CHAPTER

3 Fractions

Learning objectives
3.1 Fraction 3.2 Conversion of Fractions
3.3 Equivalent Fractions 3.4 Reducing Fractions in its Simplest Form
3.5 Addition and Subtraction of Like Fractions 3.6 Addition and Subtraction of Unlike Fractions
3.7 Comparison of Fractions 3.8 Fraction of a Number

3.1 FRACTION
When a thing is divided into equal parts, then each part is called a fraction of the whole.

→ Whole (1) Olympiad Bite

A fraction always means numerator is divided by
 
denominator.
The denominator of a fraction tells us the total number
 
→ Divided whole into 3 equal parts of equal parts into which the figure is divided.
The numerator of a fraction tells us the number of
 
parts chosen.
1
→ Each part is of the whole
3

1 → Numerator
Here 1 ÷ 3 =
3 → Denominator

Proper Fractions
A fraction in which numerator is less than the denominator is called proper fraction.
4
For example : is a proper fraction.
7
Improper Fractions
A fraction in which numerator is more than the denominator is called improper fraction.
For example : 7 is an improper fraction.
4
Like Fractions
Fractions having same denominators are called like fractions.
3 4
For example : and are like fractions.
11 11
Unlike Fractions
Fractions having different denominators are called unlike fractions. (It doesn't matter what the
numerator is)
4 5
For example : and are unlike fractions.
9 13
Unit Fractions
A fraction with numerator as 1 is called unit fraction.
1 1
For example : and are unit fractions.
5 7

Mixed Fractions
Mixed fraction is a combination of a whole number and a proper fraction.

3 2 2 2
For example : Fraction of shaded part of is + = 1 + = 1 is a mixed fraction.
3 3 3 3

3.2 CONVERSION OF FRACTIONS

Conversion of Mixed Fractions into Improper Fractions

4
Convert 3 into improper fraction.
7
h Multiply the whole part by denominator, i.e., 3 × 7 = 21

h Add numerator to the result of previous step, i.e., 21 + 4 = 25

h Thesum obtained will be the numerator for the required improper fraction, whereas denominator
remains same.
4 7 × 3 + 4 25
So, 3 = =
7 7 7

Conversion of Improper Fractions into Mixed Fractions

13
Convert into mixed fraction.
5
h Divide the numerator by the denominator, i.e., divide 13 by 5, by long-division method.

2 Quotient
5 1 3
– 1 0
3 Remainder
h Write down the whole number (Quotient) result.
h Use the remainder as the new numerator over the denominator. This is the fraction part of the
mixed fraction.
13 3
So, =2
5 5
 SELF TEST - 1
3 3 4
1. 4 is same as ________. (C) (D)
8 7 7
32
(A) 35 (B)
8 8 4. Which of the following is a unit fraction?
35 32 5 1
(C) (D) (A) (B)
3 3 7 8
2. Which of the following will come in place
of (?)? 3 2
(C) (D)
4 5
41 ?
=4 5. The given figure shows ______ as shaded
9 9
(A) 2 (B) 4 fraction.
(C) 5 (D) 1
3. What fraction of the given figure is unshaded?

1 3
(A) 2 (B)
2
2 4
3 1
(A) 5 (B)
3 (C) 1 (D)
2
8 8 4 4

3.3 EQUIVALENT FRACTIONS

1
In the following collections, (one-third) of each collection is shaded. The fraction represented by
3
shaded portion of each collection is given below the figure.

1 2 4
3 6 12
The shaded portion of the figures are equal. So, the fractions which indicate the shaded portions should
also be equal, i.e., 1 = 2 = 4 and so on.
3 6 12
These fractions are called equivalent (equal) fractions.

To find a fraction equivalent to a given fraction

h To get a fraction equivalent to a given fraction, we multiply or divide the numerator and the
denominator of the given fraction by the same non-zero number.

For example : Find two equivalent fractions of 6 .

