Samacheer Kalvi Class 8 Maths Solutions Chapter 1 Numbers Exercise 1.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.2

Question 1. Fill in the blanks:
(i) The value of \( \frac{-5}{12}+\frac{7}{15} = \_\_\_\_\_\_\_\_. \)
(ii) The value of \( \left(\frac{-3}{6}\right) \times\left(\frac{18}{-9}\right) = \_\_\_\_\_\_\_\_. \)
(iii) The value of \( \left(\frac{-15}{23}\right) \div\left(\frac{30}{-46}\right) = \_\_\_\_\_\_\_\_. \)
(iv) The rational number \( \_\_\_\_\_\_\_\_ \) does not have a reciprocal.
(v) The multiplicative inverse of -1 is \( \_\_\_\_\_\_\_\_. \)
Answer:
(i) \( \frac{1}{20} \)
(ii) 1
(iii) 1
(iv) 0
(v) -1
In simple words: These questions test basic operations with rational numbers and properties like reciprocals and multiplicative inverses. Remember that adding fractions needs a common bottom number, while multiplying just needs you to multiply tops and bottoms.

🎯 Exam Tip: For fill-in-the-blanks, always perform the calculation carefully or recall the definition before writing your final answer to avoid common errors.

Question 2. Say True or False
(i) All rational numbers have an additive inverse.
(ii) The rational numbers that are equal to their additive inverses are 0 and -1.
(iii) The additive inverse of \( \frac{-11}{-17} \) is \( \frac{11}{17} \)
(iv) The rational number which is its own reciprocal is -1.
(v) The multiplicative inverse exists for all rational numbers.
Answer:
(i) True
(ii) False
(iii) False
(iv) True
(v) False
In simple words: We are looking at properties of rational numbers. The additive inverse of a number makes it zero when added, and the multiplicative inverse makes it one when multiplied. Not all numbers have a multiplicative inverse (like zero).

🎯 Exam Tip: Understand the exact definitions of additive inverse and multiplicative inverse. Remember that 0 does not have a multiplicative inverse, and \( \frac{-11}{-17} \) is the same as \( \frac{11}{17} \).

Question 3. Find the sum
(i) \( \frac{7}{5}+\frac{3}{5} \)
(ii) \( \frac{7}{5}+\frac{5}{7} \)
(iii) \( \frac{6}{5}+\left(\frac{-14}{15}\right) \)
(iv) \( -4 \frac{2}{3}+7 \frac{5}{12} \)
Answer:
(i) \( \frac{7}{5}+\frac{3}{5} = \frac{7+3}{5} = \frac{10}{5} = 2 \)
(ii) \( \frac{7}{5}+\frac{5}{7} = \frac{7 \times 7 + 5 \times 5}{35} = \frac{49+25}{35} = \frac{74}{35} \)
(iii) \( \frac{6}{5}+\left(\frac{-14}{15}\right) = \frac{6 \times 3 + (-14)}{15} = \frac{18+(-14)}{15} = \frac{4}{15} \)
(iv) \( -4 \frac{2}{3}+7 \frac{5}{12} = -\frac{14}{3}+\frac{89}{12} \)
\( \implies = \frac{-14 \times 4+89}{12} = \frac{-56+89}{12} = \frac{33}{12} = \frac{11}{4} \)
In simple words: When adding fractions, if the bottom numbers are the same, just add the top numbers. If the bottom numbers are different, you need to find a common bottom number before adding. Mixed numbers should be changed into improper fractions first.

🎯 Exam Tip: Always simplify fractions to their lowest terms after finding the sum. Pay close attention to negative signs in your calculations.

Question 4. Subtract \( \frac{-8}{44} \) from \( \frac{-17}{11} \)
Answer:
\( \frac{-17}{11} - \left(\frac{-8}{44}\right) = \frac{-17}{11} + \frac{8}{44} \)
\( \implies = \frac{-17 \times 4 + 8}{44} = \frac{-68+8}{44} \)
\( \implies = \frac{-60}{44} = \frac{-15}{11} \)
In simple words: To subtract fractions, first change "subtract a negative" into "add a positive". Then, find a common bottom number for both fractions. Finally, subtract the top numbers and simplify the result.

