7.25

https://drive.google.com/file/d/1zvtbLCs_iwYPBpmet3LX6NkMrJ4kpVPH/view?usp=drivesdkhttps://drive.google.com/file/d/1zvtbLCs_iwYPBpmet3LX6NkMrJ4kpVPH/view?usp=drivesdkhttps://drive.google.com/file/d/1zvtbLCs_iwYPBpmet3LX6NkMrJ4kpVPH/view?usp=drivesdk

7.24

https://drive.google.com/file/d/1yrZt82Bi5zRIEwbvqrkxw3dKWjK5NC2U/view?usp=drivesdkhttps://drive.google.com/file/d/1yrZt82Bi5zRIEwbvqrkxw3dKWjK5NC2U/view?usp=drivesdk

7.23

https://drive.google.com/file/d/1ydnffCR1C1pZd2jIxEwnURKWajw6uE_g/view?usp=drivesdkhttps://drive.google.com/file/d/1ydnffCR1C1pZd2jIxEwnURKWajw6uE_g/view?usp=drivesdk

Exercise 1.2(cbse)

 Question 1:

Express each number as product of its prime factors: 

Answer 1:

(i) 140=2x2x5x7=2²x5x7

(ii) 156=2x2x3x13=2²x3x13

(iii) 3825=3x3x5x5x17-3²x5²x17

(iv)5005= 5x7x11x13

(v)7429 =17x19x23

Question 2:

Find the LCM and HCF of the following pairs of integers and verify that

LCM X HCF = product of the two numbers.

Answer:2

(1) 26 and 91

26=2x13

91=7x13

HCF= 13 (highest common factor)

LCM=2x7x13=182 (every common number)

Product of the two numbers = 26x91=2366

HCFxLCM =13x182=2366

Hence, product of two numbers =HCF x LCM

Verified.

(ii)510 and 92

510= 2x3x5x17

92=2x2x23

HCF=2

LCM 2x2x3x5x17x23-23460

Product of the two numbers = 510x92=46920

HCFx LCM=2x23460 =46920

Hence, product of two numbers = HCF x LCM

Verified

(iii) 336 and 54

336= 2x2x2x2x3x7 336=2¹x3x7

54=2x3x3x3

54=2×3³

HCF=2×3=6

LCM=2¹x3¹x7=3024

Product of the two numbers 336x54=18144

HCF×LCM=6×3024= 18144

HENCE, HCF×LCM= Product of two numbers verified

Exercise 1.1(cbse 10th)

 Question 1:

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

(i) 135 and 225 (ii) 196 and 38220

Answer 1:

(i) 135 and 225

Since 225 135, we apply the division lemma to 225 and 135 to obtain 

225= 135 x 1 +90

Since remainder 90not equal to 0 , we apply the division lemma to 135 and 90 to

obtain

135 =90 x 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the

division lemma to obtain

90 =2 x 45 +0

Since the remainder is zero, the process stops. Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to

obtain

38220= 196 x 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

 867= 255 x 3 + 102

Since remainder 102 is equal to 0, we apply the division lemma to 255 and 102

to obtain

255=102×2+51

We consider the new divisor and new remainder 51 apply the division lemma to obtain 

102=51×2+0

Since the remainder is 0 the process stops

Since the divisor at this stage is 51

Therefore HCF of 867 and 255 is 51.

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