NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 03 The World of Numbers

Exercise Set 3.1

Question 1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
Answer: The merchant gets 15 ingots for every 2 bags of spices. First, find how many ingots he gets per bag - that's 15 divided by 2, which is 7.5 ingots. Now multiply this by 12 bags: 12 times 7.5 equals 90. So the merchant will take home 90 copper ingots.
In simple words: If 2 bags give 15 ingots, then 12 bags give 90 ingots.

Exam Tip: Set up a proportion or find the rate per bag first - this method works for all exchange problems.

Question 2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
Answer: These numbers are all prime numbers - numbers that can only be divided by 1 and themselves. The next three prime numbers after 19 are 23, 29, and 31.
In simple words: The pattern is prime numbers. They only have two factors: 1 and the number itself.

Exam Tip: Always identify what property the numbers share before finding the next ones in the pattern.

Question 3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.
Answer: Natural numbers are not closed under subtraction. Closure means the result must also be a natural number. For example, 5 - 3 gives 2, which is a natural number. However, 3 - 5 gives -2, which is not a natural number because it is negative. Since subtraction can produce negative numbers, the set of natural numbers does not have the closure property for subtraction.
In simple words: Natural numbers work fine for addition, but subtraction can give negative answers, which are not natural numbers.

Exam Tip: To show closure fails, you only need one counterexample where the operation gives a result outside the set.

Question 4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Answer: Each finger except the thumb has 3 joints. Since you have 4 usable fingers on one hand, the total number of joints is 4 times 3, which equals 12. You can count up to 12 using one hand this way. This naturally leads to a base-12 counting system. Since counting reaches 12 on one hand, ancient peoples developed the duodecimal (base-12) system for their calculations.
In simple words: One hand has 12 finger joints total, so people used base-12 counting instead of base-10.

Exam Tip: Explain both the calculation and its historical significance - show how practical methods led to different number systems.

Exercise Set 3.2

Question 1. The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?
Answer: The starting temperature is 4°C. When it drops by 15°C, you subtract: 4 minus 15 equals -11°C. Therefore, the midnight temperature is -11°C.
In simple words: Start at 4°C and go down 15 degrees, which takes you to -11°C.

Exam Tip: "Drops by" means subtract; "rises by" means add. Pay careful attention to these directional words.

Question 2. A spice trader takes a loan (debt) of Rs.850. The next day, he makes a profit (fortune) of Rs.1,200. The following week, he incurs a loss of Rs.450. Write this sequence as an equation using integers and calculate his final financial standing.
Answer: Represent the debt as -Rs.850, the profit as +Rs.1200, and the loss as -Rs.450. The equation becomes: -Rs.850 + Rs.1200 - Rs.450. Working through step by step: -850 plus 1200 gives 350, then subtract 450 to get -100. His final financial standing is -Rs.100, meaning he still owes Rs.100.
In simple words: Debts and losses are negative; profits and gains are positive. Add them all together to find the final result.

Exam Tip: Always assign the correct signs - debt and loss are negative, profit and gain are positive - before doing the arithmetic.

Question 3. Calculate the following using Brahmagupta's laws:
(i) (−12) × 5
(ii) (−8) × (−7)
(iii) 0 − (−14)
(iv) (−20) ÷ 4
Answer:
(i) According to Brahmagupta's rules, a negative number times a positive number gives a negative number (debt times fortune equals debt). So (−12) × 5 = −60.
(ii) A negative times a negative gives a positive (debt times debt equals fortune). So (−8) × (−7) = 56.
(iii) Zero minus a negative equals zero plus that positive number (removing a debt increases your wealth). So 0 − (−14) = 0 + 14 = 14.
(iv) A negative divided by a positive gives a negative (debt divided by fortune equals debt). So (−20) ÷ 4 = −5.
In simple words: Brahmagupta thought of negatives as debts and positives as fortunes. Debt times fortune stays debt; debt times debt becomes fortune; removing debt adds wealth.

Exam Tip: Remember the sign rules: negative and negative make positive; negative and positive make negative. The magnitude is found by ordinary arithmetic.

Question 4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 − (−5) = 15).
Answer: Imagine you have Rs.10. A negative amount represents a debt owed. So −Rs.5 means you owe Rs.5 to someone. Now, 10 − (−5) means you remove or cancel a debt of Rs.5. When your debt is removed, your money goes up by Rs.5. Therefore, 10 − (−5) = 10 + 5 = 15. Removing a debt has the same effect as gaining money.
In simple words: Canceling a debt is like getting extra money, so subtracting a negative is the same as adding.

