GSEB Class 7 Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.1

 

Question 1. Solve:
(i) \( 2 - \frac { 3 }{ 5 } \)
(ii) \( 4 + \frac { 7 }{ 8 } \)
(iii) \( \frac { 3 }{ 5 } + \frac { 2 }{ 7 } \)
(iv) \( \frac {9}{ 11 } – \frac { 4 }{ 15 } \)
(v) \( \frac { 7 }{ 10 } + \frac { 2 }{ 5 } + \frac { 3 }{ 2 } \)
(vi) \( 2\frac { 2 }{ 3 } + 3\frac { 1 }{ 2 } \)
(vii) \( 8\frac {1}{2 } - 3\frac { 5 }{ 8 } \)
Answer:
(i) \( 2 - \frac { 3 }{ 5 } = \frac { 2 }{ 1 } - \frac { 3 }{ 5 } \)
\( = \frac { 2(5) - 1(3) }{ 5 } \)
\( = \frac { 10- 3 }{5} \)
\( = \frac { 7 }{ 5 } \)
(ii) \( 4 + \frac { 7 }{ 8 } \)
\( = \frac { 4(8) + 1(7) }{ 8 } \)
\( = \frac { 32 + 7 }{ 8 } \)
\( = \frac { 39 }{8} \) or \( 4\frac { 7 }{ 8 } \)
(iii) \( \frac { 3 }{ 5 } + \frac { 2 }{ 7 } \)
\( = \frac { 7(3) + 5(2) }{ 35 } \)
\( = \frac { 21 + 10 }{ 35 } \)
\( = \frac { 31 }{ 35 } \)
(iv) \( \frac { 9 }{ 11 } – \frac { 4 }{ 15 } \)
\( = \frac { 135 – 44}{ 165 } \)
\( = \frac { 91 }{ 165 } \)
(v) \( \frac { 7 }{ 10 } + \frac { 2 }{ 5 } + \frac { 3 }{ 2 } \)
\( = \frac { 1(7) + 2(2)+5(3) }{ 10} \) [LCM of 2, 10 and 5 is 10]
\( = \frac { 7 + 4+15 }{ 10 } \)
\( = \frac { 26 }{ 10 } \)
\( = \frac { 13 }{ 5 } \)
(vi) \( 2\frac { 2 }{ 3 } + 3\frac { 1 }{ 2 } \)
\( = \frac { 8 }{ 3 } + \frac { 7 }{ 2 } \)
\( = \frac {2(8) + 3(7) }{ 6 } \)
\( = \frac { 16 + 21 }{ 6 } \) [LCM of 3 and 2 is 6]
\( = \frac { 37 }{6} = 6\frac { 1 }{ 6 } \)
(vii) \( 8\frac { 1 }{ 2 } – 3\frac { 5 }{ 8 } \)
\( = \frac { 17 }{ 2 } – \frac { 29 }{ 8 } \)
\( = \frac { 4(17) – 1(29) }{ 8 } \)
\( = \frac { 68-29 }{ 8 } \)
\( = \frac { 39 }{ 8 } = 4\frac {7}{8} \)
In simple words: To solve these fraction problems, first find a common bottom number for all fractions. Then, add or subtract the top numbers as needed. Sometimes, you may need to convert mixed numbers into simple fractions first. Finally, simplify your answer if possible.

Exam Tip: Always convert mixed fractions to improper fractions before performing addition or subtraction. Remember to find the Least Common Multiple (LCM) of the denominators to easily combine fractions.

 

Question 2. Arrange the following in descending order:
(i) \( \frac { 2 }{ 9 }, \frac { 2 }{ 3 }, \frac { 8 }{ 21 } \)
(ii) \( \frac { 1 }{ 5 }, \frac { 3 }{7}, \frac { 7 }{ 10 } \)
Answer:
(i) To arrange \( \frac {2}{9}, \frac { 2 }{ 3 }, \frac { 8 }{ 21 } \) in descending order, we first determine the LCM of the denominators 9, 3, and 21.
The LCM calculation is shown below:

The LCM of 9, 3, 21 is \( 3 \times 3 \times 7 = 63 \).
Now, we convert each fraction to have a common denominator of 63:
\( \frac { 2 }{ 9 } = \frac { 2\times7 }{ 9\times7 } = \frac { 14 }{ 63 } \)
\( \frac { 2 }{ 3 } = \frac { 2\times21 }{ 3\times21 } = \frac { 42 }{ 63 } \)
\( \frac { 8 }{ 21 } = \frac { 8\times3 }{ 21\times3 } = \frac { 24 }{ 63 } \)
Since, \( 42 > 24 > 14 \), the equivalent fractions in descending order are \( \frac { 42 }{ 63 } > \frac { 24 }{ 63 } > \frac { 14 }{ 63 } \).
Thus, the original fractions arranged in descending order are \( \frac { 2 }{ 3 }, \frac { 8 }{ 21 }, \frac { 2 }{9} \).

