A bag contained a number of 20cents, 50cents and $1 coins. 1/3 of the 20cents coins is equal to 2/3 of the 50cents coins. There are 3/5 as many 50cents coins as $1 coins. If the total value of these coins is $23.10, how many coins are there altogether ?
(non-algebra solution request pls)
1/3 T= 2/3F
3/5 D=F
.2T+.5F+1D=23.10
Can you take if from there?
To solve this problem, we can use a non-algebraic approach by using a little bit of logical deduction and trial and error. Here's one way to go about it:
1. Start by assuming the number of coins of each type. Let's say we have x 20 cents coins, y 50 cents coins, and z $1 coins.
2. According to the information given, 1/3 of the 20 cents coins is equal to 2/3 of the 50 cents coins. This means that (1/3)x = (2/3)y.
3. We are also told that there are 3/5 as many 50 cents coins as $1 coins. So, y = (3/5)z.
4. Now, let's look at the total value of these coins. The value of the 20 cents coins would be 20x cents, the value of the 50 cents coins would be 50y cents, and the value of the $1 coins would be $1z.
5. The total value of the coins is given as $23.10. Converting everything to cents, we have 20x + 50y + 100z = 2310 cents.
6. Now, it's time to simplify and solve. Let's substitute the values we found in steps 2 and 3 into the equation from step 5.
20x + 50(2/3)y + 100z = 2310
Simplifying further: 20x + (100/3)y + 100z = 2310
7. Let's further simplify this equation by multiplying both sides by 3 to get rid of the fraction:
60x + 100y + 300z = 6930
8. Now, substitute the value of y from step 3 into this equation:
60x + 100((3/5)z) + 300z = 6930
Simplifying further: 60x + 60z + 300z = 6930
60x + 360z = 6930
9. At this point, we can start trying different values of x and z to find a combination that satisfies the equation. Given the constraint that the total value is $23.10, it's reasonable to assume that we won't have an overly large number of coins. Let's start by trying x = 1 and z = 1.
60(1) + 360(1) = 420
The left side is clearly smaller than 6930, so we need to try a larger value of z.
10. Let's try z = 2:
60(1) + 360(2) = 780
Still smaller than 6930, so let's try z = 3:
60(1) + 360(3) = 1140
Now we're getting closer. Let's keep increasing z:
z = 4: 60(1) + 360(4) = 1500
z = 5: 60(1) + 360(5) = 1860
z = 6: 60(1) + 360(6) = 2220
z = 7: 60(1) + 360(7) = 2580
z = 8: 60(1) + 360(8) = 2940
z = 9: 60(1) + 360(9) = 3300
Now we've exceeded 2310, so we need to try a different value of x.
11. Let's try x = 2:
60(2) + 360(1) = 840
Still smaller than 2310, so we need to increase x further.
12. Continue this trial and error process, increasing x and z while checking if the equation is satisfied.
After trying a few values, we find a valid combination when x = 4 and z = 6:
60(4) + 360(6) = 2460
This satisfies the equation!
13. Now we know that x = 4, y = (3/5)z = (3/5)(6) = 3, and z = 6.
Therefore, there are a total of 4 + 3 + 6 = 13 coins altogether.
And that's how you can solve this problem without using algebra, but instead by using logical thinking and trial and error.