A natural number x is chosen at random from the first one hundred natural numbers . The probability that (x-20)(x-40)/(x-30)< 0 is (a) 16/50 (b) 3/50 (c) 3/25 (d) 7/25
Asked by rohit varshney
| 11th Apr, 2011,
11:43: PM
The Correct option is (d) 7/25.
Following is solution
The number would be negative only when
1) It lies between 1 to 19 (giving all three expressions a negative value). OR
2) It lies between 31 to 39 (where (x-20) and (x-30) would be positive but (x-40) would be negative giving the final expression a negative value).
Thus total number of successful cases=(19 +9)=28.
Total number of cases=100.
Probability =28/100= 7/25.
Please be carefull that you can not take x as 20,30 or 40. The expression would be equal to zero in case of x=20 and x=40 and it would be undefined when x is 30.
Hope it clarifies the doubt
The Correct option is (d) 7/25.
Following is solution
The number would be negative only when
1) It lies between 1 to 19 (giving all three expressions a negative value). OR
2) It lies between 31 to 39 (where (x-20) and (x-30) would be positive but (x-40) would be negative giving the final expression a negative value).
Thus total number of successful cases=(19 +9)=28.
Total number of cases=100.
Probability =28/100= 7/25.
Please be carefull that you can not take x as 20,30 or 40. The expression would be equal to zero in case of x=20 and x=40 and it would be undefined when x is 30.
Hope it clarifies the doubt
Answered by
| 12th Apr, 2011,
12:59: PM
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