Question
Present average age of 4 friends: βAβ, βBβ,
βCβ and βDβ is 24 years. If their ages are in the arithmetic progression in the same order and βDβ is 4 years older than βBβ, then find the sum of present ages of βAβ and βCβ.Solution
Common difference between ages = (4/2) = 2 years. So, let present ages of βAβ, βBβ, βCβ and βDβ be βxβ, (x + 2) years, (x + 4) years and (x + 6) years, respectively According to question: x + x + 2 + x + 4 + x + 6 = 96 Or, 4x = 84 Or, x = 21 years (Age of A) Present age of βCβ = x + 4 = 21 + 4 = 25 years Required sum = 21 + 25 = 46 years
49.99% of 639.99 + 159.98% of 49.99 = ?2
(5.013 β 30.04) = ? + 11.98% of 4799.98
(51.99Β² - 19.05Β² )Γ· ? = 14.11Β² - 140.33
125.9% Γ· 9.05 x 99.98 = ? - 69.97 Γ β324.02 Γ· 5.98
What approximate value will replace the question mark (?) in the following?
29.99...
(18.21)Β² - (12.9)Β² = 20% of 649.9 - ? + 400.033
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
39.9% of 1720 + 80.2% of 630 = 89.9% of 1280 + ?
(11.75)2 - 49.99% of 120 - ? = (8.23)2Β