online user counter

MATHMAA

# KVPY2013solution5

KVPY2013solution5 contains the solutions of the questions 10.

10) In the figure given below, ABCDEF is a regular hexagon of side length 1, AFPS and ABQR are squares . Then the ratio Area(APQ)/Area(SRP) equals

A) $$\frac{\sqrt{2}+1}{2}$$  B) $$\sqrt{2}$$    C)$$\frac{3\sqrt{3}}{4}$$  D) 2

Solution :

From the figure we know the values AB = BC = CD = DE = EF = AF =1 and ABQR, AFPS are squares then AB = BQ = QR = AR = 1 = AF = FP = PS = AS and AQ, AP are diagonals of length $$\sqrt{2}$$ each.

In a regular polygon of sides n , each interior angle = $$\frac{(n-2)*\Pi}{n}$$ .

In a regular hexagon  each interior angle = (6-2)* 180/6  = 120 degrees.

$$\angle FAB = 120^{\circ}$$

$$\angle RAB = 90^{\circ}=\angle FAS$$ as ABQR and AFPS are squares.

$$\angle FAB = \angle RAB + \angle FAS - \angle RAS$$

120 = 90 + 90 - $$\angle RAS$$

$$\angle RAS = 60^{\circ}$$ means triangle RAS is equvilateral triangle of side 1.

SR = 1.

Similarly we prove PS = 1 .

$$\angle RAS = \angle RAQ + \angle PAS - \angle PAQ$$

60 = 45 + 45-$$\angle PAQ$$

$$\angle PAQ = 30^{\circ}$$

$$\angle ASR = 60^{\circ} and \angle ASP = 90^{\circ}$$ then $$\angle RSP= 30$$

$$\frac{ar\triangle APQ}{ar\triangle SRP}$$=$$\frac{\frac{1}{2}AP*AQ*\sin (30^{\circ})}{\frac{1}{2}*RS*PS*\sin 30^{\circ}}$$

= $$\frac{\sqrt{2}*\sqrt{2}}{1*1}$$ = 2

Hence option D is the correct answer.

11) A person X is running around a circular track completing one round every 40 seconds. Another person Y running in the opposite direction meets X every 15 seconds. The time, expressed in seconds, taken by Y to complete one round is
A) 12.5     B) 24       C) 25      D) 55

Solution :

X takes 40 seconds to complete one round. In 15 seconds it completes 15/40 =3/8 part of the circle.

Y meets X in the opposite direct in every 15 seconds.

If X completes 3/8 part in 15 seconds then Y completes 5/8 parts in the opposite direction.

Time takes to complete 1 part (round the circle) by Y = $$\frac{15}{\frac{5}{8}}$$

= $$\frac{15*8}{3}$$ =24 seconds

Therefore Y takes 24 seconds to complete one round.

Hence the option B is the correct answer.

12) The least positive integer n for which
$$\sqrt{n+1}$$ - $$\sqrt{n-1}$$ < 0.2 is
A)  24    B) 25    C)  26    D)  27

Solution :

Rationalize left side, we get

$$\frac{2}{\sqrt{n+1}+\sqrt{n-1}}< 0.2$$

$$\sqrt{n+1}+\sqrt{n-1}$$> 10

$$\sqrt{n+1}$$ > 10 - $$\sqrt{n-1}$$

squaring on both sides we get,

$$\sqrt{n-1}$$ >4.9

again squaring on both sides we get,

n> $$4.9^{2} +1$$ =25.01

So the least positive integer n = 26   (since we can estimate this by taking 5^2+1)

Therefore the option is (C).

To go back to the previous solution click Back and for further solutions click Next.