c
Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:
Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.
Now you have 4 litres of water in jug A.
let no of boys in group is = x
then total sum = 30 * x
after joining one more boys with a weight of 35 kg the total sum is = 30x + 35
after joining the new student the weight will increase 1 kg
so total sum
30x + 35 = 31(x+1)
x = 4
Ans is 2(Prime numbers)
Sum of 5 consecutive nos is 35
so X + (X+1) + (X+2) + (X+3)+ (X+4) = 35
5X + 10 = 35
X = 5
So the 5 consecutive numbers are : 5, 6, 7, 8, 9
The Prime numbers are 5 and 7
2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
Which means the train will travel 180 m in 6 seconds so
We need to convert this into minutes
6×10 = 60s= 1 minute.
180×10 = 1800 m train will travel 1.8 km in 1 minute.
1.8×60 = 108 kmph
Answer is 274 .
Let the required term be x.
Pattern :- 1) Find the difference between consecutive terms as :
2 – 1 = 1
4 – 2 = 2
13 – 4 = 9
31 – 13 = 18
112 – 31 = 81
2) Now write the above numbers obtained by differences in series as :
1, 2, 9, 18, 81, y
(Suppose y be next term after 81.)
3) Look the difference series carefully and it follows the below pattern as :
(9 * 2)/1 = 18
(18 * 9)/2 = 81
(81 * 18)/9 = 162 = y (the next term after 81)
Thus the required answer is :
x = 112 + y
x = 112 +162 =274
C. 20
150m
4
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
450÷100×30=135
450-135=315
Ans: 315
C