1 A B C together can do a piece of work in 10 days All the three started workingat it

Ans1 (A+B+C) do 1 work in 10 days.
So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5
Remaining work=1-2/5=3/5
(B+C) take 10 more days to complete 3/5 work.
So( B+C)'s 1 day work=3/50
Now A'S 1 day work=(A+B+C)'s 1 day work-(B+C)'s 1 day work=1/10-3/50=1/25
A does 1/25 work in in 1 day
Therefore 1 work in 25 days.(Ans)
8 years agoHelpfull: Yes(35) No(2)

2) d=s*t
45mins=3/4 hr and 48mins=4/5 hr
s*(3/4)=(s-5)*(4/5)
3s/4=(4s-20)/5
15s=16s-80
s=80 km.
8 years agoHelpfull: Yes(27) No(16)
S*T1=(S-5)8T2
S*45=(S-5)48 SO S=80 PUT DIS IN FST EQN N DIVIDE BY 60
SO THE ANS IS 60..
9 years agoHelpfull: Yes(15) No(3)
2. 60 k.m will be the ans .make a eqn with previous speed*time takin in hour = (speed-5)*time taking in hour .,.the eqn was posssible as the ditance it is going to n fro is same
9 years agoHelpfull: Yes(14) No(6)
s1/s2=t2/t1;
s1/s1-5=48/45
from this equn s1=80kmph
s=d/t
d=s*t
=80*45/60=60km
8 years agoHelpfull: Yes(9) No(1)
1) 25 days
8 years agoHelpfull: Yes(4) No(1)
d=s*t
d= s1*(45/60) that is s1*(3/4)
d= s2*(48/60) that is s2*(4/5)
s1*(3/4)=s2*(4/5)
s1*(3/4)=(s1-5)*(4/5) that is speed reduced by 5kmph
s1=80 kmph
d=s.t
d=80*(3/4)= 60 km
8 years agoHelpfull: Yes(3) No(0)
S * (45/60) = (S-5) * (48/60)
on simplifying s=80 kmphr i.e speed
distance = 80*(45/60) or 75*(48/60) = 60km
8 years agoHelpfull: Yes(2) No(0)
(1) 25 days
(2) 80km
8 years agoHelpfull: Yes(1) No(1)
2) s=d/t
=> (s1/s2)=(d1/d2)*(t2/t1)
=>( x / (x-5) )=(d / d) * ( 48 / 45 )
=>( 4x-20 ) *4 = 15x
=>x=80
6 years agoHelpfull: Yes(1) No(0)
speed(s)=distance(d)/time(t)...............here distance constant
s*t=constant
s1*t1=s2*t2
speed time
x 45
x-5 48
x*45=(x-5)*48

3x=240

x=80km
8 years agoHelpfull: Yes(0) No(5)
25 DAYS WOULD BE THE ANSWER
6 years agoHelpfull: Yes(0) No(0)
1. A+B+C's 1 day work 1/10
B+C's 1 day work=3/50
A's 1 day work=1/10-3/50=1/25
A can complete work in 25 days
2.
60x/45=60x/48+5
4x/3=5x/4+5
16x=15x+60
x=60 km
6 years agoHelpfull: Yes(0) No(0)
given 1/a+1/b+1/c=1/10
so 1/b+1/c=1/10-1/a
now a work for 4 days and b nd c work for 14 days now
4/a+14/b+14/c=1
4/a+14(1/10-1/a) =1
by solving it we got a=25 days
6 years agoHelpfull: Yes(0) No(0)
1
A+B+C CAN COMPLETE IN 10 DAYS
LET'S TAKE WORK TO BE 10 UNIT THEN THEY TOGHTER COMPLETE 1 UNIT A DAY
IN 4 DAYS THEY COMPLETE 4 UNIT I.E 6 UNIT IS REMAINING . NOW B+C CAN COMPLETE THIS 6 UNIT IN 10 DAYS i.e 6/10 UNIT A DAY
A'S 1 DAY WORK IS 1-6/10=4/10 unit
HENCE A WILL COMPLETE WORK IN = TOTAL WORK/ONE DAY WORK
=10unit/4/10unit/DAY
=10*10/4
=25 DAYS
6 years agoHelpfull: Yes(0) No(0)
2. speed=distance/time
lets assume distance=x;
change the time in hrs=t1=(45/60)=3/4,t2=(48/60)=4/5
speed of train reduced=distance/t1-distance/t2=(how much speed reduced)=5
x/t1-x/t2=5
4x/3-5x/4=5 do LCM
16x-15x=60
x=60
4 years agoHelpfull: Yes(0) No(0)

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