SNAP 2015 Question Paper

This can be answered either by using averages concept or alligations concept.

Alligations :$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$  where q1 and q2 are the number of innings in two group's , p1 and p2 are there respective averages(average runs) , p is the overall average.

Here q1=15 , q2=1 , p1=32 , p= 34 (since increased by 2 runs) 

On solving the equation you get p2=64

The least number that is divisible by 14,16,18 will be the LCM of three which is 1008 but it is not a perfect square .

$$1008=2^4\cdot3^2\cdot7^{ }$$ To make it a perfect square you need to multiply it by 7.

1008*7=7056

All of them will clap together after LCM(20,30,40,50) minutes = 600 minutes=10hrs

Therefore they will clap together again at 8pm.

Let 100x be the total numbers of voters in the voters list. => 90x voters voted

A=1.4B  (since A received 40% more votes than B)

C=1.2B

Given A-C = 4000

1.4B-1.2B=4000  => B=20000 => A=24000 and C = 28000

So total voter who voted =A+B+C= 72000 which is 90% of the total list.

So when 90%= 72000 , 100%= 80000.

Let total number of students=100x

10x=students who cleared the cut off in all the sections

5x=students who  cleared none of the sections.

Remaining = 85x

Out of these 85x , 30% cleared only 1st section , 20% cleared only 2nd section, 30% cleared only section 3 

Together they constitute 60% of 85x

Remaining= 40% of 85x =1020

On solving x=30

Total students=100x= 100*30=3000

This can be solved in an easier manner by using allegations concept.

$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$ where p1 ,p2 are profit percentages ,   q1,q2 are the number of articles , p is the overall profit percentage.

Here q1=3/5  ,  q2 =2/5 , p1=20 , p2=0

On solving you get p as 12

 P=50% and D=10%

The trader earns a revenue of Rs 54*20 after selling the items. = Rs 1080.

This is equivalent to a 50 percent profit.

Hence the cost price of all the items is Rs 1080*2/3 = Rs 720.

The revenue of Rs 1080 was earned after a discount of 10 percent.

If K is the marked price: K*(9/10) = K = Rs 1200.

Hence the profit percentage if no discount is provided :

$$\frac{\left(1200-720\right)}{720}\cdot100\ =\ 66.66\%$$

Final volume of Milk = Initial volume of milk$$\left(1-\frac{y}{x}\right)^n$$ 

where y=quantity of milk removed in each iteration , x= total quantity of the mixture and n=number of times the process being repeated 

Final volume of milk = $$50\cdot\left(1-\frac{5}{50}\right)^4$$

= 32.805

For the 1st drop the distance travelled by the ball = 200m

After 1st drop it rebounces to a height of 4/5 * 200 and then falls from that height . The total distance travelled in this case would be $$2\cdot\frac{4}{5}\cdot200$$

Similarly the total distance travelled in next case would be $$2\cdot\frac{4}{5}\cdot\frac{4}{5}\cdot200$$

So total distance = 

200 + $$2\cdot200\cdot\left(\frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^{3\ }+....\right)$$

200+400$$\left(\frac{4}{\frac{5}{1-\frac{4}{5}}}\right)$$    = 1800m

$$\frac{12}{2}\left(16+93\right)$$The numbers will be of the form 7k+2 where k is an whole number.

The smallest two digit number is when k=2 which is 16 and the largest 2 digit number is 93 k= 13

So sum = 16+23+.... +93  which are in AP. 

Sum to n terms of an AP= n/2(a+l), where n=number of terms , a=1st term ,l=last term

Here n=12 , a=16 , l=93

Hence sum = 654