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Working with Fractions - Some Problems Involving Fractions

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A fraction is a number representing part of a whole. It is written in the form ab\frac{a}{b}, where aa is the numerator and bb is the non-zero denominator.

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The operator 'of' represents multiplication in fractional word problems. For example, 12\frac{1}{2} of 2020 means 12×20\frac{1}{2} \times 20.

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To add or subtract fractions with different denominators, first find the Least Common Multiple (LCM) of the denominators to convert them into like fractions.

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To multiply two fractions, multiply their numerators and denominators separately: ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}.

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The reciprocal of a non-zero fraction ab\frac{a}{b} is ba\frac{b}{a}. The product of a fraction and its reciprocal is always 11.

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To divide one fraction by another, multiply the first fraction by the reciprocal of the second: ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}.

📐Formulae

Product of Fractions=Product of NumeratorsProduct of Denominators\text{Product of Fractions} = \frac{\text{Product of Numerators}}{\text{Product of Denominators}}

ab÷cd=ab×dc=adbc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}

Reciprocal of xy=yx\text{Reciprocal of } \frac{x}{y} = \frac{y}{x}

Value of ’of’ operator: 1n of X=1n×X\text{Value of 'of' operator: } \frac{1}{n} \text{ of } X = \frac{1}{n} \times X

💡Examples

Problem 1:

Lipika reads a book for 1341 \frac{3}{4} hours every day. She reads the entire book in 66 days. How many hours in all were required by her to read the book?

Solution:

Time spent every day = 1341 \frac{3}{4} hours = 74\frac{7}{4} hours. Number of days = 66. Total hours = 74×6=7×64=424=212=1012\frac{7}{4} \times 6 = \frac{7 \times 6}{4} = \frac{42}{4} = \frac{21}{2} = 10 \frac{1}{2} hours.

Explanation:

Since the time spent per day is constant, we multiply the fraction representing daily hours by the total number of days. Convert the mixed fraction to an improper fraction before multiplication.

Problem 2:

A rectangular sheet of paper is 121212 \frac{1}{2} cm long and 102310 \frac{2}{3} cm wide. Find its perimeter.

Solution:

Length (ll) = 1212=25212 \frac{1}{2} = \frac{25}{2} cm. Breadth (bb) = 1023=32310 \frac{2}{3} = \frac{32}{3} cm. Perimeter = 2×(l+b)2 \times (l + b) Perimeter = 2×(252+323)2 \times (\frac{25}{2} + \frac{32}{3}) LCM of 22 and 33 is 66. Perimeter = 2×(25×36+32×26)=2×(75+646)2 \times (\frac{25 \times 3}{6} + \frac{32 \times 2}{6}) = 2 \times (\frac{75 + 64}{6}) Perimeter = 2×1396=1393=46132 \times \frac{139}{6} = \frac{139}{3} = 46 \frac{1}{3} cm.

Explanation:

To find the perimeter, we add the length and breadth first. Since they have different denominators, we use the LCM (66) to find equivalent fractions, sum them, and then multiply by 22.

Problem 3:

A wire of length 1212 meters is cut into pieces of equal length. If each piece is 2142 \frac{1}{4} meters long, how many pieces are there? Also, find the remaining length if we subtract 55 meters from the original length.

Solution:

Total length = 1212 m. Length of one piece = 214=942 \frac{1}{4} = \frac{9}{4} m. Number of pieces = 12÷94=12×49=489=163=51312 \div \frac{9}{4} = 12 \times \frac{4}{9} = \frac{48}{9} = \frac{16}{3} = 5 \frac{1}{3} pieces. Remaining length calculation after cutting 55 meters: 12−57\begin{array}{r} 12 \\ - 5 \\ \hline 7 \end{array} Remaining length = 77 m.

Explanation:

To find the number of pieces, we divide the total length by the length of one piece (multiply by the reciprocal). For the subtraction part, we use vertical arithmetic for the whole numbers.