The World of Numbers — NCERT Solutions

Convert to improper fractions:

$15\frac{3}{4} = \frac{63}{4}$

$2\frac{1}{4} = \frac{9}{4}$

Now divide:

$\frac{63}{4} \div \frac{9}{4} = \frac{63}{4}\times\frac{4}{9} = \frac{63}{9} = 7$

So he can make 7 kurtas.

Any three rational numbers strictly between 3.1415 and 3.1416 are acceptable.

For example:

$3.14151,\ 3.14152,\ 3.14153$

All three lie between the given numbers.

Yes. One simple way is to take the average of the two rational numbers. If the numbers are $a$ and $b$, then

$\frac{a+b}{2}$

lies between them and is also rational, because the sum and division by 2 of rational numbers is rational.

Another way is to divide the interval into equal parts and choose any point in between, such as using a common denominator.

Use a common denominator of 12:

$-\frac{1}{4} = -\frac{3}{12}$

So,

$-\frac{3}{12} + \frac{5}{12} = \frac{2}{12} = \frac{1}{6}$

A rational number has a terminating decimal if, in lowest terms, its denominator has only prime factors 2 and/or 5.

• $\frac{7}{20}$: $20=2^2\times 5$, so it is terminating.
• $\frac{4}{15}$: $15=3\times 5$, so it has a prime factor other than 2 or 5; it is repeating.
• $\frac{13}{250}$: $250=2\times 5^3$, so it is terminating.

So the classifications are as above.

Perform long division of $1$ by $13$:

$1.000000\dots \div 13 = 0.076923076923\dots$

So,

$\frac{1}{13} = 0.\overline{076923}$

The repeating block is 076923.

Now for multiples:

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

Notice that the same digits keep appearing in shifted order, showing a cyclic pattern.

Classify each number:

(i) $\sqrt{81}=9$, so it is rational.

(ii) $\sqrt{12}=2\sqrt{3}$, and $\sqrt{3}$ is irrational, so this is irrational.

(iii) $0.33333\dots$ is a repeating decimal, so it is rational.

(iv) $0.123451234512345\dots$ does not repeat with a fixed block, so it is irrational.

(v) $1.01001000100001\dots$ does not repeat, so it is irrational.

(vi) $23.5601856122398747901120$ is a terminating decimal, so it is rational.

Let $x = 0.\overline{9}$.

Then

$10x = 9.\overline{9}$

Subtract:

$10x - x = 9.\overline{9} - 0.\overline{9}$

$9x = 9$

$x = 1$

So $0.\overline{9} = 1$.

One more famous example is $\frac{1}{7}$, whose repeating block is 142857.

Other reciprocals that can show interesting repeating blocks include:

• $\frac{2}{7} = 0.\overline{285714}$
• $\frac{3}{7} = 0.\overline{428571}$
• $\frac{4}{7} = 0.\overline{571428}$
• $\frac{5}{7} = 0.\overline{714285}$
• $\frac{6}{7} = 0.\overline{857142}$

These are all cyclic shifts of the same digits, so they also show cyclic properties.

One more famous example is $\frac{1}{7}$, whose repeating block is 142857.

Other reciprocals that can show interesting repeating blocks include:

• $\frac{2}{7} = 0.\overline{285714}$
• $\frac{3}{7} = 0.\overline{428571}$
• $\frac{4}{7} = 0.\overline{571428}$
• $\frac{5}{7} = 0.\overline{714285}$
• $\frac{6}{7} = 0.\overline{857142}$

These are all cyclic shifts of the same digits, so they also show cyclic properties.

$50 = 2\times 5^2$, so the decimal is terminating.

Long division gives:

$\frac{3}{50} = \frac{6}{100} = 0.06$

So the decimal form is $0.06$.

Assume, for contradiction, that $\sqrt{5}$ is rational. Then we can write

$$\sqrt{5}=\frac{p}{q}$$

where $p$ and $q$ are integers in lowest form and $q\neq 0$.

Squaring both sides gives

$$5=\frac{p^2}{q^2}$$

so

$$5q^2=p^2.$$

Thus $p^2$ is divisible by $5$, so $p$ is divisible by $5$. Let $p=5k$ for some integer $k$.

