Question 12 Marks
Find the greatest possible length which can be used to measure exactly the length $7m, 3m\ 85cm$ and $12m\ 95cm$.
Answer$7m = 700cm$
$3m\ 85cm = 385cm$
$12m\ 95cm = 1295cm$
$HCF$ of $700, 385$ and $1295$ is $35$.
So to measure exactly, the greatest length we can use is $35cm$.
View full question & answer→Question 22 Marks
Give an example of two irrationals whose sum is rational.
AnswerConsider the irrational numbers, $\big(5+\sqrt7\big)$ and $\big(5-\sqrt7\big)$
$\big(5+\sqrt7\big)+\big(5-\sqrt7\big)=10$
which is rational number.
View full question & answer→Question 32 Marks
The $HCF$ of two numbers is $23$ and their $LCM$ is $1449$. If one of the numbers is $161$, find the other.
Answer$H.C.F = 23; L.C.M. = 1449$
For any two numbers $a$ and $b$, we have
$a \times b = L.C.M. \times H.C.F.$
$\therefore\text{b}=\frac{\text{LCM}\times\text{HCF}}{\text{a}}$
$\Rightarrow\text{b}=\frac{1449\times23}{161}=207$
View full question & answer→Question 42 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$23, 31$
Answer$23$ and $31$ are prime numbers, so
$HCF (23, 31) = 1$
$LCM (23, 31) = 23 \times 31 = 713$
$HCF \times LCM = 1 \times 713$
$23 \times 31 = 713$
$\Rightarrow HCF \times LCM =$ product of given numbers
Hence verified.
View full question & answer→Question 52 Marks
Find the simplest form of:
$\frac{368}{496}$
AnswerPrime factorisation of $368$ and $496$ is:
$ 368=2^4 \times 23 $
$ 496=2^4 \times 31 $
Therefore, $\frac{368}{496}=\frac{2^4\times23}{2^4\times31}=\frac{23}{31}$
Thus, simplest form of $\frac{368}{496}$ is $\frac{23}{31}.$
View full question & answer→Question 62 Marks
Short-Answer Questions:
Express $0.\bar4$ as a rational number in simplest form.
AnswerLet $\text{x}=0.\bar4$ then,
$\text{x}=0.4444\dots\ \ \dots(\text{i})$
$\therefore\text{10x}=4.444\dots\ \ \dots(\text{ii})$
On subtracting $(i)$ from $(ii)$, we get
$\text{9x}=4$
$\Rightarrow\text{x}=\frac{4}{9}$
Hence, $0.\bar4=\frac{4}{9}$
View full question & answer→Question 72 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$24, 36, 40$
Answer$ 24=2 \times 2 \times 2 \times 3=2^3 \times 3 $
$ 36=2 \times 2 \times 3 \times 4=2^2 \times 3^2 $
$ 40=2 \times 2 \times 2 \times 5=2^3 \times 5 $
$ \operatorname{HCF}(24,36,40)=2^2=4 $
$ \operatorname{LCM}(24,36,40)=2^3 \times 3^2 \times 5=360$
View full question & answer→Question 82 Marks
Very-Short-Answer Questions:
The $LCM$ of two numbers is $1200$. Show that the $HCF$ of these numbers cannot be $500$. Why?
AnswerGiven that the $LCM$ is $1200$
We know that, the $LCM$ is always a multiple of the $HCF$.
Multiples of $500$ are $500, 1000, 1500, 2000,$ and so on...
Clearly, $1200$ is not a multiple of $500$, and so the $HCF$ of those two numbers cannot be $500$.
View full question & answer→Question 92 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$12, 15, 21$.
Answer$ 12=2 \times 2 \times 3=2^2 \times 3 $
$ 15=3 \times 5 $
$ 21=3 \times 7 $
$ \operatorname{HCF}(12,15,21)=3 $
$ \operatorname{LCM}(12,15,21)=2^2 \times 3 \times 5 \times 7=420$
View full question & answer→Question 102 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$36, 84$.
