Let required amount of coal be y metric tonnes
9 engines consumes 24 metric tonnes of coal in 8 hours a day.
More engines, more amount of coal (direct proportion)
If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)
More hours, more amount of coal(direct proportion)
Now we have , 9 × ( 1 / 3 ) × 8 × y = 8 × ( 1 / 4 ) × 13 × 24
? 3 × 8 × y = 8 × 6 × 13
? 3 × y = 6 × 13 ? y = 2 x 13 = 26 metric tonnes
? Speed = Distance / Time = (300/15)
= 20 m/sec
= (20 x 18) / 5 = 72 km/hr
Ratio of the areas = area of original square / area of new square
= [ d2 / 2 ] / [ (2d)2 / 2 ] = 1/4
? New area becomes 4 fold.
So, 24 is wrong, it should be 8 (48/6 = 8).
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.
Let the number be x,
Then, 4x/5 - 2x/3 = 8
? (12x - 10x) / 15 = 8
? 2x = 120
? x = 60
39°
Share of Rakesh : Share of Dinesh : Share of Mahesh
= 5000 : 8000 : 12000 = 5 : 8 : 12
Total earned profit = ? 12500
? Share of Dinesh in profit
= [8/(5 + 8 + 12)] x 12500 = (8/25) x 12500
= 8 x 500 = ? 4000
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴ Required number = (90 x 4) + 4 = 364.
⟹ log (33 ) = 1.431
⟹ 3 log 3 = 1.431
⟹ log 3 = 0.477
∴ log 9 = log(32 ) = 2 log 3 = (2 x 0.477) = 0.954.
∴ Required number of ways | = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) | |||||||||||||
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= (45 + 18 + 1) | ||||||||||||||
= 64. |
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