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Maths 3 Marks Question

Real Numbers · CBSE STD 10 (CBSE - English Medium). 16 questions.

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Question 13 Marks
Use Euclid’s division algorithm to find the HCF of 441, 567, 693.
Answer
Let a = 693, b = 567 and c = 441 By Euclid's division algorithmes,
By Euclid's division algorithms,
a = bq + r .....(i)
[$\because$ dividend = divisor × quotient + remainder]
First we take, a = 693 and b = 567 and find their HCF.
693 = 567 × 1 + 126
567 = 126 × 4 + 63
126 = 63 × 2 + 0
$\therefore$ HCF (693, 567) = 63
Now, we take c = 441 and say d = 63, then find their HCF.
Again, using Euclid's division algorithm,
c = dq + r
⇒ 441 + 63 × 7 + 0
$\therefore$ HCF (693, 567, 441) = 63
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Question 23 Marks
Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.
Answer
By Euclid's division lemma, $b=a q+r$
Where $a, b, q, r$ are +ve integers and here $a=3$ then $b=3 q+r$ then $0 \leq r<3$ or $r=0,1,2$, so $b$ become $b=3 q, 3 q+1,3 q+2$,
$b=3 q$
$\Rightarrow(b)^2=(3 q)^2$
$\Rightarrow b^2=3.3 q^2=3 m \text { where, } 3 q^2=m$
So, as $b^2$ is perfect square so 3 m will also be perfect square.
$\text { When } r=1, b=3 q+1$
$\Rightarrow(b)^2=(3 q+1)^2$
$\Rightarrow b^2=9 q^2+1+2 \times 3 q$
$\Rightarrow b^2=3\left[3 q^2+q\right]+1$
$\Rightarrow b^2=3 m+1 \text { and } m=3 q^2+2 q$
So, $b^2$ is perfect square or a number of the form $3 m+1$ is perfect square.
$\text { When } r=2, b=3 q+2$
$\Rightarrow b^2=9 q^2+4+2 \cdot 3 q \cdot 2$
$=9 q^2+3+3 \times 4 q+1$
$=3\left[3 q^2+1+4 q\right]+1$
$\Rightarrow b^2=3 m+1$
Again, a number of the form $3 m+1$ is perfect square,
Hence, a number of the form $(3 m+2)$ can never be perfect square.
But a number of the form $3 m$, and $3 m+1$ are perfect squares.
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Question 33 Marks
A positive integer is of the form $3q + 1$, q being a natural number. Can you write its square in any form other than $3\ m + 1$, i.e., 3m or $3\ m + 2$ for some integer m? Justify your answer
Answer
By Euclid's division lemma,
b = aq + r where b, q, r are natural number and a = 3
$\therefore\text{b}=3\text{q}+\text{r where }0\leq\text{ r}<3\text{ so r}=0,1,2,$
At $r = 0, b = 3q$
$\Rightarrow b^2= (3q)^2 = 3.3q^2$
$\Rightarrow b^2 = 3m$, where $m = 3q^2$​​​​​​​
So, a number of the form 3m is prefect square.
At $r = 1, b = 3q + 1$
$\Rightarrow b^2 = (3q + 1)^2$
$\Rightarrow b^2 = 9q^2 + 1 + 6q$
$\Rightarrow b^2 = 3[3q^2 = 2q] + 1$
$\Rightarrow b^2 = 3m + 1$, where $m = 3q^2 + 2q$
So, a number of the form (3m + 1) is also perfect square.
At $r = 2, b = 3q + 2$
$\Rightarrow b^2 = (3q)^2 + (2)^2 + 2(3q)(2)$
$= 9q^2 + 4 + 3 \times 4q$
$= 9q^2 + 3 + 3 + \times 4q + 1 = 3[3q^2 + 1 + 4q] + 1$
$\Rightarrow b^2 = 3m + 1$, where $m = 3q^2 + 4q + 1$
Hence, a perfect square will be of the form 3m and (3m + 1) for m being a natural number.
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Question 43 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 x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, $x^2 + y^2 = (2m + 1)^2 + (2m + 3)^2$
$= 4m^2 + 1 + 4m + 4m^2 + 9 + 12m$ [$\because$ $(a + b)^2 = a^2 + 2ab + b^2]$
$= 8m^2 + 16m + 10 = even$
$= 2(4m^2 + 8m + 5) of 4(2m^2 + 4m + 2) + 1$
Hence, $x^2 + y^2$​​​​​​​ is even for every positive integer m but not divisible by 4.
