Solve the problem 8/7(6/7+(-4/5))
Answers
Answer:
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.
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.
Question 7:
Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat?
Who had the larger share? By how much?
Answer:
The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu = 1 – 3/5 = (5 - 3)/5 = 2/5
Comparing the parts of an apple eaten by both Ritu and Somu: 3/5 > 2/5
Larger share will be more by 3/5 – 2/5 = 1/5
Thus Ritu’s part is 1/5 more than Somu’s part.
Question 8:
Michael finished colouring a picture in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour.
Who worked longer? By what fraction was it longer?
Answer:
Time taken by Michael to colour the picture = 7/12 hour
Time taken by Vaibhav to colour the picture = 3/4 hour
Converting both fractions in like fractions, 7/12 and (3 *3)/(4 * 3) = 9/12
Here, 7/12 < 9/12
=> 7/12 < 3/4
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by 3/4 - 7/12 = 9/12 – 7/12
= (9 - 7)/12
= 2/12 = 1/6 hours
Thus, Vaibhav took 1/6 hour more than Michael.
hope it is helpful for you dear