+76 1.) Last summer, Ben sold lemonade. On a certain day, he observed that the money he had , consisted of 25¢ coins and $5.00 coins. In fact, 9 less than twice the number of 25¢ coins was the number of $5.00 coins. The total value of money is $129.25. How many coins of each type were there?
2.) Mister Adam sells tapioca pearls in 4 kilo and 3 kilo containers. With much demand for the tapioca pearls, he sold 30 containers weighing 300 kilos. How many of each type did he sell?
3.) A bank teller counted $100 and $50 bills. This totalled to 117 bills to worth $9950. How many bills of each type were there?
4.) Mel is an enterprising woman. She sews quilled bags and sells them at $220 or $150 depending on the size. Today, of the 15 bags she brought, 3 were unsold and this gave her $2290. How many of each kind was she able to sell?
5.) In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.59. People attended the activity and $12,150 was raised. How many couples attended?
1.) Last summer, Ben sold lemonade. On a certain day, he observed that the money he had , consisted of 25¢ coins and $5.00 coins. In fact, 9 less than twice the number of 25¢ coins was the number of $5.00 coins. The total value of money is $129.25. How many coins of each type were there?
Let q=number of .25c
Let d=number of $5
2q - 9=d, and,
.25q + 5d=129.25
Solve the following system:
{2 q-9 = d
5 d+0.25 q = 129.25
In the first equation, look to solve for d:
{2 q-9 = d
5 d+0.25 q = 129.25
2 q-9 = d is equivalent to d = 2 q-9:
{d = 2 q-9
5 d+0.25 q = 129.25
Substitute d = 2 q-9 into the second equation:
{d = 2 q-9
0.25 q+5 (2 q-9) = 129.25
0.25 q+5 (2 q-9) = (10 q-45)+0.25 q = 10.25 q-45:
{d = 2 q-9
10.25 q-45 = 129.25
In the second equation, look to solve for q:
{d = 2 q-9
10.25 q-45 = 129.25
10.25 q-45 = (41 q)/4-45 and 129.25 = 517/4:
(41 q)/4-45 = 517/4
Add 45 to both sides:
{d = 2 q-9
(41 q)/4 = 697/4
Multiply both sides by 4/41:
{d = 2 q-9
q = 17
Substitute q = 17 into the first equation:
Answer: | d=25 and q=17
2.) Mister Adam sells tapioca pearls in 4 kilo and 3 kilo containers. With much demand for the tapioca pearls, he sold 30 containers weighing 300 kilos. How many of each type did he sell?
I think there maybe something wrong with your question: he cannot sell 30 containers weighing 300 kilos, because 300/30=10 kilos. But he only has cans that weigh 4 and 3 kilos!. So, even if he sold 30 cans of 4-kilos=120 kilos????????????.
x = number of 25 cent coins, then according to the problem ( x (2) -9 ) is defined as the number of 5 dollar coins
so the value of 25 cent coins is x (25) cents and the value of 5 dollar coins is 500 (x(2) - 9) cents and these two added together = 12925 cents
(everything needs to be expressed in cents....makes 'cents' , huh? Ha)
so: x(25) + ( 500(x(2)-9)) = 12925 now solve for 'x' (the number of 25 cent coins)
25x + 1000x - 4500 =12925
1025 x = 17425
x = 17 = number of 25 cent coins Two times this and subtract 9 to get 5 dollar coins = 2(17)-9 = 25 5 dollar coins
~jc
3.) A bank teller counted $100 and $50 bills. This totalled to 117 bills to worth $9950. How many bills of each type were there?
Let the number of $50=f
Let the number of $100=h,
f + h=117 and,
50f + 100h=9950
Solve the following system:
{f+h = 117 | (equation 1)
50 f+100 h = 9950 | (equation 2)
Swap equation 1 with equation 2:
{50 f+100 h = 9950 | (equation 1)
f+h = 117 | (equation 2)
Subtract 1/50 × (equation 1) from equation 2:
{50 f+100 h = 9950 | (equation 1)
0 f-h = -82 | (equation 2)
Divide equation 1 by 50:
{f+2 h = 199 | (equation 1)
0 f-h = -82 | (equation 2)
Multiply equation 2 by -1:
{f+2 h = 199 | (equation 1)
0 f+h = 82 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{f+0 h = 35 | (equation 1)
0 f+h = 82 | (equation 2)
Collect results:
Answer: | f=35 and h=82
For Q # 3
Let x = 100 bills the rest are 50's or 117 - x is the number of 50s
the value of the 100's is x (100) the value of the 50's is 50 (117-x) when these values are added they total 9950
x (100) + 50 (117-x) = 9950 SOlve this for x
100x + 5850 - 50x = 9950
50x = 4100
x = 82 The number of 100s 117- x = 117 - 82 = 35 50s
~jc
For Q #5
x = number of couples Uh-OH I think you neglected to tell us how many people attended !
How many people attended ?
~jc
4.) Mel is an enterprising woman. She sews quilled bags and sells them at $220 or $150 depending on the size. Today, of the 15 bags she brought, 3 were unsold and this gave her $2290. How many of each kind was she able to sell?
Let the number of bags of $220=t
Let the number of bags of $150=h
h + t=12, and,
220t + 150h=2290
Solve the following system:
{h+t = 12 | (equation 1)
150 h+220 t = 2290 | (equation 2)
Swap equation 1 with equation 2:
{150 h+220 t = 2290 | (equation 1)
h+t = 12 | (equation 2)
Subtract 1/150 × (equation 1) from equation 2:
{150 h+220 t = 2290 | (equation 1)
0 h-(7 t)/15 = (-49)/15 | (equation 2)
Divide equation 1 by 10:
{15 h+22 t = 229 | (equation 1)
0 h-(7 t)/15 = -49/15 | (equation 2)
Multiply equation 2 by -15/7:
{15 h+22 t = 229 | (equation 1)
0 h+t = 7 | (equation 2)
Subtract 22 × (equation 2) from equation 1:
{15 h+0 t = 75 | (equation 1)
0 h+t = 7 | (equation 2)
Divide equation 1 by 15:
{h+0 t = 5 | (equation 1)
0 h+t = 7 | (equation 2)
Collect results:
Answer: | h=5 and t=7
