Class 8 Maths Rational Numbers Worksheet - Free Printable
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Step-by-step solution for: Class 8 Maths Rational Numbers Worksheet
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Step-by-step solution for: Class 8 Maths Rational Numbers Worksheet
Absolutely! Let’s solve each problem step by step and explain the reasoning clearly.
---
1. Fill in the boxes:
> $\frac{6}{8} = \frac{-18}{\square} = \frac{30}{\square}$
We need to find two missing denominators such that all three fractions are equivalent.
First, simplify $\frac{6}{8}$:
$$
\frac{6}{8} = \frac{3}{4}
$$
So we want:
$$
\frac{-18}{x} = \frac{3}{4} \quad \text{and} \quad \frac{30}{y} = \frac{3}{4}
$$
---
Solve for first box ($\frac{-18}{x} = \frac{3}{4}$):
Cross-multiply:
$$
-18 \cdot 4 = 3 \cdot x \\
-72 = 3x \\
x = -24
$$
✔ First box: -24
---
Solve for second box ($\frac{30}{y} = \frac{3}{4}$):
Cross-multiply:
$$
30 \cdot 4 = 3 \cdot y \\
120 = 3y \\
y = 40
$$
✔ Second box: 40
---
Final Answer for #1:
> $\frac{6}{8} = \frac{-18}{\boxed{-24}} = \frac{30}{\boxed{40}}$
---
2. Locate the points on a number line:
> $\frac{3}{6}, -\frac{1}{6}, 1, -\frac{5}{6}, 0$
First, simplify where possible:
- $\frac{3}{6} = \frac{1}{2} = 0.5$
- $-\frac{1}{6} \approx -0.166...$
- $1 = 1.0$
- $-\frac{5}{6} \approx -0.833...$
- $0 = 0$
Now, arrange them from least to greatest:
$$
-\frac{5}{6} < -\frac{1}{6} < 0 < \frac{3}{6} < 1
$$
On a number line, mark these positions approximately:
```
<---|-----|-----|-----|-----|-----|-----|-----|-----|-----|--->
-1 -5/6 -1/6 0 1/2 1
```
You would place dots at:
- $-\frac{5}{6}$ (just left of -0.8)
- $-\frac{1}{6}$ (just left of 0)
- $0$
- $\frac{1}{2}$ (midway between 0 and 1)
- $1$
✔ You can sketch this on paper or visualize it as above.
---
3. Determine which of the two rational numbers is greater:
> $-\frac{7}{12}, -\frac{8}{23}$
Both are negative. The *greater* number is the one *closer to zero* (less negative).
Compare their absolute values:
- $|\frac{7}{12}| = \frac{7}{12} \approx 0.583$
- $|\frac{8}{23}| \approx 0.348$
Since $0.348 < 0.583$, then $-\frac{8}{23}$ is *closer to zero* than $-\frac{7}{12}$.
Therefore:
✔ $-\frac{8}{23} > -\frac{7}{12}$
---
4. Arrange the following in descending order:
> $-\frac{5}{4}, \frac{2}{3}, -\frac{7}{10}, \frac{4}{5}$
Descending order = from largest to smallest.
Convert to decimals for easier comparison:
- $-\frac{5}{4} = -1.25$
- $\frac{2}{3} \approx 0.666...$
- $-\frac{7}{10} = -0.7$
- $\frac{4}{5} = 0.8$
Now order from largest to smallest:
$$
0.8 > 0.666... > -0.7 > -1.25
$$
Which corresponds to:
✔ $\frac{4}{5} > \frac{2}{3} > -\frac{7}{10} > -\frac{5}{4}$
---
5. Verify that $a + (b + c) = (a + b) + c$ using:
> $a = \frac{1}{2}, b = \frac{3}{4}, c = -\frac{2}{3}$
This is verifying the Associative Property of Addition.
---
Left Side: $a + (b + c)$
First compute $b + c$:
$$
\frac{3}{4} + (-\frac{2}{3}) = \frac{3}{4} - \frac{2}{3}
$$
Find common denominator (LCM of 4 and 3 is 12):
$$
= \frac{9}{12} - \frac{8}{12} = \frac{1}{12}
$$
Now add $a$:
$$
\frac{1}{2} + \frac{1}{12} = \frac{6}{12} + \frac{1}{12} = \frac{7}{12}
$$
✔ Left side = $\frac{7}{12}$
---
Right Side: $(a + b) + c$
First compute $a + b$:
$$
\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}
$$
Now add $c$:
$$
\frac{5}{4} + (-\frac{2}{3}) = \frac{5}{4} - \frac{2}{3}
$$
Common denominator (LCM of 4 and 3 is 12):
$$
= \frac{15}{12} - \frac{8}{12} = \frac{7}{12}
$$
✔ Right side = $\frac{7}{12}$
---
Conclusion: Both sides equal $\frac{7}{12}$ → Associative property verified.
