Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
09 Aug 2021, 04:15
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
87% (01:51) correct
13%
(02:25)
wrong
based on 139
sessions
History
Date
Time
Result
Not Attempted Yet
The highest common factor and the least common multiple of two positive integers are 7 and 140, respectively. If the numbers are between 20 and 45, then what is the sum of the numbers ?
(A) 77
(B) 70
(C) 63
(D) 56
(E) 54
(A) 77
(B) 70
(C) 63
(D) 56
(E) 54
09 Aug 2021, 04:55
Kudos
Bookmarks
Solution:
Since the HCF of \(2\) numbers \(= 7\).
We can assume the \(2\) numbers to be \(7x\) and \(7y\). Where \(x\) and \(y\) are co-primes.
Now we know the product of 2 number = product of their LCM and HCF. Thus we can write:
\(⇒ 7x\times 7y=140\times 7\)
\(⇒ 49xy=140\times 7\)
\(⇒ xy=20\)
So the value of \(x\) and \(y\) can be \((1,20)\) or \((4,5)\)
Since \(x\) and \(y\) has to be between \(25\) and \(45\), \(x=4\) and \(y=5\)
And the 2 number \(= 7\times 4=28\) and \(7\times 5=35\)
Thus their sum \(= 28+35=63\)
Hence the right answer is Option C.
Since the HCF of \(2\) numbers \(= 7\).
We can assume the \(2\) numbers to be \(7x\) and \(7y\). Where \(x\) and \(y\) are co-primes.
Now we know the product of 2 number = product of their LCM and HCF. Thus we can write:
\(⇒ 7x\times 7y=140\times 7\)
\(⇒ 49xy=140\times 7\)
\(⇒ xy=20\)
So the value of \(x\) and \(y\) can be \((1,20)\) or \((4,5)\)
Since \(x\) and \(y\) has to be between \(25\) and \(45\), \(x=4\) and \(y=5\)
And the 2 number \(= 7\times 4=28\) and \(7\times 5=35\)
Thus their sum \(= 28+35=63\)
Hence the right answer is Option C.
ash124
Joined: 21 Jan 2018
Last visit: 06 Apr 2025
Posts: 107
Own Kudos:
Given Kudos: 64
Status:Improving everyday
Location: India
Schools: ISB '23 Judge '23 Said '23 Sloan '24 Fuqua '24 LBS '24 CEIBS '24 INSEAD Jan 22 intake Yale '24
GMAT 1: 660 Q49 V31
GPA: 3.23
WE:Engineering (Aerospace and Defense)
Products:
09 Aug 2021, 05:28
Kudos
Bookmarks
Understand the Question:
1) Two numbers' (x & y) HCF & LCM are 7 and 140 respectively.
2) 20 < x, y < 45
Find x+ y?
We know from the above statements:
HCF*LCM = x*y
7*140 = x*y
7 *7*2^2*5 = x*y
We also know HCF is 7 ,means Number are multiple of 7
In Range 20 - 45 : numbers can be 21 , 28, 35, 42
21 & 42 can't be the pair because it will in-validate HCF(7) as HCF will be 21.
7 *7*2^2*5 = x*y
we need multiple of 5 & 7 to satisfy above equation:
and only 35 fulfills the criteria.
We identify the first number as 35.
Next we need multiple of 4 & 7
And only 28 fulfills the criteria.
We identify the first number as 28.
Add them and the answer will be 63
1) Two numbers' (x & y) HCF & LCM are 7 and 140 respectively.
2) 20 < x, y < 45
Find x+ y?
We know from the above statements:
HCF*LCM = x*y
7*140 = x*y
7 *7*2^2*5 = x*y
We also know HCF is 7 ,means Number are multiple of 7
In Range 20 - 45 : numbers can be 21 , 28, 35, 42
21 & 42 can't be the pair because it will in-validate HCF(7) as HCF will be 21.
7 *7*2^2*5 = x*y
we need multiple of 5 & 7 to satisfy above equation:
and only 35 fulfills the criteria.
We identify the first number as 35.
Next we need multiple of 4 & 7
And only 28 fulfills the criteria.
We identify the first number as 28.
Add them and the answer will be 63