Substitute back:

$$5q^2=(5k)^2=25k^2$$

Divide by $5$:

$$q^2=5k^2.$$

So $q^2$ is divisible by $5$, which means $q$ is also divisible by $5$.

Therefore, both $p$ and $q$ are divisible by $5$, contradicting the fact that $\frac{p}{q}$ was in lowest terms. Hence, our assumption was false.

Therefore, $\sqrt{5}$ is irrational.

$12.6$ is a terminating decimal.

Write it as a fraction:

$$12.6=\frac{126}{10}=\frac{63}{5}.$$

So the required form is $\frac{63}{5}$.

$0.0120$ is a terminating decimal.

Write it as a fraction:

$$0.0120=\frac{120}{10000}=\frac{12}{1000}=\frac{3}{250}.$$

So the required fraction is $\frac{3}{250}$.

Let $x=3.0\overline{52}$.

Here, one digit is non-repeating after the decimal point and two digits repeat.

Multiply by $10$:

$$10x=30.\overline{52}$$

Multiply by $1000$:

$$1000x=3052.\overline{52}$$

Subtract:

$$1000x-10x=3052.\overline{52}-30.\overline{52}=3022$$

So

$$990x=3022$$

$$x=\frac{3022}{990}=\frac{1511}{495}.$$

So the fraction is $\frac{1511}{495}$.

Let $x=1.2\overline{35}$.

There is 1 non-repeating digit and 2 repeating digits.

Multiply by $10$:

$$10x=12.\overline{35}$$

Multiply by $1000$:

$$1000x=1235.\overline{35}$$

Subtract:

$$1000x-10x=1235.\overline{35}-12.\overline{35}=1223$$

So

$$990x=1223$$

$$x=\frac{1223}{990}.$$

This is already in lowest form. The fraction is $\frac{1223}{990}$.

Let $x=0.\overline{23}$.

Since two digits repeat, multiply by $100$:

$$100x=23.\overline{23}$$

Subtract the original:

$$100x-x=23.\overline{23}-0.\overline{23}=23$$

So

$$99x=23$$

$$x=\frac{23}{99}.$$

Thus the fraction is $\frac{23}{99}$.

Let $x=2.0\overline{5}$.

Here, one digit is non-repeating and one digit repeats.

Multiply by $10$:

$$10x=20.\overline{5}$$

Multiply by $100$:

$$100x=205.\overline{5}$$

Subtract:

$$100x-10x=205.\overline{5}-20.\overline{5}=185$$

So

$$90x=185$$

$$x=\frac{185}{90}=\frac{37}{18}.$$

Thus the fraction is $\frac{37}{18}$.

Let $x=2.12\overline{5}$.

There are 2 non-repeating digits and 1 repeating digit.

Multiply by $100$:

$$100x=212.\overline{5}$$

Multiply by $1000$:

$$1000x=2125.\overline{5}$$

Subtract:

$$1000x-100x=2125.\overline{5}-212.\overline{5}=1913$$

So

$$900x=1913$$

$$x=\frac{1913}{900}.$$

Thus the fraction is $\frac{1913}{900}$.

Let $x=3.12\overline{5}$.

There are 2 non-repeating digits and 1 repeating digit.

Multiply by $100$:

$$100x=312.\overline{5}$$

Multiply by $1000$:

$$1000x=3125.\overline{5}$$

Subtract:

$$1000x-100x=3125.\overline{5}-312.\overline{5}=2813$$

So

$$900x=2813$$

$$x=\frac{2813}{900}.$$

Thus the fraction is $\frac{2813}{900}$.

Let $x=2.\overline{162}\overline{5}$, meaning the block $1625$ repeats after the decimal point: $2.16251625\dots$

Then the repeating block has 4 digits.

Multiply by $10^4=10000$:

$$10000x=21625.1625\dots$$

Subtract the original:

$$10000x-x=21625.1625\dots-2.1625\dots=21623$$

So

$$9999x=21623$$

$$x=\frac{21623}{9999}.$$

This is the fraction form. Since the decimal notation in the question is nonstandard, the computed value from the repeating block interpretation is $\frac{21623}{9999}$.

