Then, woman's age = (10X + y) years; husband's age = (10y + x) years.
Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)
⟺ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⟺ 10x = 8y ⟺ x = (4/5)y
Clearly, y should be a single-digit multiple of 5, which is 5.
So, x = 4, y = 5.
Hence, woman's age = 10x + y = 45 years.
Then, we have :
d - 1 = s and 2 (s - 1) = d.
Solving these two equations, we get: d = 4, s = 3.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⟺ 3X = 60 ⟺ x = 20.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
C - 4 = 2 (A - 4) ...(ii)
Also, A + 4 = 31 or A= 31-4 = 27.
Putting A = 27 in (i), we get: B = 9.
Putting A = 27 in (ii), we get C = 50.
Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.
Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.
Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.
Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.
So, 3x - 5 = 4 (x - 5) ⟺ 3x - 5 = 4x - 20 ⟺ x = 15.
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