aryabhatta mock test

FIITJEE

ARYABHATTA MOCK TEST – IV QP CODE: 127219

Time Allowed : 2 Hours Max. Marks : 100 GENERAL INSTRUCTION 1. Do not write your name on the questionnaire.

2. Write your roll no. on each page of the questionnaire and the Answer Sheet in the space

provide

3. All the questions are compulsory

4. Read questions carefully; think twice before you write the answer. No overwriting or

cutting is allowed on the Answer Sheet. Another copy of the questionnaire or answer

sheet will not be provided.

5. Do your rough work in the space provided in the questionnaire.

6. The questionnaire contains four sections. Section A contains 10 questions on Logical

Reasoning of 1 mark each, Section B contains 20 Multiple Choice Questions of 1 marks

each, Section C contains 20 Free Response Type Questions of 2 marks each and

Section D contains 10 Free Response Type Questions of 3 marks each.

7. No working or descriptive answer of any question is to be given. Only the Answers

are to be written on the Separate Answer sheet provided to you.

8. Fill answers of question number 1 to 30 in OMR Sheet and for remaining questions

(31 to 60) write answer in the Answer Sheet attached with this paper.

9. If answer of free response type question is in decimals then write answer upto two places of

decimals.

10. Use Blue or Black pen to write the answer on the Answer Sheet.

11. Answers should be clearly written in the space provided on the Answer sheet.

12. Use of calculator is not allowed.

Name of the Candidate : ……………….………………………………….. Enrollment Number :…………………………………………………….

Section – A 1. Find the missing term.

(A) 610 (B) 660 (C) 670 (D) 690 Directions (Questions 2 – 3): Read the following information to answer these questions : Seven poles A, B, C, D, E, F and G are put in such a way that the distance between the two decreases by 1 metre. The distance between the first two poles, A and B, is 10 metres. 2. What is distance, between the first pole A and the last pole G? (A) 40 m (B) 45 m (C) 49 m (D) None of these 3. If a monkey hops from pole G to pole C, then how much distance did it cover? (A) 19 m (B) 22 m (C) 26 m (D) 32 m 4. Find the missing number in the second figure on the basis of numbers arranged in the first

figure.

(A) 30 (B) 58 (C) 160 (D) 32 5. From among the four alternatives given below, which letter replaces in the given figure the

question mark?

3 P 8

9 G 11

2 U 4

3 W 1

7 ? 18

(A) A (B) B (C) S (D) Y

space for rough work

Directions (Question 6): In the following question consists of two sets of figures. Figures 1, 2, 3 and 4 constitute the Problem Set while figure (A), (B), (C) & (D) constitute the Answer Set. There is a definite relationship between figure 1 and 2. Establish a similar relationship between figures 3 and 4 by selecting a suitable figure from the Answer Set that would replace the question mark (?) in figure 4. Problem Set Answer Set 6.

?

1 2 3 4 (A) (B) (C) (D) 7. If FAST is coded as 798 and LAST is coded as 906 then BUSY is coded as (A) 1759 (B) 1431 (C) 952 (D) 948 8. Salil travels 9km towards east, then 8km towards north, and finally travels 7km towards east.

How far is he from his original place and in which direction (approximately)? (A) 15km, north east (B) 18km, south west

(C) 15km, south west (D) 18km, north east 9. Find the mirror image of (X):

(X) (A) (B) (C) (D) 10. 3, 4, 10, 33, 136, 685, ? (A) 4016 (B) 4106 (C) 4116 (D) 4226

Section – B

11. A set of n numbers has the sum ‘S’. Each number of the set is increased by 20, then multiplied by 5, and then decreased by 20. The sum of the numbers in the new set thus obtained is:

(A) S + 20 n (B) 5 S + 80 n (C) 5 S + 40 n (D) 5 S + 100 n 12. A spherical ball is formed by melting three spherical balls of radii 1 cm, 2 cm and 3 cm.

During this process 25% of material is wasted. The radius of the new ball be? (A) 6 cm (B) 5 cm (C) 3 cm (D) 2 cm


13. An aeroplane leaves an airport and flies due north at a speed of 1000 km/h. At the same

time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/h.

After 1

12

hours, the distance between the two planes is

(A)300 61km (B) 61 300 km

(C) 36100 km (D)30 61km

14. If the quotient of 4 3 2x 11x 44x 76x 48− − − + , when divided by ( )2x 7x 12− + is Ax2 + Bx + C,

then the descending order of A, B, C is: (A) A, B, C (B) B, C, A (C) A, C, B (D) C, A, B 15. Two brands of chocolate are available in packs of 24 and 15 respectively. If I need to buy an

equal number of chocolates of both kinds, then what is the least number of boxes of each kind I would need to buy?

(A) 5, 6 (B) 5, 8 (C) 5, 4 (D) 12, 14 16. How many bricks, each measuring 25 cm × 11.25 cm × 6 cm will be needed to build a wall 8

m long, 6 m high and 22.5 cm thick? (A) 5600 (B) 6000 (C) 6400 (D) 7200 17. The edge of a cube is increased by 100%. The surface area of the cube is increased by (A) 100% (B) 200% (C) 300% (D) 400%

18. In figure ABCD and FECG are parallelograms equal in area. If ar(AQE) = 12 cm2. then ar(|| gmFGBQ) is equal to

A Q

D E C

B

G F

(A) 12 cm2 (B) 20 cm2 (C) 24 cm2 (D) 36 cm2 19. A chess board contains 64 equal squares and the area of each square is 6.25 cm2. A border

round the board is 2 cm wide. Find the length of the side of chess board. (A) 36 cm (B) 24 cm (C) 16 cm (D) 32 cm 20. Fractorise : p3(q – r)3 + q3(r – p)3 + r3(p – q)3 (A) 2pq(p + q)(q + r)(r – p) (B) 3pqr(p – q)(r – q)(r – p) (C) 2pqr(p – q)(q – r)(p – r) (D) 3pqr(p – q)(q – r)(r – p)


21. A man can row 40 km upstream and 55 km downstream in 13 hours. Also, he can row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the man in still water and speed of the current.

