GSEB Class 7 Maths Solutions Chapter 2 અપૂર્ણાંક અને દશાંશ સંખ્યાઓ Exercise 2.1

 

1. ઉકેલોઃ

 

Question 1. (i) \( 2 – \frac {3}{5} \)
Answer:
\( 2 – \frac {3}{5} \)
\( = \frac{2}{1}-\frac{3}{5} \)
\( = \frac{2 \times 5-3 \times 1}{5} \) (Because the LCM of 1 and 5 is 5)
\( = \frac{10-3}{5} \)
\( = \frac {7}{5} \)
\( = 1\frac {2}{5} \)
In simple words: To subtract these numbers, first make them both fractions. Find the smallest common multiple for the bottom numbers. Then, adjust the top numbers and do the subtraction. Convert the result back to a mixed number if needed.

Exam Tip: When dealing with whole numbers and fractions, always convert the whole number to a fraction with a denominator of 1 to simplify calculations.

 

Question 1. (ii) \( 4 + \frac {1}{2} \)
Answer:
\( 4 + \frac {1}{2} \)
\( = \frac{4}{1}+\frac{1}{2} \)
\( = \frac{4 \times 2 + 1 \times 1}{2} \) (Because the LCM of 1 and 2 is 2)
\( = \frac{8+1}{2} \)
\( = \frac{9}{2} \)
\( = 4\frac {1}{2} \)
In simple words: To add a whole number and a fraction, write the whole number as a fraction over one. Then find a common bottom number (denominator) for both. Add the top numbers, and you will get the answer as a fraction, which you can turn into a mixed number.

Exam Tip: Adding a whole number to a proper fraction often results in a mixed number where the whole number is the integer part and the fraction remains.

 

Question 1. (iii) \( \frac{3}{5}+\frac{2}{7} \)
Answer:
\( \frac{3}{5}+\frac{2}{7} \)
\( = \frac{3 \times 7+2 \times 5}{35} \) (Because the LCM of 5 and 7 is 35)
\( = \frac{21+10}{35} \)
\( = \frac {31}{35} \)
In simple words: To add these fractions, you first need a common bottom number. For 5 and 7, the lowest common multiple is 35. Change both fractions to have this new bottom number, then simply add the top numbers together.

Exam Tip: When adding fractions with different denominators, always find the Least Common Multiple (LCM) of the denominators to simplify the addition process.

 

Question 1. (iv) \( \frac{9}{11}-\frac{4}{15} \)
Answer:
\( \frac{9}{11}-\frac{4}{15} \)
\( = \frac{9 \times 15-4 \times 11}{165} \) (Because the LCM of 11 and 15 is 165)
\( = \frac{135-44}{165} \)
\( = \frac {91}{165} \)
In simple words: When subtracting fractions, you should always find a common bottom number first. For 11 and 15, the smallest common multiple is 165. Change both fractions to use this common bottom number, and then subtract the top numbers.

Exam Tip: Remember to multiply both the numerator and denominator by the same factor when finding a common denominator to maintain the fraction's value.

 

Question 1. (v) \( \frac{7}{10}+\frac{2}{5}+\frac{3}{2} \)
Answer:
\( \frac{7}{10}+\frac{2}{5}+\frac{3}{2} \)
\( = \frac{7 \times 1+2 \times 2+3 \times 5}{10} \) (Because the LCM of 10, 5 and 2 is 10)
\( = \frac{7+4+15}{10} \)
\( = \frac {26}{10} \)
\( = \frac {13}{5} \)
\( = 2\frac {3}{5} \)
In simple words: To add three fractions, find the smallest common bottom number for all of them. For 10, 5, and 2, this is 10. Change each fraction so they all have the same bottom number. Then, add all the top numbers together and simplify your final answer.

Exam Tip: Always simplify your final fractional answer to its lowest terms or convert it to a mixed number if it's an improper fraction.

 

Question 1. (vi) \( 2 \frac{2}{3}+3 \frac{1}{2} \)
Answer:
\( 2 \frac{2}{3}+3 \frac{1}{2} \)
\( = \frac{8}{3}+\frac{7}{2} \)
\( = \frac{8 \times 2+7 \times 3}{6} \) (Because the LCM of 3 and 2 is 6)
\( = \frac {16+21}{6} \)
\( = \frac {37}{6} \)
\( = 6\frac {1}{6} \)
In simple words: To add mixed numbers, first turn them into improper fractions. Then, find a common bottom number for these fractions. Add the top numbers together and simplify the result, possibly converting it back into a mixed number.

