Verify : 6/13 x (-2/5) + 1/5 x 6/13 = 6/13 ( -2/5+1/5)
Answers
Answer:
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.
Question 6:
Salil wants to put a picture in a frame. The picture is 7 cm wide. To fit in the frame the picture cannot be more than 7 cm wide.
How much should the picture be trimmed?
Answer:
Given, the width of the picture = 7 cm = 38/5 cm
and the width of the picture frame = 7 cm = 73/10
Therefore, the picture should be trimmed = 38/5 – 73/10
= (38 * 2 – 73 * 1)/10
= (76 – 73)/10
= 3/10
Thus, the picture should be trimmed by 3/10 cm.
The great majority of lichens have ascomycetes as the mycobiont. Of the 18,000 or so known lichens fewer than 50 have basidiomycete mycobionts and such lichens are referred to as basidiolichens .