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Bunuel
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The HCF of two numbers is 11 and their LCM is 7700. If one of these nu 22 Apr 2021, 04:15
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C
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5% (low)

Question Stats:

91% (01:40) correct 9% (01:54) wrong based on 131 sessions

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The HCF of two numbers is 11 and their LCM is 7700. If one of these numbers is 275, then the other one is:

(A) 279
(B) 283
(C) 308
(D) 318
(E) 338
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C
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ManyataM
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The HCF of two numbers is 11 and their LCM is 7700. If one of these nu 22 Apr 2021, 04:24
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Bunuel
The HCF of two numbers is 11 and their LCM is 7700. If one of these numbers is 275, then the other one is:

(A) 279
(B) 283
(C) 308
(D) 318
(E) 338
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Answer is option C
The product of 2 numbers = Lcm * Hcf of those two numbers

7700*11 = x*275
x = 308

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Rajaking
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The HCF of two numbers is 11 and their LCM is 7700. If one of these nu 22 Apr 2021, 05:01
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(C) 308

Product of two numbers = HCF X LCM

275 * X = 11 * 7700

X = 308

Cross-check - For 275 and 308, the HCF is 11 [as 11 is the only common factor] and LCM is 7700.
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ruis
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The HCF of two numbers is 11 and their LCM is 7700. If one of these nu 08 Jan 2024, 12:11
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My approach was:
1) Prime decompose the numbers:
7700 = 11*2^2*5^2*7
275 = 11*5^2

2) Find the numbers among the answers choices that have at least one 7 and at least two 2´s when prime decomposed
Option C is 7*2^2*11

Please correct me if necessary
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The HCF of two numbers is 11 and their LCM is 7700. If one of these nu 10 Jan 2025, 04:12
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10-15 sec solution.
Given, the product of LCM and HCF = product of numbers. Looking at LCM, we have 4 as a factor. Given 275 is not even, the other number must be a factor of 4. Using divisibility rules of 4 and looking at the last 2 digits, only C (08) works
Bunuel
The HCF of two numbers is 11 and their LCM is 7700. If one of these numbers is 275, then the other one is:

(A) 279
(B) 283
(C) 308
(D) 318
(E) 338
Show more
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