A. 978626
Percentage of people passed = (passed in eng) + (passed in math) – ( passed in both)
Which is 90% of students passed and 10% failed
10% of X = 40
X = 400
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
x+y+z
—– = 6800
3
x>=6400
6400 + y + z = 20400
y + z = 14000
to get the greatest of y and z, lets assume y = 6400
so, z = 7600
so ANS is 7600
Collective performance
Postive attitude towards working
Shaping team player
Turning individuals into team players
11 14
12 15
13 16
=======
14 17
the answers is 14
21
40
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
512
Rs. 1500
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
let the for digit number be = pqrs
p=q/3 ==>q = 3p
r=p+q=p+3p =4p
s=3q = 3(3P)=9p
number:
p 3p 4p 9p
let p=1 answer is 1349
if it 2 answer 2 6 8 18
so it becomes five digit number so correct answer is 1349
area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2
In a cube all the diagonal and sides are equal, we can go diagonally.