CBSE Class 6 Mathematics Fractions Worksheet Set 07

Question 1. Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Answer: Natural numbers from 2 to 12 are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Total number of natural numbers = 11
Prime numbers among them are: 2, 3, 5, 7, 11
Total number of prime numbers = 5
Required fraction = \(\frac{5}{11}\)
In simple words: First write down all numbers from 2 to 12, count how many are prime, and put that number over the total count.

Exam Tip: Remember that 1 is neither prime nor composite, and 2 is the only even prime number.

Question 2. Reduce the following fractions to its lowest form
a) \(\frac{48}{60}\)
b) \(\frac{126}{90}\)
Answer:
a) \(\frac{48}{60}\):
Find the H.C.F of 48 and 60, which is 12.
Divide both numerator and denominator by 12:
\(\frac{48 \div 12}{60 \div 12} = \frac{4}{5}\)
b) \(\frac{126}{90}\):
Find the H.C.F of 126 and 90, which is 18.
Divide both numerator and denominator by 18:
\(\frac{126 \div 18}{90 \div 18} = \frac{7}{5}\)
In simple words: Divide the top and bottom numbers by their highest common factor until they cannot be divided any further.

Exam Tip: You can also simplify step-by-step by dividing by common factors like 2, 3, or 6 until no common factor remains except 1.

Question 3. Match the following
Column I:
a) \(\frac{13}{7} - \frac{6}{7}\)
b) \(1 + \frac{3}{8}\)
c) \(1\frac{1}{4} - \frac{2}{3}\)
d) \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{2}{3}\)
e) \(7\frac{3}{4} - 5\frac{2}{6}\)
f) \(\frac{15}{21} + \frac{6}{7} - \frac{1}{3}\)

Column II:
a) \(\frac{26}{21}\)
b) \(\frac{5}{12}\)
c) \(2\frac{5}{12}\)
d) \(\frac{7}{12}\)
e) \(\frac{11}{8}\)
f) 1
Answer:
Let us solve each expression in Column I to find its correct match in Column II:

a) \(\frac{13}{7} - \frac{6}{7} = \frac{13 - 6}{7} = \frac{7}{7} = 1\) (Matches with f)
b) \(1 + \frac{3}{8} = \frac{8}{8} + \frac{3}{8} = \frac{11}{8}\) (Matches with e)
c) \(1\frac{1}{4} - \frac{2}{3} = \frac{5}{4} - \frac{2}{3}\)
LCM of 4 and 3 is 12.
\(= \frac{15}{12} - \frac{8}{12} = \frac{7}{12}\) (Matches with d)
d) \(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{2}{3}\)
LCM of 2, 3, 4 is 12.
\(= \frac{6}{12} + \frac{4}{12} + \frac{3}{12} - \frac{8}{12} = \frac{6 + 4 + 3 - 8}{12} = \frac{5}{12}\) (Matches with b)
e) \(7\frac{3}{4} - 5\frac{2}{6} = \frac{31}{4} - \frac{32}{6}\)
LCM of 4 and 6 is 12.
\(= \frac{93}{12} - \frac{64}{12} = \frac{29}{12} = 2\frac{5}{12}\) (Matches with c)
f) \(\frac{15}{21} + \frac{6}{7} - \frac{1}{3} = \frac{5}{7} + \frac{6}{7} - \frac{1}{3} = \frac{11}{7} - \frac{1}{3}\)
LCM of 7 and 3 is 21.
\(= \frac{33}{21} - \frac{7}{21} = \frac{26}{21}\) (Matches with a)

Correct Pairwise Matching:
- (a) matches with (f)
- (b) matches with (e)
- (c) matches with (d)
- (d) matches with (b)
- (e) matches with (c)
- (f) matches with (a)
In simple words: Simplify each fraction expression on the left by finding common denominators and matching the final result to the correct option on the right.

Exam Tip: Always find the LCM of denominators before adding or subtracting unlike fractions.

Question 4. If the H.C.F of two numbers is 23 and L.C.M is 1449. One of the numbers is 161, find the other number.
Answer:
We know the relation between H.C.F, L.C.M, and two numbers:
\(\text{First Number} \times \text{Second Number} = \text{H.C.F} \times \text{L.C.M}\)
Given:
H.C.F = 23
L.C.M = 1449
First Number = 161
Let the other number be \(x\).
\(161 \times x = 23 \times 1449\)
\(x = \frac{23 \times 1449}{161}\)
Since \(161 = 23 \times 7\):
\(x = \frac{1449}{7}\)
\(x = 207\)
The other number is 207.
In simple words: Multiply the H.C.F and L.C.M together, then divide the result by the given number to find the missing number.

Exam Tip: Memorize the formula Product of two numbers = HCF × LCM, as it is frequently tested in exams.