NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | नए पैटर्न पर आधारित एनसीईआरटी समाधान कक्षा 10वी के अध्याय 1 प्रश्नावली 1.2 के सभी प्रश्नों का हल अंग्रेजी में | Exercise 1.2 Class 10 Maths Solutions In English on New Pattern | Class 10th Math Chapter 1 Real Numbers Ex. 1.2 Solutions In Hindi & English | Ex 1.2 class 10 math solution with Video and pdf download in Hindi and English |Ex 1.2 class 10 | Exercise 1.2 Class 10 Maths Question 3 | Class 10 Maths Chapter 1 Exercise 1.2 Question 2 | Class 10 Maths Chapter 1 Exercise 1.2 Solutions | क्लास 10 के अध्याय 1.2 का समाधान का पीडीएफ और विडियो इंग्लिश और हिंदी में | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो हिंदी और इंग्लिश में | Exercise 1.2 Class 10 Maths Question 1,2,3,4,5,6 and 7 Solution |
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| NCERT New Pattern Solutions Class 10th Chapter 1 Exercise 1.2 |
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 1 |
Chapter 1- Exercise 1.2
1. Express the following numbers as a product of prime factors:-
i) 140
Solution: (i) 140 = 2×2×5×7 = 22×5×7
ii) 156
Solution:- 156 = 2×2×3×13 = 22×3×13
iii) 3825
Solution:- 3825 = 3×3×5×5×17 = 32×52×17
iv) 5005
Solution:- 5005 = 5×7×11×13
v) 7429
Solution:- 7429 = 17×19×23
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 2 |
2. Find the HCF and LCM of the following pairs of integers and check that the product of the two numbers = HCF × LCM.
i) 26 and 91
Solution:-
26 = 2×13
91 = 7×13
HCF =13
LCM = 2×7×13 =182
Product of the two numbers = 26×91 = 2366
HCF×LCM = 13×182 = 2366
Hence, product of two numbers = HCF×LCM
ii) 510 and 92
Solution:-
510 = 2×3×5×17
92 = 2×2×23
HCF = 2
LCM= 2×2×3×5×17×23 = 23460
Product of the two numbers = 510×92 = 46920
HCF×LCM = 2×23460
HCF×LCM = 46920
Hence, product of two numbers = HCF×LCM
iii) 336 and 54
Solution:-
336 = 2×2×2×2×3×7
336 = 2⁴×3×7
54 = 2×3×3×3
54 =2×3³
HCF = 2×3 = 6
LCM = 24×33×7 = 3024
Product of the two numbers = 336×54 = 18144
HCF×LCM = 6×3024 = 18144
Hence, product of two numbers = HCF×LCM
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 3 |
3. Find the HCF and LCM of the following integers by the prime factorization method.
Prime factors of any number is the representation of a number as a product of prime numbers it is composed of
for example:- Prime factors of 20 = 2 × 2 × 5, HCF = Highest common factor = The product of the factors that are common to the numbers LCF = Least Common Factor = Product of all the factors of numbers without duplicating the factor.
i) 12, 15 and 21
Solution:- let us write prime factors of the given numbers :-
12 = 2×2×3
= 2²×3
15 = 3×5
21 = 3×7
Only 3 is common in all the three numbers, therefore
HCF = 3
As 3 is common in all three numbers, it will be taken as 1 time in the product of calculating LCM
LCM = 2² ×3×5×7 = 420
ii) 17, 23 and 29
Solution:- Let us write the prime factors of the given numbers:-
17 = 1×17
23 = 1×23
29 = 1×29
As only 1 is common from all the three factors
HCF = 1
As nothing except 1 is common from all three numbers, simply multiplying them will give LCM of numbers.
LCM = 17×23×29 = 11339
iii) 8, 9 and 25
Let us write the prime factors of the given numbers:-
8=2×2×2
9 = 3×3
25 = 5×5
As nothing is common in the numbers
HCF = 1
As nothing is common in factors of numbers, numbers are simply multiplied to obtain LCM.
