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Naddika [18.5K]
3 years ago
7

High grade steel consists of 85% iron and 15% magnese. Low grade steel consists of 67% iron and 33% mag ese. NASA orders 500 ton

s of steel and specifies that it must be in the proportion 80% iron and 20% steel. How many tons of high grade and low grade steel must you melt together to create steel that matches NASA’s requirements?
Mathematics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

361\dfrac{1}{9} tons of high steel and 138\dfrac{8}{9} tons of low steel

Step-by-step explanation:

Let x be the number of tons of high grade steel and y be the number of tons ow low grade steel needed.

In x tons of high grade steel there are

0.85x tons of iron

0.15x tons of magnese

In y tons of low grade steel there are

0.67y tons of iron

0.33y tons of magnese

NASA orders 500 tons of steel, so

x+y=500

and specifies that it must be in the proportion 80% iron and 20% magnese, so

0.85x+0.67y=500\cdot 0.8\\ \\0.85x+0.67y=400

From the first equation,

x=500-y

Substitute it into the second equation:

0.85(500-y)+0.67y=400\\ \\425-0.85y+0.67y=400\\ \\-0.18y=-25\\ \\y=\dfrac{2,500}{18}=138\dfrac{8}{9}\ tons\\ \\x=361\dfrac{1}{9}\ tons

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There are 2504 computer science students at a school. Of these, 1876 have taken a course in Java, 999 have taken a course in Lin
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Answer:

492

Step-by-step explanation:

As per the given data of the question:

Total number of students = 2504

Number of students in Java (J) = 1876

Number of students in Linux (L) = 999

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J∩L = 876

L∩C = 231

C∩J = 290

L∩J∩C = 189

Now according to Venn-diagram as drawn below:

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= J∩L - L∩J∩C

= 876 - 189

= 687

Number of students haven taken courses in Java and C both only

= C∩J - L∩J∩C

= 290 - 189

= 101

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= L∩C - L∩J∩C

= 231 - 189

= 42

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