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
Suppose they meet after y hours .
Then, 21y - 16y = 60
? y = 12
?required distance = (16 x 12 + 21 x 12) km.
= 444 km.
Let Reena?s present age be 'K' yrs.
Given:{(K+3) x 4} - {(K-3) x 3)}
=> K = 21 yrs
So her present age is 21 yrs.
Then, her age after 1 year is 22 yrs.
Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25 => x = 5
Hence, B's age = 2x = 10 years.
suppose the ages of Sunil and Anil are 8x yr and 7x yr, respectively.
After 6 yr, = 136x + 102 = 133x + 114
136x ? 133x = 114 ? 102
3x = 12 X = 12/3 = 4
Age of Sunil = 8x = 8 × 4 = 32 yr
Present age Anil = 7x = 7 × 4 = 28 yr
Required difference = (32 ? 28) = 4 yr
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