XAT 2016 - QUANT

AB = AC = CD, => $$\angle CAD = \angle CDA = 20^{\circ}$$

and $$\angle ABC = \angle ACB$$

In $$\triangle$$ ACD

=> $$\angle ACD + \angle CAD + \angle CDA = 180^{\circ}$$

=> $$\angle ACD = 180^{\circ} - 20^{\circ} - 20^{\circ} = 140^{\circ}$$

=> $$\angle ACB = 180^{\circ} - 140^{\circ} = 40^{\circ} = \angle ABC$$

Similarly, In $$\triangle$$ ABC

=> $$\angle BAC = 180^{\circ} - 40^{\circ} - 40^{\circ} = 100^{\circ}$$

$$\therefore \angle BAD = 100^{\circ} + 20^{\circ} = 120^{\circ}$$

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

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.

AB + BC + CD = 70 ------------(i)

AB + BD = 45 ------------------(ii)

AC + CD = 30 ------------------(iii)

AC + CB + BD = 65 ------------(iv)

Adding (i) & (iv)

=> AB + BC + CD + AC + CB + BD = 70 + 65

(AB + BD) + 2 BC + (AC + CD) = 135

From (ii) & (iii)

=> 45 + 2 BC + 30 = 135

=> 2 BC = 135 - 75 = 60

=> BC = $$\frac{60}{2} = 30$$ min

10 hour analog clock is used. 

Time in mirror = 3 hours 42 minutes and 20 seconds

=> Actual time = 10 - (3 hrs 42 min 20  sec)

= 6 hrs 17 min 20 sec

$$\therefore$$ Time after 5 minutes = 6 hrs 22 min 20 sec

|a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10

Expression : $$[abc - (a + b + c)]$$

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if $$a = -10 , b = -9 , c = 8$$

=> Max value = $$[(-10 \times -9 \times 8) - (-10 -9 + 8)]$$

= $$720 + 11 = 731$$

When a square sheet is folded in half, its area is also halved. Now, an isosceles right triangle is formed after folded. For 2nd & 3rd folds, again areas halved each time and isosceles right triangle is formed

Equal sides = 10 cm

Area of last such triangle = $$\frac{1}{2} \times 10 \times 10 = 50 cm^2$$

$$\therefore$$ Area of original square = $$50 \times 2 \times 2 \times 2$$

= $$400 cm^2$$

Let radius of incircle = $$r$$, => Radius of circumcircle = $$2r$$

Difference in area = $$\pi [(2r)^2 - (r)^2] = 2156$$

=> $$3 \times \frac{22}{7} \times r^2 = 2156$$

=> $$r^2 = \frac{2156 \times 7}{66}$$

=> $$r = \sqrt{\frac{686}{3}}$$

Now, height of equilateral triangle = $$3 r = \frac{\sqrt{3}}{2} a$$    (where $$a$$ is side of triangle)

=> $$3 \times \sqrt{\frac{686}{3}} = \frac{\sqrt{3}}{2} a$$

=> $$a = 2 \sqrt{686}$$

$$\therefore$$ Area of triangle = $$\frac{\sqrt{3}}{4} a^2$$

= $$\frac{\sqrt{3}}{4} \times 4 \times 686 = 686 \sqrt{3} cm^2$$

If the object does not change direction at point C then it will be at D above the person. But since it is given that object changed direction and continued flying upwards, thus object would reach point E. 

=> CG is perpendicular bisector to AB => BG = 300 m

In $$\triangle$$ BCG

=> $$cos 30 = \frac{BG}{BC}$$

=> $$\frac{\sqrt{3}}{2} = \frac{300}{BC}$$

=> $$BC = \frac{600}{\sqrt{3}} = 200 \sqrt{3}$$

Now, according to statement I, time can increase only when the angle with ground level will increase.

But, according to statement II, if CD = $$200 \sqrt{3}$$ (= BC), then at constant speed, it will not take any additional time and thus there should not be any increase in angle.

But in first statement direction has been changed whereas in second statement direction has not been changed. 

So both the statements are inconsistent with each other. 