Exam Tip: Use real-world scenarios like debts or temperature to make abstract rules concrete and memorable.

Exercise Set 3.3

Question 1. Prove that the following rational numbers are equal:
(i) 2/3 and 4/6
(ii) 5/4 and 10/8
(iii) -3/5 and -6/10
(iv) 9/3 and 3
Answer:
(i) To show two rational numbers are equal, simplify both or compare them. The first fraction 2/3 is already simplified. The second fraction 4/6 can be reduced by dividing both numerator and denominator by 2, giving 2/3. Since both equal 2/3, they are equal.
(ii) The first fraction 5/4 stays as is. For the second, divide both parts of 10/8 by 2 to get 5/4. Both equal 5/4, so they are equal.
(iii) The first fraction -3/5 stays as is. For the second, divide both parts of -6/10 by 2 to get -3/5. Both equal -3/5, so they are equal.
(iv) The fraction 9/3 simplifies to 3 by dividing 9 by 3. So 9/3 and 3 are equal.
In simple words: Two fractions are equal if they simplify to the same thing. Divide top and bottom by the same number to reduce a fraction.

Exam Tip: Always reduce fractions to simplest form - divide by the greatest common factor of the numerator and denominator.

Question 2. Find the sum:
(i) 2/5 + 3/10
(ii) 7/12 + 5/8
(iii) – 4/7 + 3/14
Answer:
(i) To add fractions, use a common denominator. The LCM of 5 and 10 is 10. Change 2/5 to 4/10 by multiplying both parts by 2. Now add: 4/10 + 3/10 = 7/10.
(ii) The LCM of 12 and 8 is 24. Change 7/12 to 14/24 and 5/8 to 15/24. Now add: 14/24 + 15/24 = 29/24.
(iii) The LCM of 7 and 14 is 14. Change -4/7 to -8/14. Now add: -8/14 + 3/14 = -5/14.
In simple words: Find the LCM of the denominators, convert both fractions, then add the numerators.

Exam Tip: Always use the smallest common denominator (LCM) to keep numbers manageable - it makes the final answer easier to check.

Question 3. Find the difference:
(i) 5/6 – 1/4
(ii) 11/8 – 3/4
(iii) -7/9 – (-2/3)
Answer:
(i) The LCM of 6 and 4 is 12. Change 5/6 to 10/12 and 1/4 to 3/12. Now subtract: 10/12 - 3/12 = 7/12.
(ii) The LCM of 8 and 4 is 8. Keep 11/8 and change 3/4 to 6/8. Now subtract: 11/8 - 6/8 = 5/8.
(iii) Rewrite -7/9 - (-2/3) as -7/9 + 2/3 (subtracting a negative becomes adding). The LCM of 9 and 3 is 9. Change 2/3 to 6/9. Now add: -7/9 + 6/9 = -1/9.
In simple words: Make the denominators match, then subtract the numerators. Remember that subtracting a negative is the same as adding.

Exam Tip: Convert subtraction of negatives to addition first - this prevents sign errors in the working.

Question 4. Find the product:
(i) 2/3 × 3/10
(ii) 7/11 × 5/8
(iii) -4/7 × 5/14
Answer:
(i) Multiply the numerators together and the denominators together: (2 × 3)/(3 × 10) = 6/30. Reduce by dividing both by 6 to get 1/5.
(ii) Multiply straight across: (7 × 5)/(11 × 8) = 35/88. This is already in simplest form.
(iii) Multiply: (-4 × 5)/(7 × 14) = -20/98. Reduce by dividing both by 2 to get -10/49.
In simple words: For fractions, multiply the tops together and the bottoms together. Then reduce the result if possible.

Exam Tip: Look for common factors to cancel before multiplying - this keeps the numbers smaller and easier to work with.

Question 5. Find the quotient:
(i) 2/3 ÷ 3/10
(ii) 7/11 ÷ 5/8
(iii) -4/7 ÷ 5/14
Answer:
(i) To divide by a fraction, multiply by its reciprocal. The reciprocal of 3/10 is 10/3. So: 2/3 × 10/3 = (2 × 10)/(3 × 3) = 20/9.
(ii) The reciprocal of 5/8 is 8/5. So: 7/11 × 8/5 = (7 × 8)/(11 × 5) = 56/55.
(iii) The reciprocal of 5/14 is 14/5. So: -4/7 × 14/5 = (-4 × 14)/(7 × 5) = -56/35. Reduce by dividing by 7 to get -8/5.
In simple words: Division by a fraction means multiply by the flipped version (reciprocal) of that fraction.

