International Maths Olympiad Practice Book

8/12/2019 International Maths Olympiad Practice Book

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INSTANT

WORK BOOK

International Mathematics Olympiad 7


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Class 7 3

1. (C) : Positive integers are greater than negative

integers.2. (D) : 179 is a prime number.3. (C) : 4 9 = 36

3 12 = 3612

72 36 36 =

1 36 = 364. (D)

5. (D)

6. (D) : (5) (4) (3) (2) (1) 0 + 0 (1) (2) (3) (4) (5) = 0 + 0 = 0

7. (D)

8. (B) : 10 + (3) + 4 = 10 3 + 4 = 9\ 1 + A + (6) = 9 A = 9 1 + 6 = 4and 0 + (2) + B = 9 B = 9 + 2 = 7

9. (C) : |7| + |5| |7| |3| = 7 + 5 (7) (3) = 7 + 5 7 3 = 2

10. (B) : Distance covered in rst 1 sec. = +5 cm. Distance covered in next 1 sec. = 2 cm.\ Distance covered in 2 sec. = 5 2 = 3 cm.

\ Time taken to cover 57 cm. =23

57

sec.

= 38 sec.

\ Time taken to cover 60 cm. = (38 + 1) sec.

= 39 sec.11. (A) : Capacity of tank = 500 litres.

Rate at which water is owing out = 9 litres every hour. Amount of water owing out in 1 hr. = 9 litres. Amount of water owing out in 10 hrs = 90 litres.\ Amount of water left after 10 hrs. = (500 90) litres = 410 litres.

12. (D) : LCM of 7, 3, 3, 9 = 63

\

47

99

3663

13

2121

2163

23

2121

4263

59

77

3563

= = = =, , ,

The ascending order is,2163

3563

3663

4263

, , ,

i.e.,1

3

5

9

4

7

2

3, , ,

Now, the average of59

and 47

is 12

59

47

+

=12

35 3663

71126

+ =

Chapter-1 : NUMBER SYSTEM

13. (C) : 0 + 0 = 0

14. (D)

15. (D) : = = 3

8 243

833 24

X X

= 924 24

X

X = 9

16. (C) : ( ) ( ) ( )( ) ( ) ( )4 9 252 3 5

= 2 3 5 = 3017. (C)

18. (A)

19. (B) : No. of plants in 1 row = Total no of plantsNo of rows

..

=63021

30=

20. (C) : Sum of even numbers between 10 and 20 = 12 + 14 + 16 + 18 = 60 and sum of odd numbers between 10 and 20 = 11 + 13 + 15 + 17 + 19 = 75\ Difference = 75 60 = 15

21. (D)

22. (D) : 19 (4 + (2)) = 19 4 + (19) (2)Associative Property.

23. (A)

24. (D) : Let the other number be x .

Then, x

= 4

39

16

x =

916

34

x =2764

25. (B)

26. (B)

27. (A)

28. (A) : 199 is an odd number. Therefore, the product havenegative sign.

29. (A)

30. (A) : 222 1

342 56 8 9 108 + +( ){ }+

= 222 1

342 56 17 108 + { } +

= 222 27 108 222 1 35 +[ ] = = 87


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IMO Work Book Solutions4

1. (B) : 34

52

7

104

8

15

19

5

27

10

68

15+ + = + +

LCM of 5, 10, 15 = 30

\

195

2710

6815

11430

8130

13630

33130

+ + = + + =

2. (B) : 1 210

1 1

51

15

65

+ = + = =

3. (B) : Required fraction =38

4. (A) : The rational number lying between17

and18

= 12

17

18

+

=12

1556

15112

=\ = 112

5. (C) : Total no. of boxes = 25 No. of shaded boxes = 9

\ Fraction =9

25

6. (D) : Total no. of squares = 18 Let, total no. of shaded squares be x .

According to question,

79 18

= x

x = 14Out of these 14 squares 6 are shaded. So, we have toshade 8 more squares.

7. (B)

8. (B) : = = = 95 20

27 45ab c

= = = 95

44

3620 20

36a

a

= = = 9

533

2715

2715

bb

= = = 2595

55

4525

45c

c

9. (C) :1

1 1

2 1

3

1

1 16 1

3

1

1 3

7+

+

=+ +

=+

=1

7 37

710+ =

Chapter-2 : FRACTIONS AND DECIMALS

10. (C)

11. (D) : 9 720

18720

9 35= = .

12. (B) : 1 1

31

14

1 1

51

16

1 1

........

n

=3 1

34 1

45 1

56 1

61

........ n

n

=23

34

45

56

1 2 ( )

=........ nn n

13. (C) : Total no. of animals = 192. Fraction of cattle =

716

\ Number of cattles =7

16192 12 7 84 = =

Out of these 84 cattles23

are dairy cows.

\ No. of dairy cows =23

84 56 =14. (B)

15. (B) : Let, Toms have ` x

Then, Jasmines have`

(41 x ).Now,

14

2 1

741of x x = + ( )

14

2 41

7 7 x

x = +

14

17

14 417

x x + = +

7 428

557

+

= x

x = ` 20\ Tom have ` 20

16. (C) : x % of 24 = 64

or x 100

24 64 =

x =64 100

24800

3 =

x = 26623

17. (B) : 0.00639 0.213

=0 00639

0 213100000100000

..

=639

213000 03= .


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Class 7 5

18. (A) : Total time = 5 14

214

hrs = hrs.

Time for English and Mathematics = 234

114

hrs hrs= .

\ Time for other subjects =214

114

104

=

=52

212

hrs hrs= .19. (D)20. (B)

21. (C) : Total runs = 321 No. of wickets = 15

\ Average score per wicket =32115

21 4= .

22. (D) :

+

+234

134

234

134

......... upto times30

= 30 234

134

= 30 11

474

+

= 30 4

430 =

23. (C) : 1 1

21

13

1 1

41

110

........

=12

23

34

89

910

........ = 110

24. (A) : Total length of rope = 30 m.

Length of piece to be cut = 334

154

m m= .

\ Number of pieces =30154

8=

25. (C) : 0.231 = 0 231 1000

1000231

1000. =

26. (B) : 2 3 2 3 2 3 11 1 1 1 ( ) = ( )

= 2 3 1 2 3 5 1

51 1 1 ( )[ ] = +( ) = ( ) =

27. (A) :6

1638

=

28. (D) : 913

534

283

234

= =112 69

12

=4312

3 712

=

29. (B) : 1 1

1 1

1 1

6

1 1

1 16 1

6

+ +

= + +

= 1 1

1 6

5

1 15 6

5

1 511

++

= + + = + =11 5

111611

+ =

30. (D) : 452

238

378

132

198

318

+ = +

=52 19 31

8648

8 + = =

vvv


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1. (C) : Let the one part be x .

Then, the other part will be 184 x .Now,

13

17

184 8 = ( ) + x x

x x 3

1847 7

8= +

x x 3 7

1847

8+ = +

or 1021

2407

x =

x =240 21

7 10

x = 72\ First part = 72

Second part = 184 72 = 112

2. (D) :2

1 1

11

1 x

x x

++

=

2

1 111

1 2

1 11

x

x x x

x

x + +

= +

=

2

21

x x =

2x = 2 x

3x = 2 x =23

3. (A) : Let the numbers be x , x + 1, x + 2 and x + 3.We have, x + (x + 1) + ( x + 2) + ( x + 3) = 70 4x + 6 = 70 4x = 64 x = 16\ Greatest integer = x + 3 = 16 + 3 = 19

4. (C) : Let each nephew receives = ` x Then, each daughter receives = ` (4x ) and each son receives = ` (5x )Now, 2 x + 4 4 x + 5 5 x = ` 8600or 2 x + 16 x + 25 x = ` 8600 43 x = ` 8600 x = ` 200\ Each daughter receives ` (4 200) = ` 800

5. (A) : Let the number be x . Then, acc. to question,5 4 20

810

x +( ) =

or 5( x + 4) 20 = 80

Chapter-3 : SIMPLE EQUATIONS

5x + 20 20 = 80

5x = 80 x = 16

6. (B) : Let the marks in English be x . Then, the marks in Mathematics = 240 x .Now, we are given that,

13

240 1

230( ) = + x x

803 2

30 = + x x

x x

3 280 30+ =

56

50 x =

x = 60\ Marks in English = 60.

7. (A)

8. (D) : Fraction of women workers =13

So, fraction of men workers =23

Now,12

13of are married women =

16 are married

13

16

of

women have children =1

18 have children.

\

Women having no children =13

118

518

= ...(i)

Also, 34

23

of

are married men =12 are marrie

23

12

of

men have children =13

have children.

