2 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 12 Marks

Check whether$ 6^n$ can end with the digit $0$ for any natural number n.

Answer

If the number $6^{ n }$ ends with the digit zero, then it is divisibli by 5 . Therefore the prime factorization of $6^{ n }$ contains the prime 5 . This is not possible because the only prime in the factorisation of $6^{ n }$ is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorisation of $6^n$. Hence it is very clear that there is no value of $n$ in natural numbers for which $6^{ n }$ ends with the digit zero.

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Question 22 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{987}{10500}$

Answer

The given number is
$\frac{987}{10500}=\frac{987\div21}{10500\div21}=\frac{47}{500}$
Now, $500 = 2^2 \times 5^3$
The denominator can be written in the form of $2^m \times 5^n.$
So, the given number has a terminating decimal expansion.
$\frac{987}{10500}=\frac{47}{500}=\frac{47}{5^3\times2^2}\times\frac{2}{2}$ $\frac{94}{5^3\times2^3}=\frac{94}{1000}=0.094$

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Question 32 Marks

Find the LCM and HCF of the following integer by applying the prime factorisation method.
17, 23 and 29

Answer

17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
LCM of 17, 23, & 29
$\begin{array}{c|c} 17 & 17-23-29 \\ \hline 23 & 1-23-29\\\hline29 & 1-1-29\\\hline & 1-1-1\end{array}$
So LCM of 17, 23 & 29 is 17 × 23 × 29
= 11339
From the above, HCF (17, 23, 29) = 1 and LCM (17, 23, 29) = 11339

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Question 42 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
475 and 495

Answer

We need to find H.C.F. of 475 and 495.
By applying Euclid’s Division lemma
495 = 475 × 1 + 20.
Since remainder ≠ 0, apply division lemma on 475 and remainder 20.
475 = 20 × 23 + 15.
Since remainder ≠ 0, apply division lemma on 20 and remainder 15.
20 = 15 × 1 + 5.
Since remainder ≠ 0, apply division lemma on15 and remainder 5.
15 = 5 × 3 + 0.
Therefore, H.C.F. of 475 and 495 = 5.

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Question 52 Marks

What is a composite number?

Answer

A composite number is a positive integer which has a divisor other than one or itself.
In other words a composite number is any positive integer greater than one that is not a prime number.

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Question 62 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{125}{441}$

Answer

The given number is $\frac{125}{441}$
Here, $441 = 3^2 \times 7^2 441 = 32 \times 72$ and none of $3$ and $7$ is a factor of $125.$
So, the given number is in its simplest form.
Now, $441 = 3^2 \times 7^2$ is not of the form $2^m \times 5^n.$
So, the given number has a non-terminating repeating decimal expansion.

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Question 72 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m \times 5^n,$ where, m, n are non-negative integers.
$\frac{14588}{625}$

Answer

The given number is $\frac{14588}{625}.$
Clearly, $625 = 5^4$ is of the form $2^m \times 5^n,$ where $m = 0$ and $n = 4.$
So, the given number has terminating decimal expansion.
$\therefore\ \frac{14588}{625}=\frac{14588\times2^4}{2^4\times5^4}=\frac{14588\times16}{(2\times5)^4}$ $=\frac{233408}{(10)^4}=\frac{233408}{10000}=23.3408$

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Question 82 Marks

Find the $LCM$ and $HCF$ of the following integer by applying the prime factorisation method.
$8, 9$ and $25$

Answer

$8 = 2^3 \times 1$
$9 = 3^2 \times 1$
$25 = 5^2 \times 1$
LCM of $8, 9, 25$
$\begin{array}{c|c} 2 & 8-9-25 \\ \hline 2 & 4-9-25\\\hline2 & 2-9-25\\\hline 3 & 1-9-25\\\hline3 & 1-3-25\\\hline 5 & 1-1-25\\\hline5&1-1-5\\\hline & 1-1-1\end{array}$
So $LCM$ of $8, 9, 25$ is $= 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$
$= 1800$
From the above $HCF (8, 9, 25) = 1$ and $LCM (8, 9, 25) = 1800.$

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Question 92 Marks

If the prime factorization of a natural number n is $2^3 \times 3^2 \times 5^2 \times 7$, write the number of consecutive zeros in n.