Answer$ 36=2 \times 2 \times 3 \times 3=2^2 \times 3 $
$ 84=2 \times 2 \times 3 \times 7=2^2 \times 3 \times 7 $
$ \operatorname{HCF}(36,84)=2^2 \times 3=12 $
$ \operatorname{LCM}(36,84)=2^2 \times 3^2 \times 7=252 $
$ H C F \times \operatorname{LCM}=3024 $
$ 36 \times 84=3024$
$\Rightarrow HCF \times LCM =$ product of given numbers
Hence verified.
View full question & answer→Question 112 Marks
Very-Short-Answer Questions:
If the product of two numbers is $1050$ and their $HCF$ is $25$, find their $LCM$.
Answer$HCF \times LCM =$ Product of the two numbers
$\Rightarrow 25 \times LCM = 1050$
$\Rightarrow LCM = 42$
View full question & answer→Question 122 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$21, 28, 36, 45$.
Answer$ 21=3 \times 7 $
$ 28=2 \times 2 \times 7=2^2 \times 7 $
$ 36=2 \times 2 \times 3 \times 3=2^2 \times 3^2 $
$ 45=3 \times 3 \times 5=3^2 \times 5 $
$ \operatorname{HCF}(21,28,36,45)=1 $
$ \operatorname{LCM}(21,28,36,45)=2^2 \times 3^2 \times 5 \times 7=1260$
View full question & answer→Question 132 Marks
Find the least number which should be added to $2497$ so that the sum is exactly divisible by $5, 6, 4$ and $3$.
Answer$L.C.M.$ of $5, 6, 4$ and $3 = 60$.
On dividing $2497$ by $60$, the remainder is $37$.
$\therefore$ $L.C.M.$ of $5, 6, 4$ and $3 = 60$. Number to be added $= (60 - 37) = 23$.
View full question & answer→Question 142 Marks
Find the $HCF$ and $LCM$ of $12, 15, 18, 27$.
AnswerTo find the $HCF$, of $12, 15, 18, 27$
we will find the prime factorisation of each number.
$ 12=2^2 \times 3 $
$ 15=3 \times 5 $
$ 18=2 \times 3^2 $
$ 27=3^3$
$\text { So, the HCF }=3$
$\mathrm{LCM}=2^2 \times 3^3 \times 5=540$
View full question & answer→Question 152 Marks
Find the largest number which divides $320$ and $457$ leaving remainders $5$ and $7$ respectively.
AnswerSubtracting $5$ and $7$ from $320$ and $457$ respectively:
$320 - 5 = 315,$
$457 - 7 = 450$
Let us now find the $HCF$ of $315$ and $405$ through prime factorization:
$\begin{array}{c|c} 3 & 315 \\ \hline 3 & 105\\ \hline5&35\\ \hline&7 \end{array}$ $\begin{array}{c|c} 2 & 450 \\ \hline 3 & 225\\ \hline3&75\\ \hline 5&25 \\ \hline&5 \end{array}$
$315=3 \times 3 \times 5 \times 7 $
$=3^2 \times 5 \times 7$
$450=2 \times 3 \times 3 \times 5 \times 5$
$=2 \times 3^2 \times 5^2$
$\therefore \text { H.C.F. of } 315 \text { and } 450 \text { is } 3^2 \times 5=9 \times 5=45$
$\therefore$ The required number is $45$
View full question & answer→Question 162 Marks
Find the simplest form of:
$\frac{1095}{1168}$
AnswerPrime factorisation of $1095$ and $1168$ is:
$1095 = 3 × 5 × 73$
$1168 = 2^4× 73$
Therefore, $\frac{1095}{1168}=\frac{3\times\text{5}\times73}{2^4\times73}=\frac{15}{16}$
Thus, simplest form of $\frac{1095}{1168}$ is $\frac{15}{16}.$
View full question & answer→Question 172 Marks
Three pieces of timber $42m$, $49m$ and $63m$ long have to be divided into planks of the same length. What is the greatest possible length of each plank?
