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Maths 5 Marks Questions

Real Numbers · CBSE STD 10 (CBSE - English Medium). 20 questions.

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Question 15 Marks
Show that any number of the form $4^{ n }, n \in N$ can never end with the digit 0 .
Answer
We have $4^n=(2 \times 2)^n=2^n \times 2^n$
The prime factors of $4^n$ are $2^n$
For any $n$, the number $2^n$ would with a zero digit if number is divisible by 5
So, the prime factorisation should contain at least one prime factor 5
But $4 n=2^n \times 2^n$ which contains the prime factors 2
So, by the uniqueness of the fundamental theorem, no other prime factorisation of $4^n$ exists.
Thus, any number of the form $4^n$ can never end with the digit 0
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Question 25 Marks
Use Euclid's algorithm to find HCF of 1190 and 1445. express the HCF in the form 1190m + 1445n.
Answer
Using Euclid's algorithm
HCF of (1190 and 1445).
1445 = 1190 × 1 + 255
1190 = 255 × 4 + 170
255 = 170 × 1 + 85
170 = 85 × 2 + 0
HCF = 85
Now,
85 = 255 - 170
= (1445 - 1190 × 1) - (1190 - 225 × 4)
= (1445 - 1190 × 1) - (1150 - (1445 - 1190 × 1) × 4)
= 1445 - 190 × 1 - 1190 + 1445 × 4 - 1190 × 4
= 1445 × 5 - 1190 × 6
= 85
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Question 35 Marks
The traffic lights at three different road crossing change after every 48 secons, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m. then at what time will they again change simultaneously?
Answer
Let us find the LCM of 48, 72 and 108 through prime factorisation:
$\begin{array}{c|c} 2 & 48 \\ \hline 2 & 24\\ \hline2&12\\ \hline2&6\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 72 \\ \hline 2 & 36\\ \hline2&18\\ \hline3&9\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 108 \\ \hline 2 & 54\\ \hline3&27\\ \hline3&9\\ \hline&3 \end{array}$
$48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$
LCM of $48,72,108$ is $2^4 \times 3^3$
= 16 × 27sec
= 432sec
=7min 12sec
Three bells toll together after 7min 12sec.
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Question 45 Marks
Find the leash number of square tiles required to pave the ceiling of a room 15m 17cm long and 9m 2cm broad.
Answer
Length of room = 15m 17cm= 1500cm + 17cm = 1517cm
Breadth of room = 9m 2cm
= 900cm + 2cm = 902cm
Now, H.C.F. of 1517 and 902 is 41
Thus, area of each tile = (41 × 41) = 1681 sq. cm
Hence, required number of tiles $=\frac{1517\times902}{1681}=814$
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Question 55 Marks
Three measuring rods are 64cm, 80cm and 96cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.
Answer
Let us find the LCM of 64, 80 and 96 through prime factorization:
$\begin{array}{c|c} 2 & 64 \\ \hline 2 & 32\\ \hline2&16\\ \hline2&8\\ \hline2&4\\\hline&2 \end{array}$ $\begin{array}{c|c} 2 & 80 \\ \hline 2 & 40\\ \hline2&20\\ \hline2&10\\ \hline&5 \end{array}$ $\begin{array}{c|c} 2 & 96 \\ \hline 2 & 48\\ \hline2&24\\ \hline2&12\\ \hline2&6\\\hline&3 \end{array}$
$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$
$80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5$
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
L.C.M. of 64, 80 and 96
$= 2^6 \times 5 \times 3 = 64 \times 15$
= 960cm = 9.6m
Therefore, the least length of the cloth that can be measured an exact number of times by the rods of 64cm, 80cm and 96cm = 9.6m
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Question 65 Marks
In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.
Answer
To find the minimum number of rooms required first find the maximum number of participants that can be accomodated in each room such that the number of participants in each room is same.
This can be determined by finding the HCF of 60, 84 and 108.
