Question
A = 1200 – X where X is a real number. A > 0 and A is
divisible by 3, 5, 7 and 11. The sum of the two largest possible value of X is:Solution
Two largest Possible value of X = 45 and (-1110) [ Both are real numbers ] For X = 45, A = 1155 [Divisible by 3,5,7,11] For X = (-1110), A = 2310 [ Divisible by 3,5,7,11 ] Both of the above values of X satisfy the given conditions. Therefore, the sum of two largest possible values of X for which A is divisible by the given numbers (A > 0) will be 45 + (-1110) = (-1065) ∴ The largest possible value of X is -1065 for which the given condition is satisfied.
1885 ÷ 64.98 + 7.29 + ? = 69.09
212 + 14 × 23 – 28 × 15 = ? Â
(22² × 8²) ÷ (92.4 ÷ 4.2) =? × 32
567-4824 ÷ 134 =? × 9
Determine the value of 'p' in the expression.
28 ÷ 22p + 1 = 43Â
What will come in place of (?) in the given expression.
(15) ² - (13) ² = ?? = 6.25% of 240 + 25 2 + 17 2 – 16 × 17
35% of 840 + 162 = ? – 25% × 300
(7/5) × (3/4) × (5/9) × (6/7) × 3112 = ?
1024 ÷ 16 + 800 ÷ √64 + ? = 200 * 2