3 Marks Question - Maths STD 10 Questions - Vidyadip

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35 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Prove that if $x$ and $y$ are both odd positive integers then $x^2 + y^2$ is even but not divisible by $4$.

Answer

Let the two odd positive numbers be $x = 2k + 1$ and $y = 2p + 1$
Hence $x^2+ y^2 = (2k + 1)^2 + (2p + 1)^2$
$= 4k^2 + 4k + 1 + 4p^2 + 4p + 1$
$= 4k^2 + 4p^2+ 4k + 4p + 2$
$= 4(k^2 + p^2+ k + p) + 2$
Clearly notice that the sum of square is even the number is not divisible by $4$
Hence if $x$ and $y$ are odd positive integers, then $x^2+ y^2$ is even but not divisible by $4$

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Question 23 Marks

Express the following as a fraction in simplest form:
$2.2\bar4$

Answer

Let $\text{x}=2.2\bar4$
$\therefore\text{x}=2.2444\dots(1)$
$\text{10x}=22.2444\dots(2)$
$\text{100x}=224.444\dots(3)$
On subtracting equation (2) from (3), we get
$\text{90x}=202$
$\Rightarrow\text{x}=\frac{202}{90}$
$=\frac{101}{45}$
Hence, $2.24=\frac{\overline{101}}{45}$

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Question 33 Marks

Express the following as a fraction in simplest form:
$2.\bar4$

Answer

Let $\text{x}=2. \bar4$
$\therefore\text{x}=2.444\dots(1)$
$\text{10x}=24.444\dots(2)$
On subtracting equation (1) from (2), we get
$\text{9x}=22$
$\Rightarrow\text{x}=\frac{22}{9}$
$\therefore2.4=\frac{\overline{22}}{9}$

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Question 43 Marks

Using prime factorization, find the $HCF$ and $LCM$ of:
$396, 1080$

Answer

$396 = 2 \times 2 \times 3 \times 3 \times 11 $
$= 2^2 \times 3^2\times 11 1080 $
$= 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 $
$= 2^3 \times 3^3 \times 5$
HCF $(396, 1080)$
$ = 2^2 \times 3^2$
$=36$LCM $(396, 1080) $
$= 2^3 \times 3^3 \times 11 \times 5$
$ =11880$
HCF $\times $ LCM
$= 427680 144 \times 198$
$ = 427680$
$\Rightarrow $ HCF $\times $ LCM = product of given numbers Hence verified.

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Question 53 Marks

Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Answer

Let 'a' be a given positive integer.
On dividing 'a' by 6, let 'q' be the quotient and 'r' be the remainder.
Then, by Euclid's algorithm, we have
a = 6q + r, where 0 ≤ r < 6
⇒ a = 6q + r, where r = 0, 1, 2, 3, 4, 5, 6
⇒ a = 6q or a + 6q + 1 or a + 6q + 2 or a + 6q + 3 or a + 6q + 4 or a + 6q + 5
But, a = 6q, a + 6q + 2, a + 6q + 4 gives even values of 'a'.
Thus, when 'a' is odd, it is of the form 6q + 1, 6q + 3 or 6q + 5 for some integer 'q'.

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Question 63 Marks

Express the following as a fraction in simplest form:
$0.1\bar2$

Answer

Let $\text{x}=0.1\bar2$
$\therefore\text{x}=0.1222\dots(1)$
$\text{10x}=12.222\dots(2)$
On subtracting equation (1) from (2), we get
$\text{9x}=12$
$\Rightarrow\text{x}=\frac{12}{9}$
$\therefore0.12=\frac{\overline{12}}{9}$

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Question 73 Marks

Find the smallest number which when divided by $28$ and $32$ leaves remainders $8$ and $12$ respectively.

Answer

The smallest number which when divided by $28$ and $32$ can be determined by finding the LCM of $28$ and $32$
$28 = 2^2 \times 7$
$32 = 2^5$
LCM $= 2^5 \times 7$
$= 224$
The smallest number that when divided by $28$ and $32$ leaves a reminder $8$ and $12$
$= 224 - 8 - 12 = 204$

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Question 83 Marks

Short-Answer Questions:
Express $0.\overline{23}$ as a rational number in simplest form.

