International Maths Olympiad Practice Book

International Mathematics Olympiad

7

WORK BOOK

INSTANT

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Chapter-1 : Number system 1. (C) : Positive integers are greater than negative integers. 2. (D) : 179 is a prime number.

−3 X = 8 −24

⇒

−3 3 − X × = 8 3 24

−9 − X = 24 24 ⇒ X = 9 ⇒

1 × −72 = −36 ≠ 36 2 1 × 36 = 36

16. (C) :

4. (D)

17. (C)

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 = –4 and 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 first 1 sec. = +5 cm. Distance covered in next 1 sec. = –2 cm. \ Distance covered in 2 sec. = 5 – 2 = 3 cm. 2  \ Time taken to cover 57 cm. =  × 57 sec. 3 = 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 flowing out = 9 litres every hour. ⇒ Amount of water flowing out in 1 hr. = 9 litres. Amount of water flowing 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 4 9 36 1 21 21 2 21 42 5 7 35 \ × = , × = , × = , × = 7 9 63 3 21 63 3 21 63 9 7 63 21 35 36 42 , , , The ascending order is, 63 63 63 63 1 5 4 2 i.e., , , , 3 9 7 3 5 4 1 5 4 Now, the average of and is  +  9 7 2 9 7  1  35 + 36  71 = =   2  63  126 Class 7

14. (D) 15. (D) :

3. (C) : –4 × –9 = 36 –3 × –12 = 36



13. (C) : 0 + 0 = 0

( −4) × ( −9) × ( −25) = 2 × 3 × 5 = 30 ( −2) × ( −3) × ( −5)

18. (A) Total no. of plants No. of rows 630 = 30 = 21 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

19. (B) : No. of plants in 1 row =

21. (D) 22. (D) : –19 × (4 + (–2)) = –19 × 4 + (–19) × (–2) Associative Property. 23. (A) 24. (D) : Let the other number be x.  −4  −9 Then, x ×   =  3  16 ⇒ x = ⇒ x =

−9 3 × 16 −4 27 64

25. (B) 26. (B) 27. (A) 28. (A) : 199 is an odd number. Therefore, the product have negative sign. 29. (A) 1  30. (A) : 222 −  {42 + (56 − 8 + 9)} + 108 3  1  = 222 −  {42 + 56 − 17} + 108 3  = 222 − [27 + 108] = 222 − 135 = 87

3

Chapter-2 : Fractions and Decimals 1. (B) : 3

4 7 8 19 27 68 +2 +4 = + + 5 10 15 5 10 15

LCM of 5, 10, 15 = 30 19 27 68 114 81 136 331 + + = + + = 30 30 30 30 5 10 15

\

2 1 1 6 = 1+ = 1 = 10 5 5 5

2. (B) : 1 +

3. (B) : Required fraction =

3 8

1 1 4. (A) : The rational number lying between and 8 7 1  1 1 =  +  2 7 8 = \

1  15  15 = 2  56  112 = 112

5. (C) : Total no. of boxes = 25 No. of shaded boxes = 9 9 25 6. (D) : Total no. of squares = 18 Let, total no. of shaded squares be x. According to question, \ Fraction =

7 x = 9 18 ⇒ x = 14 Out of these 14 squares 6 are shaded. So, we have to shade 8 more squares. 7. (B)

11. (D) : 9

7 187 = = 9.35 20 20

1   12. (B) :  1 −   1 −  3 

−9 a 27 −45 = = = 5 20 b c

−9 4 −36 a × = = 5 4 20 20



 3 − 1  4 − 1  5 − 1  6 − 1 ........  n − 1 =     3   4   5   6  n 



=

2 3 4 5 ........ (n − 1) 2 × × × × × = 3 4 5 6 n n

13. (C) : Total no. of animals = 192. 7 16 7 × 192 = 12 × 7 = 84 \ Number of cattles = 16 2 Out of these 84 cattles are dairy cows. 3 2 \ No. of dairy cows = × 84 = 56 3 14. (B) Fraction of cattle =

15. (B) : Let, Tom’s have ` x Then, Jasmine’s have `(41 – x). 1 1 Now, of x = 2 + ( 41 − x ) 4 7 ⇒

1 41 x x = 2+ − 4 7 7

1 1 14 + 41 x+ x= 4 7 7

⇒

55  7 + 4   x = 28 7 ⇒ x = ` 20 \ Tom have ` 20

⇒ a = −36

16. (C) : x % of 24 = 64 x × 24 = 64 100

−9 −3 27 27 × = = ⇒ b = −15 b 5 −3 −15

or

−9 5 −45 −45 × = = ⇒ c = 25 25 c 5 5

⇒ x =

9. (C) :

1  1  1 1    1 −   1 −  ........  1 −  n 4 5 6

⇒

8. (B) :

1 1+



4

10. (C)

1 1 2+ 3

=

1 1+

1 6 +1 3

=

=

1 1+

3 7

1 7 = 7 + 3 10 7

64 × 100 800 = 24 3

2 3 17. (B) : 0.00639 ÷ 0.213 ⇒ x = 266



=

=

0.00639 100000 × 0.213 100000 639 = 0.03 21300

IMO Work Book Solutions

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

1 21 18. (A) : Total time = 5 hrs = hrs. 4 4

3 15 m. Length of piece to be cut = 3 m = 4 4 30 = 8 \ Number of pieces = 15 4 1000 231 = 25. (C) : 0.231 = 0.231 × 1000 1000

3 11 Time for English and Mathematics = 2 hrs = hrs. 4 4 21 11 10 − = \ Time for other subjects = 4 4 4 5 1 = hrs = 2 hrs. 2 2 19. (D)

26. (B) : 2 − 3 (2 − 3)−1 

20. (B)



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

3



3 



3



3 

22. (D) :  −2  −  −1   +  −2  −  −1   + ......... upto 30 times 4 4   4 4  

 3   3   = 30 ×  −2  −  −1    4   4  



 −11 7  = 30 ×  +   4 4



= 30 ×

−4 = −30 4

1  1   23. (C) :  1 −   1 −   1 −    2 3 

=

1  ........  1   1 −  4 10

1 1 2 3 ........ 8 9 × × × × × = 10 2 3 4 9 10

−1

−1 = 2 − 3 ( −1) 

= (2 + 3 )

−1

= (5 )

−1

−1

=

1 5

6 3 = 16 8 112 − 69 1 3 28 23 − 28. (D) : 9 − 5 = = 12 3 4 3 4 43 7 =3 = 12 12 27. (A) :

321 = 21.4 \ Average score per wicket = 15 

= [2 − 3 × ( −1)]

−1

29. (B) : 1 +

1 1

= 1+

1

1 1+ 1 6 −1 1− 6 6 1 1 5 11 + 5 16 = 1+ = 1+ = = 1 + = 6 5+6 11 11 11 1+ 5 5 5 3 7 13 19 31 30. (D) : 4 − 2 + 3 = − + 2 8 8 2 8 8

=

1+

52 − 19 + 31 64 = =8 8 8

vvv

Class 7

5

Chapter-3 : Simple Equations 1. (C) : Let the one part be x. Then, the other part will be 184 – x. Now,

1 1 × x = (184 − x ) + 8 3 7

x 184 x = − +8 ⇒ 7 7 3 x x 184 + = +8 7 3 7

⇒

or

10 x 240 = 21 7

⇒ x =

240 × 21 7 × 10

⇒ x = 72 Second part = 184 – 72 = 112

1+

2x 1 1+

=1

x 1− x

2x 2x =1 ⇒ =1 1 1+ 1− x 1+ 1− x + x 1− x 2x ⇒ = 1 ⇒ 2x = 2 – x 2−x 2 ⇒ 3x = 2 ⇒ x = 3 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 × 4x + 5 × 5x = ` 8600 or 2x + 16x + 25x = ` 8600 ⇒ 43x = ` 8600 ⇒ x = ` 200 \ Each daughter receives `(4 × 200) = ` 800 5. (A) : Let the number be x. Then, acc. to question, 5 ( x + 4) − 20 = 10 8 or 5(x + 4) – 20 = 80

