NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
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Exercise 1.1
Question 1:
Express each number as a product of its prime factors:
निम्नलिखित संख्याओं को अभाज्य गुणनखंड के रूप में व्यक्त कीजिये :
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Answer
(i) 140 = 2 × 2 × 5 × 7 = 22× 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22× 13 × 3
(iii) 3825 = 3 × 3 × 5 × 5 × 17= 32× 52× 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 2:
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
पूर्णांकों के निम्नलिखित युग्मों के LCM and HCF ज्ञात कीजिए तथा इसकी जाँच कीजिए कि दो संख्याओं का गुणनफल = LCM × HCF है|
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Answer
(i) 26 = 2 \(\times\) 13
and 91 = 7 \(\times\) 13
HCF = 13
LCM \(= \frac{Product \,of\, two \,numbers}{HCF}\)
\(\quad \quad = \frac{26 \times 91}{13} = 2 \times 91 = 182\)
Verification
product of 26 and 91 = 26 × 91 = 2366
And product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91.
(ii) 510 = 2\(\times\) 3\(\times\)5\(\times\)17
and 92 = 2\(\times\) 2\(\times\)23
HCF = 2
LCM \(= \frac{Product \,of\, two \,numbers}{HCF}\)
\(\quad \quad = \frac{510 \times 92}{2} = 510 \times 46 =23460\)
Verification
product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92.
(iii) 336 = 2 \(\times\) 2\(\times\) 2\(\times\) 2\(\times\) 3\(\times\) 7
and 54 = 2 \(\times\) 3\(\times\) 3\(\times\) 3
HCF = 2\(\times\) 3 = 6
LCM \(= \frac{Product \,of\, two \,numbers}{HCF}\)
\(\quad \quad = \frac{336 \times 541}{6} = 2 \times 91 = 3024\)
Verification
product of 336 and 54 = 336 × 54 = 18,144
And product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 3:
Find the LCM and HCF of the following integers by applying the prime factorisation method.
अभाज्य गुणनखंड विधि द्वारा निम्नलिखित पूर्णांकों के LCM और HCF ज्ञात कीजिए |
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Answer
(i) 12, 15 and 21
\(\quad\) 12=2×2×3
\(\quad\) 15=5×3
\(\quad\) 21=7×3
Therefore,
HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
\(\quad\) 17=17×1
\(\quad\) 23=23×1
\(\quad\) 29=29×1
Therefore,
HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
\(\quad\) 8=2×2×2×1
\(\quad\) 9=3×3×1
\(\quad\) 25=5×5×1
Therefore,
HCF(8,9,25)=1
LCM(8,9,25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).
HCF (306, 657) = 9, दिया है | LCM (306, 657) ज्ञात कीजिए |
Answer
HCF×LCM=Product of the two given numbers
\(\quad\) 9 × LCM = 306 × 657
LCM = (306×657)/9 = 22338
Hence, LCM(306,657) = 22338
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 5:
Check whether 6n can end with the digit 0 for any natural number n.
जाँच कीजिए कि क्या किसी प्राकृत संख्या n के लिए संख्या 6n अंक 0 पर समाप्त हो सकती है |
Answer
If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.
So, value 6n should be divisible by 2 and 5 both.
6n is divisible by 2 but not divisible by 5 as the prime factors of 6 are 2 and 3.
So, it can not end with 0.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 6:
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
व्याख्या कीजिए 7 × 11 × 13 + 13 और 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 भाज्य संख्या क्यों है ?
Answer
7 × 11 × 13 + 13 = 13 (7 × 11 + 1) = 13(77 + 1 ) = 13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
And,
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 +1) = 5(1008 + 1) = 5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 7:
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
किसी खेल के मैदान के चारों ओर एक वृत्ताकार पथ है। इस मैदान का एक चक्कर लगाने में सोनिया को 18 मिनट लगते हैं, जबकि इसी मैदान का एक चक्कर लगाने में रवि को 12 मिनट लगते हैं। मान लीजिए वे दोनों एक ही स्थान और एक ही समय पर चलना प्रारंभ करके एक ही दिशा में चलते हैं। कितने समय बाद वे पुनः प्रांरभिक स्थान पर मिलेंगे?
Answer
LCM of 18 and 12
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Exercise 1.2
Question 1:
Prove that √5 is irrational.
सिद्ध कीजिए कि √5 एक अपरिमेय संख्या है |
Answer
Let us take √5 as rational number
let, \(\sqrt{5} = \frac{a}{b}\) where, a and b are co-prime.
\(\Rightarrow\) b√5 = a
Squaring on both sides, we get
5b2 = a2 …(i)
Therefore, 5 divides a2.
Also, 5 divides a.
Putting a = 5c in (i), we get
5b2 = (5c)2
⇒ 5b2 = 25c2
⇒ b2 = 5c2
Similarly, 5 divides b2.
Also, 5 divides b.
Therefore, a and b have at least 5 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
So our assumption is wrong that 2 is rational.
So, we conclude that 2 is irrational.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 2:
Prove that 3 + 2 √5 is irrational.
सिद्ध कीजिए कि 3 + 2√5 एक अपरिमेय संख्या है |
Answer
Let us take 3 + 2√5 as rational number
let, 3 + 2\(\sqrt{5} = \frac{a}{b}\) where, a and b are co-prime.
\(\Rightarrow 2\sqrt{5} = \frac{a}{b} – 3\)
\(\Rightarrow \sqrt{5} = \frac{1}{2}(\frac{a}{b} – 3)\)
since, a and b are integers,
\(\frac{1}{2}(\frac{a}{b} – 3)\) is a rational number.
Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 is irrational.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question 3:
Prove that the following are irrationals :
सिद्ध कीजिये कि निम्नलिखित संख्याएँ अपरिमेय हैं :
(i) \(\frac{1}{\sqrt{2}}\) (ii) 7√5 (iii) 6 + √2
Answer
Let us take \(\frac{1}{\sqrt{2}}\) as rational number
let, \(\frac{1}{\sqrt{2}} = \frac{a}{b}\) where, a and b are co-prime.
\(\Rightarrow\) b = a√2
Squaring on both sides, we get
b2 = 2a2 …(i)
Therefore, 2 divides b2.
Also, 2 divides b.
Putting b = 2c in (i), we get
(2c)2 = 2a2
⇒ 4c2 = 2a2
⇒ 2c2 = a2
Similarly, 2 divides a2.
Also, 2 divides a.
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
So our assumption is wrong that \(\frac{1}{\sqrt{2}}\) is rational.
So, we conclude that \(\frac{1}{\sqrt{2}}\) is irrational.
(ii) Let us take 7√5 as rational number
let, 7\(\sqrt{5} = \frac{a}{b}\) where, a and b are co-prime.
\(\Rightarrow \sqrt{5} = \frac{a}{7b}\)
since, a and b are integers,
\(\frac{a}{7b}\) is a rational number.
Therefore, √5 is also a rational number. But this contradicts the fact that 7√5 is irrational.
So, we conclude that 7√5 is irrational.
(iii) Let us take 6 + √2 as rational number
let, 6 + \(\sqrt{2} = \frac{a}{b}\) where, a and b are co-prime.
\(\Rightarrow \sqrt{2} = \frac{a}{b} – 6\)
since, a and b are integers,
\(\frac{a}{b} – 6\) is a rational number.
Therefore, √2 is also a rational number. But this contradicts the fact that 6 + √2 is irrational.
So, we conclude that 6 + √2 is irrational.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