Exam Tip: Always flip and multiply for division - never forget the reciprocal step or your answer will be inverted.

Question 6. Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3
Answer: Start with the left side (LHS). First, add inside the brackets: 1/2 = 2/4, so 2/4 + 3/4 = 5/4. Then multiply: 5/4 × 8/3 = (5 × 8)/(4 × 3) = 40/12, which reduces to 10/3. Now work on the right side (RHS). Calculate 1/2 × 8/3 = 8/6 = 4/3. Calculate 3/4 × 8/3 = 24/12 = 6/3 = 2. Wait, let me recalculate: 3/4 × 8/3 = (3 × 8)/(4 × 3) = 24/12 = 2. Actually, 24/12 = 2, but we need 6/3 for the same denominator as 4/3. So 24/12 simplifies to 2, which is 6/3. Then 4/3 + 6/3 = 10/3. Both sides equal 10/3, so the equation is proven true. This shows the distributive property works with fractions.
In simple words: You can distribute multiplication over addition for fractions, just like with whole numbers - both ways give the same answer.

Exam Tip: Show all steps clearly: simplify inside brackets first on LHS, then distribute on RHS, then compare both final answers.

Question 7. Simplify the following using the distributive property: (7/9)(6/7 − 3/4).
Answer: Apply the distributive property by multiplying 7/9 by each term: 7/9 × 6/7 - 7/9 × 3/4. For the first product: (7 × 6)/(9 × 7) = 42/63 = 6/9 = 2/3. For the second product: (7 × 3)/(9 × 4) = 21/36 = 7/12. Now subtract: 2/3 - 7/12. Find a common denominator of 12: 2/3 = 8/12. So 8/12 - 7/12 = 1/12.
In simple words: Multiply each term in the parentheses separately, then combine the results.

Exam Tip: Check if numerators and denominators have common factors you can cancel before doing the final subtraction.

Question 8. Find the rational number x such that: (5/6)(x + 3/5) = (5/6)x + 1/2
Answer: Expand the left side by distributing 5/6: (5/6)x + (5/6 × 3/5) = (5/6)x + 1/2. Simplify the second term on the left: 5/6 × 3/5 = (5 × 3)/(6 × 5) = 15/30 = 1/2. So the left side becomes (5/6)x + 1/2, which exactly matches the right side. This equation is true for every rational number x, not just one specific value. Therefore, x can be any rational number.
In simple words: When you expand and simplify, both sides become identical. This means the equation works for all values of x.

Exam Tip: If after simplifying the equation is always true (like 1/2 = 1/2), then the variable can be any number from the given set.

Exercise Set 3.4

Question 1. Represent the rational numbers 2/3, -5/4 and 1½ on a single number line.
Answer: First, convert 1½ to an improper fraction: 1½ = 3/2. Next, change all fractions to decimals to see where they sit: -5/4 = -1.25, 2/3 ≈ 0.67, and 3/2 = 1.5. On the number line, -5/4 lies between -2 and -1 (closer to -1), 2/3 lies between 0 and 1 (closer to 1), and 3/2 lies between 1 and 2 (closer to 2). The order from left to right is: -5/4, then 2/3, then 3/2. Mark these points accordingly on the line.
In simple words: Convert to decimals to find approximate positions, then mark each point on the number line in the correct order.

Exam Tip: Always convert to decimals first if the fractions have different denominators - this makes ordering them much easier.

Question 2. Find three distinct rational numbers that lie strictly between -1/2 and 1/4.
Answer: Change both numbers to a common denominator of 4: -1/2 = -2/4. Now both numbers are -2/4 and 1/4. Look for numbers with denominator 4 that fall strictly between them: -1/4 (between -2/4 and 1/4), 0/4 = 0 (between -2/4 and 1/4), and 1/8 (which also falls between -2/4 and 1/4). Note: There are infinitely many rational numbers between any two rational numbers, not just three.
In simple words: Find a common denominator, then pick numbers with that denominator that sit between the two given numbers.

Exam Tip: Remember that there are always infinitely many rational numbers between two given rationals - this is a key property of the rational number system.

Question 3. Simplify the expression: (-1/4) + (5/12)
Answer: Use a common denominator of 12. Change -1/4 to -3/12 by multiplying both parts by 3. Now add: -3/12 + 5/12 = 2/12. Reduce by dividing both by 2 to get 1/6.
In simple words: Find the LCM of the denominators, convert both fractions, add them, then simplify.

Exam Tip: Always reduce your final answer to simplest form by dividing numerator and denominator by their GCD.

