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
Take one screw out from the rest of the three tyres and put the screws in the fourth tyre. Now all tyres will be held by three screws each.
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
5 (3+2)
if he wins then it becomes 4 double than 2
and if looses then opponent becomes 3 which is equal.
4, 5, 15, 49, 201, 1011, 6073
Correct option is A)
2nd term =(1stterm×1+2)=(4×1+2)=6
3rd term =(2ndterm×2+3)=(6×2+3)=15
4th term =(3rdterm×3+4)=(15×3+4)=49
5th term =(4thterm×4+5)=(49×4+5)=201 and so on
∴5 is wrong.
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
on the earth
a = 2
d = 5 – 2 = 3
Then 12th term is :
This in arithmetic progression
L = a + (12 – 1) d
L = 2 + 11 * 3
L = 2 + 33 = 35
R>P>Q>T
V>S>P
so the lightest is T.
b