Example 1:
Assign, A=20, B=10, C=5, D=5(Because C is equal to D as
given), E=1.
A/B = 20/10 = 2. So A/B = 2
A/C = 20/5 = 4. So A/C = 4
A/E = 20/1 = 20. So A/E = 20
Therefore “A/E is Greatest”
Example 2:
Assign, A=100, B=50, C=20, D=20(Because C is equal to D as
given), E=10.
A/B = 100/50 = 2. So A/B = 2
A/C = 100/20 = 5. So A/C = 5
A/E = 100/10 = 10. So A/E = 10
Therefore “A/E is Greatest”
(10×1000) / (2×3.1416×1.75) = 909 times (about)
cousin, daughter of maternal uncle
I am Srilaxmi. I born and bought-up in WARANGAL. My father Agriculture cum politician and my Mom homemaker. I blessed with three brothers and one sister. My elder brother was married and working as a TA in MPDO office at Warangal. My first younger brother was also married and working with Sushee Infra Pvt Ltd as a Asset Manager at Hyderabad. Younger brother and sister are into their studies. Coming to me I completed my MBA from ICFAI university in 2008 and worked as a Audit assistant in accounting firm for two years. I had actively participated in bank audits i.e. is in CBI (Concurrent audit) and SBI(Statutory audit) during my services. Later I pursued CS Executive and attempted for five times then I give up to clear my M.Com. At present I was studying postal studies of CS Executive programme. My hobbies are reading books and listening to music.
6
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
740
Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.
Dedication ,determination and discipline will be the great asset of company.
125
12.9
2+3+5+7+11+13+17+19+23+29=129
129/10=12.9