(A) 8 km/hr, 5 km/hr (B) 8 km/hr, 3 km/hr (C) 5 km/hr, 3 km/hr (D) 9 km/hr, 6 km/hr 22. The figure shows a rectangle ABCD with a semi-circle and a

circle inscribed inside it as shown. What is the ratio of the area of the circle to that of the semi-circle?

(A) ( )2

2 1− (B) ( )2

2 2 1−

(C) ( )2

2 1 / 2− (D) none of these

A B

C D

23. A point P is selected inside an equilateral triangle. If sum of lengths of perpendicular

dropped on to sides from P is 2019, thenlength of altitude of triangle

2019

, is equal to

(A) 1 (B) 2 (C) 3 (D) None of these 24. In the figure, AB and CD are two diameters of a circle (with

centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded

region. (Use 22

7 = )

(A) 65.50 cm2 (B) 66.50 cm2 (C) 67.50 cm2 (D) 68.50 cm2

B

D O

A

C

25. If ( )2 4 21f x 1 x 7x k− = − + and ( )3 6 3

2f x 2 x 9x k ,− = − + then the value of (k2 – k1) is

(A) 6 (B) 7 (C) 8 (D) 9 26. A pharmaceutical company manufactures 6000 strips of prescribed diabetic drugs for

Rs.800000 every month. In July 2019, the company supplied 600 strips of free medicines to

doctors at various hospitals. Of the remaining medicines, it was able to sell 4

th5

of the strips

at 25% discount and the balance at the printed price of Rs.250. Assuming vender’s discount at the rate of a uniform 30% of the total revenue, the approximate percentage profit/loss of the pharmaceutical company in July 2019 is

(A) 5.5% (profit) (B) 4% (loss) (C) 5.5% (loss) (D) None of these 27. In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the

fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

(A) 20 min (B) 15 min (C) 10 min (D) 5 min


28. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is:

(A) 4

(B) 1

2

−

(C) 5

(D) 2 1−

29. An egg seller sells his eggs only in packs of 3 eggs, 6 eggs, 9 eggs, 12 eggs etc., but the

rate is not necessarily uniform. One day Raju (which is not the same egg seller) purchased at the rate of 3 eggs for a rupee and the next hour he purchased equal number of eggs at the rate of 6 eggs for a rupee. Next day he sold all the eggs at the rate of 9 eggs for Rs.2. What is his percentage profit or loss?

(A) 10% loss (B) 11.11% loss (C) 3% loss (D) 2.5% profit 30. Pranjul purchased a hotel worth Rs.10 lakh and Sara purchased a car worth Rs.16 lakh. The

value of hotel every year increases by 20% of the previous value and the value of car every year depreciates by 25%. What is the difference between the price of hotel and car after 3 years?

(A) Rs.925000 (B) Rs.10,53,000 (C) remains constant (D) can’t be determined

Section – C

31. ABCD is a square. A circle is inscribed in the square. Also taking

A, B, C, D (the vertices of square) as the centres of four quadrants, drawn inside the circle, which are touching each other on the mid-points of the sides of square. Area of square is 4 cm2

and area of the shaded region is ( ) 22 K cm − then K = ?

A

D

B

C

32. A, B and C three weavers have to supply an order of 100 shawls. A can weave a shawl in 2

hours, B in 3 hours and C in 4 hours respectively. It is known that even being a joint contract each one weaves his own shawl completely i.e., no other weaver help to the rest weavers. In how many hours they will complete the order irrespective of day or night?

33. In a certain examination paper, there are n questions. For J = 1, 2, ….. n, there are 2n – j

students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is:

34. In the given figure, P is a point on AB such that AP : PB

= 4 : 3. PQ is parallel to AC and QD is parallel to CP. In

ARC, ARC = 90°, and in PQS, PSQ = 90°. The length of QS is 6cms. What is ratio AP : PD?

A P D

B

Q

S

R

C


35. The average of 4 distinct prime numbers a, b, c, d is 35, where a < b < c < d. a and d are equidistant from 36 and b and c are equidistant from 34 and a, b are equidistant from 30 and c and d are equidistant from 40. The difference between a and d is

36. f is a function for which f(1) = 1 and f(x) = 2x + f(x – 1) for each natural number x 2. Find f(31).

37. If a, b and c are 3 consecutive integers between – 10 to + 10 (both inclusive), how many

integer values are possible for the expression ( )

3 3 3

2

a b c 3abc

a b c

+ + +

+ +?

38. Points A and B are 90 km apart from each other on a highway. A car starts from A (towards

B) and another from B at the same time. If they go in the same direction they meet in 9 hours

and if they go in opposite direction they meet in 9

7hours. Speed of car starting from point A

is

39. If 3 1

x ,2

+= then find the value of 4x3 + 2x2 – 8x + 2.

40. In a triangle ABC, AD, BE and CF are the internal angle bisectors of the angles A, B and C

respectively. If D divides BC in the ratio 1 : 2 and E divides AC in the ratio 3 : 4, What is the ratio in which F divides AB?

41. Two triangles are similar but not congruent and the lengths of the sides of first triangle are

6cm, 11 cm, 12 cm. The sides of second also have integral length and one of them is congruent to a side of a first, what is perimeter of second triangle?

42. How many integers ‘x’ in {1, 2, 3, …, 99, 100} are there such that x2 + x3 is the square of an

integer? 43. If f(x) be a quadratic polynomial such that f(2) = –3, and f(–2) = 21. Then, the absolute value

(modulas) of coefficient of ‘x’ in f(x) is_____? 44. A and B together can finish a work in 15 days. A and C take 2 days more to complete the

same work than that of B and C. A, B and C together complete the work in 8 days. In how many days will A finish it separately?

45. If 2 2 2 2 2 2

1 1 1 1 1 11 1 ... 1

1 2 2 3 2007 2008+ + + + + + + + + =

1K

K− then K = ?

