NCERT Solutions: The World of Numbers

Exercise Set 3.1 (Page 43)

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?

Ans: Rate: 15 ingots per 2 bags $\Rightarrow {\displaystyle \frac{15}{2}}$ ingots per bag.
For 12 bags: $\text{Ingots}=12\times {\displaystyle \frac{15}{2}}={\displaystyle \frac{180}{2}}=\mathbf{90}$ copper ingots.
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.
Ans: 11, 13, 17, 19 are all prime numbers – natural numbers greater than 1 divisible only by 1 and themselves. They are specifically the prime numbers between 10 and 20.

Next three primes after 19: 23, 29, 31.
(Verification: 20 = 2×10; 21 = 3×7; 22 = 2×11; 23 is prime; 24, 25, 26, 27, 28 are composite; 29 is prime; 30 is composite; 31 is prime.)
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.
Ans: No, natural numbers are not closed under subtraction.
Counter-examples:

• $3-5=-2$. Here $3,5\in \mathbb{N}$ but $-2\mathrm{\notin }\mathbb{N}$.
• $7-7=0$. Here $7\in \mathbb{N}$ but $0\mathrm{\notin }\mathbb{N}$.

Since the result of subtracting a larger natural number from a smaller one (or equal ones) lies outside $\mathbb{N}$, natural numbers are not closed under subtraction.*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?

Ans: One hand has 4 fingers (excluding thumb). Each finger has 3 joints. Using the thumb to point to each joint:$$4\text{ fingers}\times 3\text{ joints each}=\mathbf{12}\text{ counts on one hand}$$

This allows counting up to 12 on a single hand. This directly gives rise to base-12 (duodecimal) counting – the reason we have 12 inches in a foot, 12 months in a year, and 24 (= 2×12) hours in a day. Unlike base-10 (counting on fingertips), base-12 is more divisible (12 is divisible by 1, 2, 3, 4, 6, 12), making it practical for trade and measurement.

Think and Reflect (Page 46)

Q. Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt. (Hint: If someone takes away (-) four of your debts that are each worth ₹3 (that is, -3), you are effectively ₹12 richer! Therefore, (-3) × (-4) = +12.)
Ans: If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt. For example, if someone takes away (-) four of your debts each worth ₹3 (i.e., -3), you become ₹12 richer. Therefore, $(-3)\times (-4)=+12$.
In general: removing a debt = gaining a fortune, hence $(-)\times (-)=(+)$.

Exercise Set 3.2 (Page 46)

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?Ans: Noon temperature: $+4\mathrm{°}C$
Drop: $15\mathrm{°}C$
$$\text{Midnight temperature}=4-15=-\mathbf{11}\mathbf{°}\mathbf{C}$$
The midnight temperature is $-11\mathrm{°}C$ (11 degrees below zero).
Question 2: A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.
Ans: Using Brahmagupta’s convention (fortune = positive, debt = negative):
$$\text{Final standing}=(-850)+1200+(-450)$$$$=-850+1200-450$$$$=1200-1300=-\mathbf{\text{₹}}\mathbf{100}$$
The trader ends with a net debt of ₹100.
Question 3: Calculate the following using Brahmagupta’s laws: (i) (-12) × 5   (ii) (-8) × (-7)   (iii) 0 – (-14)   (iv) (-20) ÷ 4
Ans:

• (i) $(-12)\times 5=-\mathbf{60}$   [Debt × Fortune = Debt]
• (ii) $(-8)\times (-7)=+\mathbf{56}$   [Debt × Debt = Fortune]
• (iii) $0-(-14)=0+14=\mathbf{14}$   [Subtracting a negative = adding a positive]
• (iv) $(-20)÷4=-\mathbf{5}$   [Debt ÷ Fortune = Debt]

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).

Ans:

Real-world example: Suppose you have ₹10 in hand. Someone cancels (removes) a debt of ₹5 that you owe them. Cancelling a debt of ₹5 is the same as gaining ₹5.
$$10-(-5)=10+5=15$$
In Brahmagupta’s terms: removing (subtracting) a debt (negative) is equivalent to gaining a fortune (adding a positive). Hence subtracting a negative number = adding its positive counterpart.

