Exercise - 2.1
1. Solve:
i) $\text{2-}\frac{\text{3}}{\text{5}}$
Ans: The solution is given as,
$2-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}=1\frac{2}{5}$
ii) $\text{4+}\frac{\text{7}}{\text{8}}$
Ans: The solution is given as,
$4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=4\frac{7}{8}$
ii) $\frac{\text{3}}{\text{5}}\text{+}\frac{\text{2}}{\text{7}}$
Ans: The solution is given as,
$\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\frac{31}{35}$
iii) $\frac{\text{9}}{\text{11}}\text{-}\frac{\text{4}}{\text{15}}$
Ans: The solution is given as,
$\frac{9}{11}-\frac{4}{15}=\frac{135-44}{165}=\frac{91}{165}$
(v) $\frac{\text{7}}{\text{10}}\text{+}\frac{\text{2}}{\text{5}}\text{+}\frac{\text{3}}{\text{2}}$
Ans: The solution is given as,
$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$
(vi) $\text{2}\frac{\text{2}}{\text{3}}\text{+3}\frac{\text{1}}{\text{2}}$
Ans: The solution is given as, $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$
(vii) $\text{8}\frac{\text{1}}{\text{2}}\text{-3}\frac{\text{5}}{\text{8}}$
Ans: The solution is given as, $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$
2. Arrange the following in descending order:
(i) $\frac{\text{2}}{\text{9}}\text{,}\frac{\text{2}}{\text{3}}\text{,}\frac{\text{8}}{\text{21}}$
Ans: Converting into the fractions with same denominator,
$\frac{2}{9},\frac{2}{3},\frac{8}{21}$ => $\frac{14}{63},\frac{42}{63},\frac{24}{63}$
Arranging the terms in descending order, $\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$
Converting the fraction into simplest form,
$\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$
(ii) $\frac{\text{1}}{\text{5}}\text{,}\frac{\text{3}}{\text{7}}\text{,}\frac{\text{7}}{\text{10}}$
Ans: Converting into the fractions with same denominator, $\frac{1}{5},\frac{3}{7},\frac{7}{10}\,\,\,\,\,\Rightarrow \,\,\,\,\frac{14}{70},\frac{30}{70},\frac{49}{70}$
Arranging the term in descending order,
$\Rightarrow \,\,\frac{49}{70}>\,\frac{30}{70}>\,\frac{14}{70}$
Converting the fraction into simplest form,
$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$
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?
$\left( \text{Along the first row}\frac{\text{4}}{\text{11}}\text{+}\frac{\text{9}}{\text{11}}\text{+}\frac{\text{2}}{\text{11}}\text{+=}\frac{\text{15}}{\text{11}} \right)$
Ans: If the sum of fractions in each row, in each column and along the diagonals is same then it is a magic square.
Calculating the sum of,
first row $=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$
second row \[=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}=\frac{15}{11}\]
third row $=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}=\frac{15}{11}$
first column $=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}=\frac{15}{11}$
second column $=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}=\frac{15}{11}$
third column $=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}=\frac{15}{11}$
first diagonal along top left to bottom right $=\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}=\frac{15}{11}$
second diagonal along top right to bottom left $=\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2+5+8}{11}=\frac{15}{11}$
Observe that the sum of fractions in each row, in each column and along the diagonals is same,
Hence, it’s a magic square.
4. A rectangular sheet of paper is $\text{12}\frac{\text{1}}{\text{2}}$ cm long and $\text{10}\frac{\text{2}}{\text{3}}$ cm wide. Find its Perimeter?
Ans: Given: A rectangular sheet of paper has
\[\text{Length =12}\frac{1}{2}\] cm
\[\text{Breadth =10}\frac{2}{3}\] cm
\[\text{Perimeter of rectangle = 2 (length + breadth)}\]
$=2\left( 12\frac{1}{2}+10\frac{2}{3} \right)=2\left( \frac{25}{2}+\frac{32}{3} \right)$
$=2\left( \frac{25\times 3+32\times 2}{6} \right)=2\left( \frac{75+64}{6} \right)$
$=2\times \frac{139}{6}=\frac{139}{3}=46\frac{1}{3}$ cm
Hence, the perimeter of the rectangular sheet is $46\frac{1}{3}$ cm.
5. Find the perimeter of
(i) $\text{ }\!\!\Delta\!\!\text{ ABE}$,
Ans: Given: In $\text{ }\!\!\Delta\!\!\text{ ABE,}\,\text{AB=}\frac{\text{5}}{\text{2}}\text{cm,}\,\text{BE=2}\frac{\text{3}}{\text{4}}\text{cm,}\,\text{AE=3}\frac{\text{3}}{\text{5}}\text{cm}$
The perimeter of triangle is equal to the sum of all the sides of the triangle.