\ Men having no children =23

13

13 = ...(ii)

\ Total no. of workers having no children =13

518

+

[from (i) & (ii)]

=1118

9. (B) : Let the number of working days be x Then, the number of days he is idle = 60 xAccording to question, 20 x 3(60 x ) = 280

or 20 x 180 + 3 x = 280 23 x = 460 x = 20.\ Working days = 20 and Idle days = 40


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Class 7 7

10. (B) : Let the number of notes of ` 100 denomination be x .Then, number of notes of ` 50 denomination = 85 x Now, we are given that, 100 x + 50 (85 x ) = 5000or 100x + 4250 50 x = 5000 50 x = 750 x = 15\ Number of notes of ` 50 denomination = 85 15 = 70\ Amount required = 70 50

= ` 350011. (D) : Let the units digit be x . Then,

tens digit = 2x\ The number = 20 x + x = 21 x .Now, the number after interchanging the digits = 10 x + 2 x = 12 x According to question,

12x = 21 x 36or, 36 = 21 x 12 x 9x = 36 x = 4\ The original number = 84.

12. (C) : p {4 (2 8 4)} = 8 p { 4 (2 2)} = 8 p (4 0} = 8 p + 4 = 8 p = 4

13. (A) : Let the amount be ` x . Total number of boys = 14

\ Share of 1 boy = ` x 14

Also, when total number of boys = 18

Share of 1 boy = ` x 18


x x 14 18

80= +

x x 14 18

80 =

9 7126

80x = x

x = ` 5040

14. (C) : 25

5 1 3

51 x +( ) + =

2 2

535

1 x + + =

2

55 1 x + =

2x + 1 = 1 2x = 0 x = 0

15. (C)

16. (B) : Let the number be x. Then,

23

12

17

x + x + x + x = 37

or, 28 21 642

37 x x x x + + =+ 42

or, 97 x = 37 42

x =1554

97

x = 16 297

17. (C) : Let rst angle be x Then, other will be (44 + x )Now,

x + (44 + x ) = 180or, 2 x = 136

x = 68\ First angle = 68 and second angle = (68 + 44) = 112.

18. (C) : 34

7 1 2 1

232

x ( )

= +x x

x

214

34

4 12

32

x x x

x

+

= +

214

34

5 12

32

x x

x = +

21

4

5

2

3

2

3

4

1

2 x x + x =

21 10 4

4 x x x

=6 3 2

4+

or, 7 x = 7 x = 1

19. (A) : Number of legs of 50 hens = 2 50 = 100 Number of heads of 50 hens = 1 50 = 50 Number of legs of 45 goats = 4 45 = 180 Number of heads of 45 goats = 1 45 = 45And, Number of legs of 8 camels = 8 4 = 32 Number of heads of 8 camels = 8 1 = 8 Let, the number of keepers be x .Then, Total number of feet = 224 + Total no. of heads 100 + 180 + 32 + 2 x = 224 + 50 + 45 + 8 + x or, 2 x x = 224 + 103 312 x = 15\ Number of keepers = 15

20. (A) : From the options we see that the given conditionis satis ed by 1 st option. Hence, the answer is 3, 4.

21. (D) : Let, the ages of A and B be 5 x and 3 x . After 6years, their ages will be 5 x + 6 and 3 x + 6.According to question,

5 63 6

75

x x

++ =


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or, 5 (5 x + 6) = 7 (3 x + 6) 25 x + 30 = 21 x + 42 25 x 21 x = 42 30 4x = 12 x = 3\ The present age of A = 5 3 = 15 years and the present age of B = 3 3 = 9 years

Also, the sum of their present ages = (15 + 9) years = 24 years.

22. (C) : The cost of rst 1000 copies = 1000 x and the cost of after 1000 copies = ( z 1000) y .\ The total cost = 1000 x + (z 1000) y

= 1000 ( x y ) + zy .23. (A) : Let, the required number be x .

Then, the boy had to multiply with 25, i.e., 25 x = 25 x \ The correct answer = 25 x ...(1)But, he multiply with 52 then, the answer becomes 52 x .According to question, 52 x = 324 + 25 x or, 52 x 25 x = 324 27 x = 324 x = 12

24. (C)

25. (D) : Let the maximum marks be x .According to question, 35% of x = 80 + 60

or35

100 x = 140

x =140 100

35

x = 400

26. (C) : Average = 13

i.e.,11 12 13 14

513+ + + + = x

50 + x = 13 5 x = 15

27. (D) : Let C gets be ` x .Then, D gets = ` x B gets = ` (125 + x )and A gets = ` (125 + x + x ) = ` (125 + 2 x )Now, 125 + 2 x + 125 + x + x + x = 750or, 5 x + 250 = 750 5x = 500 x = ` 100\ As share = ` (125 + 2 100) = ` (125 + 200) = ` 325

28. (C) : Let the value of rst prize be ` x

Then, the value of second prize = `34

x

and the value of third prize = `12

34

x = `38

x

Now,

x x + + =34

38 2550x

or,8 6 3

8 x x x + +

= 2550

17 x = 2550 8 x = 1200\ The value of rst prize = ` 1200

29. (C) : Let the share of each member from rest of themembers be x .Then, Fathers share = 3 share of each member

14

= 3x

x =1

12

Also, if father took 14

of the cake, 34

of the cake wasleft.Now,

112

part taken by 1 member

34

part taken by 12 34

members

= 9 members

\ Total no. of members = (9 + 1) = 10.

30. (A) : Let the present age of B be x years.Then, the present age of A = 2 x years.After 30 years, their ages will be ( x + 30) and (2 + 30)According to question,

112 30 2 30 x x +( ) = +( )

or32

30 2 30 x x +( ) = +( )

32

2 30 45 x x = x = 30

\ As present age = 60 years and Bs age = 30years.

vvv


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Class 7 9

1. (A) : Since PQ || ST

\ PQS = QST (Alternate angles) QST = 98 QSA + AST = 98 AST = 70Now, AB is a straight line.\ AST + TSB = 180 (linear pair) TSB = 110

2. (C) : EAC is a straight line.\ EAD + DAC = 180 (linear pair) DAC = 98

Also, AC = AD\ ADC = ACDNow, In D ADC, DAC + ADC + ACD = 180 2 ACD = 180 98 = 82 ACD = 41Now, AB || CD\ ACD = CAB (Alternate angles) CAB = 41In D ACB,

y + BCA + CAB = 180 y = 75

3. (A) : a + b = 180 (Interior consecutive angles)4. (A)

5. (C) : In Rhombus ABCD, D + A = 180 A = 180 120 A = 60 = C (Opp. angles are equal)Now, CA bisects C as we know the digonals of arhombus bisect the angles.\ DCA = 30Also, in square CGEF. GCA = 90 y + DCA = 90 y = 60

6. (C) : Let angles be 7 x and 11 x . Then, 7x + 11 x = 180 18 x = 180 x = 10\ The angles are 70 and 110.

7. (B) : As the sum of all the angles around a point is360.

Chapter-4 : PRACTICAL GEOMETRY & LINESAND ANGLES

8. (B) : Draw a line EG

passing through Fand parallel to AB.

Now, AB || CDand AB || EG EG || CD.Let, CFE = 1 and EFA = 2.Since, CD || EG\ 1 = FCD (Alternate angles) 1 = 58 ...(i)Also, FGBA is a ||gm.\ FAB + DBA = 180 (interior consecutive angles) FAB = 68and 2 = FAB (Alternate angles) 2 = 68 ...(ii)Adding (i) and (ii) we get, 1 + 2 = 58 + 68 CFA = 126

9. (A) : It is given that, AC = 13 cm.Let, BC = 12 cm.Then, by Pythagoras theorem,

we have, AB2 + BC 2 = AC2

AB2 + (12) 2 = (13) 2

AB2 = 169 144 = 25 AB = 5 cm.

10. (B) : In DAEB EAB + B + AEB = 180 EAB + 90 + 74 = 180 EAB = 16 ...(i)Now, DAB = 90 FAB = 90 20 = 70 ...(ii)From eqn. (i) and (ii) FAE + EAB = 70 FAE = 70 16 = 54Now, In DFAG,

x + FAG + FGA = 180 x + 2 54 = 180 x = 72Now, AGE is a straight line.\ AGF + FGE = 180 FGE = 126 GEB + y = 126 y = 52\ x + y = 72 + 52 = 124


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11. (D) : Let the angle be x .Then, supplement of x = 180 x

Now, we are given that, x = 2(180 x ) 45 x + 2 x = 360 45 3x = 315 x = 105\ The angle is 105.