46. If the area of the shaded region in the diagram shown is K(–1) where the given triangle is isosceles with vertices of bases lying on axis of the radius perpendicular to the diameter of the two small semicircles, then K = ?

8 O 8


47. Two circles each with radius 10 cm are placed so that they are tangent to each other and straight line. A small circle is nested between them so that is tangent to larger circle and line. What is radius of smaller circle?

48. In a right angle triangle ABC, right angled at B. If AB : BC = 1 : 2. A perpendicular is dropped

from B to AC which meets AC at D. Then AD : DC = ? 49. Let n = 640640640643, what is the remainder when n2 is divided by 8. 50. The height of a cone is 40 cm. A small cone is cut at the top by a plane parallel to the base.

If the volume of the small cone be 1

64of the volume of the given cone, the height above the

base where the section is made is equal to 6 k, then find ‘k’.

Section – D

51. What is the area of the largest rectangle that can be inscribed in a right angled triangle with

legs of length 3 and 4? 52. In the given figure a circle of radius 4 cm is inscribed in a triangle which touches the side at

D, E, F respectively such that BD = 6 cm and DC = 8 cm. Then find the value of 1

16(AB +

AC).

B C

E

A

O

D

F

53. Places P1 and P2 are 250 km apart from each other on a national highway. A car starts from

P1 and another from P2 at the same time with their uniform speeds respectively. If they go in the same direction then they meet in 5 hours and if they go in opposite directions they meet

in 25

13hours. Then, ( )

1of sum of their speeds ?

8=

54. A circle is inscribed in a triangle whose sides are 40, 40 and 48 cm respectively. A smaller

circle touching two equal sides of the triangle and to the first circle. Find the radius of smaller circle.


55. ABCD is a quadrilateral such that AD = 9 cm, BC = 13 cm and DAB = BCD = 90°. P and Q are two points on AB and CD respectively, such that DQ : BP = 1 : 2 and DQ is an integer. How many values can DQ take, for which the maximum possible area of the quadrilateral PBQD is 150 sq. cm?

56. The centre of a circle inside a triangle is at a distance of 625 cm from each of the vertices of

the triangle. If the diameter of the circle is 350 cm and the circle is touching only two sides of the triangle, then find the area of the triangle.

57. A bouncing tennis ball is dropped from a height of 32 m. The ball rebounds each time to a

height equal to half the height of the previous bounce. The approximate distance traveled by the ball when it hits the ground for the eleventh time, is

58. The student fest in an Engineering College is to be held in one month’s time and no

sponsorship has yet been arranged by the students. Finally, the General Secretary (GS) of the student body took the initiative and decided to go alone for sponsorship collection. Infact, he is the only student doing the fund raising job on the first day. However, seeing his enthusiasm, other students also joined him as follows: on the second day, 2 more students join him; on the third day, 3 more students join the group of the previous day and so on. In this manner, the sponsorship collection is completed in exactly 20 days. If an MBA student, is twice as efficient as an engineering student then the number of days which 11 MBA students would take to do the same activity, is

59. Three horses are grazing within a semi-circular field. In

the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi circles with centres at P and R respectively, and S is the centre of the circle touching the two semi circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi circles and the horses tied at S can graze within the circle centred at S. The percentage of the area of the semi circle with diameter AB that cannot be grazed by the horses is :

A

S

O B R P

60. In the following figure, the diameter of the circle is 3 cm. AB and

MN are two diameters such that MN is perpendicular to AB. In addition CG is perpendicular to AB such that AE : EB = 1 : 2, and DF is perpendicular to MN such that NL : LM = 1 : 2 and length

of DH is ( )K 2 1

2

− then K = ?

A

D F

B

N G

C M

O

L H

E


FIITJEE Aryabhatta Mock Test – IV

Answer Sheet for Q. No. 31 to 60

Name : ______________ Batch Code : ___________ Enrollment No: ______________________________ GENERAL INSTRUCTION 1. No overwriting or cutting is allowed on the Answer Sheet. Another copy of the answer sheet

will not be provided. 2. Use Blue or Black pens to write the answer on the Answer Sheet. 3. Answers should be clearly written in the space provided on the Answer sheet. 4. If Answer is in decimals then write answer upto two places of decimal.

Q. No. Answer Q. No. Answer

31. 46.

32. 47.

33. 48.

34. 49.

35. 50.

36. 51.

37. 52.

38. 53.

39. 54.

40. 55.

41. 56.

42. 57.

43. 58.

44. 59.

45. 60.

FIITJEE

ARYABHATTA MOCK TEST – IV

QP CODE: 127219

ANSWERS

Section – A 1. D 2. B 3. C 4. B 5. B 6. A 7. B 8. D 9. A 10. C

Section – B

11. B 12. C 13. A 14. A 15. B 16. C 17. C 18. C 19. B 20. D 21. B 22. D 23. A 24. B 25. C 26. C 27. C 28. B 29. B 30. B

Section – C 31. 4 32. 93 33. 12 34. 7 : 3 35. 14 36. 991 37. 2 38. 40 km/hr

39. 5 40. 3 : 2 41. 58 42. 9 43. 6 44. 40 45. 2008 46. 16 47. 2.5 48. 1: 4 49. 1 50. 5

Section – D

51. 3 52. 1.75 53. 16.25 54. 3 55. 9 56. 387072 cm2 57. 96 m 58. 70 59. 13.8% 60. 2

SOLUTIONS

Section – A

1. We have : ( )2 2 2 21 5 4 3 10 510;+ + + =

( )2 2 2 23 4 6 2 10 650;+ + + =

So, missing number = ( )2 2 2 21 2 8 0 10 690+ + + =

2. The arrangement of poles is as under :

B G C D E

6 m 10 m 9 m A

8 m 7 m F

5 m

Distance between A and G = ( )10 9 8 7 6 5 45m+ + + + + =

3. Required distance 8 7 6 5= + + + 26m.=

4.

option 2 is the correct answer. 5. 3 + 8 = 11, and P is 11th letter from right end. 9 + 11 = 20, and G is 20th letter from right end. Similarly, 7 + 18 = 25, and B is 25th letter from right end. 6. As per observation.

7. FAST = ( )2 2 2 26 1 19 20 798+ + + =

LAST = ( )2 2 212 1 19 20 906+ + + =

BUSY = ( )2 2 2 22 21 19 25 1431+ + + =

8.

9

7 B

7

8

9 A Let A and B be the initial and the final points respectively in the journey. Salil is towards the north east with respect to the original place.

AB = ( ) ( ) ( )2 2

16 8 18km approx+ =

9. Mirror reflection 10. (3 x 1) + 1 = 4; (4 x 2) + 2 = 10; (10 x 3) + 3 = 33 ; …

(685 x 6) + 6 = 4116

Section – B 11. Let S = a1 + a2 + a3 + … + an

20n + S = (a1 + 20) + (a2 + 20) + (a3 + 20) + ….. + (an + 20)

5(20n + S) = 5(a1 + 20) + 5(a2 + 20) + 5(a3 + 20) + ….. + 5(an + 20)

5(20n + S) – 20n = {5(a1 + 20) – 20} + {5(a2 + 20) – 20} + {5(a3 + 20) – 20} + ….. + {5(an + 20) – 20}

Thus, required sum = 5(20n + S) – 20n = 5S + 80n

12. Vol. of new ball ( )3 3 3

1 2 3

3 4r r r

4 3= + +

( )3 3 31 2 3= + +

( )1 8 27= + +

= 36 cm3

34r 36

3 =

3 36 3r 27

4

= =

3r 27 3 cm = =

13. Let the first aeroplane lies in the direction OA and second in the direction OB.

A

O B

1500 km

1800 km Distance travelled by the first plane

= speed time = 1000 3

2= 1500 km

Distance travelled by the second plane

= 1200 3

2= 1800 km

Distance between the plane

= ( ) ( )2 2

1500 1800 2250000 3240000+ = +

= 5490000 300 61= km.

14.

2

2 4 3 2

4 3 2

3 2

3 2

2

2

x 4x 84

x 7x 12 x 11x 44x 76x 48

x 7x 12x

4x 56x 76x 48

4x 28x 48x

84x 28x 48

84x 588x 1008

616x 1056

− −

− + − − − +

− +− + −

− − − +

− + −+ − +

− − +

− + −+ − +

− +

Comparing, A = 1, B = –4, C = –84

Descending order is A, B, C 15. L.C.M. of 24 and 15 is 120

Number of boxes required120

524

= boxes for 1st brand.

Number of boxes required120

815

= boxes for 2nd brand.

16. Volume of the brick = 25 × 11.25 × 6 = 1687.5 cm3

For the wall, l = 8 m = 800 cm, h = 6 m = 600 cm, b = 22.5 cm Volume of the wall = (800 × 600 × 22.5) cm3 = 10800000 cm3

Volume of the wall

Number of bricksVolume of the brick

=

3

3

10800000 cm6400

1687.5 cm

= =

17. Let ‘a’ be the edge of a cube. Area = 6a2 = 6A New area = 6 × (2a)2 = 24A

Increased 18A

% 100 % 300%6A

= =

18. ar(|| gmABCD) = ar(|| gmFECG)

ar(|| gmABCD) – ar(|| gmBQEC) = ar(|| gmFECG) – ar(|| gmBQEC)

ar(|| gmADEQ) = ar(|| gmFGBQ)

Now, ar(|| gmAQED) = ar(ADE) + a1(AQE) = 12 + 12 = 24 cm2 So, ar(|| gmFGBQ) = 24 cm2

19. Let the length of the side of the chess board be x cm.

Then, Area of 64 squares = (x – 4)2

(x – 4)2 = 64 × 6.25

2 2x 8x 16 400 x 8x 384 0 − + = − − =

2x 24x 16x 384 0 − + − =

( )( )x 24 x 16 0 x 24 cm − + = =

Chess board

2cm 2cm

2cm

2cm

x cm

20. We have, p3(q – r)3 + q3(r – p)3 + r3(p – q)3

( ) ( ) ( ) 3 3 3

p q r q r p r p q= − − + −

Let ( ) ( ) ( )p q r a,q r p b and r p q c− = − = − =

Then, ( ) ( ) ( )a b c p q r q r p r p q+ + = − + − + −

a b c pq pr qr qp rp rq 0 + + = − + − + − =

3 3 3a b c 3abc + + =

( ) ( ) ( ) ( ) ( ) ( )3 3 3

p q r q r p r p q 3p q r q r p r p q − + − + − = − − −

( ) ( ) ( ) ( ) ( ) ( )3 3 33 3 3p q r q r p r p q 3pqr p q q r r p − + − + − = − − −

21. Let rate of man in still water = x km/hr and rate of current = y km/hr. Speed in upstream

= (x – y) km/hr and speed in downstream = (x + y) km/hr.

40 55 30 4u

13 and 10x y x y x y x y

+ = + =− + − +

Let 1 1

u and vx y x y

= =− +

So equations becomes 40u + 55v = 13 and 30u + 44v = 10

40u + 55v – 13 = 0 and 30u + 44v – 10 = 0 By cross multiplication, we get

( ) ( ) ( ) ( )

u v 1

55 10 44 13 40 10 30 13 44 40 55 30

−= =

− − − − − − −

u v 1

22 10 110 = =

22 1 10 1

u and v110 5 110 11

= = = =

So, 1 1

u x y 55 x y

= = − =−

…(i)

and 1 1

v x y 1110 x y

= = + =+

…(ii)

Adding (1) & (2), we get 2x = 16 x = 8 and putting x = 8 in (1), we get x = 3.

