challenging questions for various math … questions for various math competitions 1. ... each of...

MATHEMATICS–X CHALLENGING QUESTIONS 445

CHALLENGING QUESTIONS FOR VARIOUSMATH COMPETITIONS

1. Show that there are infinitely many positive primes.2. Solve the following equation :

2 25 8 3 2 53 2

x x x x xx x

3. Prove that the equations 2 (p + 2) x + py + 5 = 0 and px – 3 (p + 2) y – 8 = 0 cannot have an infinite numberof solutions for any real values of p.

4. The speed of a train is reduced to 56 of its original speed because of some defect in its engine, after it

covers a distance of 240 km with a uniform speed. As a result, the train reaches its destination late by 24minutes. If this defect in the engine would occur after covering a distance of 30 km more, then the trainwould have reached its destination 18 minutes late. Find the speed of the train and the distance of thejourney.

5. Solve the following system of equation in x and y :(a) ax + by = 1 (b) (a – b) x + (a + b)y = a2 – 2ab – b2

2 2

2abbx aya b

(a + b) (x + y) = a2 + b2

6. Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1.7. If and are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate :

(a) 2 2 (b) 3 3 (c)

(d) 3 3

1 1

(e) 4 4

8. Solve the following equation for x :

(a) 1 ; 2

12 122

x x

x

(b) 1 12525

xx

(c) 5 8 2 17x x x (d) 21 12 3 8 0 ; 0x x x

x x

(e) 13 4 13 2 140x x (f) 4 23 1 3 1 116 40 9 0 ;

2 1 2 1 2x x xx x

9. Find the least and the greatest values of 2

2

14 92 3

x xx x

for all real value of x.

10. If the sum of the roots of the equation 1 1 p

a x x b

is zero, show that the product of the roots is

2 21 ( ).2

a b AMIT B

AJAJ

446 CHALLENGING QUESTIONS MATHEMATICS–X

11. Show that the roots of the equation : ( )( ) ( )( ) ( )( ) 0x a x b x b x c x c x a are always realand these cannot be equal unless a = b = c.

12. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simulta-neously, fill the pool in the same time during which the pool is filled by the third pipe alone. The secondpipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find thetime required by each pipe to fill the pool seperately.

13. The product of two numbers is 63. When their sum is added to the sum of their squares, we get 146. Findthe numbers.

14. The vertices of a triangle ABC are A(5, 6), B(a, 3) and C(8, b). Find the values of a and b so that the sideAB is parallel to y-axis and BC is parallel to the x-axis. Also find ABC.

15. The co-ordinates of the opposite vertices of a square are (0, 1) and (4, 3). Find the co-ordinates of theremaining vertices.

16. An equilateral triangle has one vertex at (3, 4) and another at (–2, 3). Find the coordinates of the thirdvertex.

17. If the vertices of a triangle have integral coordinates, prove that the triangle cannot be equilateral.

18. If 1 1n n

n n

a ba b

is the arithmetic mean between a and b, find the value of n.

19. If p2, q2, r2 are in A.P., prove that 1 1 1, ,

q r r p p q are also in A.P..

20. If the mth term of an A.P. is 1n

and the nth term is 1m

, show that the sum of mn terms is 1 ( 1).2

mn

21. The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that

( ) ( ) ( ) 0a b cq r r p p qp q r

.

22. Each of the A.P.’s 2, 4, 6, 8,.... and 3, 6, 9, 12,.... is continued to 200 terms. How many terms of these twoA.P.’s are identical?

23. The interior angles of a polygon are in A.P. The smallest angle is 120° and the common difference is 5°.Find the number of sides of the polygon.

24. The sum of n terms of two A.P.’s are in the ratio (2n – 1) : (3n + 5). Find the ratio of their 30th terms.25. Two cars start together in the same direction from the same place. The first goes with uniform speed of

10 km/h. The second goes at a speed of 8 km/h in the first hour and increases the speed by 12

km in each

succeeding hour. After how many hours will the second car overtake the first car if both cars go non-stop?

26. In the given figure, the side AC of ABC is produced to G so that CG = 12

AC. If D is the midpoint of BC,

GD is produced to meet AB in E and DF is drawn parallel to AB, prove that DE = 13 GE.

A

EF

C

GDB

AMIT BAJA

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27. Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum ofthe squares of the medians of the triangle.

28. In trapezium ABCD, AB || DC and AB = 12

DC. EF drawn parallel to AB cuts AD in F and BC in E such that

BE 3 .EC 4

Diagonal DB intersects EF at G. Prove that AB 7 .FE 10

29. If is an acute angle and tan + cot = 2, find the value of tan10 + cot10 .

30. Find an acute angle , when cos sin 1 3cos sin 1 3

.

31. If cosec – sin = m and sec – cos = n, prove that (m2n)2/3 + (mn2)2/3 = 1.32. An equilateral triangle is inscribed in a circle of radius 4 cm. Find its side.

33. If a sin + b cos = c, then show that a cos – b sin 2 2 2 .a b c

34. Prove that : 2 2 2 2 1sin 5 sin 10 ... sin 85 sin 90 9 .2

35. If x and y are two unequal real positive quantities, show that the equations (i) 2

2 ( )sin4

x yxy

and

(ii) 1cos xx

are both impossible.

36. If (1 + sin x) (1 + sin y) (1 + sin z) = (1 – sin x) (1 – sin y) (1 – sin z), prove that each is equal to± cos x cos y cos z.

37. Two stations due south of a leaning tower which leans towards the north are at distances a and b fromits foot. If be the elvations of the top of the tower from these stations, prove that its inclination to

the horizontal is given by cot cotcot .b ab a

38. From a window, h metre high above the ground, of a house in a street, the angles of elevation anddepression of the top and the foot of another house on the opposite sides of the street are and respectively. Show that the height of the opposite house is h (1 + tan cot ).

39. At the foot of a mountain the elevation of its summit is 45°, after ascending 1000 m towards the mountainup a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

40. A ladder rests against a wall at an angle to the horizontal. Its foot is pulled away from the wall througha distance a so that it slides a distance b down the wall making an angle with the horizontal. Show that

cos cossin sin

ab

.

41. If the diameter of the cross-section of a wire is decreased by 5%, how much percent will the length beincreased so that the volume remains the same?

42. The three vertices of a rhombus lie on a circle with centre O. If O is the fourth vertex of the rhombus andif the area of the rhombus is 50 3 cm2, find the radius of the circle.AMIT B

AJAJ


43. In the given figure, a semi-circle is drawn with line-segment PR as a diameter. Q is the mid-point of theline-segment PR. Two semi-circles with line-segments PQ and QR as diameters are drawn. A circle isdrawn which touches the three semi-circles. If PR = 24 cm, find the area of the shaded region.

P B Q C R

44. A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20cm so as to cover the whole surface. Find the length and weight of the wire assuming the specific gravityto be 8.88 gm/cm3.

45. A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered intothe water and its size is such that when it touches the sides, it is just immersed as shown in the givenfigure. What fraction of water overflows?

6 cm

8 cm

O

46. The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that thevolumes of the three portions starting from the top are in the ratio 1 : 7 : 19.

47. A solid metallic cylinder with base radius 6 cm and height 14 cm is given. On one side of the cylinder, ahemisphere shape of solid and on the other side, a conical shape of solid are removed. If the base radiusof the hemisphere is equal to the radius of the cylinder and the base radius of the cone is 3 cm and itsheight is 5 cm, find the volume and the total surface area of the remaining solid.

14 c

m

6 cm

5 cm

3 cm

AMIT BAJA

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48. The following table shows the distribution of the life-time of 350 radio tubes.

Life-Time (in hrs.) 300-400 400-500 500-600 600-700 700-800 800-900 900-1000No. of tubes 6 18 73 165 62 22 4

Assum that the class frequencies are uniformly distributed within the corresponding classes, find thepercentage of tubes that have life time :

(a) greater than 760 hours.

(b) less than 530 hours.

49. If 1 4x and 2 6y , find the probability that 5.x y

50. Using the digits 2, 3, 5 and 6 only once, all possible 4-digit numbers are formed. A number is selected atrandom. What is the probability that the selected number is exactly divisible by 5?

ANSWERS

2. x = –1 4. speed of train = 60 km/hr, distance = 360 km

5. (a) 2 2 2 2,a bx ya b a b

(b) 2, abx a b y

a b

6. a = 1, b = 7

7. (a) 2

2

2b aca

(b) 3

3

3abc ba

(c) 2 2b ac

ac

(d) 3

3

3abc bc

(e) 2 2 2 2

4

( 2 ) 2b ac a ca

8. (a) x = 1, 1 (b) 125 or25

x (c) x = 9 (d) 11, 2,2

x (e) x = 2 (f) 1 1 3, or12 8 4

x

9. least value = – 5 and greatest value = 4 12. 15 hr, 10 hr, 6 hr

13.17 37

2

and 17 37

2

or 7 and 9 14. a = 5, b = 3; ABC = 90°

15. (3, 0) and (1, 4) 16. 1 3 7 5 3 1 3 7 5 3, or ,

2 2 2 2

18. n = 0 22. n = 66 23. n = 9 24. 9 : 14

25. t = 9 hours 29. 2 30. = 60° 32. 4 3 cm

39. 1365 m 41. 10.8% nearly 42. r = 10 cm

43. 220 cm 44. 1200 cm, 426.24 2 gm 45. 3 : 8

47. 3 2345 cm , (267 3 34) cm 48. (a) 14.57% (b) 13.14%

49.56 50.

14

AMIT BAJA

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HINTS/SOLUTIONS TO CHALLENGING QUESTIONS1. Let if possible, there are finite number of positive primes p1, p2, p3, ...., pn such that p1 < p2 < p3 < ... < pn.

Let p = 1 + p1 p2 p3....pn. Clearly, p1 p2 p3....pn is divisible by each of p1, p2, p3,...., pn. p = 1 + p1 p2 p3....pnis not divisible by any one of p1, p2, p3, ...., pn. p is a prime or it has prime divisors other than p1, p2, p3,...., pn which means there exists a positiveprime different from p1, p2, p3, ...., pn, which contradicts that there are finite number of positive primes.Hence, the number of positive primes is infinite.

2. Given equation can be rewritten as :

2 25 8 3 53 2

x x x xx xx x

Taking LCM and simplifying gives

2 8 3 3 5

3 2x xx x

2 6 2 3 6 3 53 2

x xx x

2( 3) 2 3( 2) 3 5

3 2x xx x

2 32 3 5

3 2x x

2 3 2 30

3 2 3 2x x x x

– = 1 Ans.x

3. For infinite number of solutions,

2( 2) 53( 2) 8

p pp p

from the first two ratios, we get 2

2 26( 2) 6 62 2

p pp pp p

, which is not pos-

sible. Hence, p is not real and so the two equations cannot have infinite number of solutions.

4. Let original speed of the train is x km/h and distance be y km. Then, the scheduled time, yt hrx

...(1)

By given first condition, we have the total time taken 240 24056

y hrx x

.

Also, this distance would have been covered by the train in 24 2 .60 5

t hr t hr

hence, we have : 240 240 2 25 5 56

y ytx xx

[ using (1)]AMIT B

AJAJ


Simplifying, we get 2x – y + 240 = 0 ...(2)Again, in second case,

Total time 270 270 18 35 60 106

y hr t hr t hrx x

310

y hrx

Simplifying, we get 3x – 2y + 540 = 0 ...(3)Solving (2) and (3), we get x = 60, y = 360.Hence, speed of train = 60 km/hr and distance = 360 km.

5. (a) 1 0ax by

2 2

2 0abbx aya b

By cross multiplication, we have

2 2 2 2

12 2( 1) ( 1)

x yab ab a a b bb a a b

a b a b

3 2 2 3 2 2

2 2 2 2

1x ya ab a b b a ba b a b

2 2 2 2

2 2 2 2 2 2 2 2

( ) 1 ( ) 1anda a b b a bx ya b a b a b a b

2 2 2 2and a bx y

a b a b(b) Given system of equations may be written as

(a – b) x + (a + b) y – (a2 – 2ab – b2) = 0(a + b) x + (a + b) y – (a2 + b2) = 0

Now use cross-multiplication to get x = a + b and 2aby

a b

6. Here, remainder = 0.

On dividing, we get

Quotient = x2 + x + 7 and remainder = x (a – 1) + (b – 7).

Remainder = 0 x (a – 1) + (b – 7) = 0 . x + 0

1 0a and b – 7 = 0

a = 1 and b = 7.

+ + 8 + + + + 7x x x ax b x x4 3 2 2x2 +1

x x ax b3 2 + 7 + +

x x4 2 + – –

x x3 +

) (

7 + ( – 1) + x x a b2

7 + 7x2

x a b ( – 1) + ( – 7)

– –

– –AMIT BAJA

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7. here, ,b ca a

(a) 2 2

2 2 22

2( ) 2 2b c b aca a a

(b) 3 3 3( ) 3 ( ). Now put values of + and .

(c) 2 2 2( ) 2 .

Now put values of + and .

(d) 3 3 3

3 3 3 3

1 1 ( ) 3 ( ) .( ) ( )

Now put values of + and .

(e) 4 4 2 2 2 2 2 2 2 2( ) 2 [( ) 2 ] 2( ) . Now put values of + and .

8. (a) Given equation 1 1 1

1 2 22 2 22(2 ) 1 4 2 1 3 22

x x xx x

x x xx

23 2 3 2 2 1 02(3 2 ) (2 ) 4 3

x xx x x xx x x

x = 1, 1

(b) 1 1 1 125 2525 25

x xx x

x = 25 or 1=25

x

(c) Squaring both sides, 5 8 2 ( 5)( 8) 2 17x x x x x

( 5)( 8) 2.x x Again squaring, we get

2( 5)( 8) 4 13 36 0 4 or 9.x x x x x x

but x = 4 do not satisfy the given equation. x = 9.(d) From the given equation, we have

2

2

1 1 12 2. . 3 8 0x x xx xx

22

1 12 4 3 8 0x xxx

21 1 12 2. . 4 3 8 0x x x

x x x

21 12 4 4 3 8 0x x

x x

21 12 3 0x x

x x

AMIT BAJA

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Let 21 32 3 0 0 or .2

x y y y y yx

1 0xx

or 1 3

2x

x

x2 – 1 = 0 or 2x2 – 3x – 2 = 0

x = ± 1 or x = 2, –12

Ans.

(e) From the given equation, we have

2 13 (2 ) 13 2 140 0x x

3 × 22x + 2 – 13 × 2x – 140 = 0 3 × 22x × 22 – 13 × 2x – 140 = 0 12 × 22x – 13 × 2x – 140 = 0

Let 22 12 13 140 0x y y y

12y2 + 35y – 48 y – 140 = 0 (12y + 35) (y – 4) = 0

35 or 4

12y y

352 or 2 4

12x x

2x = 22 x = 2 ( as 2x > 0 for all real x)

( f ) Let 23 1

2 1x yx

Given equation reduces to 16 y2 – 40 y + 9 = 0

(4y – 1) (4y – 9) = 0

1 9or4 4

y y

2 23 1 1 3 1 9or

2 1 4 2 1 4x xx x

3 1 1 3 1 3or2 1 2 2 1 2

x xx x

1 1 3= – , or Ans.

12 8 4x

AMIT BAJA

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9. Let 2

2

14 92 3

x x yx x

Cross multiplying and rearranging the terms, we get

2 (1 ) (14 2 ) (9 3 ) 0x y x y y ..(1)

here, discriminant, D = (14 – 2y)2 – 4 (1 – y) ( 9 – 3y) = 160 – 8y2 – 8y

28( 20)y y

8( 5)( 4)y y

Now, since x is real, hence two roots of equation (1) are real. Hence, D 0.

–8 (y + 5) (y – 4) 0 (y + 5) (y – 4) 0 either y + 5 0 and y – 4 0 ...(2)

or 5 0y and 4 0y ...(3)

from eqn. (2), 5y and 4y 5 4y

from eqn. (3), 5y and 4,y which is not possible.

5 4.y

least value = – 5 and greatest value = 4.

10.1 1 p

a x x b

( ) ( ) ( )( )x b a x p a x x b

simplifying we get

2 (2 ) 0px x ap bp abp a b

Let are roots of this equation.

then, 2 ( ) 2ap bp a b pp p

Since 0 ( ) 2 0a b p 2p

a b

...(1)

Now, ( )( ).2

abp a b a b a bab ab a bp p

[ using (1)]

2 21 ( )2

a b AMIT B

AJAJ


11. Simplifying the given equation, we get23 2( ) ( ) 0x a b c x ab bc ca

discriminant, 2[ 2( )] 4 3 ( )d a b c ab bc ca

2 2 24( )a b c ab bc ca

2 2 22(2 2 2 2 2 2 )a b c ab bc ca

2 2 2 2 2 22[( 2 ) ( 2 ) ( 2 )]a b ab b c bc c a ca

2 2 22[( ) ( ) ( ) ] 0a b b c c a ,

being the sum of three perfect squares. Hence the roots are real.

Now, 2 2 22[( ) ( ) ( ) ]a b b c c a will vanish only when a = b = c.

So, the roots will be equal, when a = b = c.12. Let V be the volume of the pool and x the number of hours required by the second pipe alone to fill the

pool. Parts of the pool filled by the first, second and third pipes in one hour are respectively

, and .5 4

V V Vx x x

Let the time taken by first and second pipes to fill the pool simultaneously be t hours, which is same timetaken by third pipe.

. .5 4

V V Vt tx x x

1 1 1

5 4x x x

Simplifying, we get 2 8 20 0x x x = 10, –2. x = 10 ( x can’t be negative) Required timing for first, second and third pipe are 15 hours, 10 hours and 6 hours respectively.

13. Let the numbers be x and 63x

. Then,

2263 63 146x x

x x

263 63 632. . 146x x x

x x x

263 63 272 0x x

x x

Let 63 .x yx

Then

y2 + y – 272 = 0 y = – 17 or y = 16

AMIT BAJA

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263 17 3717 17 63 0 .2

x x x xx

and, 263 16 16 63 0 9 or 7.x x x xx

Required numbers are –17 + 37 –17 – 37and or 7 and 9.2 2

Ans.

14. ABC is clearly a right-angled triangle with ABC = 90°. The abscissae of A and B will be equal, sinceAB is parallel to the y-axis and the ordinates of B and C will be equal, since BC is parallel to the x-axis.

15. Here, AB2 = (x – 0)2 + (y – 1)2 = x2 + y2 – 2y + 1BC2 = (x – 4)2 + (y – 3)2 = x2 + y2 – 8x – 6y + 25AC2 = (4 – 0)2 + (3 – 1)2 = 20.

Since, ABCD is a square, we have AB = BC AB2 = BC2

x2 + y2 – 2y + 1 = x2 + y2 – 8x – 6y + 25 y = 6 – 2x ...(1)Now, B = 90° AB2 + BC2 = AC2

x2 + y2 – 2y + 1 + x2 + y2 – 8x – 6y + 25 = 20

D C(4, 3)

A(0, 1) B ( )x, y

x2 + y2 – 4x – 4y + 3 = 0 x2 + (6 – 2x)2 – 4x – 4 (6 – 2x) + 3 = 0 [ using eqn. (1)] x2 – 4x + 3 = 0 x = 3 or x = 1when x = 3, from eqn. (1), y = 0when x = 1, from eqn. (1), y = 4 co-ordinates of the other vertices are (3, 0) and (1, 4). Ans.

16. Here, BC = AB = AC BC2 = AB2 = AC2

Now, BC2 = (–2 – 3)2 + (3 – 4)2 = 26

A( )x, y

B(3, 4) C(–2, 3)AB2 = (x – 3)2 + (y – 4)2

AC2 = (x + 2)2 + (y – 3)2

BC2 = AB2 (x – 3)2 + (y – 4)2 = 26 x2 + y2 – 6x – 8y – 1 = 0 ...(1)BC2 = AB2 (x + 2)2 + (y – 3)2 = 26 x2 + y2 + 4x – 6y – 13 = 0 ...(2)

Subtracting (1) from (2), 10 x + 2y – 12 = 0 y = 6 – 5 x10 x + 2y – 12 = 0 y = 6 – 5x

Using value of y in eqn. (1), we get

2 2(6 5 ) 6 8(6 5 ) 1 0x x x x

2 1 32 2 1 02

x x x

AMIT BAJA

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1 3 7 5 36 5

2 2y

The third vertex has the coordinates as

1 + 3 7 – 5 3 1 – 3 7 + 5 3, or ,2 2 2 2

Ans.

17. Let A (x1, y1), B(x2, y2) and C(x3, y3) be the vertices of a triangle ABC where xi, yi, ; i = 1, 2, 3 are integers.

Then, area of ABC is given by 12

|x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

= A rational numberIf possible, let ABC be an equilateral. Then its area is given by

23 3(side)4 4

(A positive integer)

= An irrational number.This is a contradiction to the fact that the area is a rational number. Hence, the triangle cannot beequilateral.

18. According to question, 1 1

2

n n

n n

a b a ba b

1 12( ) ( )( )n n n na b a b a b

Simplifying, we get

( )( ) 0 0 asn n n na b a b a b a b

0

1n n

n n a a aa bb b b

n = 0 Ans.

19. Since p2, q2, r2 are in A.P., we have

2 2 22q p r ...(1)

Now, 1 1 1, andq r r p p q

will be in A.P..

if 2 1 1

r p q r p q

i.e. if 2 2( )( )

p q rr p q r p q

i.e. if 2 (q + r) (p + q) = (r + p) (p + 2q + r)Simplifying, we geti.e. if p2 + r2 = 2q2, which is true by eqn. (1).

AMIT BAJA

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20. Let a be the first term and d be the common difference of the given A.P. Then,

1 1( 1)ma a m dn n

...(1)

and1 1( 1)na a n dm m

...(2)

Subtracting eqn. (2) from eqn. (1), we get

1 1 1( ) ( ) m nm n d m n d dn m mn mn

Putting 1dmn

in eqn. (1), we get 1amn

Now, [2 ( 1) ]2mn

mnS a mn d

Using 1a dmn

in Smn, we get 1 ( 1)2mnS mn

21. Let A be the first term and D be the common difference of the given A.P. Then,

2[2 ( 1) ] [2 ( 1) ]2p aa A p D A p D

p ...(1)

2[2 ( 1) ] [2 ( 1) ]2q bb A q D A q D

q ...(2)

2[2 ( 1) ] [2 ( 1) ]2r cc A r D A r D

r ...(3)

multiplying eqn. (1), (2) and (3) by (q – r), (r – p) and (p – q) respectively and adding, we get the desiredresult.

22. For the first A.P. 2, 4, 6,.... we have a200 = 2 + (200 – 1) × 2 = 400.

For the second A.P. 3, 6, 9, .... we have a200 = 3 + (200 – 1) × 3 = 600.

Observe that the identical terms in the two A.P.’s will be upto 400 only, since 400 < 600.

Clearly, the identical terms of two A.P.’s will be the multiples of 2 × 3 = 6. Hence, terms will be 6, 12, 18,....

Let nth term of this A.P. be less than or equal to 400.

i.e. 400na 6 + (n – 1) × 6 400 266 .3

n

Since n must be a positive integer, n = 66.AMIT B

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23. We know sum of all the interior angles of a polygon of n sides is (n – 2) × 180°. Clearly, here a = 120°,d = 5°.

[2 ( 1) ]2nnS a n d

[2 120 ( 1) 5 ]2n n

[235 5 ]2n n

according to given question, [235 5 ] ( 2) 1802n n n

Simplifying, we get2 25 144 0 16 or 9.n n n n

when n = 16, the 16th angle = 120° + (16 – 1) × 5° = 195° > 180°, which is not possible. n = 9.24. Let a1 and a2 be the first terms and d1 and d2 be the common differences of the two A.P.’s. Let S1 and S2

be the sums of the two A.P.’s respectively.

Then, 1 1 1[2 ( 1) ]2nS a n d and 2 2 2[2 ( 1) ]

2nS a n d

1 1 1

2 2 2

2 ( 1) 2 12 ( 1) 3 5

S a n d nS a n d n

...[given] ...(1)

Now, ratio of the 30th terms of the two A.P.’s

1 1 1 1 1 1

2 2 2 2 2 2

(30 1) 29 2 58(30 1) 29 2 58

a d a d a da d a d a d

...(2)

from (1) and (2), we observe 1 58 59.n n

Putting n = 59 in eqn. (1), we get

1 1

2 2

2 58 2 59 1 92 58 3 59 5 14

a da d

= 9 : 14 Ans.

