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Svetlanka [38]
3 years ago
14

Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than

Maria? (Show your work please!)
Mathematics
1 answer:
liraira [26]3 years ago
6 0

Given:

Atul has \dfrac{2}{3} lb of candy.

Jose has \dfrac{3}{5} lb of candy.

Maria has \dfrac{1}{2} lb less than Jose.

To find:

How many more pounds of candy does Atul have than Maria?

Solution:

Since, Maria has \dfrac{1}{2} lb less than Jose, therefore

Maria has = \dfrac{3}{5}-\dfrac{1}{2} lb

Maria has = \dfrac{6-5}{10} lb

Maria has = \dfrac{1}{10} lb

Difference between the candies Atul and Maria have.

Difference = \dfrac{2}{3}-\dfrac{1}{10} lb

Difference = \dfrac{20-3}{30} lb

Difference = \dfrac{17}{30} lb

Therefore,  Atul have \dfrac{17}{30} lb of candy more than Maria.

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Equation is 4b + 5m = €10.90, and values of b and m are €1.35 and €1.1 respectively.

Step-by-step explanation:

  • Step 1: Find equation for Thursday's sales.

Given b = price of a loaf of bread and m = price of one liter of milk

Total sales = €10.90

Equation ⇒ 4b + 5m = €10.90

  • Step 2: Find the values of b and m.

3b + 2m = 6.25 ----- (1)

4b + 5m = 10.90 ---- (2)

Multiply eq(1) by 5 and eq(2) by 2 to make coefficients of m equal.

15b + 10m = 31.25

8b + 10m = 21.8

Subtract eq(2) from eq(1)

⇒ 7b = 9.45

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Substitute for b and find m using eq(1).

⇒ 3 × 1.35 + 2m = 6.25

⇒ 4.05 + 2m = 6.25

⇒ 2m = 2.2

∴∴ m = 2.2/2 = €1.1

8 0
3 years ago
I need help with this problem
-BARSIC- [3]
Regular is
$97.99 after sale price is 83.30
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Answer:  101 revolutions

Step-by-step explanation:

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So whole ml's


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2              15            15/30=1/2
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5                6             6/30=1/5
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10               3              3/30=1/10
15              2              2/30=1/15
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