Question 4. A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?
Answer: Convert the mixed numbers to improper fractions: 15¾ = 63/4 metres and 2¼ = 9/4 metres per kurta. To find the number of kurtas, divide the total amount by the amount per kurta: (63/4) ÷ (9/4). To divide fractions, multiply by the reciprocal: (63/4) × (4/9) = (63 × 4)/(4 × 9) = 63/9 = 7. Therefore, the tailor can make exactly 7 kurtas.
In simple words: Divide the total fabric by the amount needed per garment to find how many can be made.

Exam Tip: Always convert mixed numbers to improper fractions before dividing - this prevents mistakes in the calculation.

Question 5. Find three rational numbers between 3.1415 and 3.1416.
Answer: Add extra decimal places between the two numbers: 3.1415 < 3.14151 < 3.14152 < 3.14153 < 3.1416. Three rational numbers that work are 3.14151, 3.14152, and 3.14153. Note: There are infinitely many rational numbers between any two given numbers - you could also use 3.141511, 3.141512, 3.141513, and so on.
In simple words: Insert more decimal places between the two given numbers to find rational numbers between them.

Exam Tip: When dealing with decimals, adding more decimal places is the easiest way to find numbers in between.

Question 6. Can you think of other way(s) to find a rational number between any two rational numbers?
Answer: Yes, there are several methods: First, take the average of the two numbers - if a and b are two rationals, then (a + b)/2 lies between them. Second, convert both to the same denominator and pick any fraction with that denominator that sits between them. Third, convert to decimal form and add more decimal places to create numbers in between. All three methods work because the set of rational numbers is dense, meaning there are infinitely many rationals between any two given rationals.
In simple words: You can use averages, common denominators, or decimal expansion - all three ways will give you rational numbers in between.

Exam Tip: The average method (a + b)/2 is often the quickest and easiest for finding just one number in between.

Exercise Set 3.5

Question 1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
Answer: A fraction in lowest terms has a terminating decimal if and only if the denominator contains only 2 and 5 as prime factors. If the denominator has any other prime factor, the decimal will repeat.
For 7/20: The prime factorization is 20 = 2² × 5. Since only 2 and 5 appear, this fraction terminates. By long division: 7/20 = 0.35 (terminating).
For 4/15: The prime factorization is 15 = 3 × 5. Since 3 is a prime factor (not 2 or 5), this fraction repeats. By long division: 4/15 = 0.2666... (repeating).
For 13/250: The prime factorization is 250 = 2 × 5³. Since only 2 and 5 appear, this fraction terminates. By long division: 13/250 = 0.052 (terminating).
In simple words: If the denominator has only 2s and 5s as factors, the decimal stops. If it has any other prime, the decimal repeats forever.

Exam Tip: Always factor the denominator first - this tells you whether the decimal terminates or repeats without needing to divide.

Question 2. Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?
Answer: Performing long division for 1/13: 1/13 = 0.076923076923..., so the repeating block is 076923. Now calculate the other fractions: 2/13 = 0.153846153846... with repeating block 153846; 3/13 = 0.230769230769... with repeating block 230769; 4/13 = 0.307692307692... with repeating block 307692; 5/13 = 0.384615384615... with repeating block 384615; 6/13 = 0.461538461538... with repeating block 461538. Notice that each repeating block is a cyclic shift of the same digits - they are rearrangements of the block 076923. This shows a beautiful cyclic property: the blocks cycle through different starting points but contain the same digits. Therefore, 1/13 produces a cyclic number.
In simple words: Each fraction 1/13, 2/13, 3/13 gives the same digits in a different cyclic order - they are cyclic shifts of one another.

Exam Tip: When working with fractions of the same denominator, look for patterns in the repeating blocks - they often have beautiful cyclic or symmetric properties.

Question 3. Classify the following numbers as rational or irrational:
(i) √81
(ii) √12
(iii) 0.33333...
(iv) 0.123451234512345...
(v) 1.01001000100001...
(vi) 23.560185612239874790120
Answer:
(i) √81 = 9, which can be written as 9/1. Since it can be expressed as a fraction of two integers, it is rational.
(ii) √12 = 2√3. Since √3 is irrational (it cannot be expressed as a ratio of integers), √12 is also irrational.
(iii) 0.33333... has the repeating block 3. A repeating decimal is rational. You can write it as the fraction 1/3.
(iv) 0.123451234512345... has the repeating block 12345. Since it repeats, it is rational. By algebraic manipulation: let x = 0.123451234512345..., then 100000x = 12345.123451234..., so 100000x - x = 12345, giving 99999x = 12345. Therefore x = 12345/99999 = 4115/33333 (after reducing).
(v) 1.01001000100001... is non-terminating and non-repeating. The number of zeros between the 1s keeps increasing - it never settles into a repeating pattern. Therefore, it is irrational.
(vi) 23.560185612239874790120 is a terminating decimal (it ends at the last digit). Every terminating decimal is rational. It equals 23560185612239874790120 / 1000000000000000000000.
In simple words: Rational numbers can be fractions or either terminating or repeating decimals. Irrational numbers are non-terminating and non-repeating.

