
By: drstevewu (offline) Wednesday, August 21 2013 @ 11:14 PM CDT (Read 587 times)



drstevewu 
Hi, wanting to see if there is another working for the following question:
Mr Tan bought a total of 100 fruits consisting of oranges, pineapples and pears for $74. Each orange cost 55 cents and each pear cost 15 cents more than an orange. The cost of each pineapple is $1.20. If she bought an equal number of oranges and pears, how many pears did she buy?
Thank you.
Cheers,
Steve

Junior
Registered: 04/20/10 Posts: 31





By: gerchan (offline) Thursday, August 22 2013 @ 12:22 AM CDT



gerchan 
I used algebra method:
let y be the number of oranges or pears
ley x be the number of pineapples
0.55 + 0.70 = 1.25
x + 2y = 100  (1)
1.2x + 1.25y = 74 (2)
Equation (1) multiply by 1.2
1.2x +2.4y = 120 (3)
subtract (2) from (3) you will get:
1.15y = 46
y = 40
therefore, the number of pears is 40.

Newbie
Registered: 02/23/10 Posts: 3





By: echeewh (offline) Thursday, August 22 2013 @ 03:10 AM CDT



echeewh 
Hi <drstevewu>
Following is my Alternative worked solution:
<Units of Fruits>
Or : P
1 : 1
Or + P = 2u
PA = 100  2u
<Cost>
Or = 55
P = 55 + 15 = 70
PA = 120
Using <Number x Value> method, we have ...
(1u × 55) + (1u × 70) + [(100  2u) × 120] = 7400
55u + 70u + 12000  240u = 7400
240u  55u  70u = 12000  7400
115u > 4600
1u > 40
Number of Pears (P) = 1u > 40
========
Trust this helps.
Do let me know again if there's further clarifications.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627