🎯 Exam Tip: Remember that subtracting a negative number is the same as adding a positive number. Always simplify the resulting fraction to its simplest form.

Question 5. Evaluate
(i) \( \frac{9}{132} \times \frac{-11}{3} \)
(ii) \( \frac{-7}{27} \times \frac{24}{-35} \)
Answer:
(i) \( \frac{9}{132} \times \frac{-11}{3} \)
\( \implies = \frac{\overset{3}{\cancel{9}}}{\underset{12}{\cancel{132}}} \times \frac{\overset{-1}{\cancel{-11}}}{\underset{1}{\cancel{3}}} \)
\( \implies = \frac{3 \times (-1)}{12 \times 1} = \frac{-3}{12} = \frac{-1}{4} \)
(ii) \( \frac{-7}{27} \times \frac{24}{-35} \)
\( \implies = \frac{\overset{1}{\cancel{-7}}}{\underset{9}{\cancel{27}}} \times \frac{\overset{8}{\cancel{24}}}{\underset{5}{\cancel{-35}}} \)
\( \implies = \frac{1 \times 8}{9 \times 5} = \frac{8}{45} \)
In simple words: To multiply fractions, you can cross-cancel numbers diagonally first to make them smaller. Then, multiply the top numbers together and the bottom numbers together. Make sure to simplify the final answer.

🎯 Exam Tip: Look for common factors in the numerators and denominators to simplify before multiplying. This makes the calculation much easier and reduces the chance of errors.

Question 6. Divide
(i) \( \frac{-21}{5} \) by \( \frac{-7}{-10} \)
(ii) \( \frac{-3}{13} \) by -3
(iii) -2 by \( \frac{-6}{15} \)
Answer:
(i) \( \frac{-21}{5} \div \frac{-7}{-10} \)
\( \implies = \frac{-21}{5} \div \frac{7}{10} \) (Since \( \frac{-7}{-10} = \frac{7}{10} \))
\( \implies = \frac{-21}{5} \times \frac{10}{7} \)
\( \implies = \frac{\overset{-3}{\cancel{-21}}}{\underset{1}{\cancel{5}}} \times \frac{\overset{2}{\cancel{10}}}{\underset{1}{\cancel{7}}} \)
\( \implies = \frac{-3 \times 2}{1 \times 1} = -6 \)
(ii) \( \frac{-3}{13} \div (-3) \)
\( \implies = \frac{-3}{13} \times \frac{1}{-3} \)
\( \implies = \frac{\overset{1}{\cancel{-3}}}{13} \times \frac{1}{\underset{1}{\cancel{-3}}} \)
\( \implies = \frac{1 \times 1}{13 \times 1} = \frac{1}{13} \)
(iii) \( -2 \div \frac{-6}{15} \)
\( \implies = -2 \times \frac{15}{-6} \)
\( \implies = \frac{\overset{1}{\cancel{-2}}}{1} \times \frac{15}{\underset{3}{\cancel{-6}}} \)
\( \implies = \frac{1 \times 15}{1 \times 3} = \frac{15}{3} = 5 \)
In simple words: When dividing fractions, flip the second fraction upside down (find its reciprocal) and then multiply it by the first fraction. Always simplify the result. Remember that a negative divided by a negative gives a positive.

🎯 Exam Tip: Changing division into multiplication by the reciprocal is a crucial step. Pay careful attention to the signs of the numbers to ensure the correct final sign.