Think and Reflect (Page 47)

Q. Can you explain why we need q≠0 in the definition of a rational number?
Ans: Division by zero is undefined in mathematics. If $q=0$, then $\frac{p}{q}$ would mean dividing $p$ by 0. For instance, $\frac{5}{0}$ has no defined value because no number multiplied by 0 gives 5. Hence we must require $q\mathrm{\ne }0$ for $\frac{p}{q}$ to be a meaningful number.

Think and Reflect (Page 49)

Question 1: While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?
Answer: Find the LCM of the denominators and convert the fractions into equivalent fractions with the same denominator.
Question 2: Verify the distributive law for rational numbers.
Answer: Take 1/2, 3/4 and 2/5
LHS: 1/2 × (3/4 + 2/5) = 1/2 × (15/20 + 8/20) = 1/2 × 23/20 = 23/40
RHS: (1/2 × 3/4) + (1/2 × 2/5) = 3/8 + 2/10 = 15/40 + 8/40 = 23/40
Since LHS = RHS, the distributive law is verified.

Exercise Set 3.3 (Page 49)

Question 1: Prove that the following rational numbers are equal: (i) 23 and 46   (ii) 54 and 108   (iii) −35 and −610   (iv) 93 and 3
Ans:
Using the equality rule: $\frac{a}{b}=\frac{c}{d}\iff ad=bc$.

• (i) $\frac{2}{3}$ vs $\frac{4}{6}$: $2\times 6=12$ and $3\times 4=12$. Since $12=12$, they are equal. 
• (ii) $\frac{5}{4}$ vs $\frac{10}{8}$: $5\times 8=40$ and $4\times 10=40$. Since $40=40$, they are equal. 
• (iii) $-\frac{3}{5}$ vs $-\frac{6}{10}$: $(-3)\times 10=-30$ and $5\times (-6)=-30$. Since $-30=-30$, they are equal. 
• (iv) $\frac{9}{3}$ vs $3=\frac{3}{1}$: $9\times 1=9$ and $3\times 3=9$. Since $9=9$, they are equal. 

Question 2: Find the sum: (i) 25+310   (ii) 712+58   (iii) −47+314

Ans:

• (i) LCM of 5 and 10 = 10: ${\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{10}}={\displaystyle \frac{4}{10}}+{\displaystyle \frac{3}{10}}={\displaystyle \frac{7}{10}}$
• (ii) LCM of 12 and 8 = 24: ${\displaystyle \frac{7}{12}}+{\displaystyle \frac{5}{8}}={\displaystyle \frac{14}{24}}+{\displaystyle \frac{15}{24}}={\displaystyle \frac{29}{24}}$
• (iii) LCM of 7 and 14 = 14: $-{\displaystyle \frac{4}{7}}+{\displaystyle \frac{3}{14}}=-{\displaystyle \frac{8}{14}}+{\displaystyle \frac{3}{14}}=-{\displaystyle \frac{5}{14}}$

Question 3: Find the difference: (i) 56−14   (ii) 118−34   (iii) −79−(−23)

Ans:

• (i) LCM of 6 and 4 = 12: ${\displaystyle \frac{5}{6}}-{\displaystyle \frac{1}{4}}={\displaystyle \frac{10}{12}}-{\displaystyle \frac{3}{12}}={\displaystyle \frac{7}{12}}$
• (ii) LCM of 8 and 4 = 8: ${\displaystyle \frac{11}{8}}-{\displaystyle \frac{3}{4}}={\displaystyle \frac{11}{8}}-{\displaystyle \frac{6}{8}}={\displaystyle \frac{5}{8}}$
• (iii) $-{\displaystyle \frac{7}{9}}-(-{\displaystyle \frac{2}{3}})=-{\displaystyle \frac{7}{9}}+{\displaystyle \frac{2}{3}}$. LCM = 9: $=-{\displaystyle \frac{7}{9}}+{\displaystyle \frac{6}{9}}=-{\displaystyle \frac{1}{9}}$

Question 4: Find the product: (i) 23×310   (ii) 711×58   (iii) −47×514

Ans:

• (i) ${\displaystyle \frac{2}{3}}\times {\displaystyle \frac{3}{10}}={\displaystyle \frac{6}{30}}={\displaystyle \frac{1}{5}}$
• (ii) ${\displaystyle \frac{7}{11}}\times {\displaystyle \frac{5}{8}}={\displaystyle \frac{35}{88}}$
• (iii) $-{\displaystyle \frac{4}{7}}\times {\displaystyle \frac{5}{14}}=-{\displaystyle \frac{20}{98}}=-{\displaystyle \frac{10}{49}}$