6

6. (B) : Let the marks in English be x. Then, the marks in Mathematics = 240 – x. Now, we are given that, 1 (240 − x ) = 1 x + 30 3 2 x x ⇒ 80 − = + 30 3 2 x x + = 80 − 30 ⇒ 3 2



5x = 50 6 ⇒ x = 60 \ Marks in English = 60. ⇒

\ First part = 72

2. (D) :

⇒ 5x + 20 – 20 = 80 ⇒ 5x = 80 ⇒ x = 16

7. (A) 8. (D) : Fraction of women workers = So, fraction of men workers =

2 3

1 3

1  1 1 Now,  of  are married women = are married  2 3 6  1 1 1  of  women have children = have children. 3 6 18 1 1 5 − = 3 18 18 1  3 2 are marrie Also,  of  are married men = 2  4 3 \

Women having no children =

...(i)

 2 1 1  of  men have children = have children. 3 2 3 2 1 1 \ Men having no children = − = ...(ii) 3 3 3 1 5 \ Total no. of workers having no children = + 3 18 [from (i) & (ii)]

=

9. (B) : Let the number of working days be x Then, the number of days he is idle = 60 – x According to question, 20x – 3(60 – x) = 280 or 20x – 180 + 3x = 280 ⇒ 23x = 460 ⇒ x = 20. \ Working days = 20 and Idle days = 40

11 18

IMO Work Book Solutions

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, 100x + 50 (85 – x) = 5000 or 100x + 4250 – 50x = 5000 ⇒ 50x = 750 ⇒ x = 15 \ Number of notes of ` 50 denomination = 85 – 15 = 70 \ Amount required = 70 × 50 = ` 3500 11. (D) : Let the unit’s digit be x. Then, ten’s digit = 2x \ The number = 20x + x = 21x. Now, the number after interchanging the digits = 10x + 2x = 12x According to question, 12x = 21x – 36 or, 36 = 21x – 12x ⇒ 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

15. (C) 16. (B) : Let the number be x. Then, 2 1 1 x + x + x + x = 37 3 2 7 28 x + 21x + 6 x + 42 x = 37 42 or, 97x = 37 × 42 or,

⇒ x =

1554 97

⇒ x = 16

2 97

17. (C) : Let first angle be x° Then, other will be (44 + x)° Now, x° + (44 + x)° = 180° or, 2x = 136° ⇒ x = 68° \ First angle = 68° and second angle = (68 + 44)° = 112°. 3 1− x 3  18. (C) : (7 x − 1) −  2x −  = x +  4 2 2 3   4x − 1+ x  3  21 ⇒  x −  −   = x +  4   4 2 2 ⇒

21 3 5x − 1 3 x− − = x+ 4 4 2 2

⇒

21 5x 3 3 1 x− −x= + − 4 2 2 4 2

 x \ Share of 1 boy = `   14

⇒

6+3−2 21x − 10 x − 4 x = 4 4

Also, when total number of boys = 18

or, 7x = 7 ⇒ x = 1

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

 x Share of 1 boy = `    18  According to question, x x = + 80 14 18 x x − = 80 14 18

⇒

9x − 7 x = 80 126 ⇒ x = ` 5040 ⇒

14. (C) :

2 (5 x + 1) + 3 = 1 5 5

⇒ 2 x +

2 3 + =1 5 5

⇒ 2 x + 5 = 1 5 ⇒ 2x + 1 = 1 ⇒ 2x = 0 ⇒ x = 0 Class 7

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 = 45 And, 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 + 2x = 224 + 50 + 45 + 8 + x or, 2x – x = 224 + 103 – 312 ⇒ x = 15 \ Number of keepers = 15 20. (A) : From the options we see that the given condition is satisfied by 1st option. Hence, the answer is 3, 4. 21. (D) : Let, the ages of A and B be 5x and 3x. After 6 years, their ages will be 5x + 6 and 3x + 6. According to question, 5x + 6 7 = 3 x + 6 5

7

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

or, 5 (5x + 6) = 7 (3x + 6) ⇒ 25x + 30 = 21x + 42 ⇒ 25x – 21x = 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.

3  Then, the value of second prize = `  x  4 3  1 3  and the value of third prize = `  × x  = `  x  8 2 4 Now, or,

22. (C) : The cost of first 1000 copies = 1000x 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 = 25x \ The correct answer = 25x ...(1) But, he multiply with 52 then, the answer becomes 52x. According to question, 52x = 324 + 25x or, 52x – 25x = 324 ⇒ 27x = 324 ⇒ x = 12

⇒



1 part taken by 1 member 12

⇒

3 3  part taken by  12 ×  members  4 4



⇒ x = 400

⇒ 50 + x = 13 × 5 ⇒ x = 15

1 12

1 3 of the cake, of the cake was Also, if father took 4 4 left. Now,

140 × 100 35

11 + 12 + 13 + 14 + x = 13 i.e., 5

1 = 3x 4

⇒ x =

35 x = 140 100

26. (C) : Average = 13

8x + 6x + 3x = 2550 8

29. (C) : Let the share of each member from rest of the members be x. Then, Father’s share = 3 × share of each member

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

⇒ x =

3 3 x + x = 2550 4 8

⇒ 17x = 2550 × 8 ⇒ x = 1200 \ The value of first prize = ` 1200

24. (C)

or

x+

= 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 = ‘2x’ years. After 30 years, their ages will be (x + 30) and (2 + 30) According to question, 1 1 ( x + 30) = (2 x + 30) 2

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

or

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

3 x − 2 x = 30 − 45 2 ⇒ x = 30 ⇒

\ A’s present age = 60 years and B’s age = 30 years.

vvv

8

IMO Work Book Solutions

Chapter-4 : Practical Geometry & lines and angles 1. (A) : Since PQ || ST \ ∠PQS = ∠QST (Alternate angles) ⇒ ∠QST = 98° ⇒ ∠QSA + ∠AST = 98° ⇒ ∠AST = 70° Now, 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 = ∠ACD Now, In D ADC, ∠DAC + ∠ADC + ∠ACD = 180° ⇒ 2 ∠ACD = 180° – 98° = 82° ⇒ ∠ACD = 41° Now, AB || CD \ ∠ACD = ∠CAB (Alternate angles) ⇒ ∠CAB = 41° In 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 a rhombus bisect the angles. \ ∠DCA = 30° Also, in square CGEF. ∠GCA = 90° ⇒ y + ∠DCA = 90° ⇒ y = 60° 6. (C) : Let angles be 7x and 11x. Then, 7x + 11x = 180° ⇒ 18x = 180° ⇒ x = 10° \ The angles are 70° and 110°. 7. (B) : As the sum of all the angles around a point is 360°.

Class 7

8. (B) : Draw a line EG passing through F and parallel to AB. Now, AB || CD and 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 = 68° and ∠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 + BC2 = 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° = 54° Now, In DFAG, x + ∠FAG + ∠FGA = 180° ⇒ x + 2 × 54° = 180° ⇒ x = 72° Now, AGE is a straight line. \ ∠AGF + ∠FGE = 180° ⇒ ∠FGE = 126° ⇒ ∠GEB + y = 126° ⇒ y = 52° \ x + y = 72° + 52° = 124°

9

11. (D) : Let the angle be x°. Then, supplement of x = 180° – x Now, we are given that, x = 2(180 – x)° – 45° ⇒ x + 2x = 360° – 45° ⇒ 3x = 315° ⇒ x = 105° \ The angle is 105°. 12. (D) : Since GHCI || EABF. \ ∠HCB = ∠CBF (Alternate angles) ⇒ ∠HCB = 66° Since, HC || AB and AH || CB. \ ABCH is a ||gm. \ ∠HCB + ∠AHC = 180° (Interior consecutive angles) ⇒ ∠AHC = 114° But, 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 = 50° Now, AHB is a straight line. \ ∠AHC + ∠CHE + ∠EHG + ∠GHB = 180° ⇒ 72° + 50° + 42° + x° = 180° ⇒ x = 16° 18. (D) : Vertically opp. angles. 20. (B) : 45° is the only angle which is its own complementary angle. 21. (D) 22. (C) : ABCD is a rhombus. \ AB = AD. ⇒ ∠ADB = ∠ABD (angle opp. to equal sides are equal) ⇒ ∠ABD = 27° Now, ABE is a straight line. \ ∠ABD + ∠DBE = 180° ⇒ ∠DBE = 153° 23. (A) : ∠ABE = ∠CBF (Vertically opp. angles) ⇒ ∠CBF = 48°