(ii) To arrange \( \frac { 1 }{ 5 }, \frac { 3 }{ 7 }, \frac { 7 }{ 10 } \) in descending order, we find the LCM of the denominators 5, 7, and 10.
The LCM calculation is shown below:
The LCM of 5, 7, and 10 is \( 5 \times 2 \times 7 = 70 \).
Now, we convert each fraction to have a common denominator of 70:
\( \frac { 1 }{ 5 } = \frac { 1\times14 }{ 5\times14 } = \frac { 14 }{ 70 } \)
\( \frac { 3 }{ 7 } = \frac { 3\times10 }{ 7\times10 } = \frac { 30 }{ 70 } \)
\( \frac { 7 }{ 10 } = \frac { 7\times7 }{ 10\times7 } = \frac { 49 }{ 70 } \)
Since, \( 49 > 30 > 14 \), the equivalent fractions in descending order are \( \frac { 49 }{ 70 } > \frac { 30 }{ 70 } > \frac { 14 }{ 70 } \).
Thus, the original fractions arranged in descending order are \( \frac { 7 }{ 10 }, \frac { 3 }{7}, \frac { 1 }{ 5 } \).
In simple words: To put fractions in order, first find the smallest common bottom number for all of them. Change each fraction so they all have this same bottom number. Then, simply look at the top numbers to see which is biggest, and arrange them from largest to smallest.

Exam Tip: Always find the LCM of the denominators to convert fractions into equivalent forms with a common denominator. This makes comparison much easier. When writing the final answer, remember to use the original fractions, not their equivalent forms.

 

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?

Answer: We need to calculate the sum of numbers in each row, each column, and both diagonals to check if they are all identical.
Along the first row:
\( \frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{4+9+2}{11} = \frac{15}{11} \)
Along the second row:
\( \frac{3}{11} + \frac{5}{11} + \frac{7}{11} = \frac{3+5+7}{11} = \frac{15}{11} \)
Along the third row:
\( \frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{8+1+6}{11} = \frac{15}{11} \)
Along the first column:
\( \frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{4+3+8}{11} = \frac{15}{11} \)
Along the second column:
\( \frac{9}{11} + \frac{5}{11} + \frac{1}{11} = \frac{9+5+1}{11} = \frac{15}{11} \)
Along the third column:
\( \frac{2}{11} + \frac{7}{11} + \frac{6}{11} = \frac{2+7+6}{11} = \frac{15}{11} \)
Along the first diagonal (top-left to bottom-right):
\( \frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{4+5+6}{11} = \frac{15}{11} \)
Along the second diagonal (top-right to bottom-left):
\( \frac{8}{11} + \frac{5}{11} + \frac{2}{11} = \frac{8+5+2}{11} = \frac{15}{11} \)
Since the sum of numbers in each case is the same \( \left( \frac{15}{11} \right) \), it is a magic square.
In simple words: We added up all the numbers in each line going across, down, and diagonally. Because every single total came out to be exactly the same, this square is indeed a magic square.

Exam Tip: To verify if a square is "magic," carefully calculate the sum of all rows, all columns, and both diagonals. All these sums must be identical for it to be a magic square.

 

Question 4. A rectangular sheet of paper is \( 12\frac {1}{ 2 } \) cm long and \( 10 \frac { 2 }{ 3 } \) cm wide. Find its perimeter.
Answer: First, we convert the given mixed fractions to improper fractions.
Length \( = 12\frac { 1 }{ 2 } \) cm \( = \frac { (12 \times 2) + 1 }{ 2 } \) cm \( = \frac { 25 }{ 2 } \) cm
Width \( = 10\frac {2}{3} \) cm \( = \frac { (10 \times 3) + 2 }{ 3 } \) cm \( = \frac { 32 }{ 3 } \) cm
The formula for the perimeter of a rectangle is \( 2 \times (\text{Length + Width}) \).
So, the perimeter of the rectangular piece of paper is:
\( = 2 \left[ \frac {25}{ 2 } + \frac {32}{ 3 } \right] \) cm
To add the fractions, we find the LCM of the denominators 2 and 3, which is 6.
\( = 2 \left[ \frac { (25 \times 3) + (32 \times 2) }{6} \right] \) cm
\( = 2 \left[ \frac { 75+64 }{6} \right] \) cm
\( = 2 \left[ \frac { 139 }{ 6 } \right] \) cm
\( = \frac { 2 \times 139 }{ 6 } \) cm
\( = \frac { 139 }{ 3 } \) cm
Finally, we convert the improper fraction back to a mixed fraction:
\( = 46 \frac { 1 }{ 3 } \) cm
In simple words: We changed the mixed numbers for length and width into simple fractions. Then, we used the formula for a rectangle's perimeter: two times the sum of its length and width. We found a common bottom number to add the fractions, and then multiplied by two. The final answer is given as a mixed number.