According to the given figure, $\text{ }\!\!\Delta\!\!\text{ ABE=AB+BE+AE}$
$=\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}=\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$
$=\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20}$ cm
Hence, the perimeter of $\text{ }\!\!\Delta\!\!\text{ ABE}$ is $8\frac{17}{20}$ cm.
(ii) The rectangle $\text{BCDE}$ in this figure. Whose perimeter is greater?
Ans: Given: In rectangle $\text{BCDE,}$ $\text{BE=2}\frac{\text{3}}{\text{4}}\text{cm,}\,\text{ED=}\frac{\text{7}}{\text{6}}\text{cm}$
The perimeter of the rectangle is given by,
\[\text{Perimeter of rectangle = 2 (length + breadth)}\]
$=2\left( 2\frac{3}{4}+\frac{7}{6} \right)=2\left( \frac{11}{4}+\frac{7}{6} \right)$
$=2\left( \frac{33+14}{12} \right)=\frac{47}{6}=7\frac{5}{6}\text{cm}$
Hence, the perimeter of rectangle $\text{BCDE}$ is $7\frac{5}{6}$ cm.
Compare the perimeter of triangle with the perimeter of rectangle,
$\text{8}\frac{\text{17}}{\text{20}}\text{cm }>\text{ 7}\frac{\text{5}}{\text{6}}\text{cm}$
Hence, the perimeter of triangle \[\text{ABE}\] is greater as compared to the perimeter of rectangle $\text{BCDE}$
6. Salil wants to put a picture in a frame. The picture is $\text{7}\frac{\text{3}}{\text{5}}$ cm wide. To fit in the frame the picture cannot be more than $\text{7}\frac{\text{3}}{\text{10}}$ cm wide. How much should the picture be trimmed?
Ans: Given: The width of the picture$=7\frac{3}{5}$ cm and
The width of picture frame$=7\frac{3}{10}$ cm
The picture should be
trimmed$=7\frac{3}{5}-7\frac{3}{10}=\frac{38}{5}-\frac{73}{10}$ $=\frac{76-73}{10}=\frac{3}{10}$ cm
Hence, the original picture should be trimmed by $\frac{3}{10}$ cm.
7. Ritu ate $\frac{\text{3}}{\text{5}}$ 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: Given: The part of an apple eaten by Ritu $=\frac{3}{5}$
The part of an apple eaten by Somu $=1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}$
Compare the parts of apple eaten by Ritu and Somu,
$\frac{3}{5}>\frac{2}{5}$
Observe that Ritu’s part is larger than Somu’s part.
Also, the larger share is more by $\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$ part as compared to the smaller part.
Hence, Ritu’s part is $\frac{1}{5}$ more as compared to Somu’s part.
8. Michael finished colouring a picture in $\frac{\text{7}}{\text{12}}$ hour. Vaibhav finished colouring the same picture in $\frac{\text{3}}{\text{4}}$ hour. Who worked longer? By what fraction was it longer?
Ans: Given: Time taken by Michael for coloring the picture $=\frac{7}{12}$ hour and
Time taken by Vaibhav for coloring the picture $=\frac{3}{4}$ hour
Convert both fractions in the fractions such that their denominator is same, $\frac{7}{12}$ and
$\frac{3\times 3}{4\times 3}=\frac{9}{12}$
Observe that,
$\frac{7}{12}<\frac{9}{12}\Rightarrow \frac{7}{12}<\frac{3}{4}$
Hence, Vaibhav worked for a longer time.
Calculate the time for which Vaibhav worked longer,
$\frac{3}{4}-\frac{7}{12}=\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}$ hour
Therefore, Vaibhav took $\frac{1}{6}$ hour more than Michael.
NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.1
Suppose you have a Christmas Cake. You planned to eat it alone. Let us assume that the size of the cake is 1 unit.
However, your friend arrives and now you have to share your cake equally. What will you do? You will cut the cake into two pieces. The amount and size of the cake remain the same but you now have two parts - one is for you and one is for your friend. So how will you determine how much units of cake you have for yourself and for your friend? You will divide the amount - 1 Unit by 2. This will look like this -
So by dividing the cake into 2 equal halves you and your friends celebrate Christmas. Thus you make two parts of the whole cake. This making of parts of a whole is called fraction. Fraction tells us the number of parts of a whole. In this case, we assumed that the whole cake is 1 Unit in size or amount.