10

[EC = EB] 24. (D) : ∠EBC = ∠ECB ⇒ ∠ECB = 62° In D EBC, ∠ECB + ∠EBC + ∠CEB = 180° ⇒ ∠CEB = 56° ∠CEB = ∠DEA (vertically opp. angles) \ ∠DEA = 56° Now, In D DEA 38° + 56° + y = 180° ⇒ y = 86° 25. (D) : Since D BFC is an equilateral triangle. \ ∠FBC = 60° But, ∠ABC = 90° (angle of a square) ⇒ ∠ABF + ∠FBC = 90° ⇒ ∠ABF = 30° In D AFB, ∠BAF = 60° But, ∠BAD = 90° ⇒ ∠BAF + ∠FAD = 90° ⇒ ∠FAD = 30° And ∠EAF = 2 × ∠FAD = 2 × 30° = 60° 26. (A) : Let the interior angles be 2x and 5x. Then, 2x + 5x = 70° (Exterior angle property) ⇒ 7x = 70° ⇒ x = 10° \ The angles are 20°, 50°, 110° 27. (B) 28. (D)

19. (B)

Also, ∠FCB = 48°

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

[FC = FB]

29. (B) : FG || CBH \ ∠FGC = ∠GCB (Alternate angles) ⇒ ∠GCB = 70° And GCA is a straight line. \ ∠GCB + ∠BCA = 180° ⇒ ∠BCA = 110° In D ABC, 110° + ∠CBA + ∠CAB = 180°

[ CB = CA ] ⇒ 2 ∠CBA = 70° ⇒ ∠CBA = 35° = ∠CAB.  CBH is a straight line. \ y = 180° – 35° = 145° Also, y = x + ∠CAB = x + 35° (alternate angles) ⇒ x = 110° IMO Work Book Solutions

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

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

vvv

Class 7

11

Chapter-5 : The Triangle and Its Properties & Congurence of Triangles 1. (C) : 132 = 122 + 52 ⇒ This triangle satisfies 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 Now, BC is a straight line. \ ∠ADB + ∠ADC = 180° ⇒ 2∠ADC = 180° ⇒ ∠ADC = 90°

(by CPCT)

4. (A) : In D CED, CE = ED \ ∠EDC = ∠ECD ⇒ ∠ECD = 28° Also, ∠ECD = ∠BCA (vertically opp. angles) ⇒ ∠BCA = 28° In DBCA, 62° + 28° + ∠BAC = 180° ⇒ ∠BAC = 90° Now, BAF is a straight line. \ ∠BAC + ∠CAF = 180° ⇒ y = 90° 5. (C) 6. (A) : Since ABC is an equilateral triangle. \ ∠A = ∠B = ∠C = 60° [ DA =DB] And, ∠DBA = ∠DAB = (60 – x)° In D DAB, ∠DBA + ∠DAB + ∠ADB = 180° ⇒ 2(60 – x)° + 88° = 180° ⇒ 2(60 – x)° = 92° ⇒ 60° – x = 46° ⇒ x = 14°

12

A

7. (C) : B

8. (D)

D

C

9. (A) 10. (B) : Let the third side of triangle be 3 cm. Then, 3 + 3 = 6 < 8 So, it is not possible because we know that the sum of two sides will be greater than the third side of the triangle. So, the third side is 8 cm. 11. (B) : Let the angles are x, 2x and x. Then, x + 2x + x = 180° ⇒ 4x = 180° ⇒ x = 45° \ The greatest angle = 2 × 45° = 90° 12. (B) : It is given that, AB = BC and AC2 = 100 cm2 By using pythagoras theorem, AC2 = AB2 + BC2 ⇒ 100 = 2AB2 ⇒ AB2 = 50 ⇒ AB = 5 2 cm i.e., AB = BC = 5 2 cm A 13. (D) : B

14. (D)

D

C

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

IMO Work Book Solutions

\ 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

[EB = CD] ⇒ AE + EB = 18 m ⇒ AE = 5 m In D AED, AE2 + ED2 = AD2 (By Pythagoras theorem) 2 2 2 or, x = (5) + (12) = (25 + 144) cm2 = 169 cm2 ⇒ x = 13 cm. 17. (A) 18. (C) : We have,

1 hx ...(i) 2 (where h is the height of the altitude) And, Area of square CDEF = x × x ...(ii) From (i) and (ii) Area of D ABC =

x × x = ⇒ h = 2x

1 ×h×x 2

19. (A) : Since PQ = PR ⇒ ∠PQR = ∠PRQ ...(i) Now, let ∠P be x. Then, ∠Q = 2x (given) and ∠R = 2x {from (i)} Now, ∠P + ∠Q + ∠R = 180° x + 2x + 2x = 180° ⇒ 5x = 180° ⇒ x = 36° \ ∠Q = 2 × 36° = 72° 20. (A) : In D FGC, ∠GCF = 92° (given) As we know, ∠CGF = 60° (angle of equilateral triangle) \ x + 60° + 92° = 180° Class 7

⇒ x = 28° Now, In D BCF, ∠CBF = 60° ∠FCB = 180° – 92° (linear pair) ⇒ ∠FCB = 88° \ ∠BFC + 88° + 60° = 180° ⇒ ∠BFC = 32° And, ∠CFE = 90° ⇒ y + 32° = 90° ⇒ y = 58° \ y – 2x = 58° – 2 × 28° = 58° – 56° = 2° 21. (B) : In DADC and DCBA AD = BC (given) ∠DAC = ∠BCA AC = AC \ DADC @ DCBA \ DC = AB

[ AD || BC] (common) (By SAS) (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 = 60° Now, In D GCF, 60° + 72° + x = 180° ⇒ x = 48° 23. (A) : ∠FCA = ∠BFD (corresponding angles) ⇒ x = 51° and ∠CAB = ∠AOD (Alternate angles) ⇒ ∠AOD = 83° Also, ∠FOB = ∠AOD (Vertically opp. angles) ⇒ ∠FOB = 83° Now, In D FOB y = 51° + 83° (Exterior angle property) ⇒ y = 134°  x + y = 51° + 134° = 185° 24. (B) 25. (D) : By RHS. 26. (C) : It is given that, D ABC is a right angled triangle at A. C D B

A

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

13

29. (B) : We have, 82 + 62 = 100 = 102 So, it is a right angled triangle.



= (2AD)2 – AD2 = 4AD2 – AD2 = 3AD2 27. (A) : Since, AB = AC ⇒

\ Area of D ABC =

1 1 AB = AC 2 2

⇒ BF = EC Also, AB = AC ⇒ ∠B = ∠C In D BEC and D CFB EC = FB ∠B = ∠C BC = BC \ D BEC @ D CFB ⇒ BE = CF

1 × AB × BC 2

1 × 8 × 6 = 24 cm2 2 30. (A) : Draw AD perpendicular bisector to BC. Then, BD = DC = 5 cm.

