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A frog falls into a well and is trying to escape. Each day he is able to climb up 10% of the total night in the morning, but at night he falls 5% of the total height. When will the frog be able to come out?
(a) 10th day
(b) 11th day
(c) 19th day
(d) 20th day
Read Solution (Total 4)
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- (c) 19th day
Let the total height be 'x'
Resultant climb in a day= 0.1x - 0.05x= 0.05x
Upto 18 days, total climbing=18*0.05x=0.9x
So on 19th day, frog will cover 0.9x+0.1x= x and will come out - 8 years agoHelpfull: Yes(5) No(0)
- 19 th day.
Because it every day it gains 5 % .....After 18 days it will be at 90 %.....
On 19th morning it Reaches 100 %.. - 8 years agoHelpfull: Yes(2) No(0)
- 19
10% forward and 5% backward then the net movement is 10-5 = 5%
frog climbs 90% in 18 days (18 * 5%)
so on the next day he will climb 10% to make it 100% - 6 years agoHelpfull: Yes(0) No(0)
- Let the total height of the well be x.
The frog climbs 10% of the height each day.
10% of x = x/10.
The frog slips 5% of the height in the night.
5% of x/10 = x/200.
Percentage of the distance covered by the frog in one day = x/20-x/200 = 19x/200.
1 day - 19x/200
? - x
? * 19x/200= x
? = 200/19 = 10.5
Therefore, the frog will be out of the well on the 11th day. - 3 years agoHelpfull: Yes(0) No(0)
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