$60 = 2^2 \times 3 \times 5$
$84 = 2^2 \times 3 \times 7$
$108 = 2^2 \times 3^2$
$HCF = 2^2 \times 3$
$= 12$
So, the minimum number of rooms required
$=\frac{\text{Total number of participants}}{12}$
$=\frac{60+84+108}{12}$
$=21$
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Question 75 Marks
Find the largest number which divides 546 and 764, leaving remainders 6 and 8 respectively.
Answer
546 and 764 are divided by the largest number leaving remainers 6 and 8 respectively.
$546-6=540$
$764-8=756$
So, 540 and 756 are exactly divisible by the required number.
Thus, the required number is the HCF of 540 and 756
$540=2^2 \times 3^3 \times 5$
$756=2^2 \times 3^3 \times 7$
$\operatorname{HCF}(540,756)=2^2 \times 3^3=108$
which is the required number.
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Question 85 Marks
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together?
Answer
2 = 2 × 1
4 = 2 × 2
6 = 2 × 3
8 = 2 × 2 × 2
10 = 2 × 5
12 = 2 × 2 × 3
L.C.M. of 2, 4, 6, 8, 10, 12 minutes
= 2 × 2 × 2 × 3 × 5 = 120 minutes
= 2 hours
After every 2 hours they toll together,
Required number of times $=\Big(\frac{30}{2}+1\Big)$
$$= 16 times
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Question 95 Marks
Find the HCF and LCM of $\frac{8}{9},\frac{10}{27}$ and $\frac{16}{81}.$
Answer
$\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{HCF of the numerators}}{\text{LCM of the denominators}}$
$=\frac{\text{HCF}(8,10,16)}{\text{LCM}(9,27,81)}$
and
$\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{LCM of the numerators}}{\text{HCF of the denominators}}$
$=\frac{\text{LCM}(8,10,16)}{\text{HCF}(9,27,81)}$
Consider,
$8 = 2^3$
$10 = 2 \times 5$
$16 = 2^4$
So, the $\operatorname{HCF}(8,10,16)=2$ and $\operatorname{LCM}(8,10,16)=2^4 \times 5=80$
$9 = 3^2$
$27 = 3^3$
$81 = 3^4$
So, the $\operatorname{HCF}(9,27,81)=3^2=9$ and $\operatorname{LCM}(9,27,81)=3^4=81$
$\Rightarrow\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{2}{81}$ and $\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{80}{9}$
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Question 105 Marks
An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?
Answer
Interval of beeping together = LCM (60 seconds, 62 seconds)
The prime factorization of 60 and 62:
60 = 30 × 2, 62 = 31 × 2
$\therefore$ L.C.M of 60 and 62 is 30 31 × 2 = 1860 sec = 31min
$\therefore$ electronic device will beep after every 31 minutes
After 10 a.m., it will beep at 10hrs 31 minutes
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Question 115 Marks
Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.
Answer
Total pens = 1001
Total pencils = 910
We need to find maximum no.of students among whom 1001 pens and 910 pencils can be distributed in such a way that each students get same no.of pens and pencils.
Then we need to find HCF of 1001 and 910
Prime factorization of,
1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13
HCF=product of commom prime factor of least power
HCF = 7 × 13 = 91
Here HCF of 1001 and 910 is 91.
Hence among 91 students 1001 pens and 910 pencils can be distributed such that each student get same no. of pens and pencils.
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Question 125 Marks
Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.
Answer
Let 'a' be a given positive od integer.
On dividing 'a' by 4, let q be the quotient and r be the remainder.
Then, by euclid's algorithm, we have
a = 4q + r, where 0 ≤ r < 4
⇒ a = 4q + r, where r = 0, 1, 2, 3
⇒ a = 4q or a = 4q + 1 or a = 4q + 2 or a = 4q + 3
But, a = 4q and a = 4q + 2 = 2(2q + 1) are clearly even.
Thus, when 'a' is odd, it is of the form:
a = (4q + 1) or (4q + 3) for some integer q.
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Question 135 Marks
Show that $\big(4+3\sqrt2\big)$ is irrational.
Answer
If possible, let $\big(4+3\sqrt2\big)$ be rational.
Then 4 and $3\sqrt2$ are rational.
