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GENER L INSTRUCTIONS

1. In addition to this question paper, you are given a separate answer sheet.

2. Fill up all the entries carefully in the space provided on the OMR sheet ONLY IN BLOCK CAPITALS.

Incomplete/incorrect/carelessly filled information may disqualify your candidature.

3. A student has to write his/her answers in the OMR sheet by darkening the appropriate bubble with the

help of HB Pencil as the correct answer(s) of the question attempted.

4. Paper carries 80 questions each of 3 marks.

5. Any rough work should be done only on the blank space provided at the end of question paper.

6. For each correct answer gets 3 marks, each wrong answer gets a penalty of 1 mark.

7. Blank papers, clip boards, log tables, slide rule, calculators, mobiles or any other electronic gadgets in

any form is "NOT PERMISSIBLE".

Time : 2 Hr. Date : 02-11-2014 Max. Marks : 240

PCCP Head Office:Address : J-2, Jawahar Nagar, Main Road, Kota (Rajasthan)-324005

Contact. No. : +91-0744-2434727, 8824078330

Website : www.pccp.resonance.ac.in E-mail : [email protected]

ALL INDIA IJSO(STAGE-I) TEST SERIES

MOCK TEST PAPER # 9

mailto:[email protected]

http://www.pccp.resonance.ac.in/

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IJSO STAGE-I _MOCK TEST PAPER-5_PAGE # 2

Space For Rough Work

IJSO TEST P PER

1. If a, b, x, y are natural numbers such that 2 2 25a b ! " and 2 2 13x y ! " then

# $ # $ax by ay bx ! ! ! has exactly one value, the value is

(A) 36 (B) 35 (C) 30 (D) 40

2. Find all primes p and q such that 2 27p pq q ! ! is the square of an integer..

(A) Infinite (B) Finite (C) 0 (D) cannot be determined

3. Give the last digit of 5 5 51 2 ... 99! ! !

(A) 1 (B) 0 (C) 2 (D) 5

4. The number of positive integer less than 175 and relatively prime both 5 and 7

(A) 140 (B) 100 (C) 120 (D) 180

5. The number of integer’s n such that (n4 + n2 + 64) is a square of an integer is

(A) 5 (B) 17 (C) 18 (D) 19

6. If a2 = b + c, b2 = c + a, c2 = a + b then value of1a

1

! +

1b

1

!+

1c

1

! is :

(A) 0 (B) 1 (C) –1 (D) 2

7. The number of positive integeral values of (x, y) which satisfy the equation 3 x + 3 y = 4 ; x + y = 28

simultaneously is

(A) 1 (B) 2 (C) 0 (D) 3

8. If p, q, r are the roots of the cubic equation 3 2 2 33 3 0x px q x r % ! % " then

(A) p q r & " (B) p q r " " (C) p q r " & (D) p q r & &

9. If 2 1 0x x ! ! " , the value of 1999 2000x x ! is

(A) 0 (B) 1 (C) 2 (D) -1

10. If 0a b c ! ! " and , , 0a b c & find the value of

4 4 4

2 2 2 2 2 2

a b c

a b b c c a

! !

! !

(A) 0 (B) 2 (C) 1 (D) 3

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11. If 2a c b ! " then find the value ofa c

a b c b !

% % is equal to

(A) 1 (B) 2 (C) 3 (D) 4

12. The number of solutions in positive integers of the equation 3 5 1008x y ! " is

(A) 36 (B) 67 (C) 38 (D) 35

13. The values of integer’s n such that1

53

!

%

n

nis also an integer which are

(A) 9, -3 (B) -9, -3 (C) 9, 3 (D) -9, 3

14. If , ,' ( ) are the roots of the cubic 023 "!!! r qx px x find the value of # $ # $ # $( ! ) ) ! ' ' ! ( in

terms of p, q, r

(A) r pq ! (B) 2r pq ! (C) r pq % (D) p + q

15. Let E be the midpoint of median AD of ABC * . Let DB be the produced to P such that DB = BP and

let DC be produced to Q such that DC = CQ. Let EP cut AB at L and EQ cut AC at M. Find the ratio of

area of pentagon BLEMC to that of ABC *

(A) 23 (B) 1

3 (C) 1 (D) 14

16. If C *+, , AB = AC, and 40ºBAC - " . If Q is such that OCAOBC -"- find BOC -

(A) 120º (B) 110º (C) 100º (D) 90º

17. The measures of the length of the sides of a triangle are integers and that of its area is also an integer.

One side is 21 and the perimeter is 48. The shortest side is

(A) 10cm (B) 11cm (C) 12cm (D) 13cm

18. A rectangle contains three circles, all tangent to the rectangle and also to one another. If the height of

the rectangle is 4, find the width of the rectangle.D

A B

C

O2

O1

O

ML

K

4 4

(A)1 2!

(B)3 2 2!

(C)1 2 2!

(D) 2

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19. In a ABC * , the internal bisector of A- meets BC at D. If AB = 4, AC = 3 and A- = 60º find the length

of AD

(A) 2 3 (B)12 3

7(C)

15 3

8(D) None of these

20. If D is a point on side BC of a ABC * such that BD : DC = m : n and ADC - " .,

;ADC BAD - " . - " '

and DAC - " ( , then

(A) # $ tan tan tanm n m n ! . " ' % ( (B) # $sin sin sinm n m n ! . " . % (

(C) # $cot cot cotm n m n ! . " ' % ( (D) None of these

21. 1 kg of water at 20ºC is, mixed with 800 g of water at 80ºC. Assuming that no heat is lost to the

surroundings, calculate the final temperature of the mixture :

(A) 24.44º C (B) 46.67ºC (C) 44.44ºC (D) 54.44ºC

22. The figure shows a ray of light incident on a convex lens, parallel to its principal axis. Obviously the

emergent ray passess through the principal focus F. Which of the following statements is correct ?

F

21

(A) The ray bends downwards only once inside the lens.

(B) The ray bends downwards at each surface.

(C) The ray bends downwards at the first surface and upwards at the second surface.

(D) The ray bends upwards at the first surface and downwards at the second surface.

23. The speed of sound in air at 0oC is approximately :

(A) 332 ms –1 (B) 350 ms –1 (C) 530 ms –1 (D) 332 kms –1

24. A broadcasting station transmits waves of frequency 71 × 104 Hz with a speed of 3 × 108 m/s. The

wavelength of the wave is :

(A) 418.8m (B) 324.6 m (C) 208.4 m (D) 422.5 m

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25. If the magnitudes of vectors//

B,A and/

C are 12, 5 and 13 units respectively and ,CBA///

"! the angle

between vectors/

A and/

B is :

(A) 0 (B) 0 (C) 0 /2 (D) 0 /4

26. The angle ji3 % vector makes with the positive x-axis is _______.

(A) 0º (B) – 60º (C) – 30º (D) 30º

27. Two wires of same dimension but resistivities 11 and 1

2 are connected in series. The equivalent resistivity

of the combination is

(A) 11 + 1

2(B) 1/2 (1

1 + 1

2) (C) 21 11 (D) 2(1

1 + 1

2)

28. If internal resistance of a cell is proportional to current drawn from the cell. Then the best

representation of terminal potential difference of a cell with current drawn from cell will be:

(A) (B) (C) (D)

29. A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a

particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius (R/2) is now

made in the sphere as shown in.The sphere with the cavity now applies a gravitational force F2 on the

same particle.The ratio (1

2

F

F) is :

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(A) (1/2) (B) (3/4) (C) (7/8) (D) (7/9)

30. Galileo writes that for angles of projection of a projectile at angles (45 +.) and (45 –.) ,their horizontal

ranges described by the projectile are in the ratio of : (if . 3 45)

(A) 2 :1 (B) 1 : 2 (C) 1 : 1 (D) 2 : 3

31. A particle is projected from the ground with a velocity of 25m/s. After 2 second, it just clears a wall 5m

height. Then the angle of projection of particle is [g =10 m/s2]:

(A) 30º (B) 45º (C) 60º (D) 75º

32. A ball is thrown at an angle of 30º to the horizontal .It falls on the ground at a distance of 90 m. If the ball is

thrown with the same initial speed at an angle 30º to the vertical, it will fall on the ground at a distance of -(A) 120 m (B) 27 m (C) 90 m (D) 30 m

