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What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?
- a)9
- b)8
- c)5
- d)4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 =...
5P9 + 3R7 + 2Q8 = 1114
We can see that, the sum of last digits i.e. (5 + 3 + 2) = 10 and 1 is added as carry
And 2 is added as carry in the sum of middle digits of the numbers (P + Q + R)
⇒ (P + Q + R + 2) will be 11 only.
⇒ P + Q + R = 9
⇒ P + Q + R = 9
For maximum value of Q, values of P and R should be minimum
If P = R = 0 then Q = 9
∴ Maximum value of Q = 9
If P = R = 0 then Q = 9
∴ Maximum value of Q = 9
Most Upvoted Answer
What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 =...
Maximum value of Q in the equation 5P9 3R7 2Q8 = 1114 is to be found.
Solution:
The given equation is 5P9 3R7 2Q8 = 1114.
We know that the maximum value of any digit in a place value system is one less than the base of the system.
In this case, the base is 10 (since we use 10 digits in the decimal system). Therefore, the maximum value of any digit in this system is 9.
To find the maximum value of Q, we need to assume that the other digits (P and R) take their maximum values.
So, we can assume that P = 9 and R = 7. Substituting these values in the equation, we get:
5(9) 3(7) 2Q8 = 1114
45 + 21 + 2Q8 = 1114
2Q8 = 1114 - 66
2Q8 = 1048
Dividing both sides by 2, we get:
Q8 = 524
Since the maximum value of any digit in the decimal system is 9, we can conclude that the maximum value of Q is 9.
Therefore, the correct option is (A) 9.
Solution:
The given equation is 5P9 3R7 2Q8 = 1114.
We know that the maximum value of any digit in a place value system is one less than the base of the system.
In this case, the base is 10 (since we use 10 digits in the decimal system). Therefore, the maximum value of any digit in this system is 9.
To find the maximum value of Q, we need to assume that the other digits (P and R) take their maximum values.
So, we can assume that P = 9 and R = 7. Substituting these values in the equation, we get:
5(9) 3(7) 2Q8 = 1114
45 + 21 + 2Q8 = 1114
2Q8 = 1114 - 66
2Q8 = 1048
Dividing both sides by 2, we get:
Q8 = 524
Since the maximum value of any digit in the decimal system is 9, we can conclude that the maximum value of Q is 9.
Therefore, the correct option is (A) 9.
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Community Answer
What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 =...
5(9)+3(7)2Q8=1114
45+21+2Q8=1114
2Q8=1114-66
2Q8=1048
dividing both side by 2,we get
Q8=524
Q=9
45+21+2Q8=1114
2Q8=1114-66
2Q8=1048
dividing both side by 2,we get
Q8=524
Q=9
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What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer? for Defence 2025 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer?.
What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer? for Defence 2025 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What would be the maximum value of Q in the equation 5P9 + 3R7 + 2Q8 = 1114?a)9b)8c)5d)4Correct answer is option 'A'. Can you explain this answer?.
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