12. (D) : Since GHCI || EABF.\ HCB = CBF (Alternate angles) HCB = 66Since, HC || AB and AH || CB.\ ABCH is a ||gm.\ HCB + AHC = 180 (Interior consecutive angles) AHC = 114But, x = AHC (Vertically opp. angles) x = 114

Also, x = KDH (Alternate angles) 114 = 48 + y y = 66 \ x y = 114 66 = 48

13. (B)

14. (B)

15. (A) :

16. (B) : Since 3 + 4 = 7 < 8.17. (B) : FHD = CHE (Vertically opp. angles)

CHE = 50Now, AHB is a straight line.\ AHC + CHE + EHG + GHB = 180 72 + 50 + 42 + x = 180 x = 16

18. (D) : Vertically opp. angles.19. (B)

20. (B) : 45 is the only angle which is its own complementaryangle.

21. (D)

22. (C) : ABCD is a rhombus.\ AB = AD. ADB = ABD (angle opp. to equal sides are equal) ABD = 27Now, ABE is a straight line.\ ABD + DBE = 180 DBE = 153

23. (A) : ABE = CBF (Vertically opp. angles) CBF = 48

Also, FCB = 48 FC FB=[ ]

Now, ABE = BCG (corresponding angles)\ BCG = 48As we know that, x + BCG + FCB = 360 x = 360 96 x = 264

24. (D) : EBC = ECB EC EB=[ ] ECB = 62In D EBC, ECB + EBC + CEB = 180 CEB = 56 CEB = DEA (vertically opp. angles)\ DEA = 56Now, In D DEA 38 + 56 + y = 180 y = 86

25. (D) : Since D BFC is an equilateral triangle.\ FBC = 60But, ABC = 90 (angle of a square) ABF + FBC = 90 ABF = 30In D AFB, BAF = 60But, BAD = 90 BAF + FAD = 90 FAD = 30And EAF = 2 FAD = 2 30 = 60

26. (A) : Let the interior angles be 2 x and 5 x .Then, 2x + 5 x = 70 (Exterior angle property) 7x = 70 x = 10\ The angles are 20, 50, 110

27. (B)

28. (D)

29. (B) : FG || CBH\ FGC = GCB (Alternate angles) GCB = 70And GCA is a straight line.\ GCB + BCA = 180 BCA = 110In D ABC, 110 + CBA + CAB = 180 2 CBA = 70 CB CA=[ ] CBA = 35 = CAB.

CBH is a straight line.\ y = 180 35 = 145Also, y = x + CAB = x + 35 (alternate angles) x = 110


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Class 7 11

30. (D) : It is given that, XZ || UV and ST || XY.Now, UVY = XZY (corresponding angles) XZY = 72

In D SZT, 53 + 72 + STZ = 180 STZ = 55Also, STZ = XYZ (corresponding angles) XYZ = 55

vvv


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1. (C) : 13 2 = 12 2 + 5 2

This triangle satis es Pythagoras theorem.\ It is a right angle triangle.

2. (A) : It is given that, AB + BC = 10 cm ...(i) BC + CA = 12 cm ...(ii) CA + AB = 16 cm ...(iii)Adding (i), (ii) and (iii); we get, 2(AB + BC + CA) = 10 + 12 + 16 AB + BC + CA = 19 cm.

3. (C) : In D ADB and D ADC

AB = AC (given) ABD = ACD

BD = DC (given)\ D ADB @ D ADC (By SAS)\ ADB = ADC (by CPCT)Now, BC is a straight line.\ ADB + ADC = 180 2ADC = 180 ADC = 90

4. (A) : In D CED,

CE = ED\ EDC = ECD ECD = 28

Also, ECD = BCA (vertically opp. angles) BCA = 28In D BCA, 62 + 28 + BAC = 180 BAC = 90Now, BAF is a straight line.\

BAC +

CAF = 180 y = 90

5. (C)

6. (A) : Since ABC is an equilateral triangle.\ A = B = C = 60And, DBA = DAB = (60 x ) DA=DB[ ]In D DAB, DBA + DAB + ADB = 180 2(60 x ) + 88 = 180 2(60 x ) = 92

60 x = 46 x = 14

Chapter-5 : THE TRIANGLE AND ITSPROPERTIES & CONGURENCE OF TRIANGLES

7. (C) :

A

B CD

8. (D)

9. (A)

10. (B) : Let the third side of triangle be 3 cm.Then, 3 + 3 = 6 < 8So, it is not possible because we know that the sumof two sides will be greater than the third side of thetriangle. So, the third side is 8 cm.

11. (B) : Let the angles are x , 2x and x .Then, x + 2 x + x = 180 4x = 180 x = 45\ The greatest angle = 2 45 = 90

12. (B) : It is given that, AB = BC and AC 2 = 100 cm 2

By using pythagoras theorem,AC2 = AB2 + BC2

100 = 2AB 2

AB2 = 50 AB = 5 2 cm

i.e., AB = BC = 5 2 cm

13. (D) :

A

B CD

14. (D)

15. (C) : In D AEB, A = DAE + BAD A = 60 + 90 A = 150And, AE = AB ABE = AEB\ A + ABE + AEB = 180 2AEB = 30 AEB = 15Now, E = 60 DEF = 60 15 = 45


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Class 7 13

\ In D EFD, DEF + EDF + EFD = 180 45 + 60 + y = 180 y = 75

16. (C) : Let AB and CD be the two chimneys.Let the distance between their tops be x m.

Then, AB = 18 m AE + EB = 18 m EB CD=[ ] AE = 5 mIn D AED, AE2 + ED 2 = AD2 (By Pythagoras theorem)or, x 2 = (5) 2 + (12) 2 = (25 + 144) cm 2 = 169 cm 2

x = 13 cm.17. (A)

18. (C) : We have,

Area of D ABC =12

hx ...(i)

(where h is the height of the altitude)And, Area of square CDEF = x x ...(ii)From (i) and (ii)

x x =12

h x h = 2 x

19. (A) : Since PQ = PR PQR = PRQ ...(i)Now, let P be x .Then, Q = 2 x (given)and R = 2 x {from (i)}Now,

P + Q + R = 180 x + 2 x + 2x = 180 5x = 180 x = 36

\ Q = 2 36 = 7220. (A) : In D FGC, GCF = 92 (given)

As we know, CGF = 60 (angle of equilateral triangle)\ x + 60 + 92 = 180

x = 28Now, In D BCF, CBF = 60

FCB = 180 92 (linear pair) FCB = 88\ BFC + 88 + 60 = 180 BFC = 32And, CFE = 90 y + 32 = 90 y = 58 \ y 2 x = 58 2 28 = 58 56 = 2

21. (B) : In DADC and D CBAAD = BC (given)DAC = BCA [ AD || BC]AC = AC (common)\ DADC @ DCBA (By SAS)\ DC = AB (By CPCT)

22. (B) : We know that, IFH = 60 (angle of equilateral triangle)and DCB = 72 (opposite angles of a rhombus are equal)Now, DCB = GCE = 72 (vertically opp. angles)Also, IFH = GFC = 60 (vertically opp. angles)and, GFC = FCE = 60 (Alternate angles) y = 60Now, In D GCF, 60 + 72 + x = 180 x = 48

23. (A) : FCA = BFD (corresponding angles) x = 51and CAB = AOD (Alternate angles) AOD = 83Also, FOB = AOD (Vertically opp. angles) FOB = 83Now, In D FOB y = 51 + 83 (Exterior angle property) y = 134

x + y = 51 + 134 = 18524. (B)

25. (D) : By RHS.26. (C) : It is given that, D ABC is a right angled triangle at

A.C

AB

D

Now, In D ABD AD2 + BA2 = BD2 ...(i) [Pythagoras theorem]And In D ABC AC2 + AB2 = BC 2 ...(ii) [Pythagoras theorem]Subtracting eqn. (i) from (ii) we get, BC2 BD 2 = AC2 AD2


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= (2AD)2 AD2

= 4AD2 AD2

= 3AD2

27. (A) : Since, AB = AC

12

12

AB AC=

BF = ECAlso, AB = AC B = CIn D BEC and D CFB EC = FB (proved above) B = C (proved above) BC = BC (common)\ D BEC @ D CFB (by SAS) BE = CF (by CPCT)

28. (C)

29. (B) : We have,82 + 6 2 = 100 = 10 2

So, it is a right angled triangle.

\ Area of D ABC =12

AB BC

=12

8 6 24 2 = cm

30. (A) : Draw AD perpendicularbisector to BC.Then, BD = DC = 5 cm.