If $$(a + b)^{(a + b)}$$ is divisible by 500, 

$$500 = 2^2 \times 5^3$$

=> Least value of $$a + b = 2 \times 5 = 10$$

For least $$a$$ and $$b$$, let $$a = 1$$

=> $$b = 10 - 1 = 9$$

$$\therefore$$ Min $$(a \times b) = 1 \times 9 = 9$$

Let walking speed = $$a$$

=> Driving speed = $$8a$$ 

Let the distance between temple and office = x

Acc. to ques,

=> $$\frac{x}{a} = \frac{1}{a} + \frac{x + 1}{8 a}$$

=> $$8x = 8 + x + 1$$

=> $$x = \frac{9}{7}$$

Since a,b,c are consecutive integers

=> $$a = b-1$$ and $$c = b+1$$

Expression : $$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$

= $$\frac{(b - 1)^3 + b^3 + (b + 1)^3 + 3 (b - 1) b (b + 1)}{(b - 1 + b + b + 1)^2}$$

= $$\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 - 3b}{9 b^2}$$

= $$\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$$

Putting different values of b from - 10 to 10, we can verify that only - 1 and 1 satisfies to get integer values for the expression.

Ans - (C)

Let P and Q be the centres of the circles with arcs x and y respectively.

Thus, $$\angle APB = 90$$ and $$\angle AQB = 60$$

Also, length of arc $$y = 10 \pi$$ cm

=> $$\frac{\theta}{360} \times 2 \pi r = 10 \pi$$

=> $$\frac{1}{6} \times 2 \times r = 10$$

=> $$r = AQ = 10 \times 3 = 30$$ cm

=> AB = 30 ($$\because \triangle$$ AQB is equilateral triangle)

Also, $$\triangle$$ APB is right isosceles triangle, => $$AP = \frac{30}{\sqrt{2}}$$ 

$$\therefore$$ Arc length = $$x = \frac{90}{360} \times 2 \pi \times \frac{30}{\sqrt{2}}$$

= $$\frac{15 \pi}{\sqrt{2}}$$

Let $$DQ = x$$, => $$BP = 2x$$

Acc. to ques,

=> $$ar (\triangle BPD) + ar (\triangle BQD) \leq ar (PBQD)$$

=> $$(\frac{1}{2} \times AD \times BP) + (\frac{1}{2} \times BC \times QD) \leq 150$$

=> $$(\frac{1}{2} \times 9 \times 2x) + (\frac{1}{2} \times 13 \times x) \leq 150$$

=> $$31x \leq 300$$ => $$x \leq \frac{300}{31}$$

=> $$x \leq 9.68$$

Thus, for $$x$$ to be an integer and positive, 9 different values (1 to 9) are possible. 

Since the population of the different areas is unknown, None of the two statements can be inferred.
Hence D is the correct answer.

Since there is no data available about the population of the different areas, all the other options except C can be inferred. Option C cannot be inferred as it talks about the ratio of the customers.
Hence C is the correct answer.

Option  A,B ,C,D talk about the increase or decrease in the number of shops with respect to each other. Since the total number of shops can be different in two different years, hence they cannot be definitely concluded.

Option E compares the numbers of shops of two types in the same year. Since the percentage is given and total number of shops is same, hence E can be concluded. Hence E is the correct answer.

Let's tabulate the information given in the graph, 

For a salary range we can get the highest and the lowest value of the range but cannot get the distribution of individual student. 

In the year 2008 we have a complete salary range(76-100) inside it and some students of the range 51-75. 

In the year 2009 some students of the range 51-75 and some students of the range 75-100. 

In the year 2010 some students of the range 51-75 and some students of the range 75-100. 

In the year 2012 we have a complete salary range 51-75 inside it and some students of the range (76-100) . 

Since we cannot determine the exact students and cannot infer a particular year, Hence answer is E i.e cannot be determined.

Let's tabulate the information given in the graph, 

In all the options since 2009 is given 

Lets check if 2009 is satisfying the given criteria 

Median salary of 2009 = 24 l 

Avg salary of 2008 = 16 l

So Median salary of 2009 is between 40 to 50% higher than 2008 avg salary.

Hence options 1,3,4,5 can be negated . 

Option B is the correct answer

Let's tabulate the information given in the graph, 

The number of years cannot be calculated with precision as the distribution pattern is unknown .

Hence E is the correct answer

Let's tabulate the information given in the graph, 

It is clear from the table that the  mean salary after discarding the unemployed students is greater than the mean salary for the years 2008,2010,2013

Hence A is the correct answer.