8 is the answer
b’coz no two have the same age
8+6+5+4+3=26
a) who started with small amount of money?
b) who started with greatest amount of money?
c) what amount did B have?
status after 3 games
A-40
B-40
c-80
d-16
status after 2 games
A-20
B-20
C-40
D-96
status after 1 game
A-10
B-10
C-108
D-48
status after 0 games
A-5
B-93
C-54
D-24
so answers are
a)A
b)B
c)93
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
The number of ways of selecting a group of eight is
5 men and 3 women=5C5×6C3 =20
4 men and 4 women=5C4×6C4 =75
3 men and 5 women=5C3×6C5=60
2 men and 6 women=5C2×6C6=10
Thus the total possible cases is 20+75+60+10=165.
This is more logical …
Let the 1st flag 1 placed at the origin ….
in crossing 8 flags he traveled 7 distances….
s=d/t
=7/8
time for 4 flags t=(d/s)=4/(7/8)=(4*8)/7=4.5714285714285714285714285714286
saying 20%alcohol in 15 litre solution.
It means (20/100)*15=3 litres
Now 5 litre of water is added not alcohol.
So it means only the volume of solution is increases by 5 not alcohol.
It means 15+5=20
Now % of alcohol is
(3/20)*100=15%.
15/46
Lets assume total LCM(5,8) = 40units.
As, 5 men or 8 women do equal amount of work in a day,
1 Man does 8units/day and 1 Woman does 5units/day.
3M and 5W in 10 days do (3*8 + 5*5)*10 = 490units
To do 490 units in 14 days, number of Women required = 490/(14*5) = 7
10%=394;50%=1970;250%=9850;30%=1182;
250%+30%=9850+1182=11032