ncert solutions class 10 maths chapter 1 · solution : we know, hcf x lcm=product of two numbers...

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QUESTION-1 Use Euclid's di vision algorithm to find the HCF of: i)135 and 225 ii)196 and 38220 iii)867 and 225 Solution: i) We start with the larger number i.e 225 By Euclid's division algorithm.we have 225=1X 135+90 135=1 x90+45 90=2x45+0 Hence, HCF(225, 135)=HCF(135,90)=HCF(90,45)=45 Therefore.the HCF of 135 and 225 is 45 ii) We start with the larger number i.e 38220 By Euclid's division algorithm.we have 38220=196 X 195+0 196=196X 1+0 Hence , HCF( 196,38220)=196 Therefore.the HCF of 196 and 38220 is 196 iii) We start with the larger number i.e 867 By Euclid's division algorithm.we have 867=225X 3+102 225=102X 2+51 102=51 x2+0 Hence, HCF(867,225)=HCF(225,102)=HCF( 102,51)=51 Therefore.the HCF of 867 and 225 is 51 Question 2: Show that any positive odd i nteger is of the form 6q+1,6q+3 & 6q+5 where q is some integer. NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 - REAL NUMBERS

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Page 1: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the

QUESTION-1

Use Euclid's division algorithm to find the HCF of:

i)135 and 225

ii)196 and 38220

iii)867 and 225

Solution:

i) We start with the larger number i.e 225

By Euclid's division algorithm.we have

225= 1 X 135+90

135=1 x90+45

90=2x45+0

Hence, HCF(225, 135)=HCF(135,90)=HCF(90,45)=45

Therefore.the HCF of 135 and 225 is 45

ii) We start with the larger number i.e 38220

By Euclid's division algorithm.we have

38220= 196 X 195+0

196=196X 1 +0

Hence , HCF(196,38220)=196

Therefore.the HCF of 196 and 38220 is 196

iii) We start with the larger number i.e 867

By Euclid's division algorithm.we have

867=225 X 3+102

225=102 X 2+51

102=51 x2+0

Hence, HCF(867,225)=HCF(225, 102)=HCF(102,51 )=51

Therefore.the HCF of 867 and 225 is 51

Question 2:

Show that any positive odd integer is of the form 6q+1,6q+3 & 6q+5

where q is some integer.

NCERT SOLUTIONS CLASS 10 MATHS

CHAPTER 1 - REAL NUMBERS

Page 2: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 3: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 4: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 5: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 6: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 7: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 8: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 9: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 10: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 11: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the
Page 12: NCERT SOLUTIONS CLASS 10 MATHS CHAPTER 1 · Solution : We know, HCF x LCM=Product of two numbers i.e 9xLCM=306x657 306x657 LCM: Question 10: 22338 Check whether 611 can end with the