For every 2 bags, he receives 15 ingots. Since $12 \div 2=6$, 12 bags make 6 such groups. Therefore the number of ingots is $6\times 15=90$.
He will leave with $90$ copper ingots.
Each of $11,13,17,19$ has exactly two positive factors: 1 and itself. Continuing the prime-number pattern after 19 gives $23,29,31$.
They are prime numbers between 10 and 20. The next three primes after 19 are $23, 29, 31$.
A set is closed under an operation only if applying the operation to any two members always gives a member of the same set. Subtraction of natural numbers can give a negative number or 0, so natural numbers are not closed under subtraction.
No. Natural numbers are not closed under subtraction. For example, $5-2=3$ is natural, but $2-5=-3$ is not a natural number. Also, $4-4=0$, and 0 is not a natural number in the usual NCERT convention $\mathbb{N}=\{1,2,3,\ldots\}$.
The thumb acts as the pointer. The other four fingers each have 3 countable joints, so the total is $3+3+3+3=12$. Because one full hand-count gives 12, larger counts can be grouped by twelves.
Using the thumb to count the three joints on each of the four fingers, one can count $4\times 3=12$ positions on one hand. This naturally supports counting in groups of 12, which is the idea behind base-12 systems.
A drop of $15$ °C from $4$ °C gives $4-15=-11$. Therefore the temperature is $-11$ °C.
The midnight temperature is $-11$ °C.
Represent debt/loss as negative and profit as positive. Then the final amount is $-850+1200-450=350-450=-100$. A negative result means he is still in debt by `100.
The integer equation is $-850+1200-450=-100$. His final standing is a debt of `100.
A negative times a positive is negative, so $(-12)\times 5=-60$. A negative times a negative is positive, so $(-8)\times(-7)=56$. Subtracting a negative adds the positive: $0-(-14)=14$. A negative divided by a positive is negative: $(-20)\div 4=-5$.
(i) $-60$
(ii) $56$
(iii) $14$
(iv) $-5$
Think of $-5$ as owing `5. Subtracting $-5$ means taking away that debt. Taking away a debt improves the balance by the same amount, so it has the same effect as adding $+5$.
A negative amount can represent debt. If you have `10 and a debt of `5 is removed, your net amount increases by `5. Thus $10-(-5)$ means removing a debt of 5, which gives $10+5=15$.
Reduce each fraction by dividing numerator and denominator by their common factor: $4/6$ by 2, $10/8$ by 2, and $-6/10$ by 2. Also $9\div 3=3$. Hence each pair represents the same rational number.
(i) $\dfrac{4}{6}=\dfrac{2}{3}$
(ii) $\dfrac{10}{8}=\dfrac{5}{4}$
(iii) $-\dfrac{6}{10}=-\dfrac{3}{5}$
(iv) $\dfrac{9}{3}=3$
(i) $\dfrac{2}{5}+\dfrac{3}{10}=\dfrac{4}{10}+\dfrac{3}{10}=\dfrac{7}{10}$. (ii) $\dfrac{7}{12}+\dfrac{5}{8}=\dfrac{14}{24}+\dfrac{15}{24}=\dfrac{29}{24}$. (iii) $-\dfrac{4}{7}+\dfrac{3}{14}=-\dfrac{8}{14}+\dfrac{3}{14}=-\dfrac{5}{14}$.
(i) $\dfrac{7}{10}$
(ii) $\dfrac{29}{24}$
(iii) $-\dfrac{5}{14}$
(i) $\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}$. (ii) $\dfrac{11}{8}-\dfrac{3}{4}=\dfrac{11}{8}-\dfrac{6}{8}=\dfrac{5}{8}$. (iii) $-\dfrac{7}{9}-\left(-\dfrac{2}{3}\right)=-\dfrac{7}{9}+\dfrac{6}{9}=-\dfrac{1}{9}$.
(i) $\dfrac{7}{12}$
(ii) $\dfrac{5}{8}$
(iii) $-\dfrac{1}{9}$
Multiply numerators and denominators, then simplify: $\dfrac{2}{3}\times\dfrac{3}{10}=\dfrac{6}{30}=\dfrac{1}{5}$; $\dfrac{7}{11}\times\dfrac{5}{8}=\dfrac{35}{88}$; $-\dfrac{4}{7}\times\dfrac{5}{14}=-\dfrac{20}{98}=-\dfrac{10}{49}$.