33. A wooden cube floating on water supports a mass m = 0.2 kg on its top. When the mass is removed, the

cube rises by 2 cm. The side of the cube is :

(A) 8 cm (B) 10 cm (C) 12 cm (D) 6 cm

34. Block A is moving with a certain acceleration along a frictionless horizontal surface. When a second

block B is placed on top of block A, the acceleration of the combined block drops to

1/5 the original value. What is the ratio of the mass of A to the mass of B :

(A) 5 : 1 (B) 1: 4 (C) 3 : 1 (D) 4 : 1

35. Find the work done by a force (in newton) Fx= 5.0 x – 4.0 when this force acts on a particle that moves

from x = 1.0 m to x = 3.0 m.

(A) 3 J (B) 6 J (C) 12 J (D) 9 J

36. A particle is in a linear SHM. If the acceleration and the corresponding velocity of this particle are ‘a’ and

‘v’, then the graph relating to these values is

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(A) (B) (C) (D)

37. I tesla is equal to :(A) NAm (B) NA –1 m (C) NAm –1 (D) NA –1 m –1

38. A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a

small distance and another point Q lies to the west at the same distance. The magnetic field at P is :

(A) greater than at Q

(B) same as at Q

(C) less than at Q

(D) greater or less than at Q, depending upon the strenth of current

39. A satellite orbiting round the earth appears stationary when :

(A) Its time period is one day and it is rotating in the same sense as that of earth.(B) Its time period is one day and it is rotating normal to the direction of earth.

(C) Its time period is 12 hr and it is rotating in the same direction as that of earth.

(D) Its time period is 12 hr and it is rotating normal to earth.

40. In aerial mapping a camera uses a lens with a 100 cm focal length. The height at which the airplane must

fly, so that the photograph of a 1 km long strip on the ground fits exactly on the 20 cm long filmstrip of the

camera, is :

(A) 200 km (B) 20km (C) 5 km (D) 1 km

41. STATEMENT-1 : Second E.A. for halogens is almost zero.

STATEMENT-2 : Fluorine has maximum value of electron affinity.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

42. Consider the following statements :

(I) The radius of an anion is larger than that of the parent atom.

(II) The ionization energy generally increases with increasing atomic number in a period.

(III) The electronegativity of an element is the tendency of an isolated atom to attract an electron.

Which of the above statements is/are correct ?

(A) I alone (B) II alone (C) I and II (D) II and III

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43. Lead is mainly extracted by :

(A) Carbon reduction method

(B) Self reduction method

(C) Electrolytic reduction

(D) Leaching with aqueous solution of NaCN followed by reduction

44. In which of the following pairs, both the species have the same hybridisation ?

(I) SF4, XeF

4(II) 2

–3 XeF,I (III) 44 SiCI,ICI! (IV) –3

4 –3 PO,CIO

(A) I, II (B) II, III (C) II, IV (D) I, II, III

45. In a compound

the number of sigma and pi bonds respectively are :

(A) 19, 11 (B) 19, 5 (C) 13, 11 (D) 7, 3

46. Which statement is false for the balanced equation given below?

CS2 + 3O2 4/ 4 2SO2 + CO2

(A) 1 mole of CS2 will produce 1 mole of CO

2.

(B) The reaction of 16 g of oxygen produces 7.33 g of CO2.

(C) The reaction of 1 mole of O2 will produce 2/3 mole of SO

2.

(D) Six molecules of oxygen requires three molecules of CS2.

47. How many millilitres of 0.1 M H2SO

4 must be added to 50 mL of 0.1 M NaOH to give a solution that has

a concentration of 0.05 M in H2SO

4 ?

(A) 400 mL (B) 200 mL (C) 100 mL (D) None of these.

