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shtirl [24]
3 years ago
6

A set of problems on Algebra, Geometry, and Combinatorics consists of 52, 70, and 88 problems. If 2 more problems from each cate

gory were added to the set, what is the largest number of tests that can be made containing an equal amount of problems from each category?
Mathematics
2 answers:
Talja [164]3 years ago
6 0
Anwser: 36
explanation: i take geometry so ik
MatroZZZ [7]3 years ago
4 0

Answer:

The largest number of tests that can be made is 36

Step-by-step explanation:

Let us list the information

  • Algebra consists of 52 problems
  • Geometry consists of 70 problems
  • Combinatorics consists of 88 problems
  • 2 more problems from each category were added to the set

That means each category will increase by 2

∴ Algebra = 52 + 2 = 54

∴ Geometry = 70 + 2 = 72

∴ Combinatorics = 88 + 2 = 90

We need to find the largest number of test that can be made containing an equal amount of problem

∵ 54 = 6 × 9

∵ 72 = 6 × 12

∵ 90 = 6 × 15

6 is the number of problem in each test because we need an equal amount of problems in each category and the other numbers are the number of tests in each category

∴ The largest number of Algebra tests is 9

∴ The largest number of Geometry tests is 12

∴ The largest number of Combinatorics test is 15

Add them to find the largest number of tests that can be made

∵ 9 + 12 + 15 = 36

∴ The largest number of tests that can be made is 36

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5 0
3 years ago
Read 2 more answers
A stone is a British unit of weight equivalent to 14 pounds. After losing 1 1/2 stone. Jim is now 85% of his original weight. Wh
evablogger [386]

Answer:

Jim's current weight = 119 pounds

Step-by-step explanation:

1 Stone = 14 pounds

1\frac{1}{2} stone = 1.5 stone

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Jim lost 21 pounds

Let X be Jim's Original Weight

Y be his present weight

As per given statement in the Question:

After losing 1.5 stones (21 pounds of weight) Jim now weighs Y

Present weight = original weight - 21

Y = X -21     <u>                                                    Equation 1</u>

Also Current Weight = 85 % (Original weight)

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Y=\frac{85X}{100}

put in Equation 1

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85X = (X-21) 100

85 X = 100 X -2100

or

2100 = 100 X - 85X

2100 = 15X

or

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X=\frac{2100}{15}

X= 140 pounds ( Original Weight)

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Y = X -21

Y = 140 -21

Y = 119 pounds (Current Weight)

5 0
3 years ago
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