Question 5: Find the quotient: (i) 23÷310   (ii) 711÷58   (iii) −47÷514

Ans:

• (i) ${\displaystyle \frac{2}{3}}÷{\displaystyle \frac{3}{10}}={\displaystyle \frac{2}{3}}\times {\displaystyle \frac{10}{3}}={\displaystyle \frac{20}{9}}$
• (ii) ${\displaystyle \frac{7}{11}}÷{\displaystyle \frac{5}{8}}={\displaystyle \frac{7}{11}}\times {\displaystyle \frac{8}{5}}={\displaystyle \frac{56}{55}}$
• (iii) $-{\displaystyle \frac{4}{7}}÷{\displaystyle \frac{5}{14}}=-{\displaystyle \frac{4}{7}}\times {\displaystyle \frac{14}{5}}=-{\displaystyle \frac{56}{35}}=-{\displaystyle \frac{8}{5}}$

Question 6: Show that: (12+34)×83=12×83+34×83.

Ans:

LHS: $(\frac{1}{2}+\frac{3}{4})\times \frac{8}{3}=\frac{5}{4}\times \frac{8}{3}=\frac{40}{12}=\frac{10}{3}$
RHS: $\frac{1}{2}\times \frac{8}{3}+\frac{3}{4}\times \frac{8}{3}=\frac{4}{3}+\frac{6}{3}=\frac{10}{3}$
Since LHS = RHS = $\frac{10}{3}$, the distributive law is verified. 
Question 7: Simplify the following using the distributive property: 79(67−34).
Ans:
$$\frac{7}{9}(\frac{6}{7}-\frac{3}{4})=\frac{7}{9}\times \frac{6}{7}-\frac{7}{9}\times \frac{3}{4}=\frac{6}{9}-\frac{7}{12}=\frac{2}{3}-\frac{7}{12}$$
LCM of 3 and 12 = 12: $={\displaystyle \frac{8}{12}}-{\displaystyle \frac{7}{12}}={\displaystyle \frac{\mathbf{1}}{\mathbf{12}}}$
Question 8: Find the rational number x such that: 56(x+35)=56x+12.
Ans: Expand the LHS using distributivity:
$$\frac{5}{6}x+\frac{5}{6}\times \frac{3}{5}=\frac{5}{6}x+\frac{1}{2}$$
This equals the RHS for any value of $x$ – the equation is an identity (true for all rational $x$).

Think and Reflect (Page 51)

Try and represent 8/5 and -7/4 on a number line.

Exercise Set 3.4 (Page 52)

Question 1: Represent the rational numbers 23, −54 and 112 on a single number line.
Ans:

Fraction	Decimal	Position
-5/4 (green)	-1.25	Between -2 and -1, exactly 3/4 of the way from -2
2/3 (blue)	≈0.667	Between 0 and 1, exactly 2/3 of the way from 0
1½ (red)	1.5	Between 1 and 2, exactly halfway (1 + 1/2)

Question 2: Find three distinct rational numbers that lie strictly between −12 and 14.

Ans: Convert to common denominator 20: $-\frac{1}{2}=-\frac{10}{20}$ and $\frac{1}{4}=\frac{5}{20}$.

Three rational numbers: $-\frac{\mathbf{7}}{\mathbf{20}},\mathbf{0},\frac{\mathbf{3}}{\mathbf{20}}$.
Verification: $-\frac{10}{20}<-\frac{7}{20}<0<\frac{3}{20}<\frac{5}{20}$. 
Question 3: Simplify the expression: (−14)+(512).
Ans:
LCM of 4 and 12 = 12:
$$-\frac{1}{4}+\frac{5}{12}=-\frac{3}{12}+\frac{5}{12}=\frac{2}{12}=\frac{\mathbf{1}}{\mathbf{6}}$$
Question 4: A tailor has 1534 metres of fine silk. If making one kurta requires 214 metres of silk, exactly how many kurtas can he make?Ans: Convert to improper fractions: $15\frac{3}{4}=\frac{63}{4}$ m and $2\frac{1}{4}=\frac{9}{4}$ m.
$$\text{Number of kurtas}=\frac{63}{4}÷\frac{9}{4}=\frac{63}{4}\times \frac{4}{9}=\frac{63}{9}=\mathbf{7}$$
The tailor can make exactly 7 kurtas.
Question 5: Find three rational numbers between 3.1415 and 3.1416.
Ans:
Three examples: $\mathbf{3.14151},\mathbf{3.14153},\mathbf{3.14158}$
In fractional form: $\frac{314151}{100000}$, $\frac{314153}{100000}$, $\frac{314158}{100000}$.
*Question 6: Can you think of other way(s) to find a rational number between any two rational numbers?
Ans: Apart from the averaging method $\frac{a+b}{2}$, two other methods:

