XAT 2011 Question Paper

Using properties of secant, $$PS \times ST = QS \times SR$$ -------------Eqn(I)

Also, for two numbers, $$PS$$ and $$ST$$, we know that harmonic mean is less than geometric mean.

=> $$\frac{2}{\frac{1}{PS} + \frac{1}{ST}} < \sqrt{PS \times ST}$$

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}}$$

Using Eqn(I)

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}}$$ ------Eqn(II)

Also, for two numbers, $$QS$$ and $$SR$$, geometric mean is less than arithmetic mean.

=> $$\sqrt{QS \times SR} < \frac{QS + SR}{2}$$

=> $$\frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$$  $$(\because QS + SR = QR)$$

Multiplying both sides by $$2$$

=> $$\frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$$ ----------Eqn(III)

From eqn(II) and (III)

$$\therefore \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$$

To find the minimum or maximum value, we need to find first derivative and put it equal to zero

Expression : $$f(x) = 21 sin x + 72 cos x$$

=> $$f'(x) = 21 cos x - 72 sin x = 0$$

=> $$21 cos x = 72 sin x$$

=> $$\frac{sin x}{cos x} = tan x = \frac{21}{72} = \frac{7}{24}$$

=> $$sin x = \frac{7}{\sqrt{7^2 + 24^2}} = \frac{7}{\sqrt{49 + 576}}$$

=> $$sin x = \frac{7}{\sqrt{625}} = \frac{7}{25}$$

Similarly, $$ cos x = \frac{24}{25}$$

Putting it in the original expression, we get :

=> $$f(x) = (21 \times \frac{7}{25}) + (72 \times \frac{24}{25})$$

= $$\frac{147}{25} + \frac{1728}{25} = \frac{1728 + 147}{25}$$

= $$\frac{1875}{25} = 75$$

Shortcut Method : Maximum value of $$a sin x + b cos x = \sqrt{a^2 + b^2}$$

and minimum value = $$- \sqrt{a^2 + b^2}$$

=> Max value of $$21 sin x + 72 cos x = \sqrt{(21)^2 + (72)^2}$$

= $$\sqrt{441 + 5184} = \sqrt{5625}$$

= $$75$$

For northern neighbourhood,

Probability that there is no major crime = $$(1 - 0.418) = 0.582$$

Probability that there is no minor crime = $$(1 - 0.612) = 0.388$$

For southern neighbourhood,

Probability that there is no major crime = $$(1 - 0.355) = 0.645$$

Probability that there is no minor crime = $$(1 - 0.520) = 0.480$$

$$\therefore$$ Probability that no crime of either type is committed in either neighbourhood on any given day

= $$0.582 \times 0.388 \times 0.645 \times 0.480$$

= $$0.069$$

KL = lighthouse

BA = initial position man of man and BC = shadow

After moving 300 m east, DE = new position of man and EF = shadow

Given : AB = DE = 6 m

BC = 24 m and EF = 30 m and BE = 300 m

$$\triangle$$ LBE is right angled triangle (sea level).

To find : KL = ?

Solution : In $$\triangle$$ KLF and $$\triangle$$ DEF

=> $$\angle KLF = \angle DEF = 90$$

$$\angle KFL = \angle DFE$$ (common angle)

=> $$\triangle KLF \sim \triangle DEF$$

=> $$\frac{KL}{DE} = \frac{LF}{EF}$$ -----------Eqn(I)

Similarly, $$\triangle KLC \sim \triangle ABC$$

=> $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)

From eqn (I) and (II), and using AB = DE

=> $$\frac{LC}{BC} = \frac{LF}{EF}$$

=> $$\frac{LC}{24} = \frac{LF}{30}$$

=> $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$

If, LC is 4 part $$\equiv$$ LF is 5 part

=> $$LB = 4x$$ and $$LE = 5x$$

$$\because$$ $$\triangle$$ LBE is right angled triangle

=> $$(LE)^2 - (LB)^2 = (BE)^2$$

=> $$25X^2 - 16X^2 = 90000$$

=> $$x^2 = \frac{90000}{9} = 10000$$

=> $$x = \sqrt{10000} = 100$$

=> $$LB = 400$$ and $$LE = 500$$

=> $$LC = LB + BC = 400 + 24 = 424$$

Now, using Eqn (II), we get : 

=> $$KL = \frac{LC}{BC} \times AB$$

= $$\frac{424}{24} \times 6 = \frac{424}{4}$$

= $$106 m$$

KL = lighthouse

BA = initial position man of man and BC = shadow

After moving 300 m east, DE = new position of man and EF = shadow

Given : AB = DE = 6 m

BC = 24 m and EF = 30 m and BE = 300 m

$$\triangle$$ LBE is right angled triangle (sea level).

To find : LE = ?

