Lessonly Interview Questions, Process, and Tips

Let the total rotations be x:
So the distance covered by both the wheels will be same,
9 * x = 7 * (x + 10)
x = 35 rotations
distance travelled = 9 * 35 = 7 * (35 + 10) = 315 feet

5

1

125

.

I’m interested in this job because I can see that, in this role, my skills could help solve this problem within your company. I also see an opportunity for me to learn and grow these skills, so we both would benefit personally, professionally, and financially.

A. 80 : 81 : 84

The answer probably lies in finding the in-centre of the
traingle. Bisect all the angles of the triangles and the
point where these angle bisectors meet gives u the point P

The probability will be
white ball = 5/9
red ball = 4/9
The answer can be 5/9 and 4/9 or 1

1500

According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300​−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.

Kerala

3×8=24=2xt ===> t=12hours

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros