Let the speed of first lady be (v + 2)km/h and speed of second lady will be v km/h.
Total distance = 24 km
For first lady, Speed = Distance/Time
v + 2 = 24/t ...(i)
Let time taken by first lady to cover distance of 24 km be t h.
Then, time taken by second lady to cover same distance will be(t + 1)h.
For second lady, Speed =Distance/time
v = 24/(t + 1) ...(ii)
Form Eqs.(i) and (ii), we get
24/t - 2 = 24/(t + 1)
? (24 - 2t)/t = 24/(t + 1)
? (24 - 2t)(t + 1) = 24t
? 24t - 2t2 + 24 - 2t = 24t
? 2t2 + 2t - 24 = 0
? t2 + t - 12 = 0
? t2 + 4t - 3t - 12 = 0
? t(t + 4) -3(t + 4) = 0
? (t - 3)(t + 4) = 0
? t = 3, -4
? t = 3 (t ?- 4)
? The distance travelled by the first lady in one hour
= 24/t = 24/3 = 8 km
And distance travelled by the second lady in one hour
= 24/t + 1
= 24/3 + 1 = 6 km
? 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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