TCS Company Numerical Ability Algebra

A bus starts at 6:00 pm. from starting point at d speed if 18m/s. reached its destination. Waited for
40minutes. And again returned back at the speed of 28m/s. Find the time to reach the destination ?

Read Solution (Total 2)

7:52 pm as { s=d/t ;; 18=d/x=>d=18x ;; s=d/t ;; 28=d/(x-40)=>d=28(x-40) ;;
18x=28(x-40)
18 x = 28 x -28*40
28*40=28x-18x
1120=10x
112=x and distance is 18*112=2016 or 28(112-40)=2016
so after 112 minutes it reaches the destination means = 6:00 + 112 min => 7:52 min }
9 years agoHelpfull: Yes(15) No(8)

7.12pm will b the ans

s1/s2=t1/t2
or, 18/28=t2/t2+40*60
t2=4320 sec
and 4320 sec = 72 mins

72 mins after 6.00 pm will b 7:12pm
9 years agoHelpfull: Yes(0) No(4)

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