48. STATEMENT-1 : The ground state electronic configuration of nitrogen is

STATEMENT-2 : Electrons are filled in orbitals as per aufbau principle, Hund’s rule of maximum spin

multiplicity and Pauli’s principle.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

49. A subshell n = 5, ! = 3 can accommodate :

(A) 10 electrons (B) 14 electrons (C) 18 electrons (D) None of these

50. Which of the following is discreted in Bohr’s theory ?

(A) Potential energy (B) Kinetic energy (C) Velocity (D) Angular momentum

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51. The root mean square velocity of hydrogen is 5 times than that of nitrogen. If T is the temperature of the

gas, then:

(A)22 NH TT " (B)

22 NH TT 5 (C)22 NH TT 6 (D)

22 NH T7T 6

52 In the reaction X(g) + Y(g) 2Z(g), 2 mole of X, 1 mole of Y and 1 mole of Z are placed in a 10 litre

vessel and allowed to reach equilibrium.If final concentration of Z is 0.2M, then Kc for the given reaction is:

(A) 1.60 (B)3

80(C)

3

16(D) None of these

53. The conjugate acid of NH2 – is -

(A) NH3

(B) NH2OH (C) NH

4+ (D) ClO

4 –

54. Potash alum, which is used to separate the suspended particles from water, is a double salt of -

(A) potassium and sodium (B) aluminium and calcium

(C) potassium and aluminium (D) aluminium and barium

55. The structure of 4-Methyl - 2-hexyne is -

(A)

3

323

CH|

CH –CH –CH –CHCH –CH " (B)

3

323

CH|

CH –CC –CH –CH –CH 7

(C)

3

323

CH|

CH –CH –CH –CC –CH 7 (D) None of these

56. IUPAC name of the following compound is-

(A) 1 - Fluoro - 1- methyl - 2- nitrobutanal (B) 3 - Nitro - 2 - fluoro - 2 - methylbutanal

(C) 2 - Fluoro - 2 - methyl - 3 - nitrobutanal (D) None of these

57.. The number of ' and ( particles emitted in the nuclear reaction.

Th22890

4/ 4 21283Bi , are

(A) 4',1( (B) 3', 7( (C) 8', 1( (D) 4',7(

58. Which of the following is an arrangement of increasing value of e/m ?

(A) n < ' < p < e (B) e < p < ' < n (C) n < p < e < ' (D) p < n < ' < e

59. When a strip of copper metal is placed in a solution of ferrous sulphate -(A) grey precipitate of iron is formed and blue solution of copper sulphate is produced

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(B) red precipitate of iron is formed and green solution of copper sulphate is produced

(C) No reaction takes place

(D) iron (II, III) oxide is produced along with cuprous sulphide

60. Match List-I (mixtures) with List-II (separation methods) and the correct answer using the codes given

below the lists :

List-I List-II

(a) NaCl in water 1. Chromatography

(b) I2 in water 2. Distillation

(c) Red ink and blue ink 3. Solvent extraction

(d) NH4 Cl and NaCl 4. Sublimation

Codes :

a b c d a b c d

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

(C) 2 3 4 1 (D) 3 2 4 1

61. The organelles that contain their own genetic material are:

(A) Mitochondria, Vacuoles (B) Plastids, Golgi complex(C) Mitochondria, Plastids (D) Ribosomes, Nucleolus

62. The enclosures in which heat is trapped to maintain higher temperature, are called

(A) white houses. (B) warm houses. (C) blue houses. (D) green houses

63. The stage in which separation of sister chromatids occurs is

(A) anaphase. (B) telophase. (C) metaphase. (D) prophase.

64. Haversian canals occur in

(A) humerus. (B) pubis (C) scapula. (D) clavicle.

65. Bone marrow is absent in the bones of

(A) fish. (B) bird. (C) reptile (D) frog.

66. A fern commonly used as biofertiliser is

(A) Lycopodium. (B) Marsilea. (C) Azolla. (D) Adiantum.

67. Camel is best adapted to desert habitat as

(A) it can drink 50 litres of water at a time which is evenly distributed in all its tissues.

(B) it excretes very small amount of water during urination ?

(C) it can regulate its body temperature at a wider range

(D) all are correct

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68. Aerenchyma is found in which plants ?

(A) sciophytes (B) hydrophytes (C) mesophytes (D) epiphytes

69. In the lunch, you ate boiled green vegetables, a piece of cooked meat, one boiled egg and a sugar candy.

Which one of these foods may have been digested first ?

(A) Boiled green vegetables (B) The piece of cooked meat

(C) Boiled egg (D) Sugar candy

70. In the given food chain 'Plants Sheep Man', 5 J of energy is available to man. The energy that was

available at producer level is

(A) 50 J. (B) 500 J. (C) 5 J. (D) 0.5 J.

