XAT 2017 Question Paper

$$f(x) = ax+b$$.
$$f(f(x)) = f(ax+b) = a(ax+b)+b = a^2x + ab+b$$
We have been given that $$f(f(x)=9x+8$$
$$9x+8=a^2x+ab+b$$
Equating the co-efficient of $$x$$, we get, $$a^2=9$$. Therefore, $$a$$ can be $$3$$ or $$-3$$. But, it has been given that $$a$$ is a positive real number. 
Equating the constants, we get, $$ab+b=8$$
If we substitute $$a=3$$, we get, $$3b+b=8$$
$$4b=8$$
$$b=2$$
Therefore, $$a+b=3+2=5$$
Therefore, option C is the right answer. 

Both of them row the boat for the same time. Therefore, we can ignore the time taken to row the boat and add it to the final answer. 
Arup will walk 2 km in 2/4 = 30 minutes.
Swarup will walk 2 km in 2/5 = 24 minutes.

Therefore, Swarup will start driving 30-24 = 6 minutes earlier than Arup. 
In 6 minutes, Swarup will gain a lead of 6*40/60 = 4 km. 
Arup drives at 50 kmph (i.e, 10 kmph faster than Swarup). Therefore, Arup will cover the 4 km advantage that Swarup has in 4/10*60 = 24 minutes. 
Therefore, the time after which both of them will meet is 30 (to walk) + 30 (to row the boat) + 24 = 1 hour 24 minutes after Arup starts. 
Since both of them start at 8 AM, they will meet again at 9:24 AM. 
Therefore, option D is the right answer. 

Let the age of the father be 'f', mother be 'm', Hari be 'h', Chari be 'c'. It has been given that Chari is 4 years older than Gouri. Therefore, the age of Gouri is c-4. 

When Chari was born, the sum of the ages of Hari, his father and mother was 70 years. 
If Chari's age is 'c' now, then Chari's father's age when Chari was born would have been 'f-c' (i.e, Current age - the number of years that has passed after Chari's birth). The same holds true for all the family members. 

=> f - c + m - c + h - c = 70
f+m+h-3c = 70 -------------(1)

The sum of the ages of the 4 family members when Gouri was born was twice the sum of the ages of the father and mother at the time of Hari's birth. 

=> f-(c-4) + m -(c-4) + h -(c-4) + c - (c-4) = 2(f-h+m-h)
=> f + m + h + c - 4(c-4) = 2f + 2m - 4h
f+m+h-3c + 16 = 2f+2m-4h
Substituting (1), we get,
70 +16 = 2f+2m-4h
43 + 2h= f+m ---------------(2)
Substituting (2) in (1), we get, 
43 + 2h + h - 3c = 70
3h - 3c = 27
=> h-c = 9
Therefore, the difference between the age of Hari and Chari is 9 years. Therefore, option E is the right answer.

Both Amit and Ben scored 35 marks. 4 marks are awarded for a correct answer and 1 mark is deducted for an incorrect answer. 
Let the number of questions that Amit got right be 'x'. 
=> 4x -(10-x) = 35
5x = 45
x = 9.
Therefore, Amit must have made only 1 mistake. The same must have been the case with Ben too. 
The responses given by the 3 persons are as follows:


AMIT     T T F F T T F T T F
BENN    T T T F F T F T T F 
CHITRA T T T T F F T F T T

Amit and Ben have given different responses for question 3 and question 5. Therefore, one of them must be wrong in each of these questions. Also, Amit must have given the correct answer for one of these 2 questions and Ben must have answered the other one correct, since both of them got 9 questions correct. 

Let us assume Amit has given an incorrect response for question 3 and Ben has given an incorrect response for question 5. 
In this case, Chitra's would have given 4 correct responses (questions 1,2,3, and 9). Chitra's score would have been 4*4 - 6 = 10.

Let us assume Amit has given an incorrect response for question 5 and Ben has given an incorrect response for question 3. 
In this case, Chitra's would have given 4 correct responses (questions 1,2,5, and 9). Chitra's score would have been 4*4 - 6 = 10.

Therefore, Chitra's score should have been 10 and hence, option A is the right answer. 

Let us note down the information given:

No person can major in all 3 subjects. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. There are 60 students in total.

It has been given that average marks scored by students majoring in Maths in English is 45. 
Average marks scored by students majoring in Biology in English is 60.
But the combined average marks scored by students majoring in Maths and Biology in English is 50.
=> $$\dfrac{45*(a+6) + 60*(c+15)}{a+6+c+15} = 50$$
$$45a+270+60c+900 = 50a+50c+1050$$
$$5a=10c+120$$
$$a=2c+24$$
To maximize b, we have to minimize 'a' and 'c'. The least value that 'c' can take is 0.
The corresponding value of a is 24. 
24+6+b+15 = 60
=> b = 15.
Therefore, the maximum number of students who could have majored only in physics is 15. Therefore,option D is the right answer. 