Exam Tip: Check: does it terminate (rational), repeat (rational), or neither (irrational)? This is the fastest way to classify.

Question 4. The number 0.9̅ (which means 0.99999...) is a rational number. Using algebra (let x = 0.9̅, multiply by 10, and subtract), explain why 0.9̅ is exactly equal to 1.
Answer: Let x = 0.99999... Multiply both sides by 10: 10x = 9.99999... Subtract the original equation from this new one: 10x - x = 9.99999... - 0.99999... This gives 9x = 9, so x = 1. But we started with x = 0.99999..., so therefore 0.99999... = 1. The surprising result shows that the repeating decimal 0.9̅ is exactly equal to the whole number 1, not just approximately.
In simple words: When you subtract to get rid of the repeating part, you find that 0.99999... actually equals 1.

Exam Tip: This shows that one number can have two different decimal representations - a key insight into decimal notation.

Question 5. We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.
Answer: Calculate the decimals for the reciprocals of various numbers. For 1/7 = 0.142857142857..., the repeating block is 142857, which is cyclic. Other examples with cyclic repeating blocks are: 1/17 = 0.0588235294117647... (repeating block: 0588235294117647); 1/19 = 0.052631578947368421... (repeating block: 052631578947368421). These repeating blocks show cyclic properties - if you shift the digits, you get the decimal expansion of another fraction with the same denominator. Therefore, numbers like 7, 17, and 19 have the property that their reciprocals produce cyclic repeating decimals.
In simple words: Prime numbers like 7, 17, and 19 often produce cyclic repeating patterns in their reciprocals' decimal expansions.

Exam Tip: Cyclic numbers are special - their reciprocals show beautiful mathematical patterns that connect number theory, decimals, and group theory.

End of Chapter Exercises

Question 1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(i) 3/50
(ii) 2/9
Answer:
(i) Divide 3 by 50. The calculation gives 3/50 = 0.06, which stops after two decimal places. Therefore, 3/50 is a terminating decimal.
(ii) Divide 2 by 9. The calculation gives 2/9 = 0.2222..., in which the digit 2 repeats endlessly. This is written as 0.2̅. Therefore, 2/9 is a non-terminating repeating decimal.
In simple words: Some divisions finish with a remainder of zero (terminating), while others cycle through the same digits forever (repeating).

Exam Tip: Use long division and watch what happens to the remainder - if it becomes zero, the decimal terminates; if it repeats, the decimal repeats.

Question 2. Prove that √5 is an irrational number.
Answer: Use proof by contradiction. Suppose √5 is rational. Then it can be written as p/q, where p and q are integers, q ≠ 0, and p/q is in lowest terms (meaning p and q share no common factor except 1). Square both sides: 5 = p²/q², so p² = 5q². This means p² is divisible by 5, which implies p is also divisible by 5 (since if p were not divisible by 5, then p² would not be either). Write p = 5k for some integer k. Substitute into p² = 5q²: (5k)² = 5q², which gives 25k² = 5q², so 5k² = q². This means q² is divisible by 5, so q is also divisible by 5. Now both p and q are divisible by 5. But this contradicts our assumption that p and q have no common factor. Therefore, our supposition was wrong, and √5 must be irrational.
In simple words: Assume it is rational, then prove this leads to a contradiction. Since the assumption fails, it must be irrational.

Exam Tip: Proof by contradiction is powerful - assume the opposite of what you want to prove, then show this leads to an impossibility.