Solution : In $$\triangle$$ KLF and $$\triangle$$ DEF

=> $$\angle KLF = \angle DEF = 90$$

$$\angle KFL = \angle DFE$$ (common angle)

=> $$\triangle KLF \sim \triangle DEF$$

=> $$\frac{KL}{DE} = \frac{LF}{EF}$$ -----------Eqn(I)

Similarly, $$\triangle KLC \sim \triangle ABC$$

=> $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)

From eqn (I) and (II), and using AB = DE

=> $$\frac{LC}{BC} = \frac{LF}{EF}$$

=> $$\frac{LC}{24} = \frac{LF}{30}$$

=> $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$

If, LC is 4 part $$\equiv$$ LF is 5 part

=> $$LB = 4x$$ and $$LE = 5x$$

$$\because$$ $$\triangle$$ LBE is right angled triangle

=> $$(LE)^2 - (LB)^2 = (BE)^2$$

=> $$25^2 - 16X^2 = 90000$$

=> $$x^2 = \frac{90000}{9} = 10000$$

=> $$x = \sqrt{10000} = 100$$

$$\therefore LE = 5 \times 100 = 500 m$$

Original position of the ladder = AC = $$25$$ ft

Base = BC = $$7$$ ft

=> In $$\triangle$$ ABC, using Pythagoras theorem

=> $$(AB)^2 = (AC)^2 - (BC)^2$$

=> $$(AB)^2 = (25)^2 - (7)^2$$

=> $$AB = \sqrt{625 - 49} = \sqrt{576}$$

=> $$AB = 24$$ ft

After drawing out the base, the new position of ladder = ED = $$25$$ ft

and $$AD = x$$ and $$CE = 2x$$

To find : CE = ?

Solution : In $$\triangle$$ DBE

$$DB = (24 - x)$$ and $$BE = (7 + 2x)$$

=> $$(DB)^2 + (BE)^2 = (ED)^2$$

=> $$(24 - x)^2 + (7 + 2x)^2 = (25)^2$$

=> $$(576 - 48x + x^2) + (49 + 28x + 4x^2) = 625$$

=> $$625 - 20x + 5x^2 = 625$$

=> $$5x^2 = 20x$$

=> $$x = \frac{20}{5} = 4$$

$$\therefore$$ $$CE = 2 \times 4 = 8$$ ft

None of the options include 8 in the interval.

A logarithm function of the form $$log_a b$$ is true, iff $$b > 0$$

and $$c = a^b$$ can be written as = $$log_a c = b$$

We have, $$f(x) =log_{7}({ log_{3}(log_{5}(20x-x^{2}-91 )))}$$

=> $$log_{3}(log_{5}(20x - x^{2} - 91 )) > 0$$

=> $$log_{5}(20x - x^{2} - 91 ) > (3)^0$$

=> $$log_{5}(20x - x^{2} - 91 ) > 1$$

=> $$20x - x^{2} - 91 > (5)^1$$

=> $$x^2 - 20x + 96 < 0$$

=> $$(x - 8) (x - 12) < 0$$

=> $$8 < x < 12$$

$$\therefore$$ Domain of $$f(x)$$ = $$(8,12)$$

As the mechanic has decided to check two machines, thus if he identifies either both defective machines or  both non defective, then the probability that he is able to catch the last bus is sum of the two. Thus, the possible outcomes are :

(i) : Both are defective machines, probability = $$(\frac{1}{2})(\frac{1}{3}) = \frac{1}{6}$$

(ii) : First is defective and second is non defective, probability = $$(\frac{1}{2})(\frac{2}{3}) = \frac{1}{3}$$

(iii) : First is non defective and second is defective, probability = $$(\frac{1}{2})(\frac{2}{3}) = \frac{1}{3}$$

(iv) : Both are non defective machines, probability = $$(\frac{1}{2})(\frac{1}{3}) = \frac{1}{6}$$

In the first and last cases, the mechanic would have identified the defective machines in time to catch the bus.

$$\therefore$$ Probability that he is able to catch the last bus = $$\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$$

=> Ans - (D)

Number of teams in the bottom group $$= 10$$
Let the total number of teams in top group = $$n$$
Total number of teams $$ = 10 + n$$

=> Number of matches played amongst the bottom group teams = $$^{10}C_2$$
= $$\frac{10 \times 9}{1 \times 2} = 45$$

Number of points bottom group teams get playing amongst themselves $$= 45\cdot1=45$$

Let the total number of teams in top group = $$n$$

wkt, "Each of the bottom ten teams earned half of their total points against the other nine teams in the bottom ten"
i.e. they get half the total points by playing amongst themselves and the other half of total points by playing with the top group teams.

Since they got 45points playing amongst themselves, the bottom teams get 45 points from their matches against top group teams, => 45 out of $$10 n$$ points
Total points by matches between top teams and bottom teams $$= 10\cdot{n}\cdot1 =10n$$ points

Number of points that top group teams get from matches playing amongst themselves = $$^nC_2$$

Number of points that top group gets against the bottom group = $$10n - 45$$

wkt, "Exactly half of the points earned by each team were earned in games against the ten teams which finished at the bottom of the table.". Therefore the top n teams obtained half points by playing amongst themselves and half the points by playing against bottom 10 teams.

=> $$^nC_2 = 10n - 45$$

=> $$n (n - 1) = 20n - 90$$

=> $$n^2 - 21n + 90 = 0$$

=> $$(n - 6) (n - 15) = 0$$

If, $$n = 6$$, top group would get = $$C^n_2 + 10n - 45$$

= $$C^6_2 + 60 - 45 = 30$$

Average points per game = $$\frac{30}{6} = 5$$

Bottom teams will get on an average = $$\frac{45 + 45}{10} = 9$$

This is not possible.

=> $$n = 15$$

$$\therefore$$ Total number of teams = $$15 + 10 = 25$$

Given : PQ = 10 , QT = 8 , RS = 7

Now, TQ*TP=TR*TS (Secant from same external point)

8(PQ+QT)=TR(TR+RS)

8*18=TR(TR+7)

TR=9

TS=9+7=16

$$\therefore ar (\triangle PTS) = \frac{1}{2} \times PT \times TS \times sin60 $$

= $$\frac{1}{4} \times (16 \sqrt{3}) \times 18$$

= $$72 \sqrt{3}$$