Question 7. Find (a + b) ÷ (a – b) if
(i) \( a = \frac{1}{2}, b = \frac{2}{3} \)
(ii) \( a = \frac{-3}{5}, b = \frac{2}{15} \)
Answer:
(i) If \( a = \frac{1}{2}, b = \frac{2}{3} \)
First, find \( a+b \):
\( a+b = \frac{1}{2}+\frac{2}{3} = \frac{1 \times 3 + 2 \times 2}{6} = \frac{3+4}{6} = \frac{7}{6} \)
Next, find \( a-b \):
\( a-b = \frac{1}{2}-\frac{2}{3} = \frac{1 \times 3 - 2 \times 2}{6} = \frac{3-4}{6} = \frac{-1}{6} \)
Finally, find \( (a+b) \div (a-b) \):
\( (a+b) \div (a-b) = \frac{7}{6} \div \left(\frac{-1}{6}\right) = \frac{7}{6} \times \frac{6}{-1} = \frac{7 \times 1}{1 \times -1} = -7 \)
(ii) If \( a = \frac{-3}{5}, b = \frac{2}{15} \)
First, find \( a+b \):
\( a+b = \frac{-3}{5}+\frac{2}{15} = \frac{-3 \times 3 + 2}{15} = \frac{-9+2}{15} = \frac{-7}{15} \)
Next, find \( a-b \):
\( a-b = \frac{-3}{5}-\frac{2}{15} = \frac{-3 \times 3 - 2}{15} = \frac{-9-2}{15} = \frac{-11}{15} \)
Finally, find \( (a+b) \div (a-b) \):
\( (a+b) \div (a-b) = \frac{-7}{15} \div \left(\frac{-11}{15}\right) = \frac{-7}{15} \times \frac{15}{-11} = \frac{-7 \times 1}{-1 \times 11} = \frac{7}{11} \)
In simple words: To solve this, first add the two numbers, then subtract the second number from the first. After that, divide the sum by the difference. Remember to simplify all fractions at each step.

🎯 Exam Tip: Break down the problem into smaller, manageable steps: calculate (a+b), then (a-b), and finally perform the division. This helps prevent errors in complex fraction operations.

Question 8. Simplify \( \frac{1}{2}+\left(\frac{3}{2}-\frac{2}{5}\right) \div \frac{3}{10} \times 3 \) and show that it is a rational number between 11 and 12.
Answer:
We need to simplify the expression: \( \frac{1}{2}+\left(\frac{3}{2}-\frac{2}{5}\right) \div \frac{3}{10} \times 3 \)
First, calculate the expression inside the parenthesis:
\( \left(\frac{3}{2}-\frac{2}{5}\right) = \left(\frac{3 \times 5 - 2 \times 2}{10}\right) = \left(\frac{15-4}{10}\right) = \frac{11}{10} \)
Now substitute this back into the main expression:
\( \frac{1}{2} + \frac{11}{10} \div \frac{3}{10} \times 3 \)
Next, perform the division:
\( \frac{11}{10} \div \frac{3}{10} = \frac{11}{10} \times \frac{10}{3} = \frac{11}{3} \)
Now, the expression is:
\( \frac{1}{2} + \frac{11}{3} \times 3 \)
Next, perform the multiplication:
\( \frac{11}{3} \times 3 = 11 \)
Finally, perform the addition:
\( \frac{1}{2} + 11 = 11\frac{1}{2} = \frac{23}{2} \)
The simplified value is \( \frac{23}{2} \).
Since \( 11 = \frac{22}{2} \) and \( 12 = \frac{24}{2} \), we can see that \( \frac{23}{2} \) is indeed between 11 and 12. This result confirms it is a rational number because it can be written as a fraction of two integers.
In simple words: Follow the order of operations (BODMAS/PEMDAS): first brackets, then division, then multiplication, and finally addition. The answer should be a fraction, and we need to check if it sits between 11 and 12.

🎯 Exam Tip: Always follow the order of operations (PEMDAS/BODMAS) strictly. Perform calculations within parentheses first, then exponents, multiplication and division (from left to right), and finally addition and subtraction (from left to right).