9
6 × 2 12 6÷3 2
= and =
9 × 2 18 9÷3 3
 12 2 6
So, and are equivalent fractions of .
18 3 9
3.4 REDUCING FRACTIONS IN ITS SIMPLEST FORM
h A fraction is said to be in its lowest/simplest form, if HCF of its numerator and denominator is 1.
h To convert a fraction into simplest form, divide the numerator and denominator by their HCF.
20
For example : Reduce to its lowest form.
25
Factors of 20 are 1 , 2, 4, 5 , 10 and 20.
Factors of 25 are 1 , 5 and 25.
Common factors are 1, 5. So, HCF is 5.
20 ÷ 5
\ Divide numerator and denominator of the given fraction by 5 i.e.,= = 4.
25 ÷ 5 5
4 20
Hence, is the simplest form of .
5 25

3.5 ADDITION AND SUBTRACTION OF LIKE FRACTIONS

Addition of Like Fractions
Sum of numerators
Addition of like fractions =
Common denominator
1 2 1+ 2 3
For example : + = =
5 5 5 5

Subtraction of Like Fractions

Difference of numerators
Subtraction of like fractions =
Common denominator
For example : 6 − 2 = 6 − 2 = 4
11 11 11 11

3.6 ADDITION AND SUBTRACTION OF UNLIKE FRACTIONS

Addition of Unlike Fractions

h To add unlike fractions convert them into equivalent fractions with same denominator and then
add.
1 2
For example : Add and .
4 3
h Convert fractions into equivalent fractions
1× 3 3 2×4 8
= and =
4 × 3 12 3 × 4 12
h Add the equivalent fractions
3 8 3 + 8 11
+ = =
12 12 12 12
So, 1 + 2 = 11
4 3 12
Subtraction of Unlike Fractions
h To subtract unlike fractions convert them into equivalent fractions with same denominator and
subtract.
5 1
For example : −
9 6
h Convert fractions into equivalent fractions

5 × 2 = 10 and 1 × 3 = 3
9 × 2 18 6 × 3 18
h Subtract
these equivalent fractions
10 3 10 − 3 7
− = =
18 18 18 18
5 1 7
So, − =
9 6 18
3.7 COMPARISON OF FRACTIONS

Comparison of Like Fractions

If the denominators of given fractions are same, then the fraction with greater numerator is greater.
5 2 1 3
For example : > and <
8 8 7 7
Comparison of Unlike Fractions
h If the numerators of two given unlike fractions are same, then the fraction with greater denominator
is smaller than the fraction with smaller denominator.
3 3 1 1
For example : > and <
8 11 6 5
h If the numerators of two given unlike fractions is different, then we convert the fractions into
equivalent fractions of same denominator and then compare.
2 3
For example : Compare and
7 5
LCM of 7 and 5 = 35
2 3 10  2 × 5 21  3×7 
Equivalent fractions of and with the denominator 35 are  i.e.,  and  i.e., 
respectively. 7 5 35  7 × 5 35  5×7 

10 21 2 3
Thus, < or <
35 35 7 5
3.8 FRACTION OF A NUMBER Olympiad Bite
Let’s learn how to find the fraction of a collection or a number
with the help of an example. 4 4
of 15 is same as × 15 = 12
3 5 5
For example : Find of 25 candies.
5
3
is same as 3 parts out of 5 equal parts.
5
Distribute 25 candies into 5 equal parts.

