Ques:- A starts from a place at 11.00 am and travels at a speed of 4 kmph , B starts at 1.00 pm and travels with speeds of 1 kmph for 1 hr , 2 kmph for the next hr , 3 kmph for the next hr and so on. At wht time will B catch up with A ?
Recent Answer : Added by DK BOSS On 2021-07-17 06:18:40:
the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;
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Recent Answer : Added by padmavathy On 2022-08-14 16:42:03:
8
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A. Rs. 9432
B. Rs. 8568
C. Rs. 8712
D. Rs. 8352
Recent Answer : Added by narendra On 2022-09-18 09:03:23:
c
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Recent Answer : Added by Rama On 2022-08-14 16:28:56:
198.20
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Recent Answer : Added by Admin On 2020-05-17 12:00:01:
75%
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Recent Answer : Added by Admin On 2020-05-17 11:59:36:
let original radius=100
original surface area lets say it x =4 * (22/7) * 100 *100
new radius=100 + 50% of 100=150
new surface area lets say it y = 4 * (22/7)* 150* 150
increase in surface area=((y-x)/x) * 100
i.e.
((4*(22/7)*150*150 – 4*(22/7)*100 *100) / (4*(22/7)*100*100))*100
=125
so the answer is 125%.
hope this helps you.
Ques:- A bus started from the bus stand at 8 Am and after staying 30 minutes at a destination return back to the bus stand. The Destination is 27 miles from the bus stand . The Speed of the bus is 18mph . In the return journey the bus travells with 50% fast speed.At what time it is return to the bus stand ?
Recent Answer : Added by Admin On 2020-05-17 12:00:34:
Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
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Recent Answer : Added by Mayuri On 2021-07-08 16:28:56:
the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;