$\Rightarrow4+3\sqrt2-4$ is rational $\dots(\because$ diference of two rationals is rational$)$
$\Rightarrow3\sqrt2$ is rational
$\Rightarrow\sqrt2$ is rational $\dots(\because$ 3 is rational$)$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\big(4+3\sqrt2\big)$ is rational.
Hence, $\big(4+3\sqrt2\big)$ is irrational.
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Question 145 Marks
Prove that $\frac{2}{\sqrt7}$ is an irrational number.
HINT: $\frac{2}{\sqrt7}=\Big(\frac{2}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}\Big)=\frac{2}{7}.\sqrt7$
Answer
$\frac{2}{\sqrt{7}}=\left(\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\right)=\frac{2}{7} \cdot \sqrt{7}$ Let $\frac{2}{7} \sqrt{7}$ is a rational number.
$\therefore \frac{2}{7} \sqrt{7}=\frac{ p }{ q }$, where p and q are some integers and
$\operatorname{HCF}(p, q)=1 \ldots(1) \Rightarrow 2 \sqrt{7} q=7 p$
$\Rightarrow(2 \sqrt{7} q)^2=(7 p)^2$
$\Rightarrow 7\left(4 q^2\right)=49 p^2$
$\Rightarrow 4 q^2=7 p^2 \Rightarrow q^2$ is divisible by 7
$\Rightarrow$ $q$ is divisible by $7 \ldots$ (2) Let $q=7 m$, where $m$ is some integer. $\therefore 2 \sqrt{7} q=2 p$
$\Rightarrow[2 \sqrt{7} q (7 m)]^2=7 p ^2$
$\Rightarrow 343\left(4 m^2\right)=49 p ^2$
$\Rightarrow 7\left(4 m^2\right)= p ^2 \Rightarrow p ^2$ is divisible by 7
$\Rightarrow p$ is divisible by 7
From (2) and (3), 7 is a common factor of both $p$ and $q$, which contradicts (1).
Hence, our assumption is wrong. Thus, $\frac{2}{\sqrt{7}}$ is irrational.
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Question 155 Marks
Prove that $\big(4-5\sqrt2\big)$ is an irrational number.
Answer
Let $\text{x}=4-5\sqrt2$ be a rational number.
$\text{x}=4-5\sqrt2$
$\Rightarrow\text{x}^2=\big(4-5\sqrt2\big)^2$
$\Rightarrow\text{x}^2=(4)^2+\big(2\sqrt3\big)^2-2(4)\big(5\sqrt2\big)$
$\Rightarrow\text{x}^2=16+50-40\sqrt2$
$\Rightarrow\text{x}^2-66=-40\sqrt2$
$\Rightarrow\frac{66-\text{x}^2}{40}=\sqrt2$
Since x is a rational number, $x ^2$ is also a rational number. $\Rightarrow 66- x ^2$ is a rational number
$\Rightarrow\frac{66-\text{x}^2}{40}$ is a rational number
$\Rightarrow\sqrt2$ is a rational number
But $\sqrt2$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(4-5\sqrt2\big)$ is an irrational number.
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Question 165 Marks
Prove that $\big(5-2\sqrt3\big)$ is an irrational number.
Answer
Let $\text{x}=5-2\sqrt3$ be a rational number.
$\text{x}=5-2\sqrt3$
$\Rightarrow\text{x}^2=\big(5-2\sqrt3\big)^2$
$\Rightarrow\text{x}^2=(5)^2+\big(2\sqrt3\big)^2-2(5)\big(2\sqrt3\big)$
$\Rightarrow\text{x}^2=25+12-20\sqrt3$
$\Rightarrow\text{x}^2-37=-20\sqrt3$
$\Rightarrow\frac{37-\text{x}^2}{20}=\sqrt3$
Since x is a rational number, $x ^2$ is also a rational number.
$\Rightarrow 37-x^2$ is a rational number
$\Rightarrow \frac{37-x^2}{20}$ is a rational number
$\Rightarrow \sqrt{3}$ is a rational number
But $\sqrt{3}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $(5-2 \sqrt{3})$ is an irrational number.
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Question 175 Marks
Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.