• Common denominator method: Write both fractions with the same denominator $q$. If the numerators differ by more than 1, any integer numerator between them gives a valid rational.
• Decimal method: Write both numbers as decimals. Insert a decimal number between them. Convert back to $\frac{p}{q}$ form.

Think and Reflect (Page 53)

Q. Can 2 be written as a rational number pq?
Ans: No. By proof by contradiction (Hippasus, c. 400 BCE), assuming $\sqrt{2}=\frac{p}{q}$ in lowest terms leads to both $p$ and $q$ being even, contradicting the assumption that they are co-prime. Therefore $\sqrt{2}$ is irrational and cannot be expressed as $\frac{p}{q}$.

Think and Reflect (Page 55)

Q. Try to prove the irrationality of 3 using proof by contradiction. Will the same approach work for 5, 7, or 10?
Ans: Proof: √3 is irrational

1. Assume $\sqrt{3}=\frac{p}{q}$, $\gcd (p,q)=1$.
2. Then $p^2=3q^2$, so $3\mid p^2$. Since 3 is prime, $3\mid p$. Write $p=3k$.
3. Then $3q^2=9k^2\Rightarrow q^2=3k^2$, so $3\mid q^2\Rightarrow 3\mid q$.
4. Both $p$ and $q$ divisible by 3 – contradiction. ∎

The same approach works for $\sqrt{5}$ and $\sqrt{7}$ (primes). For $\sqrt{10}$: assume $\sqrt{10}=\frac{p}{q}$, then $p^2=10q^2$. We can show $2\mid p$ (write $p=2k$), then $10q^2=4k^2\Rightarrow 5q^2=2k^2$, leading to contradictions with parity or divisibility. So yes, $\sqrt{10}$ is also irrational.
Question: We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
Answer: Lines of irrational length can be obtained using geometric constructions, such as the diagonal of a square. For example, a square with side 1 unit has a diagonal of length √2, which is irrational.

Think and Reflect (Page 56)

Question: Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
Answer:
To construct √3, first construct √2 using a right triangle with sides 1 and 1. Then take √2 as one side and 1 as the other side of a new right triangle; its hypotenuse will be √3.
To construct √5, take √4 (=2) as one side and 1 as the other side of a right triangle; the hypotenuse will be √5.
In general, to construct √n, use the previous length √(n-1) as one side and 1 as the other side of a right triangle. The hypotenuse gives √n.

Think and Reflect (Page 57)

Question: Try to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?
Answer:
10/3 = 3.333… (3 repeats)
11/12 = 0.91666… (6 repeats)
Observation:
Both are non-terminating decimals, but the digits repeat. Hence, rational numbers can have repeating decimal expansions.Think and Reflect (Page 58)Question: The decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?Answer: A decimal terminates when the fraction can be written with a denominator that is a power of 10, i.e., $10^k=2^k\times 5^k$ for some integer $k$. If $q=2^m\times 5^n$, we can multiply numerator and denominator by the appropriate power of 2 or 5 to make the denominator $2^{\max (m,n)}\times 5^{\max (m,n)}=10^{\max (m,n)}$. This gives a terminating decimal. If $q$ has any prime factor other than 2 or 5, this manoeuvre is impossible and the decimal repeats.

Exercise Set 3.5 (Page 61)

Question 1: Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 720, 415 and 13250. Then check your answers by explicitly performing the long divisions.