=

In D ABD

(proved above) (proved above) (common) (by SAS) (by CPCT)

AB2 – BD2 = AD2 ⇒ AD2 = 100 – 25 = 75 cm2 ⇒ AD = 5 3 cm

28. (C)

Now, Area of D ABC =

1 × 10 × 5 3 = 25 3 cm2 2

vvv

14

IMO Work Book Solutions

Chapter-6 : Comparing Quantities 1. (C) : C.P. of 11 oranges = ` 10  10  \ C.P. of 1 orange = `    11  Also, S.P. of 10 oranges = ` 11  11  \ S.P. of 1 orange = `    10  Now, Profit = S.P. – C.P. =

11 10  21  − = `   10  10 11

21 \ Profit % = 110 × 100 = 21% 10 11

2. (c) : Let the total no. of votes be x. Then, according to question, 30% of x + 15000 = 70% of x 30 70 x + 15000 = x 100 100  70 − 30  x or 15000 =   100  15000 × 100 ⇒ x = = 37500 40 \ Total no. of votes = 37500 ⇒

Number of votes of winning candidate = 70% of 37500 70 × 37500 = 26250 = 100 3. (C) : Let the salary in 1999 = ` x Then, salary in 2000 = ` [x + 20% of x] 6 x 5 36 6  6 Also, salary in 2001 = `  x + 20% of x  = ` x 25 5  5

= `

But, salary in 2001 = ` 26640 \ 36 x = ` 26640 25  26640 × 25  ⇒ x = `  = ` 18500  36 \ 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, 20x + (36 – x) × 10 = 6.60 × 100 or 20x + 360 – 10x = 660 ⇒ 10x = 300 ⇒ x = 30 5. (A) : Let the cost price be ` x. Then, gain × 100 gain % = C.P. Class 7

 100 × 5   100 × S.P.  or, C.P. =  =    100 + 20   gain% + 100  500 = 4.16 = 120 Now, cost price of 10 eggs = 4.16 i.e., Number of eggs in ` 4.16 = 10 Number of eggs in ` 1 =

10 4.16

10 × 5 = 12 4.16 6. (A) : Let the numbers be x and 2x. According to question, \ Number of eggs in ` 5 =

x+7 3 = 2 x + 7 5 ⇒ 5(x + 7) = 3(2x + 7) ⇒ 5x + 35 = 6x + 21 ⇒ x = 14 \ The greater number = 2 × 14 = 28. 7. (B) : In 60 days 210 men complete 1 work  1 In 1 day they complete   work  60  1  1 × 12 =   work  5 60 4 Now, work is left for (210 + 70) workers 5 = 280 workers Let the no. of days be x.

In 12 days they complete

Then, 280 × x =

4 × 210 × 60 5

4 × 210 × 12 = 36 days. 280 8. (C) : Let the third number be 100. Then, first number = 100 – 20% of 100 = 80 and second number = 75. \ Difference of first and second = 5. ⇒ x =

5 1 × 100 = 6 % 80 4 9. (A) : Let the number be x. According to question, x – 4 = 80% of x \ Required % =

⇒ x – 4 = or, x −

80 4 ×x = x 100 5

4 x=4 5

x =4 5 ⇒ x = 20 ⇒

15

10. (B) : Selling price of 1 horse = ` 4000 Profit % = 25. 100 × S.P. 100 × 4000 = = ` 3200 100 + Profit% 125 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) = ` 4800 Now, Cost price of 2nd horse = ` 4800 Selling price = ` 4000 \ Loss = ` 800 \ C.P. =

800 2 × 100 = 16 % 4800 3 11. (D) : If selling price = ` 144 Let the cost price of jug be ` x. Loss % =

1 x of x = 7 7 Loss = C.P. – S.P

Then, Loss = Now,

x = x − 144 7 6x or 144 = 7 ⇒ x = ` 168. Since cost price of jug = `168. Now, selling price = ` 189 \ Profit = `(189 – 168) = ` 21 ⇒

21 × 100 = 12.5% 168 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, Profit = ` (100 – 80) = ` 20 \ Profit% =

20 × 100 = 25% 80 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 \ Profit = ` (192.50 – 175) = ` 17.50 And, Profit % =

17.50 × 100 = 10% 175 14. (C) : In 9 days P can complete 1 work Profit % =

 1 \ In 1 day P can complete   work.  9 Now, Q is 50% more efficient than P \ In 1 day Q can complete

16

=

1 1 + 50% of 9 9

1 1 3 1 + = = 9 18 18 6 \ No. of days in which Q can complete the work

=



=

15. (B)

1 = 6 days. 1 6

16. (C) : Let the selling price of shirts be ` x. 1 25 % Profit = 12 % = 2 2 100 × x 200 x 8  = \ Cost price = = ` x  9  25  225   100 +  2 \ Profit = S.P. – C.P. 8 1  = x − x =` x  9 9 Now, cost price of pants = Selling price of shirts = ` x \ Selling price = Profit =

120 × x x (100 + 20) 6  = ` x  = 5  100 100

6 1  x − x = ` x  5  5

1 1 x + x = 700 5 9 14 x or, = 700 45 ⇒ x = 2250.

Now,

8  \ Cost price of shirts = `  × 2250 = ` 2000 9  17. (B) : 1 day = 24 hrs. 1 hr = 60 min. \ 1 day = (60 × 24) min = 1440 min. 36 × 100 = 2.5% 1440 18. (D) : Selling price of article = ` 450 Loss % = 20. Required % =

\ Cost price =

100 × S.P.

(100 − Loss%)

100 × 450 100 × 450 = =` 562.50 (100 − 20) 80 Now, if he makes a profit of 20% then,

=

(562.50) × (100 + 20)

=` 675 100 19. (A) : Let the two cars starts from the point P. Now, In D PST, PS2 + ST2 = PT2 ⇒ PT2 = (62 + 82) = 100 ⇒ PT = 10 miles Similarly, PR = 10 miles. \ Distance between car A and car B is (10 + 10) miles = 20 miles S.P. =

IMO Work Book Solutions

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

26. (B) : The present value of machine = ` 100,000 The value is depreciates 5% every year. \ The value of machine after 1 year = 100000 – 5% of 100000 = 100000 – 5000 = ` 95000 And the value of machine after 2 years = 95000 – 5% of 95000 = 95000 – 4750 = ` 90,250.

2 2  Then, Alok have to save ` of x  = ` x 5  5 2 x 5 × 100 = 40% \ Required % = x 21. (D) : Let the number of boys be x. 2 Then, the number of girls = 14 %of x 7 100 1 x x× = = 7 100 7 Now, total no. of students = 560.

27. (A) : Let the number be x. Then, 200% of x = 20 200 × x = 20 100 ⇒ x = 10

⇒

x \ x + = 560 7 8x = 560 or, 7 ⇒ x = 490

50 × 10 = 5 100 28. (C) : Let the marked price be ` 100 Then, Selling price = ` (100 – 10% of 100) = ` 90. Profit % = 5 Now, 50% of 10 =

And no. of girls = (560 – 490) = 70. 22. (B) : Cost price of radio = ` 600 Profit % = 25 \ Selling price =

600 × (100 + 25) = ` 750 100   S.P. =  C.P. × (100 + g%)    100  

23. (B) : Total no. of people = 12000 No. of people wearing same colour = 4800 4800 × 100 = 40% 12000 24. (B) : Let the maximum marks be x. According to question, 36% of x = 113 + 85 \ Required % =

36 × x = 198 100 198 × 100 ⇒ x = 36 ⇒ x = 550 \ Maximum marks = 550 ⇒

100 × 90 \ C.P. = = ` 85.71 105 Now, M.P. – C.P. = `(100 – 85.71) = ` 14.29 2 14.29 × 100 = 16 % 3 85.71 29. (C) : Let the cost price be ` 100. Then, Marked price = 100 + 10% of 100 = ` 110 And, selling price = 110 – 10% of 110 = ` 99 \ This type of deal bears loss. \ Required % =

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) = ` 0.8x Now, Amount of milk purchased in ` 0.8x = 1 litre  1  × x  litres \ Amount of milk purchased in ` x =   0 .8 x

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

=



= 1.25 litres

 1.25 − 1 \ Required % =   × 100 = 25% 1 

78 × 1200 = ` 936.00 100 vvv

Class 7

17

Chapter-7 : Algebraic Expressions −2 2   8   −1  1. (A) : ( −5 x 2 y ) ×  xy z ×  xyz 2  ×  z   4   3   15 −2