In D ABD

AB2 BD2 = AD2

AD2 = 100 25 = 75 cm 2

AD = 5 3 cm

Now, Area of D ABC =12

10 5 3 25 3 2 = cm

vvv


15/32

Class 7 15

Chapter-6 : COMPARING QUANTITIES

1. (C) : C.P. of 11 oranges = ` 10

\ C.P. of 1 orange = ` 1011

Also, S.P. of 10 oranges = ` 11

\ S.P. of 1 orange = ` 1110

Now, Pro t = S.P. C.P. = = `1110

1011

2110

\ Pro t % =

211101011

100 21 = %

2. (C) : Let the total no. of votes be x .Then, according to question, 30% of x + 15000 = 70% of x

30

10015000

70100

x x + =

or 15000 =70 30

100

x

x =15000 100

40

= 37500

\ Total no. of votes = 37500

Number of votes of winning candidate = 70% of 37500

=70

10037500 26250 =

3. (C) : Let the salary in 1999 = ` x Then, salary in 2000 = ` [x + 20% of x ]

= ` 65

x

Also, salary in 2001 = ` 65

20 6

5 x x + % of = ` 36

25 x

But, salary in 2001 = ` 26640

\ 3625

x = 26640`

x = ` 26640 25

36

= ` 18500\ Salary in 1999 = ` 18500

4. (C) : Let the number of coins of 20 p be x Then, the number of coins of 10 p = (36 x )According to question, 20 x + (36 x ) 10 = 6.60 100or 20 x + 360 10 x = 660

10 x = 300 x = 30

5. (A) : Let the cost price be ` x . Then,

gain % =gainCP. .

100

or, C.P. =

100

100

+

S P

gain

. .

% =

100 5

100 20

+

=500120

4 16= .Now, cost price of 10 eggs = 4.16i.e., Number of eggs in ` 4.16 = 10

Number of eggs in ` 1 =10

4 16.

\ Number of eggs in ` 5 =10

4 165 12

. =

6. (A) : Let the numbers be x and 2 x .


x x

++ =

72 7

35

5(x + 7) = 3(2 x + 7) 5x + 35 = 6 x + 21 x = 14\ The greater number = 2 14 = 28.

7. (B) : In 60 days 210 men complete 1 work

In 1 day they complete1

60

work

In 12 days they complete 160

12 15

= work

Now, 45

work is left for (210 + 70) workers = 280 workersLet the no. of days be x .

Then, 280 x =45

210 60

x =4 210 12

280

= 36 days.

8. (C) : Let the third number be 100.

Then, rst number = 100 20% of 100 = 80and second number = 75.\ Difference of rst and second = 5.

\ Required % =5

80100 6

14

= %

9. (A) : Let the number be x .According to question, x 4 = 80% of x

x 4 =80

100 x = 4

5 x

or, x x 45 = 4

x 5

4= x = 20


16/32


10. (B) : Selling price of 1 horse = ` 4000 Pro t % = 25.

\ C.P. =100

100100 4000

1253200

+ = =SP

ofit. .

Pr %`

Since the man having neither loss nor gain.\ Cost price of second horse = S.P. of 2 horses C.P. of 1 horse

= ` (8000 3200) = ` 4800Now, Cost price of 2 nd horse = ` 4800 Selling price = ` 4000\ Loss = ` 800

Loss % =800

4800100 16

23

= %

11. (D) : If selling price = ` 144Let the cost price of jug be ` x .

Then, Loss =1

7 7of x

x =

Now, Loss = C.P. S.P

x

x 7

144=

or 144 =67 x

x = ` 168.Since cost price of jug = ` 168.Now, selling price = ` 189\ Pro t = ` (189 168) = ` 21

\ Pro t% =

21

168 100 12 5 = . %12. (A) : Let cost price of 25 articles be ` 100

\ Cost price of 1 article = ` (100 25) = ` 4\ Cost price of 20 articles = ` 80.And, selling price of 20 articles = cost price of 25 articles\ Selling price of 20 articles = ` 100.Now, Pro t = ` (100 80) = ` 20

And, Pro t % =2080

100 = 25%

13. (B) : Cost price of 70 kg rice = ` 175 Selling price of 1 kg rice = ` 2.75\ Selling price of 70 kg rice = ` (2.75 70) = ` 192.50\ Pro t = ` (192.50 175) = ` 17.50

Pro t % =17 50175

100 10.

% =

14. (C) : In 9 days P can complete 1 work

\ In 1 day P can complete19

work.

Now, Q is 50% more ef cient than P\ In 1 day Q can complete

=19

50 1

9+ % of

=19

118

318

16

+ = =\ No. of days in which Q can complete the work

=116

= 6 days.

15. (B)

16. (C) : Let the selling price of shirts be ` x .

Pro t = 1212

252

% %=

\ Cost price =100

100 25

2

200225

+

= x x = ` 8

9 x

\ Pro t = S.P. C.P.

= x x x

89

19

= `

Now, cost price of pants = Selling price of shirts = ` x

\ Selling price = x 100 20100+( ) = 120100 65 = x x

`

Pro t =65

15

x x x =

`

Now, 15

19

700 x x + =

or, 1445

700 x =

x = 2250.\ Cost price of shirts = ` `8

92250 2000

=

17. (B) : 1 day = 24 hrs. 1 hr = 60 min.\ 1 day = (60 24) min = 1440 min.

Required % =36

1440100 2 5 = . %

18. (D) : Selling price of article = ` 450 Loss % = 20.

\ Cost price =100

100

( )S P

Loss. .

%

=100 450

100 20

100 450

80

562 50

( ) =

= .

Now, if he makes a pro t of 20% then,

S.P. =562 50 100 20

100675

.( ) +( ) =

19. (A) : Let the two cars starts from the point P.Now, In D PST, PS 2 + ST 2 = PT 2

PT 2 = (6 2 + 8 2) = 100 PT = 10 milesSimilarly,

PR = 10 miles.\ Distance between car A and car B is

(10 + 10) miles = 20 miles


17/32

Class 7 17

20. (C) : Let the total allowance be ` x .

Then, Alok have to save ` `25

25

of x x

=

\ Required % =

25 100 40

x

x = %

21. (D) : Let the number of boys be x .

Then, the number of girls = 1427 %of x

=100

71

100 x = x

7Now, total no. of students = 560.

\ x x +

7 = 560

or,87

560 x =

x = 490

And no. of girls = (560 490) = 70.

22. (B) : Cost price of radio =` 600 Pro t % = 25

\ Selling price =600 100 25

100 +( )

= ` 750SP

CP g. .

. . ( %)=

+100100

23. (B) : Total no. of people = 12000No. of people wearing same colour = 4800

\ Required % =4800

12000100 40 = %

24. (B) : Let the maximum marks be x .According to question, 36% of x = 113 + 85

36

100 x =198

x =198 100

36

x = 550\ Maximum marks = 550

25. (C) : Earning of Niharika = ` 1200. Payable amount = 78% of earnings

=78

1001200 = ` 936.00

26. (B) : The present value of machine = ` 100,000The value is depreciates 5% every year.\ The value of machine after 1 year = 100000 5% of 100000 = 100000 5000 = ` 95000And the value of machine after 2 years = 95000 5% of 95000

= 95000 4750 = ` 90,250.27. (A) : Let the number be x .

Then, 200% of x = 20

200100

20 =x x = 10

Now, 50% of 10 =50

10010 5 =

28. (C) : Let the marked price be ` 100Then,

Selling price =`

(100 10% of 100) =`

90. Pro t % = 5

\ C.P. =100 90

105

= ` 85.71Now,

M.P. C.P. = ` (100 85.71) = ` 14.29

\ Required % =14 2985 71

100..

= 1623

%

29. (C) : Let the cost price be ` 100.Then, Marked price = 100 + 10% of 100 = ` 110And, selling price = 110 10% of 110 = ` 99 \ This type of deal bears loss.