Ans:

• $\frac{7}{20}$: $20=2^2\times 5$. Only 2 and 5 → Terminating. $\frac{7}{20}=\mathbf{0.35}$
• $\frac{4}{15}$: $15=3\times 5$. Has prime factor 3 → Non-terminating repeating. $\frac{4}{15}=\mathbf{0.2}\bar{\mathbf{6}}$
• $\frac{13}{250}$: $250=2\times 5^3$. Only 2 and 5 → Terminating. $\frac{13}{250}=\mathbf{0.052}$

Question 2: Perform the long division for 113. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213? Now compute 313, 413, etc. What do you notice?

Ans:

$\frac{1}{13}=0.\overline{076923}$ (repeating block: 076923, period 6)

• $\frac{1}{13}=0.\overline{076923}$
• $\frac{2}{13}=0.\overline{153846}$
• $\frac{3}{13}=0.\overline{230769}$
• $\frac{4}{13}=0.\overline{307692}$
• $\frac{5}{13}=0.\overline{384615}$
• $\frac{6}{13}=0.\overline{461538}$

Observation: The decimal expansions use two cyclic sequences – 076923 and 153846 – which are cyclic rotations of each other. Unlike $\frac{1}{7}$, the cycle for $\frac{1}{13}$ splits into two groups.
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…

Ans:

• (i) $\sqrt{81}=9$ → Rational.
• (ii) $\sqrt{12}=2\sqrt{3}$. Since $\sqrt{3}$ is irrational → Irrational.
• (iii) $0.\bar{3}=\frac{1}{3}$ → Rational.
• (iv) $0.\overline{12345}$ – the block repeats → Rational.
• (v) $1.01001000100001\dots$ – each group of zeros increases; no block repeats → Irrational.
• (vi) If the decimal terminates at the last digit shown → Rational.

Question 4: The number 0.9‾ (which means 0.99999…) is a rational number. Using algebra, explain why 0.9‾ is exactly equal to 1.

Ans:

1. Let $x=0.\bar{9}=0.99999\dots$
2. Multiply both sides by 10: $10x=9.99999\dots$
3. Subtract: $10x-x=9.99999\dots -0.99999\dots$
4. $9x=9$, therefore $x=1$.

Hence $0.\bar{9}=1$ exactly. This is not an approximation – they represent the same real number.*Question 5: We have seen that the repeating block of 17 is a cyclic number. Try to find more numbers (n) whose reciprocals (1n) produce decimals with repeating blocks that are cyclic.

Ans:

A number $n$ gives a full-period cyclic number if $n$ is a prime and 10 is a primitive root modulo $n$. These are called full-reptend primes.

• $n=7$: period 6 → 142857 (cyclic)
• $n=17$: period 16 → 0588235294117647 (cyclic)
• $n=19$: period 18 → 052631578947368421 (cyclic)
• $n=23$: period 22 → cyclic

Think and Reflect (Page 64)

Question: Consider this puzzle: What is the square root of -1? We know that 1 × 1 = 1. We also know that (-1) × (-1) = 1. There is no real number that, when multiplied by itself, results in a negative number. Thus, -1 cannot exist on the number line. Explain.
Answer: The square of any real number is always non-negative. Since both 1 × 1 and (-1) × (-1) give 1, no real number squared can give -1. Therefore, √-1 is not a real number and cannot be represented on the number line.

End-of-Chapter Exercises (Page 63)

Question 1: Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal: (i) 350   (ii) 29

Ans:

• (i) $\frac{3}{50}$: $50=2\times 5^2$ → terminating. $\frac{3}{50}=\frac{6}{100}=\mathbf{0.06}$
• (ii) $\frac{2}{9}$: $9=3^2$ → non-terminating repeating. $=\mathbf{0.}\bar{\mathbf{2}}$

Question 2: Prove that 5 is an irrational number.

Ans:

Proof: √5 is irrational (by contradiction)

1. Assume $\sqrt{5}=\frac{p}{q}$, where $\gcd (p,q)=1$ and $q\mathrm{\ne }0$.
2. Squaring: $p^2=5q^2$. So $5\mid p^2$. Since 5 is prime, $5\mid p$. Write $p=5k$.
3. Substituting: $5q^2=25k^2\Rightarrow q^2=5k^2$, so $5\mid q$.
4. Both $p$ and $q$ divisible by 5, contradicting $\gcd (p,q)=1$.
   Therefore $\sqrt{5}$ is irrational.  

