A thin lens of refractive index 1.5 has a focal length 15 cm in air. When the lens is placed in a medium of refractive index 4/3, what is the new focal length?
This is the correct answer.
Mathematics: If the cost of 18 Apple and 45 Mangoes is Rs. 981, then calculate the cost of 10 Apple and 25 Mangoes?
545 is answer!Take cost of one apple as x andCost of one mango is y18 x+90 y=891=>2x+5y=10910x+25y=?=>109×5=545
If A and B are sets,where [math]n(AUB)=36, n(A∩B)=16, n(A-B)=15[/math] then find n(B)?
The question has already been answered correctly. Let me try a different take on this.Being sure of whats what can help you figure this solution easily...:)n(AUB) means the the number of elements in both A & B. And that includes the ones in common too.n(A∩B) means the ones that in common.n(A-B) refers to the elements that belong to A alone. That means the elements in A excluding the common elements.n(B) is the number of elements that belong to B and that includes the common ones too.Now coming to your question.The total number of elements that can fit in both the sets is 36.While the number of common elements is 16.And also given that n(A-B)=15.Now according to the definitions above....n(A) should include the ones that belong exclusively to A alone and also the ones in common.So,n(A)=n(A-B)+n(A∩B) =15+16 =31Therefore, n(A) is 31.Now when n(AUB) is 36 and n(A) is 31, it is clear that the elements exclusive to B alone is 5, i.e.36-31. Now n(B)=n(A∩B)+n(B-A)Therefore, n(B) is 16+5 = 21.n(B)=21.
When eggs in basket were divided by 2, 3, 4, 5, and 6, there was always an egg left over. When taken out in groups of 7, the basket got emptied. How many eggs did it contain?
First of all, if the eggs left after divided by 7 is 0 then the number must be a multiple of 7.If divided by 2,3,4,5,6 one egg is left so we know that the actual number of eggs must have 1 or 6 at its ones place because then only it’ll give one after being divided by 5. But if it’ll have 6 at its ones place then it doesn’t leave one behind as a number with 6 at its ones place is divisible by 2. So the multiples of 7 having 6 at their ones place are eliminated.We’re just left with the multiples of 7 having 1 at ones place.So we’ll write down all the numbers having 1 at ones place.21, 91, 161, 231, 301…Now by Hit and Trial method we’ll check if they can be divided by those five numbers giving one. Remember, that we’ll move from small number to larger number when dividing.21 is a multiple of 3. So it doesn’t give 1 when divided. Thus, eliminated.91 gives 1 when divided with 2 and 3 but not with 4. Thus, eliminated.161 divides by 3 giving 2. So it is also eliminated.231 divides by 3. So it is eliminated.301 divides by 2 leaving 1 behind, with 3,4,5,6 and leaving 1 behind. So the required number is 301.So, the minimum number of eggs basket must contain in order to satisfy this need is 301.Thank You