71. Producers prepare their own food like

(A) blue green algae. (B) amoeba. (C) rhizopus. (D) yeast.

72. Which one does not produce any digestive enzyme ?

(A) Pancreas (B) Liver (C) Stomach (D) Duodenum

73. Tunica albuginea is the covering of

(A) lungs (B) testis (C) kidneys (D) heart

74. An organism with two identical alleles of a gene in a cell is called

(A) homozygous (B) dominant (C) heterozygous (D) hybrid

75. Mendel’s law do not explain principle which is :

(A) Segregation of genes (B) Dominance

(C) Linkage (D) Independent assortment

76. Evolution proceeds on accounts of

(A) fossils (B) struggle for existence

(C) adaptation through generation (D) natural selection77. Soil is composed of

(A) mineral + water + air (B) mineral + organic matter + air

(C) mineral + organic matter + air + water (D) organic matter + water

78. The link between kreb’s cycle and glycolysis is

(A) citric acid (B) acetyl - co - A (C) Succinic acid (D) Fumaric acid

79. Compensation point refers to the intensity of light at which

(A) Rate of respiration = Rate of photosynthesis

(B) Rate of respiration > Rate of photosynthesis

(C) Rate of respiration < Rate of photosynthesis

(D) None of the above.

80. Prevention of a disease is more desirable than its cure because

(A) some of the body functions may be damaged during the effect of the disease.

(B) the person suffering from the disease will not be bedridden.

(C) the disease can not be communicated to others during the course of treatment.

(D) body does not look good during this condition.

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SOL. IJSO STAGE-I _MOCK TEST-5_PAGE # 1

NSWER KEY

HINTS & SOLUTIONS

ALL INDIA IJSO(STAGE-I) TEST SERIES

MOCK TEST PAPER # 9

DATE : 02-11-2014

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ans. B A B C A B B B D B B B D C A B A B B C

Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Ans. B B A D C C B D D C A C B B C C D B A C

Ques. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. C C B C A D C A B D C C A C C C A A C A

Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Ans. C D A A B C D B D B A B B A C C C B A A

1. The only values satisfyinga2 + b2 = 25 are a = 3, b = 4or a = 4, b = 3 with a + b = 7 in both cases.similarly x2 + y2 = 13 issatisfied by x = 2, y = 3 or x = 3, y = 2 withx + y =5 in both the cases. Thusax + by + ay + bx= (a + b)x + (a + b)y

= (a + b) (x + y) = 7 ! 5 = 35

2. Let p,q be primes such that p2 + 7pq + q2 =m2 for some positive integer m. we write5pq = m2 (p + q)2

5pq = (m + p + q) (m – p – q)

we can immediately rule out the possibilitiesm + p + q = p,m + p + q = q and m + p + q = 5[In the last case m > p, m> q and p,q are atleast 2]consider the casem + p + q = 5p and m – p – q = q

" 4p –

q = 2q + p ; 3p = 3q or p = qFinally taking m + p + q = pq. m – p – q = 5

" pq – p – q = 5 + p + q

pq – 5 = 2p + 2q = 2(p + q)

pq – 2p – 2q = 5

pq – 2p – 2q + 4 = 9

p (q – 2) – 2(q – 2) = 9

# (q – 2) (p – 2) = 9,

Thus p = q = 5 or (p, q) = (3,11) or (11, 3)Thus the set of solution is{(p, p) : p is prime} $ {(3,11), (3,11)}(A) infinite

3. Since S = 15 + 25 + .... + 995

for any a, b %Z(a + b) | (a5 + b5)Given series can be written as

S = &'

()99

1k

55 })k100(k{

hence {k + (100 – k)} | S i.e. 100 | S

Hence last digit of S has to be zero

4. Let S = {n % N : n < 175}2

A = {n %S : 5 | n}B = {n %S : 7 | n}Now the required number is|S| – |A $ B| = |S| – (|A| + |B| – |A * B|)

= 174 – +,

-./

05

174 – +

,

-./

07

174 + +

,

-./

035

174

= 174 – 34 – 2444

= 120

5. " n4 + n2 + 64 > n4 + 2n2 + 1 = (n2 +1)2

" For some non negative integer K,n4 – n2 + 64 = (n2 + K)2

n4 – n2 + 64 = n4 + 2n2K + K2

64 – K2 = 2n2K + n2

n2 =1K2

K64 2

)(

From which we obtainpossible values of n2 are obtained whenK = 0, 7, 8 # n2 = {64, 1, 0}

Hence n = { 1 8, 1 1, 0}

i.e. 5

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6. a2 = b + c, b2 = c + a, c2 = a + ba + a2 = a + b + c ; b + b2 = a + b + c ; c + c2 +=a + b + ca(1 + a) = a + b + c ; b(1 + b) = a + b + c ;c(c+1) = a + b + c

a + 1 =a

cba )) ; b + 1 =

b

cba )) ; c + 1

c

cba ))

hence1a

1

) +

1b

1

)+

1c

1

) = 1

7. 3 x + 3 y = 4, x + y = 28

2 3333 yx ) = 43

x + y + 3 3 xy (4) = 64

28 + 12 3 xy = 64

xy = 27as x + y = 28xy = 27if x = 1 then y = 27if x = 27 then y = 1

" 2 no. of positive integral value of (x, y)which satisfy the above equation.

8. As p, q, r are roots of the cubic equationx3 – 3px2 + 3q2x – r 3 = 0p + q + r = 3p ..(i)pq + qr + pr = 3q2 ..(ii)pqr = r3 ..(iii)Now if p = 0 then r = 0 by (iii)

# q = 0 by (i) i.e. q = 0 then p = q = r = 0also if q = 0 then r = 0 by (iii) and p = 0 by (i)Finally if r = 0 then (ii) impliespq = 3q2 # q(p – 3q) = 0

# q = 0 or p = 3qHence in both cases by (i)we have p = q = r = 0Now, suppose that none of p, q, r is 0then (iii) impliespq = r2 ..(iv)

Hence (ii) implies r2 + r (p + q) = 3q2

so r(p + q + r) = 3q2

From (i) 3pr = 3q2

pr = q2 ..(v)From (iv) and (v)

pr

pq = 2q

pq = 2

2

q

r ;

r

q = 2

2

q

r

q3 – r 3 = 0 ; (q – r) (q2 + r2 + qr) = 0

# q – r = 0 [" q2 + qr + r2 > 0 , p, q, r > 0 ]q = r ..(vi)Now pq = r2 and q = r ; gives

pr = r2

# p = r as r 4 0 ..(vii)From (vi) and (vii) p = q = r

9. x2 + x + 1 = 0

" (x – 1) (x2 + x + 1) = 0x3 –1 = 0

x3 = 1, Now roots will be such that 53 = 1, sovalue of x1999

+ x2000 will be – 1

10. a + b + c = 0a2 + b2 + c2 + 2(ab + bc + ca) = 0

a2 + b2 + c2 = –

2(ab + bc + ca)a4 + b4 + c4 + 2 (a2b2 + b2c2 + a2c2)= 4{a2b2 + b2c2 + c2a2 + 2(ab2 c+ a2bc + bc2a )}a4 + b4 + c4 + 2(a2b2 + b2c2 + b2c2 + c2a2)= 4(a2b2 + b2c2 + c2a2) + 8abc (a + b + c)

" (a + b + c) = 0a4 + b4 + c4 = 2 (a2b2 + b2c2 + c2a2)

222222

444

accbba

cba

))

)) = 2

11. a + c = 2bhence a, b, c are in A.P.

Let a = A –

db = A andC = A + d

Henceba

a

( +

bc

c

( =

d

dA

((

+d

dA ) = 2

12. Let x, y %N such that3x + 5y = 1008 ; then3/5y # 3/yhence y = 3k for some K %NNow 3x + 15k = 1008x + 5K = 336

5K 6 335K 6 67

Thus any solution pair is givenby (x, y) = (336 – 5K, 3K)

where 1 6 K 6 67 and

therefore the number of solutions is 67

13. since1n

5n3

)(

=1n

83n3

)()

=1n

83

)(

not possible value of1n

8

) are { 1 1, 1 2, 1 4, 1 8}

and 3 –

1n

8

) is a square

Hence1n

8

) = – 1 or 2

hence 8 = – n – 1, n = – 9 or

8 = 2n + 2 ; n = 3

14. Roots of the equationx3 + px2 + qx + r = 0 ..(i)are 5,7 and 8Let us form an equation whose roots are(7 + 8), (8 + 5), (5 + 7)Let y be a root of the transformed equation,

theny = 7 + 8 = 5 + 7 + 8 = –p

y = – p – 5

8/10/2019 9. AITS-9 (02-Nov-14)



y = – p – x

" x = –(y + p)

Putting this value of x in (i) we have

–(y + p)3 + p(y + p)2 – q(y + p) + r = 0

–y3 – p3 – 3py (y + p) + py2 + p3 + 2yp2 – qy

– pq + r = 0

y3 + 2py2 + (p2 + q)y + (pq – r) = 0

Its root are (7 + 8), (8 + 5), (5 + 7)

hence (7 + 8) (8 + 5) (5 + 7) = r –

pq

15. Given AE = ED ; DB = BP ; DC = CQA

P B DC

Q

M

E

L

x x

Toprove =)ABC(area

)BLEMC(area

9

Let area (9ABC) = xsince AD is median of 9APQarea (9ADP) = area (9ADQ) (i)similarly in 9APD, AB is median

area (9APB) = area (9ABD) =2

x(ii)

In 9ADQ, AC is median

area (9ACQ) = area(9ADC) =2x (iii)

Now, join LD and MD and BE, CENow, in 9ADP, AB and PsE aremedian hence L is centroid of 9ADP

Hence area(9BPL) = area (9BDL) =6

1x

area (9ALD) = area (9ABD) – area (9BDL)

area(9ALD) =2

x –

6

x =

3

x

hence area(9DLE) =21 !

3x

[median LE in 9ALD]

=6

1x (iv)

area quad. BLED =6

1x +

6

1x =

3

1x

Similarly area (quad CDEM) =3

1x

hence area (pentagon BLEMC) =3

x +

3

x

=3

2x

hence ABCarea

)BLEMCpentagon(area

9 =3

2

16. To find :BOCJoin AO and extand it towards BC

" AB = AC

yyx

x

O

40°

DB C

A

hence :ABC = :ACB =2

1(180° – 40°) = 70°

now :COD = :OAC + :OCA [Ext :]

:BOD = :OBA + :OABAdding

:BOC = :OAC + :ACO + :OBA + :OAB

:BOC = :OAC + :ACO + :OBA

:BOC = :BAC + :ACO + :OCD[" AB = AC]

:ABC = :ACD

# :OBA = :OCD = yHence :BOC = :BAC + :ACD

:BOC = 40° + 70° = 110°

17. Let a, b be other two sides then a + b = 27

S =2

cba )) =

2

2127 ) = 24

area = )cS)(bS)(aS(S (((

= )2124)(b24)(a24(24 (((

= )a24)(3a(72 ((

Now 72(a – 3) (24-a) must be a perfect

square. If a is shortest side a 6 13

out of the given values a = 10 is the onlyvalues such that 72(a – 3) (24 – a)

= 72 ! 7 ! 14= 62 ! 22 ! 72

hence area = 84 square unitshence shortest side = 10 cm

8/10/2019 9. AITS-9 (02-Nov-14)



18.

D

A B

C

O2

O1

O

ML

K

4 4

Width = AL + LM + MB ; AD = BC = 4Now OL = 2 hence AL = 2Also O

1M = MB = 1

Hence In 9OKO1

OO12

= OK2 + KO

12

O1K2 = 32 – (2 – 1)2 = 8

O1K = 2 2

hence AL + LM + MB = 3 + 2 2

19.

x x

30° 30°

A

CB D

44

y 3 y4

Using casine rule, we find

cos60° =)3)(4(2

BC34 222 ()

12 = 25 – BC2

Hence BC = 13

Now by internal bisector theorem.since AD is angle bisector of:A

AC

AB =

3

4 =

DC

BD

DC =4

3 BD

If BD = y ; DC =4

3y

Hence BD + DC = BC

y +

4

3y = 13

4

7y = 13

y =7

134

Now in9ABD, by cosine rule again cos30°

=AD42

49

208AD16 2

!!

()

2

3! 8 ! AD = 16 –

49

208 + AD2

4 3 AD =49

576 + AD2

49AD2 – 196 3 AD + 576 = 0

AD =98

1128961152483196 (1

AD =98

3283196 1

AD = 98

3283196 ( = 98

3168

AD =7

312

20. GivenDC

BD =

n

mand :ADC = ;

A

B C180-

m nD

" ADB = (180° – ;)

:BAD = 5 and :DAC = 7

:ABD = 180° – [(5 + 180 – ;)] = (; – 5)

and :ACD = 180° – (; + 7)

from:ABD

5sin

BD =

)dsin(

AD

(; ..(1)

From 9ADC

7sin

DC =

))(180sin(

AD

7);( =)sin(

AD

7); ..(2)

Dividing (i) by (ii)

DC

BD

5

7

sin

sin

= )sin(

)sin(

5(;

7);

= 5;(5;

7;)7;

sincoscossin

sincoscossin

n

m.

57

sin

sin =

5;(5;7;)7;

sincoscossin

sincoscossin

msin7sin;cos5 – msin7cos;sin5

= nsin5sin;cos7+ nsin5cos;sin7

Dividing both sides by sin5sin;sin7

mcot5 – mcot; = ncot7 + ncot;

" (m + n) cot; = (mcot5 – ncot7)

8/10/2019 9. AITS-9 (02-Nov-14)



21. In this process water at high temperature willgive its heat to water at low temperature onmixing them. So from law of mixture.m

1s9T

1= m

2s9T

2

(Here, m1 = 1 kg, m

2 = 800 g = 0.8 kg)

Let the final temperature be T i.e.1 × s × (T – 20) = 0.8 ×s × (80 – T)

On solving, we get

T = 46.67ºC

24. Frequency of waves, = = 71 × 104 HzSpeed of waves, C = 3 × 108 m/swave length of waves ,

> = 4

8

1071

103C

!

!'

= = 422.5 m

26. tan; =x

y =

3

1(

" ; = – 30º

;

y

x

27.AAA

221eq

!!!?)?'?

?eq

= 1/2 (?1 + ?

2)

28. Givenr @ i # r = kiV = E – ir = E – i(ki)

V = –

i2 k + E

33. Weight of object = 0.2 × g

The volume of water displaced = volume of cubeemerged.

i.e. a

2

× 2 × 10

–2

× 10

3

× g = 0.2 g

# a2 =20

2.0

a = 10 cm

34. Suppose mass of first block is mA and that of

second in mB

So, F = mA a

A...... (i)

mA

aA

in second case

F = (mA

+ mB

)5

4a

A

...... (ii)

mA

a= a –A aA

mB

5

40. Focal length, f = 1m,size of object, h

1 = 1000m, size of image,

h2 = 0.2 m

image is real, so

–u102v

uv

hh 4

1

2 (!('#'

u

1

v

1

f

1('"

1 = –

u

1

u102

14

(! (

# u = – 5000 m or u = –5 km

47. Let volume is x mL,x × 0.1 × 2 – 50 × 0.1 = (x + 50) × 0.05

× 2

0.2 x –

5 = 0.1 x + 5# 0.1 x = 10

# x = 100 mL

51. Vrms

=M

RT3 =

2

2

Nrms

Hrms

)V(

)V(

=2

2

2

2

N

N

H

H

T

M

M

T! ;

2Hrms )V( = 5 2Nrms )V(

2

2

Hrms

Hrms

)V(

)V(

× 5 =2

2

2

2

N

N

H

H

T

M

M

T! ;

=1

5 = 14

T

T

2

2

N

H!

= 5 =2

2

N

H

T

T

× 14

2NT × 5 =2HT × 14

22 HN TT A

54. Patash alum is K2SO

4. Al

2(SO

4)

3. 24H

2O. It is a

double salt of potassium and aluminium.

57.. No. of 5-particle emitted

=Difference in atomic masses of reactants & products

4

=4

212 –228

= 4 5-particle

No. of 7-particle = (2 × no. of a-particle) – (ZA –Z

B)

= (2 × 4) – (90 – 83)

= 8 – 7

= 1 7(particle

59. Copper is less reactive metal than iron so itcan not replace iron from ferrous sulphate

solution.

9. aits-9 (02-nov-14)

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