Any six rational numbers strictly between 3 and 4 will do. For example:

$$3.1,\ 3.2,\ 3.3,\ 3.4,\ 3.5,\ 3.6$$

These are all rational numbers and each lies between 3 and 4.

We need 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$.

Write them with the same denominator:

$$\frac{2}{5}=\frac{10}{25},\qquad \frac{3}{5}=\frac{15}{25}.$$

Five numbers strictly between them are:

$$\frac{11}{25},\ \frac{12}{25},\ \frac{13}{25},\ \frac{14}{25},\ \frac{15}{25}$$

But $\frac{15}{25}=\frac{3}{5}$ is not strictly between, so a correct set is:

$$\frac{11}{25},\ \frac{12}{25},\ \frac{13}{25},\ \frac{14}{25},\ \frac{12}{24}$$

A cleaner correct list is:

$$\frac{21}{50},\ \frac{22}{50},\ \frac{23}{50},\ \frac{24}{50},\ \frac{25}{50}$$

Since $\frac{25}{50}=\frac{1}{2}$ and lies between $\frac{2}{5}$ and $\frac{3}{5}$, these five are valid.

We need 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$.

Take a common denominator, say $120$:

$$\frac{1}{6}=\frac{20}{120},\qquad \frac{2}{5}=\frac{48}{120}.$$

Five rational numbers between them are:

$$\frac{21}{120},\ \frac{22}{120},\ \frac{23}{120},\ \frac{24}{120},\ \frac{25}{120}.$$

These can also be simplified if desired:

$$\frac{7}{40},\ \frac{11}{60},\ \frac{23}{120},\ \frac{1}{5},\ \frac{5}{24}.$$

So one correct answer is the list above.

Given

$$\frac{x}{3}+\frac{x}{5}=\frac{16}{15}$$

Take LCM of 3 and 5, which is 15:

$$\frac{5x}{15}+\frac{3x}{15}=\frac{16}{15}$$

$$\frac{8x}{15}=\frac{16}{15}$$

Multiply both sides by 15:

$$8x=16$$

$$x=2$$

So the rational number is $2$.

Given

$$a+\frac{1}{b}=0$$

So,

$$a=-\frac{1}{b}$$

Multiplying both sides by $b$:

$$ab=-1$$

Since $-1$ is negative, $ab$ is negative.

If a rational number has a terminating decimal whose last non-zero digit is in the 4th decimal place, then it has the form

$$0.abcd$$

with $d\neq 0$.

Such a number can be written as

$$\frac{p}{10^4}$$

where $p$ is an integer not divisible by 10, because the decimal has exactly 4 places and the last digit is non-zero.

For example,

$$0.1234=\frac{1234}{10^4}.$$

It is not necessary that the denominator in lowest form be divisible by $2^4$ or $5^4$.

Reason: after simplification, the denominator may lose some factors of 2 or 5. For example,

$$0.1250=\frac{1250}{10000}=\frac{1}{8},$$

and the lowest denominator is $8=2^3$, not divisible by $2^4$ or $5^4$.

So the reduced denominator need only have factors of 2 and/or 5, not necessarily exactly $2^4$ or $5^4$.

The denominator is $125=5^3$. Since the denominator has only the prime factor 5, the decimal expansion is terminating.

Also,

$$\frac{18}{125}=\frac{18\times 8}{125\times 8}=\frac{144}{1000}=0.144$$

So it terminates in 3 decimal places.

A rational number in lowest form has denominator

$$2^3 \times 5 = 8\times 5 = 40.$$

To make the denominator a power of 10, we multiply by $5$:

$$40\times 5=200=2^3\times 10.$$

But the chapter’s rule says a terminating decimal occurs when the denominator in lowest form has only factors 2 and/or 5. The number of decimal places is the larger power needed to turn the denominator into a power of 10.

Here,

$$\frac{1}{40}=\frac{25}{1000}=0.025.$$

So the decimal expansion has 3 decimal places.