8

−1

× ×  × (x2 × x × x ) × (y × y × y ) × =  −5 × 3 15 4  (z × z2 × z ) −4 4 4 4 x y z = 9

Bring all the like terms together.  a 2 2a 2   b 3 3b 3   −c 3 4c 3  + + a2  +  + + b3  +  − + c3     3   4  2 3 4 5

 3a 2 + 4a 2 + 6a 2   4b3 + 9b3 + 12b3   +  +  6 12

= 

 −5c 3 − 16c 3 + 20c 3    20

=

13 2 25 3 1 3 a + b − c 6 12 20

3. (A) 2

1 1   4. (C) :  2 x + −  2x −    3y  3 y 

2

1 1  1 1  =  2 x + + 2x − 2x + − 2x +    3y 3y   3y 3 y 

{ a2 − b2 = (a + b ) (a − b )}

= 4 x ×  5. (A) :  x +  Now,

2 8x = 3y 3y

2

7. (D)

2

=

p 2 267 p + 18 − 2 40

8. (C) : Let us bring the like terms together. 2y y y   −5 2 2 2   − −  + y + y + y2 +  2y +    3 3 3 3 3

18

9. (C) : Sum = (8a – 6a2 + 9) + (–10a – 8 + 8a2) = 2a2 – 2a + 1 and Difference = –3 – 2a2 + 2a – 1 = –2a2 + 2a – 4 10. (B) : When x = 15 and y = 3, we have, 9 × (15)2 + 49 × (3)2 – 42 × 15 × 3 = 2025 + 441 – 1890 = 576 −b , we get, a 2 b2 b2  −b   −b  +c a  + b  + c = a × 2 −  a   a  a a b2 b2 − +c = c = a a 12. (D) : At t = 4, we have, d = –16 (4)2 + 1000 = –256 + 1000 = 744

11. (C) : At x =

13. (D) : 3x4 – 12y4 = 3{x4 – 4y4} = 3[(x2)2 – (2y2)2] = 3{(x2 – 2y2) (x2 + 2y2)} 2

1 1   =  x +  − 4 = 144 – 4 = 140 x x 1   \  x −  = 140  x 2 24 15  p  4p   5p 6. (D) :  − − 6 = − 3 ×  p− p + 18   5   8 2 5 8

= 2y + 4

 a 2 − b 2 = (a + b )(a − b )

1  = 12 x

  x − 

 5y 3 − y 3 − 4y 3    + 4 2

2

 a 2 b3 c 3   2a 2 3b3 4c 3  2 3 3 2. (C) :  + −  + + −  + (a + b + c )  2 3 4  3 4 5 



2 2 2  6 y + 2y − y − y   −5 y + 2y + 3 y  + =     + 3 3

 5 3 y3  − 2y 3  + 4  y −  2 2

1 1 1  2 14. (D) :  3 x − − 2 × 3x ×  = 9 x + 2  2x 2x 4x  (a − b )2 =    a 2 + b 2 − 2ab  2 ⇒ (6)2 = 9 x + 2 ⇒ 9 x +

1

1 4x 2

−3

= 36 + 3 = 39 4x 2 15. (A) : Required expression = (3x2 – 4y2 + 5xy + 20) – (–x2 – y2 + 6xy + 20) = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20 = 4x2 – 3y2 – xy 16. (A) : x – x8 + x2 – 1.7x10 + 1.4 x8 – 7.8x2 + 4 – 9x = 4 + (x – 9x) + (x2 – 7.8x2) + (–x8 + 1.4x8) – 1.7 x10 = 4 – 8x – 6.8x2 + 0.4x8 – 1.7x10 17. (C) 18. (C) : 4st (s – t) – 6s2 (t – t2) – 3t2 (2s2 – s) + 2st (s – t) = 4s2t – 4st2 – 6s2t + 6s2t2 – 6t2s2 + 3st2 + 2s2t – 2st2 IMO Work Book Solutions

= (4s2t – 6s2t + 2s2t) + (–4st2 + 3st2 – 2st2) = –3st2

= a4 + 4a2b2 + b4 – a4 + 8a2b2 – b4 = 12a2b2 24. (C) : (2l – 3m)2 = (–1)2 ⇒ 4l2 + 9m2 – 12lm = 1 ⇒ 4l2 + 9m2 – 12 × 20 = 1 ⇒ 4l2 + 9m2 = 241

19. (C) : (a3 – 2a2 + 4a – 5) – (–a3 – 8a + 2a2 + 5) = a3 – 2a2 + 4a – 5 + a3 + 8a – 2a2 – 5 = 2a3 – 4a2 + 12a – 10 20. (D) : We have, x = 5 and y = x + 7 = 5 + 7 = 12. Now,

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

2 2 x 2 + y 2 = (5) + (12) = 25 + 144



(2)3

=

169 = 13

27. (A)

(2)2

21. (B) : 2 × –2× +2–a=5 or 16 – 8 + 2 – a = 5 ⇒ 10 – a = 5 ⇒ a = 5  1  1   2 1  22. (C) :  x −   x +    x + 2  x x x   1 1  2  2  =  x − 2   x + 2   x  x 

28. (B) : If 4l2 + ( k + 10)ml + 25m2 is a perfect square. Then, 4l2 + (k + 10)ml + 25m2 must be equal to 4l2 + 2 × 2l × 5m + 25m2 ⇒ k + 10 = 20 ⇒ k = 10  (a − b ) (a + b )  

= a2 − b2

1  =  x 4 − 4   x  23. (A) : Required expression = (a4 + 4a2b2 + b4) – (a4 – 8a2b2 + b4)

 

29. (D) x 3 = y 4  6 1 7 Then,  +  = = 1  7 7 7

30. (C) :

vvv

Class 7

19

Chapter-8: Exponents and Powers 1. (A) : 23 + 23 + 23 + 23 = 4 × (23) = 22 × 23 = 25 −1  −1

 2. (C) : 6 −1 +  3    2  

=

{ } {} 1 2 + 6 3

x

5

 5  125   125  3. (C) :  =  ×  2  8   8   5 ⇒    2

3×5

 5 ×   2

3x

15 + 3 x

⇒ y = 1 Now, (0.25)y = (0.25)1 = 0.25

 5 =   2

−1

5 6

=

−1

=

 5 12. (B) :    3

6 5

 5 ⇒    3

18

18

18

5

 1  −  8 4. (D) : ÷ 4  1  1  −   −  4 2  1 =  −  2

5−4

3

 1  1 ÷  ×   2  2 3

 3  p   2 5. (A) :   =   ÷    2    q 3  p \    q

−10

= (1)

−10

−3

−2

1

1

 1  1 =  −  ÷   = –1  2  2 3

3

 2  2 =   ÷  =1  3  3

=1

−3 −3  −3  6. (A) :  1  −  1   ÷  1   2   4  3 

{

}

3 3 3 = (3) − (2) ÷ ( 4) 19 = {27 − 8} ÷ 64 = 64 7. (A)

8/3

÷3

2

)×3

3/2

=



 = 3 3



13 / 6 2/3+3/ 2 = (3 ) = (3 )

8   − 2

× 33 / 2

9. (D) : (–8)5 + (–8)5 = 2 × (–8)5 = 2 × (–2)15 = (–1)15 × (2)16 = –(2)16 = –(4)8 3y

 1 11. (B) :   = 0.008  5 ⇒ (0.2)3y = (0.2)3

20

 5 =   3

 5 =   3

5 5 ⇒   =    3  3 or 8 + x = 6 ⇒ x = –2

8+ x

8+ x

8+ x

13. (A)  a −2 × b −3  14. (D) :  −3 = (a −2+ 3 × b −3 + 4 ) = a × b = ab  a × b −4   2  −2  15. (A) :   −1     2   

−1

{}

 1 =  4

−2  −1

 

−1

= [16]

=

1 16

16. (D) : 3n = 81 Squaring both sides, we get, 3n = (81)2 ⇒ 3n = (3)4×2 ⇒ 3n = 38 ⇒ n = 8 17. (D) : 8x–1 = 2x+3 ⇒ (2)3(x – 1) = 2x+3 On comparing, 3(x – 1) = x + 3 ⇒ 3x – 3 = x + 3 ⇒ 2x = 6 ⇒ x = 3 3