Answer

Given that prime factorization of a natural number is,
$n = 2^3 \times 3^2 \times 5^2 \times 7$
$n = 8 \times 9 \times 25 \times 7$
$n = 72 \times 175$
$n = 12600$

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Question 102 Marks

Has the rational number $\frac{441}{2^2\times5^7\times7^2}$ terminating or a non terminating decimal representation?

Answer

Given rational number is $\frac{441}{2^2\times5^7\times7^2}$
The denominator $(2^2 \times 5^7 \times 7^2)$ of given rational number is not in the from of $(2^m \times 5^n).$
So, this is a non-terminating decimal representation.

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Question 112 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
18 and 24

Answer

We need to find H.C.F. of 18 and 24.
By applying division lemma
24 = 18 × 1 + 6
Since remainder ≠ 0, apply division lemma on divisor 18 and remainder 6.
18 = 6 × 3 + 0
Therefore, H.C.F. of 18 and 24 is 6

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Question 122 Marks

Prove that the product of two consecutive positive integers is divisible by 2.

Answer

Let, $(n-1)$ and $n$ be two consecutive positive integers
$\therefore$ Their product $= n(n - 1)$
$= n^2 - n$
We know that any positive integer is of the form $2 q$ or $2 q+1$, for some integer $q$ When $n=2 q$, we have
$n^2 - n = (2q)^2 - 2q$
$= 4q^2 - 2q$
$2(2q - 1)$
Then $n ^2- n$ is divisible by 2 .
When $n=2 q+1$, we have
$n^2 - n = (2q + 1)^2 - (2q + 1)$
$= 4q^2 + 4q + 1 - 2q - 1$
$= 4q^2 + 2q$
$= 2(2q + 1)$
Then $n ^2- n$ is divisible by 2 .
Hence the product of two consecutive positive integers is divisible by 2 .

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Question 132 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{35}{50}$

Answer

$\frac{35}{50}=\frac{7}{10}=\frac{7}{2\times5}$
So, the simplest from of the given rational number $\frac{35}{50}$ is $\frac{7}{10}$ and determination 10 is of the form$ (2^m \times 5^n)$, where m and n are non-nagative integer.
Thus, $\frac{35}{50}$ has a terminating decimal expansion.

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Question 142 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{23}{8}$

Answer

$8 = 2^3$
The denominator is of the form $2^m \times 5^n.$
Hence, the decimal expansion of $\frac{23}{6}$ is terminating.

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Question 152 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
70 and 30

Answer

Prime factors of 70 and 30 are
70 = 2 × 5 × 7
30 = 2 × 3 × 5
HCF (70, 30) = Product of the smallest power of each common prime factor in the number
= 2 × 5
= 10

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Question 162 Marks

State Euclid's division lemma.

Answer

Euclid’s Division Lemma:
Let a and b be any two positive integers.
Then, there exist unique integers q and r such that
a = bq + r, 0 ≤ r < b
If b|a then r = 0
Otherwise, r satisfies the stronger inequality 0 < r < b.

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Question 172 Marks

The HCF to two numbers is 16 and their product is 3072. Find their LCM.

Answer

Given: HCF of two numbers is 16. If product of numbers is 3072
To Find: L.C.M of numbers
L.C.M × H.C.F = First Number × Second Number
L.C.M × 16 = 3072
L.C.M $= \frac{3072}{16}$
L.C.M = 192

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Question 182 Marks

What can you say about the prime factorisations of the 43 of the following rationals:
$43.\overline{123456789}$

Answer

We have
$43.\overline{123456789}$
$43.\overline{123456789}$ has non-terminating decimal expansion.
So its denominator has factors other than 2 or 5.