Question 3. Convert the following decimal numbers in the form of p/q:
(i) 12.6
(ii) 0.0120
(iii) 3.05̅2̅
(iv) 1.23̅5̅
(v) 0.2̅3̅
(vi) 2.05̅
(vii) 2.125̅
(viii) 3.125̅
(ix) 2.1̅6̅2̅5̅
Answer:
(i) Write 12.6 as a fraction: 12.6 = 126/10. Reduce by dividing both by 2: 126/10 = 63/5.
(ii) Write 0.0120 as a fraction: 0.0120 = 120/10000. Reduce by dividing both by 40: 120/10000 = 3/250.
(iii) Let x = 3.052525252... Then 10x = 30.52525... and 1000x = 3052.52525... Subtract: 1000x - 10x = 3052.525... - 30.525..., giving 990x = 3022. So x = 3022/990. Reduce by dividing by 2: x = 1511/495.
(iv) Let x = 1.235353... Then 10x = 12.353... and 1000x = 1235.353... Subtract: 1000x - 10x = 1235.353... - 12.353..., giving 990x = 1223. So x = 1223/990.
(v) Let x = 0.232323... Then 100x = 23.232323... Subtract: 100x - x = 23.232... - 0.232..., giving 99x = 23. So x = 23/99.
(vi) Let x = 2.05555... Then 10x = 20.5555... and 100x = 205.5555... Subtract: 100x - 10x = 205.555... - 20.555..., giving 90x = 185. So x = 185/90. Reduce by dividing by 5: x = 37/18.
(vii) Let x = 2.125555... Then 100x = 212.5555... and 1000x = 2125.5555... Subtract: 1000x - 100x = 2125.555... - 212.555..., giving 900x = 1913. So x = 1913/900.
(viii) Let x = 3.125555... Then 100x = 312.5555... and 1000x = 3125.5555... Subtract: 1000x - 100x = 3125.555... - 312.555..., giving 900x = 2813. So x = 2813/900.
(ix) Let x = 2.162516251625... Then 10000x = 21625.162516... Subtract: 10000x - x = 21625.1625... - 2.1625..., giving 9999x = 21623. So x = 21623/9999.
In simple words: For terminating decimals, write as a fraction with a power of 10 in the denominator. For repeating decimals, multiply by a suitable power of 10 to shift the decimal point, then subtract to eliminate the repeating part.

Exam Tip: Choose the power of 10 based on the length of the repeating block - multiply by 10^n where n is the number of repeating digits.

Question 4. Locate the following rational numbers on the number line:
(i) 0.532
(ii) 1.15̅
Answer:
(i) The decimal 0.532 lies between 0 and 1. More precisely, 0.532 = 532/1000, so it is slightly past the 0.53 mark. It would be positioned between 0.53 and 0.54, very close to 0.53.
(ii) The decimal 1.15̅ = 1.15555... lies between 1 and 2. More precisely, 1.15 < 1.15555... < 1.16, so it sits between 1.15 and 1.16, just slightly past the 1.15 mark.
In simple words: Find the integer interval, then estimate where within that interval the decimal falls based on the first few digits.

Exam Tip: For marking decimals on a number line, convert to approximate position between two consecutive integers first.

Question 5. Find 6 rational numbers between 3 and 4.
Answer: Express both whole numbers with a common denominator: 3 = 21/7 and 4 = 28/7. Now list six fractions with denominator 7 that fall between these: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7. Each of these lies strictly between 21/7 and 28/7, so each lies between 3 and 4. These six rational numbers work: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.
In simple words: Use a common denominator and pick integers for the numerator that are between the original numerators.

Exam Tip: The more you multiply the original numbers, the more fractions you can fit between them - choose a denominator large enough to get the number you need.

Question 6. Find 5 rational numbers between 2/5 and 3/5.
Answer: Convert to a larger common denominator to create space for five fractions: 2/5 = 20/50 and 3/5 = 30/50. Now list five fractions with denominator 50 that fall strictly between these: 21/50, 22/50, 23/50, 24/50, 25/50. Each of these lies strictly between 20/50 and 30/50, so each lies between 2/5 and 3/5. These five rational numbers work: 21/50, 22/50, 23/50, 24/50, 25/50.
In simple words: Multiply both fractions by a larger number to get a bigger denominator with more space between them for finding intermediate fractions.

Exam Tip: If you need n rational numbers between two given fractions, use a denominator that creates at least n + 1 gaps between the numerators.

Question 7. Find 5 rational numbers between 1/6 and 2/5.
Answer: First, find a common denominator for 1/6 and 2/5. The LCM of 6 and 5 is 30: 1/6 = 5/30 and 2/5 = 12/30. Now list five fractions with denominator 30 that fall strictly between these: 6/30, 7/30, 8/30, 9/30, 10/30. These all lie strictly between 5/30 and 12/30. In simplest form, they are: 1/5, 7/30, 4/15, 3/10, 1/3. These five rational numbers work: 6/30, 7/30, 8/30, 9/30, 10/30 (or in reduced form: 1/5, 7/30, 4/15, 3/10, 1/3).
In simple words: Convert to a common denominator, then pick five numerators that sit between the original numerators.

