Dice
O
a)0
b)206
c)0
d)250
e)39
f)92
P Q R S
TP -> UQ -> RV -> SW
Answers-
U sits opposite to S
Q sits in between S and R (after interchange)
Rs. 1500
Phillips
To solve this problem, we can break it down into steps:
Step 1: Determine the individual rates of work for A, B, and C.
If A needs 8 days to finish the task, then their work rate is 1/8 of the task per day.
If B needs 12 days to finish the task, then their work rate is 1/12 of the task per day.
If C needs 16 days to finish the task, then their work rate is 1/16 of the task per day.
Step 2: Calculate the combined work rate of A and B.
If A works for 2 days, their contribution will be 2 * (1/8) = 1/4 of the task completed.
If B works until 25% of the job is left for C, then they will complete 75% of the task.
Step 3: Calculate the time it takes for B to complete 75% of the task.
Since B’s work rate is 1/12 of the task per day, it will take B (75%)/(1/12) = 9 days to complete 75% of the task.
Step 4: Calculate the remaining work for C.
If B completes 75% of the task, then the remaining work for C is 100% – 75% = 25% of the task.
Step 5: Calculate the time it takes for C to complete the remaining work.
Since C’s work rate is 1/16 of the task per day, it will take C (25%)/(1/16) = 4 days to complete the remaining 25% of the task.
Step 6: Calculate the total time required.
A worked for 2 days, B worked for 9 days, and C worked for 4 days, totaling 2 + 9 + 4 = 15 days.
Therefore, it will take a total of 15 days for A to work for 2 days, B to work until 25% of the job is left, and C to complete the remaining work.
a/g
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
If the bell rung when they r started is also counted then it
will be 5 otherwise 4
( d ) Cannot be determined
a
The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.
Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?