Question 9. Simplify
(i) \( \left[\frac{11}{8} \times\left(\frac{-6}{33}\right)\right]+\left[\frac{1}{3}+\left(\frac{3}{5} \div \frac{9}{20}\right)\right]-\left[\frac{4}{7} \times \frac{-7}{5}\right] \)
(ii) \( \left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right]-\left[\frac{3}{4} \times \frac{4}{3}\right]+\left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right] \)
Answer:
(i) We simplify each bracket separately:
First bracket: \( \left[\frac{11}{8} \times\left(\frac{-6}{33}\right)\right] = \left[\frac{\overset{1}{\cancel{11}}}{\underset{4}{\cancel{8}}} \times \frac{\overset{-3}{\cancel{-6}}}{\underset{3}{\cancel{33}}}\right] = \left[\frac{1 \times (-3)}{4 \times 3}\right] = \left[\frac{-3}{12}\right] = \frac{-1}{4} \)
Second bracket: \( \left[\frac{1}{3}+\left(\frac{3}{5} \div \frac{9}{20}\right)\right] \)
Inside, division first: \( \frac{3}{5} \div \frac{9}{20} = \frac{3}{5} \times \frac{20}{9} = \frac{\overset{1}{\cancel{3}}}{\underset{1}{\cancel{5}}} \times \frac{\overset{4}{\cancel{20}}}{\underset{3}{\cancel{9}}} = \frac{1 \times 4}{1 \times 3} = \frac{4}{3} \)
Then, add to \( \frac{1}{3} \): \( \frac{1}{3}+\frac{4}{3} = \frac{1+4}{3} = \frac{5}{3} \)
Third bracket: \( \left[\frac{4}{7} \times \frac{-7}{5}\right] = \left[\frac{\overset{4}{\cancel{4}}}{\underset{1}{\cancel{7}}} \times \frac{\overset{-1}{\cancel{-7}}}{\underset{5}{\cancel{5}}}\right] = \left[\frac{4 \times (-1)}{1 \times 5}\right] = \frac{-4}{5} \)
Now combine the simplified brackets:
\( \frac{-1}{4} + \frac{5}{3} - \left(\frac{-4}{5}\right) = \frac{-1}{4} + \frac{5}{3} + \frac{4}{5} \)
Find the LCM of 4, 3, 5 which is 60.
\( = \frac{-1 \times 15}{60} + \frac{5 \times 20}{60} + \frac{4 \times 12}{60} \)
\( = \frac{-15}{60} + \frac{100}{60} + \frac{48}{60} = \frac{-15+100+48}{60} = \frac{133}{60} \)
(ii) We simplify each bracket separately:
First bracket: \( \left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right] = \left[\frac{4}{3} \times \frac{-7}{8}\right] = \left[\frac{\overset{1}{\cancel{4}}}{3} \times \frac{-7}{\underset{2}{\cancel{8}}}\right] = \frac{1 \times (-7)}{3 \times 2} = \frac{-7}{6} \)
Second bracket: \( \left[\frac{3}{4} \times \frac{4}{3}\right] = \left[\frac{\overset{1}{\cancel{3}}}{\underset{1}{\cancel{4}}} \times \frac{\overset{1}{\cancel{4}}}{\underset{1}{\cancel{3}}}\right] = 1 \)
Third bracket: \( \left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right] = \left[\frac{\overset{1}{\cancel{4}}}{3} \times \frac{-1}{\underset{1}{\cancel{4}}}\right] = \frac{1 \times (-1)}{3 \times 1} = \frac{-1}{3} \)
Now combine the simplified brackets:
\( \frac{-7}{6} - 1 + \left(\frac{-1}{3}\right) = \frac{-7}{6} - 1 - \frac{1}{3} \)
Find the LCM of 6 and 3 which is 6.
\( = \frac{-7}{6} - \frac{1 \times 6}{6} - \frac{1 \times 2}{6} = \frac{-7-6-2}{6} = \frac{-15}{6} = \frac{-5}{2} \)
In simple words: For these long problems, tackle one small part at a time. First, solve everything inside the parentheses. Remember to change division into multiplication by flipping the fraction. Combine the results carefully, paying attention to positive and negative signs.

🎯 Exam Tip: When simplifying expressions with multiple operations and brackets, follow the BODMAS/PEMDAS rule carefully. Simplify each bracket individually before combining them with addition or subtraction.

Question 10. A student had multiplied a number by \( \frac{4}{3} \) instead of dividing it by \( \frac{4}{3} \) and got 70 more than the correct answer. Find the number.
Answer:
Let the unknown number be 'a'.
The correct operation should have been dividing 'a' by \( \frac{4}{3} \). This is \( a \div \frac{4}{3} = a \times \frac{3}{4} = \frac{3a}{4} \).
The student did multiply 'a' by \( \frac{4}{3} \). This is \( a \times \frac{4}{3} = \frac{4a}{3} \).
The problem states that the student's answer was 70 more than the correct answer. So, we can write the equation:
\( \frac{4a}{3} - \frac{3a}{4} = 70 \)
To solve for 'a', find a common denominator for the fractions, which is 12.
\( \frac{4a \times 4}{12} - \frac{3a \times 3}{12} = 70 \)
\( \frac{16a}{12} - \frac{9a}{12} = 70 \)
\( \frac{16a - 9a}{12} = 70 \)
\( \frac{7a}{12} = 70 \)
To find 'a', multiply both sides by 12 and divide by 7:
\( 7a = 70 \times 12 \)
\( a = \frac{70 \times 12}{7} \)
\( a = 10 \times 12 \)
\( a = 120 \)
Thus, the number is 120.
In simple words: We set up an equation where the wrong answer minus the correct answer equals 70. Then, we solved this equation to find the mystery number. It helps to write division as multiplication by the reciprocal.

