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
In 3 hours thinker candle becomes half
In the same time thinner candle becomes one fourth. Now the
required condition is satisfied. So 3 Hours
A. Rs 140
7+8+11=26
B: 7/26*520= 140
The first number is 36
Step-by-step explanation:
let their common factor be ‘x’
hence,
the ratio be 3x/5x
hence, by condition;
(3x-9)/(5x-9)=9/17 ch upu
9(5x-9)=17(3x-9)
45x-81=51x-153
153-81=51x-45x
72=6x
x=72/6
x=12
the first number is 3x=3×12=36
2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
A parabola is obtained as the intersection of a cone with a
plane
9 years
Bcz 3×3=9
5×3=15
7×3=21
Total = 45
Avg of this 45/3 = 15
let the no be x
then given that the difference between the number and 3/5th
of the number is 50
therefor , x-3/5x=50
2x=250
x=125 is the answer
6,36
Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
19=
13
240+x
⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.
11
to reach land
first time it travels 8 mtrs
next time (half=4) 4+4 mtrs (up&down)
next time (half=4) 2+2 mtrs (up&down)
next time (half=4) 1+1 mtrs (up&down)
and so on.
Altogether,close to 24 mtrs.
So we consider the 2nd statement first. We can form an equation out of it.
14x-6=13y+3=9z+3
Using this, we can understand that the multiple of 14 and the multiple of 13 and 9 must have a difference of 9. The easiest way to ensure that is multiplying it by 9
14*9=126
13*9=117
If the 5th farmer gives 3 apples to the 4th farmer, they would have 123 and 120 apples respectively. However, we also know that the 2nd farmer has 117 apples (13*9=117, and this is a multiple of 9) if the 5th farmer gives 3 apples too the 2nd farmer, the 3rd, 4th and 5th farmers would have 120 apples each.
Now that we got 120, we should check if the first part of the question makes sense along with it. The equation would be
7a+1=11b-1=120
We know that 11*11=121 and 7*17=119. When we add 1 to 119 and subtract 1 from 121, we get 120 for each. In this way, all the farmers have 120 apples each.
Therefore, the 3rd farmer had a yield of 11 per tree and the 4th farmer had a yield of 9 per tree.
325
30/100*142.85 = approx. 100
so ans is 42.85
10 x 27=270