Question
A bag contains 10 red gems, 12 yellow gems and 6 green
gems. 3 gems are drawn randomly. What is the probability that the gems drawn contain gems of different colours?Solution
Total number of gems = 10 + 12 + 6 = 28 Let S be the sample space. Then, n(S) = number of ways of drawing 3 gems out of 28 = 28C3 = (28 x 27 x 26)/ (3 x 2 x 1) = 3276 Let E= event of drawing 3 gems of the different colour n(E) = 10C1 x 12C1 x 6C1 = 10 x 12 x 6 = 720 :. P(E) = (n(E))/(n(S)) = 720/3276 = 20/91
(22 × 52 ) + 4 × 6 = ? - √324
What should come in place of (?) question mark in the given expression.
 (25% of 320) + (3/8 of 400) − 30 = ?
(5832)1/3  × 10.11 × 11.97 ÷ 16.32 = ? + 45.022
82% of 400 + √(?) = 130% of 600 - 85% of 400
If (x + 1/x) = 5, then value of x3 + 1/x3 is:
Simplify: (1 ÷ 0.08)
What should come in place of (?) question mark in the given expression.
{ (144 ÷ 12) × 5 } − (18 ÷ 3) = ?
Simplify the following expressions and choose the correct option.
(3/4 of 256) + (2/5 of 150) - (72 ÷ 7)
464 + 181 +? = (154 × 25) - (15) 2 Â
15% of 1800 + 22 = ?Â