Answer

Let $\text{x}=0.\overline{23}$ then,
$\text{x}=0.232323\dots\ \ \dots(\text{i})$
$\therefore\text{100x}=23.2323\dots\ \ \dots(\text{ii})$
On subtracting (i) from (ii), we get
$\text{99x}=23$
$\Rightarrow\text{x}=\frac{23}{99}$
Hence, $0.\overline{23}=\frac{23}{99}$

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Question 93 Marks

Using Euclid's algortihm, find the HCF of:
960 and 1575

Answer

Dividing 1575 by 960, we get
Quotient = 1, Remainder = 615
$\therefore$ 1575 = 960 × 1 + 615
Dividing 960 by 615, we get
Quotient = 1, Remainder = 345
$\therefore$ 960 = 615 × 1 + 345
Dividing 615 by 345
Quotient = 1, Remainder = 270
$\therefore$ 615 = 345 × 1 + 270
Dividing 345 by 270, we get
Quotient = 1, Remainder = 75
$\therefore$ 345 = 270 × 1 + 75
Dividing 270 by 75, we get
Quotient = 3, Remainder =45
$\therefore$ 270 = 75 × 3 + 45
Dividing 75 by 45, we get
Quotient = 1, Remainder = 30
$\therefore$ 75 = 45 × 1 + 30
Dividing 45 by 30, we get
Remainder = 15, Quotient = 1
$\therefore$ 45 = 30 × 1 + 15
Dividing 30 by 15, we get
Quotient = 2, Remainder = 0
$\therefore$ H.C.F. of 1575 and 960 is 15

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Question 103 Marks

Express the following as a fraction in simplest form:
$0.\bar8$

Answer

Let
$\text{x}=0. \bar8$
$\therefore\text{x}=0.888\dots(1)$
$\text{10x}=8.888\dots(2)$
On subtracting equation (1) from (2), we get
$\text{9x}=8$
$\Rightarrow\text{x}=\frac{8}{9}$
$\therefore0.8=\frac{\bar8}{9}$

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Question 113 Marks

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Answer

Let 'a' be a given positive odd integer.
On dividing 'a' by 4, let 'q' be the quotient and 'r' be the remainder.
Then, by Euclid's algorithm, we have
a = 4m + r, where 0 ≤ r < 4
⇒ a = 4m + r, where r = 0, 1, 2, 3
⇒ a = 4m or a = 4m + 1 or a = 4m +2 or a = 4m + 3
But, a = 4m and a = 4m + 2 = 2(2m +1) are clearly even.
Thus, when 'a' is odd, it is of the form a = (4m + 1) or (4m + 3) for some integer m.

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Question 123 Marks

Give prime factorisation of $4620$.

Answer

$4620 = 2 \times 2 \times 3 \times 5 \times 7 \times 11$
$= 2^2 \times 3 \times 5 \times 7 \times 11$ is the prime factorisation of $4620$

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Question 133 Marks

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

Answer

Let the two numbers be a and b.
HCF × LCM = ab
⇒ 145 × 2175 = 725 × b
⇒ b = 435
So, the other number is 435

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Question 143 Marks

Using Euclid's algortihm, find the HCF of:
504 and 1188

Answer

On dividing 1188 by 504, we get
Quotient = 2, Remainder = 180
$\therefore$ 1188 = 504 × 2+ 180
Dividing 504 by 180
Quotient = 2, remainder = 144
$\therefore$ 504 = 180 × 2 + 144
Dividing 180 by 144, we get
Quotient = 1, Remainder = 36
Dividing 144 by 36
Quotient = 4, Remainder = 0
$\therefore$ H.C.F. of 1188 and 504 is 36

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Question 153 Marks

Find the HCF of $1008$ and $1080$ by prime factorization method.

Answer

$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7$
$= 2^4 \times 3^2 \times 7$
$1080 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$
So, the HCF $= 2^3 \times 3^2 = 72$

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Question 163 Marks

The HCF of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.

Answer

Let the number be a and 81
HCF × LCM = product of the two numbers
27 × 162 = 81a
a = 54
So, the other number is 54

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Question 173 Marks

Express the following as a fraction in simplest form:
$0.\overline{24}$

Answer

Let $\text{x}=0. \overline{24}$
$\therefore\text{x}=0.2424\dots(1)$
$\text{100x}=24.2424\dots(2)$
On subtracting equation (1) from (2), we get
$\text{99x}=24$
$\Rightarrow\text{x}=\frac{8}{33}$
$\therefore0.24=\frac{\bar{8}}{33}$

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Question 183 Marks

Short-Answer Questions:
Explain why 0.15015001500015... is an irrational number.

Answer

The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational numbers.
Clearly, 0.15015001500015... is a non-terminating and non-repeating decimal.
Hence, it is irrational.

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Question 193 Marks

Using prime factorization, find the HCF and LCM of:
$144, 198$

Answer

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$= 2^4 \times 3^2 198 $
$= 2 \times 3 \times 3 \times 11$
$ = 2 \times 3^2 \times 11$
HCF $(144, 198) = 2 \times 3^2 = 18$
LCM $(144, 198) = 2^4 \times 3^2 \times 11 = 1584$
HCF $\times $ LCM = 28512 144 × 198 = 28512
$\Rightarrow $ HCF $\times $ LCM = product of given numbers Hence verified.