30. (B) : Let the cost of milk of 1 litre be ` x .After decreasing the cost of 1 litre milk will be` (x 20% of x ) = ` 0.8 x Now,Amount of milk purchased in ` 0.8 x = 1 litre

\ Amount of milk purchased in ` x =1

0 8. x x

litres

= 1.25 litres

\ Required % =1 25 1

1100 25

.%

=

vvv


18/32


1. (A) :( )

5

2

3

8

15

1

42 2 2 x y xy z xyz z

=

( ) ( )

5 2

38

151

42 2

2

x x x y y y

z z z

( )

=49

4 4 4 x y z

2. (C) : a b c a b c a b c 2 3 3 2 3 3

2 3 3

2 3 42

33

44

5+

+ +

+ + +( )

Bring all the like terms together.

a aa

b bb

c c c

2 22

3 33

3 33

22

3 33

4 44

5+ +

+ + +

+

+

=3 4 6

64 9 12

12

5 16 2020

2 2 2 3 3 3

3 3 3

a a a b b b

c c c

+ +

+

+ +

+

+

=136

2512

120

2 3 3a b c +

3. (A)

4. (C) : 2 1

32

13

2 2

x y

x y

+

= 2 13

2 13

2 13

2 13

x y

x y

x y

x y

+ +

+ +

a b a b a b2 2 = +( ) ( ){ }

= 4 2

383

x y

x y

=

5. (A) : x x

+

=1

12

Now,

x x

x x

= +

1 14

2 2

= 144 4 = 140

\ x x

=1

140

6. (D) :45

3 5

86

2245

158

182 p p p

p p

= +

= p

p2

226740

18 +7. (D)

8. (C) : Let us bring the like terms together.

2 2

3 3 35

323

52 2

2

2 2 2

3 3 3

y y y y

y y y

y y y

+

+ + +

+

+ 44

Chapter-7 : ALGEBRAIC EXPRESSIONS

=6 2

3

5 2 3

3

5 42

4

2 2 2

3 3 3

y y y y y y y

y y y

+

+

+ +

+

+

= 2y + 49. (C) : Sum = (8 a 6 a 2 + 9) + (10 a 8 + 8 a 2)

= 2a 2 2 a + 1and Difference = 3 2 a 2 + 2 a 1 = 2 a 2 + 2 a 4

10. (B) : When x = 15 and y = 3, we have, 9 (15) 2 + 49 (3) 2 42 15 3 = 2025 + 441 1890 = 576

11. (C) : At x =ba

, we get,

a ba

b ba

c

+

+2

= a b

a

ba

c +2

2

2

=ba

ba

c c 2 2

+ = 12. (D) : At t = 4, we have,

d = 16 (4) 2 + 1000= 256 + 1000

= 74413. (D) : 3x 4 12 y 4 = 3{x 4 4 y 4}

= 3[(x 2)2 (2 y 2)2] = 3{(x 2 2 y 2) (x 2 + 2 y 2)}

a b a b a b2 2 = + ( )( )

14. (D) : 3 1

29

1

42 3

12

22

2 x x x

x x

x

= +

a b

a b ab

( ) =+

2

2 2 2

(6)2 = 9 1

432 2 x x

+

9 1

42

2 x x + = 36 + 3 = 39

15. (A) : Required expression= (3x 2 4 y 2 + 5xy + 20) ( x 2 y 2 + 6xy + 20)

= 3x 2 4 y 2 + 5xy + 20 + x 2 + y 2 6 xy 20 = 4x 2 3 y 2 xy

16. (A) : x x 8 + x 2 1.7 x 10 + 1.4 x 8 7.8 x 2 + 4 9 x = 4 + ( x 9x ) + (x 2 7.8 x 2) + (x 8 + 1.4 x 8) 1.7 x 10

= 4 8 x 6.8 x 2 + 0.4 x 8 1.7 x 1017. (C)

18. (C) : 4st (s t ) 6 s 2 (t t 2) 3 t 2 (2s 2 s ) + 2 st (s t ) = 4s 2t 4 st 2 6 s 2t + 6s 2t 2 6 t 2s 2 + 3st 2 + 2s 2t 2 st 2


19/32

Class 7 19

= (4 s 2t 6 s 2t + 2 s 2t ) + (4 st 2 + 3 st 2 2 st 2) = 3 st 2

19. (C) : (a 3 2 a 2 + 4 a 5) ( a 3 8 a + 2a 2 + 5) = a 3 2 a 2 + 4a 5 + a 3 + 8a 2 a 2 5 = 2 a 3 4 a 2 + 12 a 10

20. (D) : We have, x = 5 and y = x + 7 = 5 + 7 = 12.Now,

x y 2 2 2 25 12 25 144+ = ( ) + ( ) = +

= 169 = 1321. (B) : 2 (2) 3 2 (2) 2 + 2 a = 5

or 16 8 + 2 a = 5 10 a = 5 a = 5

22. (C) : x x x

x x

x

+

+

1 1 122

= x x

x x

22

22

1 1

+

a b a b

a b

( ) +( )= 2 2

= x x

44

1

23. (A) : Required expression

= (a 4 + 4a 2b 2 + b 4) (a 4 8 a 2b 2 + b 4)

= a 4 + 4 a 2b 2 + b 4 a 4 + 8 a 2b 2 b 4

= 12 a 2b 2 24. (C) : (2l 3 m )2 = (1) 2

4 l 2 + 9m 2 12 lm = 1 4 l 2 + 9m 2 12 20 = 1 4 l 2 + 9m 2 = 241

25. (D)

26. (B) : [(2.3) (1) 5 (0.5) 2] [(1.2) (1) 2 (0.5) 4] = (2.3 0.25) (1.2 0.0625) = 0.043125

27. (A)

28. (B) : If 4l 2 + ( k + 10) ml + 25 m 2 is a perfect square.Then, 4 l 2 + (k + 10) ml + 25 m 2 must be equal to

4l 2 + 2 2 l 5m + 25 m 2 k + 10 = 20 k = 10

29. (D)30. (C) :

x y

= 34

Then, 67

17

77

1+

= =

vvv


20/32


1. (A) : 23 + 2 3 + 2 3 + 2 3 = 4 (2 3)

= 2 2 2 3 = 2 5

2. (C) : 6 3

216

23

56

65

11 1 1 1

+

= +{ } ={ } =3. (C) :

1258

1258

52

5 18

=

x

52

52

52

3 5 3 18

=

x

5

2

5

2

15 3 18

=

+ x

On comparing the powers on both sides, we get,15 + 3 x = 18

3x = 18 15 = 3 x = 1

4. (D) :

1212

1814

5

4

=

1

2

1

2

1

2

5 4 3 2

=

1

2

1

2

1 1

= 1

5. (A) : pq

=

=

=2

332

23

23

13 3 3 3

\ pq

= ( ) =

10101 1

6. (A) : 13

12

14

3 3 3

= 3 2 43 3 3( ) ( ){ } ( )

= 27 8 64 19

64{ } =

7. (A)

8. (B) : 9 27 3 3 3 34 3 2 3 3 2 2 4 3 3 2 3 3 2/ / / / / /( ) = ( )

= 3 3 38 3 2 3 2/ /( )

= 3 383

2 3 2

/

= 3 2 3 3 2( ) +/ / = 3

13 6( ) / 9. (D) : (8) 5 + (8) 5 = 2 (8) 5 = 2 (2) 15

= (1) 15 (2) 16 = (2) 16 = (4) 8

10. (B)

11. (B) :15

0 0083

=y

.

(0.2) 3y = (0.2) 3

Chapter-8: EXPONENTS AND POWERS

y = 1

Now, (0.25) y = (0.25) 1 = 0.25

12. (B) :53

53

53

5 11 8

=

+ x

53

53

5 11 8

=

+ + x

53

53

6 8

=

+ x

or 8 + x = 6 x = 2

13. (A)

14. (D) : a b

a ba b

+ +

=

( )2 3

3 42 3 3 4 = a b = ab

15. (A) :

= { } = [ ] =

12

14

16 1

16

2 2 1

2 11

16. (D) : 3 81n =Squaring both sides, we get, 3n = (81) 2

3n = (3) 42

3n = 3

8

n = 817. (D) : 8x 1 = 2 x +3

(2)3(x 1) = 2 x +3 On comparing, 3(x 1) = x + 3 3x 3 = x + 3 2x = 6 x = 3

18. (A) :

= 1

27

1

19683

3

19. (B) : 34

14

43

4 8

33

8

1 1 1 1 1

= { } = { } =

20. (A) :32

2432

3

23

23

273 5 5

5

3 5 5 35

3

=

=

=

= / /

88

21. (A) : 75

75

75

3 2 8

=

+ x

7

5

7

5

3 2 8

=

+ + x

x + 5 = 8 x = 3


21/32

Class 7 21

22. (C) : 32

32

32

2 5 8

=

+a

32

32

5 2 8

=

+ +a

or a + 7 = 8 a = 1

23. (A) : Let the number be x . Then, (8) 1 x = (10) 1

=

1

81

10 x

x =

8

10

x =45

24. (A) : Let the number be x . Then, (15) 1 x = (5) 1

=

1

151

5 x

( ) =1

155 x

x =13

3 1=

25. (C) : a b b a = 3 7 7 3 = 2187 343 = 184426. (C) : 3x = 500

3x 2 =3

3

50092

x

=

27. (A) : 2325

12

2 3

4

0 53

2

= 1 1 (2) 5 2 3 3 2 (2) 4 = 9 (2) 5 + 3 4

= 9 (2) 6 =964

28. (B) : Let the number be x . Then, x 4 3 = 64 x 4 3 = 4 3

x =4

4

3

3

x = 4 3 + 3

x = 4 6

x = 2 12

29. (A) :4 162

12

212

1 2 2 1 2

( )( )

= ( ){ } /

/

= 4 42

1

2( ) =

30. (C) :35

35

53

3 6 1 2

=

x

35

35

3 6 1 2

=

( ) x

35

35

3 1 2

=

( ) x

or 3 = (1 2 x )

1 2 x = 3 2x = 2 x = 1

vvv


22/32


1. (C) : Area of trapezium ABCD

= 12 (Sum of parallel sides) height

=12

15 24 10 +( )

=390

22

cm = 195 cm2

2. (A) : 1 decameter = 10 m.\ 2.4 decameter = 24 m.Now,

Area of ||gm = Base Height 576 = 24 PT PT = 24 m.