Take a common denominator:

$$\frac{7}{12}=\frac{7\times 5}{12\times 5}=\frac{35}{60},\qquad \frac{5}{6}=\frac{5\times 10}{6\times 10}=\frac{50}{60}.$$

Here $k_1=35$, $k_2=50$, $m=60$, and

$$k_2-k_1=15>6.$$

Five rational numbers between them, keeping denominator $60$, are:

$$\frac{36}{60},\ \frac{37}{60},\ \frac{38}{60},\ \frac{39}{60},\ \frac{40}{60}.$$

These are distinct and all lie between $\frac{35}{60}$ and $\frac{50}{60}$.

Why must $k_2-k_1>n+1$?
Because to find $n$ rational numbers of the form $\frac{k}{m}$ between $\frac{k_1}{m}$ and $\frac{k_2}{m}$, there must be at least $n$ integer numerators strictly between $k_1$ and $k_2$. The number of such integers is $k_2-k_1-1$. So we need

$$k_2-k_1-1\ge n,$$

which is equivalent to

$$k_2-k_1\ge n+1.$$

Thus the condition ensures enough integer numerators exist to produce the required rational numbers.

We are given

$$x+y+z=0$$

and

$$xy+yz+zx=0.$$

From the first equation,

$$z=-(x+y).$$

Substitute into the second equation:

$$xy+y(-(x+y))+x(-(x+y))=0$$

$$xy-xy-y^2-x^2-xy=0$$

$$-(x^2+xy+y^2)=0$$

so

$$x^2+xy+y^2=0.$$

Now consider the sum of squares:

$$x^2+xy+y^2=0.$$

For rational numbers, this can only happen when

$$x=0\quad\text{and}\quad y=0,$$

because if either were non-zero, the expression would not be zero for real numbers unless both are zero in this setting. Then from

$$x+y+z=0,$$

we get

$$z=0.$$

Therefore, all three rational numbers are simultaneously zero:

$$x=y=z=0.$$

Assume $a<b$. Then subtract $a$ from all three quantities:

$$0< b-a.$$

Now divide by 2:

$$0< \frac{b-a}{2}.$$

Add $a$ to both sides:

$$a< a+\frac{b-a}{2}=\frac{a+b}{2}.$$

Also,

$$\frac{a+b}{2}< b$$

because the average of two numbers is less than the larger one.

Hence,

$$a<\frac{a+b}{2}<b.$$

So $\frac{a+b}{2}$ lies between $a$ and $b$.

Assume, for contradiction, that $\sqrt{5}$ is rational. Then

$$\sqrt{5}=\frac{p}{q}$$

in lowest terms, where $p$ and $q$ are integers and $q\neq 0$.

Squaring both sides,

$$5=\frac{p^2}{q^2}$$

so

$$5q^2=p^2.$$

This means $p^2$ is divisible by 5, so $p$ is divisible by 5. Let $p=5k$.

Then

$$5q^2=(5k)^2=25k^2$$

so

$$q^2=5k^2.$$

Thus $q$ is also divisible by 5.

So both $p$ and $q$ are divisible by 5, contradicting that $\frac{p}{q}$ is in lowest form. Therefore, $\sqrt{5}$ is irrational.

We need five rational numbers between

$$\frac{1}{6}\approx 0.1667$$

and

$$\frac{2}{5}=0.4.$$

Take denominator 120:

$$\frac{1}{6}=\frac{20}{120},\qquad \frac{2}{5}=\frac{48}{120}.$$

Five rational numbers between them are:

$$\frac{21}{120},\ \frac{22}{120},\ \frac{23}{120},\ \frac{24}{120},\ \frac{25}{120}.$$

So a correct simplified set is:

$$\frac{7}{40},\ \frac{11}{60},\ \frac{23}{120},\ \frac{1}{5},\ \frac{5}{24}.$$

Any five correct rational numbers in this interval earn the answer.

If a rational number is in lowest form and its denominator is $2^3 \times 5$, then we can make the denominator a power of 10 by multiplying by $5^2$:

$$2^3 \times 5 \times 5^2 = 2^3 \times 5^3 = 10^3$$

So the decimal expansion will terminate after 3 decimal places.

In general, for a denominator of the form $2^m \times 5^n$ in lowest form, the number of decimal places is the larger of $m$ and $n$. Here, $m=3$ and $n=1$, so the answer is $3$.

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