Speed of man in still water = 8 km/hr and speed of current = 3 km/hr 22. If the radius of the semi circle is R and that of

the circle is r, then from the figure given below

EB =R 2 R r r 2= + +

r 2 1

3 2 2R 2 1

−= = −

+

( )22 2

2

r r2 2 3 2 2 2[17 12 2]

RR

2

= = − = −

A B

C D E

2R

23. let ABC be an equilateral triangle with side a, i.e. AB

= BC = AC = a Given, PL + PM + PN = 2019

Now, area of ABC = ar(PBC) + ar(CPA) +

ar(APB)

=1 1 1 1

a AD BC PL PM AC AB PN2 2 2 2 = + +

1 1 1 1

a AD a PL a PM a PN2 2 2 2 = + +

A

M

B

N

C D L

P

( )1 1

a AD a PL PM PN2 2 = + +

AD = PL + PM + PN

AD = 2019

Length of altitude AD

2019 2019= =

20191

2019=

24. O is the centre of the circle, OA = 7 cm

AB = 2OA = 2 × 7 = 14 OC = OA = 7 cm

AB and CD are perpendicular to each other

OC ⊥ AB

Area ABC = 21 1AB OC 14cm 7cm 49 cm

2 2 = =

Again OD = OA = 7 cm

B

D O

A

C

Radius of the small circle = ( )1 1 7

OD 7 cm2 2 2

= =

Area of the small circle = 2 222 7 7 11 7 77cm cm

7 2 2 2 2

= =

Radius of the big circle =14

cm 7cm2

=

Area of the semi-circle OACB = 21 227 7 cm

2 7

= 11 7 cm2 = 77 cm2 Now, area of the shaded region = [Area of the small circle] + [Area of the big semi-circle OABC] – [Area of ABC]

= 2 2 2 277 77 154 98cm 77cm 49cm cm

2 2

+ −+ − =

= 2 2 2231 98 133cm cm 66.50 cm

2 2

−= =

25. f(x2 – 1) = x4 – 7x2 + k1

f(0) = 1 – 7 + k1 = k1 – 6 [putting x2 = 1] and f(x3 – 2) = x6 – 9x3 + k2

f(0) = 4 – 18 + k2 = k2 – 14 [putting x3 = 2] Now, f(0) = k1 – 6 = k2 – 14

k2 – k1 = 8

26. Total revenue 1 4

5400 250 5400 0.75 250 10800005 5

= + =

Net revenue after 30% discount to vendors = 0.7 × 1080000 = 756000

Loss = 800000 – 756000 = 44000

Loss percent 44000

100 5.5%800000

= =

27. Let F and S denote the faster and slower runner. Since the ratio of speeds of slower is to faster runner is 1 : 2, hence when S completes one round, F completes 2 round of the circular track. Thus in 5 minutes (when they meet for the first time) S has completed one round (of 1000

meter) and F has completed two rounds and hence traversed 2000 m. Thus F needs 5 × 2 = 10 minutes to traverse 200 × 2 = 4000 m.

28. AM = BM = BN = AN = 1 cm and AMBN is a square.

P B A

N

M

The area of shaded region = area of sector AMN – area of AMN

( )2 90 1 1 1

1360 2 4 2

= − = −

But since we have just equal area common inside the circle with centre B.

A

N

M

Total area common to both the circles 1

2 14 2 2

= − = −

29. CP of one egg (in first case) 1

33.33 paise3

= =

CP of one egg (in second case) 1

16.66 paise6

= =

Average CP of one egg ( )33.33 16.66

25 paise2

+= =

SP of one egg 200

9= [Re. 1 = 100 paise]

profit-loss

20025

9 10025

−

= CP SP

loss% 100CP

− =

= 11.11% loss 30. Worth of hotel after 3 years = 1000000 (1.2)3 = 1728000

Worth of car after 3 years =

33

16000004

= Rs.6,75,000 So, the difference in their worth (pertaining to hotel and car) is = 1728000 – 675000 = 10,53,000

Section – C

31. Area of region x = Area of square – Area of inscribed circle = (4 – )

x x

x x

y

Fig.(i)

x x

x x

Fig.(ii)

y

Fig.(iii) Area of region y = Area of square – 4 (area of quadrant)

( ) ( )21

4 4 1 44

= − = −

Required area (of shaded region) = Area of square – [Area of region x + Area of region y]

= 4 – [4 – + 4 – ] = 2 − 4 K = 4 32. LCM of 2, 3 and 4 = 12 In 12 hours A will make 6 shawls In 12 hours B will make 4 shawls In 12 hours A will make 3 shawls Total shawls = 13 i.e., in 12 hours they will weave 13 shawls so, in 84 hours they will weave 91 shawls Now, in 9 hours A will make 4 shawls in 9 hours B will make 3 shawls in 9 hours C will make 2 shawls Total shawls = 9 So, they will complete 100 shawls in 93 hours 33. No. of students who answered 1 or more questions wrongly = 2n – 1 No. of students who answered 2 or more questions wrongly = 2n – 2 No. of students who answered 3 or more questions wrongly = 2n – 3

No. of students who answered n questions wrongly = 2n – n = 1 Hence total number of wrong answers

n 1 n 2 n 3 2 12 2 2 .....2 2 1− − −= + + + + +

( )n

n1 2 1

4095 2 12 1

−= = −

−

2n = 4096 = 212 n = 12 34.

A P D

B

Q

C

P D

B

Q

Given that AP : PB = 4x : 3x

Since ABC PBQ (AC || PQ, B is common)

PD : DB = 4y : 3y ( DQ || PC) (As P divides AB in the ratio 4 : 3 so D divides PB in the same ratio 4 : 3) But PB = PD + DB

3x = 4y + 3y = 7y

x 7

y 3 =

we are required to find AP 4x x 7

PD 4y y 3= = =

35. Given that a, b, c and d are primes such that a < b < c < d. Also, a < 36 < d …(i) b < 34 < c …(ii) a < 30 < b …(iii) c < 40 < d …(iv) Combining equations (ii) and (iv), we get that 34 < c < 40.

