Solution NCERT Exercise 2.1 part2
Fractions and Decimals 7th Math
Solution NCERT Exercise 2.1 part2
Question (3) In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?
(Along the first row `4/11+9/11+2/11` `=15/11`)
Solution :
Sum along the second row
`=3/11+5/11+7/11`
`=(3+5+7)/11 = 15/11`
Sum along the third row
`=8/11+1/11+6/11`
`=(8+1+6)/11 = 15/11`
Sum along the first column
`=4/11+3/11+8/11`
`=(4+3+8)/11=15/11`
Sum along the second column
`=9/11+5/11+1/11`
`=(9+5+1)/11=15/11`
Sum along the third column
`=2/11+7/11+6/11`
`=(2+7+6)/11 = 15/11`
Sum along the one of the diagonals
`=4/11+5/11+6/11`
`=(4+5+6)/11 = 15/11`
Sum along the second diagonal
`= 2/11+5/11+8/11`
`= (2+5+8)/11=15/11`
Since sum of the numbers in each row, in each column and along the diagonal is the same, thus this is a magic square. Answer
Question (4) A rectangular sheet of paper is `12 1/2` cm long and `10 2/3` cm wide. Find its perimeter.
Solution :
Given, length of paper `=12 1/2` cm
`=25/2` cm
And, width of paper `=10 2/3` cm
`=32/3` cm
Thus, perimeter = ?
We know that, perimeter of a rectangle = 2 (length + width)
Thus, perimeter of given sheet of paper `= 2( 25/2+32/3)`
`=2(((25xx3)+(32xx2))/6)`
`= 2xx(75+64)/6`
`= cancel2xx139/(cancel2 xx3)`
`=139/3`
`=46 1/3` cm
Thus, perimeter of given sheet of paper `=46 1/3` cm Answer
Question (5) Find the perimeter of (i) `Delta` ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution :
Given, sides of triangle ABE = `5/2` cm, `2 3/4` cm and `3 3/5` cm
And, length of rectangle BCDE = `2 3/4` cm
And width of rectangle BCDE `= 7/6` cm
Thus, perimeter of triangle and reactangle = ?
And bigger perimeter = ?
Calculation of perimeter of triangle ABE
We know that, perimeter of a triangle = sum of all the three sides
`= 5/2+2 3/4 + 3 3/5` cm
`=5/2+11/4+18/5` cm
`=(5xx10)+(11xx5)+(18xx4)/20` cm
`= (50+55+72)/20` cm
`= 177/20` cm
`=8 17/20` cm
Thus, perimeter of triangle ABE `=8 17/20` cm
Calculation of perimeter of rectangle BCED
We know that, perimeter of a rectangle = 2 (length + width)
Thus, perimeter of rectangle BCDE `=2(2 3/4 + 7/6)` cm
`= 2 (11/4+7/6)` cm
`= 2 (((11xx3)+(7xx2))/12)` cm
`=2((33+14)/(12))` cm
`=cancel2 xx47/(cancel2 xx6)` cm
`=47/6` cm
`=7 5/6` cm
Thus, perimeter of triangle `=8 17/20` cm and perimeter of rectangle `=7 5/6` cm. And triangle has bigger perimeter. Answer
Question (6) Sahil wants to put a picture in a frame. The picture is `7 3/5` cm wide. To fit in the frame the picture cannot be more than `7 3/10` cm wide. How much should the picture be trimmed?
Solution :
Given, width of picture `=7 5/6` cm
`=47/6` cm
And width of frame `=7 3/10` cm
`=73/10` cm
Thus, width of picture trimmed to fit in the frame = ?
The width of picture should be trimmed to fit in the frame = width of picture – width of frame
`=47/6-73/10` cm
`=((47xx5)-(73xx3))/30` cm
`=(235-219)/30` cm
`= 16/30` cm
Thus, width of picture to be trimmed to fit in the frame `=16/30` cm Answer
Question (7) Ritu ate `3/5` part of an apple and the remaining apple was eater by her brother Somu. How much part of the apple did Somu eat? Whos had the larger share? By how much?
Solution :
Given, Share of apple eaten by Ritu = `3/5` part
Thus, part of apple eaten by Somu = ?
Who ate larger part and by how much ?
The part of apple eaten by Somu `=1-3/5`
`=(5-3)/5 = 2/5` part
Thus Somu ate `2/5` part of apple
Since, `3/5 > 2/5`, thus Ritu ate larger part.
Ritu ate how much large part than that of Somu
= Part of apple ate by Ritu – Part of apple ate by Somu
`=3/5-2/5`
`=(3-2)/5=1/5` part
Thus, Somu ate `2/5` part of apple. Ritu ate larger part of apple. Ritu ate `1/5` more part than Somu. Answer
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?
Solution :
Given, Time of working by Michael `=7/12` hour
Time of working by Vaibhav `=3/4` hour
Thus, longer work = By whom?
How much longer ?
Time by Michael and Vaibhav
`=7/12, 3/4`
`=7/12, (3xx3)/(4xx3)`
`=7/12, 9/12`
Clearly, `7/12 < 9/12`
Thus, `3/4` is greater.
Thus, Vaibhav worked for longer.
Now, Vaibhava worked for how much longer `=3/4-7/12` hour
`=((3xx3)-7)/12` hour
`=(9-7)/12` hour
`=2/12` hour
`=1/6` hour
Thus, Vaibhav worked longer for `1/6` hour. Answer
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