Question 3: Convert the following decimal numbers in the form of pq: (i) 12.6   (ii) 0.0120   (iii) 3.052‾   (iv) 1.235‾   (v) 0.23‾   (vi) 2.05‾   (vii) 2.125‾   (viii) 3.125‾   (ix) 2.1625‾

Ans:

• (i) $12.6=\frac{126}{10}=\frac{\mathbf{63}}{\mathbf{5}}$
• (ii) $0.0120=\frac{120}{10000}=\frac{\mathbf{3}}{\mathbf{250}}$
• (iii) $3.0\overline{52}$: $990x=3022\Rightarrow x=\frac{\mathbf{1511}}{\mathbf{495}}$
• (iv) $1.2\overline{35}$: $990x=1223\Rightarrow x=\frac{\mathbf{1223}}{\mathbf{990}}$
• (v) $0.\overline{23}$: $99x=23\Rightarrow x=\frac{\mathbf{23}}{\mathbf{99}}$
• (vi) $2.0\bar{5}$: $90x=185\Rightarrow x=\frac{\mathbf{37}}{\mathbf{18}}$
• (vii) $2.12\bar{5}$: $900x=1913\Rightarrow x=\frac{\mathbf{1913}}{\mathbf{900}}$
• (viii) $3.12\bar{5}$: $900x=2813\Rightarrow x=\frac{\mathbf{2813}}{\mathbf{900}}$
• (ix) $2.1\overline{625}$: $9990x=21604\Rightarrow x=\frac{\mathbf{10802}}{\mathbf{4995}}$

Question 4: Locate the following rational numbers on the number line: (i) 0.532   (ii) 1.15‾

Ans:

• (i) $0.532=\frac{133}{250}$. Lies between 0 and 1, at 532 thousandths.
• (ii) $1.1\bar{5}$: Let $x=1.1\bar{5}$. $90x=104\Rightarrow x=\frac{52}{45}\approx 1.156$. Lies between 1.1 and 1.2.

Question 5: Find 6 rational numbers between 3 and 4.

Ans: Multiply by 7: $3=\frac{21}{7}$ and $4=\frac{28}{7}$.

Six rational numbers: ${\displaystyle \frac{\mathbf{22}}{\mathbf{7}}},{\displaystyle \frac{\mathbf{23}}{\mathbf{7}}},{\displaystyle \frac{\mathbf{24}}{\mathbf{7}}},{\displaystyle \frac{\mathbf{25}}{\mathbf{7}}},{\displaystyle \frac{\mathbf{26}}{\mathbf{7}}},{\displaystyle \frac{\mathbf{27}}{\mathbf{7}}}$.
Question 6: Find 5 rational numbers between 25 and 35.
Ans: Multiply by 6: $\frac{2}{5}=\frac{12}{30}$ and $\frac{3}{5}=\frac{18}{30}$.
Five rationals: ${\displaystyle \frac{\mathbf{13}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{14}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{15}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{16}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{17}}{\mathbf{30}}}$.Question 7: Find 5 rational numbers between 16 and 25.

Ans: LCM of 6 and 5 is 30: $\frac{1}{6}=\frac{5}{30}$ and $\frac{2}{5}=\frac{12}{30}$.

Five rationals: ${\displaystyle \frac{\mathbf{6}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{7}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{8}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{9}}{\mathbf{30}}},{\displaystyle \frac{\mathbf{10}}{\mathbf{30}}}$
Simplified: ${\displaystyle \frac{1}{5}},{\displaystyle \frac{7}{30}},{\displaystyle \frac{4}{15}},{\displaystyle \frac{3}{10}},{\displaystyle \frac{1}{3}}$.

Question 8:  If x3+x5=1615, find the rational number x.

Ans: $$\frac{x}{3}+\frac{x}{5}=\frac{16}{15}⟹\frac{5x+3x}{15}=\frac{16}{15}⟹8x=16⟹\mathbf{x}=\mathbf{2}$$

Question 9: Let a and b be two non-zero rational numbers such that a+1b=0. Without assigning any numerical values, determine whether ab is positive or negative.

