international maths olympiad practice book
TRANSCRIPT
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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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IMO Work Book Solutions6
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
According to question,
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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IMO Work Book Solutions8
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.
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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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IMO Work Book Solutions10
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
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IMO Work Book Solutions12
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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IMO Work Book Solutions14
= (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
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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 .
According to question,
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
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IMO Work Book Solutions16
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
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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
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IMO Work Book Solutions18
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
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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
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IMO Work Book Solutions20
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
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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
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IMO Work Book Solutions22
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
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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) :
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IMO Work Book Solutions24
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)
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IMO Work Book Solutions26
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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IMO Work Book Solutions28
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.
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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= :
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IMO Work Book Solutions30
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
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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 `
According to question,
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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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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