25. Suppose the second car overtakes the first car after t hours. Then, the two cars travel the same distancein t hours.Distance travelled by the first car in t hours = 10 t km.Distance travelled by the second car in t hours

= sum of t terms of an A.P. with a = 8, 1 .2

d

12 8 ( 1) ( 31)2 2 4t tt t

according to question, 10 ( 31)4tt t

t2 – 9 t = 0 = t = 0 or t = 9 t = 9 hours ( t 0)AMIT B

AJAJ


26. Clearly AF = FC = 1 AC2

...(1)

also, GFD ~ GAE (AA similarity)

GF GD FDGA GE AE

GD GFGE GA

...(2)

Now, GD = GE – DE

GF = CG + CF = 12

AC + 12

AC = AC

and, GA = CG + CA = 12

AC + AC = 32

AC

Hence, from eqn. (2), we have

GE DE AC 23GE 3AC2

DE 2 DE 2 11 1GE 3 GE 3 3

1DE GE3

27. According to Apollonius theorem, if AD is a median of ABC, then

AB2 + AC2 = 12

BC2 + 2AD2. ...(1)

By symmetry, we get AB2 + BC2 = 12

AC2 + 2.BE2 ...(2)

and, AC2 + BC2 = 12

AB2 + 2.CF2 ...(3)

adding (1), (2) and (3), we get

2 (AB2 + BC2 + AC2) = 12

(AB2 + BC2 + AC2) + 2 (AD2 + BE2 + CF2)

32

(AB2 + BC2 + AC2) = 2 (AD2 + BE2 + CF2)

3 (AB2 + BC2 + AC2) = 4 (AD2 + BE2 + CF2)AMIT B

AJAJ


28. Clearly, DFG ~ DAB (AA similarity)

DF FGDA AB

...(1)

In trapezium ABCD, EF || AB || DC

AF BEDF EC

A B

CD

EF G2 1

11

AF 3 AF 31 1DF 4 DF 4

BE 3 (given)EC 4

AF+DF 7 AD 7 DF 4

DF 4 DF 4 AD 7 ...(2)

from (1) and (2), FG 4 4FG ABAB 7 7

...(3)

Again, BEG ~ BCD ( AA similarity)

BE EGBC CD

3 EG7 CD

BE 3 EC 4 EC 4 BC 71 1EC 4 BE 3 BE 3 BE 3

EG = 37

CD = 37

× 2AB [ CD = 2AB (given)]

EG = 67 AB ...(4)

adding (3) and (4), we get

4 6FG EG AB AB7 7

10 AB 7EF AB7 FE 10

29. We have, tan + cot = 2

21tan 2 tan 1 2 tantan

tan2 – 2 tan + 1 = 0 (tan – 1)2 = 0

tan – 1= 0 tan = 1 = tan 45° = 45°

tan10 + cot10 = tan1045° + cot1045° = (1)10 + (1)10 = 1 + 1 = 2. Ans.AMIT B

AJAJ


30.cos sin 1 3 .cos sin 1 3

cos sin1 3cos

cos sin 1 3cos

( dividing numerator and denominator by cos )

1 tan 1 31 tan 1 3

tan = 3 (on comparing two sides)

tan = tan 60° = 60° Ans.31. We have, cosec – sin = m and sec – cos = n

1 1sin and cos

sin cosm n

2 21 sin 1 cosand

sin cosm n

2 2cos sinand

sin cosm n

Now, substitute the values of m and n in LHS to get the desired result.32. Here, OA = OB = OC = 4 cm. Draw OD BC and OE AC. Then OD bisects BC and OE bisects AC.

Hence, BD = DC = CE = EA ( BC = CA)Clearly, ODC OEC (RHS congruence rule) OCD = OCE (cpct)

CB

A

OE

D30°30°

But DCE = 60° OCD = OCE = 30°i.e. OC is the bisector of BCA.Similarly, OB and OA are bisectors of ABC and BAC respectively.

Now, In OBD, BD BD 3cos30 .OB 4 2

BD 2 3

BC = 2BD = 4 3 cm.33. Consider, (a sin + b cos )2 + (a cos – b sin )2

= a2 sin2 + b2 cos2 + 2ab sin cos + a2 cos2 + b2 sin2 – 2ab sin cos = a2 (sin2 + cos2) + b2 (sin2 + cos2) = a2 + b2

c2 + (a cos – b sin )2 = a2 + b2 ( sin cos )a b c

(a cos – b sin )2 = a2 + b2 – c2

a cos – b sin = 2 2 2a b c

AMIT BAJA

J


34. LHS = sin25° + sin210° + ... + sin2 40° + 2

21 sin (90 40 )2

+ .... + sin2 (90° – 5°) + 12

2 2 2 2 2 1sin 5 sin 10 .... sin 40 cos 40 ... cos 5 12

2 2 2 2 2 2 3(sin 5 cos 5 ) (sin 10 cos 10 ) ... (sin 40 cos 40 )2

8 terms

3 3 11 1 ..... 1 8 92 2 2

35. (i) we have, 2 2( ) 4 ( ) 0x y xy x y if x and y are unequal.