Ans:

From $a+\frac{1}{b}=0$, we get $a=-\frac{1}{b}$.
Therefore: $ab=(-\frac{1}{b})\cdot b=-1$.
Conclusion: $ab=-1$, which is negative.
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 p104. Is it necessary that the denominator in lowest form is divisible by 2⁴ or 5⁴?
Ans: A decimal terminating at the 4th decimal place has the form $0.d_1{d}_2{d}_3{d}_4$ which equals $\frac{p}{10^4}$ where $p$ is the 4-digit integer. Since the last digit $d_4\mathrm{\ne }0$, $p$ is not divisible by 10.
Is divisibility by 24 or 54 necessary? No – not necessarily both. For example $\frac{5}{10^4}=\frac{1}{2^4\times 5^3}$, which is divisible by $2^4$ but not $5^4$. The denominator in lowest form has the form $2^m\times 5^n$ with $\max (m,n)=4$, but both exponents need not equal 4.
Question 11: Without performing division, determine whether the decimal expansion of 18125 is terminating or non-terminating. If it terminates, state the number of decimal places.
Ans:
$125=5^3$. Only 5 as prime factor → Terminating.
$$\frac{18}{125}=\frac{18\times 2^3}{5^3\times 2^3}=\frac{144}{1000}=0.144$$
Terminates after 3 decimal places.
Question 12: A rational number in its lowest form has denominator 23×5. How many decimal places will its decimal expansion have?
Ans: We need equal powers of 2 and 5. Multiply by $5^2$: denominator becomes $2^3\times 5^3=10^3$.
The decimal expansion will have 3 decimal places (= $\max (3,1)=3$).

*Question 13: Let a=712 and b=56. Express both in the form k1m and k2m where k2−k1>6. Write exactly five distinct rational numbers lying between a and b. Explain why the condition k2−k1>n+1 is necessary to find n rational numbers between them.

Ans:

LCM of 12 and 6 = 12: $a=\frac{7}{12}$, $b=\frac{10}{12}$. Here $k_2-k_1=3$, which is not $>6$. Take $m=84$: $a=\frac{49}{84}$, $b=\frac{70}{84}$. Now $k_2-k_1=21>6$. 
Five rational numbers: $\frac{\mathbf{51}}{\mathbf{84}},\frac{\mathbf{55}}{\mathbf{84}},\frac{\mathbf{59}}{\mathbf{84}},\frac{\mathbf{63}}{\mathbf{84}},\frac{\mathbf{67}}{\mathbf{84}}$

Why k2−k1>n+1? We need $n$ integers strictly between $k_1$ and $k_2$. This requires $k_1+n<k_2$, i.e., $k_2-k_1>n$. More precisely, $k_2-k_1\ge n+1$ ensures at least $n$ integers exist strictly between the endpoints.

*Question 14: Three rational numbers x, y, z satisfy x+y+z=0 and xy+yz+zx=0. Show that all three must simultaneously be zero.

Ans: Proof

Using the identity: $(x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$
Substituting $x+y+z=0$ and $xy+yz+zx=0$:
$$0=x^2+y^2+z^2+0⟹x^2+y^2+z^2=0$$
Since $x^2,y^2,z^2\ge 0$ and their sum is 0, each must individually be 0.
$$x^2=0,y^2=0,z^2=0⟹x=y=z=0.\mathrm{■}$$

*Question 15: Show that the rational number (a+b)2 lies between the rational numbers a and b.

Ans:
Proof
Assume $a<b$.

Step 1: Adding $a$ to both sides of $a<b$: $2a<a+b\Rightarrow a<\frac{a+b}{2}$.
Step 2: Adding $b$ to both sides of $a<b$: $a+b<2b\Rightarrow \frac{a+b}{2}<b$.
Combining: $a<{\displaystyle \frac{a+b}{2}}<b$.  
Question 16: Find the lengths of the hypotenuses of all the right triangles in the square root spiral .

Ans:

Each triangle has one leg equal to the hypotenuse of the previous triangle and the other leg = 1.

Triangle	Legs	Hypotenuse
1st	1, 1	$\sqrt{2}$
2nd	$\sqrt{2}$, 1	$\sqrt{3}$
3rd	$\sqrt{3}$, 1	$\sqrt{4}=2$
4th	2, 1	$\sqrt{5}$
5th	$\sqrt{5}$, 1	$\sqrt{6}$
$n$th	$\sqrt{n}$, 1	$\sqrt{n+1}$

The hypotenuse of the $n$th triangle is $\sqrt{\mathbf{n}+\mathbf{1}}$.