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Mary worked from Monday to Saturday last week and bought lunch in the company’s restaurant
every day.
Her lunches cost a different amount every day.
Each lunch cost a multiple of 20¢.
The most expensive lunch of the week cost $7.60 and the cheapest one cost $4.40.
Friday’s lunch cost exactly 11⁄2 times as much as Wednesday’s lunch.
Tuesday’s lunch cost $1.20 more than Thursday’s lunch.
(a) What is the minimum total that Mary’s six lunches last week could have cost?
(b) What is the maximum total that Mary’s six lunches last week could have cost?

Respuesta :

Answer:

25%. 4.80 divided by 1.20 = 4 x 100 = 40o%

fichoh

Using logical and arithmetic principles, the minimum and maximum total Mary could spend is $31.20 and $40.40 respectively.

  • Cheapest lunch fee = $4.40 = 440¢
  • Most expensive = $7.60 = 760¢

Minimum total cost of lunch :

Lunch cost will increase steadily by multiple of 20¢ from the cheapest unless given a stated constraint

Monday = $4.80

Tuesday = $4.40 + 1.20 + 0.2 = $5.80

Wednesday = $4.40

Thursday = $5.80 - $1.20 = $4.60

Friday = 3/2($4.40) = $6.60

Saturday = $5.00

Total = $(4.40 + 5.80 + 4.80 + 4.60 + 6.60 + 5.00) = $31.20

Maximum Total :

Lunch cost will decrease steadily by multiple of 20¢ from the most expensive unless given a stated constraint

Monday = $7.40

Tuesday = $7.60 + 1.20 = $6.40

Wednesday = $4.80

Thursday = $7.60

Friday = 3/2($4.80) = $7.20

Saturday = $7.00

Total = $(7.40 + 6.40 + 4.80 + 7.60 + 7.20 + 7.00) = $40.40

Therefore, the minimum and maximum total Mary could spend is $31.20 and $40.40 respectively.

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