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Question 192 Marks

Find the least number that is divisible by ail the numbers between 1 to 10 (both inclusive).

Answer

The required least number which is divisible by 1 to 10 will be the L.C.M. of 1 to 10.
L.C.M. of 1 to 10.
$\begin{array}{c|c} 2 & 1,2,3,4,5,6,7,8,9,10 \\ \hline 2 & 1,1,3,2,5,3,7,4,9,5\\ \hline 3 & 1,1,3,1,5,3,7,2,9,5\\ \hline 5 & 1,1,1,1,5,1,7,2,3,6\\ \hline &1,1,1,1,1,1,7,2,3,1 \end{array}$
$= 2 \times 2\times 3 \times 5\times 7 \times 2 \times 3 = 2520$

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Question 202 Marks

Complete the missing entries in the following factor tree.

Answer

We need to fill the values for a and b in the following factor tree:

It is clear from the factor tree above that b = 3 × 7 b = 21 Also, a = 2 × b a = 2 × 21 a = 42 Thus, the missing entries are 21 and 42.

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Question 212 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m \times 5^n,$ where, m, n are non-negative integers.
$\frac{7}{80}$

Answer

$\frac{7}{80}=\frac{7}{2^4\times5^1}$
($\because$ $80 = 16 \times 5 = 2 \times 2 \times 2 \times 2 \times 5 = 24 \times 51)$
{Multiplying and dividing by $5^3$}
$=\frac{875}{10000}=0.0875$

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Question 222 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{77}{210}$

Answer

The given number is $\frac{77}{210}$ and $HCF(77, 210) = 7.$
$\therefore\ \frac{77}{210}=\frac{77\div7}{210\div7}=\frac{11}{30}$
Here, $\frac{11}{30}$ is in its simplest form.
Here, $\frac{77}{210}$ is in its simplest form.
Now, $30 = 2 \times 3 \times 5$ is not of the form $2^m \times 5^n.$
So, the given number has a non-terminating repeating decimal expansion.

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Question 232 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m \times 5^n$, where, m, n are non-negative integers.
$\frac{3}{8}$

Answer

The given number is $\frac{3}{8}.$
$\frac{3}{8} = \frac{3}{2^{3}} = \frac{3}{2^{3} \times 5^{3}}$
$= \frac{3\times 5^{3}}{2^{3}\times 5^{3}} = \frac{3\times 125}{(10)3} = \frac{375}{1000} = 0.375$
Thus, the decimal expansion of given rational number is 0.375

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Question 242 Marks

Define $HCF$ of two positive integers and find the $HCF$ of the following pairs of numbers:
$100$ and $190$

Answer

Prime factors of $100$ and $190$ are
$100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$
$190 = 2 \times 5 \times 19 = 2 \times 5 \times 19$
HCF (100, 190) = Product of the smallest power of each common prime factor in the number
$= 2 \times 5$
$= 10$

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Question 252 Marks

The decimal expression of the rational number $\frac{43}{2^4\times5^3}$ will terminate after how many places of decimals.

Answer

The denominator of $\frac{43}{2^4\times5^3}$ is $2^4 \times 5^3$ which is in the form of $2^m \times 5^n$ where m and n are positive integers $\frac{43}{2^4\times5^3}$ has terminating decimals.
The decimal expansion of $\frac{43}{2^4\times5^3}$ terminates after 4 (the highest power is 4) decimal places.

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Question 262 Marks

Write whether $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$ on simplification gives a rational or an irrational number.

Answer

Let us simplify $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$
$\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}=\frac{6\sqrt{5}+6\sqrt{5}}{2\sqrt{5}}$
$=\frac{12\sqrt{5}}{2\sqrt{5}}$
$=6$
6 is rational number.

