Question

How many 3-digit even numbers can you form such that if one of the digits is 5, the following digit must be 7?
(a) 5
(b) 405
(c) 365
(d) 495
Please explain the answer in A very easy manner. And if bored of studies, go visit SSA MUSIC POINT, only on YouTube for exclusive rap songs.​

Answer 1

Answer:

Firstly, we must consider that an even number is divisible by 2 and has 0, 2, 4, 6 or 8 as its ones digit.

Now, the required number would be either ABC, 57C or A57 (because 7 must fall after 5).

So, out of the 900 three digit numbers, 450 even numbers, eliminate the A5C(A57) types: 150, 152, 154, 156, 158, 250, 252, 254, 256, 258, 350, 352, 354, 356, 358, 450, 452, 454, 456, 458, 650, 652, 654, 656, 658, 750, 752, 754, 756, 758, 850, 852, 854, 856, 858, 950, 952, 954, 956, 958 [40 numbers]

This has been done because in such form of number where 5 lies in middle, the next (ones) digit would be 7, which won't be an even number.

Now, eliminate all even numbers of the 5BC type, which are 50 in number and add only the 57C numbers: 570, 572, 574, 576, 578 (5 numbers).

So total number of three-digit when numbers, in which if 5 appears is followed by 7, is

450-40-50+5

=365

Step-by-step explanation:

answer is 365

mark me brainlest