SNAP 2008 Question Paper

Gas usage per hour of the larger burner = $$\ \frac{\ 14.4}{\ \ 80}$$

Gas usage per hour of the smaller burner = $$\ \frac{\ 14.4}{\ \ 104}$$

The required percentage = $$\ \dfrac{\ \ \frac{14.4\ }{80}-\ \frac{\ 14.4}{104}}{\ \frac{14.4\ }{80}}$$ = 1- $$\ \frac{\ \ 80}{\ 104}$$ = $$\ \frac{\ \ 24}{\ 104}$$ = 23.07

If $$\ \frac{\ a}{\ \ b}$$ is less than 1, then $$\ \frac{\ a+7}{\ \ b+7}$$ > $$\ \frac{\ a}{\ \ b}$$ given that a,b>0

Hence, if I is given. II or III are redundant.

In statistics, the coefficient of variation (COV) is a simple measure of relative event dispersion. It is equal to the ratio between the standard deviation and the mean. The most common use of the COV is to compare relative risk, although it can be applied to any type of quantitative likelihood or probability distribution.

There is another use and meaning of the COV. When interpreting mathematical models, COV is calculated as the ratio between root mean squared error and the mean of a separate dependent variable. This type of COV analysis is less common, but it can be constructive when determining if a model is a good fit for a specific task or type of analysis. It is used to measure disparity and consistency.

Assuming the cyclist speed without the wind = a  and the wind speed = b

Then $$\ \frac{\ 1km}{a+b}$$ = 3   => a+b = $$\ \frac{\ 1}{3}$$

$$\ \frac{\ 1km}{a-b}$$ = 4  => a-b = $$\ \frac{\ 1}{4}$$

Adding both the equations, we get 2a = $$\ \frac{\ 1}{3}$$ + $$\ \frac{\ 1}{4}$$

=> a = $$\ \frac{\ 7}{24}$$  

Time taken to cover 1 km = $$\ \frac{\ 1\ km}{a}$$ = $$\ \dfrac{\ 1\ km}{\ \frac{\ 7}{24}}$$ = $$\ \frac{\ 24}{\ \ 7}$$ = $$3\frac{3}{7}$$

Assuming the total capital = 6 units and the total period = 12 months

Investment of A = 6*$$\ \frac{\ 1}{2}$$ = 3 units

Investment period of A = 12*$$\ \frac{\ 1}{3}$$ = 4 month

Investment of A = 6*$$\ \frac{\ 1}{3}$$ = 2 units

Investment period of A = 12*$$\ \frac{\ 1}{4}$$ = 3 months

Investment of A = 6*$$\ \frac{\ 1}{6}$$ = 1unit

Investment period of A = 12 months

Then the profit will be divided in the ratio (3*4):(2*3):(1*12) = 12:6:12 = 2:1:2

The total number of combinations = 10*10*10  = 1000    (There are 10-digits 0 to 9 on each ring)

Since there is only 1 right combination, the number of cases in which the lock cannot be opened = 1000-1=999

Consider the number of first-class tickets =a and the second-class ticket =b

10a+3b=110

a+b=18

Solving two equations, we get a=8, b=10

After interchanging the number of tickets, the new cost = 10*10+3*8 =124

Since the clock is losing time, the original time will be more than 1:45 PM and options A,B, C are less than 1:45 PM. Hence D is the answer.

After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19

Let the C.P. per 1000 gm of the food grain for the merchant be x.

Let us now evaluate the options one by one.

Option A says the profit is straightaway 30%.

In option B, since the weight is reduced by 15%, he will be able to cheat by selling 850 grams instead of 1000 grams. 

So, his effective C.P. in this case will be 0.85x

Also, the S.P. is increased by 15% and so the S.P. will be 1.15x

Profit in this case= $$\ \frac{\ SP-CP}{CP}\cdot100$$= $$\ \frac{\ 1.15x-0.85x}{0.85x}\cdot100$$= $$\ \frac{\ 0.3x}{0.85x}\cdot100$$=35.29%.

In option C, the shopkeeper cheats by selling 700 grams instead of 1000 grams. 

So, effective CP for the shopkeeper= 0.7x

The SP remains the same as original CP as nothing is mentioned about the change. So, SP=x

Profit in this case= $$\ \frac{\ SP-CP}{CP}\cdot100$$= $$\ \frac{\ x-0.7x}{0.7x}\cdot100$$= $$\ \frac{\ 0.3x}{0.7x}\cdot100$$=42.8%

In Option D, if he mixes 30% impurities, for 1000 grams of food grain, he will be able to sell 1300 grams of food grains.

So, Effective CP remains the same=x

Effective SP= 1.3x

Profit in this case= $$\ \frac{\ SP-CP}{CP}\cdot100$$= $$\ \frac{\ 1.3x-x}{x}\cdot100$$= $$\ \frac{\ 1.3x}{x}\cdot100$$=30%

We can see that the profit is maximum in the third case, and hence, Option C is correct.