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Question 53 Marks
Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.
Answer
$ \frac{987}{10500}=\frac{3\times7\times47}{2^2\times3^1\times5^3\times7^1}=\frac{47}{2^2\times5^3}$
As denominator has prime factors only in 2 and 5 so number $\frac{987}{10500}$ is terminating decimal.
$\frac{47}{2^3\times5^3}\times2=\frac{94}{1000}=0.094$
$\begin{array}{c|c} 3 & 987 \\ \hline 7 & 329 \\ \hline & 47 \end{array}$ $\begin{array}{c|c} 5 & 10500 \\ \hline 3 & 2100 \\ \hline 7 & 700 \\ \hline 5 & 100 \\ \hline 5 & 20 \\ \hline 2 & 4 \\ \hline & 2 \end{array}$
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Question 63 Marks
Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Answer
Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively.Thus, after subtracting these remainders from the numbers.
We have the numbers, 1251 - 1 = 1250, 9377 - 2 = 9375 and 15628 - 3 = 15625 which is divisible by the required number.
Now, required number = HCF of 1250, 9375 and 15625 [for the largest number]
By Euclid's division algorithm,
a = bq + r .....(i)
[$\because$ dividend = divisor × quotient + remainder]
For largest number, put a = 15625 and b = 9375
15625 = 9375 × 1 + 6250 [from Eq. (i)]
⇒ 9375 = 6250 × 1 + 3150
⇒ 6250 = 3125 × 2 + 0
$\therefore$ HCF (15625, 9375) = 3125
Now, we take c = 1250 and d = 3125, then again using Euclid's division algorithm,
d = cq + r [from Eq. (i)]
⇒ 3125 = 1250 × 2 + 625
⇒ 1250 = 625 × 2 + 0
$\therefore$ HCF (1250, 9375, 15625) = 625
Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.
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Question 73 Marks
On a morning walk, three persons step off together and their steps measure 40cm, 42cm and 45cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Answer
We have to find the LCM of 40cm, 42cm and 45cm to get the required minimum distance.
For this, 40 = 2 × 2 × 2 × 5,
42 = 2 × 3 × 7
and 45 = 3 × 3 × 5
$\therefore$ LCM (40, 42, 45) = 2 × 3 × 5 × 2 × 2 × 3 × 7
= 30 × 12 × 7 = 210 × 12
= 2520
Minimum distance eacg should walk 2520cm. So that, can cover the same distance in complete steps.
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Question 83 Marks
Prove that $\sqrt{3}+\sqrt{5}$ is irrational.
Answer
Let us suppose $\sqrt{3}+\sqrt{5}$ is rational Let $\sqrt{3}+\sqrt{5}=\text{a},$ where a is rational. $\therefore\ \sqrt{3}=\text{a}-\sqrt{5}$ On squaring both sides, we get $(\sqrt{3})^2+(\text{a}-\sqrt{5})^2$ $\Rightarrow\ 3=\text{a}^2+5-2\text{a}\sqrt{5}\ \ \big[\because(\text{a}-\text{b})^2=\text{a}^2 +\text{b}^2-2\text{ab}\big]$ $\Rightarrow\ 2\text{a}\sqrt{5}=\text{a}^2+2$ $\therefore\ \sqrt{5}=\frac{\text{a}^2+2}{2\text{a}}$ which is contradiction.As the right hand side is rational number while V5 is irrational. Since, 3 and 5 are prime number. Hence, $\sqrt{3}+\sqrt{5}$ is irrational.
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Question 93 Marks
The numbers $525$ and $3000$ are both divisible only by $3, 5, 15, 25$ and $75$. What is $HCF (525, 3000)$? Justify your answer.
Answer
The numebr 525 and $3000$ both are divisible by $3, 5, 15, 25$, and $75$. So, highest common factor out of $3, 5, 15, 25$ and $75$ is $75$ of $HCF$ of ($525, 3000$) is $75.$
Verification: $525 = 5 \times 5 \times 3 \times 7 = 3 \times 5^2 \times 7^1$
$3000 = 23 \times 53 \times 31 = 2^3 \times 3^1 \times 5^3$
$HCF = 3^1 \times 5^2 = 75$
Hence, verified.
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Question 103 Marks
"The product of three consecutive positive integers is divisible by 6". Is this statement true or false"? Justify your answer.
Answer
Yes, the given statement is true.
Three consecutive positive integers are n, (n + 1), (n + 2). Out of 3 consecutive integers, one will be even and other will be divisible by 3.