For smaller values of 'x', cos 'x' will be close to 1 and sin x will be close to 0. 
Sin 30° = 1/2 = 0.5
As we move closer to 0, the value of cos x increases and sin x decreases. Therefore, when we move from a larger value of x to a smaller value, the value of the expression cos x - sin x will increase. 

Let us evaluate the value of the expression at x= 30° (since we know the values of cos x° and sin x° at this point) and then logically deduce the range of the value of the expression. 

At x = 30°, cos x = $$\frac{\sqrt{3}}{2}$$ and sin x = $$\frac{1}{2}$$
Therefore, cos x° - sin x° would have been 1.732/2 - 0.5 = 0.866 - 0.5 = 0.366

sin 30° + cos x - sin x will be 0.866

At x = 0, sin 30 + cos x - sin x will be 0.5 + 1 - 0 = 1.5.
At x = 5°, the value of the expression sin 30 + cos x - sin x  will be slightly less than 1.5. 
Therefore, we can infer that the upper limit of the expression will be greater than 1. None of the given limits include values greater than 1. Therefore, option E is the right answer. 

Alternate solution:

The value of sin 15° and cos 15° can be found out using the identities sin (A-B) = sin A cos B - cos A sin B and cos (A-B) = cos A cos B + sin A sin B.
Substituting A = 45° and B = 15° in these expressions, we get, 
sin 15° = 0.2588
cos 15° = 0.9659

sin 30° + cos 15° - sin 15° = 0.5 + 0.9659 - 0.2588 = 1.2071.
None of the given options capture this value in the range. Therefore, option E is the right answer.

Volume of the pyramid = $$200$$ cubic cm. 
The volume of  a pyramid is usually a third of the volume of a cuboid of the same height. 
Therefore, a square cuboid of the height of the pyramid will have a volume of $$600$$ cubic cm.

We know that the height is $$13$$ cm.
Area of the base square* height = $$600$$ cm.
=> Area of the base square = $$\frac{600}{13}$$ cm$$^2$$.
Side of the base square = $$\sqrt{\frac{600}{13}}$$ cm.

Length of diagonal of the base square = $$\sqrt{\frac{600}{13}}*\sqrt{2}$$

Now, the height of slant edge of the pyramid can be found out by using the Pythagoras theorem. 
Length of half the diagonal of the base square will form one of the sides and the height of the pyramid will form the other side. The slant height of the pyramid will be the hypotenuse of the right-angled triangle. 

Height of slant edge = $$\sqrt{13^2 + \frac{1}{4}*2*\frac{600}{13}}$$
                   = $$\sqrt{169 + \frac{600}{26}}$$
                   = $$\sqrt{\frac{4394+600}{26}}$$
                   = $$\sqrt{ \frac{4994}{26}}$$
                   =$$\sqrt{192.07}$$
                   =$$13.85$$ cm

The nearest integer is $$14$$. Therefore, option C is the right answer.

A die is rolled twice.
The number of combinations that can occur = 6*6 = 36.
We have to find the probability of the second roll being higher than the first. 
If we select 2 numbers out of the 6 and arrange them in ascending order, then we will obtain the scenario in which the number obtained in the second roll will be greater than the number obtained in the first roll. 

2 numbers out of 6 numbers can be selected in 6C2 = 15 ways. The numbers can be arranged in ascending order in only one way.
Therefore, the required probability is 15/36. 

Therefore, option C is the right answer.

We have to employ the pigeon hole principle to solve this problem.
The maximum number of students who can be picked from each department such that 5 students are not selected from the same department is 4. 
Therefore, after 4 students from each department are selected (i.e., 4*5 = 20 students in total), the 21st student selected will be the fifth student to be selected from one of the 5 departments. Therefore, 20+1 = 21 students should be selected in total to ensure that at least five students from one of the departments is selected. Therefore, option C is the right answer.

It has been given that $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. 

$$p,q,r$$ and $$s$$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $$0$$.

$$11^{p + 7}$$:

This expression must be made a perfect cube. The nearest perfect cube is $$11^9$$. Therefore, the least value that $$p$$ can take is $$9-7=2$$.

$$7^{q - 2}$$

The least value that $$q$$ can take is 2. If $$q=2$$, then the value of the expression $$7^{q-2}$$ will become $$7^0=1$$, without preventing the product from becoming a perfect cube. 

$$5^{r+1}$$:

The least value that $$r$$ can take is $$2$$.

$$3^{s})$$:

The least value that $$s$$ can take is $$3$$. 

Therefore, the least value of the expression $$p + q + r + s$$ is $$2+2+2+3=9$$.
Therefore, option E is the right answer.