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matrenka [14]
3 years ago
15

A collection of dimes and quarters is worth $9.55. If the quarters were dimes and the dimes were quarters, the total value would

be 7.60. Find the number of each coin.
Mathematics
1 answer:
kvasek [131]3 years ago
5 0

Number of dimes are 18 and number of quarters are 31

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

A collection of dimes and quarters is worth $9.55

value of 1 dime x number of dimes + value of 1 dime x number of quarters = 9.55

0.10d + 0.25q = 9.55 ---------- eqn 1

If the quarters were dimes and the dimes were quarters, the total value would be 7.60

quarters were dimes means , q = d

dimes were quarters means d = q

0.25d + 0.10q = 7.60 ----- eqn 2

Let us solve eqn 1 and eqn 2 to find "d" and "q"

Multiply eqn 1 by 2.5

0.25d + 0.625q = 23.875 ---- eqn 3

Subtract eqn 2 from eqn 3

0.25d + 0.625q = 23.875

0.25d + 0.10q = 7.60

( - ) ----------------------

0.525q = 16.275

q = 31

Substitute q = 31 in eqn 1

0.10d + 0.25q = 9.55

0.10d + 0.25(31) = 9.55

0.10d + 7.75 = 9.55

0.10d = 1.8

d = 18

Thus dimes are 18 and number of quarters are 31

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A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t
MakcuM [25]

Answer:

165.

Step-by-step explanation:

Since repetition isn't allowed, there would be 11 choices for the first donut, (11 - 1) = 10 choices for the second donut, and (11 - 2) = 9 choices for the third donut. If the order in which donuts are placed in the bag matters, there would be 11 \times 10 \times 9 unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type 3 \times 2 \times 1 = 6 times.

For example, if a bag includes donut of type x, y, and z, the count 11 \times 10 \times 9 would include the following 3 \times 2 \times 1 arrangements:

  • xyz.
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  • yzx.
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Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count 11 \times 10 \times 9 by 3 \times 2 \times 1 = 6 to find the actual number of donut combinations:

\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of 3 objects from a set of 11 distinct objects:

\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

5 0
2 years ago
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