algebra equation rather dan backsolve

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divya23: Master | Next Rank: 500 Posts; Posts: 113; Joined: Mon Apr 26, 2010 8:58 amFollowed by: 14 members; Thanked: 2 times

by divya23 » Wed Jun 01, 2011 9:56 pm

A certain tree was first planned it was 4 feet tall then height incrsd by constt rate for next 6yrs at the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr,by how many inches did the hite of tree inc each yr?
1.3/10
2.2/5
3.1/2
4.2/3
5.6/5
[spoiler]OA = 2/3[/spoiler]
[spoiler]
plz tell d correct method instead of backsolving[/spoiler]

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Source: Beat The GMAT — Problem Solving | Wed Jun 01, 2011 9:56 pm

GMATToppers: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Tue May 24, 2011 4:53 amFollowed by: 1 member; Location: Bangalore; Thanked: 6 times

by GMATToppers » Wed Jun 01, 2011 10:24 pm

divya23 wrote:A certain tree was first planned it was 4 feet tall then height incrsd by constt rate for next 6yrs at the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr,by how many inches did the hite of tree inc each yr?
1.3/10
2.2/5
3.1/2
4.2/3
5.6/5
The algebraic solution is as follows
4+6R=the height at the end of 6 years since 4 was already there and we added R units for 6 years
Here R is the amount of growth each year
The translation of the line "the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr"
the height at the end of 6 years i have explained it above.
was means equal to '='
1/5 is 20%
taller than= add 100%
4 year height=4+4R
so
4+6R=(20%+100%)4+4R
4+6R=(120/100)4+4R
4+6R=1.2(4+4R)
R=2/3

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Wed Jun 01, 2011 10:49 pm

let constant growth rate=r
then after x years, length = 4+xt
also 4+6t = (1 + 1/5)*(4+4t)
=> 20+30t = 24+24t =>t=2/3
Thus each year growth = 2/3 inches.
IMO D

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Ozlemg: Master | Next Rank: 500 Posts; Posts: 407; Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members; Thanked: 25 times

by Ozlemg » Thu Jun 02, 2011 2:41 am

IMO D

x is the constant growth
At the end of 1st year, its length is : 4+x
At the end of 2nd year, its length is : 4+x+x
At the end of 3rd year, its length is : 4+x+x+x
At the end of 4th year, its length is : 4+x+x+x+x
At the end of 5th year, its length is : 4+x+x+x+x+x
At the end of 6th year, its length is : 4+x+x+x+x+x+x
In the Question it is mentioned that "at the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr" --> if we convert the wording to numbers, it is :

(4+4x)*(1+1/5)=4+6x
1,2x=0,8--> 2/3

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SoCan: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Sun May 15, 2011 9:19 amFollowed by: 1 member; Thanked: 18 times

by SoCan » Thu Jun 02, 2011 5:38 am

divya23 wrote:A certain tree was first planned it was 4 feet tall then height incrsd by constt rate for next 6yrs at the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr,by how many inches did the hite of tree inc each yr?
1.3/10
2.2/5
3.1/2
4.2/3
5.6/5
[spoiler]OA = 2/3[/spoiler]
[spoiler]
plz tell d correct method instead of backsolving[/spoiler]

The question is written incorrectly - it asks for how many inches the tree grew but it gives the answers in feet. If you want the answer in feet, it's 2/3, but if you want the answer in inches, it's 8 (12 inches per foot - although the GMAT would provide this information on a real question).

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Thu Jun 02, 2011 5:53 am

missed the growth in inches part. As no other unit was given, just assumed that all of them were same....

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