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Romashka [77]
2 years ago
8

Pixel goes shopping with one hundred dollars and ends up buying three different types of items: A, B, and C. The unit price of A

is five dollars, B is 25 cents, and C is five cents. Pixel comes home and finds that all the money had been spent, but discovers that the total number of items is exactly equal to100. How many items of each kind did Pixel buy? No fractions allowed. (Spend exactly 100 dollars and buy a total of 100 items)
Mathematics
1 answer:
mixas84 [53]2 years ago
5 0

Answer:

  • A - 16, B - 79, C - 5

Step-by-step explanation:

<u>Given</u>

  • A costs 5$
  • B costs  $0.25
  • C costs $0.05
  • A + B + C = 100
  • 5A + 0.25B + 0.05C = 100 ⇒ 100A + 5B + C = 2000

<u>Subtract the first equation from the second:</u>

  • 99A + 4B = 1900

<u>We have a restriction in numbers:</u>

  • 0 < A, B, C < 100

<u>Since 4B can be max 400, we get 99A > 1500  and 99A < 1900 so:</u>

  • A > 1500/99 > 15

and

  • A < 1900/99 < 19

A can be 16 or 18 as it should be even number to get B the whole number.

<u>If A = 16:</u>

  • 99*16 + 4B = 1900
  • 4B = 1900 - 1584
  • 4B = 316
  • B = 316/4 = 79

<u>If A = 18:</u>

  • 99*18 + 4B = 1900
  • 4B = 1900 - 1782
  • 4B = 118
  • B = 118/4 = 29.5, discarded as decimal

<u>So we get:</u>

  • A = 16, B = 79 and C = 100 - (16 + 79) = 5

<u>Let's proof the numbers:</u>

  • 16*5 + 79*0.25 + 5*0.05 = 100, confirmed

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Your answer would be 24, so you got it right.

This is because you need to find the lowest common multiple of 6 and 8, that is, the lowest number that divides by both 6 and 8.

We can do this by either writing out the multiples of 6 and 8 until you find a common one, or you can find the prime factors of 6 and 8 and multiply them:

6-> 3 × 2
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Step-by-step explanation:

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Step-by-step explanation:

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