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Question 203 Marks

Short-Answer Questions:
Write a rational number between $\sqrt3$ and 2.

Answer

We have, $\sqrt{3}=1.732\dots$
Since 1.732... < 1.8 < 2
So, 1.8 is a rational number that lies between $\sqrt3$ and 2

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Question 213 Marks

Show that every positive integer is either even or odd.

Answer

Let 'a' be a given positive integer.
On dividing 'a' by 2, let q be the quotient and r be the remainder.
Then, by Euclid's algorithm, we have
a = 2q + r, where 0 ≤ r < 2
⇒ a = 2q + r, where r = 0, 1
⇒ a = 2q or a = 2q + 1
When a = 2q for some integer q, then clearly 'a' is even.
When a = 2q + 1 for some integer q, then clearly 'a' is odd.
Thus, every positive integer is either even or odd.

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Question 223 Marks

Find the least number which when divided by $35, 56$ and $91$ leaves the same remainder $7$ in each case.

Answer

The smallest number which when divided by $35, 56$ and $91$ can be determined by finding the LCM of $35, 56$ and $91$
$35 = 5 \times 7$
$56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7$
$91 = 7 \times 13$
LCM $= 2^3 \times 5 \times 7 \times 13$
$= 3640$
The smallest number that when divided by $35, 56, 91$ leaves a reminder $7$ in each case
$= 3640 + 7 = 3547$

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Question 233 Marks

Examine whether $\frac{17}{30}$ is a terminating decimal.

Answer

A number is a terminating decimal, if the denominator is of the form $2^m \times 5^n$, where m and n are non-negative integers.
$\frac{17}{30}=\frac{17}{2\times3\times5}$
Clearly, $\frac{17}{30}$ is a not a terminating decimal, since its denominator is not of the form $2^m \times 5^n$
It has a factor of $3$ present too.

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Question 243 Marks

Using prime factorization, find the HCF and LCM of:
$1152, 1664$

Answer

$1152=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$
$=2^7 \times 3^2$
$1664=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=2^7$
$=128$
$\operatorname{HCF}(1152,1664)$
$=2^7$
$=128$
LCM $(1152,1664)$
$=2^7 \times 3^2 \times 13$
$=14976$
$\text { HCF } \times \text { LCM }=19169281152 \times 1664$
$=1916928$
$\Rightarrow H C F \times \text { LCM }$
$=$ product of given numbers Hence verified.

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Question 253 Marks

Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

Answer

LCM of 20, 25, 35, and 40 = 1400
Now,
20 - 6 = 14
25 - 6 = 19
35 - 6 = 29
40 - 6 = 34
So, (1400 - 6 = 1394) will have remainder of 14, 19, 29, and 34 when divided by 20, 25, 35, and 40 resp.
= 1394

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Question 263 Marks

Short-Answer Questions:
Explain why $3.\overline{1416}$ is a rational number.

Answer

The number of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0$ are called rational numbers.
Let $\text{x}=3.\overline{1416}$
⇒ x = 3.141614161416... ...(i)
Since there are four repeating digits,
We multiply by 1000
⇒ 1000x = 31416.14161416... ...(ii)
Subtracting (i) from (ii), we get
999x = 31416
$\Rightarrow\text{x}=\frac{31416}{999}$
Which is of the form $\frac{\text{p}}{\text{q}}.$
So, $3.\overline{1416}$ is a rational number.

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Question 273 Marks

The HCF of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.

Answer

Let the number be a and 81
HCF × LCM = product of the two numbers
27 × 162 = 81a
a = 54
So, the other number is 54

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Question 283 Marks

Find the smallest number which when increased by $17$ is exactly divisible by both $468$ and $520$.

Answer

The smallest number which when increased by $17$ is exactly divisible by both $520$ and $468$ is obtained by subtracting $17$ from the $LCM$ of $520$ and $468$
$468 = 2^2 \times 3^2 \times 13$
$520 = 2^3 \times 5 \times 13$
$LCM = 2^3 \times 3^2 \times 5 \times 13$
$= 4680$
Smallest number which when increased by $17$ is exactly divisible by both $520$ and $468 = 4680 - 17 = 4663$

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Question 293 Marks

For any positive integer $n$, prove that $n^3-n$ is divisible by $6$ .

