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In the equation `A+B+C+D+E=FG`. Where `FG` is the two digit number whose value is `10F+G` and letters `A,B,C,D,E,F` and `G` each represents different and `FG` is as large as possible If a five digit number is made using the digits `A,B,C,D,E` without repetition tenA. the probability that the number is divisible by 11 is `2/5`B. the probability that the number is divisible by 11 is `1/5`C. the probability that the number is divisible 4 is `1/4`D. the probability that the number is divisible by 4 is `2/5` |
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Answer» Correct Answer - C `9,8,6,5,4` (`A` and `B`)………`9+8+6+5+4=32` `|(9+8)-(6+5+4)|=2` not divisible by 11 `|(9+6)-(8+5+4)|=3` not divisible by 11 `|(9+5)-(8+6+4)|=4` not divisible by 11 `|(9+4)-(8+6+5)|=6` not divisible by 11 `|(8+6)-(9+5+4)|=4` not divisible by 11 `|(8+5)-(9+6+4)|=6` not divisible by 11 `|(8+4)-(9+6+5)|=8` not divisible by 11 `|(6+5)-(9+8+4)|=10` not divisible by 11 `|(6+4)-(9+8+4)|=12` not divisible by 11 `|(5+4)-(9+8+6)|=14` not divisible by 11 (C) .......... `ul4, ul8` `ul5, ul6` `ul6, ul4` `ul8, ul4` `ul9, ul6` Probability `=(5xx3!)/(5!)=1/4` |
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