Exam Tip: Remember the formula for the perimeter of a rectangle. Always convert mixed numbers to improper fractions before performing calculations involving addition or subtraction, and then convert back to mixed numbers for the final answer if appropriate.

 

Question 5. Find the perimeter of (i) triangle ABE and (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Answer:
(i) Perimeter of \( \triangle \text{ABE} = \text{AB + BE + AE} \)
Given sides for \( \triangle \text{ABE} \):
AB \( = \frac{5}{2} \) cm
BE \( = 2\frac{1}{4} \) cm \( = \frac{9}{4} \) cm
AE \( = 3\frac{3}{5} \) cm \( = \frac{18}{5} \) cm
Perimeter \( = \frac{5}{2} + \frac{9}{4} + \frac{18}{5} \) cm
The LCM of 2, 4, and 5 is 20.
Perimeter \( = \left[ \frac{ (5 \times 10) + (9 \times 5) + (18 \times 4) }{ 20 } \right] \) cm
Perimeter \( = \left[ \frac { 50+45+72 }{ 20 } \right] \) cm
Perimeter \( = \frac { 167 }{ 20 } \) cm \( = 8\frac{7}{20} \) cm

(ii) Perimeter of rectangle BCDE \( = 2 [\text{BE + DE}] \)
Given sides for rectangle BCDE from the solution context:
BE \( = 2\frac{3}{4} \) cm \( = \frac{11}{4} \) cm
DE \( = \frac{7}{6} \) cm
Perimeter \( = 2 \left[ 2\frac{3}{4} + \frac{7}{6} \right] \) cm
Perimeter \( = 2 \left[ \frac{11}{4} + \frac{7}{6} \right] \) cm
The LCM of 4 and 6 is 12.
Perimeter \( = 2 \left[ \frac{ (11 \times 3) + (7 \times 2) }{ 12 } \right] \) cm
Perimeter \( = 2 \left[ \frac{ 33+14 }{ 12 } \right] \) cm
Perimeter \( = 2 \left[ \frac{ 47 }{ 12 } \right] \) cm
Perimeter \( = \frac{ 47 }{ 6 } \) cm \( = 7\frac{5}{6} \) cm

Now, we compare the perimeters to find which is greater: \( \frac{167}{20} \) cm and \( \frac{47}{6} \) cm.
To compare these, we find the LCM of 20 and 6, which is 60.
\( \frac{167}{20} = \frac{167 \times 3}{20 \times 3} = \frac{501}{60} \)
\( \frac{47}{6} = \frac{47 \times 10}{6 \times 10} = \frac{470}{60} \)
Since \( 501 > 470 \), we know that \( \frac{501}{60} > \frac{470}{60} \).
This implies \( \frac{167}{20} > \frac{47}{6} \).
Thus, the perimeter of \( \triangle \text{ABE} \) is greater.
In simple words: We calculated the distance around the triangle and the distance around the rectangle separately. For each shape, we added all its side lengths. To compare them, we changed both answers to fractions with the same bottom number. We found that the triangle's perimeter was larger.

Exam Tip: Be very careful when identifying the dimensions for each shape, especially when shapes share sides or are part of a larger figure. Convert all measurements to a common fraction type (improper fractions) before adding, and then convert to a common denominator to compare.

 

Question 6. Salil wants to put a picture in a frame. The picture is \( 7\frac { 3 }{ 5 } \)cm wide. To fit in the frame the picture cannot be more than \( 7\frac { 3 }{ 10 } \)cm wide. How much should the picture be trimmed?
Answer: First, we convert the mixed fractions to improper fractions.
Actual width of the picture \( = 7\frac { 3 }{ 5 } \) cm \( = \frac { (7 \times 5) + 3 }{ 5 } \) cm \( = \frac { 38 }{ 5 } \) cm
Required width (maximum allowed by frame) \( = 7\frac { 3 }{ 10 } \) cm \( = \frac { (7 \times 10) + 3 }{ 10 } \) cm \( = \frac { 73 }{ 10 } \) cm
The amount the picture needs to be trimmed is the difference between its actual width and the required width.
Amount to trim \( = \left[ \frac {38 }{ 5 } – \frac { 73 }{ 10 } \right] \) cm
To subtract these fractions, we find the LCM of the denominators 5 and 10, which is 10.
Amount to trim \( = \left[ \frac { (38 \times 2) – (73 \times 1) }{ 10 } \right] \) cm
Amount to trim \( = \frac { 76-73 }{ 10 } \) cm
Amount to trim \( = \frac { 3 }{ 10 } \) cm
Thus, the picture should be trimmed by \( \frac { 3 }{ 10 } \) cm.
In simple words: We changed the picture's size and the frame's maximum size into simple fractions. Then, we subtracted the frame's size from the picture's size to find out how much extra width needed to be cut off. The picture needs to be made smaller by a small amount.

Exam Tip: For problems involving 'trimming' or 'fitting,' always calculate the difference between the larger and smaller dimensions. Ensure all mixed numbers are converted to improper fractions for precise calculations.

 

Question 7. Ritu ate \( \frac { 3 }{ 5 } \) 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: Ritu ate \( \frac { 3 }{ 5 } \) part of an apple.
The whole apple is represented by 1 (or \( \frac{5}{5} \)).
Somu ate the remaining part, which is \( 1 - \frac { 3 }{ 5 } = \frac { 5 }{ 5 } - \frac { 3 }{ 5 } = \frac { 2 }{ 5 } \) part.
To determine who had the larger share, we compare \( \frac { 3 }{ 5 } \) (Ritu's share) and \( \frac { 2 }{ 5 } \) (Somu's share).
Since the denominators are the same, we compare the numerators: \( 3 > 2 \).
\( \implies \) Therefore, \( \frac { 3 }{ 5 } > \frac { 2 }{ 5 } \).
So, Ritu had the larger share.
To find out by how much Ritu's share was larger, we subtract Somu's share from Ritu's share:
\( \frac { 3 }{ 5 } – \frac { 2 }{ 5 } = \frac { 3-2 }{ 5 } = \frac { 1 }{ 5 } \)
Thus, Ritu had the larger share by \( \frac { 1 }{ 5 } \) part of an apple.
In simple words: Ritu ate part of an apple, and her brother Somu ate what was left. To find Somu's share, we subtracted Ritu's part from the whole apple. We then compared their shares and found that Ritu ate more. Finally, we calculated the difference to see how much more she ate.

Exam Tip: When dealing with parts of a whole, remember that the entire item is represented by 1. For fractions with the same denominator, the fraction with the larger numerator is the greater fraction.

 

Question 8. Michael finished colouring a picture in \( \frac {7}{12} \) hour. Vaibhav finished colouring the same picture in \( \frac { 3 }{ 4 } \) hour. Who worked longer? By what fraction was it longer?
Answer: Michael took \( \frac {7}{12} \) hour to colour a picture.
Vaibhav took \( \frac { 3 }{ 4 } \) hour to colour the same picture.
To compare their times, we need to express the fractions with a common denominator. The LCM of 12 and 4 is 12.
Michael's time is already \( \frac { 7 }{ 12 } \) hour.
Vaibhav's time: \( \frac { 3 }{ 4 } = \frac { 3 \times 3 }{ 4 \times 3 } = \frac { 9 }{ 12 } \) hour.
Now we compare the times: \( \frac { 7 }{ 12 } \) and \( \frac { 9 }{ 12 } \).
Since \( 9 > 7 \), we conclude that \( \frac { 9 }{ 12 } > \frac { 7 }{ 12 } \).
This means Vaibhav worked for a longer duration.
To find out by what fraction Vaibhav worked longer, we subtract Michael's time from Vaibhav's time:
\( \frac { 9 }{ 12 } – \frac { 7 }{ 12 } = \frac { 9-7 }{ 12 } = \frac { 2 }{12} \)
We can simplify this fraction: \( \frac { 2 }{ 12 } = \frac { 1 }{ 6 } \).
Thus, Vaibhav worked longer by \( \frac { 1 }{ 6 } \) of an hour.
In simple words: Michael and Vaibhav colored the same picture, but took different amounts of time. We changed their times into fractions that had the same bottom number. We saw that Vaibhav spent more time. Then, we subtracted the smaller time from the larger time to find out how much longer Vaibhav worked.

Exam Tip: When comparing quantities expressed as fractions, always convert them to equivalent fractions with a common denominator first. This makes direct comparison straightforward. Remember to simplify your final fractional answer to its lowest terms.