C. 21000
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
444km
Since the speed diff is 5km/h
It takes 12 hrs to cover 60 extra kms
Therefore in 12 hrs the total dist covered is
Relative speed×time=(21+16)×12=444 kms
8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8
B
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 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
True
C
1. How many such numbers can be formed ?
2. How many such numbers are divisible by 4 ?
a printer produced 176400 lines in 420 minutes
it produces 176400/420 lines per minute = 420
answer is : 420 lines per minutes
w(drown)