XAT 2015 - QUANT

The series is an A.P. with common difference, $$d = -66 - (-64) = -2$$

First term, $$a = -64$$ and last term $$a_n = -100$$

nth term of the series, $$a_n = a + (n - 1) d$$

=> $$-100 = -64 + (n - 1) (-2)$$

=> $$n - 1 = \frac{-36}{-2} = 18$$

=> $$n = 18 + 1 = 19$$

$$\therefore$$ Sum = $$\frac{n}{2} (a + a_n)$$

= $$\frac{19}{2} \times (-64 -100) = \frac{19}{2} \times (-164)$$

= $$19 \times (-82) = -1558$$

Volume of Cylinder = $$\pi r^2 h = \pi \times 7^2 \times 10 = 490\pi$$

Now, The solid metal cylinder is re-cast into two cones in the proportion 3 : 4 i.e. the volumes of cone 1 and cone 2 is

$$210 \pi$$ and $$280 \pi$$ respectively.

So, flat Surface area of cylinder before melting = $$2 \pi r^2 = 2 \pi \times 7^2 = 98 \pi$$

Volume of cone 1 = $$\frac{1}{3} \pi r_1^2 h = 210 \pi$$

=> $$r_1^2 = \frac{210 \times 3}{10} = 63$$

Volume of cone 2 = $$\frac{1}{3} \pi r_2^2 h = 280 \pi$$

=> $$r_2^2 = \frac{280 \times 3}{10} = 84$$

Flat surface area of cones = $$\pi r_1^2 + \pi r_2^2$$

= $$\pi (63 + 84) = 147 \pi$$

$$\therefore$$ Percentage change in surface area = $$\frac{147 \pi - 98 \pi}{98 \pi} \times 100$$

= $$\frac{1}{2} \times 100 = 50 \%$$

Let Manufacturing Cost of the product = $$Rs. 100$$

=> Maximum Retail Price(MRP) = $$100 + \frac{55}{100} \times 100 = Rs. 155$$

Retailer gives 10% discount on MRP

=> Retailer's selling price = $$155 - \frac{10}{100} \times 155 = Rs. 139.5$$

It is given that the retailer earned 23% profit on his purchase price, say $$Rs. x$$

=> $$\frac{123 x}{100} = 139.5$$

=> $$x = \frac{13950}{123} = 113.41$$

Now, the purchase price of retailer = $$x$$ = selling price of Manufacturer

$$\therefore$$ Profit earned by Manufacturer = $$113.41 - 100 = 13.41$$

$$\approx 13 \%$$

The probability that his friend receives the gift in time will be when his friend receives even one gift.

That can be calculated as the probability of his friend receiving at least one gift.

The probability that none of the retailers sends in time = $$(1 - 0.6) \times (1 - 0.8) \times (1 - 0.9) \times (1 - 0.5)$$

= $$0.4 \times 0.2 \times 0.1 \times 0.5 = 0.004$$

$$\therefore$$ Probability of his receiving at least one gift = $$1 - 0.004 = 0.996$$

Area of given figure = 144 sq meter

It is given that BCE becomes square when we will unfold it, so to find the complete area of the figure shown as dotted after unfolding we need to add the area of triangle BCE.

Thus, BC = CE = 6 m

=> Area of $$\triangle$$ BCE = $$\frac{1}{2} \times 6 \times 6 = 18$$ sq meter

$$\therefore$$ Final area of whole figure = 144 + 18 = 162 square meter.

When x  = -3, y = -10
This is satisfied only in option D.

Hence, option D is the correct answer.

Let the quantities of the chemicals X and Y, mixed to produce product M be $$5c$$ and $$4c$$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $$\frac{3}{4} \times 5c = \frac{15 c}{4}$$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $$\frac{2}{3} \times 4c = \frac{8 c}{3}$$

Quantity of B in M = $$\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $$9c = 864$$ => $$c = \frac{864}{9} = 96$$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $$\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$$

$$\therefore$$ Quantity of water added to M = $$1232 - 864 = 368$$ units

Perimeter of square ABCD = 200 ft

=> AB = $$\frac{200}{4} = 50$$ ft

=> $$DB = \sqrt{50^2 + 50^2} = 50 \sqrt{2}$$ ft

=> $$BO = r = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2}$$ ft

Width of the road = BX = $$7 \sqrt{2}$$ ft

=> $$BX = R = 25 \sqrt{2} + 7 \sqrt{2} = 32 \sqrt{2}$$

Area of bigger circle = $$\pi R^2 = \pi (32 \sqrt{2})^2 = 2048 \pi$$ sq. ft

Area of smaller circle = $$\pi r^2 = \pi (25 \sqrt{2})^2 = 1250 \pi$$ sq. ft

=> Area of road = $$2048 \pi - 1250 \pi = 798 \times \frac{22}{7} = 2508$$ sq. ft

But we have to calculate cost of construction of 50% road.