Exam Tip: When adding mixed numbers, converting them to improper fractions before finding a common denominator can prevent errors, especially with larger numbers.

 

Question 1. (vii) \( 8 \frac{1}{2}-3 \frac{5}{8} \)
Answer:
\( 8 \frac{1}{2}-3 \frac{5}{8} \)
\( = \frac{17}{2}-\frac{29}{8} \)
\( = \frac{17 \times 4-29 \times 1}{8} \) (Because the LCM of 2 and 8 is 8)
\( = \frac{68-29}{8} \)
\( = \frac {39}{8} \)
\( = 4\frac {7}{8} \)
In simple words: To subtract mixed numbers, first change them into improper fractions. Then, find the smallest common bottom number for both fractions. Subtract the top numbers and simplify your final answer, converting it back to a mixed number if appropriate.

Exam Tip: Always ensure that when converting mixed numbers to improper fractions, you multiply the whole number by the denominator and add the numerator.

 

2. નીચેનાને ઊતરતા ક્રમમાં ગોઠવોઃ

 

Question 2. (i) \( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \)
Answer:
To arrange fractions in descending order, we need to find a common denominator.
The denominators are 9, 3, and 21.
Let's find the LCM of 9, 3, and 21.

The LCM of 9, 3, and 21 is \( 3 \times 3 \times 7 = 63 \).
Now, we convert each fraction to have a denominator of 63:
\( \frac {2}{9} = \frac{2 \times 7}{9 \times 7} = \frac {14}{63} \)
\( \frac {2}{3} = \frac{2 \times 21}{3 \times 21} = \frac {42}{63} \)
\( \frac {8}{21} = \frac{8 \times 3}{21 \times 3} = \frac {24}{63} \)
Now, we compare the numerators: \( 42 > 24 > 14 \).
Therefore, \( \frac{42}{63}>\frac{24}{63}>\frac{14}{63} \).
So, in descending order, the fractions are: \( \frac{2}{3}, \frac{8}{21}, \frac{2}{9} \).
In simple words: To put fractions in order, make their bottom numbers the same by finding the smallest common multiple. Once they all have the same bottom number, simply compare their top numbers to see which is largest or smallest.

Exam Tip: When arranging fractions, converting them to equivalent fractions with a common denominator is the most reliable method for comparison.

 

Question 2. (ii) \( \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \)
Answer:
To arrange fractions in descending order, we need to find a common denominator.
The denominators are 5, 7, and 10.
Let's find the LCM of 5, 7, and 10.

The LCM of 5, 7, and 10 is \( 2 \times 5 \times 7 = 70 \).
Now, we convert each fraction to have a denominator of 70:
\( \frac {1}{5} = \frac{1 \times 14}{5 \times 14} = \frac {14}{70} \)
\( \frac {3}{7} = \frac{3 \times 10}{7 \times 10} = \frac {30}{70} \)
\( \frac {7}{10} = \frac{7 \times 7}{10 \times 7} = \frac {49}{70} \)
Now, we compare the numerators: \( 49 > 30 > 14 \).
Therefore, \( \frac{49}{70}>\frac{30}{70}>\frac{14}{70} \).
So, in descending order, the fractions are: \( \frac{7}{10}, \frac{3}{7}, \frac{1}{5} \).
In simple words: To arrange fractions, first find the smallest common bottom number (LCM) for all of them. Change each fraction so they all share this common bottom number. Then, compare the top numbers to place them in descending order, from largest to smallest.

Exam Tip: Always double-check your LCM calculation, as an incorrect LCM will lead to wrong equivalent fractions and an incorrect ordering.

 

Question 3. “જાદુઈ ચોરસ”માં દરેક આડી હરોળ, ઊભી હરોળ અને ત્રાંસી હરોળની સંખ્યાઓનો સરવાળો સમાન આવે છે. શું આ એક જાદુઈ ચોરસ છે?
Answer:
The given square is:

Let's calculate the sum of each row, column, and diagonal.
**Sum of rows:**
First row sum: \( \frac{4}{11}+\frac{9}{11}+\frac{2}{11} = \frac{4+9+2}{11} = \frac{15}{11} \)
Second row sum: \( \frac{3}{11}+\frac{5}{11}+\frac{7}{11} = \frac{3+5+7}{11} = \frac{15}{11} \)
Third row sum: \( \frac{8}{11}+\frac{1}{11}+\frac{6}{11} = \frac{8+1+6}{11} = \frac{15}{11} \)
**Sum of columns:**
First column sum: \( \frac{4}{11}+\frac{3}{11}+\frac{8}{11} = \frac{4+3+8}{11} = \frac{15}{11} \)
Second column sum: \( \frac{9}{11}+\frac{5}{11}+\frac{1}{11} = \frac{9+5+1}{11} = \frac{15}{11} \)
Third column sum: \( \frac{2}{11}+\frac{7}{11}+\frac{6}{11} = \frac{2+7+6}{11} = \frac{15}{11} \)
**Sum of diagonals:**
First diagonal sum: \( \frac{4}{11}+\frac{5}{11}+\frac{6}{11} = \frac{4+5+6}{11} = \frac{15}{11} \)
Second diagonal sum: \( \frac{2}{11}+\frac{5}{11}+\frac{8}{11} = \frac{2+5+8}{11} = \frac{15}{11} \)
Here, the sum of each row, each column, and each diagonal is the same, \( \frac{15}{11} \). Yes, this is a magic square.
In simple words: To check if a square is "magic," add up all the numbers in each horizontal row, vertical column, and corner-to-corner diagonal. If all these sums are exactly the same, then it is indeed a magic square.

Exam Tip: For a magic square problem, methodically calculate all row, column, and diagonal sums. Even if a few match, one mismatch means it is not a magic square.

 

Question 4. એક લંબચોરસ કાગળની લંબાઈ \( 12\frac {1}{2} \) સેમી અને પહોળાઈ \( 10\frac {2}{3} \) સેમી છે. તેની પરિમિતિ શોધો.
Answer:
Given:
Length of the rectangular paper \( = 12\frac {1}{2} \) cm \( = \frac{(12 \times 2) + 1}{2} \) cm \( = \frac{25}{2} \) cm
Width of the rectangular paper \( = 10\frac {2}{3} \) cm \( = \frac{(10 \times 3) + 2}{3} \) cm \( = \frac{32}{3} \) cm
The perimeter of a rectangle \( = 2 \times (\text{length} + \text{width}) \)
\( = 2 \left[ \frac{25}{2} + \frac{32}{3} \right] \) cm
\( = 2 \left[ \frac{25 \times 3 + 32 \times 2}{6} \right] \) cm (Because the LCM of 2 and 3 is 6)
\( = 2 \left[ \frac{75 + 64}{6} \right] \) cm
\( = 2 \left[ \frac{139}{6} \right] \) cm
\( = \frac{139}{3} \) cm
\( = 46\frac {1}{3} \) cm
Thus, the perimeter of the rectangular paper is \( 46\frac {1}{3} \) cm.
In simple words: To find the distance around a rectangular paper, first change the given length and width into simple fractions. Then, use the formula for a rectangle's perimeter, which is twice the sum of its length and width. Make sure you find a common bottom number before adding the fractions.

Exam Tip: Always remember the formula for the perimeter of a rectangle: \( P = 2(l+w) \). Convert mixed numbers to improper fractions first for easier calculation.

 

