SNAP 2008 Question Paper

As the graph representing production line 1 is showing a steady increase, the given graphs represent cumulative productions. The graph representing production line 2 is showing no change from the 4th month to 6th month so It was the period of breakdown.
As the graph representing production line 2 is not following any pattern, we cannot quantify the loss of production as the production in those months cannot be determined.
So Only A can be answered

Let n pages be missing starting from the p th page.

So,$$p+(p+1)+(p+2)\dots..+[p+(n−2)]+[p+(n−1)]=9808$$

$$9808=\frac{n}{2}[2\times p+(n−1)\times1]$$
$$n^2+(2p-1)n−19616=0$$
Now both p and n are natural numbers 
Now possible cases 
$$n=19616,\ p=−9807$$
$$n=−19616,\ p=9808$$
$$n=613,\ p=−29$$
$$n=−613,\ p=29$$
$$n=1,\ p=9808$$
$$n=−1,\ p=−9807$$
$$n=32,\ p=291$$
$$n=−32,\ p=−290$$

But since p and n has to be natural numbers, so all solutions except 5 and 7 are discarded.

Also, it is mentioned in the question that "a number of consecutive pages are missing". Thus, case 5 can be discarded(i.e. n = 1 and p = 9808)

So, the correct option is C

We have the series as : 2,3,13,37,86,167,288
Now taking difference between 2 consecutive terms :
3-2 =1 
13-3 =10
37-13=24
86-37 = 49
167-86 = 81
288-167 = 121
Now if you observe the differences are squares of consecutive odd numbers except 13-3 and 37-13 and so we can say in place of 13 , it should have been 12 
Hence 13 is the wrong number .

Assume the speed in km/hr be a and b.

Then, the distance travelled after half an hour = $$\ \frac{\ a}{2}$$ and $$\ \frac{\ b}{2}$$

The distance between them = $$\ \ \ \frac{\ 1}{2}\sqrt{\ a^2+b^2}$$  = 17

=> $$\sqrt{\ a^2+b^2}$$ = 34

=> $$a^2+b^2$$ = 1156  .....(1)

After 45 minutes, the difference of the distance travelled = $$\ \frac{\ 3}{4}\left(a-b\right)$$ = 10.5

=> a-b = 14   => $$a^2+b^2-2ab\ $$ = 196.......(2)

From 1 and 2, 2ab = 960 

$$a^2+b^2$$ +2ab = 1156+960= 2116

=> a+b=46

Hence a=30 and b=16

Tabulating the percentage, we get

From the table, in case of Y, the percentage is 50% and rest of the hotels have less than 50%. Hence C is the answer.

Using $$a_n$$ = a+(n-1)d

=> $$a_57$$=a+56d

From 1, the value of d can be obtained.

From 2, an equation in a and d is obtained. 

Hence, both 1 and 2 are necessary to calculate the value of $$a_57$$.

The cost of levelling and turfing a square field at Rs. 160 per hectare is Rs. 2624.40.

Hence, the area = $$\ \frac{\ 2624.40}{ 160}$$ = 16.4025 * 10000 square metres             (1 Hectare = 10000 square metres)

Assume the length of the side=a, the area, a$$^2$$ = 16.4025 * 10000 

=> a = 405 m

Then the perimeter = 4a = 405*4

The cost of surrounding = 405*4*25 paise = Rs 405

Assume the height of the tower = x and the shadow formed when the inclination = 60 is y. 

In the given figure, tan60= $$\ \frac{\ x}{y}$$...............(1)

tan30=$$\ \frac{\ x}{y+60}$$...................(2)

Dividing 1 by 2, we get, $$\ \frac{\ y}{y+60}$$ = 3

=> y=30  =>x=y$$\sqrt{\ 3}$$ = 30$$\times\ $$1.732=51.96 m 

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400    => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

The viewers in the city C watch less than one movie a week = 13600

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400 => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

E has the highest number of viewers who watch less than one movie a week.