International Mathematics Olympiad
WORK BOOK
6
INSTANT
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Chapter - 1: KNOWING OUR NUMBERS 1.
(B) : In the given number 605, 116, 218 ; 5 is at Million’s place. Therefore place value of 5 = 5 × 10 6 = 5 million.
2.
(C)
3.
(B) : For the formation of greatest 4 digit number using different digit we have 4 places. Thousand’s place, Hundred’s place, Ten’s place, One’s place 9 ® The largest digit at thousand’s place 8 ® The second largest digit at hundred’s place 6 ® Given that 6 is at ten’s place 7 ® Third largest digit at unit’s place \ Number is 9867.
4.
(A) : Mustard oil in a bottle = 920 ml Mustard oil in 25 bottles = 25 × 920 ml = 23000 ml 23000 1 l = 23 l l , then 23000 ml = 1000 1000 (B) : Difference of 500 and 200 = 500 – 200 = 300 Since C = 100, then CCC = 300.
Since 1 ml =
5. 6.
(B) : Roman numeral starts from I.
7.
(D) : Expanded form of 37034 = 3 × 10000 + 7 × 1000 + 0 × 100 + 3 × 10 + 4 × 1 = 30000 + 7000 + 0 + 30 + 4.
8.
(D) : 10 = 10 + 9 = X + IX = XIX.
9.
(D) : Capacity is not measured in centimetres as it is measuring unit of length.
10. (C) : 498 = (500 – 100) + (100 – 10) + 8 = CDXCVIII. 11. (B) : To write a number in which the smaller digit placed before the greater digit, we subtract the value of the small digit from that of the greater digit. 12. (C) : XX = 10 + 10 = 20, XXXVI = 10 + 10 + 10 + 6 = 36, V = 5 Ascending order = V, XX, XXXVI. 13. (D) : CDI + V + C + X = ((500 – 100) + 1) + 5 + 100 + 10 = (400 + 1) + 5 + 100 + 10 = 516. 14. (C) : Underlined digit is at the hundred’s place.
Class 6
15. (A) : Using the place value chart, we may write
Ten Crores Crores 3
1
Ten Lakhs Ten Thousands Hundreds Tens Lakhs Thousands 5
5
3
2
8
Ones
0
0
Hence the standard numeral = 31,55,32,800. 16. (B) : Greatest 7 digit number = 9999999 Smallest 5 digit number = 10000 Difference between them = 9999999 – 10000 = 9989999. 17. (B) : Tens place comes on immediate right of the hundreds place. 18. (C) : Roman Numbers
I
V
X
L
C
D
M
HinduArabic Numerals
1
5
10
50
100
500
1000
Since C is lying between two roman numerals greater than itself therefore C is subtracted from digit at right MCDXVIII = 1000 + (500 – 100) + 10 + 8 = 1000 + 400 + 18 = 1418. 19. (B) : Million period in international number system is given below :
Hundred Ten Millions Millions Millions 8
10
7
10
Ten Crores Crores
6
10
Ten Lakhs
20. (D) : Largest number using 4, 0, 3, 7 = 7430 Smallest number using 4, 0, 3, 7 = 3047. 21. (C) : The face value of a digit in a numeral is its own value, at whatever place it may be. Therefore face value of 3 in 31005660 is 3. 22. (C) : Beginning from the right, a comma is put after the first three digits and then after two digits. 23. (C) : Sugar distributed among 12 persons = 55 kg 200 g = (55000 + 200)g = 55200 g \ Sugar received by 1 person = 55200 ÷ 12 = 4600 g = 4 kg 600 g. 24. (B) : Place value and face value are always equal at one’s place.
IMO Work Book Solutions
25. (A) : In international system 315233729 can be written as Three hundred fifteen million, two hundred thirty three thousand seven hundred twenty nine. 3 1 5 2 3 3 7 2 9 Units Tens Hundreds Thousands Ten Thousands Hundred Thousands Millions Ten Millions Hundred Millions
26. (B) 27. (B) : Since X is lying between C and L therefore X is subtracted from L \ CCCXLVII = 100 + 100 + 100 + (50 – 10) + 7 = 300 + 47 = 347. 28. (B) : Since it is given that digit cannot be repeated therefore three digit number are 350, 305, 550, 503 i.e. 4 numbers. 29. (A) : Number of screws produced in a day = 2825 Number of screws produced in 30 days = 2825 × 30 = 84750 Number of screws 5 dealers got = 84750 Number of screws each dealer got = 84750 ¸ 5 = 16950 . 30. (A) : Number name for 900800700 is ninety crores eight lakhs and seven hundred. 9 0 0 8 0 0 7 0 0 Units Tens Hundreds Thousands Ten Thousands * Lakhs Ten lakhs Crores Ten crores
vvv
Class 6
Chapter - 2 : NUMBER SYSTEM 1.
(C) : 2 is the only even prime number.
2.
(D) : Definition
Example
(A) Closure property : A set is closed under an operation when the result of the operation on any two members in the set is also in that set.
Let a = 1, b = 2, c = 4 (A) a ÷ b = 1 ÷ 2 = 0.5 (not a whole number) \ Closure property is not satisfied under division (B) Commutative property : An operation is (B) a ÷ b = 1 ÷ 2 = 0.5 commutative if you can change the order b ÷ a = 2 ÷ 1 = 2 without affecting the result. \ Commutative property is not satisfied under division. 1 (C) Associative property : If a, b, c are three (C) (a ÷ b) ÷ c = (1 ÷ 2) ÷ 4 = 8 whole numbers then it should satisfy a ÷ (b ÷ c) = 1 ÷ (2 ÷ 4) = 2 (a * b) * c = a * (b * c) under the operation * \ (a ÷ b) ÷ c ¹ a ÷ (b ÷ c) \ Associative property is not satisfied under division. 3. 4. 5.
(B) : As explained in Q. 2, Associative property is used between three numbers. (C) : Remainder r can be zero or less than divisor. (B) : Whole number : Starts with 0 and goes to infinity. Natural number : Starts with 1 and goes to infinity. ‘0’ is the least whole number and there in no largest definite whole number.
6.
(B) : As explained in Q. 3, the given equation satisfies the associative property.
7.
(B) : a and b are whole numbers and a > b and c ¹ 0, c > 0 so a × c > b × c. For example : a = 2 and b = 1 as a > b Let c = 3 ¹ 0 and 3 > 0, then a × c = 6 > 1 × 3 = 3 = b × c.
8.
(C) : Face value of 5 in 165,234 = 5 Face value of 9 in 842,928 = 9 Difference between them = 9 – 5 = 4.
9.
(D) : Let a, b be two whole numbers (A) Closure property : a, b are two whole numbers, but a – b may not be a whole number. (B) Commutative property : a, b are two whole numbers but a – b ¹ b – a (C) Associative property : a, b and c are three whole numbers but (a – b) – c ¹ a – (b – c).
IMO Work Book Solutions
10. (B) : Number of red balls in basket = 222 \ 222 ÷ 6 = 37 groups of 6 balls are replaced by 12 white balls 37 times. \ Number of white balls = 37 × 12 = 444 11. (A) : Multiplicative Identity : The result of multiplying a number by 1 is equal to the number itself, then 1 is the identity element with respect to multiplication. It is also called the multiplicative identity. 12. (D) : 82 × 1 = 82. 13. (A) : Commutative law : If a and b be two whole numbers then, a – b = b – a. It is possible when a = b. 14. (D) 15. (C) : If a × b = 0 then either a = 0 or b = 0. 16. (D) : In option (D) it should be reciprocal of 0 which is not defined. 17. (A) : On the number line negative integers are on the left of 0 and positive integers are on the right of 0. Hence positive integers > zero > negative integer. 18. (D) : Let a and b be two integers. Let a = –6, a × b = –48 Þ –6 × b = –48 Þ b = 8 19. (A) : (–7 + (–2)) = (–7 –2) = –9 Integer 2 more than –9 = –9 + 2 = –7. 20. (A) : Positive sign preceding the bracket does not change the sign of term when bracket is removed. 21. (B) : L.H.S. = (–25) – (–42) – (–27) = – 25 + 42 + 27 = 44 R.H.S. = (–42) – (–25) + (–22) = – 42 + 25 – 22 = – 39 \ (–25) – (–42) – (–27) > (–42) – (–25) + (–22). 22. (B) : Square of any –ve integer is positive. 23. (B) : Negative sign preceding the bracket changes the sign of terms when bracket is removed. 24. (A) : 3 + 4 – 2 changes to 3 × 4 ¸ 2 = 3 × 2 = 6. 25. (C) : We first remove the ( ), then { } and lastly [ ]. N A
26. (D) : W
3 km E 8 – 3 = 5 km
8 km
\
Final position is 5 km towards south.
B S
27. (B) : 0 is greater than negative integer and less than positive integer. If a > b, then –a 37, then –99 > > > > > 54 54 54 54 2 9 18 27
9.
(C) : (A) and (B) are false as fractions with equivalent numerator and denominator are called like fractions. (D) is also false as fraction with numerator less than denominator is called proper fraction.
10. (D) : According to given question
Þ
1 2 2 4 5 5 5 1 + ? + = + + = + + 15 15 15 15 15 15 15 15 3 11 11 3 8 +?= Þ ? = = 15 15 15 15 15
(On subtracting
3 from the both sides) 15
11. (A) : In figure (I) 5 squares are shaded out of 9 and in figure (II) 6 squares are shaded out of 9. Hence the given fractions will be
5 6 5 6 11 and \ Sum = + = 9 9 9 9 9
+ Fig. I
Fig. II
12. (C) : Number of children in the group = 7 (3 girls, 4 boys) Fraction of girls among 7 children =
3 . 7
IMO Work Book Solutions
13. (C) : Figures
Shaded fraction 4 1 = (4 parts are shaded out of 8) 8 2 1 (1 part is shaded out of 2) 2 1 (1 part is shaded out of 4) 4 3 1 = (3 parts are shaded out of 6) 2 6
14. (C) : Number of total marbles in bag = 3 + 1 + 6 + 2 = 12 Number of red marbles in bag = 3 Fraction of red marbles in bag =
3 12
1 2 5 < < [Numerators are in ascending order and denominators are same] 8 8 8 1 4 3 3 1 2 4 2 1 < > ; > < ; > > 2 2 2 4 4 4 16 16 16
15. (A) :
16. (D) 17. (D) : 3
é 3 ì 1 æ 1 1 ö üù 1 - ê1 + í2 - ç1 - ÷ ýú 12 ë 4 î 2 è 2 3 ø þû
37 é 7 ì 5 -ê +í 12 ë 4 î 2 37 é 7 ì 5 = +í 12 êë 4 î 2 37 é 7 ì 5 +í = 12 êë 4 î 2
=
=
æ 3 1 ö üù çè - ÷ø ýú 2 3 þû 3 ´ 3 - 1 ´ 2 üù ýú 6 þû 7 üù ý 6 þúû
37 é 7 3 ´ 5 - 7 ù + ú 12 êë 4 6 û
37 é 7 8 ù 37 é 3 ´ 7 + 8 ´ 2 ù + = ú 12 êë 4 6 úû 12 êë 12 û 37 37 37 - 3 7 0 = = = = 0 12 12 12 12
=
18. (B) : Let total number of pages in book = x Piyush read = Pages left = 80
Class 6
3 3 x of x = 5 5
(Expanding the mixed fractions) (L.C.M. (2, 3) = 6)
(L.C.M. (2, 6) = 6) (L.C.M. (4, 6) = 12)
3 x 80 3 x + 80 ´ 5 3 x + 400 + Þ x = = 5 1 5 5 Þ 5x = 3x + 400 (Multiply by 5 on both sides) Þ 2x = 400 (Subtract 3x on both sides) Þ x = 200 (Divide by 2 on both sides)
Hence x =
19. (D) : Shaded fraction shown in Model I =
9 12
Model I
Model II 3 3 2 6 Shaded fraction shown in Model II = = ´ = 6 6 2 12 9 6 > And 12 12
20. (D) : 6 more equal sized pieces would be needed to make a whole for the given figure. 1 2 3
6 5 4
21. (D) :
Figure
Shaded Fraction
Unshaded fraction
A
1 4
3 4
Unshaded fraction of Figure A = Shaded fraction of figure B Total parts in figure B = 16 3 x x x 12 3 4 = ´ Þ = Let the shaded part of figure B is x then, = Þ Þ x = 12 4 16 16 16 16 4 4 22. (B) :
So,
Figure
Equal parts
Shaded parts
36
18
12
x
Fractional number 18 1 = 36 2 x 12
x 1 x 1 6 x 6 = Þ = ´ Þ = Þ x = 6 12 2 12 2 6 12 12
IMO Work Book Solutions
23. (C) : Alphabets K E S H A V
Number of straight lines = 3 = 4 S = 0 = 3 = 3 = 2
Þ Fraction of alphabets made of 3 straight lines =
3 . 6
24. (D) : Only alphabet S is made of semicircles Alphabets S
Number of semicircle = 2
Þ Fraction of alphabet made of semicircles =
1 . 6
25. (A) : 1 metre = 100 cm, Given length = 10 cm 10 1 = \ Required fraction = 100 10 26. (A) 27. (B) :
Figure
Shaded fraction
6 ; 6 parts are shaded out of 15 parts 15
2 ; 2 parts are shaded out of 5 parts 5
Hence
6 6¸3 2 = = 15 15 ¸ 3 5
3 8 3 ö æ 28. (C) : Reena eats ç1 + ÷ chocolate bar = 1 = è ø 5 5 5 5 x y = = 7 35 49 5 x = 7 35
29. (A) : Þ
Class 6
[Given] [Take first two fractions]
5 ´ 35 7 Þ x = 5 × 5 = 25
Þ
x=
Now take
5 y = 7 49
5 ´ 49 7 Þ y = 5 × 7 = 35
Þ
[Multiplying 35 both sides]
y =
[Multiplying 49 both sides]
30. (A) : Quantity of milk bought = 7 Milk consumed = 5 Milk left =
1 litres 2
=
15 litres 2
3 23 litres = litres 4 4
15 23 15 ´ 2 - 23 30 - 23 7 3 = = = = 1 litres 2 4 4 4 4 4
vvv
IMO Work Book Solutions
Chapter - 7 : DECIMALS 1.
(B) : Weight of boy = 56.74 kg Weight of father = 1.5 times heavier than son = 1.5 × 56.74 kg = 85.11 kg
2.
(B) : 0.111 =
1 9 Multiply by 7 both sides, we get (0.111) ´ 7 =
1 7 ´ 7 Þ 0.777 = 9 9
3.
(D) : 0.005