Question 5. આપેલ આકૃતિમાં (i) AABE (ii) લંબચોરસ BCDE ની પરિમિતિ શોધો. કોની પરિમિતિ વધારે છે?
Answer:
(i) Perimeter of triangle ABE.
From the figure, the sides of \( \triangle ABE \) are:
AB \( = \frac{5}{2} \) cm
BE \( = 2\frac{3}{4} \) cm \( = \frac{(2 \times 4) + 3}{4} \) cm \( = \frac{11}{4} \) cm
AE \( = 3\frac{3}{5} \) cm \( = \frac{(3 \times 5) + 3}{5} \) cm \( = \frac{18}{5} \) cm
Perimeter of \( \triangle ABE = AB + BE + AE \)
\( = \frac{5}{2} + \frac{11}{4} + \frac{18}{5} \)
To add these fractions, we find the LCM of the denominators 2, 4, and 5.
The LCM of 2, 4, and 5 is 20.
\( = \frac{5 \times 10}{2 \times 10} + \frac{11 \times 5}{4 \times 5} + \frac{18 \times 4}{5 \times 4} \)
\( = \frac{50}{20} + \frac{55}{20} + \frac{72}{20} \)
\( = \frac{50+55+72}{20} \)
\( = \frac{177}{20} \)
\( = 8\frac{17}{20} \) cm
Thus, the perimeter of \( \triangle ABE \) is \( 8\frac{17}{20} \) cm.
(ii) Perimeter of rectangle BCDE.
From the figure, the sides of rectangle BCDE are:
BE \( = 2\frac{3}{4} \) cm \( = \frac{11}{4} \) cm
DE \( = \frac{7}{6} \) cm
Perimeter of rectangle BCDE \( = 2 \times (\text{BE} + \text{DE}) \)
\( = 2 \left[ \frac{11}{4} + \frac{7}{6} \right] \)
To add these fractions, we find the LCM of the denominators 4 and 6.
The LCM of 4 and 6 is 12.
\( = 2 \left[ \frac{11 \times 3}{4 \times 3} + \frac{7 \times 2}{6 \times 2} \right] \)
\( = 2 \left[ \frac{33}{12} + \frac{14}{12} \right] \)
\( = 2 \left[ \frac{33+14}{12} \right] \)
\( = 2 \left[ \frac{47}{12} \right] \)
\( = \frac{47}{6} \)
\( = 7\frac{5}{6} \) cm
Thus, the perimeter of rectangle BCDE is \( 7\frac{5}{6} \) cm.

Now, we compare the perimeters to find which is greater:
Perimeter of \( \triangle ABE = 8\frac{17}{20} \) cm
Perimeter of rectangle BCDE \( = 7\frac{5}{6} \) cm
To compare \( 8\frac{17}{20} \) and \( 7\frac{5}{6} \), we can first compare the whole number parts: 8 and 7. Since 8 is greater than 7, \( 8\frac{17}{20} \) is greater than \( 7\frac{5}{6} \).
Therefore, the perimeter of \( \triangle ABE \) is greater than the perimeter of rectangle BCDE.
In simple words: First, find the distance around the triangle by adding its three sides, making sure to use a common bottom number for the fractions. Next, find the distance around the rectangle using its length and width, again finding a common bottom number. Finally, compare these two distances to see which one is longer.

Exam Tip: When comparing mixed numbers, always start by comparing the whole number part first. If the whole numbers are different, you don't need to compare the fractional parts.

 

Question 6. સલીલ એક ચિત્રને ફ્રેમમાં મૂકવા માંગે છે. ચિત્રની પહોળાઈ \( 7\frac {3}{5} \) સેમી છે. ફ્રેમમાં વ્યવસ્થિત લગાવવા માટે ચિત્રની પહોળાઈ \( 7\frac {3}{10} \) સેમીથી વધુ ન હોવી જોઈએ. ચિત્રને કેટલું કાપવું પડશે?
Answer:
Given:
Width of the picture \( = 7\frac{3}{5} \) cm \( = \frac{(7 \times 5) + 3}{5} \) cm \( = \frac{38}{5} \) cm
Required width of the picture to fit in the frame \( = 7\frac{3}{10} \) cm \( = \frac{(7 \times 10) + 3}{10} \) cm \( = \frac{73}{10} \) cm
Since the actual picture width \( (\frac{38}{5} \text{ cm}) \) is greater than the required width \( (\frac{73}{10} \text{ cm}) \), the picture needs to be trimmed.
To find out how much to trim, we subtract the required width from the actual width:
Amount to be trimmed \( = \frac{38}{5} - \frac{73}{10} \)
To subtract these fractions, we find the LCM of the denominators 5 and 10.
The LCM of 5 and 10 is 10.
Amount to be trimmed \( = \frac{38 \times 2 - 73 \times 1}{10} \)
\( = \frac{76-73}{10} \)
\( = \frac{3}{10} \) cm
Thus, to fit the picture properly, \( \frac{3}{10} \) cm of the picture must be trimmed.
In simple words: Salil needs to trim the picture. First, turn the picture's actual width and the frame's maximum width into fractions. Then, subtract the frame's width from the picture's width. This will tell you how much of the picture needs to be cut off.