Answer

$n^3-n=n\left(n^2-1\right)=n(n-1)(n+1)$
Whenever a number is divided by 3 , the remainder obtained is either $0$ or $1$ or $2$ .
$\therefore n=3 p$ or $3 p+1$ or $3 p+2$, where $p$ is some integer.
If $n =3 p$, then n is divisible by $3$ .
If $n=3 p+1$, then $n-1=3 p+1-1=3 p$ is divisible by $3$ .
If $n=3 p+2$, then $n+1=3 p+2+1=3 p+3=3(p+1)$ is divisible by $3$ .
So, we can say that one of the numbers among $n, n-1$ and $n+1$ is always divisible by $3$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by 3 .
Similarly, whenever a number is divided $2$ , the remainder obtained is $0$ or $1$ .
$\therefore n=2 q$ or $2 q+1$, where $q$ is some integer.
If $n=2 q$, then $n$ is divisible by $2$ .
If $n=2 q+1$, then $n-1=2 q+1-1=2 q$ is divisible by $2$ and $n+1=2 q+1+1=2 q+2=2(q+1)$ is divisible by
$2$.
So, we can say that one of the numbers among $n , n -1$ and $n +1$ is always divisible by $2$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by $2$ .
Since, $n(n-1)(n+1)$ is divisible by $2$ and $3$ .
$\therefore n(n-1)(n+1)=n^3-n$ is divisible by $6$ . (If a number is divisible by both $2$ and $3$ , then it is divisible by $6$ )

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Question 303 Marks

Short-Answer Questions:
Show that $\frac{\sqrt2}{3}$ is irrational.

Answer

If possible, let $\frac{\sqrt2}{3}$ be rational.
$\Rightarrow\frac{1}{3}\sqrt2$ is rational.
Now, 3 is rational, $\frac{1}{3}\sqrt2$ is rational.
$\Rightarrow\Big(3\times\frac{1}{3}\sqrt2\Big)$ is rational.
$\Rightarrow\sqrt2$ is rational.
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\frac{\sqrt2}{3}$ is rational.
Hence, $\frac{\sqrt2}{3}$ is irrational.

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Question 313 Marks

Using prime factorization, find the HCF and LCM of:
96, 404

Answer

96 = 2 × 2 × 2 × 2 × 2 × 3
404 = 2 × 2 × 101
LCM (96 and 404) = 25 × 3 × 101 = 9696
HCF (96 and 404)= 22=4
Verification:
HCF × LCM = 9696 × 4 = 38784
product of given numbers = 96 × 404 = 38784
⇒ HCF × LCM = Product of given numbers verified.

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Question 323 Marks

Using Euclid's algortihm, find the HCF of:
405 and 2520

Answer

On dividing 2520 by 405, we get
Quotient = 6, Remainder = 90
$\therefore$ 2520 = (405 × 6) + 90
Dividing 405 by 90, we get
Quotient = 4,
Remainder = 45
$\therefore$ 405 = 90 × 4 + 45
Dividing 90 by 45
Quotient = 2, Remainder = 0
$\therefore$ 90 = 45 × 2
$\therefore$ H.C.F. of 405 and 2520 is 45

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Question 333 Marks

Express the following as a fraction in simplest form:
$0.\overline{365}$

Answer

Let $\text{x}=0.\overline{365}$
$\therefore\text{x}=0.3656565\dots(1)$
$\text{10x}=3.656565\dots(2)$
$\text{1000x}=365.656565\dots(3)$
On subtracting equation (2) from (3), we get
$\text{990x}=362$
$\Rightarrow\text{x}=\frac{362}{990}$
$=\frac{181}{495}$
Hence, $0.365=\frac{\overline{181}}{495}$

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Question 343 Marks

Very-Short-Answer Questions:
Simplify: $\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}$

Answer

$\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}$
$=\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}\times\frac{2\sqrt5}{2\sqrt5}$ ...(Rationlising the denominator)
$=\frac{2\sqrt5\big(2\sqrt{45}+3\sqrt{20}\big)}{20}$
$=\frac{4\sqrt{45\times5}+6\sqrt{20\times5}}{20}$
$=\frac{4\sqrt{3^2\times5^2}+6\sqrt{2^2\times5^2}}{20}$
$=\frac{4(3\times5)+6(2\times5)}{20}$
$=\frac{60+60}{20}$
$=6$

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Question 353 Marks

Three sets of English, Mathematics and Science books containing $336, 240$ and $96$ books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there?

Answer

Let us find the HCF of 336, 240 and 96 through prime factorization:
$\begin{array}{c|c} 2 & 336 \\ \hline 2 & 168\\ \hline2&84\\ \hline2&42\\ \hline3&21\\\hline&7 \end{array}$ $\begin{array}{c|c} 2 & 240 \\ \hline 2 & 120\\ \hline2&60\\ \hline2&30\\ \hline3&15\\\hline&5 \end{array}$ $\begin{array}{c|c} 2 & 96 \\ \hline 2 & 48\\ \hline2&24\\ \hline2&12\\ \hline2&6\\\hline&3 \end{array}$
$336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7$
$240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^4 \times 3 \times 5$
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
Each stack of book will contain 48 books
Number of stacks of the same height
$=\frac{240}{48}+\frac{336}{48}+\frac{96}{48}$
$=5+7+2=14$

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