Answer$HCF$ of $42, 49, 63 = 7$
$1^{st}$ plank i.e $\frac{42}{7}=6$
$2^{nd}$ plank i.e $\frac{49}{7}=7$
$3^{rd}$ plank i.e $\frac{63}{7}=9$
Planks formed are $= 6 + 7 + 9$
$= 22$
View full question & answer→Question 182 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$8, 9, 25$
Answer$ 8=2 \times 2 \times 2=2^3 $
$ 9=3 \times 3=3^2 $
$ 25=5 \times 5=5^2 $
$ \operatorname{HCF}(8,9,25)=1 $
$ \operatorname{LCM}(8,9,25)=2^3 \times 3^2 \times 5^2=1800$
View full question & answer→Question 192 Marks
Find the largest number which divides $438$ and $606$, leaving remainder $6$ in each case.
AnswerLargest number which divides $438$ and $606$, leaving remainder $6$ is actually the largest number which divides $438 - 6 = 432$ and $606 - 6 = 600$, leaving remainder $0$.
Therefore, $HCF$ of $432$ and $600$ gives the largest number.
Now, prime factors of $432$ and $600$ are:
$432=2^4 \times 3^3$
$ 600=2^3 \times 3 \times 5^2$
$HCF =$ product of smallest power of each common prime factor in the numbers $= 2^3× 3 = 24$
Thus, the largest number which divides $438$ and $606$, leaving remainder $6$ is $24$.
View full question & answer→Question 202 Marks
Very-Short-Answer Questions:
If $a$ and $b$ are relatively prime, what is their $LCM?$
AnswerIf $a$ and $b$ are relatively prime, it means they have no common multiple other than their product.
So, $LCM(a, b) = ab$.
View full question & answer→Question 212 Marks
Find the greatest number that will divide $43, 91$ and $183$ so as to leave the same remainder in each case.
AnswerWe have to find Greatest Factor
In this case, we have to find $HCF$ with remainder (No mention of remainder in question)
Step:
- Find the Differences of numbers
- Get the $HCF$ (that differences)
We have here $43, 91$ and $183$
- So differences are
$183 - 91 = 92,$
$183 - 43 = 140,$
$91 - 43 = 48.$
Now,
- $HCF (48, 92$ and $140)$
As
$48 = 2 \times 2 \times 2 \times 2 \times 3,$
$92 = 2 \times 2 \times 23,$
$140 = 2 \times 2 \times 5 \times 7$
$HCF = 2 \times 2 = 4$
And $4$ is the required number. View full question & answer→Question 222 Marks
Find the largest four-digits number which when divided by $4, 7$ and $13$ leaves a reminder of $3$ in each case.
AnswerSo $LCM (4, 7, 13) = 364$
Largest $4$ digit number $= 9999$
On dividng $9999$ by $364$ we get reaminder as $171$
So $9999 - 171 = 9828 + 3 = 9831$
Therefore $9831$ is the number.
View full question & answer→Question 232 Marks
By what number should $1365$ be divided to get $31$ as quotient and $32$ as remainder?
AnswerBy Euclid's Division Algorithm, we have:
Dividend = (divisor $\times $ quotient) + remainder
$1365 = ($divisor $\times$ $31) + 32$
$\frac{1365-32}{31}=\text{divisor}$
$\Rightarrow\frac{1331}{31}=\text{divisor}$
$\therefore\text{Divisor}=43$
View full question & answer→Question 242 Marks
Find the greatest number of four digits which is exactly divisible by $15, 24$ and $36$.
AnswerThe greatest number four digit number = 9999
$ 15=3 \times 5 $
$ 24=2^3 \times 3 $
$ 36=2^2 \times 3^2 $
$ L C M=2^3 \times 3^2 \times 5 $
$ =360$
On dividing $9999$ by $360$, remainder $= 279$
$\therefore$ the required number $= 9999 - 279 = 9720$.
View full question & answer→Question 252 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$17, 23, 29$
Answer$17, 23$ and $29$ are prime numbers,
$HCF (17, 23, 29) = 1$
$LCM (17, 23, 29) = 17 × 23 × 29 = 11339$
View full question & answer→Question 262 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$30, 72, 432$
Answer$ 30=2 \times 3 \times 5 $
$ 72=2 \times 2 \times 2 \times 3 \times 3=2^3 \times 3^2 $
$ 432=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3=2^4 \times 3^3 $
$ \operatorname{HCF}(30,72,432)=2 \times 3=6 $
$ \operatorname{LCM}(30,72,432)=2^4 \times 3^3 \times 5=2160$
View full question & answer→