XAT 2015 Question Paper

Using distance formula,

$$CX = \sqrt{(17.5 - 5.5)^2 + (23.5 - 7.5)^2} = \sqrt{12^2 + 16^2}$$

= $$\sqrt{144 + 256} = \sqrt{400} = 20$$

=> $$AC = 2 \times CX = 40$$

$$BX = \sqrt{(17.5 - 13.5)^2 + (23.5 - 16)^2} = \sqrt{4^2 + 7.5^2}$$

= $$\sqrt{16 + 56.25} = \sqrt{72.25} = 8.5$$

=> $$BD = 2 \times BX = 17$$

$$f(x^2 - 1) = x^4 - 7x^2 + k_1$$

Put $$x^2 = 1$$ to make it 0

=> $$f(0) = (1)^2 - 7(1) + k_1 = k_1 - 6$$ --------(i)

Also, $$f(x^3 - 2) = x^6 - 9x^3 +k_2$$

Put $$x^3 = 2$$

=> $$f(0) = (2)^2 - 9(2) + k_2 = k_2 - 14$$ -----------(ii)

Equating (i) & (ii), we get :

=> $$k_1 - 6 = k_2 - 14$$

=> $$k_2 - k_1 = 14 - 6 = 8$$

Let the principal amount = $$P$$ and rate of interest = $$r \%$$

Interest accumulated from 2004 to 2007 is Rs.10,000 and from 2004 to 2010 is Rs.25,000

Using, $$C.I. = P[(1 + \frac{R}{100})^T - 1]$$

=> $$P[(1 + \frac{r}{100})^3 - 1] = 10,000$$ ----------Eqn(I)

and $$P[(1 + \frac{r}{100})^6 - 1] = 25,000$$ -----------Eqn(II)

Dividing eqn(II) from (I), we get : 

=> $$\frac{P[(1 + \frac{r}{100})^6 - 1]}{P[(1 + \frac{r}{100})^3 - 1]} = \frac{5}{2}$$

Let $$(1 + \frac{r}{100})^3 = x$$

=> $$\frac{x^2 - 1}{x - 1} = \frac{5}{2}$$

=> $$2x^2 - 5x + 3 = 0$$

=> $$(2x - 3) (x - 1) = 0$$

=> $$x = \frac{3}{2} , 1$$    $$(x \neq 1)$$ because then, r = 0

=> $$(1 + \frac{r}{100})^3 = \frac{3}{2}$$

Substituting it in eqn(I)

=> $$P[\frac{3}{2} - 1] = 10,000$$

=> $$P = 10,000 \times 2 = 20,000$$

Minimum tax paid would be = (15*0)+(6*1500*0.05)+(3*3000*0.10)= 450+900 = 1350

Maximum tax paid will be = (15*0) + (12*1500*0.05) + (9*3000*0.10) +(5*5000*0.15) = 0+900+2700+3750 = 7350

Since we have approximate the value so the actual minimum tax will be greater than 1350 and actual maximum tax will be less than 7350

Expression : $$\frac{a + b + c + d}{a + b + c - d}$$

To maximize the above expression, we have to minimize the denominator

Minimum value of the denominator = 1

So we can make $$a + b + c = 26$$ and $$d = 25$$   (as maximizing d will give denominator the least value).

So required maximum value = $$\frac{a + b + c + d}{a + b + c - d}$$

= $$\frac{26 + 25}{26 - 25} = 51$$

L.C.M. of 3,4,5,6 = 60

Number is of the form = $$60 k_1 + 2$$ -------------(i)

When divided by 11, it leaves 0 remainder so number will also be of the form = $$11 k_2$$ ---------(ii)

Hence equating (i) and (ii), we get,

$$60k_1 + 2 = 11k_2$$

$$60k_1 - 11k_2 = -2$$ or $$11k_2 - 60k_1 = 2$$ -----------(iii)

It means $$60k_1$$ will leave remainder 9 when divide by 11.

Lets consider values for 60k1, if k1=1, 60k1=60, reminder is 60mod11=5

120mod11 will be 5+5=10, 180mod11 will be 5+5+5=15, since 15>11, reminder will be 15-11=4

240mod11 reminder will be 4+5=9

$$\therefore$$ By remainder root $$\frac{4 k_1}{11}$$ should leave remainder as 9 or -2

=> Possible values of $$K_1 = 4, 15, 26, 37, 48, 59$$ (As 11 and 60 are co-prime)

$$\therefore$$ Required value = $$60 \times 59 + 2 = 3540 + 2 = 3542$$

Alternatively,
L.C.M. of 3,4,5,6 = 60
As the number 60k+2 is divisible by 11, 60k leaves a reminder of 9
60mod11=5, 120mod11=10, 180mod11=4, 240mod11=9
Hence the first number where both conditions are satisfied as 242.
As 60 and 11 are co-prime, the next number where this is true is 242+60*11
Hence, the numbers are in the form 242+660k
For 6th number, k=5   =>  3300+242=3542

It is evident that, 1 wrong 2 marks question would result in 2.33 deduction from the total (As negative in 2 marks question is 1/3 of a mark)

1 wrong of 1 mark question lead to deduction of 1.25 marks

1 unattempted of 1 mark question lead to deduction of 1.5 marks

1 unattempted of 2 marks question lead to deduction of 3 marks

$$\therefore$$ Rank of student who scores 27.5 = 5

Expression : $$f (x + a) = f (a \times x)$$

Also, $$f(1) = 4$$

Now, $$f(1003) = f(1002 + 1) = f(1002 \times 1) = f(1002)$$

Similarly, $$f(1002) = f(1001) = f(1000) = ......... = f(1) = 4$$

$$\therefore f(1003) = k = 4$$

Scenario I : Devanand's position after $$t$$ hours is $$(50 - 3t)$$ km west  of Pradeep's house, while Pradeep's position is $$4t$$ km south of his own house.

If $$d$$ is the distance between them, then

=> $$d^2 = (50 - 3t)^2 + (4t)^2$$

=> $$d^2 = 2500 - 300t + 25t^2$$

=> $$d^2 = 25 (t^2 - 12t + 36) + 1600$$

=> $$d^2 = 25 (t - 6)^2 + 1600$$

Thus, minimum distance is 40 km after 6 hours.

Thus, scenario I is possible

Scenario II & III are not possible as minimum distance in that case would be 50 km as after that distance will keep on increasing between the two.

Median of 11 integers is 15, => In ascending order 6th integer = 15

=> Numbers = 6,8,12,13,14,15,20,22

Statement I : Average of four smallest = 6 + 8 + 12 + 13

= $$\frac{39}{4} = 9.75$$

It is given that, avg of 4 largest - avg of 4 smallest = 13.25

=> Average of 4 largest = 13.25 + 9.75 = 23

=> Sum of 4 largest numbers = 23 * 4 = 92

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement I is sufficient.

Statement II : Sum of 11 integers = 11 * 16 = 176

Sum of given 8 integers = 6+8+12+13+14+15+20+22 = 110

Sum of remaining numbers = 176 - 110 = 66

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement II is sufficient.

$$\therefore$$ Either statement I or II is sufficient.