2( ) 4x y xy

2( ) 1

4x y

xy

i.e. 2sin 1 which is impossible,

Since, 0 sin 1 when 0 90 .

(ii) We have, 21 1xx

x x

Now, 2 2

2 2 1 1 1 3( 1) 1 1 02 4 2 4

x x x x x x

2

2 11 1 cos 1,xx xx

which is impossible,

Since, 0 cos 1 when 0 90 36. We have (1 + sin x) (1 + sin y) (1 + sin z) = (1 – sin x) (1 – sin y) (1 – sin z)

multiplying both sides by (1 – sin x) (1 – sin y) (1 – sin z)(1 – sin2 x) (1 – sin2y) (1 – sin2z) = [(1 – sin x) (1 – sin y) (1 – sin z)]2

cos2x . cos2y cos2z = [(1 – sin x) (1 – sin y) (1 – sin z)]2

(1 – sin x) (1 – sin y) (1 – sin z) = ± cos x cos y cos z. (1 – sin x) (1 – sin y) (1 – sin z) = (1 + sin x) (1 + sin y) (1 + sin z) = ± cos x cos y cos z.

37. Let AB be the leaning tower and let C and D be two given stations.

In AEB, we have tan coth x hx

...(1)

x

aa

bD C EA

B

hAMIT BAJA

J


In CEB, we have tan coth x h aa x

...(2)

In DEB, we have tan coth x h bb x

...(3)

from (1) and (2), we get cot cotcot cot

ah h a h

...(4)

from (1) and (3), we get cot cotcot cot

bh h b h

...(5)

from (4) and (5), we get cot cot cot cota b

rearrange and simplify to get cot cotcot b ab a

38. Let B be the window h m above the ground. D, the top and C, the foot of a house.

In DBE, we have tan cotx y xy

...(1)

In BAC, we have tan coth y hy

...(2)

D

C

EB

A y

h m

x m

from (1) and (2), we get

cot cot tan cotx h x h ...(3)

Height of the house = DC = DE + EC = (x + h) m

= (h tan cot + h) = h (1 + tan tan ).39. Let C be the foot and A be the summit of the mountain ABC such that BCA = 45° BAC = 45°

AB = BC = hDraw DE BC and DF AB.

In CE 3DEC, cos30 CE 1000CD 2

m 500 3 m ...(1)

Also, DE 1sin 30 DE 1000 500 m BFCD 2

A

F

BEC

D 60°

30°45° 1000 m

h mNow, AF = h – BF = (h – 500) m

In AFD, DF AF 500cot 60 DFAF 3 3

h

BC = CE + BE = CE + DF = 500 500 100033 3

h h AMIT B

AJAJ


Now, In ABC, AB tan 45BC

1000 1000 3 1BC ( 3 1) 1000

3 3 1 3 1hh h h

1000 2.73 m2

h = 1365 m Ans.

40. Let AB be the ladder leaning against vertical wall AC at an angle with the horizontal. When its foot Bis pulled away from the wall at B, then its top A slides down the wall to A so that BB = a and AA = b,ABC = and ABC = .

ll

A

CxaB B

y

A

Let AB = AB = l.

In ABC, we have sin b yl

...(1)

cos xl

...(2)

In ABC, we have sin yl

...(3)

cos x al

...(4)

from (1) and (3), we have

sin sin y b y bl l l

...(5)

from (2) and (4), we have cos cos x x a al l l

...(6)

dividing (6) by (5), we get cos cos cos cossin sin sin sin

a l al b b

41. Let d be the original diameter and l be the original length of the wire.

Then, the original volume of the wire 2

4d l .

Let the length be increased by x%. Then, the new length

100100 100lx xl l

b

AMIT BAJA

J


Then, the new volume 219 100

20 100xd l

according to given question, 2 219 100

20 100 4x dd l l

Simplifying, we get 3900 10.8361

x

required percent is 10.8% nearly.

42. Clearly, OA = AB = OB = OC = BC = r, the rhombus is divided into two equilateral triangles OAB and OBC,each of side r cm, by the diagonal OB.

Now, area of 2 23OAB cm4

r

Area of rhombus OABC = 2 × 3

4 r2 cm2 =

32

r2 cm2.

O r

rrr

r

A

BC

2 23 50 3 1002

r r r = 10 cm

43. here, radius of the semi-circle PEQ= The radius of the semi-circle QFR

24 cm 6 cm4

.

Also, ABQ ACQ (SSS congruence rule) AQB = AQC (cpct)also, AQB + AQC = 180° AQB = AQC = 90°Now, In ABQ, AB2 = AQ2 + BQ2

(r + 6)2 = (12 – r)2 + 62 ( AB = AE + EB = r + 6, AQ = DQ – AD = 12 – r) r = 4 cm radius of the circle with centre at A is 4 cm.

