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Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are.
  • a)
    (20,36)
  • b)
    (25,31)
  • c)
    (24,32)
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Divide 56 into two parts such that three times the first part exceeds ...
let one part be x
so the other part is 56-x
ATQ,
3x-(56-x/3)=48
(9x-56+x)/3=48
10x-56=144
10x=200
x=20
so one part is 20 and the other part is 36.
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Most Upvoted Answer
Divide 56 into two parts such that three times the first part exceeds ...
Given information:
- Let the two parts be x and y.
- We are given that three times the first part exceeds one third of the second by 48.
- This can be represented as 3x = (1/3)y + 48.

Equation:
- From the given information, the equation can be written as:
3x = (1/3)y + 48 -> 9x = y + 144 -> y = 9x - 144.

Dividing 56 into two parts:
- Since we need to divide 56 into two parts, we know that x + y = 56.
- Substituting the value of y from the equation above into this equation:
x + 9x - 144 = 56 -> 10x = 200 -> x = 20.

Finding the second part:
- Now that we have found the value of x, we can find the value of y using y = 9x - 144.
- y = 9(20) - 144 = 180 - 144 = 36.

Final answer:
- Therefore, the two parts are 20 and 36.
- Hence, the correct answer is option 'A' (20, 36).
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Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are.a)(20,36)b)(25,31)c)(24,32)d)none of theseCorrect answer is option 'A'. Can you explain this answer?
Question Description
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