If a and b are positive integers such that a/(4b) = 6.35, which of the

Given, \(\frac{a}{4b} = 6.35\)
--> a = 4b*6.35 = 25.4b
So, 4a = 4*(25.4*b) = 101.6*b
--> 4a = 2b(50.8) = 2b(50 + 0.8) = 2b*50 + 1.6b
--> 4a = 2b*50 + 1.6b

Comparing with the formula, Dividend = Divisor*Quotient + Remainder
--> Remainder = 1.6b = \(\frac{8}{5}b\)
So, remainder should be a multiple of 8
--> Only possible value is 40

IMO Option C

Pls Hit kudos if you like the solution

Hi

I used the below approach and I failed to get the answer.

a/4b = 6.35

So the remainder would be 0.35 which means R/4b = 35/100

When simplified we get R= 7 and 4b = 20 and automatically a becomes 127. Which means the reminder will be a multiple of 7.

When we substitute a and b, 4a/2b becomes 508/10.

Now all the choices are multiples of 10 and the remainder should be a multiple of 7.

Bunuel, chetan2u, can you please help where am I going wrong.

Posted from my mobile device

Hi Harsh9676,

You got 4a/2b = 508/10.
It’s absolutely correct till here.

Now as you understand, ratio only gives the most simplified values
E.g: If x/y = 2/3
Possible values of (x,y) = (2,3) or (4,6) or (20,30) or (-2,-3) so on

So possible values of 4a/2b can be
508/10 —> Remainder = 8
or
508*2/10*2 = 1016/20 —> Remainder = 16
or
508*3/10*3 = 1524/30 —> Remainder = 24
so on ...

As you can see, all possible remainders are multiples of 8 and only 40 among given options is the ONLY multiple of 8
—> Answer = 40

Hope I’m clear!

Understood.

Didn't know how I missed that.
Thanks @Dilesh4096

Posted from my mobile device

a/4b = 6.35

a/4b = 6 + 0.35

Remainder = 35/100

a / 4b = 35 / 4*25

It means that a is a multiple of 35 and b is a multiple of 25.

a/b = 35 / 25 = 7/5.

Now required work :

4a / 2b = 4*7 / 5*2 = 28/10

In this case our remainder is 8.

Therefore whatever remainder we will get must be multiple of 8.

Option C.

We got 4a/2b = 508/10.

But 508/10 = 254/5 -> Remainder = 4, not multiple of 8.

What is wrong in my reasoning?

Thanks in advance

a/4b = 6.35

a/b = 127/5 = 25.4.
Now, for a and b to be integers, a must be a multiple of 127 and b a multiple of 5.

2a/b = 50.8 = 50+0.8, remainder is 0.8
0.8 can be written as 4/5, 8/10, 12/15, 16/20, 20/25, and so on including 40/50. Here, numerator is always a multiple of 4.

So, IMO, both 20 and 40 can be the remainders.

Hi Nicooo13,

Here dividend is 4a and divisor is 2b

We have 4a/2b = 508/10 or 254/5
—> 4a = 2b(254/5)
—> 4a = 2b(250/5 + 4/5)
—> 4a = 2b*50 + 8b/5
So, quotient = 50 and remainder = 8b/5

—> Remainder only can take multiples of 8 in any way we approach the question
—> Only Possible value = 40

Hope I’m clear!

Posted from my mobile device

Ah okay, understood, many thanks Dillesh4096 !

Since a/4b = 6.35, a = (6.35)(4b) = (25.4)b. Multiplying each side by 5, we get 5a = 127b. Since a and b are integers, the smallest possible value of a is 127 and the smallest possible value of b is 5. Since 4a = 4(127) = 508 divided by 2b = 2(5) = 10 produces a remainder of 8; for any possible value of a and b, division of 4a by 2b will produce a remainder that is a multiple of 8. Of the given answer choices, only 40 is a multiple of 8.

Answer: C
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I think there is a really simple way to solve this question:

Given: a/4b = 6.35

1. Multiply by 8 both sides: 4a/2b = 50.8

2. The remainder is a multiple of 8. only 40 meets this criterion.

experts what's wrong with below?
a/4b = 6+35/100 - eq 1 (we are concerned with 35/100 only)
we need to find remainder of 4a/2b = 2a/b
divide eq 1 by 4
a/b= 35/25 (since we are only concerned with the later part)
2a/b= 70/25 = 14/5 remainder = 4 therefore answer should be 20 and 80
ScottTargetTestPrep Dillesh4096

Hi Kritisood,

For the highlighted part above, Remainder of \(\frac{4a}{2b}\) may not be same as that of \(\frac{2a}{b}\)
{E.g: remainder of \(\frac{4*3}{2*4} = \frac{12}{8}\) is \(4\), whereas remainder of \(\frac{2*3}{4} = \frac{6}{4}\) is \(2\).

Also, Always write remainder formula in the form "DIVIDEND = DIVISOR*QUOTIENT + REMAINDER"
--> 4a = 2b*QUOTIENT + REMAINDER

a/4b = 6+35/100
--> \(a = 24b + \frac{35}{25}b\)
--> \(a = 24b + \frac{7}{5}b\)
--> \(4a = 96b + \frac{28}{5}b = 96b + \frac{20 + 8}{5}b\)
--> \(4a = 96b + 4b + \frac{8}{5}b\)
--> \(4a = 2b*50 + \frac{8}{5}b\)
--> Remainder = \(\frac{8}{5}b\) or a multiple of 8

So, Only 40 --> Option C

Hope I'm Clear

Solution:

As a matter of fact, your solution correctly identifies that the remainder when 2a is divided by b must be a multiple of 4; however, as it is noted in the above comments, the remainder when 2a is divided by b is not the same as the remainder when 4a is divided by 2b.

Let's explore what happens to the remainders when we multiply both the dividend and the divisor by 2:

The remainder when 5 is divided by 3 is 2. Multiplying 5 and 3 by 2, we obtain 10 and 6. The remainder when 10 is divided by 6 is 4. As we can see, the remainder was also multiplied by 2.

Similarly, if we divide 8 by 5, the remainder is 3; and if we divide 8 * 2 = 16 by 5 * 2 = 10, the remainder is 6 (which is 3 * 2).

You can find more examples. Applying the above to our situation, if the remainder when 2a is divided by b must be a multiple of 4, then the remainder when 4a is divided by 2b must be a multiple of 4 * 2 = 8.
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I solved this with a similar approach, but you have to consider that the remainder must be able to be expressed as a numerator divided by 2b. If the numerator is 20, then 20/2b reduces to 10/b, and the numerator is no longer a multiple of 4. Only 40 can be divided by 2 and still be a multiple of 4.