XAT 2016 Question Paper

From the figure, $$a + 22 + 25 = 77$$

=> $$a = 77 - 47 = 30$$

Similarly, $$c = 55 - 22 = 33$$

and $$b = 50 - 25 = 25$$

It is given that 40 visitors did not opt for ride A and B, or both

=> $$b + d = 40$$

=> $$d = 40 - 25 = 15$$

$$\therefore$$ Total number of people who visited with the entry pass on that day

= $$a + b + c + d + 22 + 25$$

= $$30 + 25 + 33 + 15 + 47 = 150$$

AB = 54 cm and $$\triangle$$ANM , $$\triangle$$OCP , $$\triangle$$OPX are equilateral triangles.

=> MN = MR = NO = OP = PQ = QR = $$\frac{54}{3} = 18$$ cm

Thus,  MNOPQRM is a regular hexagon with side 18 cm

$$\therefore$$ Area of  MNOPQRM = $$\frac{3 \sqrt{3}}{2} (side)^2$$

= $$\frac{3 \sqrt{3}}{2} \times 18^2 = 486 \sqrt{3} cm^2$$

Area of rectangular floor = $$9.5 \times 11.5$$

= $$109.25 m^2$$

Now, cost of covering $$109 m^2$$ (with 1x1 tiles) = $$109 \times 100$$ = Rs. 10,900

Cost of covering $$0.25 m^2$$ (with 0.5 m square tile) = Rs. 30

$$\therefore$$ Total cost = $$10,900 + 30 = Rs. 10,930$$

Let A,B,C,D,E,F represents Anita, Biplove, Cheryl, Danish, Emily and Feroze respectively.

Statement I : It is not sufficient as nothing is mentioned about D and F.

Statement II : E = F, thus E and F cannot score the lowest.

Also, C scored higher than at least 2 and B scored more than D while A scored the highest.

The only person left with the lowest marks can be D.

Thus, statement II alone is sufficient.

Let number of apples = $$x$$ and oranges = $$y$$

=> $$23x + 10y = 653$$   $$(x > y)$$

Since, 653 has last digit 3, which is possible when 23 is multiplies by 1,11,21,31 and so on.

Also, $$x > y$$ => $$x = 21$$ and $$y = 17$$

=> C.P. of 1 apple = $$\frac{100}{115} \times 23 = 20$$

C.P. of 1 orange = $$\frac{100}{125} \times 10 = 8$$

=> Total C.P. = $$(21 \times 20) + (17 \times 8) = 420 + 136 = 556$$

$$\therefore$$ Profit % = $$\frac{653 - 556}{556} \times 100 = 17.4 \%$$

Set :  {$$1, 3, 3^{2}, 3^{3},…...,3^{100}$$}

Clearly, this set is a G.P. with common ratio, $$r = 3$$

Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

Number of terms = 101

Last term = $$3^{100}$$

Sum of remaining terms = $$\frac{1 (3^{100} - 1)}{3 - 1}$$

= $$\frac{3^{100} - 1}{2}$$

$$\therefore$$ Required ratio = $$\frac{3^{100}}{\frac{3^{100} - 1}{2}}$$

= $$\frac{3^{100} \times 2}{3^{100} - 1}$$

$$\approx \frac{3^{100} \times 2}{3^{100}} = 2$$

Expression : $$f(x) = 2x + f(x - 1)$$ and $$f(1) = 1$$

Putting, $$x = 2$$

=> $$f(2) = 2 \times 2 + f(1) = 4 + 1 = 5$$

and $$f(3) = 2 \times 3 + f(2) = 6 + 5 = 11$$

Similarly, $$f(4) = 2 \times 4 + f(3) = 8 + 11 = 19$$

The pattern followed is : $$f(n) = n^2 + (n - 1)$$

Now, $$n = 31$$

$$\therefore f(31) = 31^2 + (31 - 1) = 961 + 30 = 991$$

Converting numbers in decimal system

=> $$2061_B = 2B^3 + 6B + 1$$

and $$601_B = 6B^2 + 1$$

Acc. to ques, => $$2B^3 + 6B + 1 + 6B^2 + 1 = 432$$

Solving above equation, we get : $$B = 5$$

$$\therefore 1010_5$$ in decimal system = $$1(5)^3 + 1(5) = 130$$

Note :-  This question is technically wrong as we cannot have digit 6 in Base 5. 

M inlet pipe can fill $$(\frac{M}{8})^{th}$$ part of the tank in 1 hour.

Similarly, N outlet pipes can empty $$(\frac{N}{12})^{th}$$ part of the tank in 1 hour.

=> $$\frac{M}{8} - \frac{N}{12} = \frac{1}{6}$$

=> $$6 M - 4 N = 8$$

=> $$M = \frac{4 + 2 N}{3}$$

If, N = 1, => M = 2 => $$M : N = 2 : 1$$

If, N = 4, => M = 4 => $$M : N = 1 : 1$$

Thus, Ans - (E)

XYZ running cost = Rs 400/student

Each student pays Rs 2000

Profit = 2000 -  400 = 1600

ABC receives 60% of profit, i.e. 1600*0.6 = Rs 960

XYZ begins to make profit from first student itself.

For ABC to recover its investment, number of students should be 100000/960 = 104.16

Therefore, partnership needs at least 105 students to reach break even point. 

The answer is option C.