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

Redeem

I don't understand! Problem solving

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Tue Aug 05, 2008 7:15 am

I don't understand! Problem solving

by GMAT_Tomorrow » Tue Aug 05, 2008 7:17 am
Hi... I can't seem to figure out how to answer this question correctly:

Mark bought a set of 6 different flower pots of different sizes at a total cost of $8.25. Each pot cost $0.25 more than the one below it in size. What was the cost, in dollars, of the largest pot?

Please help :?
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 46
Joined: Mon Jul 07, 2008 12:10 pm
Location: Atlanta
Thanked: 5 times
Followed by:1 members

by lachlanc » Tue Aug 05, 2008 7:32 am
Here's how to do it -

let the most expensive pot = x

8.25 = x + (x-0.25) + (x-0.5) + (x-0.75) + (x-1) + (x-1.25)

8.25 = 6x - 3.75

6x = 12

x = $2
Top
Senior | Next Rank: 100 Posts
Posts: 30
Joined: Mon Jul 07, 2008 3:38 pm
Location: Atlanta
Thanked: 6 times

by Travis_1234567890 » Tue Aug 05, 2008 7:41 am
Here is a slightly different way to handle it...


6 pots totaling $8.25, each pot is bigger than the next, the prices increase by 0.25 for each pot

P = the smallest pot (or the base price, depends on how you want to look at it)
i=increase (we know this is 0.25)

This can be expressed as…

P + (P + i) + (P + 2i) + (P + 3i) + (P + 4i) + (P + 5i) = 8.25

You can simplify into

6p + 15i = 8.25

We know i=0.25, so we can substitute

6p + (15*0.25) = 8.25

6p + 3.75 = 8.25

6p = 4.5

p = .75

So, the 6th pot = 0.75 + (5 * .25) = $2.00
Top
Senior | Next Rank: 100 Posts
Posts: 58
Joined: Fri Jul 18, 2008 12:26 pm
Thanked: 2 times

by malolakrupa » Tue Aug 05, 2008 9:32 am
This could also be treated as a problem on arithmetic progression

Sum of n pots = S = 8.25 = 33/4
no. of pots = n = 6
common difference = 0.25 = 1/4
cost of smallest pot = a1

S = (n/2)*[2a1 + (n-1)*d]

substituting the values

33/4 = (6/3)*[2a1 + {(6-1)*1/4}]

Solving the above equation gives a1 = 3/4

a6 = a1 + (6-1)*1/4 = 3/4 + 5/4 = $2
Top
Post new topic Post Reply
• Page 1 of 1