Question

3/10 + 2/10+ 7/10 =1 and 1/5​

Answer 1

Answer:

=(3+ 2+ 7)/10

= 12/10

= 6/5

Answer 2

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Answer:

Exercise 2.1

Question 1:

Solve:

(i) 2 – 3/5 (ii) 4 + 7/8 (iii) 3/5 + 2/7 (iv) 9/11 – 4/15 (v) 7/10 + 2/5 + 3/2

(vi) 2 + 3 (vii) 8 - 3

Answer:

(i) 2 – 3/5 = 2/1 – 3/5

= (2 * 5 – 3 * 1)/5 [LCM (1, 5) = 5]

= (10 - 3)/5

= 7/5

= 1

(ii) 4 + 7/8 = 4/1 + 7/8

= (4 * 8 + 7)/8 [LCM (1, 8) = 8]

= (32 + 7)/8

= 39/8

= 4

(iii) 3/5 + 2/7 = (3 * 7 + 2 * 5)/35 [LCM (7, 5) = 35]

= (21 + 10)/35

= 31/35

(iv) 9/11 – 4/15 = (9 * 15 – 4 * 11)/165 [LCM (11, 15) = 165]

= (135 - 44)/165

= 91/165

(v) 7/10 + 2/5 + 3/2 = (7 * 1 + 2 * 2 + 3 * 5)/10

= (7 + 4 + 15)/10

= 26/10 [26 and 10 are divided by 2]

= 13/5

= 2

(vi) 2 + 3 = 8/3 + 7/2

= (8 * 2 + 7 * 3)/6 [LCM (2, 3) = 6]

= (16 + 21)/6

= 37/6

= 6

(vii) 8 - 3 = 17/2 – 29/8

= (17 * 4 – 29 * 1)/8 [LCM (2, 8) = 8]

= (68 - 19)/8

= 39/8

= 4

Question 2:

Arrange the following in descending order:

(i) 2/9, 2/3, 8/21 (ii) 1/5, 3/7, 7/10

Answer:

(i) 2/9, 2/3, 8/21 = (2 * 7, 2 * 21, 8 * 3)/63 [LCM (9, 3, 21) = 63]

= (14, 42, 24)/63

= 14/63, 42/63, 24/63

Now, 42/63, 24/63, 14/63 [Arrange in descending order]

So, 2/3, 8/21, 2/9

(ii) 1/5, 3/7, 7/10 = (1 * 14, 3 * 10, 7 * 7)/70

= (14, 30, 49)/70

= 14/70, 30/70, 49/70

Now, 49/70, 30/70, 14/70 [Arrange in descending order]

So, 7/10, 3/7, 1/5

Question 3:

In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

Class_7_Maths_Fraction_&_Decimals_Representing_Fractions

(Along the first row 4/11 + 9/11 + 2/11 = 15/11)

Answer:

Sum of first row: 4/11 + 9/11 + 2/11 = 15/11 (Given)

Sum of second row: 3/11 + 5/11 + 7/11 = (3 + 5 + 7)/11 = 15/11

Sum of third row: 8/11 + 1/11 + 6/11 = (8 + 1 + 6)/11 = 15/11

Sum of first column: 4/11 + 3/11 + 8/11 = (4 + 3 + 8)/11 = 15/11

Sum of second column: 9/11 + 5/11 + 1/11 = (9 + 5 + 1)/11 = 15/11

Sum of third column: 2/11 + 7/11 + 6/11 = (2 + 7 + 6)/11 = 15/11

Sum of first diagonal (left to right): 4/11 + 5/11 + 6/11 = (4 + 5 + 6)/11 = 15/11

Sum of second diagonal (left to right): 2/11 + 5/11 + /11 = (2 + 5 + 8)/11 = 15/11

Since the sum of fractions in each row, in each column and along the diagonals is same, therefore it is a magic square.

Question 4:

A rectangular sheet of paper is 12 cm long and 10 cm wide. Find its perimeter.

Answer:

Given: The sheet of paper is in rectangular form.

Length of sheet = 12 cm = 25/2 cm

and Breadth of sheet = 10 cm = 32/3

Now, Perimeter of rectangle = 2 (length + breadth)

= 2(25/2 + 32/2)

= 2[(25 * 3 + 32 * 3)/6]

= 2[(75 + 96)/6]

= 2(139/6)

= 139/3

= 46

Thus, the perimeter of the rectangular sheet is 46 cm.

Question 5:

Find the perimeter of (i) Δ ABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Class_7_Maths_Fraction_&_Decimals_Combinationof_Triangle_&_Rectangle

Answer:

(i) In Δ ABE, AB = 5/2 cm, BE = 2 cm, AE = 3 cm

The perimeter of Δ ABE = AB + BE + AE

= 5/2 + 2 + 3

= 5/2 + 11/4 + 18/5

= (5 * 10 + 11 * 5 + 18 *4)/20

= (50 + 55 + 72)/20

= 177/20

= 8 cm

Thus, the perimeter of Δ ABE is 8 cm.

(ii) In rectangle BCDE, BE = 2 cm, ED = 7/6

Perimeter of rectangle = 2(Length + Breath)

= 2(2 + 7/6)

= 2(11/4 + 7/6)

= 2[(11 * 3 + 7 * 2)/12]

= 2[(33 + 14)/12]

= 2(47/12)

= 47/6

= 7

Thus, the perimeter of rectangle BCDE is 7 cm.

Since, 8 cm > 7 cm

Therefore, the perimeter of Δ ABE is greater than that of rectangle BCDE