So, the product of all three becomes dividsible by 6,
e.g., 13, 14, 15 here 14 is even, 15 is divisible by 3.
So, 13 × 14 × 15 is divisible by 6.
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Question 113 Marks
Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
Answer
As we Know that
HCF(a, b) × LCM(a, b) = (a × b)
18 must be factor of 380.
So, $\frac{380}{18}$ should be a natural number.
But $\frac{380}{18}$ is not a natural number of 380 is not divisible by 18.
So, 380 and 18 are not the LCM and HCF of any two number.
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Question 123 Marks
Prove that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is irrational, where p, q are primes.
Answer
Let us suppose that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is rational,
again, let $\sqrt{\text{p}}+\sqrt{\text{q}}=\text{a},$ where a is rational.
$\therefore \sqrt{\text{q}}=\text{a}-\sqrt{\text{p}}$
On squaring both sides, we get
$\text{q}=\text{a}^2+\text{p}-2\text{a}\sqrt{\text{p}}\ \big[\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab}\big]$
$\therefore\ \sqrt{\text{p}}=\frac{\text{a}^2+\text{p}-\text{q}}{2\text{a}},$ which is a contradiction as the right hand side is rational number while $\sqrt{\text{p}}$ is irrational, since p and q are prime number.
Hence, $\sqrt{\text{p}}+\sqrt{\text{q}}$ is irrational.
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Question 133 Marks
Show that $12^{ n }$ cannot end with the digit 0 or 5 for any natural number n.
Answer
If any number ends with the digit 0 of 5 it is always divisible by 5 .
If $12^n$ ends with the digit zero it must be divisible by 5 .
This is possible only if prime factorisation of $12^n$ contains the prime number 5 .
Now, $12=2 \times 2 \times 3=2^2 \times 3$
$\Rightarrow 12^n=(22 \times 3)^n=2^{2 n} \times 3^n$ [since, there is no term contains 5]
Hence, there is no value of $n \in N$ for which $12^n$ ends with digit zero or five.
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Question 143 Marks
Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.
Answer
'No'. By Euclid's division lemma, we have
dividend = divisor × quotient + remainder
a = bq + r
Let b = 4 then
a = 4q + r where q, r are positive
Since $0\leq\text{r}<4\therefore\text{ r}=0,1,2,3$
So, a become of the form, 4q, 4q + 1, 4q + 2 and 4q + 3
So, all integers can be3 represented by all 4q, 4q + 1, 4q + 2, and 4q + 3 not only by 4q + 2.
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Question 153 Marks
If n is an odd integer, then show that $n^2 - 1$ is divisible by 8.
Answer
Let $a = n ^2-1$.....(i) Given that, n is an odd integer.
$\therefore n=1,3,5$ From Eq. (i), at $n=1, a=(1)^2-1=1-1=0$,
Which is divisible by 8 . From Eq. (i), at $n=3, a=(3)^2-1=9-1=8$,
which is divisible by 8 . From Eq. (i), at $n=5, a=(5)^2-1=25-1=24=3 \times 8$,
which is divisible by 8 . From Eq. (i), at $n=7$, $a=(7)^2-1=49-1=48=6 \times 8$, which is divisible by 8 . Hence, $\left(n^2-1\right)$ is divisible by 8 , where $n$ is an odd integer.
Alternate Answer
We know that an odd integer $n$ is of the from $(4 q+1)$ or $(4 q+3)$ for some integer $q$
Case I:
when $n =4 q +1$ in this case,
we have $\left(n^2-1\right)=(4 q+1)^2-1=16 q^2+1+8 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+8 q=8 q(2 q+1)=16 q^2+8 q=8 q(2 q+1)$
Which is clearly, divisible by 8 .
Case II:
When $n=4 q+3 \ln$ this case,
we have $\left(n_2-1\right)=(4 q+3)^2-1=16 q^2+9+24 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+24 q+8=8\left(2 q^2+3 q+1\right)$
which is clearly divisible by 8 . Hence, $\left(n^2-1\right)$ is divisible by 8 .
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Question 163 Marks
Explain why 3 × 5 × 7 + 7 is a composite number.
Answer
Main Concept:A number which is not prime is composite.
3 × 5 × 7 + 7 = 7[ 3 + 5 + 1] = 7[15 + 1]
= 7 × 16 have prime factors = 7 × 2 × 2 × 2 × 2
So, number (3 × 5 × 7 + 7) is not prime hence, it is composite.
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3 Marks Question - Maths STD 10 Questions - Vidyadip