Now 3 equal parts will have

3 × 5 candies = 15 candies.
3
\ of 25 candies = 15
5

SELF TEST - 2
4 4
1. Which of the following is not equivalent to ? (C) 52 (D)
5 3
8 24
(A) (B) 30
10 30 4. The simplest form of is _______.
36
20 12
(C) (D) 5 5
25 20 (A) (B)
14 7 7 6
2. By how much is greater than ?
15 15 3 4
(C) (D)
1 1 5 7
(A) (B)
5 4 5. Compare and fill in the box.
7 3 5 3 1 3
(C) (D) + +
15 5 8 8 4 4
1 1 (A) <
3. Solve : of 125 + of 81
5 3 (B) =
7 (C) >
(A) 42 (B)
5 (D) Can't be determined
 EXERCISE
4 3 1 6. What fraction of the given figure is shaded?
1. Find the value of 3 + 4 − 1 .
5 5 5
1 1
(A) 3 (B) 2
5 5
1 1
(C) 7 (D) 4
5 5
2. Arrange the following in descending order.
9 1 5 4 10
, , , , 17 5
27 2 25 24 15 (A) (B)
25 17
10 9 5 4 1 4 5 9 1 10
(A) , , , , (B) , , , , 2 10
15 27 25 24 2 24 25 27 2 15 (C) (D)
5 21
10 1 9 5 4 4 1 10 5 9 7. Which of the following fractions represents the
(C) , , , , (D) , , , ,
15 2 27 25 24 24 2 15 25 27 shaded part of the given figures?
3. What fraction of the given figure is shaded?

1 1
(A) 1 (B) 1
4 3
2 3
(C) (D)
3 4
8. Which of the following is an improper fraction?
3 1 5 1
(A) (B) (A) (B)
4 2 12 9
1 1 3 7
(C) (D) (C) (D)
4 3 4 6
2 9. A florist has 12 roses. 2 of them are white, 3 of
4. Which one of the following fractions is less them are yellow and rest are red. What fraction of
15
4 the roses are red?
than ?
5 5 7
(A) (B)
6 2 12 12
(A) (B)
5 3 5 7
(C) (D)
1 3 7 9
(C) (D) 10. How many one-sixth makes one whole?
4 5
5. Compare and fill in the box. (A) 4 (B) 6
(C) 10 (D) 12
5 3 5 1
+ − 11. Sohan filled one-fifth of a beaker with milk
8 4 8 4 and two-fifth with water. What part of the beaker
(A) < (B) > is empty?
(C) = (D) Can’t be determined (A) 1/2 (B) 1/5
(C) 3/5 (D) 2/5
 1 17. The total number of children in a class is 40. If
12. Which of the following figure shows part the fraction of the class that was present on Monday
shaded? 3
31
is , then how many children were absent on that
40
(A) (B) day?
(A) 31 (B) 18
(C) 21 (D) 9
(C) (D) None of these
18. Aman went on a 15 days trip. He spent 7 days
in Manali and rest of the days in Shimla. What
11 5 fraction of his trip did he spend in Shimla?
13. If a − = , then a = _______.
18 18 8 7
13 4 (A) (B)
(A) (B) 7 8
18 5
8 7
8 7 (C) (D)
(C) (D) 15 15
9 18
14. Kavya bought 20 bulbs out of which 8 are good. 19. Which of the following is INCORRECT?
What fraction of the bulbs is not good? 19 3 27 1
(A) = 4 (B) = 13
2 3 4 4 2 2
(A) (B) 35 1 25 1
5 5 (C) = 11 (D) =6
2 3 3 3 4 4
(C) (D) 20. Which of the following figures has the same
3 4
15. Which of the following figures has same fraction shaded fraction as the given figure?
of unshaded part as the fraction of shaded part of
the Fig. (X)?

(A) (B)
Fig. (X)

(A) (B) (C) (D) None of these

21. What fraction of the given figure is not shaded?

(C) (D)

16. What fraction of the given shapes is circle in

the given cloud?
3 7
(A) (B)
10 10
11 13
(C) (D)
20 20
4 10 32
22. The value of + + is _______.
27 9 27
5 11 11 4
(A) (B) (A) 1 (B) 2
16 16 27 9
8 3 8 7
(C) (D) (C) 1 (D) 2
16 16 27 9
  3 1 1 4
23. If − 1 = , then find the value of . (A) (B)
5 10 10 5 5
(A) 3 (B) 6 3 2
(C) 2 (D) 7 (C) (D)
5 5
24. Which of the following does not have half of
its figure shaded? 30. Shaded fraction of – Shaded fraction

(A) (B) of is equal to ________.