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Question 272 Marks

Use Euclid's division algorithm to find the HCF of:
184, 230 and 276

Answer

Given integers are 184, 230 and 276.
Let us first find the HCF of 184 and 230 by Euclid lemma.
Clearly, 230 > 184. So, we will apply Euclid’s division lemma to 230 and 184.
230 = 184 × 1 + 46
Remainder is 46 which is a non-zero number. Now, apply Euclid’s division lemma to 184 and 46.
184 = 46 × 4 + 0
The remainder at this stage is zero. Therefore, 46 is the HCF of 230 and 184.
Now, again use Euclid’s division lemma to find the HCF of 46 and 276.
276 = 46 × 6 + 0
The remainder at this stage is zero.
Therefore, 46 is the HCF of 184, 230 and 276.

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Question 282 Marks

What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

Answer

L.C.M. of 35, 56, 91
$\begin{array}{c|c} 7 & 35, 56, 91 \\ \hline 2 & 5, 8, 13 \end{array}$
= 5 × 7 × 8 × 13 = 3640
Remainder in each case = 7
The required smallest number = 3640 + 7 = 3647

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Question 292 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
56 and 88

Answer

By applying Euclid’s division lemma
88 = 56 × 1 + 32
Since remainder ≠ 0, apply division lemma on divisor of 56 and remainder 32.
56 = 32 × 1 + 24
Since remainder ≠ 0, apply division lemma on divisor of 32 and remainder 24.
32 = 24 × 1 + 8
Since remainder ≠ 0, apply division lemma on divisor of 24 and remainder 8.
24 = 8 × 3 + 0
$\therefore$ HCF of 56 and 88 is = 8.

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Question 302 Marks

What can you say about the prime factorisations of the $43$ of the following rationals:
$43.123456789$

Answer

We have
$43.123456789=\frac{43123456789}{1000000000}$
we can write denominator 1000000000 is in the from of$ (2^m \times 5^n).$
Hence, 43.123456789 is terminating decimal, where m and n are non-negative integers.

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Question 312 Marks

Write the condition to be satisfied by q so that a rational number $\frac{\text{p}}{\text{q}}$ has a terminating decimal expansion.

Answer

Let rational number be n which is in the form of $\frac{\text{p}}{\text{q}}.$
The condition for non-terminating decimal expansion is that denominator of $\frac{ p }{ q }$ is q which is not in the form of $\left(2^m \times 5^n\right)$, where m , n are nonnegativ inteagrs.

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Question 322 Marks

Use Euclid's division algorithm to find the HCF of:
196 and 38220

Answer

Given integers are 38220 and 196.
Clearly 38220 > 196. So we will apply Euclid’s division lemma to 38220 and 196, we 38220 = (196)(195) + 0 get,
The remainder at this stage is 0. So the divisor at this stage is the H.C.F.
So the H.C.F of 38220 and 196 is 196.

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Question 332 Marks

Find the LCM and HCF of the following pairs of integers and verify that $LCM \times HCF =$ Product of the integers:
$510$ and $92$

Answer

To Find: L.C.M and H.C.F of following pairs of integers
To Verify: L.C.M $\times$ H.C.F $=$ product of the numbers
Let us first find the factors of 510 and 92
$510=2 \times 3 \times 5 \times 17$
$92=2^2 \times 23$
L.C.M of 510 , and $92=2 \times 2 \times 3 \times 5 \times 17 \times 23=23460$
H.C.F of 510 , and $92=2$
We know that,
L.C.M $\times$ H.C.F $=$ First number $\times$ Second number $\Rightarrow 23460 \times 2=510 \times 92 \Rightarrow 46920=46920$
L.C.M $\times$ H.C.F $=$ First number $\times$ Second number $\Rightarrow 23460 \times 2=510 \times 92 \Rightarrow 46920=46920$
Hence verified.