(i) $\dfrac{1}{5}$
(ii) $\dfrac{35}{88}$
(iii) $-\dfrac{10}{49}$
Divide by multiplying by the reciprocal: $\dfrac{2}{3}\div\dfrac{3}{10}=\dfrac{2}{3}\times\dfrac{10}{3}=\dfrac{20}{9}$; $\dfrac{7}{11}\div\dfrac{5}{8}=\dfrac{7}{11}\times\dfrac{8}{5}=\dfrac{56}{55}$; $-\dfrac{4}{7}\div\dfrac{5}{14}=-\dfrac{4}{7}\times\dfrac{14}{5}=-\dfrac{8}{5}$.
(i) $\dfrac{20}{9}$
(ii) $\dfrac{56}{55}$
(iii) $-\dfrac{8}{5}$
LHS $=\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\times\dfrac{8}{3}=\left(\dfrac{2}{4}+\dfrac{3}{4}\right)\times\dfrac{8}{3}=\dfrac{5}{4}\times\dfrac{8}{3}=\dfrac{10}{3}$. RHS $=\dfrac{1}{2}\times\dfrac{8}{3}+\dfrac{3}{4}\times\dfrac{8}{3}=\dfrac{4}{3}+2=\dfrac{10}{3}$. Hence LHS = RHS.
Both sides are equal to $\dfrac{10}{3}$.
Using the distributive property, $\dfrac{7}{9}\left(\dfrac{6}{7}-\dfrac{3}{4}\right)=\dfrac{7}{9}\times\dfrac{6}{7}-\dfrac{7}{9}\times\dfrac{3}{4}=\dfrac{2}{3}-\dfrac{7}{12}=\dfrac{8}{12}-\dfrac{7}{12}=\dfrac{1}{12}$.
$\dfrac{1}{12}$.
Expand the left side: $\dfrac{5}{6}\left(x+\dfrac{3}{5}\right)=\dfrac{5}{6}x+\dfrac{5}{6}\times\dfrac{3}{5}=\dfrac{5}{6}x+\dfrac{1}{2}$. This is exactly the right side, so the equation is an identity true for all rational $x$.
Every rational number $x$ satisfies the equation.
Divide the unit from 0 to 1 into 3 equal parts and mark the second part for $2/3$. Since $-5/4=-1\dfrac{1}{4}$, mark it one-fourth unit to the left of $-1$. Since $11/2=5\dfrac{1}{2}$, mark it halfway between 5 and 6.
$\dfrac{2}{3}$ lies between 0 and 1, two-thirds of the way from 0 to 1. $-\dfrac{5}{4}=-1.25$ lies between $-2$ and $-1$. $\dfrac{11}{2}=5.5$ lies halfway between 5 and 6.
The numbers satisfy $-\dfrac{1}{2}\lt -\dfrac{1}{4}\lt 0\lt \dfrac{1}{8}\lt \dfrac{1}{4}$, so all three lie strictly between the given rational numbers.
Three examples are $-\dfrac{1}{4}$, $0$ and $\dfrac{1}{8}$.
$-\dfrac{1}{4}+\dfrac{5}{12}=-\dfrac{3}{12}+\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}$.
$\dfrac{1}{6}$.
$15\dfrac{3}{4}=\dfrac{63}{4}$ and $2\dfrac{1}{4}=\dfrac{9}{4}$. Number of kurtas $=\dfrac{63}{4}\div\dfrac{9}{4}=\dfrac{63}{4}\times\dfrac{4}{9}=7$.
He can make $7$ kurtas.
Each of these terminating decimals is rational, and each is greater than $3.1415$ but less than $3.1416$.
Three examples are $3.14151$, $3.14155$ and $3.14159$.
Since rational numbers are closed under addition and division by the non-zero integer 2, $\dfrac{a+b}{2}$ is rational. Also, if $a\lt b$, then $a\lt \dfrac{a+b}{2}\lt b$. Repeating this method gives as many rational numbers as needed.
Yes. One simple method is to take the average. For any two rational numbers $a$ and $b$ with $a\lt b$, the number $\dfrac{a+b}{2}$ is rational and lies between them.
A rational number in lowest form has a terminating decimal if its denominator has only factors 2 and/or 5. Here $20=2^2\times 5$, so $7/20$ terminates. $15=3\times 5$ has factor 3, so $4/15$ repeats. $250=2\times 5^3$, so $13/250$ terminates. Long division gives $0.35$, $0.2666\ldots$ and $0.052$ respectively.
$\dfrac{7}{20}$ terminates: $0.35$. $\dfrac{4}{15}$ is non-terminating repeating: $0.2\overline{6}$. $\dfrac{13}{250}$ terminates: $0.052$.
Long division gives remainders that repeat after 6 steps for $1/13$, producing $076923$. Multiplying $1/13$ by 2, 3, 4, ... gives the decimal blocks for $2/13,3/13,4/13,\ldots$. Several are rotations of one another, showing cyclic behaviour in the repeating parts.
$\dfrac{1}{13}=0.\overline{076923}$, so the repeating block is $076923$. Also $\dfrac{2}{13}=0.\overline{153846}$, $\dfrac{3}{13}=0.\overline{230769}$ and $\dfrac{4}{13}=0.\overline{307692}$. The repeating blocks show cyclic patterns, but the multiples split into related cyclic blocks rather than one single block for every numerator.
Perfect-square roots and terminating or repeating decimals are rational. Thus $\sqrt{81}=9$, $0.333\ldots$ repeats, $0.1234512345\ldots$ repeats the block 12345, and the finite decimal in (vi) is rational. $\sqrt{12}=2\sqrt3$ is irrational because $\sqrt3$ is irrational. The decimal in (v) does not repeat a fixed block, so it is irrational.
(i) Rational: $\sqrt{81}=9=\dfrac{9}{1}$.
(ii) Irrational: $\sqrt{12}=2\sqrt{3}$.
(iii) Rational: $0.33333\ldots=\dfrac{1}{3}$.
(iv) Rational: $0.1234512345\ldots=\dfrac{12345}{99999}=\dfrac{4115}{33333}$.
(v) Irrational; the zeros increase, so there is no fixed repeating block.
(vi) Rational: $23.560185612239874790120=\dfrac{23560185612239874790120}{1000000000000000000000}$.
Let $x=0.\overline{9}=0.999\ldots$. Then $10x=9.999\ldots$. Subtracting, $10x-x=9.999\ldots-0.999\ldots=9$, so $9x=9$ and $x=1$. Therefore $0.\overline{9}=1$.
$0.\overline{9}=1$.
Such denominators are examples where the decimal repetend of $1/n$ is cyclic under multiplication by suitable integers. Checking by long division for $1/17$ and $1/19$ gives repeating blocks whose multiples produce cyclic rotations.
Examples include $n=17$ and $n=19$. For instance, $\dfrac{1}{17}=0.\overline{0588235294117647}$ and $\dfrac{1}{19}=0.\overline{052631578947368421}$; multiples of these repeating blocks cycle through rotations.
For $3/50$, make the denominator 100: $\dfrac{3}{50}=\dfrac{6}{100}=0.06$. Long division of $2\div 9$ gives 0 remainder pattern with digit 2 repeating, so $2/9=0.222\ldots=0.\overline{2}$.
(i) $\dfrac{3}{50}=0.06$
(ii) $\dfrac{2}{9}=0.\overline{2}$
Suppose $\sqrt5$ is rational, so $\sqrt5=\dfrac{a}{b}$ for coprime integers $a,b$ with $b\neq 0$. Then $a^2=5b^2$, so 5 divides $a^2$ and hence 5 divides $a$. Write $a=5c$. Then $25c^2=5b^2$, so $b^2=5c^2$, and 5 divides $b$. This contradicts $a$ and $b$ being coprime. Hence $\sqrt5$ is irrational.
Write each terminating decimal over the suitable power of 10 and reduce: $12.6=126/10=63/5$; $0.0120=120/10000=3/250$; $3.052=3052/1000=763/250$; $1.235=1235/1000=247/200$; $0.23=23/100$; $2.05=205/100=41/20$; $2.125=2125/1000=17/8$; $3.125=3125/1000=25/8$; $2.1625=21625/10000=173/80$.
(i) $\dfrac{63}{5}$
(ii) $\dfrac{3}{250}$
(iii) $\dfrac{763}{250}$
(iv) $\dfrac{247}{200}$
(v) $\dfrac{23}{100}$
(vi) $\dfrac{41}{20}$
(vii) $\dfrac{17}{8}$
(viii) $\dfrac{25}{8}$
(ix) $\dfrac{173}{80}$
To locate a terminating decimal, divide the appropriate unit interval into equal decimal parts. Mark $0.532$ as 532 thousandths from 0. Mark $1.15$ as 15 hundredths to the right of 1.
(i) $0.532=\dfrac{532}{1000}=\dfrac{133}{250}$, so it lies between $0.53$ and $0.54$.
