. 

These 5 blanks have to be filled with the remaining 5 letters in 5*4*3*2*1 = 120 ways.

C is the correct answer.