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In this article we have given NCERT Solutions for Class 10 Math Chapter 1 Exercise 1.1 with pdf. These NCERT solutions are solved by subjects experts from latest edition book.

## NCERT Textbook Class 10 Math Exercise 1.1

Question 1. Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225, (ii) 196 and 38220, (iii) 867 and 255.**Solution:**

(i) We have 135 and 225

Since, 225>135, applying Euclid’s Division Lemma to 225 and 135,

We get, 225=135X1+90

Since, remainder 90 ≠0, so applying Euclid’s division lemma to 135 and 90,

We get 135=90X1+45

Since, remainder 45≠0, so applying Euclid’s division lemma to 90 and 45 and, we get

90=45X2+0

Since remainder =0

∴HCF (135, 225) is 45.

(ii) We have 196 and 38220

Since 38220> 196, applying Euclid’s division

Lemma to 38220 and 196, we get

38220=196X195+0

Since remainder=0

∴HCF (196, 38220) is 196.

(iii) We have 867 and 255

Since 867>255, applying Euclid’s division lemma to 867 and 102,

We get 867=255X3+102

Since remainder 102=0, applying Euclid’s division lemma of 102 and 51,

we get 102=51X2+0

∴ HCF (867, 255) is 51.

Question 2. Show that any positive odd integer is of the from6q+1, or 6q+3, 6q+5, where q is form integer.

Solutions: let a be any positive odd integer and b=6, then by Euclid’s division lemma

a=6q+r, r=0,1,2,3,4,5

When ,

r=0, | a=6q ….(i) |

r=1, | a=6q+1 |

r=2, | a=6q+2 ….(ii) |

r=3, | a=6q+3 |

r=4, | a=6q+4 …..(iii) |

r=5, | a=6q+5 |

Values of a in (i), (ii) and (iii) is not possible as a is odd +ve integer.

Therefore, any +ve odd integer is of the form

(6q+1), (6q+3) or (6q+50).

Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution: Let a = 616 and b = 32

Applying Euclid’s division lemma to 616 and 32, we get

616=32×19+ 8

Now, applying Euclid’s division lemma of 32 and 8, we get

32=8×4+0

Since, remainder=0

HCF (612, 32) =8

Therefore, maximum number of columns=8.

## Class 10 Math Chapter 1 Exercise 1.1 NCERT Solution

Question 4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q+1 or 3g +2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]

Solution: Let a be ay positive integer and b=3, then by Euclid’s division lemma a=3q+r and r=0, 1, 2

When,

r=0, | a=3q ….(i) |

r=1, | a=3q+1 ….(ii) |

r=2, | a=3q+2 ….(iii) |

Squaring eqn. (i) on both side, we get

a_{2} =9q^{2}

a_{2}=3X3q^{2}

=3m, where m=3q^{2}

Squaring eqn. (iii) on Both sides, we get

a^{2} = (3q+1)^{2}

= 9q^{2} +1+6q

= 3(3q^{2} +2q)+1

= 3m+1, where m=3q^{2}

Squaring eqn. (iii) on both side

a^{2} =(3q+2)^{2}

=9q^{2} +4+12q

=9q^{2}+3+1+12q

=3(3q^{2}+1+4q) +1

=3m+1,

where m=3q^{2}+1+4q.

Question 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m + 8.

Solution: let a be any +ve integer and be=3, then by Euclid’s division Lemma a=3q+r and r=0,1,2

When,

r=0, | a=3q ….(i) |

r=1, | a=3q+1 ….(ii) |

r=2, | a=3q+2 ….(iii) |

Taking cube on both side in eqn. (i) we get

a^{3} =(3q)^{3}

=27q^{3}

=9X3q^{3}=9m, where m=3q^{3}.

Taking cube on both sides in eqn. (ii), we get

a^{3}=(3q +1)^{3}

=27q^{3 }= 1+27q^{2}+9q

=9(3q^{3}+3q^{2} +q) +1=9m+1,

Where m=3q^{3}+3q^{3}+q

Taking cube on both sides in eqn. (iii), we get

a^{3}=(3q+2)^{3}

=27q^{3}+8+54q^{2}+36q

=27q^{3}+54q^{2} +36q+8

=9(3q^{3}+6q^{2}+4q)+8

=9m+8,

Where m=3q^{3}+6q^{2}+4q.

## NCERT Class 10 Math Exercises 1.1 Pdf Download

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