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
Given in the question,
Price of 6 toys is Rs. 264.37.
? Price of 1 toys is Rs. 264.37/ 6 .
? Price of 5 toys is Rs. 264.37 x 5 / 6 .
? Price of 5 toys is Rs. 44 x 5 .
? Price of 5 toys is Rs. 220.
Ratio of the speeds of A and B = 5 : 3
Thus, in a race of 5m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
40 m will be gained by A in a race of (5/2) x 40 m = 100 m
? Winning post is 100 m away from the starting point.
Velocity of stream = (downstream velocity - upstream velocity)/2
= (60/10 - 60/30)/2
= 2 km/hr
Let their ages one years ago be 4x and 3x years.
? (4x + 2)/(3x + 2) = 5 / 4
? 4(4x + 2) = 5(3x + 2)
? x = 2
? Sum of their present ages = (4x + 1) + (3x + 1) = 16 years
Required number
= 786 x 964 = 757704
Let the C.P. be Rs. x
Then 2(69 - x) / 100 = (78 - x) / 100
? 138 - 2x = 78 - x
? x = 60
? C.P. = Rs. 60
Score of A = 100 points
Score of B = 65 points
? A can give (100 - 65 ) = 35 points to B.
Given expression = log 8 + log (1/8)
= log 8 x (1/8)
= log 1
= 0
Let N = 312 x 28
? log N = 12 x log3 + 8 x log 2
? log N = 12 x 0.47712 + 8 x 0.30103
? log N = 8.13368
? No.of digits = 8 + 1 = 9
Place value of 5 in 65231 = 5 x 1000
= 5000
Face value = 5
required product = 5000 x 5 = 25000
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