Now when your friend came, you divided the cake into 2 parts. So now you both have 2 parts of the same cake. How will you write this mathematically? You will write like this - ½
So, you have ½ unit of cake and your friend has ½ unit of cake. To tell you more simplistically - you take one of the two parts and your friend takes one of the two parts of the cake. To calculate the total unit of the cake, you would simply add the parts taken by you and your friend.
Now if 3 of your friends arrive you will divide the cake into 3 pieces. Now, you have 3 parts of a whole. Mathematically that would look like this - ⅓
3 parts of the whole (1) cake.
NCERT Solutions for Class 7 Maths
Students can also refer to the following study material for Chapter 2 of Class 7 Maths:
Proper Fractions
Fraction whose numerator (the upper portion of the fraction) is less than the denominator (the lower part) is called a proper fraction. A proper fraction represents the part of a whole. So when you divide a cake into 3 pieces and you want 2 pieces from them - you can mathematically have it - You have to take ⅔ of the cake. The whole of the cake consists of three parts and you are taking two parts of that whole.
Examples of proper fraction are ⅔, 5/10, 999/1200 etc.
Improper Fraction
A fraction whose numerator is greater than the denominator is called an improper fraction. If you divide a cake into 3 pieces, you cannot take 4 pieces out of that 3. That would mathematically look like - 4/3.
In order to solve this problem of not being able to take 4 pieces out of 3, you have to buy a whole new 3-slice cake. Now that you have 3 slices of cake from the new whole, you only need to take 1 slice from the original 3-slice cake. So mathematically it would look like
1 whole ⅓ or 1⅓.
This transformed look of an improper fraction is known as a mixed fraction. Examples of improper fractions are 5/2, 11/9, 800/700 etc.
Examples of mixed fraction are 1⅓, 100⅔, 11¾ etc.
Addition and subtraction of fractions
To add fractions with the same denominator, you just need to add the numerators and keep the denominator as it is.
Example - 5/2 + 6/2 = 12/2
If the fractions are, say 5/2 and 4/6, you first need to find the LCM of the denominators and then divide each of the denominators with the LCM. Now you have to multiply the corresponding numerator with the quotient of the earlier division. You have to repeat the process for all the fractions involved. The resulting numerator would be the addition of the results of the multiplications and the resulting denominator would be that LCM.
Example - 5/2 + 4/6
Step 1 - LCM of 6 and 2 = 6
Step 2 - 6 divided by 2 gives 3 as Quotient
Step 3 - 3*5 = 15
Step 4 - 6 divided by 6 gives 1 as Quotient
Step 5 - 1*4 = 4
Result = 15+4 =19/6
6
Same is the case for subtraction.
To Find the Ascending Order or Descending Order
Before discussing this, first, remember - If two fractions have the same denominator then the fraction with the larger numerator is considered to be the greatest. And the fraction with the smaller numerator is considered to be the lowest. Now if you want to find the ascending order in -
5/2, ¾, ⅛, 8/4
You have to first make the denominator equal. How can you do that? You can multiply these fractions with such fractions that the denominators become the same in each case. You can multiply these fractions with different fractions.
So, 5/2 * 4/4 will make it 20/8 (Which ultimately boils down to 5/2, so the value remains the same)
¾ * 2/2 will make it 6/8
⅛ has already 8 in its denominator
8/4 * 2/2 will make it 16/8
Now, you can easily find the ascending order.
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Our teachers follow NCERT guidelines while preparing the Class 7 Maths Chapter 2 Exercise 2.1 solutions. The methods used are as per the syllabus. The students first need to learn the basics. So we did not use any advanced methods to prepare the NCERT Solutions for Class 7th Maths Chapter 2 - Fractions and Decimals.
We have explained the logic behind the solutions of these problems. So, while giving an answer to Question 7 of NCERT Maths Class 7 Chapter 2 Exercise 2.1, we explained in detail how we found out who ate the larger share of the apple. We have not skipped any step. Again in Question 3 of NCERT Maths solution Class 7 Chapter 2 Exercise 2.1, we have explained in detail how the magic square satisfies all the criteria to be a magic square.
NCERT Class 7 maths book solution Chapter 2 Exercise 2.1 is devised as per the syllabus of NCERT. Our Maths Class 7 Chapter 2 Exercise 2.1 solutions PDF is written not only in a simple language but also in a visually pleasing manner. We have used adequate spacing; we did not cram up our solutions in one line. As a result, students will find it easy to understand our methods used in our NCERT Solution for Class 7 Maths Chapter 2 Exercise 2.1.
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