High School

1. Compare:
(i) \(\frac{27}{30}\) and \(\frac{41}{50}\)
(ii) \(\frac{4}{7}\) and \(\frac{3}{8}\)

2. What should be subtracted from the sum of \(\frac{3}{16}\) and \(\frac{1}{8}\) so that the difference is \(\frac{1}{16}\)?

3. A triangular sheet of paper has three sides with lengths \(\frac{4}{9}\) cm, \(\frac{2}{3}\) cm, and \(\frac{5}{18}\) cm. Find its perimeter.

4. Reduce \(\frac{1264}{1648}\) and \(\frac{624}{2266}\) to their lowest terms.

5. Rahul travels one-fourth of a distance. If the total distance is 20 km, find the remaining distance he would have traveled (in km).

6. Simplify: \(1\frac{3}{10} + 2\frac{3}{10} + 3\frac{3}{10} + 4\frac{3}{10}\)

Answer :


  1. Compare the Fractions:

    (i) [tex]\frac{27}{30}[/tex] and [tex]\frac{41}{50}[/tex]:

    To compare, find a common denominator or convert them to decimals.

    [tex]\frac{27}{30}[/tex] can be simplified to [tex]\frac{9}{10}[/tex], which is 0.9.

    [tex]\frac{41}{50}[/tex] as a decimal is 0.82.

    [tex]0.9 > 0.82[/tex], so [tex]\frac{27}{30} > \frac{41}{50}[/tex].

    (ii) [tex]\frac{4}{7}[/tex] and [tex]\frac{3}{8}[/tex]:

    Convert to decimals: [tex]\frac{4}{7} \approx 0.571[/tex] and [tex]\frac{3}{8} = 0.375[/tex].

    [tex]0.571 > 0.375[/tex], so [tex]\frac{4}{7} > \frac{3}{8}[/tex].


  2. What should be subtracted from the sum of [tex]\frac{3}{16}[/tex] and [tex]\frac{1}{8}[/tex] to get [tex]\frac{1}{16}[/tex]?

    Sum of [tex]\frac{3}{16}[/tex] and [tex]\frac{1}{8} = \frac{3}{16} + \frac{2}{16} = \frac{5}{16}[/tex].

    Let the number to be subtracted be [tex]x[/tex]:

    [tex]\frac{5}{16} - x = \frac{1}{16}[/tex]

    Solving for [tex]x[/tex]:
    [tex]x = \frac{5}{16} - \frac{1}{16} = \frac{4}{16} = \frac{1}{4}[/tex].


  3. Find the perimeter of a triangular sheet of paper with sides [tex]\frac{4}{9}[/tex] cm, [tex]\frac{2}{3}[/tex] cm, and [tex]\frac{5}{18}[/tex] cm.

    To find the perimeter, sum the lengths of the sides:

    First, convert [tex]\frac{2}{3}[/tex] to have the same denominator as the other fractions:
    [tex]\frac{2}{3} = \frac{12}{18}[/tex]

    Perimeter = [tex]\frac{4}{9} + \frac{12}{18} + \frac{5}{18}[/tex]

    Convert [tex]\frac{4}{9}[/tex] to [tex]\frac{8}{18}[/tex]:

    Perimeter = [tex]\frac{8}{18} + \frac{12}{18} + \frac{5}{18} = \frac{25}{18}[/tex] cm.


  4. Reduce [tex]\frac{1264}{1648}[/tex] and [tex]\frac{624}{2266}[/tex] to the lowest terms.

    [tex]\frac{1264}{1648}[/tex]:

    Find the GCD of 1264 and 1648, which is 384.

    [tex]\frac{1264 \div 384}{1648 \div 384} = \frac{79}{103}[/tex].

    [tex]\frac{624}{2266}[/tex]:

    Find the GCD of 624 and 2266, which is 2.

    [tex]\frac{624 \div 2}{2266 \div 2} = \frac{312}{1133}[/tex], which can be further reduced using 1.

    The fraction remains [tex]\frac{312}{1133}[/tex].


  5. Rahul travels one-fourth of a distance of 20 km. Find the remaining distance.

    Distance traveled by Rahul = [tex]\frac{1}{4} \times 20 = 5[/tex] km.

    Remaining distance = [tex]20 - 5 = 15[/tex] km.


  6. Simplify: [tex]1\frac{3}{10} + 2\frac{3}{10} + 3\frac{3}{10} + 4\frac{3}{10}[/tex]:

    First, convert each mixed number to an improper fraction:

    [tex]\frac{13}{10}, \frac{23}{10}, \frac{33}{10}, \frac{43}{10}[/tex]

    Sum them:

    [tex]\frac{13}{10} + \frac{23}{10} + \frac{33}{10} + \frac{43}{10} = \frac{112}{10} = 11\frac{2}{10} = 11\frac{1}{5}[/tex].


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