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mart [117]
3 years ago
7

Ming has 15 quarters, 30 dimes, and 48 nickels. He wants to group his money so that each group had the same number of each coin.

What is the greatest number of groups he can make? How many of each coin will be in each group? How much money will each group be worth?
Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0
Find the greatest common factor. In this case it is 3. Divide each group by 3.
15÷3=5 quarters
30÷3=10 dimes
48÷3=16 nickels
Next count them by the coin's value:
5 × 0.25 = $1.25
10 × 0.10 = $1
16 × 0.05 = $0.80
Add them up:
= $3.05 in each group and there are three groups.

So, the greatest number of groups that he can make is 3; there will be 5 quarters, 10 dimes, and 16 nickels in each group, which is worth $3.05 in each group.
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3 years ago
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A box is to be made where the material for the sides and the lid cost​ $0.20 per square foot and the cost for the bottom is ​$0.
Reika [66]

Answer:

The dimension of the box is l×w×h =  (2.274 × 2.274 × 1.934) ft³

Step-by-step explanation:

From the given information;

Let a be the cost of the box

Let b  be one side of the square base ;    &

h to be the height  of the box

We know that the volume of the box = 10 cubic feet

Then;

a²h = 10

h = 10/a²

The base = (0.65)a²

The top = (0.2)a²

The side = (0.2) a × 25/a²

= 5/a

For the four sides of the box now ;

= (0.2) 4a × 25/a²

= 0.8  × 25/a

= 20 /a

The total cost of the box is:

b = 0.65a² + 0.2a² + 20 /a

b = 0.85 a² + 20 /a

Taking differential of b with respect to a ;we have:

db/da = 1.7a - 1/a²(20) = 0

1.7 a³ - 20 = 0

1.7 a³  = 20

a³ = 20/1.7

a³ = 11.77

a = \sqrt[3]{11.77}

a = 2.274 ft

Thus; the cost for the base of the box = (0.65)a²

the cost for the base of the box =(0.65) × ( 2.274)²

the cost for the base of the box = 3.362

The top of the box = (0.2)a²

The top of the box = (0.2)× ( 2.274)²

The top of the box = 1.034

The four sides of the box = 20 /a

The four sides of the box = 20/2.274

The four sides of the box = 8.795

the total cost = b = 0.85 a² + 20 /a

the total cost = 0.85 (2.274)² + 20 /2.274

the total cost = 4.395 + 8.795

the total cost = 13.19

Recall that:

the volume of the box = 10 cubic feet

Then;

a²h = 10

h = 10/a²

h = 10/ 2.274²

h  1.934

The dimension of the box is l×w×h =  (2.274 × 2.274 × 1.934) ft³

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2 years ago
a shopkeeper sold a certain number ( a two-digit number)of toys all priced at a certain value (also a two-digit number when expr
Makovka662 [10]

The answer is 91 toys sold, make the number ab where a is the 10th digit and b is the first digit. The value is 10a + b that can expressed as 10 (3) + 4 = 34

Let the price of each item: xy

10x + y

He accidentally reversed the digits to: 10b + a toys sold at 10y + x rupees per toy. To get use the formula, he sold 10a + b toys but thought he sold 10b + a toys. The number of toys that he thought he left over was 72 items more than the actual amount of toys left over. So he sold 72 more toys than he thought:

10a + b =10b + a +72

9a = 9b + 72

a = b + 8

The only numbers that could work are a = 9 and b = 1 since a and b each have to be 1 digit numbers. He reversed the digits and thought he sold 19 toys. So the actual number of toys sold was 10a + b = 10 (9) + 1 = 91 toys sold. By checking, he sold 91 – 19 = 72 toys more than the amount that he though the sold. As a result, the number of toys he thought he left over was 72 more than the actual amount left over as was stated in the question.

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