Post Name: Officer (Cash)
Previous Vacancy was: 2246
A motorist and cyclist starts from A to B at the same time where AB = 18km. The speed of motorist is 15kmph more than the cyclist. After covering half the distance the motorist rests for 30mins and there after his speed reduces by 20%. If the motorist reaches destination B 15min earlier than that of the cyclist. Then find the speed of the cyclist.
Answer: 12 km/hr
Full Solution:
Let, speed of cyclist = x km
So, speed of motorist= x+15 km
ATQ,
9/(x+15)+1/2+9/{0.8(x+15)}= 18/x-1/4
=> 9/(x+15)+45/4(x+15)-18/x= -1/4-1/2
=> (12x+60)/x(x+15)= 1
x= 12(ans)
A boat goes 30km upstream and 44km downstream in 10 hours.In 13 hours, it can go 40km upstream and 55km downstream. Determine the speed of boat in still water and stream.
Answer: 8 km/hrs
Full Solution:
30/x-y+44/x+y=10—–1
40/x-y+ 55/x+y=13——2
From 1 and 1
x=11
Y=8
Alternative Solution:
Let speed upstream be x km/hr and speed downstream be y km/hr.
1st condition, 30/x+44/y=10…….(i)
2nd condition 40/x+55/y=13…….(ii)
(i)*4-(ii)*3 we get,
Y=11
putting the value of y in(i) we get,
x=5
so speed in still water= 11+5/2 = 8 km/hr(answer)
A and B can do a piece of work in 30 days, while B and C can do the same work 24 days, and A and C can do it in 20 days. They work together for 10 days before B and C leave. How many days more will A take to finish the work?
Answer: 18 Days
Full Solution:
2(A+B+C)=1/30+1/24+1/20=1/8
So, (A+B+C)=1/16
Work efficiency of A= 1/16-1/24=1/48
ATQ,
10/16+x/48=1
x=18
So answer is 18 days
A solution of honey and water is 28 litres, with honey and water in ratio 4:3. To this a 21 litre honey-water solution is added that has honey to water ratio as 2:1. Again a 51 litre honey-water solution that has honey to water ratio as 9:8 is added to this. After this 10 litre of the solution is replaced with pure honey. What is ratio of water to honey in the final mixture?
Answer: 387: 613
Full Solution:
1st connotation, honey=28*4/7=16L
Water=12L
2nd connotation, Honey=21*2/3=14L
Water=7L
3rd connotation, honey=51*8/17=27L
Water=24L
New solution= 16+14+27:12:7:24=57:33
After replacement=(33-33*10)/90 //{(57-57*10/90+10)}=387/613
Math No: 05
A tank is filled by three pipes with uniform flow.The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by third pipe alone. the second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time is required by the first pipe is?
Answer: 15 hours
Full Solution:
Let, First tank needs =x hour
Second tank needs=y hours
Third tank needs=z hours
Atq 1,x+y=z
Atq 2,
Y=x-5
Y=z+4
x-5=z+4
=>x-z=9
=>x=9+z
Now
1/(9+z)+1/(z+4)=1/z
=>z=6.
Required, x=9+z=9+6=15hours (answer)