🎯 Exam Tip: Clearly define your variable (e.g., 'a' for the number). Formulate the equation accurately by translating the words into mathematical expressions. Remember that "dividing by a fraction" is the same as "multiplying by its reciprocal".

Objective Type Questions

Question 11. The standard form of the sum \( \frac{3}{4}+\frac{5}{6}+\left(\frac{-7}{12}\right) \) is
(A) 1
(B) \( \frac{-1}{2} \)
(C) \( \frac{1}{12} \)
(D) \( \frac{1}{22} \)
Answer: (A) 1
In simple words: To find the sum, add all the fractions together. First, find a common bottom number for all of them. Once they all have the same bottom number, add the top numbers. Then simplify the final fraction to its smallest form.

🎯 Exam Tip: For adding fractions, the least common multiple (LCM) of the denominators makes calculations simpler. Remember to handle negative signs correctly during addition.

Question 12. \( \left(\frac{3}{4}-\frac{5}{8}\right)+\frac{1}{2} = \_\_\_\_\_\_\_\_. \)
(A) \( \frac{15}{64} \)
(B) 1
(C) \( \frac{5}{8} \)
(D) \( \frac{1}{16} \)
Answer: (C) \( \frac{5}{8} \)
In simple words: First, solve the part inside the brackets by subtracting the fractions. Then, add the result to the last fraction. Make sure all fractions have the same bottom number before adding or subtracting.

🎯 Exam Tip: Always evaluate expressions within parentheses first. Finding the LCM for denominators simplifies the process of adding and subtracting fractions.

Question 13. \( \frac{3}{4}+\left(\frac{5}{8}+\frac{1}{2}\right) = \_\_\_\_\_\_\_\_. \)
(A) \( \frac{13}{10} \)
(B) \( \frac{2}{3} \)
(D) \( \frac{5}{8} \)
Answer: (B) \( \frac{2}{3} \)
In simple words: To solve this, first add the fractions inside the brackets. After that, add that result to the first fraction. Remember to find a common bottom number each time you add fractions.

🎯 Exam Tip: Prioritize operations within parentheses. When adding fractions, always ensure they share a common denominator before summing their numerators.

Question 14. \( \frac{3}{4} \times \left(\frac{5}{8} \times \frac{1}{2}\right) = \_\_\_\_\_\_\_\_. \)
(A) \( \frac{5}{8} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{15}{32} \)
(D) \( \frac{15}{16} \)
Answer: (D) \( \frac{15}{16} \)
In simple words: First, multiply the two fractions inside the brackets. Then, multiply that result by the first fraction. When multiplying fractions, just multiply the top numbers together and the bottom numbers together.

🎯 Exam Tip: For multiplication of fractions, simplify any common factors between numerators and denominators *before* multiplying to keep numbers smaller and reduce errors.

Question 15. Which of these rational numbers have an additive inverse?
(A) 7
(B) \( \frac{-5}{7} \)
(C) 0
(D) All of the options
Answer: (D) All of the options
All rational numbers have an additive inverse.
The additive inverse of 7 is -7.
The additive inverse of \( \frac{-5}{7} \) is \( \frac{5}{7} \).
The additive inverse of 0 is 0.
In simple words: The additive inverse of a number is the number that you add to it to get zero. Every single rational number, whether positive, negative, or zero, has its own additive inverse.

🎯 Exam Tip: Understand that the additive inverse simply changes the sign of a number (e.g., for \( x \), it's \( -x \)). This property holds true for all rational numbers without exception.