Exam Tip: You can leave answers with the working denominator or reduce them - both forms are correct.

Question 8. If x/3 + x/5 = 16/15, find the rational number x.
Answer: Factor out x from the left side: x(1/3 + 1/5) = 16/15. Calculate 1/3 + 1/5 using a common denominator of 15: 1/3 = 5/15 and 1/5 = 3/15, so 1/3 + 1/5 = 8/15. Now the equation becomes x × 8/15 = 16/15. Divide both sides by 8/15, which means multiply by its reciprocal: x = (16/15) × (15/8) = (16 × 15)/(15 × 8) = 16/8 = 2. Therefore, x = 2.
In simple words: Factor out x, simplify the expression in parentheses, then divide to isolate x.

Exam Tip: Always factor out the common variable first when it appears in multiple terms - this greatly simplifies the equation.

Question 9. Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
Answer: Start with the given equation a + 1/b = 0. Rearrange to get a = -1/b. Now multiply both sides by b: ab = -1. Since the product equals -1 (which is negative), ab is negative. The justification is that -1 is always negative regardless of which specific values a and b take, as long as they satisfy the given condition.
In simple words: From the condition, you can show that ab always equals -1, which is negative.

Exam Tip: Use algebraic manipulation to express ab in terms of constants, then determine the sign without needing specific numbers.

Question 10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.
Answer: If the last non-zero digit appears in the 4th decimal place, the number has exactly 4 decimal places. Any such number can be written as p/10⁴ for some integer p. Since the last non-zero digit is in the 4th place, p cannot be divisible by 10 - if it were, the decimal would terminate earlier. Thus, p is not divisible by 10, and the number can be written as p/10⁴ as required. Now, 10⁴ = 2⁴ × 5⁴. When the fraction p/10⁴ is reduced to lowest form, common factors between p and 10⁴ cancel out. Therefore, it is not necessary that the reduced denominator is divisible by 2⁴ or 5⁴. Example: 0.1250 = 1250/10000. Reduce by dividing by 1250: 1250/10000 = 1/8 = 1/2³. The denominator 8 = 2³ is divisible by 2³ but not by 2⁴ or 5⁴.
In simple words: Express with denominator 10⁴, but when you reduce, common factors cancel and the final denominator may be much smaller.

Exam Tip: Always reduce fractions fully to lowest form - the final denominator may have fewer factors than the original denominator had.

Question 11. Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.
Answer: Check the prime factorization of the denominator 125 = 5³. Since 125 contains only 5 as a prime factor (no 2s), the decimal expansion is terminating. To find how many decimal places, express the denominator as a power of 10. Multiply numerator and denominator by 8 (since 5³ × 8 = 5³ × 2³ = 10³ = 1000): 18/125 = (18 × 8)/(125 × 8) = 144/1000 = 0.144. The decimal has 3 decimal places. Therefore, 18/125 is a terminating decimal with exactly 3 decimal places.
In simple words: If the denominator has only 2s and 5s, the decimal terminates. The number of places equals the highest power of 2 or 5 in the factorization.

Exam Tip: To find the number of decimal places, multiply to make the denominator a power of 10 - the exponent on 10 is the number of places.

Question 12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Answer: The denominator is 2³ × 5 = 8 × 5 = 40. A rational number whose denominator has the form 2ᵐ × 5ⁿ has a terminating decimal expansion. The number of decimal places is determined by the greater of m and n. Here, m = 3 (the power of 2) and n = 1 (the power of 5). Since 3 is greater than 1, the decimal expansion has 3 decimal places. Example: 1/40 = 0.025, which indeed has 3 decimal places. Therefore, the decimal expansion will have 3 decimal places.
In simple words: Count the powers of 2 and 5 separately. The number of decimal places is whichever count is larger.

Exam Tip: This rule (number of places = max(m, n) where denominator = 2ᵐ × 5ⁿ) is a quick way to answer without long division.

Question 13. Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ - k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ - k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Answer: Convert both fractions to the same denominator. Since 5/6 = 10/12, we have a = 7/12 and b = 10/12. Here k₂ - k₁ = 10 - 7 = 3, which is not greater than 6. To satisfy the condition, multiply both fractions by 3: a = 7/12 = 21/36 and b = 10/12 = 30/36. Now k₁ = 21, k₂ = 30, m = 36, and k₂ - k₁ = 30 - 21 = 9, which is greater than 6. Five distinct rational numbers with denominator 36 that lie strictly between 21/36 and 30/36 are: 22/36, 23/36, 24/36, 25/36, 26/36. Why is k₂ - k₁ > n + 1 necessary? To find n rational numbers between k₁/m and k₂/m, you need at least n integers that lie strictly between k₁ and k₂. The integers strictly between k₁ and k₂ are k₁ + 1, k₁ + 2, ..., k₂ - 1, and there are exactly (k₂ - k₁ - 1) of them. To have at least n such integers, you need k₂ - k₁ - 1 ≥ n, or equivalently, k₂ - k₁ ≥ n + 1. For safety and to ensure you have enough room, the condition is stated as k₂ - k₁ > n + 1.
In simple words: The gap between the numerators must be large enough to fit at least n integers inside it. Each integer gives you one rational number.

