BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

the candies among 7 children

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

the candies among 7 children

by sanju09 » Tue Sep 14, 2010 4:08 am
Marge has n candies, where n is an integer such that 20 < n< 50. If Marge divides the candies equally among 5 children, she will have 2 candies remaining. If she divides the candies among 6 children, she will have 1 candy remaining. How many candies will remain if she divides the candies among 7 children?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4



[spoiler]Source: https://readyforgmat.com[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Tue Sep 14, 2010 5:23 am
When Marge divides n candies among 5 children let each child gets x candies.
Then, n = 5x + 2
When Marge divides n candies among 6 children let each child gets y candies.
Then, n = 6y + 1
So, we have 5x + 2 = 6y + 1 or 6y - 5x = 1. Now choose the values of x and y so that their difference is 1, and 20 < n< 50.
Possible integer values are:
x = 7, y = 6 implies 6(6) - 5(7) = 1
n = 5(7) + 2 = 6(6) + 1 = 37

If Marge divides 37 candies among 7 children, then 2 candies will be left.

The correct answer is [spoiler](C)[/spoiler].
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 261
Joined: Wed Mar 31, 2010 8:37 pm
Location: Varanasi
Thanked: 11 times
Followed by:3 members

by ankur.agrawal » Tue Sep 14, 2010 5:28 am
Alternate way particularly for this question:

equally among 5 children = we can have 22,27,32,37,42,47 as possible values.

equally among 6 children = we can have 25,31,37,42,47 as possible values.

So only 37 is possible . So n =37.

Dividing by 7 we get 2 as remainder.

C
Top
Post new topic Post Reply
• Page 1 of 1