It means c = 37 Therefore, from equation (iv), we have d = 43 Again, combining equations (iii) and (ii), we get 30 < b < 34 It means b = 31 Therefore, from equation (iii), we have a = 29 Thus, the difference between a and d is 14. 36. Look at the following pattern f(1) = 1 = (1 × 2 – 1) or (12 + 0) f(2) = 5 = (2 × 3 – 1) or (22 + 1) f(3) = 11 = (3 × 4 – 1) or (32 + 2) f(4) = 19 = (4 × 5 – 1) or (42 + 3) f(5) = 29 = (5 × 6 – 1) or (52 + 4) f(6) = 41 = (6 × 7 – 1) or (62 + 5) ……………………………………. ……………………………………. f(31) = (31 × 32 – 1) or (312 + 30) Therefore, f(31) = 991 37. Since a, b, c are consecutive integers, so we can express them as a = (b – 1) and c = (b + 1)

and −9 b 9, as we have considered that a < b < c.

( )

( ) ( ) ( ) ( )

( )

3 333 3 3

2 2

b 1 b b 1 3 b 1 b b 1a b c 3abc

a b c b 1 b b 1

− + + + + − ++ + + =

+ + − + + +

( ) ( )

3 3 3 3 3

2 2

a b c 3abc 3b 6b 3b 3b

a b c 3b

+ + + + + − =

+ +

( )

3 3 3 3

2 2

a b c 3abc 6b 3b

9ba b c

+ + + + =

+ +

( )

3 3 3 2

2

a b c 3abc 2b 1

3ba b c

+ + + + =

+ +

The above expression will be integer only when b = 1. Wherever b is an even integer, numerator will be odd and denominator will be even. An odd numerator is not completely divisible by an even denominator.

Whenever b is an odd, except for b = 1, the above expression leaves some remainder.

Thus, the only possible values of b are −1 and 1. 38. Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be

x km/hr and that of car Y be y km/hr. Case–I: When two cars move in the same direction: Distance travelled by car X = AQ Distance travelled by car Y = BQ It is given that two cars meet in 9 hours.

Distance travelled by car Y in 9 hours = 9y km BQ = 9y

A P B Q

90 km

Clearly, AQ – BQ = AB [ AB = 90 km]

9x – 9y = 90 x – y = 10 …(i) Case–II: When two cars move in opposite direction: Suppose two cars meet at point P. Then Distance travelled by car X = AP Distance travelled by car Y = BP

In this case, two cars meet in 9

7hours

Distance travelled by X in 9

7hours =

9

7x km

9AP x

7=

Distance trvelled by car Y in 9

7hours =

9

7y km

9BP y

7=

Clearly, AP + BP = AB

( )9 9 9

x y 90 x y 907 7 7

+ = + =

x + y = 70 …(ii) Solving equations (1) and (2), we get x = 40 and y = 30 Hence, speed of car X is 40 km/hr 39. Taking 4x3 + 2x2 – 8x + 2

Substituting 3 1

x ,2

+= we get

3 2

3 1 3 1 3 14 2 8 2

2 2 2

+ + ++ − +

( ) ( )

( )3 2

3 1 3 14 3 1 2

2 2

+ += + − + +

3 3 1 3 3 9 3 1 2 3 8 3 8 4

2

+ + + + + + − − +=

10

52

= =

40. Given a triangle ABC.

B

C

E

D

A

F

1 2

3

4

AD, BE and CF are angle bisectors of angles A, B and C. From the angle bisector theorem, we have

AB 1 3 AB 3

andAC 2 6 BC 4

= = =

AB : BC : AC = 3 : 4 : 6

AF AC 6 3

BF BC 4 2= = =

F divides AB in the ratio of 3 : 2. 41.

6 11

12

x y

z

Let x, y, z be integral lengths of the second triangle.

Since, triangles are similar

x y z

6 11 12= =

Case I: If x = 6, then from Eq. (i) y = 11 and z = 12 which is not possible , since triangles are not congruent.

Case II: If x = 11, then 121

y6

= and z = 2.

Which is not possible as the sides of second are integral in length. Case III: If x = 12, then from Eq. (i) y = 22 and z = 24 Thus, the sides of triangles are 12, 22, 24.

Perimeter of second triangle is (12 + 22 + 24) = 58 cm

42. Here, ( )3 2 2x x x x 1+ = + is a square of an integer if (x + 1) is m2 for some ‘x’ in the specified

set where ‘m’ is integer.

Possible values for 2 2 2 2 2m 1 ,2 ,3 ,......,10=

x + 1 = m2 x = 0, 3, 8, ……, 99 Hence number of possible values for ‘x’ for which x2 + x3 is an integer is ‘9’. Since, 0 is not

the element of given set. 43. Since, f(x) is quadratic function

f(x) = ax2 + bx + c

f(2) = 4a + 2b + c = –3 …(i) and f(–2) = 4a – 2b + c = 21 …(ii) on subtracting Eq. (ii) from Eq. (i), we get 4b = –24 b = –6 |b| = 6

44. (A + B)’s 1 day’s work 1

15=

(A + B + C)’s 1 day’s work 1

8=

C’s 1 day’s work

1 1 15 8 7

8 15 120 120

−= − = =

Let (B + C) can complete the work in x days.

(A + C) can complete the work in (x + 2) days.

(B + C)’s 1 day’s work 1

x=

(A + C)’s 1 day’s work 1

x 2=

+

B’s 1 day’s work 1 7 120 7x

x 120 120x

−= − =

and, A’s 1 day’s work 1 7

x 2 120= −

+

( )

( )

120 7 x 2

120 x 2

− +=

+

( )

106 7x

120 x 2

−=

+

Now, A’s 1 day’s work + B’s 1 day’s work = (A + B)’s 1 day’s work

( )

106 7x 120 7x 1

120 x 2 120x 15

− − + =

+

( )

2 2106x 7x 120x 240 7x 14x

120x x 2

− + + − −

+

1

15=

214x 212x 240− + + = 8x2 + 16x

11x2 – 98x – 120 = 0

11x2 – 110x + 12x – 120 = 0

11x (x – 10) + 12 (x – 10) = 0

12

x 10,and11

= −

But no. of days cannot be negative

x = 10

A’s 1 day’s work

1 7

10 2 120= −

+

1 7

12 120= −

10 7 3 1

120 120 40

−= = =

A alone can complete the work in 40 days.

