The HCF of two number is 23 and their LCM is 1449 if one of the number is 161, find the other

Question 3 Real Numbers Exercise 1B

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Answer:

Given: HCF = 23, LCM = 1449 and one number = 161

Let the other number be m.

We know, a product of two numbers = HCF x LCM

161 x m = 23 × 1449

Simplify the above expression and find the value of m:

m = (23 x 1449)/161 = 207

Therefore, the second number is 207

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The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Text Solution

`205``208``207``209`

Answer : C

Solution : For two numbers `a` and `b,` we know that <br> ` ( axx b)` = ( HCF (a ,b) ` xx ` ( LCM (a,b)) . <br> Here a '= 161 , HCF = w23 and LCM = 1449. <br> And we have to find b. <br> ` ( 161 xxb) = ( 23xx 1449) Rightarrow b = ((23 xx 1449))/161 = 207` <br> Hence , the other number is `207. `

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Let the two numbers be a and b.Let the value of a be 161.Given: HCF = 23 and LCM = 1449We know, a × b = HCF × LCM                ⇒ 161 × b = 23 × 1449                ⇒ ∴ b =`( 23 ×1449)/161 = 33327/161 = 207`

Hence, the other number b is 207. 

Concept: Euclid’s Division Lemma

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Page 2

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find
the other.

HCF of two numbers = 145LCM of two numbers = 2175Let one of the two numbers be 725 and other be x.Using the formula, product of two numbers = HCF × LCMwe conclude that725 × x = 145 × 2175x = `(145 ×2175)/ 725`= 435

Hence, the other number is 435.

Concept: Euclid’s Division Lemma

  Is there an error in this question or solution?