How would you solve these word problems (algebra)?

Question

1. You have some nickels and dimes worth $6.25. You have three times as many nickels as dimes. How many nickels do you have?

2. Mary's age is one third of her dad's age. Seven years ago, her age was one fifth of his. How old are both now?

The next one doesn't seem to make any sense, but here it is...
3. If one machine can sort coins in 15 minutes, and a second machine can sort the same number of coins in 30 minutes, how long would it take both machines working together to do the job?

Please work out the problems instead of just saying the answer. Thanks...

Answer 1

if x= number dimes
then number of nickels = 3x
so
625 cents = x dimes + 3x nickels
and
x dimes = 10 x cents
3x nickels= 5 * 3x cents = 15 x cents
so
625 = 10x cents + 15x cents = 25x cents
625 /25 = 25x/25
25 = x
so 25 dimes + 75 nickels = 6.25

2.
3m= d (Mary's age is one third of her dad's age)
and
5(m-7)=d-7 ( Seven years ago, her age was one fifth of his )
5(m-7)=d -7
or
5(m-7)+7 =d

so combine resulting in
3m = 5(m-7)+7
3m= 5m -35 +7
3m= 5m -28
28= 5m-3m
28 = 2m
14=m
Mary is 14
dad is 42
3
1st machine does x coins / 15 Min's
2nd machine does x coins /30 Min's
then 1st can do 2x coins /30Min's
plus 2nd machine does 1x coins /30 Min's
together 3x coins /30 mins
or x coins in 10 mins
answer
10 mins

Answer 2

1. Let N=# of nickels and D = # of dimes
If they are worch $6.25 combined, then
5N + 10D = 625
If you have three times as many nickels as dimes, then N = 3D
Now substitute #D for every N in the first equation:
15D + 10D = 625
25D = 625
D = 25
so N must be 3*25 = 75
Check:
25 dimes is worth $2.50,
75 nickels is worth $3.75
The sum is indeed $6.25

2. Let M = Mary's age now
Let D = Dad's age now
M = (1/3) D
Seven years ago Mary was M-7, Dad was D-7
M-7 = (1/5)(D-7)
Again, substitute the information from the first equation into the second:
(1/3)D - 7 = (1/5)(D-7)
Multiply both sides by 15 to get rid of the fractions:
5D - 105 = 3 (D-7)
5D - 105 = 3D - 21
2D = 84
D = 42 so M = 14
Check: 7 years ago Dad was 35 and Mary was 7, which is (1/5) of Dad's age.

3. Sure it makes sense!
A sorts a number of coins in 15 minutes.
B sorts the same number of coins in 30 minutes.
Surely, the two machines working together will sort all the coins faster than either one alone.
How much faster?
Let's say the number of coins each machine sorts is x.
In 1 minute, machine A sorts x/15 coins
In 1 minute, machine B sorts x/30 coins
Together they sort x/15 + x/30 coins = x/10 coins
If they sort x/10 coins in 1 minute, they will
sort x coins in 10 minutes!
The general solution for a problem like this is:
1 / (1/A + 1/B)

Answer 3

1. how much are nickels and dimes worth ?
I'm obviously not American.


2. Mary's age is one third of her dad's age. Seven years ago, her age was one fifth of his. How old are both now?

dad = x
=> mary = 1/3 x

Seven years ago
dad = x-7
mary = 1/3 x - 7

her age was one fifth of his
=> 1/3 x - 7 = 1/5(x-7)
X by LCM etc.

3. If one machine can sort coins in 15 minutes, and a second machine can sort the same number of coins in 30 minutes, how long would it take both machines working together to do the job?

machine A can sort coins in 15 minutes &
machine B can sort coins in 30 minutes

=> in 30 min machine A can = 2 sort of coins &
=> in 30 min machine B can = 1 sort of coins

=> 3 sorts = 30 mins
=> 1 sort = 10 mins

Answer 4

Here's a more algebraic way to do 3.

Let C = number of coins, and t = time machine is running.

Then, for machine 1, total coins sorted = t(C/15).
For machine 2, total coins sorted = t(C/30).

Both machines together, total coins sorted = t(C/15) + t(C/30)
= t(3C/30) = t(C/10).

So, to determine the total time to sort C coins,
t(C/10) = C
t = 10.

Answer 5

a million. 4x - 3y = 8. -8 + 6y = sixteen. Simplify the 2nd equation: upload 8 to the two factors, giving 6y = 24. Divide the two factors by 6, giving y = 4. Use this result interior the 1st equation by substituting the y-fee: 4x - 3y = 8 will become 4x - 3(4) = 8, or 4x - 12 = 8. upload 12 to the two factors, giving 4x = 20. Divide the two factors by 4, giving x = 5. the answer factor is then (5,4). 2. permit R be the known value, and P the top type value. we are provided that P = R + 0.20. the fee of eleven gallons of known, in money, is 11R. the fee of sixteen gallons of top type is 16P = sixteen(R + 0.20). the whole is 11R + sixteen(R + 0.20) = fifty 8.fifty 5, or 11R + 16R + 3.20 = fifty 8.fifty 5, or 27R + 3.20 = fifty 8.fifty 5, or 27R = fifty 5.35, or R = $2.05. P = R + 0.20 = $2.25, answer b.

Answer 6

3(0.05x) + 0.1x = 6.26

0.15x + 0.1x = 6.25

Collect like terms

0.25x = 6.25

Divide both sides of the equatio by 0.25

0.25x / 0.26 = 6.25 / 0.25

x = 6.25/0.25

x = 25
- - - - - - -

Check

3(0.05x) + 0.1x = 6.25

3[0.05(25)] + 0.1(25) = 6.25

3[1.25] + 2.5

3.75 + 2.5 = 6.25

6.25 = 6.25

- - - - - - - --

There are 75 nickels

- - - - - - -s-

Answer 7

for #1, n=nickel, d=dime

3n+d=6.25

6.25divided 3=2.08 dimes, so you have

6.25-2.08=4.17 nickels! hope that helps!

Answer 8

problem one x(10) + 3x (5) = 625
10x + 15x = 625
25x = 625
x = 25

problem two duh uh

m = 1/3 d