2 3
(A) (B)
(C) (D) 3 5
4 1
(C) (D)
5 6
25. There were 4500 males and 2500 females in a
stadium. After 2 hours, 4 of the females left the
10
stadium. How many persons were in the stadium
after 2 hours? 17 7
31. If + + = , + =
(A) 6000 (B) 5000 12 6
(C) 1500 (D) 1800 5
and – = , then what is the value
1 12
26. Find the difference between of 328 and
1 4
of 147. of ?
7
81
(A) (B) 61 1 2
7 (A) (B)
4 3
(C) 103 (D) None of these
5 3
(C) (D)
27. Pooja had 45 cupcakes. She ate 5 of them and 3 4
gave 5 cupcakes to each of her 3 friends. Find the 32. Match the following and select the correct
fraction of cupcakes left with her. option.
4 3
(A) (B) Column A Column B
9 5
4 3 2
4 5 (p) + (i) 1
(C) (D) 3 9 5
7 9
4 8 3 7
28. Which of the following necklaces has of the (q) − (ii) 1
7 5 15 13
beads shaded? 2
7 1
(r) − (iii) 1
9 5 3
(A) (B)
12 8 26
(s) + (iv)
13 13 45
(C) (D) (A) (p) → (ii); (q) → (i); (r) → (iv); (s) → (iii)
(B) (p) → (iii); (q) → (i); (r) → (iv); (s) → (ii)
1 3 (C) (p) → (iv); (q) → (i); (r) → (ii); (s) → (iii)
29. Find the missing value in 2 + 1 = 3 + ? . (D) (p) → (iii); (q) → (ii); (r) → (i); (s) → (iv)
5 5
33. Which of the following statements is/are true? 2. Mohit has some animal cards as shown. What
(i) Two fractions are said to be equivalent, if they fraction of the cards are cat cards?
have the same denominators.
Cat Dog Horse
4 20
(ii) and are equivalent fractions.
5 25 Horse Dog Cat
(A) Only (i) (B) Only (ii)
Cat Horse Cat
(C) Both (i) and (ii) (D) Neither (i) nor (ii)
Dog Cat Horse
34. Which of the following option is INCORRECT?
1 1 5
(A) is unit as well as a proper fraction. (A) (B)
7 12 12
19 6 4
(B) is an improper fraction. (C) (D) (Level-1)
11 12 9
4 5 3. Aarav filled one-fifth of a jar with milk and
(C) and are unlike fractions. two-fifth with water. What part of the jar is empty?
7 7
1 2
2 (A) (B)
(D) 5 is a mixed fraction. 5 5
3
3 1
(C) (D)  (Level-1)
1 5 2
35. of the class IV students choose to join
7 4. Which of the following fractions represents the
1
swimming. of the remaining students choose to shaded part of the given figures?
3
join running. If there were total 126 students, then
how many –
(a) students choose to join swimming?
(b) more students choose to join running than 3 3
(A) (B)
swimming? 8 4
(a) (b) 5 1
(C) (D)  (Level-1)
(A) 16 18 4 4
(B) 16 48 5. There were 320 gems in a jar. If 150 of the gems
were red and the rest of the gems were white, then
(C) 18 42
what fraction of the gems was white?
(D) 18 18
17 15
(A) (B)
SOF IMO 2019 QUESTIONS 32 32
13 19
(C) (D)  (Level-1)
1. How many shaded triangles must be unshaded 32 32
1 6. Who among the following students has written
so that of the given figure is shaded?
4 the CORRECT statement?
Kavya : Fraction of vowels in the word EXPRESSION
4
is .
5 15 18
Shruti : Sum of 2 times of and 3 times of is 7.
12 12
(A) Only Shruti (B) Only Kavya
(A) 1 (B) 2
(C) Both Shruti and Kavya
(C) 3 (D) 4 (Level-1)
(D) Neither Shruti nor Kavya (Level-1)

