Question
The ages of 4 members of a family are in the year 2011
are βXβ,βX+6β,βX+12β and βX+18β. After some years Oldest among them was dead then average reduced by 2. After how many years from his death, the average age will same as in 2011?Solution
Average of all in 2011 = {X + (X+6) + (X+12) + (X+18)}/4 = (4X + 36)/4 = X + 9 Let the oldest die after N years, then average of three persons, {X + (X+6) + (X+12) + 3N}/3 = (X+9-2) (3X+18+3N)/3 = X+7 X+6+N = X+7 N = 1 years Let after another M years, their average will be equal to that in 2011. So, (3X+18+3N+3M)/3 = X+9 (3X + 18 + 3x1 + 3M)/3 = X+9 X + 6 +1 +M = X+9 M = 2 years
What approximate value will replace the question mark (?) in the following?
β48...
(16.16 Γ Β 31.98) + 14.15% of 249.99 = ? + 99.34
(239.89 Γ· 3.89) β (144.01 Γ· 5.73) = ?2Β
- 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.)
2.51% of 800 - 3.97% of 250 = ?
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.)...
6940 ÷ 28 ÷ 7 =?
β3601 Γ β(224) Γ· β102 = ?
1726Β 1/3 + 40% of 1849.889 + 15.12 Γ 18.25 = ?
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