The arithmetic mean of 3a and 4b is greater than 50 and a is twice b,then the smallest possible integer of a is
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Answered by
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Explanation:
[(3a+4b) / 2] >50
→3a+4b>100
→3a+ 4a/2 >100 [∴a=2b]
→3a+2a>100 →5a>100
→a>20
∴Minimum value of a =21
[(3a+4b) / 2] >50
→3a+4b>100
→3a+ 4a/2 >100 [∴a=2b]
→3a+2a>100 →5a>100
→a>20
∴Minimum value of a =21
Answered by
0
Answer:
If the arithmetic mean of 6a and 8b is greater than 100 and a is four
times of b, then the smallest possible integer value of a is
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