We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2
Let, Sum = P
Then S.I = P/2
Rate = 8%
and Time = 6 years
But P/2 = (P x 8 x 6) /100 (Not possible )
Thus, data is inadequate.
? = 6888.009 - 487.999 - 87.989
? ? ? 6888 - 488 - 88
? ? ? 6888 - 576 ? ? ? 6312 ? 6310
Let the speed in still water be x km/hr
Then, 6/x + 4 + 6/x - 4 = 2
? 6[ x - 4 + x + 4] = 2(x2 - 16)
? x2 - 6x - 16 = 0
? (x - 8) (x + 2) = 0
? x = 8 km/hr
Let the width of the room be x members
Then, its area = (4x) m2
Area of each new square room = (2x)m2
Let the side of each new room = y meters
Then, y2 = 2x
Clearly, 2x is a complete square when x=2
? y2 = 4
? y = 2 m .
12 |
|
% per annum is Rs. 25. Find the sum (Rs.). |
2 |
Average rate of interest = (100 * 750) / (5000 * 3) = 5% per annual
Investment at 3% per annual = 3 / (3 + 2) × 5000 = Rs. 3000 Investment at 8% per annual = 2 / (3 + 2) × 5000 = Rs. 2000
The difference counts only due to 40% of the profit which was distributed according to their investments.
Let total profit = R.
40% of R is distributed in the ratio, 125000 : 85000 = 25 : 17
Share of 1st partner = 40% of R x25/(25 + 17)
= 40% of 25R/42 = (40/100) x (25R/42) = 5R/21
Share of 2nd partner
= 40% of 17R/42 = (40/100) x (17R/42) = 17R/105
Now according to the question
5R/21 - 17R/105 = 600
? R(25 - 17)/105 = 600
? R = (600 x 105)/8
= ? 7875
Speed of train in m/sec = 72 km/hr x 5/18 = 20 m/sec
Distance covered in meter = Time taken in crossing the platform in sec x speed in meter = 30 sec x 20 m/sec = 600m
Now, we know , Distance covered by train in crossing the platform = Length of train + Length of Platform
Therefore, length of the train = 600m - 400m = 200m
Let the number of wickets taken before the last match = y
Bowler's overall average = total run scored / total wicket taken by bowler
Then, (12.4y + 26 ) / (y + 5) = 12
? x = 85
Then, | 120 | + | 480 | = 8 ⟹ | 1 | + | 4 | = | 1 | ....(i) |
x | y | x | y | 15 |
And, | 200 | + | 400 | = | 25 | ⟹ | 1 | + | 2 | = | 1 | ....(ii) |
x | y | 3 | x | y | 24 |
Solving (i) and (ii), we get: x = 60 and y = 80.
∴ Ratio of speeds = 60 : 80 = 3 : 4.
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