XAT 2015 Question Paper

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.