2 Marks Questions

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.

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Question 22 Marks

Give an example of two irrationals whose sum is rational.

Answer

Consider 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.

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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 × b = L.C.M. × H.C.F.
$\therefore\text{b}=\frac{\text{LCM}\times\text{HCF}}{\text{a}}$
$\Rightarrow\text{b}=\frac{1449\times23}{161}=207$

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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 × 31 = 713
HCF × LCM = 1 × 713
23 × 31 = 713
⇒ HCF × LCM = product of given numbers
Hence verified.

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Question 52 Marks

Find the simplest form of:$\frac{368}{496}$

Answer

Prime 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 $\frac{23}{31}.$

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Question 62 Marks

Short-Answer Questions:
Express $0.\bar4$ as a rational number in simplest form.

Answer

Let $\text{x}=0.\bar4$ 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, $0.\bar4=\frac{4}{9}$

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Question 72 Marks

Using prime factorization, find the HCF and LCM of:
$24, 36, 40$

Answer

$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$
$36 = 2 \times 2 \times 3 \times 4 = 2^2 \times 3^2$
$40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5$
HCF $(24, 36, 40) = 2^2 = 4$
LCM $(24, 36, 40) = 2^3 \times 3^2 \times 5 = 360$

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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?

Answer

Given 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

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Question 92 Marks

Using prime factorization, find the HCF and LCM of:
$12, 15, 21$

Answer

$12 = 2 \times 2 \times 3 = 2^2\times 3 15 = 3 \times 5 21 = 3 \times 7$
HCF $(12, 15, 21) = 3$
LCM $(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$

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Question 102 Marks

Using prime factorization, find the HCF and LCM of:
$36, 84$

Answer

$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3$
$84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$
HCF $(36, 84) = 2^2 \times 3 = 12$
LCM $(36, 84) = 2^2 \times 3^2 \times 7 = 252$
HCF $\times $ LCM $= 3024$
$36 \times 84 = 3024$
$\Rightarrow$ HCF $\times $ LCM = product of given numbers
Hence verified.

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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 × LCM = Product of the two numbers
⇒ 25 × LCM = 1050
⇒ LCM = 42

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Question 122 Marks

Using prime factorization, find the HCF and LCM of:
$21, 28, 36, 45$

Answer

$21 = 3 \times 7$
$28 = 2 \times 2 \times 7 = 2^2 \times 7$
$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$
$45 = 3 \times 3 \times 5 = 3^2 \times 5$
HCF $(21, 28, 36, 45) = 1$
LCM $(21, 28, 36, 45) = 2^2 \times 3^2 \times 5 \times 7 = 1260$

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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.

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Question 142 Marks

Find the HCF and LCM of $12, 15, 18, 27.$

Answer

To find the HCF, of $12, 15, 18, 27$
we will find the prime factorisation of each number.
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$18 = 2 \times 3^2$
$27 = 3^3$
So, the HCF $= 3$
LCM $ = 2^2\times 3^3 \times 5 = 540$

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Question 152 Marks

Find the largest number which divides $320$ and $457$ leaving remainders $5$ and $7$ respectively.

Answer

Subtracting $5$ 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:
$\begin{array}{c|c} 3 & 315 \\ \hline 3 & 105\\ \hline5&35\\ \hline&7 \end{array}$ $\begin{array}{c|c} 2 & 450 \\ \hline 3 & 225\\ \hline3&75\\ \hline 5&25 \\ \hline&5 \end{array}$
$315 = 3 \times 3 \times 5 \times 7$
$= 3^2 \times 5 \times 7$
$450 = 2 \times 3 \times 3 \times 5 \times 5$
$= 2 \times 3^2 \times 5^2$
$\therefore$ $H.C.F.$ of 315 and $450$ is $3^2 \times 5 = 9 \times 5 = 45$
$\therefore$ The required number is $45$

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Question 162 Marks

Find the simplest form of:$\frac{1095}{1168}$

Answer

Prime factorisation of $1095$ and $1168$ is:
$1095 = 3 \times 5 \times 73$
$1168 = 2^4 \times 73$
Therefore, $\frac{1095}{1168}=\frac{3\times\text{5}\times73}{2^4\times73}=\frac{15}{16}$
Thus, simplest form of $\frac{1095}{1168}$ is $\frac{15}{16}.$

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Question 172 Marks

Three pieces of timber $42\ m, 49\ m$ and $63\ m$ 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 $= 6 + 7 + 9$
$= 22$

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Question 182 Marks

Using prime factorization, find the HCF and LCM of:
$8, 9, 25$

Answer

$8 = 2 \times 2 \times 2 = 2^3 9 = 3 \times 3 = 3^2 25 = 5 \times 5 = 5^2$^
HCF $(8, 9, 25) = 1$
LCM $(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 1800$

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Question 192 Marks

Find the largest number which divides $438$ and $606$, leaving remainder $6$ in each case.

Answer

Largest 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 $= 2^3\times 3 = 24$
Thus, the largest number which divides $438$ and $606$, leaving remainder $6$ is $24.$

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Question 202 Marks

Very-Short-Answer Questions:
If a and b are relatively prime, what is their LCM?

Answer

If a and b are relatively prime, it means they have no common multiple other than their product.
So, LCM(a, b) = ab.

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Question 212 Marks

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Answer

We 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 × 2 × 2 × 2 × 3,
92 = 2 × 2 × 23,
140 = 2 × 2 × 5 × 7
HCF = 2 × 2 = 4
And 4 is the required number.

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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.

Answer

So 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.

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Question 232 Marks

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?

Answer

By Euclid's Division Algorithm, we have:
Dividend = (divisor × quotient) + remainder
1365 = (divisor × 31) + 32
$\frac{1365-32}{31}=\text{divisor}$
$\Rightarrow\frac{1331}{31}=\text{divisor}$
$\therefore\text{Divisor}=43$

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Question 242 Marks

Find the greatest number of four digits which is exactly divisible by $15, 24$ and $36.$

Answer

The greatest number four digit number $= 9999$
$15 = 3 \times 5$
$24 = 2^3 \times 3$
$36 = 2^2 \times 3^2$
LCM $= 2^3 \times 3^2 \times 5$
$= 360$
On dividing $9999$ by $360,$ remainder $= 279$
$\therefore$ the required number $= 9999 - 279 = 9720$

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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) = 1LCM (17, 23, 29) = 17 × 23 × 29 = 11339

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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 $
$= 2^4 \times 3^3$
HCF $(30, 72, 432) = 2 × 3 = 6$
LCM $(30, 72, 432) = 2^4 \times 3^3 \times 5 = 2160$

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