3. (B) : Area of shaded portion = Area of rectangle {Sumof areas of D EAF and D EBC}

= 18 10 1

26 10

12

10 8 2( ) + { }cm = [180 (30 + 40)] cm 2

= (180 70) cm 2 = 110 cm 2 4. (C) : Area of roads = [(175 5) + (225 5) (5) 2] m2

= (875 + 1125 25) m 2

= 1975 m 2

\ Cost of levelling the roads = ` (1975 3) = ` 59255. (B) : Height of minute hand = 8.4 cm

\ Radius of gure = 8.4 cm.Now, Area swept in half an hour = Area of semi-circle

=12

2 r = 12

227

8 4 8 4 2

. . cm = 110.88 cm2

6. (B) : Since there is no gain or loss in the length of thewire.\ Perimeter of square = Circumference of circle 4 side = 2 p r

4 6.25 = 2 22

7 r

r = 3.98 cm7. (A) : Length of outer rectangle = 60 + 5 + 5 = 70 m.

Breadth of outer rectangle = 30 + 5 + 5 = 40 m.\ Area of outer rectangle = (70 40) m 2 = 2800 m 2

Similarly, Area of inner rectangle = (60 30) m 2

= 1800 m2

\ Area of lawn = (2800 1800) m 2 = 1000 m 2

Chapter-9 : PERIMETER AND AREA

8. (A) : Area of ||gm ABCD = AB DL

156 = 13 DL DL = 12 cmNow, In D DLA, DL2 + LA2 = AD2 [By Pythagoras theorem] LA2 = (13) 2 (12) 2

= 169 144 = 25 = (5) 2

AL = 5 cm.9. (D) : Let the radii be 3 x and 4 x .

\ Required Ratio =2 32 4

34

=

x x

10. (B) : Area of shaded portion = Area of rectangle {Sum of areas of D DGF and D GAE}

= 30 20 1

210 15

12

10 15 2 + { }cm = [600 (75 + 75)]cm 2 = (600 150) cm 2 = 450 cm 2

11. (B) : Let the side of D ABC be 2 x .Then, AB = 2 x , BC = x In D ADB,

(2x )2 = x 2 2

6+ ( ) 4x 2 x 2 = 6

3x 2 = 6

x = 2 cm\ AB = 2 2 c m = BC = AC.Now,

Area of D ABC =12

BC AD

=12

2 2 6 2

cm

= 2 3 2cm

12. (C) : Area of oor of the room = (13 9) m 2

= 117 m 2 \ Cost of carpeting = ` (117 6.40) = ` 748.80


23/32

Class 7 23

13. (B) :

40 m

Area of shaded region =14

Areaof circle

=14

227

14 14

= 154 m 2 14. (A) : We know that the diagonals of a rhombus bisect

each other at right angles.\ BOA = 90

AB = 10 cm

BD = 16 cm BO = OD = 8 cm.Now, In D AOB, OB2 + OA2 = AB2

OA2 = (10) 2 (8) 2 = 36 OA = 6 cm.\ AC = 2(OA) = 2 6 cm = 12 cm.

15. (C) : Distance covered by insect = 2 (80 + 60) cm = (2 140) cm = 280 cm.

16. (A) : Area of shaded portion = Area of quarter of circle Area of triangle

=14

227

7 7 1

27 7 2 cm

= (38.5 24.5) cm 2

= 14 cm 2

17. (D) : Required area = 5 Area of 1 face of cube = 5 (side) 2

= {5 12 12} cm 2

= 720 cm 2

18. (A) : Diameter of cylinder = 150 cm.

\ Radius of cylinder =150

2

cm = 75 cm.Now, Length of outer edge of parapet = 660 cm.

Radius of parapet =660 72 22

Radius of parapet = 105 cm.\ Width = (105 75) cm = 30 cm.

19. (B) : Let the length of rectangle be l mThen, breadth = (23 + l )m

Now, Perimeter = 2(length + breadth) 206 = 2( l + 23 + l ) 206 = 4 l + 46 4 l = 160

l = 40 m.and breadth = (40 + 23) m = 63 m.Now, Area = (63 40) m 2 = 2520 m 2

20. (A) : Area of shaded region = Area of rectangle ABCD 4 Area of D DSR

= 30 25 4 1

2 12 5 15 2

. cm

= (750 375) cm 2 = 375 cm 2

21. (C) : Since it has 30 m barbed wire.\ 2 5 + 20 = 30

22. (C) : Rope required = 2 (10 + 8 + 5 + 9) = 2 32 = 64 m.

23. (D) : Ratio of areas =2536

r r 12

22

2536

= [where r 1 and r 2 are the radii oftwo circles]

r r 1

2

56

=

Now, Ratio of their circumferences =22

56

1

2

1

2

r r

r r

= =24. (B) : We have,

BC = 1.2 decameter = 12 m

AC = 130 dm =13010

13m m=and AB = 5 m.Now, it is clear that,

AB2 + BC 2 = AC2

So, it is a right angled triangle.

\ Area =12

12

12 5 30 2 = =BC AB m 25. (B) : We know that, the diagonals of a rhombus bisect

each of its opposite angles.\ A = 60 (given)Also, AOB = 90 [diagonals bisect each other at right

angles]\ In D AOB

AOB + BAO + ABO = 180 ABO = 60Now, In D ABD,

A = 60, B = 60\ A + B + D = 180 D = 60So, D ABD is an equilateral triangle.Hence, all the sides are equal.\ AB = BD = AD = 6 cm.

26. (B) :


24/32


Area of ||gm ABCD= Area of D ADC + Area of D ABC

=12

34 12 1

234 12 +

= (204 + 204) m 2 = 408 m 2

27. (D) : The length of the largest pole is the diagonali.e.,

Length of pole = 10 10 52 2 2( ) + ( ) + ( )

= 100 100 25+ + = 225 = 15 m.

28. (C) : Cost of cultivation of eld = ` 360 per hectare

= ` 36010000

2

perm

= ` 361000

2

perm

Now, Total cost =`

3240 Area cost per m 2 = ` 3240

Area 36

1000

= ` 3240

Area =3240 1000

3690000 2

= m Side = 300 m.\ Perimeter of eld = 4 300 = 1200 m.

\

Cost of fencing =`

1200 75100

=

`

90029. (C) : Area of shaded portion

= Area of larger semicircle Sum of areas ofsmaller semicircles

= 12

22 2 R { }r [where R radius of largercircle, r radius of

smaller circle]

=12

227

14 2 72 2 ( ) ( )

=1

2

22

7

98 154 2 = cm

30. (C) : Required area

= 3 12 5 4 5 11 3 4 5 2( ) + ( ) ( )[ ]. . . m = 73.5 m 2

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Class 7 25

1. (D)

2. (A) : 25% of 200 = 50and no. of students wants to become an actor/actress ormusician = 30 + 22 = 52.

3. (B) : First six prime numbers are 2, 3, 5, 7, 11 & 13

\ Mean =2 3 5 7 11 13

6+ + + + +

=416

6 83= .4. (D) : Let the numbers be x 1, x 2, x 3, x 4 and x 5.

According to question,Mean of numbers = 20

i.e., x x x x x 1 2 3 4 55

20+ + + + = x 1 + x 2 + x 3 + x 4 + x 5 = 100 x 1 + x 2 + x 3 + x 4 = 100 x 5 ...(i)Let x 5 be the excluded number.Then, Mean of remaining numbers = 23

i.e., x x x x 1 2 3 4

423

+ + + = 100 x 5 = 23 4 [from (i)] x 5 = 8.

