SNAP 2014 Question Paper

Midpoint of P(x1,y1) , Q(x2,y2) = C($$\frac{x1+x2}{2},\ \frac{y1+y2}{2}$$)

So $$y=\frac{4-2}{2},\ -1=\ \frac{x+4}{2}$$ 

On solving x=-6 and y=1

Odds in favour of A = 1:3, thus P(A wins) = $$\frac{1}{1+3}=\frac{1}{4}$$

Odds in favour of B= 1:4, thus P(A wins) = $$\frac{1}{1+4}=\frac{1}{5}$$

Odds in favour of C = 1:5, thus P(A wins) = $$\frac{1}{1+5}=\frac{1}{6}$$

Since deadheat is not possible only 1 will win.

P(A or B or C wins) = $$\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}$$

It is given that B pays 1.5 times that of A

Let B put N number of cows for 3 month

$$\frac{Money\ paid\ by\ B}{Money\ paid\ by\ A}=\ \frac{3}{2}\ =\ \frac{3\times\ N}{80\times\ 7}$$

$$N=280$$

$$\frac{\ 100000}{600000}\times\ 100\%\ =\ 16\ \frac{\ 2}{3}\%$$Let the Cost Price of the house be C.P

 $$\left(1+\ \frac{\ 5}{100}\right)C.P\ =\ 630000$$

$$\left(1.05\right)C.P\ =\ 630000$$

C.P = 600,000

If he sold house for 500,000; he would incur loss of 100,000

loss percent =  $$\frac{\ 100000}{600000}\times\ 100\%\ =\ 16\ \frac{\ 2}{3}\%$$

Let $$a_1,\ a_2.....\ a_{11}$$ be the 11 number

$$a_1+a_2+.....\ +a_{11}\ =\ 11\times\ 10.9\ =\ 119.9$$ .... (I)

Average of first six is 10.5 thus $$a_1+a_2+.....\ +a_6\ =\ 6\times\ 10.5\ =\ 63$$ ....(II)

Average of last six is 10.5 thus $$a_6+a_7+.....\ +a_11\ =\ 6\times\ 11.4\ =\ 68.4$$ ....(III)

(II)+(III)-(I) gives $$a_6\ =\ 11.5$$

Let 100 units of work be there.

$$\eta_A=\ \ \frac{\ 100}{25}=4units\ per\ day$$

$$\eta_B=\ \ \frac{\ 100}{20}=5units\ per\ day$$

Amount of work done when they work together for 5 days = $$5\times\ \left(\eta_A+\eta_B\right)\ =\ 5\times\ \left(5+4\right)\ =\ 45\ units$$

Remaining work = 100-45 = 55 units

Time taken by B = $$\ \frac{\ 55}{5}=11\ days$$

Let the current interest be r%

$$800+\ \frac{\ 800\times\ r\times\ 3}{100}\ =\ 956$$

r = 6.5%

Updated rate of interest = 3+6.5 = 9.5%

Sum after 3 years = $$800+\ \frac{\ 800\times\ 9.5\times\ 3}{100}\ =1028$$

Since Pradeep and his wife are going in the opposite direction, their relative velocity will be added. 

Their relative velocity = 4.5+3.75 = 8.25 km/hr = 137.5 m/min

Time taken to cover 726 meters = $$\ \frac{\ 726}{137.5}\ =\ 5.28$$ minutes

$$16a^2 - 12a$$ = $$\left(4a\right)^2-\left(2\right)\left(4a\right)\left(\ \frac{\ 3}{2}\right)$$

by adding $$\left(\ \frac{\ 3}{2}\right)^2\ $$ we can write it as $$\left(\ 4a-\frac{\ 3}{2}\right)^2\ $$

Hence $$\left(\frac{\ 3}{2}\right)^2\ =\ \ \frac{\ 9}{4}$$ is required answer

Let a and b represent salaries of A and B respectively

then $$a+b=40000$$...(I)

After spending 85% of his salary, savings of A will be = (0.15)a

After spending 95% of his salary, savings of A will be = (0.05)b

Their saving are equal, therefore (0.15)a = (0.05b) or 3a=b ...(II)

Solving eqn (I)and (II) we get a = 10,000