10. (B)

−5 +11

11

−1  −1 18. (A) :   =  27  19683

8. (B) : (94 / 3 ÷ 272 / 3 ) × 33 / 2 = (32× 4 / 3 ÷ 33 × 2 / 3 ) × 33 / 2

(3

 5 ×   3

6

 5 ⇒  5  =   2  2 On comparing the powers on both sides, we get, 15 + 3x = 18 ⇒ 3x = 18 – 15 = 3 ⇒ x = 1  1  −  2

−5

−1 −1   19. (B) :  3  −  1    4   4 

 32  20. (A) :   243 

−3 / 5

3

 25  =  5 3 

 7  7 21. (A) :   ×    5  5 3+ x + 2

−1

x +2

=

−3 / 5

 7 =   5

 7  7 ⇒   =   5  5 ⇒ x + 5 = 8 ⇒ x = 3

{ } { } 4 −4 3

 2 =   3

−1

=

5×

−3 5

−8 3

−1

 2 =   3

=

−3 8

−3

=

27 8

8

8

IMO Work Book Solutions

2

 3  3 22. (C) :   ×    2  2 ⇒

 3   2

a +5+ 2

or a + 7 = 8 ⇒ a = 1

a+5

 3 =   2

 3 =   2

8

8

23. (A) : Let the number be x. Then, (–8)–1 × x = (10)–1  1  −1 ⇒   × x =    10   8  −8  ⇒ x =    10  −4 ⇒ x = 5 24. (A) : Let the number be x. Then, (–15)–1 ÷ x = (–5)–1 −1 −1 ⇒   ÷ x =    5  15   −1 ⇒   × ( −5) = x  15 

0

2

−9 64 28. (B) : Let the number be x. Then, x × 4–3 = 64 ⇒ x × 4–3 = 43 43 ⇒ x = −3 4 ⇒ x = 43 + 3 ⇒ x = 46 ⇒ x = 212

= –9 × (2)–6 =

1

 1 2 2 29. (A) :    1/ 2 2 2 ) ( ) (   =  (16)  4

{

3

 3  3 30. (C) :   ×    5  5 3 ⇒    5

1 = 3 −1 3 25. (C) : ab – ba = 37 – 73 = 2187 – 343 = 1844

3−6

−3

−6

 3 =   5

 3  3 ⇒   =    5  5 ⇒ 1 – 2x = 3 ⇒ 2x = –2 ⇒ x = –1

3x

= 500 3 x 500 ⇒ 3x – 2 = = 9 32

1

1/ 2

}2 

2× = (4) 2 = 4



⇒ x =

26. (C) :

5

 −1  3  23  27. (A) :   ×   × 23 ×    2  4  25  = 1 × –1 × (2)–5 × 23 × 32 × (2)–4 = –9 × (2)–5 + 3 – 4

 5 =   3

1− 2 x

−(1− 2 x )

−(1− 2 x )

or –3 = –(1 – 2x)

vvv

Class 7

21

Chapter-9 : Perimeter and Area 1. (C) : Area of trapezium ABCD 1 = × (Sum of parallel sides) × height 2 1 = × (15 + 24) × 10 2  390  cm2 = 195 cm2 =   2  2. (A) : 1 decameter = 10 m. \ 2.4 decameter = 24 m. Now, Area of ||gm = Base × Height ⇒ 576 = 24 × PT ⇒ PT = 24 m.

9. (D) : Let the radii be 3x and 4x.

}

 1 1 = (18 × 10) − × 6 × 10 + × 10 × 8 2 2  = [180 – (30 + 40)] cm2 = (180 – 70) cm2 = 110 cm2

 2  cm 

4. (C) : Area of roads = [(175 × 5) + (225 × 5) – (5)2] m2 = (875 + 1125 – 25) m2 = 1975 m2 \ Cost of levelling the roads = `(1975 × 3) = ` 5925 5. (B) : Height of minute hand = 8.4 cm \ Radius of figure = 8.4 cm. Now, Area swept in half an hour = Area of semi-circle 1  1 22  × πr 2 =  × × 8.4 × 8.4 cm2 = 110.88 cm2 2 7  2 6. (B) : Since there is no gain or loss in the length of the wire. \ Perimeter of square = Circumference of circle ⇒ 4 × side = 2pr

=

22 ×r ⇒ 4 × 6.25 = 2 × 7 ⇒ r = 3.98 cm 7. (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) m2 = 2800 m2 Similarly, Area of inner rectangle = (60 × 30) m2 = 1800 m2 \ Area of lawn = (2800 – 1800) m2 = 1000 m2

22

2 × π × 3x 3 = 2 × π × 4x 4 10. (B) : Area of shaded portion = Area of rectangle – {Sum of areas of D DGF and D GAE} \ Required Ratio =

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

{

8. (A) : Area of ||gm ABCD = AB × DL ⇒ 156 = 13 × DL ⇒ DL = 12 cm Now, In D DLA, DL2 + LA2 = AD2 [By Pythagoras theorem] 2 2 2 ⇒ LA = (13) – (12) = 169 – 144 = 25 = (5)2 ⇒ AL = 5 cm.

{

}

 1 1 × 10 × 15 + × 10 × 15 = 30 × 20 − 2 2  = [600 – (75 + 75)]cm2 = (600 – 150) cm2 = 450 cm2

 2  cm 

11. (B) : Let the side of D ABC be 2x. Then, AB = 2x, BC = x In D ADB, 2 (2x)2 = x + ( 6 ) ⇒ 4x2 – x2 = 6 ⇒ 3x2 = 6 2

⇒ x =

2 cm

\ AB = 2 2 cm = BC = AC. Now, 1 × BC × AD 2 1  =  × 2 2 × 6  cm2 2 

Area of D ABC =

= 2 3 cm2

12. (C) : Area of floor of the room = (13 × 9) m2 = 117 m2 \ Cost of carpeting = `(117 × 6.40) = ` 748.80

IMO Work Book Solutions

40 m

13. (B) :

1 × Areaof circle 4 1 22 × 14 × 14 = × 4 7 = 154 m2

Area of shaded region =

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 1   1 22 × 7 × 7 − × 7 × 7 cm2 =  ×  4 7 2 = (38.5 – 24.5) cm2 = 14 cm2 17. (D) : Required area = 5 × Area of 1 face of cube = 5 × (side)2 = {5 × 12 × 12} cm2 = 720 cm2 18. (A) : Diameter of cylinder = 150 cm.  150  \ Radius of cylinder =   cm = 75 cm. 2  Now, Length of outer edge of parapet = 660 cm. 660 × 7 2 × 22 ⇒ Radius of parapet = 105 cm. \ Width = (105 – 75) cm = 30 cm. ⇒ Radius of parapet =

19. (B) : Let the length of rectangle be ‘l’ m Then, breadth = (23 + l)m Now, Perimeter = 2(length + breadth) 206 = 2(l + 23 + l) ⇒ 206 = 4l + 46 ⇒ 4l = 160 Class 7

⇒ l = 40 m. and breadth = (40 + 23) m = 63 m. Now, Area = (63 × 40) m2 = 2520 m2 20. (A) : Area of shaded region = Area of rectangle ABCD – 4 × Area of D DSR 1   =  30 × 25 − 4 × × 12.5 × 15 cm2   2 = (750 – 375) cm2 = 375 cm2 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 = πr12

⇒

πr22

=

25 36

25 36 [where r1 and r2 are the radii of two circles]

r1 5 = r2 6

⇒

2πr1 r1 5 = = Now, Ratio of their circumferences = 2πr2 r2 6 24. (B) : We have, BC = 1.2 decameter = 12 m 130 m = 13m AC = 130 dm = 10 and AB = 5 m. Now, it is clear that, AB2 + BC2 = AC2 So, it is a right angled triangle. 1 1 × BC × AB = × 12 × 5 = 30m2 2 2 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 = 60° Now, In D ABD, ∠A = 60°, ∠B = 60° \ ∠A + ∠B + ∠D = 180° ⇒ ∠D = 60° So, D ABD is an equilateral triangle. Hence, all the sides are equal. \ AB = BD = AD = 6 cm. 26. (B) : \ Area =



23

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

 36  ⇒ Area ×  = ` 3240  1000 

1 1 × 34 × 12 + × 34 × 12 2 2 = (204 + 204) m2 = 408 m2

3240 × 1000 = 90000 m2 36 ⇒ Side = 300 m. \ Perimeter of field = 4 × 300 = 1200 m.

=

⇒ Area =

27. (D) : The length of the largest pole is the diagonal i.e., Length of pole =

(10) + (10) + (5)



=

100 + 100 + 25



= 225 = 15 m.

2

2

2

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

 360  per m2 = `  10000 



 36  = ` per m2  1000 

Now, Total cost = ` 3240 ⇒ Area × cost per m2 = ` 3240

75   \ Cost of fencing = `  1200 ×  = ` 900  100  29. (C) : Area of shaded portion = Area of larger semicircle – Sum of areas of smaller semicircles 1{ 2 = πR − 2 × πr 2 } [where R → radius of larger 2 circle, r → radius of smaller circle]

1 22  2 2 × (14) − 2 × (7)  2 7 1 22 × 98 = 154 cm2 = × 2 7 30. (C) : Required area =

= [(3 × 12.5) + ( 4.5 × 11) − (3 × 4.5)] m2 = 73.5 m2

vvv

24

IMO Work Book Solutions

Chapter-10: Data handling, Symmetry and Visualising Solid Shapes = Average of 6th and 7th term

1. (D) 2. (A) : 25% of 200 = 50 and no. of students wants to become an actor/actress or musician = 30 + 22 = 52. 3. (B) : First six prime numbers are 2, 3, 5, 7, 11 & 13 2 + 3 + 5 + 7 + 11 + 13 6 41 = 6.83 = 6 4. (D) : Let the numbers be x1, x2, x3, x4 and x5. According to question, Mean of numbers = 20 x + x 2 + x3 + x 4 + x5 i.e., 1 = 20 5 ⇒ x1 + x2 + x3 + x4 + x5 = 100 ⇒ x1 + x2 + x3 + x4 = 100 – x5 ...(i) Let ‘x5’ be the excluded number. Then, Mean of remaining numbers = 23 x + x 2 + x3 + x 4 i.e., 1 = 23 4 ⇒ 100 – x5 = 23 × 4 [from (i)] ⇒ x5 = 8. \ Mean =

6. (B) : Mean = 7

 55 + 90 + 40 + 80 + 20  11. (A) : Average marks =   = 57 5 Marks obtained × 100 12. (C) : Percentage = Total marks

 55 + 90 + 40 + 80 + 20  =   × 100  500



= 57%

13. (B) : Highest marks obtained by student = 90 Lowest marks obtained by student = 20 \ Ratio = 14. (D) : U.P.

90 = 9:2 20

15. (C) : Maharashtra. 16. (D) : Total no. of heads = 59 Total no. of toss = 100 \ No. of tails = (100–59) = 41

5. (C) : We are given that, Incorrect Mean of 9 observations = 35 ⇒ Incorrect sum of 9 observations = 35 × 9 = 315 Now, 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 =

 98 + 100  198 = 99 =   =  2 2 10. (B) : Mathematics.

378 = 42 9

41 100 17. (B) : Total no. of outcomes = 2 = {H, T} No. of favourable outcomes = 1 \ Probability =

\ Probability =

1 2

14 7 = 80 40 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 18. (A) : Required Probability =

3 1 = 6 2 20. (B) : Total no. of teams = 2 \ Probability =

6+8+5+ x+4 =7 5 ⇒ x + 23 = 35 ⇒ x = 12 i.e.,

\ Probability =

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

1 2

21. (D) :

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. th

 12   12  \ Median = Average of   and  + 1  2   2 Class 7

th

term

22. (C) 23. (A) 24. (C) 25. (B)

25

26. (C) 27. (D) 28. (B)

29. (D) : The figures formed are P, 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)

vvv

26

IMO Work Book Solutions

Chapter-11: Logical and Analytical Reasoning 1. (D) : Since, 2008 is the leap year. Therefore, no. of days = 366. 364 = 52. 7 \ 364th day will be Friday. and 366th day will be Sunday. Now,

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

2. (B) : We know that, OB2 = OA2 + AB2 ⇒ OB2 = 25 km. ⇒ OB = 5 km. Now, Distance between the starting point O and final point P is (25 – 5) km = 20 km.

15. (C) 16. (C) 17. (C) : 5 + 8 = 13 13 + x = 34 ⇒ x = 21 21 + 34 = 55 18. (A) : 5 men can do 1 work in 12 days.  1 1 man can do   work in 1 day.  60   1  1 × 10 = \ The work completed in 10 days =   6 60 \ Number of men required = 6.

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

19. (D) 20. (D) : Except 14, all are divisible by 5. 21. (C)

4. (A) :

22. (D) : The word is NURSE and the middle alphabet is ‘R’. 23. (A)

5. (B) 6. (B) : We see that in the given pattern smaller figure becomes larger in next step and a new figure is added. 7. (D) 8. (D) : Let the total number of people be x. Then, Number of people who work in fields =

9. (D)

24. (D) 25. (B) : In the given pattern the number of sides in the figure is decreased by 1 in each step. 26. (A) : In the given pattern the number of squares is decreased by 1 and the number of circles is increased by 1 in each step.

1 x 2

27. (D)

1 Number of people who do not work in fields = x 2 \ Number of people working in factories

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.

1 1 1 = × x= x 2 2 4

10. (B) 11. (D) : A is B’s brother means A is the brother of B. But B is C’s sister and C is D’s father means B is D’s aunt and A is D’s uncle.

28. (B) 29. (D) 30. (C) : 5 + 5 = 10 10 + 7 = 17 17 + 9 = 26 26 + 11 = 37 37 + 13 = 50 50 + 15 = 65

vvv Class 7

27

Chapter-12: Everyday Mathematics 1. (D) : Let the number of girls and boys be 4x and 3x. Then, 3x + 4x = 21 or, 7x = 21 ⇒ x = 3 \ Number of girls = 4 × 3 = 12 2. (A) : Let the total money be x. The total money spent by Dhruv 1  7 1 x =  x + x  = 3 4 12

7 5 x= x 12 12 3. (C) : No. of green apples = 54 \ No. of red apples = 30 + 54 = 84. And total no. of apples = 54 + 84 = 138 \ Money left = x −

84 14 = 138 23 4. (B) : Let the age of Ishika be ‘x’ years. Then, the age of grandfather = ‘4x’ years According to question, x + 4x = 100 or, 5x = 100 ⇒ x = 20 \ Ishika’s age = 20 years. \ Ratio =

5. (B) : Let the no. of boys and girls be 7x and 4x. According to question, 7x – 4x = 21 or, 3x = 21 ⇒ x = 7 \ Number of boys = 7 × 7 = 49 And number of girls = 7 × 4 = 28 \ Total no. of children = 49 + 28 = 77. 6 (A) : No. of bracelets with 36 beads = 4 4 × 180 = 20 36 7. (C) : Time taken to complete 100 m. distance = 20 sec. \ Time taken to complete 400 m. distance No. of bracelets with 180 beads =



 20  × 400 sec = 80 sec. =   100 

8. (D) : Percentage of attended people 315 × 100 = 75% 420 9. (D) : No. of video-tapes checked out = Total no. of tapes – No. of tapes present = 52 – 17 = 35.