5. (C) : We are given that, Incorrect Mean of 9 observations = 35 Incorrect sum of 9 observations = 35 9 = 315Now, after detection of mistake,Correct sum of observations= Incorrect sum of observations Incorrect observation

+ Correct observation Correct sum of 9 observations = 315 18 + 81

= 378

\ Correct Mean =378

942=

6. (B) : Mean = 7

i.e.,6 8 5 4

57

+ + + + = x

x + 23 = 35 x = 12

7. (C) : Mode is that observation which have highestfrequency. Since, both 4 and 6 have highest frequency\ Option (C) is correct.

8. (A)

9. (D) : Arrange the given numbers in ascending order,75, 75, 80, 94, 96, 98, 100, 102, 180, 200, 270, 610.Now, numbers of terms = 12 which is even.

\ Median = Average of122

th

and122

1+

th

term

Chapter-10: DATA HANDLING, SYMMETRYAND VISUALISING SOLID SHAPES

= Average of 6 th and 7 th term

=98 100

2198

299+

= =

10. (B) : Mathematics.

11. (A) : Average marks =55 90 40 80 20

557+ + + +

=

12. (C) : Percentage =Marks obtained

Total marks100

=55 90 40 80 20

500100+ + + +

= 57%13. (B) : Highest marks obtained by student = 90

Lowest marks obtained by student = 20

\ Ratio =9020

9 2= : 14. (D) : U.P.15. (C) : Maharashtra.16. (D) : Total no. of heads = 59

Total no. of toss = 100\ No. of tails = (10059) = 41

\ Probability =

41100

17. (B) : Total no. of outcomes = 2 = {H, T} No. of favourable outcomes = 1

\ Probability =12

18. (A) : Required Probability =1480

740

=

19. (B) : Sample space = {1, 2, 3, 4, 5, 6}\ Total no. of outcomes = 6 No. of favourable outcomes i.e., {1, 2, 3} = 3

\ Probability = 36

12

=

20. (B) : Total no. of teams = 2

\ Probability =12

21. (D) :

22. (C)

23. (A)

24. (C)

25. (B)


26/32


26. (C)

27. (D)

28. (B)

29. (D) : The gures formed areP, Q, R, S, T, U, V, W, X,Y, ZX, XY, VW, TU, UV,XU, YV, PQ, QR, RS, QTU,RVW and so on.

30. (B)

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Class 7 27

1. (D) : Since, 2008 is the leap year. Therefore, no. of

days = 366.Now, 364

752= .

\ 364 th day will be Friday.and 366 th day will be Sunday.

2. (B) : We know that,OB2 = OA2 + AB2

OB2 = 25 km. OB = 5 km.Now, Distancebetween the startingpoint O and nal pointP is (25 5) km = 20 km.

3. (C) : We have, 18 4 + 3 2 2. After correct notations, we have, 18 + 4 3 2 2 = 18 + 12 1 = 29

4. (A) :

5. (B)

6. (B) : We see that in the given pattern smaller gurebecomes larger in next step and a new gure is added.

7. (D)

8. (D) : Let the total number of people be x.Then,

Number of people who work in elds =12

x

Number of people who do not work in elds = 12

x \ Number of people working in factories

=12

12

14

= x x 9. (D)

10. (B)

11. (D) : A is Bs brother means A is the brother of B. But Bis Cs sister and C is Ds father means B is Ds aunt andA is Ds uncle.

Chapter-11: LOGICAL AND ANALYTICAL REASONIN

12. (C) : In statement 1 and 2, the common word is apple

and the common code is 8. Also, in statement 1 and 3,the common word is bring and the common code is 6.\ The code for me is 7.

13. (C)

14. (C) : Elephants and lions are animals.15. (C)

16. (C)

17. (C) : 5 + 8 = 13 13 + x = 34 x = 21

21 + 34 = 5518. (A) : 5 men can do 1 work in 12 days.

1 man can do1

60

work in 1 day.

\ The work completed in 10 days =1

6010

16

=\ Number of men required = 6.

19. (D)

20. (D) : Except 14, all are divisible by 5.21. (C)

22. (D) : The word is NURSE and the middle alphabet isR.

23. (A)

24. (D)

25. (B) : In the given pattern the number of sides in thegure is decreased by 1 in each step.

26. (A) : In the given pattern the number of squares isdecreased by 1 and the number of circles is increasedby 1 in each step.

27. (D)

28. (B)29. (D)

30. (C) : 5 + 5 = 10 10 + 7 = 17 17 + 9 = 26 26 + 11 = 37 37 + 13 = 50 50 + 15 = 65

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28/32


1. (D) : Let the number of girls and boys be 4 x and 3 x .

Then, 3x + 4 x = 21or, 7 x = 21 x = 3\ Number of girls = 4 3 = 12

2. (A) : Let the total money be x . The total money spent by Dhruv

=13

14

712

x x x +

=

\ Money left = x x x =712

512

3. (C) : No. of green apples = 54\ No. of red apples = 30 + 54 = 84.And total no. of apples = 54 + 84 = 138

\ Ratio =84

1381423

=

4. (B) : Let the age of Ishika be x years.Then, the age of grandfather = 4 x yearsAccording to question, x + 4 x = 100or, 5 x = 100 x = 20\ Ishikas age = 20 years.

5. (B) : Let the no. of boys and girls be 7 x and 4 x .According to question, 7x 4 x = 21or, 3 x = 21 x = 7\ Number of boys = 7 7 = 49And number of girls = 7 4 = 28\ Total no. of children = 49 + 28 = 77.

6 (A) : No. of bracelets with 36 beads = 4

No. of bracelets with 180 beads =4

36180 20 =

7. (C) : Time taken to complete 100 m. distance = 20 sec.\ Time taken to complete 400 m. distance

=20

100400 80

=sec sec.

8. (D) : Percentage of attended people

=315420

100 75 = %

9. (D) : No. of video-tapes checked out = Total no. of tapes No. of tapes present

= 52 17 = 35.

Chapter-12: EVERYDAY MATHEMATICS

10. (C) : 12, 24, 36, 48, 60

+12 +12 +12 +12

11. (D) : Let the total no. of questions be x . No. of questions Monika answered correctly =

34

x

\ Percentage Required =

34 100 75

x

x = %

12. (D) : The intervals at which they toll together is 2, 4, 6,8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.\ No. of times they toll together = 15 + 1 = 16.

13. (C) : No. of people in rst group = 60

\ No. of people who board the bus from rst group

=34

60 45 = .Also, No. of people in second group = 60\ No. of people who board the bus from second

group =23

60 40 = .\ Difference = 45 40 = 5.

14. (C) : S.P. = ` 54000 gain% = 20

\ C.P. =100 54000

100 20

+ =100 54000

120

= ` 45,000

This C.P. is for the friend of the man and is the sellingprice for the man.Now,

S.P. = ` 45000 Loss% = 10

\ C.P. =100 45000

100 10100 45000

90

( ) =

= ` 50,000

\ The original cost price is ` 50,000.15. (B) : Let the cost of 1 chair be ` x .

\ The cost of 1 table = ` (40 + x )Then, we have, 3(40 + x ) + 2( x ) = 745or, 120 + 3 x + 2 x = 745 5x = 625 x = ` 125\ Cost of 1 chair = ` 125And cost of 1 table = ` (125 + 40) = ` 165.

16. (D) : For Megha, Principal = ` 7200 Rate = 5% p.a. Time = 4 years.

S.I. =7200 5 4

100

= ` 1440.


29/32

Class 7 29

For Priya, P = ` 7200 R = 5% p.a. T = 5 years.

\ S.I. =7200 5 5

100

= ` 1800.

\ Difference = ` (1800 1440) = ` 360.

17. (B) : The distance covered below sea level= (100 + 20) m. = 120 m.

He came up = +35 m.\ Required distance = (120 + 35) m. = 85 m.

18. (C) : The greatest possible size of the measuring vesselis the HCF of 1653 litres, 2261 litres and 2527 litres.Now, HCF of 1653, 2261 and 2527 is,

19 1653, 2261, 252787, 119, 133

\ HCF = 19.\ Capacity = 19 litres.

