Solutions for NCERT Class 7 th Mathematics Chapter, NCERT Class 7 th Mathematics Chapter 2- Fractions and Decimals Exercise 2.1 Solutions, Exercise 2.1 Maths Solutions, NCERT Maths solutions for class 7th, Questions and answers of NCERT class 7th Mathematics, NCERT Class 7th Mathematics Chapter 2- Fractions and Decimals Exercise 2.1 Solutions, Class 7th maths solutions.
Please find the detailed Solutions for class 7th Maths Chapter 2- Fractions and Decimals Exercise 2.1 Solutions, you can check all the solutions from here chapter wise
NCERT Class 7th Mathematics Chapter 2- Fractions and Decimals Exercise 2.1 Solutions
Exercise 2.1
1) Solve the following
(i) 2-3/5
Ans. To solve ,we have to make both numbers in fraction form
⇒2/1−3/5
Lets take LCM of 1,5 = 1*5 =5
⇒2/1∗5/5 = 10/5
Since both numbers has the denominator as 5, we can solve the equation now,
The solution is
⇒10/5−3/5=7/5
(ii) 4+7/8
Ans. To solve we have to make both numbers in fraction form
⇒4/1+7/8
Lets take LCM of 1,8 = 1*8 =8
⇒4/1∗8/8=32/8
Since both numbers has the denominator as 8, we can solve the equation now,
The solution is
⇒32/8+7/8=39/8
(iii) 3/5+2/7
Ans. Since, the numbers are in fraction form,
Lets take LCM of 5,7= 5*7=35
For first number,
⇒3/5∗7/7=21/35
For second number,
⇒2/7∗5/5=10/35
The solution is
⇒21/35+10/35=31/35
(iv) 9/11−4/15
Ans. Since ,the numbers are in fraction form
Lets take LCM of 11,15= 11*15=165
For first number,
9/11*15/15=135/165
For second number,
4/15*11/11=44/165
The solution is
135/165-44/165=91/165
(v) 7/10+2/5+3/2
Ans. Since, the numbers are in fraction form,
Lets take LCM of 10,5,2= 2*5=10
For first number,
⇒7/10∗1/1=7/10
For second number,
⇒2/5∗2/2=4/10
For third number,
⇒3/2∗5/5=15/10
The solution is
⇒7/10+4/10+15/10=26/10
(vi) 2 2/3 + 3 1/2
Ans. Since, the numbers are in mixed fraction form ,we have to convert it into fractional form
For first number in mixed fraction,
2 2/3=(3∗2)+2/3=8/3
For second number in mixed fraction,
3 1/2=(2∗3)+1/2=7/2
Lets take LCM of 3,2=3*2=6
For first number,
⇒8/3∗2/2=16/6
For second number,
⇒7/2∗3/3=21/6
The solution is
⇒16/6+21/6=37/6
(vii) 8 1/2−3 5/8
Ans. Since, the numbers are in mixed fraction form we have to convert it into fractional form
For first number in mixed fraction,
8 1/2=(2∗8)+1/2=17/2
For second number in mixed fraction,
3 5/8=(8∗3)+5/8=29/8
Lets take LCM of 2,8=2*8=8
For first number,
⇒17/2∗4/4=68/8
For second number,
⇒29/8∗1/1=29/8
The solution is
⇒68/8−29/8=39/8
2) Arrange the following numbers in descending order:
(i) 2/9, 2/3, 8/21
Ans. 2/9, 2/3, 8/21
Let’s take LCM of 9,3,21
9 = 3*3
21= 3* 7
3= 3*1
So, we can take two 3’s as a common term since we four 3’s and 7 is coming only once
Therefore ,the required LCM is= 3*3*7=63
For the first number,
⇒2/9∗7/7=14/63
For the second number,
⇒2/3∗21/21=42/63
For the third number,
⇒8/21∗3/3=24/63
Descending order means arranging the numbers from largest to smallest
So,
14/63=0.22, 42/63=0.66, 24/63=0.38
Therefore, the decreasing order of rational numbers are
42/63>24/63>14/63
i.e. 2/3>8/21>2/9
b) 1/5, 3/7, 7/10
1/5, 3/7, 7/10
Let’s take LCM of 5,7,10
7=7*1
5=5*1
10=5*2
So, we can take one 5 as a common term .Since, we have two 5’s and 7,2 is coming only once
Therefore the required LCM is= 5*7*2= 70
For the first number,
⇒1/5∗14/14=14/70
For the second number,
⇒3/7∗10/10=30/70
For the third number,
⇒7/10∗7/7=49/70
Descending order means arranging the numbers from largest to smallest
So,
14/70=0.2, 30/70=0.42, 49/70=0.70
Therefore, the decreasing order of rational numbers are
49/70>30/70>14/70
i.e. 7/10>3/7>1/5
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?
4/11 3/11 8/ 11
9/11 5/11 1/11
2/11 7/11 6/11
Ans. Yes, it is a magic square.
4. A rectangular sheet of paper is 12 1/2 cm long and 10 2/3 cm wide. Find its perimeter.
Ans. Length of rectangular sheet of paper = 12 1/2 cm
= (12*2)+1 / 2 = 25/2 cm
Breadth of rectangular sheet of paper = 10 2/3 cm
= (10*3)+2 / 3 = 32/3 cm
LCM of 2, 3 = 2*3 = 6
Length (l) = 25/2 cm = 25/2 * 3/3 = 75/6cm
Breadth (b) = 32/3 cm = 32/3 * 2/2 = 64/6 cm
Perimeter of rectangular sheet of paper = 2 (l + b)
= 2 (75/6 + 64/6) cm
= 2 (139/6) cm
= 139/3 cm
= 46 1/3 cm
5. Find the perimeters of (i) Δ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Ans. Given,
In Δ ABE,
AB = 5/2 cm
AE = 3 3/5 cm = 18/5 cm
BE = 2 3/4 cm = 11/4 cm
LCM of 2, 5, 4 = 2*2*5 = 20
AB = 5/2 cm = 5/2 * 10/10 = 50/20 cm
AE = 18/5 cm = 18/5 * 4/4 = 72/20 cm
BE = 11/4 cm = 11/4*5/5 = 55/20 cm
Perimeter of Δ ABE = AB + AE + BE
50/20 cm + 72/20 cm + 55/20 cm = 177/20 cm
= 8 17/20 cm
6. Salil 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?
Ans. Breadth of picture = 7 3/5 cm = 38/5 cm
Breadth of frame = 7 3/10 cm = 73/10 cm
LCM of 5 and 10 = 2*5 = 10
Now, to make the denominator same
38/5 cm = 38/5 * 2/2 = 76/10 cm
Breadth of picture to be trimmed = 76/10 – 73/10 = 3/10cm
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?
Ans. Part of an apple eaten by Ritu = 3/5
Part of an apple eaten by Somu = 1 – 3/5 = 5/5 – 3/5 = 2/5
Comparing the parts of apple 3/5 > 2/5
Therefore, Ritu had the larger share.
Larger share will be more by = 3/5 – 2/5 = 1/5
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?
Ans. Time taken by Michael = 7/12 hour
Time taken by Vaibhav = 3/4 hour
LCM of 12 and 4 = 2*2*3 = 12
Now, making the denominators same
3/4 = 3/4 * 3/3 = 9/12
Hence, Vaibhav worked for longer time.
Vaibhav worked longer by = 9/12 – 7/12 = 2/12= 1/6 hour.
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