45. 2 2

1 1 3 11 2

2 21 2+ + = = −

2 2 2 2

1 1 1 1 3 7 8 11 1 3

2 6 3 31 2 2 3+ + + + + = + = = −

2 2 2 2 2 2

1 1 1 1 1 1 3 7 13 15 11 1 1 4

2 6 12 4 41 2 2 3 3 4+ + + + + + + + = + + = = −

It clearly implies that

2 2 2 2 2 2

1 1 1 1 1 1 11 1 ......... 1 2008 K 2008

20081 2 2 3 2007 2008+ + + + + + + + + = − =

46. Shaded area = Area of big semicircle – (Area of 2 semi circles + Area of triangle)

4

O

8

4

2 28 2 4 1

8 4 32 16 162 2 2

= − + = − −

= 16( – 1) K = 16 47. Here, AD = AE = 10 Let CM = r(radius of small circle) AB = (10 – r), AC = (10 + r) and BC = 10 AC2 = AB2 + BC2

(10 + r)2 = (10 – r)2 + (10)2

100 + 20r + r2 = 100 – 20r + r2 + 100

40r = 100

r = 2.5 cm

B C

r

A E

D

10 –

r

10

10

10

M

48.

Given AB 1

BC 2=

AB = x, BC = 2x

ABC is right angle triangle

AC = 5x and BD ⊥ AC,

Let CD = a, AD = b and BD = h

a + b = 5x …. (i)

BDC is right angle triangle, h2 = 4x2 – a2

BDA is right angle triangle, h2 = x2 – b2 On subtracting, we get 3x2 = a2 – b2

A

5x

B

h

D

C

x

b

a

2x

3x2 = (a – b) (a + b)

3x2 = (a – b) 5x [By (i)]

( )3

x a b5

= − … (ii)

From (i) and (ii), 4x x b 1

a ,ba 45 5

= = =

49. Since 640640640000 is a multiple of 8 643 = 8k + 3

n = 8k + 3 for some positive integer k.

n2 = 64k2 + 48k + 9 = 1 + a multiple of 8. 50. Let R be the radius of the given cone, r the radius of the small cone, h be the height of the

frustum and h1 be the height of the small cone.

In figure, ONC and OMA are similar by AA similarity

11

hON NC r rh 40

OM MA 40 R R

= = ==

… (1)

We are given that

Volume of small cone 1

Volume of given cone 64=

2

1

2

1r h

131 64

R 403

=

2

2

r 1 r 140

40 R 64R

=

3 3r 1 1 r 1

R 64 4 R 4

= = =

D

D

40 cm

N

M

h

R

C

B A

h1

O

From (1) and (2) 1

1h 40 10 cm

4= =

Therefore, h = 40 – h1 = (40 – 10) cm h = 30 cm = 6 k k = 5

Section – D

51. Let BC = 3 and AB = 4 Let BE = a. BD = b Area of rectangle = ab

F(a, b) lies on x y

13 4+ =

a b1

3 4+ =

b a 3 a

1 b 44 3 3

− = − =

Area = ab =212 4a 12a 4a

a3 3

− − =

( )24 a 3a

3

− −

2 2

24 3 3a 3a

3 2 2

− − + −

24 3 9

a3 2 4

− − −

24 3

3 a3 2

− −

Where area is maximum when

23

a2

−

is least i.e.,

23 3

a 0 a2 2

− = =

Maximum value of area = 3

B C

A

E

b

x

F a

a

b

D

52. Here ABC circumscribe the circle with centre O. Also, radius = 4 cm

The sides BC, CA and AB touch the circle at D, E and F respectively.

BF = BD = 6 cm CD = CE = 8 cm AF = AE = x cm (say)

The sides of the triangle are: 14 cm, (x + 6) cm and (x + 8) cm

Perimeter of ABC = [14 + (x + 6) + (x + 8)] cm = [14 + 6 + 8 + 2x] cm = 28 + 2x cm

Semi perimeter of ABC

( )1

s 28 2x cm 14 x cm2

= + = + ,

where a = BC, b = AC, c = AB

s – b = (14 + x) – (8 + x) = 6 s – a = (14 + x) – (14) = x s – c = (14 +x) – (6 + x) = 8

Area of ABC = ( ) ( ) ( ) ( )14 x 6 x 8+ cm2

B C

E

A

O

D 6 cm 8 cm

F

6 cm 8 cm

x cm x cm

= ( )14 x 48x+ cm2 … (1)

Now, ( )1

ar OBC BC OD2

=

1

14 4 282 =

( ) ( ) ( ) 21 1ar OCA CA OE x 8 4 2x 16 cm

2 2 = = + = +

( ) ( ) ( ) 21 1ar OAB AB OF x 6 4 2x 12 cm

2 2 = = + = +

ar(ABC) = ar(OBC) + ar(OCA) + ar(OAB) = 28 cm2 + (2x + 12) cm2 + (2x + 16) cm2 = (28 + 12 + 16) + 4x cm2 = (56 + 4x) cm2 … (2) From (1) and (2), we have

56 + 4x = ( ) ( )14 x 48x 4 14 x 4 14 x 3x+ + = +

( )14 x 14 x 3x+ = +

Squaring both sides (14 +x)2 = (14 + x) 3x

196 + x2 + 28x = 42x + 3x2

2x2 + 14x – 196 = 0 x2 + 7x – 98 = 0

(x – 7) (x + 14) = 0

x – 7 = 0 x = 7 or x + 14 = 0 x = (–14)

But x = (–14) is rejected x = 7 cm Thus, AB = 6 + 7 = 13 BC = 8 + 6 = 14 cm CA = 8 + 7 = 15 cm

= 1

16(AB + AC) ( )

115 13 1.75

16= + =

53. Let A and B be the two cars. A starts from P1 with constant speed of x km/hr and B starts

from P2 with constant speed y km/hr. CaseI: When the two cars move in same directions as shown in figure, the cars meet at the

position Q.

A → B →

Q ×

P2 P1 250 km Here, P1Q = 5x km i.e., the distance travelled by car A in 5 hours with x km/hr. P2Q = 5y km i.e., the distance travelled by car B in 5 hours with y km/hr speed. We have, P1Q – P2Q = 250

5x – 5y = 250 x – y = 50 …(1) Case II: When two cars move in opposite directions as shown in figure, the cars meet at the

position R.

A →

P2

× R P1

250 km B

Here, 1

25PR x km

13= and 2

25P R y km

13=

So, P1R + P2R = 25 25

250 x y 25013 13

+ =

x + y = 130 …(2) Adding (1) and (2), we get

2x = 180 x = 90 Subtracting (1) from (2), we get

2y = 80 y = 40

Their speeds are 90 km/hr and 40 km/hr.

( )1 90 40 130

of sum of their speeds 16.258 8 8

+= = =

54. Let AB = AC = 40cm, BC = 48 cm

Altitude AL is median of BC in isosceles triangle.

BL = LC = 24 Let LM = R, OP = r and AZ = x

From ABL AL2 = AB2 – BL2 = 402 – 242 = 322

AL = 32

now, = R S

A

B C

S

N

P

Q O

L

M

Z

( )( )1 AB BC CA

BC AL ML2 2

+ + =

( ) ( )1 128

48 32 R2 2

= R = 12

BL = BN ( lengths of tangents drawn from external point are equal)

AN = 40 – 24 = 16

From ANM, AM2 = AN2 + R2 AM = 20

R + 2r + x = 20

2r + x = 8 …..(i)

Since, ASO, ANM are similar.

r r x

R R 2r x

+=

+ +

r r x

12 20

+=

20r = 12r + 12x

2r = 3x On putting the value of x in Eq. (i), we get

2r

2r 83

+ =

r = 3 55. As per the given information, we figure out that ABCD is a cyclic quadrilateral.

Area of ( ) ( )1 1

BPD BP AD 2x 9 9x2 2

= = =

Similarly, area of ( )1 13x

BQD x 132 2

= =

A C

D

Q 9

P

2x

13

B

Area of (BPD + BQD) = Area of quadrilateral PBQD 150

13x

9x 1502

+

31x 300

x 9 (since, x is an integer)

56. In ABC, the radii OE and OF are perpendicular on the respective sides AC and AB.

D B C

O F E

A

Now, the radius of ABC is 175 cm and AO = BO = CO = 625 cm.

Now, we can see that AEO, CEO, BFO and AFO are right angled congruent triangles. Therefore, using Pythagoras theorem, we get AE = 600 and CE = 600

As, ADC AEO

AC AO 600 600 625

AD AE AD 600

+ = =

( )( )1200 600

AD625

=

Similarly, we have

AC AO

CD EO=

600 600 625

CD 175

+ =

( )( )1200 175CD

625 =

( )1

Area of ABC AD BC2

=

( )1

AD 2CD AD CD2

= =

D C

y° E

x°

A

O

y°

( )( ) ( )( ) 21200 600 1200 175

387072 cm625 625

= =

57. Distance traversed till the 1st rebound = 32 = 25 m

Distance traversed between 1st rebound and 2nd rebound 5 512 2 2

2= =

Distance traversed between 2nd rebound and 3rd rebound 5 412 2

2= =

Distance traversed between 3rd rebound and 4th rebound 4 312 2

2= =

Total distance covered till the 11th rebound

= 25 + 25 + 24 + 23 + … + 2−3 + 2−4

= 25 + (25 + 24 + 23 + … + 2−3 + 2−4)

= 25 + (25 + 24 + 23 + 22 + … + 2 + 1) + 1 1 1 1

2 4 8 16

+ + +

= 32 + 63 + 1 = 96 58. Day Number of students

1 2 3 4

20

1 3 6

10

210

20 21 22

1 3 6 10 ..... 210 70 226

+ + + + + = =

From the above equation, we find that the number of student-days and man-days = 22 × 70. If MBA students are twice as efficient as Engineering students, the number of days taken by

11 students = 70. 59. Let the radius of largest sphere be AO = BO = r1

A

S

O B R P

C

D

radii of the semicircles with centres P and R = AP = PO

1rDR OR RB2

= = =

Let r2 be the radius of circle with centre S, then CS = SD = r2 Again OS = CO – CS = r1 – r2

and 12

rSR SD DR r

2= + = +

(SR)2 = (OS)2 + (OR)2

( )2 2

21 12 1 2

r rr r r

2 2

+ = − +

r1 = 3r2

Total area of semicircle with diameter 21

1AB r

2=

Total area of two semicircles with diameters AO and BO

2

1r12

2 2

=

and the area of circle with centre

22 12

rS r

3

= =

Total area that can be grazed by horses 2 2

21 11

r r 13r

4 9 36

+ =

Ungrazed area 2

2 211 1

r 13 5r r

2 36 36

= − =

Percentage area that cannot be grazed

21

21

5r

36 100 13.8%r

= =

60. AO = OB = OM = ON = 1.5 cm AE : EB = 1 : 2

A

D F

B

N G

C M

O

L H

E

AE = 1 cm, EO = 0.5 cm Similarly, NL = 1 cm and OL = 0.5 cm

( ) ( )2 2

DL DO OL = −

( ) ( )2 2

DL 1.5 0.5 = −

DL 2 cm=

( )1

DH DL HL 2 HL 0.5 cm2

= − = − =

2 2 1

DH K 22

− = =

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