If a,b are two two-digit positive integers such that a is greater than b . Then, the difference between the maximum and minimum values of (a+b)/(a-b) is
1) 197 × 67/88
2) 193× 67/88
3) 197×69/89
4) 195 × 69/89
Please tell me the answer and also tell me how you hv solved it , please do explain the formula you use .
Yashasvika
Class 8
1) 197 × 67/88
2) 193× 67/88
3) 197×69/89
4) 195 × 69/89
Please tell me the answer and also tell me how you hv solved it , please do explain the formula you use .
Yashasvika
Class 8
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Answer:
sorry!! yashasvika I didn't understand ur question will u send me the pic of it question then I will help u
Given : a and b are two two-digit positive integers with different values such that a > b
To Find : the difference between the maximum and minimum values
Solution:
Two digit numbers
10 to 99
(a + b)/(a - b) is minimum when a-b is maximum
a = 99
b = 10
(99 + 10)/(99 - 10) = 109/89
(a + b)/(a - b) is maximum when a-b is minimum
a - b is minimum
a - b = 1
(99 + 98 )/(99 - 98)
= 197/1
= 197
197 - 109/89
= 17424/89
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