Answer
on dividing 'n' by 3, let 'q' be the quotient and 'r' be the remainder.
Then, n = 3q + r, where 0 ≤ r < 3
⇒ n = 3q + r, where r = 0, 1 or 2
⇒ n = 3q or n = 3q +1 or n = 3q + 2
Case 1: If n = 3q, then 'n' is clearly divisible by 3
Case 2: If n = 3q +1, then (n + 2) = (3q + 3) = 3(q + 1), which is clearly divisible by 3
In this case, (n + 2) is divisible by 3
Case 3: If n = 3q +2, then (n + 4) = (3q + 6) = 3(q + 2), which is clearly divisible by 3
In this case, (n + 4) is divisible by 3
Hence, one and only one out of n, (n + 2) and (n + 4) is divisible by 3
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Question 185 Marks
Prove that $\sqrt3$ is an irrational number.
Answer
If possible, let $\sqrt{3}$ be rational and let its simplest form be $\frac{a}{b}$
The, $a$ and $b$ are integers having no common factor other than 1 , and $b \neq 0$
Now, $\sqrt{3}=\frac{ a ^2}{b^2}$
$\Rightarrow 3=\frac{ a ^2}{b^2} \ldots$ (On squaring both sides)
$\Rightarrow 3 b^2= a ^2$
$\Rightarrow 3$ divides $a ^2 \ldots\left[\because 3\right.$ divies $\left.3 b^2\right]$
$\Rightarrow 3$ divides a $\ldots\left[\because 3\right.$ is prime and 3 divides $a ^2 \Rightarrow 3$ divides a $]$
Let $a=3 c$ for some integers $c$.
Putting $a=3 c$ in (i), we get
$3 b^2=9 c^2$
$\Rightarrow b^2=3 c^2$
$\Rightarrow 3$ divides $b ^2 \ldots\left[\because 3\right.$ divies $\left.3 c ^2\right]$
$\Rightarrow 3$ divides $b \ldots\left[\because 3\right.$ is prime and 3 divides $a ^2 \Rightarrow 3$ divides a $]$
Thus, 3 is a common factor of $a$ and $b$
But, this contradicts the fact that $a$ and $b$ have no common factor other than 1
The contradiction arises by assuming that $\sqrt{3}$ is rational.
Hence, $\sqrt{3}$ is irrational.
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Question 195 Marks
Prove that $5\sqrt2$ is an irrational number.
Answer
Let $5\sqrt2$ is a rational number.
$\therefore5\sqrt2=\frac{\text{p}}{\text{q}},$ where p and q are some integers and HCF (p, q) = 1 ...(1)
$\Rightarrow5\sqrt2\text{q}=\text{p}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{p}^2$
$\Rightarrow2(25\text{q}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by 2
⇒ p is divisible by 2 ...(2)
Let p = 2m, where m is some integer.
$\Rightarrow5\sqrt2\text{q}=\text{2m}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{2m}^2$
$\Rightarrow2(25\text{q}^2)=\text{4m}^2$
$\Rightarrow\text{25q}^2=\text{2m}^2$
$\Rightarrow q^2$ is divisible by 2
⇒ q is divisible by 2 ...(3)
From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, $5\sqrt2$ is irrational.
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Question 205 Marks
Prove that $\big(2\sqrt3-1\big)$ is an irrational number.
Answer
Let $\text{x}=2\sqrt3-1$ be a rational number.
$\text{x}=2\sqrt3-1$
$\Rightarrow\text{x}^2=\big(2\sqrt3-1\big)^2$
$\Rightarrow\text{x}^2=\big(2\sqrt3\big)^2+1^2-2\big(2\sqrt3\big)(1)$
$\Rightarrow\text{x}^2=12+1-4\sqrt3$
$\Rightarrow\text{x}^2-13=-4\sqrt3$
$\Rightarrow\frac{13-\text{x}^2}{4}=\sqrt3$
Since x is a rational number, $x^2$ is also a rational number.
$\Rightarrow 13 − x^2$is a rational number
$\Rightarrow\frac{13-\text{x}^2}{4}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(2\sqrt3-1\big)$ is an irrational number.
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