Question
21 rotten bananas are accidentally mixed with 135 good
ones. It is not possible to just look at a banana and tell whether or not it is rotten. One banana is taken out at random from this lot. Determine the probability that the banana is taken out is a good one.Solution
Numbers of bananas = Numbers of rotten bananas + Numbers of good bananas ∴ Total number of bananas = 135 + 21 = 156 bananas P(E) = (Number of favourable outcomes) / (Total number of outcomes) P(picking a good banana) = 135/156 = 45/52Â
[(82 × 162)/12] - 28 = ?
24 × ?2 – 11 × 4 = 1900
Simplify the following expression:
  (400 +175) ² - (400 – 175) ² / (400 × 175)
√256 * 3 – 15% of 300 + ? = 150% of 160
756 + 432 – 361 + ? = 990
[(√576 × √144) ÷ √1296]2 = ? ÷ 3
136% of 560 - 2/7 of 630 + 45% of 420 =?
Simplify: 0.6 ÷ 0.04 + 0.125 × 0.8
What will come in the place of question mark (?) in the given expression?
888 + 777 - 666 = (? + 60) X 3 + 444