XAT 2011 Question Paper - QUANT

Income tax payable by Sangeeta according to her salary band = 0.2*(337425-300000) + 0.1(300000-190000) = Rs. 18485

Total education cess levied on income tax = 2 + 1 = 3% of income tax 

Therefore, the total income tax payable by Sangeeta = 1.03*18485 = 19039.55.

Hence, the required percentage = $$\dfrac{19039.55}{337425}\times 100$$ = 5.64 percent

Therefore, we can say that option A is the correct answer. 

Madan's tax deduction at source = Rs. 3,17,910

From the table we can see that one can pay this much amount of tax if he earns more than 5 lakh. 

Let say Madan's is Rs. x more than 5 lakhs. Therefore, Madan's total income = Rs. 500000+x.  

Total income tax paid by Madan = 1.03{0.3*(x) + 0.2*(200000) + 0.1(140000)}

$$\Rightarrow$$ $$\dfrac{317910}{1.03} = 0.3x + 54000$$

$$\Rightarrow$$ x = 848835.

Therefore, Madan's total income = 500000+848835 = Rs. 1348835.  Hence, option A is the correct answer. 

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)