Exam Tip: This method is powerful - by choosing a large enough denominator, you can find as many rational numbers as you want between any two given rationals.

Question 14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
Answer: Use the algebraic identity (x + y + z)² = x² + y² + z² + 2(xy + yz + zx). Substitute the given conditions: From the first equation, (x + y + z)² = 0² = 0. From the second equation, xy + yz + zx = 0. Plugging into the identity: 0 = x² + y² + z² + 2(0), which gives x² + y² + z² = 0. Now, x², y², and z² are all non-negative rational numbers (since the square of any real number is non-negative). The only way a sum of non-negative numbers can equal zero is if each term is zero. Therefore, x² = 0, y² = 0, and z² = 0. Taking square roots: x = 0, y = 0, and z = 0. Thus, all three rational numbers must be simultaneously zero.
In simple words: Use an identity to show that x² + y² + z² = 0. Since squares are non-negative, each must be zero, so all variables are zero.

Exam Tip: When a sum of non-negative quantities equals zero, each quantity must be zero - this is a powerful technique for proving equalities.

Question 15. Show that the rational number (a + b) / 2 lies between the rational numbers a and b.
Answer: Assume a < b and show that a < (a + b) / 2 < b. First, show a < (a + b) / 2: Since a < b, add a to both sides to get 2a < a + b. Divide both sides by 2: a < (a + b) / 2. Next, show (a + b) / 2 < b: Since a < b, add b to both sides to get a + b < 2b. Divide both sides by 2: (a + b) / 2 < b. Combining these results: a < (a + b) / 2 < b. Therefore, the rational number (a + b) / 2, which is the average of a and b, lies strictly between them. Note: If b < a, the same reasoning gives b < (a + b) / 2 < a. In either case, the average lies between the two numbers.
In simple words: The average of two numbers always sits between them - this is the simplest way to find a number in between any two given numbers.

Exam Tip: The average method (a + b)/2 is the quickest way to find a rational number strictly between two given rationals.

Question 16. Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
Answer: In the square root spiral, each successive right triangle is made by using one leg as 1 unit and the other leg as the hypotenuse of the triangle before it. Applying Pythagoras' theorem, we get:

First triangle: Sides = 1, 1 → Hypotenuse = \( \sqrt{1^2 + 1^2} = \sqrt{2} \)

Second triangle: Sides = \( \sqrt{2} \), 1 → Hypotenuse = \( \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3} \)

Third triangle: Sides = \( \sqrt{3} \), 1 → Hypotenuse = \( \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \)

Fourth triangle: Sides = 2, 1 → Hypotenuse = \( \sqrt{2^2 + 1^2} = \sqrt{5} \)

Fifth triangle: Sides = \( \sqrt{5} \), 1 → Hypotenuse = \( \sqrt{6} \)

Sixth triangle: Hypotenuse = \( \sqrt{7} \)

Seventh triangle: Hypotenuse = \( \sqrt{8} = 2\sqrt{2} \)

Eighth triangle: Hypotenuse = \( \sqrt{9} = 3 \)

Ninth triangle: Hypotenuse = \( \sqrt{10} \)

Tenth triangle: Hypotenuse = \( \sqrt{11} \)

The hypotenuse lengths follow a clear pattern: \( \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}, \sqrt{10}, \sqrt{11}, \ldots \), which simplifies to \( \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, 2\sqrt{2}, 3, \sqrt{10}, \sqrt{11}, \ldots \) The general pattern shows that each hypotenuse equals the square root of an integer starting from 2 and increasing by 1 for each new triangle in the spiral.
In simple words: Each triangle adds one side that is 1 unit long. The other side is the answer from the triangle before it. Using the Pythagoras rule, the hypotenuses come out to be \( \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5} \) and so on.

Exam Tip: Remember the key pattern - hypotenuses are \( \sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots \) in order. Always verify at least the first two or three calculations using the Pythagoras theorem to show you understand the method, not just the answer.