4
7. Match the figures given in Column I with their
unshaded fractions given in Column II and select
the correct option. 15
(R) (iii)
8
Column I Column II

4
(P) (i)
3 7
(S) (iv)
8

(A) (P) – (ii), (Q) – (iii), (R) – (iv), (S) – (i)

(Q) (ii) 9 (B) (P) – (iv), (Q) – (iii), (R) – (ii), (S) – (i)
8 (C) (P) – (ii), (Q) – (i), (R) – (iv), (S) – (iii)
(D) (P) – (i), (Q) – (ii), (R) – (iii), (S) – (iv)
 (Level-2)
 HINTS & EXPLANATIONS
SELF TEST - 1 19 + 23 − 6 36 1
= = =7
3 4 × 8 + 3 35 5 5 5
1. (A) : 4 = =
8 8 8 2. (C) : The given fractions are
41 5 9 1 5 4 10 1 1 1 1 2
2. (C) : =4 , , , , i.e., , , , ,
9 9 27 2 25 24 15 3 2 5 6 3
3. (A) : Total number of equal parts = 8 LCM of 2, 3, 5 and 6 is 30.
Number of unshaded parts = 5 1 1 × 10 10 1 1 × 15 15
So, = = ; = = ;
5 3 3 × 10 30 2 2 × 15 30
\ Required fraction =
8 1 1× 6 6 1 1× 5 5
4. (B) = = ; = = ;
5 5 × 6 30 6 6 × 5 30
1 1 2 2 × 10 20
5. (D) : Required fraction = 1 + 1 + = 2 = =
4 4 3 3 × 10 30
SELF TEST - 2 Clearly, 20 > 15 > 10 > 6 > 5

