7 : 3
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
find out what is the problem and taking the bad out always works.
Cousin
20- buy 15 books
15- buy 20 books
Original price is 20.
6/5
3000
22.5
If a blue stone is thrown into a red sea, several things could happen depending on the context and the properties of the stone and the sea:
Symbolically: Since blue and red are contrasting colors, the interaction of a blue stone in a red sea could be seen as a visual or metaphorical contrast. It could represent a stark difference or an unexpected element introduced into an existing situation.
Scientifically: In reality, the color of the stone and the sea would not have a direct physical impact on each other. The stone would sink or float based on its density and the water’s buoyancy. The color of the water, whether red or any other color, does not change the fundamental principles of objects interacting with liquids.
It’s important to note that red seas, in the context of bodies of water, typically do not exist naturally. The phrase “red sea” is often used metaphorically or symbolically rather than referring to an actual body of water with a red color.
38 years
Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.
x = 65(x−8)
⇒5x=6x−48
⇒x=48 years
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years
∴ Rajan’s sister’s present age = (30 + 8) years = 38 years
30/100*142.85 = approx. 100
so ans is 42.85
a:37
b:22
c:50
d:11
e:25
f:27
Yes, it is possible by using a HDMI cable
A = 20k x 2y, B = 15k x 2y, C = 20k x 1.5y => 100Ky
B contribution = (15k x 2y) /100ky
B will get 0.3 from 25k which is 7.5k = 7500
24 times