19. (B) : Let the total capacity of drum be x litres.Then, we have,

34

15 7

12 x x =

34

712

15 x x =

212

15 x =

x = 90

\ Capacity = 90 litres.20. (B) : Amount spend on education = ` (30% of 15000)

= ` `30

10015000 4500 =

21. (D) : S.P. = ` 285 Loss% = 5

\ C.P. = ` `100 285

100 5100 285

95 =

( )

= ` 300

Now, C.P. = ` 300 gain% = 15

\ S.P. =300 100 15

100+( )

= 3(115) = ` 345

22. (B) : Total no. of employees = 1600\ No. of female employees = 60% of 1600 = 960And, No. of male employees = 1600 960 = 640Out of 640 males 50% are computer literate.\ No. of males who are computer literate = 50% of 640 = 320Now, Total no. of employees who are computer literate = 62% of 1600 = 992 \ No. of males + No. of females = 992 320 + No. of females = 992 No. of females who are computer literate = 672

23. (D) : Age of Aman = 17 years. \ Age of his sister = (17 + 3) yrs. = 20 years.Now, Age of mother = (20 + 21) yrs. = 41 years.\ Age of father = (41 + 8) = 49 years.

24. (B) : Total no. of persons in the park= 1 + 1 + 1 + (2 2) + 2

= 3 + 4 + 2 = 925. (B) : The time after which they toll together is the LCM

of 36, 40 and 48.

\ LCM = 2 2 2 2 3 3 5 = 720 seconds = 12 minutes.

26. (A) : Let the original amount be ` x .Then,

Eldest sons share = ` x 2

Youngest sons share = ` `13 2 6

= x x

\ Amount left = ` x x x +

2 6 =

` x x 2

3 = `

x 3

27. (A) : Total no. of marbles = 8No. of marble marked with 2 = 1

\ Probability =18

28. (C) : S.P. = ` 3200

Pro t% = 3313

1003

=

\ C.P. = ` 100 3200

100 100

3

+

= `

100 3200 3400

=`

240029. (C) : Let the length of total journey be x km.

Then,

x =23

33 x +

or, x x =23

33

x 3

33= x = 99

\ Total distance = 99 km.30. (B) : The concentration of water in both the containers

= 80 % and 75%

\ Ratio =8075

16 15= :


30/32


1. (B) : RAGER

2. (D) : A 33. (B)

4. (D) : Z = 26 2 = 52 ACT = (1 + 3 + 20) 2 = 48Now, BAT = (2 + 1 + 20) 2 = 46

5. (C)

6. (C) : 1 0 m

25m

15mN

S

EW

P

7. (B)

8. (B)

9. (D)

10. (C) :

11. (D)

12. (B) :

13. (A) : After adding the digits the numbers will be; 8 + 5 + 3 = 16; 3 + 9 + 5 = 17; 4 + 8 + 6 = 18; 2 + 4 + 9 = 15; 4 + 9 + 7 = 20; 7 + 6 + 6 = 19; 9 + 1 + 4 = 14Arrange the above numbers in decreasing order,i.e., 20, 19, 18, 17, 16, 15, 14 497, 766, 486, 395, 853, 249, 914 \ The middle term is 395

14. (B)

15. (A)

16. (D) : The equation will be, 24 + 16 8 6 9 = 24 + 2 6 9 = 27

17. (C)

18. (A) : (3 + 27) 2 = 15 (6 + 56) 2 = 31 (9 + 81) 2 = 45

19. (C) : In winters, we use quilt and in this code language

quilt is called mosquito net.20. (D)

21. (B) : [(31 19) {5 (5 + 2 3)} of 3 + (2)] (1) = [12 1 3 + (2)] (1) = 34 (1) = 34

22. (B) : Since, D XYZ is an equilateral triangle.\ XZO = 60Now, In D XZQ ZXQ = XQZ = 30 (Angle opp. to equal sides

are equal)We know that, ZXQ + XZQ + ZQX = 180 XZQ = 180 60 = 120 XZO + OZQ = 120 OZQ = 120 60 = 60

23. (D)

24. (A) : Let the number be x. Then,

95

45of x =

x = 25Now,

15

25 5 =

25. (D) : Area of ||gm PQRS = Area of D PQR + Area of D RPS

=12

24 6 1

224 6 + = 144 cm 2

26. (C) : After 4 years, Johns sisters age = 12 + 4 = 16 years. Johns age = ( n + 16) yrs.\ Sum = ( n + 16 + 16) yrs. = ( n + 32) yrs.

27. (B) : Since CD is a straight line.\ 40 + 35 + BOC = 180 BOC = 105Now, AB is a straight line.\ y = 180 (105 + 48) = 27

28. (C) : Converting all the fractions into like fractions.\ LCM of 3, 7, 10 = 210

\ 4 703 70

280210

4 307 30

120210

7 2110 21

147210

=

=

=

; ;

The order is,

280210

147210

120210

, ,

2012 - 6 th SOF IMO


31/32

Class 7 31

\ The descending order is,4

3

7

10

4

7, ,

29. (C) : Area of square = 81 Side Side = 81 Side = 9 cm.Now, Diameter of circle = side of square = 9 cm.

\ Radius =9

2cm.

\ Circumference = 2 p r

= 2 22

7

9

2

cm = 28

2

7cm

30. (C) : It is given that, AB = AF\ AFB = ABF = p (say)Now, In D ABF, 36 + p + p = 180

p = 72Now, BC is a straight line\ BFA + AFE + EFG = 180 72 + 58 + EFG = 180 EFG = 50In D EFG, FG = FE\ FGE = FEGNow, FGE + FEG + EFG = 180 2FGE = 180 50 = 130 FGE = 65

Now, again BC is a straight line.\ FGE + x = 180 x = 180 65 = 115

31. (D) : 3889 + 12.952 x = 3854.002 3901.952 x = 3854.002 x = 47.95

32. (A) : Length of painting = (24.5 2 3) cm = 18.5 cm Breadth of painting = 6 cm.\ Area of painting = (18.5 6) cm 2 = 111 cm 2

33. (C) : 1 hour = 3600 sec.

\ Fraction = 253600

1144

=

34. (A) : Let the number required be x . Then,

1

2166of x =

x = 332

\ 30% of 332 =30

100332 99 6 = .

35. (C) : a = 40 (Alternate angles)

36. (B) : Total amount of expenditure = 100

Amount of expenditure on research & development= 5

\ Total expenditure is 20 times the expenditure onresearch and development.

37. (C)

38. (A) : Cost price of 1 chair = ` 450\ Cost price of 24 chairs = ` (450 24) = ` 10800 Selling price of 24 chairs

= ` [(16 600) + (400 8)] = ` 12800\ Proft = ` (12800 10800) = ` 2000

\ Proft % =2000

10800100 18

14

27 = %

39. (D) : Time = 2 years Rate = 12% p.a. S.I. = ` 1620Let P be the principle.

\ S.I. =P 12 2

100

1620 100 = P 12 2 P = ` 6750

40. (B)

41. (D) : 6 glasses =3

5 of jug

\ 1 glass =3

30 of jug

\ Amount left =3

5

3

30

1

2 =

42. (B) : It will be a proportion, i.e.,

4

9 324=

x

x =4 324

9

x = 144\ 144L of orange syrup is required.

43. (C) : Let the cost of skirt be ` x.

Then, the cost of bag = 55% of x =55

100

11

20 = x x `


Cost of skirt = 31.50 + Cost of bag

x = 31 50 11

20. + x

x x 1120

= 31.50

x = ` 70\ Cost of skirt = ` 70and, cost of bag = ` (70 31.50) = ` 38.5\ Total money required = ` (70 + 38.5) = ` 108.50

44. (D) : Let the no. of phonecards Sonia has be x .Then, the no. of phonecards Jasmine has = 3 x .According to question, x + 3 x = 60

4 x = 60 x = 15\ Jasmine has 45 phonecards.

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32/32

45. (A) : Distance covered in 20 min. = 18 km.\ Distance covered in 1 hr 20 min., i.e., 80 min.

=1820

80 72 = km

46. (A) : Area of path= Area of square ABCD

Area of square PQRS

= [80 80 72 72] m 2 = 1216 m 2

47. (B) : Cost of 1 litre of milk = ` 19.75\ Cost of 42 litres of milk = ` (19.75 42) = ` 829.50

48. (C) : Loss = ` 60 S.P. = ` 660\ C.P. = Loss + S.P. = ` (60 + 660) = ` 720Now,

gain % = 15

\ S.P. = ` 720 100 15

100 +( )

=`

720 115100

= ` 828

49. (D) : Let x be the distance between the buildings ABand CD.It is given that,

OB = OD = 17 m. CD = 15 m.and, AB = 8 m.In D AOB,

AB2 + AO2 = OB 2 AO2 = OB 2 AB2 = (17) 2 (8) 2 = 225\ OA = 15 m.Similarly, In D COD OC 2 = OD 2 CD 2 = (17) 2 (15) 2 = 64\ OC = 8 m.\ The distance between the buildings = (15 + 8) m = 23 m.

50. (A) : Distance between the two places = 82 + (13) = 95 m.

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International Maths Olympiad Practice Book