Exam Tip: When dealing with practical problems involving measurements, it is crucial to convert all quantities to a consistent unit or fraction type before performing calculations.

 

Question 7. રીતુએ એક સફરજનનો \( \frac {3}{5} \) ભાગ ખાધો અને બાકીનો બચેલો ભાગ એના ભાઈ સોમએ ખાધો. સફરજનનો કેટલો ભાગ સોમએ ખાધો? કોનો ભાગ વધારે હતો? કેટલો વધારે હતો?
Answer:
Part of the apple eaten by Ritu \( = \frac{3}{5} \)
Total apple represents 1 whole.
Part of the apple eaten by Somu \( = 1 - \frac{3}{5} \)
\( = \frac{5}{5} - \frac{3}{5} \)
\( = \frac{5-3}{5} \)
\( = \frac{2}{5} \)
So, Somu ate \( \frac{2}{5} \) part of the apple.

To find whose share was more, we compare Ritu's share \( (\frac{3}{5}) \) and Somu's share \( (\frac{2}{5}) \).
Since both fractions have the same denominator, we compare their numerators.
We know that \( 3 > 2 \).
Therefore, \( \frac{3}{5} > \frac{2}{5} \).
So, Ritu's share was more than Somu's share.

To find how much more Ritu's share was, we subtract Somu's share from Ritu's share:
Difference \( = \frac{3}{5} - \frac{2}{5} \)
\( = \frac{3-2}{5} \)
\( = \frac{1}{5} \)
Thus, Ritu ate \( \frac{1}{5} \) more of the apple than Somu.
In simple words: First, figure out how much apple Somu ate by subtracting Ritu's share from the whole apple. Then, compare Ritu's share and Somu's share to see who ate more. Finally, subtract the smaller share from the larger one to find out how much more one person ate.

Exam Tip: When dealing with "remaining part" problems, assume the total is 1 (or \( \frac{x}{x} \)) and subtract the known part to find the unknown part. Comparing fractions with the same denominator is as simple as comparing their numerators.

 

Question 8. મનોજે એક ચિત્રમાં રંગ પૂરવાનું કાર્ય \( \frac {7}{12} \) કલાકમાં પૂર્ણ કર્યું. વૈભવે તે જ ચિત્રમાં રંગ પૂરવાનું કાર્ય \( \frac {3}{4} \) કલાકમાં પૂર્ણ કર્યું. કોણે વધુ સમય કાર્ય કર્યું? આ સમય કેટલો વધારે હતો?
Answer:
Time taken by Manoj to paint a picture \( = \frac{7}{12} \) hour
Time taken by Vaibhav to paint the same picture \( = \frac{3}{4} \) hour
To determine who took more time, we compare \( \frac{7}{12} \) and \( \frac{3}{4} \).
We need to find a common denominator for 12 and 4. The LCM of 12 and 4 is 12.
Convert \( \frac{3}{4} \) to an equivalent fraction with a denominator of 12:
\( \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \)
Now, compare \( \frac{7}{12} \) and \( \frac{9}{12} \).
Since \( 9 > 7 \), it means \( \frac{9}{12} > \frac{7}{12} \).
Therefore, Vaibhav took more time than Manoj.

To find out how much more time Vaibhav took, we subtract Manoj's time from Vaibhav's time:
Difference in time \( = \frac{9}{12} - \frac{7}{12} \)
\( = \frac{9-7}{12} \)
\( = \frac{2}{12} \)
\( = \frac{1}{6} \) hour
Thus, Vaibhav took \( \frac{1}{6} \) hour more than Manoj to complete the work.
In simple words: First, compare the time Manoj and Vaibhav took by making sure their fractions have the same bottom number. This shows who spent more time. Then, subtract the shorter time from the longer time to figure out exactly how much more time that person worked.

Exam Tip: In comparison problems, always convert fractions to a common denominator before making a judgment to avoid errors. Simplify the final difference fraction to its lowest terms.