LCM = 2×2×2×3×3×5×5 = 1800
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 4 |
4. Given HCF (306, 657) = 9. Find the LCM of (306, 657).
Solution:- Given: HCF of (306, 657) = 9
We know that,
LCM × HCF = product of two numbers
LCM×HCF = 306×657
LCM = 306×65l / 𝐻𝐶𝐹
= 306×657 / 9
LCM = 22338
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 5 |
5. Check whether the number 6ⁿ can end with the digit 0 for some natural number n.
Solution:- We need to find can 6ⁿ end with zero
If any number has last digit 0,
Then, it should be divisible by 10
Factors of 10 = 2×5
So, Value 6ⁿ should be divisible by 2 and 5
Prime factorisation of 6ⁿ = (2×3)ⁿ
Hence, 6ⁿ is divisible by 2 but not by 5.
It can not end with 0.
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 6 |
6. Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.
Solution- Composite numbers are those numbers, which can be written in the form of the product of two or more integers, and at least one of them should not be 1
(i) 7 × 11 × 13 + 13
= (7 × 11 × 13) + (13 × 1)
Taking 13 as common, we get,
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)= 13 × 78
= 13 × 13 × 6
As the given no is a multiple of two or more integers, one of them being other than 1.
Hence, it is a composite number.
Therefore, it is a composite number.
(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= (7 × 6 × 5 × 4 × 3 × 2 × 1) + (5 × 1)
Taking 5 as common, we get,
= 5×(7×6×4×3×2×1+1)
= 5× (1008+1)
= 5×1009
As the given no is a multiple of two integers, one of them being other than 1.
Hence, it is a composite number.
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 पीडीएफ और विडियो इंग्लिश में | Exercise 1.2 Class 10 Maths Question 7 |
7. There is a circular path around a playground. Sonia takes 18 minutes to cover one round of the ground, while Ravi takes 12 minutes to cover one round of the same ground. Suppose they both start at the same place and at the same time and move in the same direction. After how much time will they meet again at the starting point.
Solution- Both of them start from the same point and start moving in same direction
Sonia takes 18 minutes and Ravi takes 12 minutes to complete the circle. After 12 minutes Ravi will be back to the starting point and Sonia must have covered
(12/18) = 2/3 of the rounds. After 12 more
minutes Ravi has completed 2 rounds and Sonia must have covered
(24/18) = 4/3 of the rounds. After 12 more minutes Ravi has completed 3 rounds and Sonia must have completed (36/18) = 2 rounds.
Hence after 36 minutes both will again meet at the starting point.
Alternate Method:-
They will meet again after LCM of both values at starting point.
18 = 2×3×3 And 12 = 2×2×3
LCM of 12 and 18 = 2×2×3×3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
Chapter 1 (Real Numbers) Exercise 1.1 Solution
नए पैटर्न पर आधारित एनसीईआरटी समाधान कक्षा 10वी के अध्याय 1 प्रश्नावली 1.2 के सभी प्रश्नों का हल अंग्रेजी में दिया गया है, यदि समाधान समझ न आए तो कृपया video की सहायता ले 🙏
NCERT New Pattern Class 10 Maths Chapter 1 Exercise 1.2 Solutions PDF In English and Hindi | कक्षा 10 गणित के लिए एनसीईआरटी समाधान हिंदी में अध्याय 1 प्रश्नावली 1.2 का पीडीएफ और विडियो इंग्लिश में | कक्षा 10 के प्रश्नावली 1.2 का समाधान अंग्रेजी में | Exercise 1.2 Class 10 Maths Solutions In English on NCERT New Pattern | Class 10th Math Chapter 1 Real Numbers Ex. 1.2 Solutions In Hindi & English | Ex 1.2 class 10 math solution with Video and pdf download in English |
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