NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 4: Fractions & Decimals - Problems and Solutions
Question 115

Question. 115

Sunita found on Monday that she had gained \(1\dfrac{1}{4}\,\text{kg}\). Earlier her weight was \(46\dfrac{3}{8}\,\text{kg}\). What was her weight on Monday?

Answer:

\(47\dfrac{5}{8}\,\text{kg}\)

Detailed Answer with Explanation:

Goal: Sunita gained weight on Monday. So, we add the gain to the earlier weight.

Given:

Earlier weight = \(46\dfrac{3}{8}\,\text{kg}\)

Gain = \(1\dfrac{1}{4}\,\text{kg}\)

Step 1: Use the same kind of fractions.

We already have eighths (\(\dfrac{\;}{8}\)) in \(46\dfrac{3}{8}\).

Change \(\dfrac{1}{4}\) into eighths: \(\dfrac{1}{4}=\dfrac{2}{8}\).

So, \(1\dfrac{1}{4}=1+\dfrac{2}{8}\).

Step 2: Add whole parts and fraction parts separately.

Whole parts: \(46+1=47\).

Fraction parts: \(\dfrac{3}{8}+\dfrac{2}{8}=\dfrac{5}{8}\).

Step 3: Combine the whole and fraction parts.

\(47\) and \(\dfrac{5}{8}\) together give \(47\dfrac{5}{8}\,\text{kg}\).

Answer: \(\boxed{47\dfrac{5}{8}\,\text{kg}}\).

(Quick check in decimals: \(46\dfrac{3}{8}=46.375\), \(1\dfrac{1}{4}=1.25\). Add: \(46.375+1.25=47.625=47\dfrac{5}{8}\).)

NCERT Exemplar Solutions Class 6 – Mathematics – Unit 4: Fractions & Decimals – Problems and Solutions | Detailed Answers