Question
Four years ago, the age of the mother was 12 times the
age of her daughter. Two years hence, the mother's age will be 4 times the daughter's age. After how many years will the mother's age be 3 times that of her daughter?Solution
Let the present age of the daughter be x years. Four years ago, the age of the daughter = (x - 4) years Four years ago, the age of the mother = 12 × (x - 4) = (12x - 48) years Two years hence, Age of the daughter = (x - 4) + 4 + 2 = (x + 2) years Age of the mother = (12x - 48) + 4 + 2 = (12x - 42) years According to the question, (x + 2) × 4 = (12x - 42)  =4x + 8 = 12x - 42 = 12x – 4x = 42 + 8 8x =50  x = 6.25 Now, the present age of the mother = (12x - 48) + 4 = 12x - 44 = 12 × 6.25 - 44 = 31 Let after y years the age of the mother will be three times that of her daughter. According to the question, (6.25 + y) × 3 = (31 + y) = 18.75 + 3y = 31 + y = 2y = 31 – 18.75  y = 12.25/2  y = 6.125=6(1/8) years. After 6(1/8) years, the mother's age will be 3 times that of her daughter.
100, 75, 59, 55, 46, 45
Find the wrong number in the given number series.
2, 6, 12, 22, 30, 428137, 5405, 3669, 2701, 2201, 1997
- Find the wrong number, in the given number series.
1, 8, 27, 65, 125, 216 147 490 707 831 895 922 930
...9Â Â Â Â 20Â Â Â Â Â 63Â Â Â Â Â Â 255Â Â Â Â Â Â 1285Â Â Â Â Â 7716
...21, 44, 78, 121, 179, 246
8Â Â Â Â 10Â Â Â Â 36Â Â Â Â Â 216Â Â Â Â Â 2592Â Â Â Â 62208
- Find the wrong number in the given number series.
60, 75, 101, 150, 210, 285 97, 98, 107, 132, 181, 264