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Question 342 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m \times 5^n$, where, m, n are non-negative integers.
$\frac{129}{2^2\times5^7}$

Answer

The given number is $\frac{129}{2^2\times5^2}.$
$\frac{129}{2^2\times5^7}=\frac{125\times2^5}{(2^7\times5^7)}=\frac{129\times32}{(10)^7}$ $=\frac{4128}{10000000}=0.0004128$
Thus, the decimal expansion of $\frac{129}{2^2\times5^7}$ is 0.0004128

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Question 352 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
32 and 54

Answer

By applying Euclid’s division lemma
54 = 32 × 1 + 22
Since remainder ≠ 0, apply division lemma on division of 32 and remainder 22.
32 = 22 × 1 + 10
Since remainder ≠ 0, apply division lemma on division of 22 and remainder 10.
22 = 10 × 2 + 2
Since remainder ≠ 0, apply division lemma on division of 10 and remainder 2.
10 = 2 × 5 [remainder 0]
Hence, HCF of 32 and 54 is 2

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Question 362 Marks

The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.

Answer

Second Number = 361
L.C.M × H.C.F = First Number × Second Number
180 × 6 = 30 × Second Number
Second Number $=\frac{180\times6}{30}=36$

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Question 372 Marks

If p and q are two prime number, then what is their HCF?

Answer

It is given that p and q are two prime numbers; we have to find their HCF.
We know that the factors of any prime number are 1 and the prime number itself.
For example, let p = 2 and q = 3
Thus, the factors are as follows
p = 2 × 1
And
q = 3 × 1
Now, the HCF of 2 and 3 is 1.
Thus the HCF of p and q is 1.

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Question 382 Marks

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer

Explain: Why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
We can see that both the numbers have common factor 7 and 1.
7 × 11 × 13 + 13 = (77 + 1) × 13
= 78 × 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 × 1) × 2
= 1008 × 5
And we know that composite numbers are those numbers which have at least one more factor other than 1.
Hence after simplification we see that both numbers are even and therefore the given two numbers are composite numbers.

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Question 392 Marks

The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.

Answer

L.C.M × H.C.F = First Number × Second Number
2175 × 145 = 725 × Second Number
$\therefore$ Second Number $= \frac{(2175 \times 145)}{725} = 435$

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Question 402 Marks

Use Euclid's division algorithm to find the HCF of:
867 and 255

Answer

867 > 255
867 = 225 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Hence, HCF of 867 and 255 is 51.

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Question 412 Marks

A rational number in its decimal expansion is$ 327.7081$. What can you say about the prime factors of q, when this number is expressed in the form $\frac{\text{p}}{\text{q}}?$ Give reasons.

Answer

327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form $2^m \times 5^n.$
Thus, $327.7081=\frac{3277081}{10000}=\frac{\text{p}}{\text{q}}\ (\text{say})$
$\therefore$ $q = 10^4 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^4 \times 5^4 = (2 \times 5)^4$
Hence, the prime factors of q is 2 and 5.

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Question 422 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
105 and 120

Answer

By applying Euclid’s division lemma
120 = 105 × 1 + 15
Since remainder ≠ 0, applying division lemma on divisor 105 and remainder 15.
105 = 15 × 7 + 0
$\therefore$ H.C.F of 105 and 120 = 15

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Question 432 Marks

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers:
26 and 91

Answer

To Find: L.C.M and H.C.F of following pairs of integers
To Verify: L.C.M × H.C.F = product of the numbers
Let us first find the factors of 26 and 91
26 = 2 × 13
91 = 7 × 13
L.C.M of 26, and 91 = 2 × 7 × 13 = 182
H.C.F of 26, and 91 = 13
We know that,
L.C.M × H.C.F = First number × Second number ⇒ 182 × 13 = 26 × 91 ⇒ 2366 = 2366
L.C.M × H.C.F = First number × Second number ⇒ 182 × 13 = 26 × 91 ⇒ 2366 = 2366
Hence verified.

