Answer is 400.(14^2=196, 16^2=256, 18^2=324, 20^2=400.)
time is inversely proportional to speed.
So if speed taken is 4/5 of usual speed, time taken will be 5/4 of usual time
and the difference between time is 10 min
So,
(5/4)t -t=10 min
usual time,t=40 min
late time=40+10=50 min
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.
In 30 mins- Thief – 20km, Owner – 0km
In 1hr – Thief – 40km, Owner – 25km
In 1.5hrs – Thief – 60km, Owner – 50km
In 2hrs – Thief – 80km, Owner – 75km
In 2.5hrs – Thief – 100km, Owner – 100km
Answer: In 2.5 hours the owner would have overtaken the Theif
30/100*142.85 = approx. 100
so ans is 42.85
The formula to find number of diagonals (D) given total number of vertices or sides (N) is
N * (N – 3)
D = ———–
2
Using the formula, we get
1325 * 2 = N * (N – 3)
N2 – 3N – 2650 = 0
Solving the quadratic equation, we get N = 53 or -50
It is obvious that answer is 53 as number of vertices can not be negative.
Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on……
Hence the series is 0, 0, 0, 2, 5, 9, 14, …….. (as diagram with 1,2 or 3 vertices will have 0 diagonals).
Using the series one can arrive to the formula given above.
24 days
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
=>800+x=653∗60=>800+x=20∗65=1300=>x=1300−800=500
Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:
D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins
117000
5! = 120
A