Exercise 2.1
1. Compare the following fractions by using the symbol > or < or = ;
Solution:
(i) 7/9 and 8/13
Let us first find the LCM of 9 and 13 .
We have
9×13 = 117
∴ L.c.m of 9 and 3 is 117.
Now , we convert the given fraction to equivalent fraction with denominator 117.
We have,
(ii)11/9 and 5/9
Let us,
Here the denominators are equal
(iii) 37/41 and 19/30
Let us first find the LCM of 41 and 30 .
We have,
LCM of 41 and 30 is 41×30 = 1230
Now we convert the given fractions to equivalent
Fraction with denominator 1230 .
We have,
(iv)17/15 and 119/105
Solution:
2. Arrange the following fractions in ascending order:
Solution:
3. Arrange the following fractions in descending order:
(i) (4/5), (7/10), (11/15), (17/20)
(ii) (2/7), *(11/35), (9/14), (13/28)
Solution:
(i) Given (4/5), (7/10), (11/17), (17/20)
Now we have to arrange these in ascending order, to arrange these in ascending order we have to make those as equivalent fractions by taking LCM’s
LCM of 5, 10, 15 and 20 is 60
Equivalent fraction are
(48/60) , (44/60), (51/60)
We know that 51> 48> 44> 42
Now arrange in ascending order
Hence (17/20) > (4/5)> (11/15) > (7/10)
(ii) Given (2/7), (11/35), (9/14), (13/28)
Now we have to arrange these in ascending order, to arrange
We have to make those as equivalent fractions by taking LCM’s.
LCM if 7, 35, 14 and 28 is 140
Equivalent fraction are
(40/140), (44/140), (95/140), (60/140)
Now arrange in ascending order
Hence (9/14)> (13/28)>(11/35) > (2/7)
4. Wrote five equivalent fraction of (3/5)
Solution:
Given (3/5)
By multiplying or dividing both the numerator and denominator so that it keeps the same value by this we can get the equivalent fraction.
Equivalent fraction are
(6/10), (9/15), (12/20), (15/25), (18/30)
5. Find the sum:
6. Find the difference of
(i) (6/7) – (9/11)
(ii) 8 – (5/9)
(iii) 9 – 5 (2/3)
(iv) 4 (3/10) – 1 (2/15)
Solution:
(i) Given (6/7) – (9/11)
To find the difference we have to make it fractions.
Taking LCM of 7 and 11 is 77.
Now converting the given fraction into equivalent fractions with denominators 77
Equivalent fractions are (66/77) and (63/77)
(6/7) – (9/11) = (66/77) – (62/77)
= (66- 63) / 77
= (3/77)
(ii) Giving 8 – (5/9)
To find the difference we have to make it equivalent fractions
Taking LCM of 1 and 9 is 9
Now converting the given fractions into equivalent fractions with denominators 9
Equivalent fractions are (72/9) and (5/9)
8 – (5/9) = (71/9) – (5/9)
= (72-5)/9
= (67/9)
(iii) Given 9 – 5 (2/3)
First convert the given mixed fraction into improper fractions .
We get 5 (2/3) = (17/3)
To find the difference we have to make it equivalent fractions
Taking LCM OF 1 and 3 is 3.
Now converting the given fractions into equivalent fractions with denominator 3
Equivalent fraction are (27/3) and (17/3)
9 – 5 (2/3) = (27/3) – (17/3)
= (10/3)
(iv) Given 4 (3/10) – 1 (2/15)
First convert the given mixed fraction into improper fractions .
We get (43/10) and (17/15)
To find the difference we have to make it equivalent fractions.
Taking LCM of 10 and 15 is 30.
Now converting the given fractions into equivalent fractions with denominators 30.
Equivalent fractions are (129/30) and (34/30)
4 (3/4) – 1 (2/15) = (43/10) – (17/15)
= (129/30) – (34/30)
= (129-34)/30
= (95/30)
(19/6)
8. Simplify:
Solution:
9. What should be added to 5 (3/7) to get 12?
Solution:
Given 5 (3/7)
First the given mixed fractions into improper fractions we get, (38/7)
Let x be the number added to (38/7) to get 12
∴ x +_ (38/7) 12
X = 12 (38/7) 12
X = 12 – (38-7)
Bu taking LCM for 7 and 1 is 7
X = (12×7-38)/7
X = (84-38)/7
X = (46/7)
Then, (46/7) is the number which is added to 5 (3/7) to get 12.
10. What should be added to 5 (4/15) to get 12(3/5) ?
Solution:
Given 5 (4/15)
First convert given mixed fraction into improper fraction we get (79/15)
Let x be the number added to (79/15) to get (63/5)
∴ x + (79/15) = (63/5)
X = (63/5) – (79/15)
By taking LCM for 15 and 5 is 15
X= (63×3-79)/15
X = (189-79)/15
X = (110/15) = (22/3)
Then, (22/3) is the number which is added to 5 (4/15) to get 12 (3/5).
13. A rectangular sheet of paper is 12 (1/2) cm long and 10 (2/3) cm wide. Find its perimeter.
Solution:
Given length of rectangular sheet of paper is 12 (1/2) i.e (25/2)
Breadth of rectangular sheet of paper is 10 (2/3) i.e (32/3)
But we know that perimeter of rectangular of = 2 (length + nreadth)
Perimeter of rectangular sheet = 2 [(25/2) + (32/3)]
LCM of 2 and 3 is 6, by taking this and simplify we get
Perimerter = 2 [(2⨯3]
= 2 [75/6) + (64/6)]
= 2 (139/6)
= (139/3)
= 46 (1/3) cm
14. In a ‘’magic square’’, the sum of the number in each row, in each column and along the diagonal is the same. Is this a magic square?
(4/11) | (9/11) |
(2/11) |
(3/11) |
(5/11) |
(7/11) |
(8/11) |
(1/11) | (6/11) |
Solution:
(4/11) | (9/11) |
(2/11) |
(3/11) |
(5/11) |
(7/11) |
(8/11) |
(1/11) |
(6/11) |
Then, the sum of number in each row, in each column and the diagonal is same and the sum is 15/11.
15. The cost of mathematics book is Rs 25 (3/4) and that of Science book is Rs 20 (1/2).
Solution:
Given the cost of mathematics book is Rs 20 (1/2) i.e (41/2)
Now taking the LCM of 2 and 4 is 4
Now we have to convert the given fractions into its equivalent fractions with denominators 4
Mathematics book cost is Rs (103/4)
Science book cost IS Rs (41⨯ 2/2 ⨯2)= (82/4)
(103 – 82)/ 4 = 21/4 = 5(1/4)
Hence the cost of mathematics book is more than cost of science book by 5 (1/4)
16. (i) Provide the number in box [ ] and also given its simplest form in each of the following:
(i) (2/3) ⨯ [ ]= (10/30)
(ii) (3/5)⨯ [ ]= (24/75)
Solution:
(i) (2/3)⨯ [5/10] = (10/30)
(ii) (3/5)⨯ [8/15]