1. (D) : (A)
8
=
8÷2 4
= i.e., 20 > 15 > 10 > 6 > 5
10 10 ÷ 2 5 30 30 30 30 30
24 24 ÷ 6 4 2 1 1 1 1
(B) ⇒ > > > >
= = 3 2 3 5 6
30 30 ÷ 6 5
So, the correct descending order is
(C) 20 = 20 ÷ 5 = 4 10 1 9 5 4
25 25 ÷ 5 5 , , , ,
15 2 27 25 24
12 12 ÷ 4 3 4
(D) = = ≠ 3. (B) : Total number of equal parts = 12
20 20 ÷ 4 5 5
Number of shaded parts = 6
14 7 14 − 7 7
2. (C) : − = = 6
15 15 15 15 \ Shaded fraction =
12
1 1 HCF of 6 and 12 = 6
3. (C) : of 125 + of 81 = 25 + 27 = 52
5 3 6 6÷6 1
Hence, the simplest form of = =
30 30 ÷ 6 5 12 12 ÷ 6 2
4. (B) : = =
36 36 ÷ 6 6
4. (B) : Required fraction = 4 − 2
5 3 5+3 8 8÷8 LCM of 5 and 15 = 15 5 15
5. (B) : + = = = =1
8 8 8 8 8÷8 4 2 4 × 3 2 × 1 12 2
So, − = − = −
and 1 + 3 = 1 + 3 = 4 = 4 ÷ 4 = 1 5 15 5 × 3 15 × 1 15 15
4 4 4 4 4÷4
= 12 − 2 = 10 = 2
5 3 1 3 15 15 3
So, + = +
8 8 4 4 5. (B) : L.C.M of 4 and 8 = 8
EXERCISE 3 3×2 6
So, = =
4 3 1 19 23 6 4 4×2 8
1. (C) : 3 + 4 − 1 = + − 5 3 5 6 5 + 6 11
5 5 5 5 5 5 Thus, + = + = =
8 4 8 8 8 8
4
 1 1× 2 2 5 11 16 16 ÷ 2 8
Also, = = ⇒ a= + = = =
4 4×2 8 18 18 18 18 ÷ 2 9
5 1 5 2 5−2 3 14. (B) : Total number of bulbs bought = 20
Thus, − = − = =
8 4 8 8 8 8 Number of good bulbs = 8
11 3 So, number of bulbs that are not good = 20 – 8 = 12
So, >
8 8 12 12 ÷ 4 3
6. (C) : Total number of equal parts = 50 So, required fraction = = =
20 20 ÷ 4 5
Number of shaded parts = 20
4 1
20 15. (C) : Shaded fraction of Fig. (X) = =
\ Shaded fraction = 8 2
50
HCF of 20 and 50 = 10 Figure in option Unshaded fraction
20 20 ÷ 10 2
Hence, simplest form of = = 3
50 50 ÷ 10 5 A
1 1 5
7. (B) : Required shaded fraction = 1 + = 1
3 3 1
8. (D) B
3
9. (B) : Total number of roses = 12
Number of white roses = 2 4 1
Number of yellow roses = 3 C =
8 2
So, number of red roses = 12 – 2 – 3 = 7
7 3
So, fraction of roses that are red = D
12 4
1
10. (B) : × 6 = 1 16. (A) : Total number of shapes = 16
6 Number of circles = 5
So, 6 one-sixths will make one whole.
5
1 So, required fraction =
11. (D) : Fraction of beaker filled with milk = 16
2 5 17. (D) : Total number of children = 40
Fraction of beaker filled with water =
5 Since, fraction of class that was present on Monday
1 2 3 31
Total fraction of beaker which is filled = + = =
5 5 5 40
3 Number of children present on Monday 31
So, fraction of beaker which is empty = 1 − ⇒ =
5 Total number of childrren 40
5 3 2
= − =
5 5 5 Number of children present on Monday 31
⇒ =
12. (A) : (A) Total number of equal parts = 6 40 40
Number of shaded parts = 2 \ Number of children present on Monday = 31
2 1
So, shaded fraction = = ⇒ Number of children absent on Monday
6 3  = 40 – 31 = 9
(B) Total number of equal parts = 6
Number of shaded parts = 3 18. (C) : Total number of days of trip = 15
3 1 Number of days spent in Manali = 7
So, shaded fraction = =
6 2 So, number of days spent in Shimla = 15 – 7 = 8
(C) Total number of equal parts = 8
Number of shaded parts = 4 8
\ Fraction of trip spent in Shimla =
4 1 15
So, shaded fraction = = 35 2
8 2 19. (C) : Option (C) is incorrect, as = 11
11 5 3 3
13. (C) : We have, a − =
18 18
 10 5 So, number of cupcakes left with her = 45 – 5 – 15 = 25
20. (B) : Shaded fraction of given figure = =
18 9 25 5
4 2 \ Fraction of cupcakes left with her = =
(A) Shaded fraction = = 45 9
6 3 3
28. (D) : (A) Shaded fraction =
5 7
(B) Shaded fraction = 5
9 (B) Shaded fraction =
7
4 1 4 1
(C) Shaded fraction = = (C) Shaded fraction = =
8 2 8 2
21. (B) : Total number of equal parts = 40 4
(D) Shaded fraction =
Total number of unshaded parts = 28 7
28 28 ÷ 4 7 1 3
\ Required fraction = = = 29. (B) : 2 + 1 = 3 + ?
40 40 ÷ 4 10 5 5
4 10 32 11 8 19
22. (B) : + + ⇒ + =3+ ? ⇒ =3+ ?
27 9 27 5 5 5
4 10 × 3 32 19
= + + ⇒ ? = −3
27 9 × 3 27 5
4 30 32 4 + 30 + 32
= + + = 19 3 × 5 19 15 4
27 27 27 27 = − = − =
5 1× 5 5 5 5
66 66 ÷ 3 22 4
= = = =2
27 27 ÷ 3 9 9 4 1 1 1
30. (D) : Required difference = − = −
 3 1 8 3 2 3
23. (D) : We have, − 1 = 1 × 3 1× 2 3 2 1
5 10 10 = − = − =
2×3 3×2 6 6 6
 13 1  1 13
i.e., − = ⇒ = + 31. (B) : We have,
5 10 10 5 10 10
 14  7 17
⇒ = ⇒ = ⇒= 7 + + = ...(1)
5 10 5 5 12
24. (D)
25. (A) : Number of females left the stadium 7
Also, + = ...(2)
4 6
= × 2500 = 1000
10 From (1) & (2), we have
So, number of females in the stadium = 2500 – 1000 7 17
= 1500 + =
6 12
So, number of persons left in the stadium after
17 7 17 7 × 2 17 14 3
2 hours = 4500 + 1500 = 6000 ⇒ = − = − = − =
12 6 12 6 × 2 12 12 12
1
26. (B) : of 328 = 328 ÷ 4 = 82
4 5
Also, – =
1 of 147 = 147 ÷ 7 = 21 12
7 3 5
\ Required difference = 82 – 21 = 61 ⇒ − =
12 12
27. (D) : Number of cupcakes Pooja had = 45
Number of cupcakes she ate = 5 5 3 8 2
Number of cupcakes she gave to her 3 friends ⇒ = + = =
12 12 12 3
= 3 × 5 = 15