Required Construction = $$\frac{2508}{2} = 1254$$ sq. ft

$$\therefore$$ Cost of 1254 ft = $$1254 \times 100 = Rs. 1,25,400$$

In $$\triangle$$ ABF

=> $$tan 30 = \frac{BF}{AB}$$

=> $$\frac{1}{\sqrt{3}} = \frac{10}{AB}$$

=> $$AB = 10 \sqrt{3}$$

Similarly, $$ED = 10 \sqrt{3}$$

Also, $$\angle ECD = \angle BCF = 60$$ (Vertically opposite angles)

In $$\triangle$$ BCF

=> $$tan 60 = \frac{BF}{BC}$$

=> $$\sqrt{3} = \frac{10}{BC}$$

=> $$BC = \frac{10}{\sqrt{3}}$$

=> Height = $$AD = AB + BC + CD = 10 \sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10 \sqrt{3}}{\sqrt{3}}$$

$$\therefore area (\triangle AED) = \frac{1}{2} \times AD \times ED$$

= $$\frac{1}{2} \times \frac{40 + 10 \sqrt{3}}{\sqrt{3}} \times 10 \sqrt{3}$$

= $$50 (\sqrt{3} + 4)$$

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.

AB + BC + CD + AD = 1120 ------------Eqn(I)

PB + BC + CD + PD = 1000 -------------Eqn(II)

Subtracting eqn(II) from (I), we get :

=> AB - PB + (AD - PD) = 120

=> AB - PB + AP = 120

=> AB + AP = 120 + PB

Now, if AB = PB, => AP = 120

=> AD = 600 and BC = 480, then AB + PB + CD = 40, which is not possible (We know that BC = PD. If BC = PD = 480, then BC+PD = 960. PB + BC + CD + PD = 1000.
=> PB+CD = 40. Therefore, AB + PB+CD should be greater than 40).

Similarly, AB = AP is also not possible. Thus $$AP = BP$$

=> $$\angle ABC = x + (180 - 2x) = (180 - x)$$

=> $$sin \angle ABC = sin (180 - x) = sin x$$

Also, perimeter of PBCD = $$10y = 1000$$ => $$y = 100$$

and perimeter of ABCD = $$AB + 10y = 1120$$ => $$AB = 120$$

Applying cosine rule in $$\triangle$$ ABP

=> $$cos x = \frac{(AB)^2 + (AP)^2 - (BP)^2}{2 AB AP}$$

=> $$cos x = \frac{(120)^2 + (100)^2 - (100)^2}{2 \times 120 \times 100}$$

=> $$cos x = \frac{120}{200} = \frac{3}{5}$$

$$\therefore sin x = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}}$$

= $$\sqrt{\frac{16}{25}} = \frac{4}{5}$$

Since the three digit number is divisible by 10, then the unit's digit is 0

Let the three digit number = $$a b 0$$

After the digits are interchanged, the new number is also divisible by 10, thus only a and b are interchanged.

=> New number = $$b a 0$$

Difference between number is divisible by 40

=> $$(100a + 10b) - (100b + 10a) = 40 k$$    (k is constant)

=> $$90a - 90b = 90 (a - b) = 40 k$$

=> k = $$\frac{9\left(a-b\right)}{4}$$

Since k is a natural number (a-b) should be a multiple of 4

If a = 9 , the values of b that satisfies the given equation are 1,5

If a = 8 , the value of b that satisfies the given equation is 4

If a = 7 , the values of b that satisfies the given equation is 3

If a = 6 , the values of b that satisfies the given equation is 2

If a = 5 , the values of b that satisfies the given equation is  1

The number could be = 510,620,730,840,950, 910

Thus, there are 6 numbers that satisfy these conditions.

If a point is equidistant from all 3 vertices, it has to be the circumcentre. The given circle with centre S is concentric and touches two sides.

As S is equidistant from 2 of the sides (say AB and AC), => It lies on angle bisector of $$\angle A$$.

=> $$\triangle ABC$$ is isosceles with AB = AC

Radius of the circle = RS = SQ = 175 cm and SA = SB = SC = 625 cm

=> $$AR = \sqrt{625^2 - 175^2} = 600$$

Let SP = x

=> $$(BP)^2 = (BA)^2 - (AP)^2 = (BS)^2 - (SP)^2$$

=> $$1200^2 - (625 + x)^2 = 625^2 - x^2$$

=>$$1200^2 - 625^2 - x^2 - 2*625x = 625^2 - x^2$$

=> $$1200^2 - 2 * 625^2 = 1250x$$

=> $$x = \frac{658750}{1250} = 527$$

=> $$BP = \sqrt{625^2 - 527^2} = 336$$

$$\therefore$$ ar $$(\triangle ABC)$$ = $$\triangle ASB + \triangle ASC + \triangle SBC$$

= $$(600 \times 175) + (600 \times 175) + (527 \times 336)$$

= $$105000 + 105000 + 177072 = 387072$$

None of the answers given are correct. The reasoning is as given below.