(ii) $1.15=\dfrac{115}{100}=\dfrac{23}{20}$, so it lies between $1.1$ and $1.2$, exactly at $1+\dfrac{3}{20}$.
These are $3.1,3.2,3.3,3.4,3.5,3.6$, and each lies strictly between 3 and 4.
Six examples are $\dfrac{31}{10}$, $\dfrac{32}{10}$, $\dfrac{33}{10}$, $\dfrac{34}{10}$, $\dfrac{35}{10}$ and $\dfrac{36}{10}$.
Write $\dfrac{2}{5}=\dfrac{20}{50}$ and $\dfrac{3}{5}=\dfrac{30}{50}$. Fractions with numerators between 20 and 30 lie between the two given numbers.
Five examples are $\dfrac{21}{50}$, $\dfrac{22}{50}$, $\dfrac{23}{50}$, $\dfrac{24}{50}$ and $\dfrac{25}{50}$.
Use denominator 30: $\dfrac{1}{6}=\dfrac{5}{30}$ and $\dfrac{2}{5}=\dfrac{12}{30}$. Therefore $6/30,7/30,8/30,9/30,10/30$ lie strictly between them.
Five examples are $\dfrac{6}{30}$, $\dfrac{7}{30}$, $\dfrac{8}{30}$, $\dfrac{9}{30}$ and $\dfrac{10}{30}$.
$\dfrac{x}{3}+\dfrac{x}{5}=\dfrac{5x+3x}{15}=\dfrac{8x}{15}$. So $\dfrac{8x}{15}=\dfrac{16}{15}$, hence $8x=16$ and $x=2$.
$x=2$.
From $a+\dfrac{1}{b}=0$, we get $a=-\dfrac{1}{b}$. Multiplying both sides by $b$ gives $ab=-1$. Therefore $ab$ is negative.
$ab$ is negative.
A terminating decimal ending at the 4th decimal place is an integer number of ten-thousandths, so it is $p/10^4$. Since the 4th decimal digit is non-zero and no later non-zero digits occur, $p$ is not divisible by 10. Now $10^4=2^4\times 5^4$. When reducing $p/10^4$, factors common with $p$ may cancel. But because $p$ is not divisible by 10, it cannot contain both a factor 2 and a factor 5 at the same time. Hence at least one full factor $2^4$ or $5^4$ remains in the denominator.
If the last non-zero digit is in the 4th decimal place, the number can be written as $\dfrac{p}{10^4}$ with $p$ not divisible by 10. In lowest form, its denominator must be divisible by at least one of $2^4$ or $5^4$.
$125=5^3$. A denominator with only factors 2 and 5 gives a terminating decimal. Since $\dfrac{18}{125}=\dfrac{18\times 8}{125\times 8}=\dfrac{144}{1000}=0.144$, it has 3 decimal places.
It is terminating and has $3$ decimal places.
The denominator is $2^3\times 5$. To make it a power of 10, multiply by $5^2$: $(2^3\times 5)\times 5^2=2^3\times 5^3=10^3$. Therefore the decimal terminates within 3 decimal places.
It will have $3$ decimal places.
Using denominator 48 gives many integer numerators strictly between 28 and 40. Any numerator from 29 to 39 gives a rational number between $a$ and $b$; choosing five of them gives the listed answers. In general, if the endpoints are $k_1/m$ and $k_2/m$, the number of integer numerators strictly between them is $k_2-k_1-1$. To find $n$ such rational numbers, we need at least $n$ integer numerators between, so $k_2-k_1-1\ge n$, i.e. $k_2-k_1\ge n+1$. The stronger condition $k_2-k_1>n+1$ also guarantees enough choices.
Take $m=48$. Then $a=\dfrac{7}{12}=\dfrac{28}{48}$ and $b=\dfrac{5}{6}=\dfrac{40}{48}$, so $k_2-k_1=12>6$. Five rational numbers between them are $\dfrac{29}{48}$, $\dfrac{30}{48}$, $\dfrac{31}{48}$, $\dfrac{32}{48}$ and $\dfrac{33}{48}$.
Since $x+y+z=0$, squaring gives $(x+y+z)^2=0$. Thus $x^2+y^2+z^2+2(xy+yz+zx)=0$. Given $xy+yz+zx=0$, we get $x^2+y^2+z^2=0$. Squares of rational numbers are non-negative, so each square must be 0. Hence $x=y=z=0$.