✔ Verified: $a + (b + c) = (a + b) + c$
---
6. Simplify:
> $\frac{-4}{11} + \frac{-2}{3} - \left(\frac{-5}{9}\right)$
First, simplify the expression:
Note: subtracting a negative is adding a positive:
$$
= \frac{-4}{11} + \frac{-2}{3} + \frac{5}{9}
$$
Now, we need a common denominator to combine these fractions.
Denominators: 11, 3, 9
Prime factors:
- 11 → 11
- 3 → 3
- 9 → 3²
LCM = 11 × 3² = 11 × 9 = 99
Convert each fraction:
- $\frac{-4}{11} = \frac{-4 \times 9}{99} = \frac{-36}{99}$
- $\frac{-2}{3} = \frac{-2 \times 33}{99} = \frac{-66}{99}$
- $\frac{5}{9} = \frac{5 \times 11}{99} = \frac{55}{99}$
Now add:
$$
\frac{-36}{99} + \frac{-66}{99} + \frac{55}{99} = \frac{-36 -66 + 55}{99} = \frac{-47}{99}
$$
✔ Final Answer: $\boxed{\frac{-47}{99}}$
---
## ✔ Final Answers Summary:
1. $\frac{6}{8} = \frac{-18}{\boxed{-24}} = \frac{30}{\boxed{40}}$
2. Number line points: $-\frac{5}{6}, -\frac{1}{6}, 0, \frac{1}{2}, 1$ (from left to right)
3. Greater number: $\boxed{-\frac{8}{23}}$
4. Descending order: $\boxed{\frac{4}{5} > \frac{2}{3} > -\frac{7}{10} > -\frac{5}{4}}$
5. Verified: $a + (b + c) = (a + b) + c = \frac{7}{12}$
6. Simplified: $\boxed{\frac{-47}{99}}$
Let me know if you’d like a visual sketch for #2 or more explanation on any part!
---
1. Fill in the boxes:
> $\frac{6}{8} = \frac{-18}{\square} = \frac{30}{\square}$
We need to find two missing denominators such that all three fractions are equivalent.
First, simplify $\frac{6}{8}$:
$$
\frac{6}{8} = \frac{3}{4}
$$
So we want:
$$
\frac{-18}{x} = \frac{3}{4} \quad \text{and} \quad \frac{30}{y} = \frac{3}{4}
$$
---
Solve for first box ($\frac{-18}{x} = \frac{3}{4}$):
Cross-multiply:
$$
-18 \cdot 4 = 3 \cdot x \\
-72 = 3x \\
x = -24
$$
✔ First box: -24
---
Solve for second box ($\frac{30}{y} = \frac{3}{4}$):
Cross-multiply:
$$
30 \cdot 4 = 3 \cdot y \\
120 = 3y \\
y = 40
$$
✔ Second box: 40
---
Final Answer for #1:
> $\frac{6}{8} = \frac{-18}{\boxed{-24}} = \frac{30}{\boxed{40}}$
---
2. Locate the points on a number line:
> $\frac{3}{6}, -\frac{1}{6}, 1, -\frac{5}{6}, 0$
First, simplify where possible:
- $\frac{3}{6} = \frac{1}{2} = 0.5$
- $-\frac{1}{6} \approx -0.166...$
- $1 = 1.0$
- $-\frac{5}{6} \approx -0.833...$
- $0 = 0$
Now, arrange them from least to greatest:
$$
-\frac{5}{6} < -\frac{1}{6} < 0 < \frac{3}{6} < 1
$$
On a number line, mark these positions approximately:
```
<---|-----|-----|-----|-----|-----|-----|-----|-----|-----|--->
-1 -5/6 -1/6 0 1/2 1
```
You would place dots at:
- $-\frac{5}{6}$ (just left of -0.8)
- $-\frac{1}{6}$ (just left of 0)
- $0$
- $\frac{1}{2}$ (midway between 0 and 1)
- $1$
✔ You can sketch this on paper or visualize it as above.