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Question 442 Marks

Find the LCM and HCF of the following integer by applying the prime factorisation method.
12, 15 and 21

Answer

12 = 2 × 2 × 3 15 = 5 × 3 21 = 7 × 3 3 is common for all, it means HCF of 12, 15 & 21 is 3 Now calculate LCM of 12, 15, 21$\begin{array}{c|c} 2 & 12-15-21 \\ \hline 2 & 6-15-21\\\hline3 & 3-15-21\\\hline 5 & 1-5-7\\\hline7 & 1-1-7\\\hline & 1-1-1\end{array}$
So LCM of 12, 15, 21 is 2 × 2 × 3 × 5 × 7 = 420 From the above, HCF (12, 15, 21) = 3 and LCM (12, 15, 21) = 420

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Question 452 Marks

What can you say about the prime factorisations of the 43 of the following rationals:
$27.\overline{142857}$

Answer

We have
$27.\overline{142857}$
We can see that it is a non-terminating repeating decimal expersion.
So, its denominator has factors others them 2 or 5.

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Question 462 Marks

State Fundamental Theorem of Arithmetic.

Answer

Fundamental Theorem of Arithmetics: Every composite number can be expressed (factorised) as a product of primes and this factorization is unique except for the order in which the prime factors occur.

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Question 472 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
155 and 1385

Answer

We need to find H.C.F. of 155 and 1385.
By applying Euclid’s Division lemma
1385 = 155 × 8 + 145.
Since remainder ≠ 0, apply division lemma on divisor 155 and remainder 145.
155 = 145 × 1 + 10.
Since remainder ≠ 0, apply division lemma on divisor 145 and remainder 10.
145 = 10 × 14 + 5.
Since remainder ≠ 0, apply division lemma on divisor 10 and remainder 5.
10 = 5 × 2 + 0.
Therefore, H.C.F. of 155 and 1385 = 5.

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Question 482 Marks

Find the $LCM$ and $HCF$ of the following pairs of integers and verify that $LCM × HCF =$ Product of the integers:
$336$ and $54$

Answer

To Find: L.C.M and H.C.F of following pairs of integers
To Verify: L.C.M × H.C.F = product of the numbers
Let us first find the factors of 336 and 54
$336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7$
$54 = 2 \times 3 \times 3 \times 3$
L.C.M of 336, and $54 = 2^4 \times 3^3 \times 7 = 3024$
H.C.F of 336, and $54 = 2 × 3 = 6$
We know that,
L.C.M × H.C.F = First number × Second number
$\Rightarrow 3024 \times 6 = 336 \times 54$
$\Rightarrow 18144 = 18144$
Hence verified.

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Question 492 Marks

If the product of two numbers is 1080 and their HCF is 30, find their LCM.

Answer

It is given that the product of two numbers is 1080.
Let the two numbers be a and b.
Therefore,
a × b = 1080
HCF is 30.
We need to find the LCM
We know that the product of two numbers is equal to the product of the HCF and LCM.
Thus,
$\text{LCM}=\frac{\text{a}\times\text{b}}{\text{HCF}}$
$\text{LCM}=\frac{1080}{30}$
$\text{LCM}=36$
Hence the LCM is 36.

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Question 502 Marks

Use Euclid's division algorithm to find the HCF of:
135 and 225

Answer

We have 225 > 135,
So, we apply the division lemma to 225 and 135 to obtain.
225 = 135 × 1 + 90
Here remainder 90 ≠ 0, we apply the division lemma again to 135 and 90 to obtain.
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45 ≠ 0, and apply the division lemma to obtain.
90 = 2 × 45 + 0
Since at this time the remainder is zero, the process is stopped.
The divisor at this stage is 45
Therefore, the HCF of 135 and 225 is 45.

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2 Marks Questions - Maths STD 10 Questions - Vidyadip