P B Q C R

D

E F

A r cm

r cmr c

m

Now, Area of the circle with centre A = (4)2 cm2 = 16 cm2

Area of the semi-circle with centre B = area of the semi-circle with centre at 2 2 21C (6) cm 18 cm2

AMIT BAJA

J


Area of the semi-circle with centre at 2 2 2Q (12) cm 72 cm2

Hence, the required area of the shaded region

= The area of the semi-circle with centre at Q – The area of the semi-circle with centre at B

– The area of the semi-circle with centre at C – The area of the circle with centre at A

= (72 – 2 × 18 – 16 ) cm2 = 20 cm2 Ans.

44. Clearly, one round of wire covers 4 mm = (0.4 cm) in thickness of the surface of the cylinder.

no. of rounds to cover 24 cm 24 600.4

Length of wire in completing one round = 2r = 2 × 10 cm = 20 cm.

Length of wire in covering the whole surface

= 20 × 60 cm = 1200 cm

Now, radius of copper wire = 2 mm = 0.2 cm

Volume of wire = (0.2)2 × 1200 cm3 = 48 2 cm3

So, weight of wire 248 8.88 gm = 426.24 2 gm Ans.

45. In 6 3 3VO A, tan sin8 4 5

In PVO, 3sin 3 cmVO 5 8r r r

r

P Q

V

OA B6 cm

r rO

8 cmV1 = Volume of sphere 3 3 34 (3) cm 36 cm3

V2 = Volume of the water = Volume of the cone 2 3 31 (6) 8 cm 96 cm3

Fraction of water that flows out 1 2V :V 36 : 96 3 : 8 Ans.

46. here, VL=LO .h

clearly, VOA ~ VOA

11

VO OA 3 2=VO O A 2 3

r h rrr h

also, VOA ~ VLC

22

VO OA 3=VL LC 3

r h rrr h

V1 = Volume of cone VCD 2

2 22

1 1 13 3 3 27

rr h h r h

O

L

V

A B

C Dr2

r1

A

h

h

Br

h

OAMIT BAJA

J


V2 = Volume of the frustum ABDC 2 21 2 1 2

1 ( )3

r r r r h

2 2 2

21 4 2 73 9 9 9 27

r r r h r h

V3 = Volume of the frustum ABBA = 2 21 1

1 ( )3

r r rr h

2 2

2 21 4 2 193 9 3 27

r rr h r h

required ratio = V1 : V2 : V3 = 1 : 7 : 19 Ans.47. Volume of cylinder = (6)2 × 14 cm3 = 504 cm3

Volume of cone which is removed 2 3 31 (3) 5 cm 15 cm3

Volume of hemi-sphere which is removed 3 3 32 (6) cm 144 cm3

Required volume of the remaining solid = (504 – 15 – 144 ) cm3 = 345 cm3.Again, CSA of the cylinder = 2 (6) × 14 cm2 = 168 cm2

CSA of the cone 2 2 2 23 3 5 cm 3 34 cm

CSA of the hemisphere = 2 (6)2 cm2 = 72 cm2

Area of the metallic portion between the two circles at the top of the cylinder= [ (6)2 – (3)2] cm2 = 27cm2

Total area of the remaining metallic portion is

2(168 3 34 72 27 ) cm 2= (267 + 3 34) π cm

48. (i) No. of tubes with the life-time greater than 760 hours

624 22 (800 760) 50.8 51,800 700

since no. of tubes can’t be fractional.

required percentage 51 100350

= 14.57

(ii) no. of tubes with life less than 530 hours

736 18 30 45.9 46

100

required percentage 46 100350

13.14AMIT B

AJAJ


49. Clearly 1 4, 2 6x y represents a rectangle with the coordinates (1, 2), (4,2 ), (1, 6), (4, 6). Let’ssexamine the graph of the total region in which the successful region is shaded.The regions we are comparing are areas.Now, P(unsuccessful outcome)

Area of unsuccessful region=Area of total region

D(1, 6) C(4, 6)

B(4, 2)A(1, 2) E(3, 2)

6

2

F(1,4)

1 4 x y+ =5

1 2 2Area of FAE 2 12=Area of ABCD 3 4 12 6

P (successful outcome) 116

56

50. Clearly unit digit can be any of the four digits given. Ten’s digit can be any of the remaining three digits(as one digit is used up at units place and repetition is not allowed). Similarly, hundreds and thousandsdigit can be any of the remaining two digits and one digit respectively. Total possible numbers formed = 4 × 3 × 2 × 1 = 24.A number is divisible by 5 if unit digit is 5.Now, out of these 24 numbers, six numbers starts with digit 2 at thousands place, out of which twonumbers have 5 at their unit place. Clearly, total number having 5 at their unit place = 2 × 3 = 6 (sixnumbers starts with digit 5 also).

Required probability 624

1 Ans.4

AMIT BAJA

J

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