The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.
B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.
The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.
The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.
Now combining the two ratios, we get A : B : C = 9 : 15 : 25.
Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x
The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.
Therefore, B's age = 15x = 15*3 = 45
0.0169 / 0.0130 = 169 / 130
= 13 / 10
Given Exp. = 4 / 7 + {(2q - p) / (2q + p)}
Dividing numerator as well as denominator by q,
Exp = 4/7 + {2-p/q) / (2 + p/q)}
= 4/7 + {(2 - 4/5) / (2 + 4/5)}
= 4/7 + 6/14
= 4/7 + 3/7
=7/7
=1.
25% of 25% = (25/100) x (25/100) = 625/10000 = 0.625
Let y% of 20 = .05
Then, (y x 20)/100 = .05
? y = .25
2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
81/3% = (25/3 x 1/100) = 1/12
.025 = (25/1000) x 100% = 2.5%
Let N% of 2/7 is 1/35
? (2/7 x N) / 100 = 1/35
? N = 1/35 x 7/2 x 100 = 10%
Let the money interest at 8% interest be ? P .
Then, the money interest at 10% interest = ?(4000 - P)
According to the question,
(P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352
? 8P + 40000 - 10P = 35200
? 40000 - 35200 = 2P
? P = 4800/2 = ? 2400
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