999000 is a multiple of 8 but not of 16. If N! is a multiple of 16, M! would also be a multiple of 16 and hence M!-N! would be a multiple of 16.

Hence, as M!-N! = 999000, it would imply that N! is a multiple of 8 and not of 16. Therefore, N is either 4 or 5. So, N! is either 24 or 120. So, it would imply that M! is either 999024 or 999120. Both of which are not factorials for any natural number. 

Hence, the given question is wrong. 

Let O be the centre of the clock. Let the person's eye be at A and the tip of minute hand at 5.00 p.m. is at P and at 5.10 p.m. at Q

AM = 1800 m and OP = OQ = $$200\sqrt{3}$$ m

In $$\triangle$$ APM

=> $$tan 30 = \frac{PM}{AM}$$

=> $$\frac{1}{\sqrt{3}} = \frac{PM}{1800}$$

=> $$PM = \frac{1800}{\sqrt{3}} = 600 \sqrt{3}$$

=> $$OM = PM - OP = 600 \sqrt{3} - 200 \sqrt{3} = 400 \sqrt{3}$$

In $$\triangle$$ OBM

=> $$tan 60 = \frac{OM}{BM}$$

=> $$\sqrt{3} = \frac{400 \sqrt{3}}{BM}$$

=> $$BM = 400$$ m

=> $$AB = AM - BM = 1800 - 400 = 1400$$ m

Time taken to reach B from A = 10 minutes = 600 sec

$$\therefore$$ Speed of the person = $$\frac{1400}{600} = \frac{7}{3}$$ m/s

= $$(\frac{7}{3} \times \frac{18}{5})$$ km/hr = $$8.4$$ km/hr

Height of cone comes down to 50%, => it becomes $$\frac{1}{2}$$

=> Volume would become $$\frac{1}{8}$$ as radius will also become half by similar triangles.

Let the capacity of cone = 24 litres

Volume of water run-off = $$24 - \frac{1}{8} \times 24 = 21$$ litres

Volume of water left in the cone = $$ \frac{1}{8} \times 24 = 3$$ litres

Pipe A's efficiency = $$\frac{24}{8} = 3$$ litres/hr

Pipe B's efficiency = $$\frac{24}{12} = 2$$ litres/hr

Pipe C's efficiency = $$\frac{24}{-4} = -6$$ litres/hr

All will run 19 hours simultaneously (going by the options)

=> Net effect = $$(3 + 2 - 6) \times 19 = -19$$ litres

This means that after 19 hours, 19 litres of water has been removed, we need to remove 2 more litres as per the requirement. Thus,  C will definitely run for another hour.

If we run A and C together for the 20th hour, net effect = $$(3 - 6) \times 1 = -3$$ litres

Run B for 30 minutes => $$2 \times \frac{1}{2} = 1$$ litres

$$\therefore$$ Volume of water removed = $$-19 -3 + 1 = -21$$ litres

Thus, Pipe B was open for 19 hours 30 minutes.

By using graphs 2A and 3A, we get the above table.

For employees 4 and 5, the training days is more than 17.

Average of the bonus of 4, 5 = $$\ \frac{\ 21+18}{2}$$= 19.5

D is the correct answer.

Using graphs 1 and 3A, we get the above table.

The effective score of the employees is greater than 7 for employees 1, 4, 5, 7.

Among them, the bonus is less than 20 lakhs for 4, 7.

A is the correct answer. 

By using the data in 2A and 3A, we get the above table.

From the above table, it is clear that for the employees 1, 2, 3, 7 the number of training days increased from Survey 1 to 2.

Out of them for the employees 2, 3 the annual bonus decreased.

B is the correct answer.

from the above table, it is clear that for the employees 1, 2, 3, 7 the number of training days increased from Survey 1  to Survey 2.

Out of which for the employees 2, 3 the effective score increased by at least 1.0

A is the correct answer.

From the above table, it is clear that the decreasing order of gain of vote share = BCEDA

D is the correct answer.

Using the data in the table, we can calculate the number of neutral votes for each party as follows:

Party B: 108128 * (100 - 30.4 - 29.7)/100 = 43143

Party C: 96620 * (100 - 32.50 - 26.60)/100 = 39517

Party D: 41524 * (100 - 30.60 - 36.10)/100 = 13827

Party E: 32724 * (100 - 21.60 - 41) /100 = 12239

Thus, we can see that Party B has the maximum number of neutral votes.

Hence, option A is the correct answer.

The table representing the Vote share of parties in the years 2000 and 2010 is as follows.

From the table, we can see that 2 and 7.5 are possible values.

Party E’s votes in 2000 are less than 2% of the total votes as given in the question. Thus, the Gain in vote share will be between 3 and 5%.

Thus, only 2.5% is the option that is not possible.

Hence, option B is the answer.

The table representing the Vote share and Tweet share in the year 2010 is as follows:

The table shows that the maximum difference between the vote share and tweet share is for “Other” parties.

Hence, option E is the correct answer.