You could just write out \(\frac{5n}{4}\) and then write out values using different values of n, until you found a remainder of 3. Then you could use that value of n to find A, and then compare it to B. Ex. if n = 3 then you have 15/4 which is equal to 3 with a remainder of 3 (mixed fraction of 3 and 3/4 expresses this). Then plug in 3 for 10n/4 in the second part and you get 30/4 = 7 R2 (same idea is expressed by mixed fraction 7 2/4). So, based on that the answer is C.

One might ask, is this always true? What about other values of n with a R2? It happens to be true, but my method doesn't show it. Brent's did. The method I used gives a reasonable guess, I think, of the right answer.

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5n = mult(4) + 3; 3 is the remainder

10 n = 5n * 2 = [mult(4) + 3] * 2 = mult(4) + 6 = mult(4) + 2

So, the remainder is always 2.

The remainder when 5n is divided by 4 is 3

Theory: Dividend = Divisor*Quotient + Remainder

5n -> Dividend
4 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> 5n = 4*a + 3 = 4a + 3

Quantity A: The remainder when 10n is divided by 4

10n = 2*5n = 2*(4a + 3) = 8a + 6 = 4*2a + 4 + 2 = 4*(2a + 1) + 2
=> 10n when divided by 4 gives 2a + 1 as quotient and 2 as remainder

Clearly, Quantity A(2) = Quantity B(2)

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Remainders

Hi Brent GreenlightTestPrep
In this type of remainder question, to clarify does it mean we should never add/combine any numbers as in below line?

"This means that 4(2k + 1) + 2 is 2 greater than a MULTIPLE OF 4"

As I have add 2 + 1 = 3 ( Remainder). Therefore A > B.