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Rs.265.80
450÷100×30=135
450-135=315
Ans: 315
Ans : – 215
Explanation :-
1^3 = 1-1=0
2^3 = 8-1 = 7
3^3 = 27-1=26
4^3 = 64 – 1 = 63
5^3 = 125-1 = 124
Hence, Ans is
6^3 = 216-1=215
the answer is (a)40
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
Creating a virtual image of the search topic
sample space=36
prob of one of the dice getting face 6
i.e n={(1,6)(2,6)(3,6)(4,6)(5,6)(6,1)(6,2)(6,3)(6,4)
(6,5)(6,6)}=11
p=n/s
p=11/36
ans:11/36
28
to arrange m objects in n places => nCm
i.e. for this example m=2(+,- sign) and n = 8(places between two number from 1 to 9)
SO, answer is : 8C2 = (8*7) / 2 = 28
60 more men
21
72
CPNCBZ