28

=

+12

10. (C) :

12,

+12 24,

+12 36,

+12 48,

60

11. (D) : Let the total no. of questions be x. 3 No. of questions Monika answered correctly = x 4 3 x \ Percentage Required = 4 × 100 = 75% 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 first group = 60 \ No. of people who board the bus from first group 3 = × 60 = 45. 4 Also, No. of people in second group = 60 \ No. of people who board the bus from second 2 group = × 60 = 40. 3 \ Difference = 45 – 40 = 5. 14. (C) : S.P. = ` 54000 gain% = 20 100 × 54000 100 × 54000 = = ` 45,000 120 100 + 20 This C.P. is for the friend of the man and is the selling price for the man. Now, S.P. = ` 45000 Loss% = 10

\ C.P. =

100 × 45000 100 × 45000 = = ` 50,000 (100 − 10) 90 \ The original cost price is ` 50,000. \ C.P. =

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) = 745 or, 120 + 3x + 2x = 745 ⇒ 5x = 625 x = ` 125 \ Cost of 1 chair = ` 125 And 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 = ` 1440. 100 IMO Work Book Solutions

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

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.

7200 × 5 × 5 = ` 1800. 100 \ Difference = ` (1800 – 1440) = ` 360.

24. (B) : Total no. of persons in the park = 1 + 1 + 1 + (2 × 2) + 2 =3+4+2=9

\ S.I. =

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.

25. (B) : The time after which they toll together is the LCM of 36, 40 and 48.

18. (C) : The greatest possible size of the measuring vessel is the HCF of 1653 litres, 2261 litres and 2527 litres. Now, HCF of 1653, 2261 and 2527 is, 19 1653, 2261, 2527 87, 119, 133

\ HCF = 19. \ Capacity = 19 litres.

19. (B) : Let the total capacity of drum be ‘x’ litres. Then, we have, 3 7 x − 15 = x 4 12 3 7 x− x = 15 ⇒ 4 12 2x ⇒ = 15 12 ⇒ x = 90

\ Capacity = 90 litres. 20. (B) : Amount spend on education = ` (30% of 15000)  30  × 15000 = ` 4500 = `  100  21. (D) : S.P. = ` 285 Loss% = 5 100 × 285  100 × 285  \ C.P. = `   = `   = ` 300 100 − 5 95 ( )   Now, C.P. = ` 300 gain% = 15 300 (100 + 15) \ S.P. = = 3(115) = ` 345 100 22. (B) : Total no. of employees = 1600 \ No. of female employees = 60% of 1600 = 960 And, No. of male employees = 1600 – 960 = 640 Out of 640 males 50% are computer literate. \ No. of males who are computer literate = 50% of 640 = 320 Now, 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 Class 7

\ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720 seconds = 12 minutes. 26. (A) : Let the original amount be ` x. Then,  x Eldest son’s share = `    2 x  1 x Youngest son’s share = `  ×  = `  3 2 6  2x  x   x x \ Amount left = `  x −  +   = `  x − = `    3 2 6   3   27. (A) : Total no. of marbles = 8 No. of marble marked with 2 = 1 \ Probability =

1 8

28. (C) : S.P. = ` 3200 1 100 = 3 3  100 × 3200  100 × 3200 × 3  \ C.P. = `  = `  100    400    100 +   3   Profit% = 33



= ` 2400

29. (C) : Let the length of total journey be x km. Then, 2 x + 33 3 2 or, x − x = 33 3 x ⇒ = 33 3 ⇒ x = 99 x =

\ Total distance = 99 km. 30. (B) : The concentration of water in both the containers = 80 % and 75% \ Ratio =

80 = 16 : 15 75

29

2012 - 6th SOF IMO 19. (C) : In winters, we use quilt and in this code language quilt is called mosquito net.

1. (B) : RAGER 2. (d) : A

3

20. (D)

3. (b)

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

4. (d) : Z = 26 × 2 = 52 ACT = (1 + 3 + 20) × 2 = 48 Now, BAT = (2 + 1 + 20) × 2 = 46 5. (C)

7. (B)

W

10m

6. (C) :

N 15m E 25m S

P

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 = 14 Arrange 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

30

22. (B) : Since, D XYZ is an equilateral triangle. \ ∠XZO = 60° Now, 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, 9 of x = 45 5 ⇒ x = 25 Now, 1 × 25 = 5 5 25. (D) : Area of ||gm PQRS = Area of D PQR + Area of D RPS 1 1 × 24 × 6 + × 24 × 6 = 144 cm2 2 2 26. (C) : After 4 years, John’s sister’s age = 12 + 4 = 16 years. John’s 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 = 105° Now, 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 × 70 280 4 × 30 120 = ; = ; 3 × 70 210 7 × 30 210 The order is, 280 147 120 , , 210 210 210 \

7 × 21 147 = 10 × 21 210

IMO Work Book Solutions

37. (C)

\ The descending order is, 4 7 4 , , 3 10 7

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 \ Profit = `(12800 – 10800) = ` 2000

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

\ Profit % =

9 cm. 2 \ Circumference = 2pr \ Radius =



2000 14 × 100 = 18 % 10800 27

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

2 22 9   = 2 × ×  cm = 28 cm  7 7 2

30. (C) : It is given that, AB = AF \ ∠AFB = ∠ABF = p (say) Now, In D ABF, 36° + p + p = 180° ⇒ p = 72° Now, BC is a straight line \ ∠BFA + ∠AFE + ∠EFG = 180° ⇒ 72° + 58° + ∠EFG = 180° ⇒ ∠EFG = 50° In D EFG, FG = FE \ ∠FGE = ∠FEG Now, ∠FGE + ∠FEG + ∠EFG = 180° ⇒ 2∠FGE = 180° – 50° = 130° ⇒ ∠FGE = 65° Now, again BC is a straight line. \ ∠FGE + x = 180° ⇒ x = 180° – 65° = 115°

P × 12 × 2 100 ⇒ 1620 × 100 = P × 12 × 2 ⇒ P = ` 6750

\ S.I. =

40. (B) 41. (D) : 6 glasses =

3 of jug 5

3 of jug 30 3 3 1 = \ Amount left = − 5 30 2 42. (B) : It will be a proportion, i.e., \ 1 glass =

4 x = 9 324 4 × 324 9 ⇒ x = 144 \ 144L of orange syrup is required. ⇒ x =

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

43. (C) : Let the cost of skirt be ` x. Then, the cost of bag = 55% of x =

32. (A) : Length of painting = (24.5 – 2 × 3) cm = 18.5 cm Breadth of painting = 6 cm. \ Area of painting = (18.5 × 6) cm2 = 111 cm2

According to question,

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

⇒ x = 31.50 +

Cost of skirt = 31.50 + Cost of bag

25 1 = 3600 144 34. (A) : Let the number required be x. Then, \ Fraction =

⇒

1 of x = 166 2 ⇒ x = 332

⇒ \ and, \

30 × 332 = 99.6 100 35. (C) : a = 40° (Alternate angles) \ 30% of 332 =

36. (B) : Total amount of expenditure = 100 Amount of expenditure on research & development =5 \ Total expenditure is 20 times the expenditure on research and development. Class 7

55 11 × x =` x 100 20

11 x 20 11 x− x = 31.50 20 x = ` 70 Cost of skirt = ` 70 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 = 3x. According to question, x + 3x = 60 ⇒ 4x = 60 ⇒ x = 15 \ Jasmine has 45 phonecards.

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45. (A) : Distance covered in 20 min. = 18 km. \ Distance covered in 1 hr 20 min., i.e., 80 min. 18 × 80 = 72 km 20 46. (A) : Area of path = Area of square ABCD – Area of square PQRS = [80 × 80 – 72 × 72] m2 = 1216 m2 =

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) = ` 720 Now, gain % = 15  720 × 115   720 × (100 + 15)  \ S.P. = `  = `   100  100   = ` 828

49. (D) : Let x be the distance between the buildings AB and CD. It is given that, OB = OD = 17 m. CD = 15 m. and, AB = 8 m. In D AOB, AB2 + AO2 = OB2 ⇒ AO2 = OB2 – AB2 = (17)2 – (8)2 = 225 \ OA = 15 m. Similarly, In D COD OC2 = OD2 – CD2 = (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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IMO Work Book Solutions