SNAP 2007 Question Paper

Let us draw the diagram of the ladder against the wall as given in the question.

Let the length of the ladder be t metres.

We can find BC, i.e the distance of the ladder from the foot of the wall using Pythagorus theorem

$$\ BC^2=\ t^2-5^2$$.,

=>$$\ BC^2=\ t^2-25$$

=> BC=$$\ BC^{ }=\ \sqrt{\ t^2-25}$$

Now, when the ladder is slips 2 m away from the wall, the ladder lies on the ground completely. This can be depicted as:

Now, BC+ BC'= t 

=> $$\ \ \sqrt{\ t^2-25}+2=t$$

=> $$\ \ \sqrt{\ t^2-25}=t-2$$

Squaring both sides,

$$\ \ t^2-25=\left(t-2\right)^2$$

=> $$\ \ t^2-25=\left(t^2-4t+4\right)$$

=>$$4t=29.\ And\ so,\ t=7.25$$

In the year 2003, Wheat: Rice=1.2 and it is given that Rice production= 4 lac tons.

Let the wheat production during 2003 be x.

So, $$\ \frac{\ x}{4}=\ \frac{\ 1.2}{1}$$

=> x= 4.8 lac tons

The number of years in which ratio is greater than 1 will be the years where wheat production was greater than the rice production.
So number of years = 3

Considering statement 1 alone, we can make a table as follow:

We get S-1220=P and S'-1400=1.32P

Since there are 3 variables, S, S' and P, we cannot get the values of the variable using the two equations alone. 

So, Statement 1 alone is not sufficient.

Using statement 2, 1.32P= 4300

=> P= 4300/1.32. But again, two variables, S and S' are unknown.

So, So, Statement 1 alone is not sufficient.

When we use both the given statements, we will have 

S-1220=4300/1.32 and 

S'-1400=4300.

Using these two equations, we can find S and S' respectively and hence answer the question asked.

Therefore, both statements are necessary to answer the question.

At t=0, speed was 40 kmph

At t=180, speed becomes 70 kmph

Therefore, increase in speed= $$\ \frac{\ 70-40}{40}.100=\ \frac{\ 30}{40}.100=75\%\ \ \ $$

Each of the options given assumes the speed of the train to vary linearly wrt time.

So, let us assume the speed of the train at time t be given by 40+ at, because at t=0, the speed was already 40 kmph.

When we put x=30, we get,

45= 40+ 30a

=> 5= 30a Or, a= $$\ \frac{\ 1}{6}\ \ \ $$

So, Speed= 40+ $$\ \frac{\ t}{6}\ \ \ $$

We will check again to verify the results. 

When t=60, Speed= 40+ $$\ \frac{\ 60}{6}\ \ \ $$= 50 kmph, which matches with the value given in the table.

So, Option D is correct

$$2\ \frac{\ 1}{2}\ $$ hours after the beginning of the time period means 120 +$$\left(2\ \cdot60\right)+\left(\frac{\ 1}{2}\cdot60\right)=\ 120+30=150\ \ \ $$ minutes.

From the table, we can infer that after 150 minutes, Speed of the train was 65 km/hour

Understanding the premise of the question, we can infer that the price of the laptop when taken individually is, let's say x and that of the printer is y. 

But, when taken together under the bundle offer, the price is lesser than (x+y), because of the discount given, let's say a%.

Statement 1 gives us the discounted price of the printer and laptop taken together, i.e. $$\ \left(x+y\right)\left(1-\ \frac{\ a}{100}\right)$$

We have three variables and cannot solve this equation.

Statement 2 gives us the value of x. But still, two variables, a and y are unknown.

Even if we use both the statements together, we will have 2 variables and only 1 equation.

Therefore, both statements are insufficient to find the value of y.

Considering statement X,

Percentage increase for Managerial candidates= $$\ \frac{\ \left(45292-19236\right)}{19236}\cdot100=\ \ \frac{\ 26056}{19236}\cdot100=135.45\%$$

Percentage increase in Technical candidates= $$\ \frac{\ \left(63298-61205\right)}{61205}\cdot100=3.3\%$$

Therefore statement X is true.

Considering statement Y,

Total registrations in 2004= 61205+19236= 80441.

Total registrations in 2005= 63298+45292= 108690.

So, growth in the overall number of registrations= $$\ \frac{\ \left(108690-80441\right)}{80441}\cdot100=35.11\%$$, which is greater than 25%.

So, both statement X and Y are true.