let the 4 digit number be ABCD.
First Digit is A :
Therefore; according to the question
A=B/3
B=3A
C=A+B=A+3A=4A
D=3B=3(3A)=9A
Since the last Digit is D and it can neither be double-digit nor 0 ;
Therefore ;
A=1;
B=3A=3(1)=3
C=A+B=1+3=4
D=9A=9(1)=9
Therefore, the 4 digit number is 1349.
X filling rate is 1/18 or 4/72 per hour
Y filling rate is 1/24 or 3/72 per hour
Combined as a pair it’s 7/72 in a complete 2 hour alternating sequence
72+ 7/72 =10 sequences = 70/72 on 20 hours, The outstanding balance is just 2/72
So at 20.5 hours with X turn filling is 100% complete with 72/72
Answer: It will take 20.5 hours to completely fill this tank.
Analysis of measured filling process, from X = 42/72 share & from Y = 30/72 share
R
T,P,R,Q,S
At 680Km
Y R G
B B G
Y G Y
16
Let n be the number of days it takes A and B, working together, to finish. And we know B=A+10 and B=3A, so:
3A=A+10
2A=10
A=5
Then B=15
So:
1/A + 1/B=1/n where n is the total amount of days. So:
1/5 + 1/15=1/n
3n+n=15
n=15/4 days
C. Monday
A year has 52 weeks and every week has 7 days.
We just add up one day to the present day of the year and that would automatically act as the day of the next year.
In case of leap years in between, one extra day gets added in the entire year making it 366 days instead of 365 days in an ordinary year and we add two days in that case.
Hint: Assume that the speed of the stream is x and the speed of the boat in still water is x. From the statement of the question form two equations in two variables x and y. This system is reducible to linear equations in two variables. Reduce the system to a system of linear equations in two variables by proper substitutions. Solve the system of equations using any one of the methods like Substitution method, elimination method, graphical method or using matrices. Hence find the value of x satisfying both the equation. The value of x will be the speed of the stream.
Complete step-by-step answer:
Let the speed of the stream be x, and the speed of the boat in still water be y.
We have the speed of the boat upstream = y-x.
Speed of the boat downstream = y+ x.
Now since it takes 14 hours to reach a place at a distance of 48 km and come back, we have the sum of the times taken to reach the place downstream and time taken to return back upstream is equal to 14.
Now, we know that time =Distance speed
Using, we get
Time taken to reach the place =48y+x and the time taken to return back =48y−x.
Hence, we have
48y+x+48y−x=14
Dividing both sides by 2, we get
24y+x+24y−x=7 —–(i)
Also, the time taken to cover 4km downstream is equal to the time taken to cover 3km upstream.
Hence, we have 4y+x=3y−x
Transposing the term on RHS to LHS, we get
4y+x−3y−x=0 ——– (ii)
Put 1y+x=t and 1y−x=u, we have
24t+24u=7 ——-(iii)4t−3u=0 ——–(iv)
Multiplying equation (iv) by 6 and subtracting from equation (iii), we get
24t−24t+24u+18u=7⇒42u=7
Dividing both sides by 42, we get
u=742=16
Substituting the value of u in equation (iv), we get
4t−3(16)=0⇒4t−12=0
Adding 12 on both sides, we get
4t=12
Dividing both sides by 4, we get
t=18
Reverting to original variables, we have
1y+x=18 and 1y−x=16
Taking reciprocals on both sides in both equations, we have
y+ x=8 ——- (v)y−x=6 ——–(vi)
Adding equation (v) and equation (vi), we get
2y=14
Dividing both sides by 2, we get
y=7.
Substituting the value of y in equation (v), we get
7+x=8
Subtracting 7 from both sides we get
x = 8-7 =1
Hence the speed of the stream is 1 km/hr.
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
Burn Rope One first, it will burn in 30 mins.Mark 30 mins and at the end of 30 mins now burn second rope from both sides (15 mins it will take ),and simultaneously burn third and last rope until it is half left.30+15= 45 , Remaining time 7.5 mins.By this time last rope is burnt half ,and now burn it from both sides as well it will exhaust in 7.5 mins. 30+15+7.5=52.5 measured.
No of men employed = N
No of days to finish the work = 9 days
No of men after increase = (N + 10)
No of days to finish the work = 6 days
Equating mandays
9N = (N+10)*6
9N — 6N = 60
3N = 60
N = 20
No of men employed = 20
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds