Question

Use the three steps to solve the problem. Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have? { a0quarters, a1dimes }

Answer 1

d = q + 10
10d + 25q = 345 cents
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Substitute for "d" and solve for "q";
10(q+10) + 25q = 345

35q + 100 = 345
35q = 245
q = 7 (# of quarters)
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Since d = q+10, d = 17 (# of dimes)
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Cheers,
Stan H.

Answer 2

Answer:

betty has 24 coins.

Step-by-step explanation:

Let Betty has number of dimes ($0.10) = D

and number of quarters ($.25) = Q

Now equations will be

D = Q + 10 ----------(1)

(0.10)D + 0.25Q = 3.45 -----------(2)

Now by putting the value of D from equation (1) to equation (2)

$(.10)(Q+10)+.25Q=3.45$

.10Q + 1 + .25Q = 3.45

.35Q + 1 = 3.45

.35Q = 2.45

Q = $\frac{2.45}{.35}=7$

from equation (1)

D = 7 + 10 = 17

So betty has (7 + 17) = 24 coins.