4
 4 3 4 × 3 3 12 3 3. (B) : Let total capacity of jar be 1.
32. (B) : (p) + = + = + 1
3 9 3×3 9 9 9 Then, fraction of jar filled with milk =
5
12 + 3 15 5 2 2
= = = =1 Fraction of jar filled with water =
9 9 3 3 5
8 3 8 × 3 3 24 3 So, fraction of jar remains empty = 1 –  1 + 2 
(q) − = − = − 5 5
5 15 5 × 3 15 15 15 3 2
24 − 3 21 7 2 = 1− =
= = = =1 5 5
15 15 5 5 4. (D) : Required difference = 3 − 4 = 3 − 1
7 1 7 × 5 1 × 9 35 9 35 − 9 26 4 8 4 2
(r) − = − = − = =
9 5 9 × 5 5 × 9 45 45 45 45 3 1× 2 3 2
= − = −
12 8 12 + 8 20 7 4 2×2 4 4
(s) + = = =1 1
13 13 13 13 13 =
4
33. (B) 34. (C) 5. (A) : Total number of gems = 320
35. (D) : Total number of students = 126 Number of red gems = 150
(a) Number of students choose to join swimming \ Number of white gems = 320 – 150 = 170
1 170 17
= of 126 = 126 ÷ 7 = 18 \ Required fraction = =
7 320 32
(b) Total number of students choose swimming = 18 6. (A) : Number of alphabets in the word
\ Remaining number of students = 126 – 18 = 108 EXPRESSION = 10
Number of students choose to join running Number of vowels (E, E, I, O) = 4
1 4 2
= of 108 = 108 ÷ 3 = 36 \ Required fraction = =
3 10 5
So, number of more students choose to join running So, Kavya has written the incorrect statement.
than swimming = 36 – 18 = 18 18
Now, 2 times of 15 + 3 times of
SOF IMO 2019 QUESTIONS 12 12
1. (D) : Total number of equal triangles = 16 15 18 30 54
= 2× +3× = +
1 12 12 12 12
Shaded fraction = 84
4 = =7
So, number of triangles shaded to get 1 shaded 12
1 4 So, Shruti has written the correct statement.
fraction = × 16 = 4
4
Number of triangles already shaded = 8 7. (A) : (P) Unshaded fraction = 2 + 2 + 5 = 9
Therefore number of triangles that must be unshaded 8 8 8 8
=8–4=4 4 6 5 15
(Q) Unshaded fraction = + + =
2. (B) : Total number of cards = 12 8 8 8 8
Number of cat cards = 5 3 3 1 7
(R) Unshaded fraction = + + =
8 8 8 8
So, required fraction = 5
12 (S) Unshaded fraction = 2 + 3 + 3 = 8 = 4
6 6 6 6 3