SNAP 2013 Question Paper

$$\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$$? 

=$$\frac{\ \log_{3^3}3^2\times\ \log_{2^4}2^6}{\log_{\left(\sqrt{\ 2}\right)^4}\sqrt{\ 2}}$$

=$$\frac{\ \ \frac{\ 2}{3}\times\ \frac{\ 6}{4}}{\ \frac{\ 1}{4}}$$

=4

D is the correct answer.

Let the number of seats in the stadium = x

Number of seats in the lower deck = a and the number of seats in the upper deck be b

x=a+b

a=$$\ \frac{\ x}{4}$$ and b = $$\ \frac{\ 3x}{4}$$

Now in the lower deck $$\ \frac{\ 4a}{5}$$ seats were sold and $$\ \frac{\ a}{5}$$ seats were unsold.

Number of total seats sold in the stadium = $$\ \frac{\ 2x}{3}$$ 

Number of unsold seats in the lower deck =  $$\ \frac{\ a}{5}$$ =  $$\ \frac{\ x}{20}$$

Number of unsold seats in the stadium =  $$\ \frac{\ x}{3}$$

Required fraction = $$\ \frac{\ \ \frac{\ x}{20}}{\ \frac{\ x}{3}}$$

= $$\ \frac{\ 3}{\ \ 20}$$

A is the correct answer.

Let t be the start time

A starts at t, while B starts at (t-2)

Total distance covered = 200 km

As they are travelling in the opposite direction,

Relative distance covered i,e 200=((200/4)*t )+((400/7)*(t-2))

t=176min i.e 2hrs 56min from start (6 am)

Therefore they meet at 8.56 am.

A is the correct answer.

Quantity of water let in by the leak in 1 hr= 15 tonnes

Quantity of water thrown out by the pumps in 1 hr = 12 tonnes

Net quantity of water filled in the ship in 1 hour =15-12=3 tonnes

60 tonnes water is filled in  = 60/3= 20 hours

Required speed = 40/20= 2kmph

Minimum distance will be LCM of 40 cm, 42 cm and 45 cm

40 = 2³ x 5

42 = 2 x 3 x 7

45 = 3² x 5

LCM = 2³ x 3² x 5 x 7 = 2520 cm

=25 m 20cm

A is the correct answer.

The least number which when divided by 6, 7, 8, 9 and 10 leaves a remainder of 4, 5, 6 7 and 8 respectively is LCM(6, 7, 8, 9, 10) -2

=2520-2= 2518

The number will be in the form of k * LCM(6, 7, 8, 9, 10) -2

When k = 3, the value of x =7558

When k = 4, the value of x =10078

So the value of x = 7558

Sum of the digits of x = 7+5+5+8= 25

A is the correct answer.

It is a decreasing AP 25,24.5,24,23.5,...

The sum will be largest, if nth term is zero then

25 + (n-1)*(-.5) = 0

=> n = 51

after 51 terms AP contains -ve terms

so, max sum will be obtained upto 50/51th term

S = $$\frac{51}{2}$$ *(25+0) = 637.5

Total members :240
Males = 2/3 (240) =160
so Females = 240-160 =80
Graduate Males = 160(0.15) =24
Non graduate males = 160-24 = 136
Graduate Females = (0.75) 80 = 60
So non graduate females = 80-60 =20
Now difference between the number of females who are non-graduates and the number of males who are graduates : 24-20 =4

Total members = 240

Number of males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160

Number of males who are non-graduates =  $$\ \frac{\ 85}{100}\times\ 160$$ = 136

Number of females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80

Number of females who are graduates = $$\ \frac{\ 3}{4}\times\ 80$$ = 60

Sum of male non graduates and female graduates = 136+60=196

D is the correct answer.

Total members in the team = 240

Males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160
Females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80

Number of females who are non graduates = $$\ \frac{\ 1}{4}\times\ 80$$ = 20

ratio between the total number of males and the number of females who are non-graduates

= 160:20

=8:1

B is the correct answer.