say the work is w and let no of days taken by b is ‘x’ which
we have to calculate
so work done by a in one day is w/6
work done by b in one day is w/x
a and b together can do work in 4 days ie=(w/6)+(w/x)=(w/4)
solving equation x=12
so no of days taken by b=12
if A is true B has to be true
Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
36
8days
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
First nos series is 7,9,11,?
ie odd number siries ie 7,9,11,13
Second number series is 16,15,14
ie 1 less the previous number 16,15,14,13
Ans —-series is 7,16,9,15,11,14,13,13
I’m the capable of what u expect from me and you should say what is the tallent u have which is used for that company…🥰
132
1/2 Hr -> 80 Cheques
Per Hour -> 80 * 2 = 160
For 7 Hours -> 160 * 7 = 1120
For 7 1/2 Hour -> 1120 + 80 = 1200
clerk can process 1200 cheques in Sever & one half an hour day.
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Let n be the number of days it takes A and B, working together, to finish. And we know B=A+10 and B=3A, so:
3A=A+10
2A=10
A=5
Then B=15
So:
1/A + 1/B=1/n where n is the total amount of days. So:
1/5 + 1/15=1/n
3n+n=15
n=15/4 days
C will get RS 95