Betty has ten more dimes than quaters. If she has $3.45, how many coins does she have?

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Answer 1

Is it safe to assume that ALL Betty's money is in dimes and quarters?

First you subtract the ten dimes. That gives you 2.45 of equal dimes and quarter. Divide that by 35 cents and you get 7 quarters and 17 dimes.

Of course if Betty also has an undisclosed number of nickles and pennies, and a ten dollar bill, it could be lots of different amounts.

Answer 2

If Betty has $3.45, the question is...is the $3.45 ALL in coins, or are there dollar bills included?

ASSUMING that the $3.45 is in fact all coins, this is how I went about figuring out how many coins Betty had in all - of course assuming that there are 10 more dimes than quarters:
We already know that the number "ten" is significant. We know that there are AT LEAST ten dimes. So, if we first consider this fact, if a dime is worth .10, .10 x 10 = $1.00. We know there is a dollar worth of dimes.

However, quarters being worth .25 could not equal .45; if two quarters were used that would equal .50 and surpass .45. So, there have to be more dimes. Question is...how many? And how many quarters? And how can the # of dimes surpass the number of quarters by 10?

IF we have a dollars worth of dimes, we can look at the .45 cents as being a quarter and 2 extra dimes already. That brings tha total to 12 dimes and 1 quarter. Not working yet because there are 11 more dimes, not 10.

Now that we've accounted for the 1st dollar (10 dimes) and the .45 in the $3.45 (1 quarter, 2 dimes), 2 dollars still needs to be accounted for AND give us exactly 10 more dimes than quarters.

Backtracking...and for reference purposes...
we have used 12 dimes so far and 1 quarter.
*Knowing 4 quarters = $1.00, we could use trial & error and say that one of the two dollars remaining is accounted for by 4 quarters bringing the total to 12 dimes and 5 quarters. BUT, accounting for the next dollar, we might say that we could use two more quarters (7 quarters) to take us to $2.50 and then 5 more dimes (.50) taking us to 17 times.

And, there we go. 17 dimes and 7 quarters (10 more dimes than quarters).

To answer the overall question of how many coins does Betty have, the answer is 24 coins.

Answer 3

I'm going to assume that her $3.45 consists solely of dimes and quarters, since no other coin was mentioned.

So...
Let x = # dimes
Let y = # quarters

From the question we know that:
0.10x + 0.25y = 3.45 (you must include the value of the coin)

We also know that:
x = 10 + y since there are 10 more dimes than quarters.

Substitute: 0.10(10+y) + 0.25y = 3.45
Simplify: 1+0.10y + 0.25y = 3.45 -> 0.35y + 1 = 3.45
Solve for y: y = (3.45 - 1)/0.35
and y = 7

Thus, Betty has 17 dimes (x = 10 + 7) and 7 quarters (y = 7) for a total of 24 coins.

Answer 4

Betty has 7 quarters and 17 dimes. 24 coins total for $3.45.

Answer 5

7 Quarters + 17 Dimes = 24 Coins

Answer 6

Easy. You know she has 10 more dimes, and as a dime is 10 cents, that takes the rest down to $2.45, which we assume is an equal number of dimes and quarters. the 2 added together is 35 cents, which divides into $2.45 7 times, so 7+7+10 = 24 coins.

Answer 7

Let x = # quarters Betty has ten more dimes than quarters. Written mathematically, #D= x + 10, and #Q= x. A dime has a value of .1 (of 1) and a quarter .25 (of 1). Now set up a second equation: .25(x) + .1(x+10) = 3.45 value of quarter times how many of them, plus value of dime times how many of them (10+x, ten more than quarters)= a total amount of $3.45 distribute and rearrange: .25x + .1x + 1 = 3.45 .35x + 1 = 3.45 .35x = 2.45 Now divide 2.45 by .35 to get X alone, and solve for X. This tells you how many quarters Betty has. If betty has ten more dimes than quarters, then X+10 should give you # dimes. Ta da.

Answer 8

Betty has 24 coins. She has 17 dimes ($1.70) and 7 quarters ($1.75) which adds to $3.45.

Two equations, two variables:

D = number of dimes
Q = number of quarters

EQ1
0.10D + 0.25Q = 3.45

EQ2
D - 10 = Q

Solving them you get D = 17 and Q = 7
Hence, Betty has 24 coins.

Answer 9

24 coins- 17 dimes and 7 quarters

Answer 10

Ummmmmmm here it goes.
OBVIOUSLY you have to have an odd number of quarters for the 5 to be in the ones place. And you know that there is 10 more dimes than quarters. By process of elimination 7 quarters= 1.75 + 17 dimes= 1.70 equals 3.45.
TADA