---
3. Determine which of the two rational numbers is greater:
> $-\frac{7}{12}, -\frac{8}{23}$
Both are negative. The *greater* number is the one *closer to zero* (less negative).
Compare their absolute values:
- $|\frac{7}{12}| = \frac{7}{12} \approx 0.583$
- $|\frac{8}{23}| \approx 0.348$
Since $0.348 < 0.583$, then $-\frac{8}{23}$ is *closer to zero* than $-\frac{7}{12}$.
Therefore:
✔ $-\frac{8}{23} > -\frac{7}{12}$
---
4. Arrange the following in descending order:
> $-\frac{5}{4}, \frac{2}{3}, -\frac{7}{10}, \frac{4}{5}$
Descending order = from largest to smallest.
Convert to decimals for easier comparison:
- $-\frac{5}{4} = -1.25$
- $\frac{2}{3} \approx 0.666...$
- $-\frac{7}{10} = -0.7$
- $\frac{4}{5} = 0.8$
Now order from largest to smallest:
$$
0.8 > 0.666... > -0.7 > -1.25
$$
Which corresponds to:
✔ $\frac{4}{5} > \frac{2}{3} > -\frac{7}{10} > -\frac{5}{4}$
---
5. Verify that $a + (b + c) = (a + b) + c$ using:
> $a = \frac{1}{2}, b = \frac{3}{4}, c = -\frac{2}{3}$
This is verifying the Associative Property of Addition.
---
Left Side: $a + (b + c)$
First compute $b + c$:
$$
\frac{3}{4} + (-\frac{2}{3}) = \frac{3}{4} - \frac{2}{3}
$$
Find common denominator (LCM of 4 and 3 is 12):
$$
= \frac{9}{12} - \frac{8}{12} = \frac{1}{12}
$$
Now add $a$:
$$
\frac{1}{2} + \frac{1}{12} = \frac{6}{12} + \frac{1}{12} = \frac{7}{12}
$$
✔ Left side = $\frac{7}{12}$
---
Right Side: $(a + b) + c$
First compute $a + b$:
$$
\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}
$$
Now add $c$:
$$
\frac{5}{4} + (-\frac{2}{3}) = \frac{5}{4} - \frac{2}{3}
$$
Common denominator (LCM of 4 and 3 is 12):
$$
= \frac{15}{12} - \frac{8}{12} = \frac{7}{12}
$$
✔ Right side = $\frac{7}{12}$
---
Conclusion: Both sides equal $\frac{7}{12}$ → Associative property verified.
✔ Verified: $a + (b + c) = (a + b) + c$
---
6. Simplify:
> $\frac{-4}{11} + \frac{-2}{3} - \left(\frac{-5}{9}\right)$
First, simplify the expression:
Note: subtracting a negative is adding a positive:
$$
= \frac{-4}{11} + \frac{-2}{3} + \frac{5}{9}
$$
Now, we need a common denominator to combine these fractions.
Denominators: 11, 3, 9
Prime factors:
- 11 → 11
- 3 → 3
- 9 → 3²
LCM = 11 × 3² = 11 × 9 = 99
Convert each fraction:
- $\frac{-4}{11} = \frac{-4 \times 9}{99} = \frac{-36}{99}$
- $\frac{-2}{3} = \frac{-2 \times 33}{99} = \frac{-66}{99}$
- $\frac{5}{9} = \frac{5 \times 11}{99} = \frac{55}{99}$
Now add:
$$
\frac{-36}{99} + \frac{-66}{99} + \frac{55}{99} = \frac{-36 -66 + 55}{99} = \frac{-47}{99}
$$
✔ Final Answer: $\boxed{\frac{-47}{99}}$
---
## ✔ Final Answers Summary:
1. $\frac{6}{8} = \frac{-18}{\boxed{-24}} = \frac{30}{\boxed{40}}$
2. Number line points: $-\frac{5}{6}, -\frac{1}{6}, 0, \frac{1}{2}, 1$ (from left to right)
3. Greater number: $\boxed{-\frac{8}{23}}$
4. Descending order: $\boxed{\frac{4}{5} > \frac{2}{3} > -\frac{7}{10} > -\frac{5}{4}}$
5. Verified: $a + (b + c) = (a + b) + c = \frac{7}{12}$
6. Simplified: $\boxed{\frac{-47}{99}}$
Let me know if you’d like a visual sketch for #2 or more explanation on any part!
Parent Tip: Review the logic above to help your child master the concept of rational number worksheet grade 8.