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A non stop bus to Amritsar overtakes an auto also moving towards Amritsar at 10 am. The bus reaches Amritsar at 12.30 pm and starts on the return journey after 1 hr. On the way back it meets the auto at 2 pm. At what time the auto will reach Amritsar? a) 2.30pm b) 3.00pm c) 3.15pm d) 3.30pm
ravi
- Jun 13th, 2007
Its simple, It takes one hour to reach and hence 2 + 1 is 3. Its meets the auto by 2 so it takes one more hour to reach hence 3 is the answer.
shivakumar007
- Aug 18th, 2006
(Sorry for the poor formatting of my First Reply)Reaches by 3:00 pmSolution:(I am not an expert)let's split up time in unit's of 0.5 hrs (ie) 1/2 hoursbus & auto met at a point Awe i''ll take dist...
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A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 70 characters. The % reduction in the number of sheets is closest to a) 20 b) 5 c) 30 d) 35
Navins
- Sep 6th, 2007
20 sheets 55 lines 65 characters?x?sheets?65 lines?70 characters=> x=(20*65*70)/65*55=25.45
sultan
- Jul 25th, 2006
20% is d ans20*55*65 = 71500 charsLet the number of sheets required be x here.Hence, x*65*70 = 71500 charsor x = 71500/65*70 = 15.71 = 16 sheets(approx)hence % reduction in sheets = 20%
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A ship started from a port and moving with I mph and another ship started from L and moving with H mph. At which place these two ships meet.
Between I and J and close to J
shikhi
- Jul 23rd, 2007
From the port to the perticular distance is L.
So from simple maths, You can say that if?(I+H) mph distance can cover in 1 hour, then l distace cancover by L/(I+H) hours. -
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In a mixture R is 2 parts, S is 1 part in order to make S to 25% of the mixture, how much R is to be added
½ part of R
santosh
- Sep 29th, 2007
Initial r=2s=1%s=(1/3)100=33.3%r=66.6to make %s=25,33.3-25=8.3 reduce%r=66.6+8.3=74.9=7575=(x)100=>x=75/100=3/4it means it is 3rd part of 4=>1 part is added
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A bus started from bus stand 8.00am and after 30min staying at destination it returned back to the bus stand. The destination is 27 miles from the bus stand the speed of the bus 50percent fast speed at what time it returns to the bus stand.
1.00am
santosh vr
- Sep 29th, 2007
Dist=27 Let ongoing speed=27m/hrtime=1hrit reaches at 9:00waits there upto 9:30dist=27speed=150%of 27=54time=27/54 =1/2hr=30minit reaches home at 10:00
Rameshsam
- Sep 6th, 2007
Hi Eswara,As you denoted if you take the intial speed as 18 means you will get an and is 1:00 pm..But you noted as 11:00 am., Can you explain me how you?got that?
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A traveler walks a certain distance. Had he gone half a kilometer an hour faster, he would have walked it in 4/5 of the time, and had he gone half a Kilometer an hour slower, he would have walked 2 ½ hr longer. What is the distance? a) 10 Km b) 15 Km c) 20 Km d) Data Insufficient
santosh vr
- Sep 29th, 2007
Let speed=stime=tspeed1=s+0.5time1=(4/5)tdist=(s+0.5)t *(0.8)speed2=s-0.5time2=t+2.5dist=(s-0.5)(t+2.5)st+2.5s-0.5t-1.25=st 2.5s-0.5t=1.25---------(1) 0.8st +0.4t=st =>0.2st=o.4t=>s=2---------(2) sub s=2
=>t=4.75/0.5=9.5dist=st=19ksrsbabu1
- Jul 22nd, 2006
since the distance travelled is same st=(s+0.5)(4/5)ton solvin u will get s=2there is one more equationst=(s-0.5)(t+5/2)substitute s=2 in the above eq and find 't' u will get s=2,t=7.5 therefore d=st=15kmans:15km
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