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 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

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