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
45
Slice the cake
1
40
40
30
60
– 10 cows
+ 20 days
if 40 cows can eat for forty days
then 20 cows can eat for 80 days
Ans- 6,6 litres
Let x Litres from can1 (where, water = x/4 litres and milk = 3x/4 litres)
Now
12-x liters from can 2 ( where, water = 6-x/2, milk = 6-x/2)
Now given
water/milk = 3/5
0.25x+6-0.5x / 0.75 +6-0.5x = 3/5
6-0.25x/6+0.25 = 3/5
Solve it you get
x = 6 litres from can1
Then 12-x= 12-6=6 litres from can2
Actual price is RS.60 and (plus)he got profit of 10%. So 60+ 10%(60) = 66
if A is true B has to be true
s=36kmph
in meter per second is=36*5/18=>10
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
tan 30=h/x
tan 15=h/100+x
so h/tan 30=(h/tan 15)-100
=33.33app
ans: 250 rs exactly
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.