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

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If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 02 May 2018, 04:08
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A
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C
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Question Stats:

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If a and b are positive integers such that a/(4b) = 6.35, which of the following could be the remainder when 4a is divided by 2b?

I. 10
II. 20
III. 40


A) I only
B) II only
C) III only
D) I and II only
E) I, II and III

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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 02 May 2018, 10:30
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a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C
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If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 15 May 2018, 07:32
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C


Vyshak
Can you please highlight more?
1) Why you didn't say that 8/10 = 4/5, so the remainder is 4?
2) Why the remainder is multiple of 8 not 8 itself (as one & only solution) as per your algebraic output?
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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 15 May 2018, 08:28
hisho wrote:
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C


Vyshak
Can you please highlight more?
1) Why you didn't say that 8/10 = 4/5, so the remainder is 4?
2) Why the remainder is multiple of 8 not 8 itself (as one & only solution) as per your algebraic output?

Replying for Vyshak (hope you dont mind :grin: )

Quote:
1) Why you didn't say that 8/10 = 4/5, so the remainder is 4?


\(\frac{4a}{2b}\) = \(\frac{508}{10}\) --> \(\frac{50 + 8}{10}\)

Here, 50 will be divided by 10 , leaving a remainder of 8

Quote:
2) Why the remainder is multiple of 8 not 8 itself (as one & only solution) as per your algebraic output?


Please remember 8 is a multiple of 8 itself as 8*1 = 8

Hope this helps..
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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 15 May 2018, 11:36
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We have numerator = 6.35 - to make this into whole number, best is to multiply by 100. Denominator multiplies b by 4; hence, set b = 25. Now we get a = 635 and b = 25.

We want 4a / 2b: 4(635) / 2(25) = 2540/50. Without reducing, we can see that 2500 is divisible by 50 which will leave a remainder of 40 so III is valid.

Let's quickly try to double it up: 5080/100 gives remainder of 80. Not a choice, but worth noting that it's a multiple of 40. Let's try the opposite, 1270 / 25. We could do the math to see we'll have a remainder of 8/10 left over; hence, form will be a multiple of 8. Neither 10 nor 20 qualify. Answer is C.
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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 17 May 2018, 03:43
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C



Hi Just wanted to know how you got 127 for the value of a
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If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 17 May 2018, 19:55
saswata4s wrote:
If a and b are positive integers such that a/(4b) = 6.35, which of the following could be the remainder when 4a is divided by 2b?

I. 10
II. 20
III. 40


A) I only
B) II only
C) III only
D) I and II only
E) I, II and III


(4a/2b)/(a/4b)=8
8*6.35=4a/2b=50.8
only remainder 40 is divisible by 8
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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 19 May 2018, 10:45
menonrit wrote:
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C



Hi Just wanted to know how you got 127 for the value of a


a/4b=6.35

a/4b=635/100

Reduce 635/100 into small fraction by dividing by 5 in both numerator and denominator

a/4b=127/20
a/b= 127*4/20

a/b=127/5

Give kudos if it helps

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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 24 May 2018, 05:44
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C


Shouldn't we reduce 8/10 further to 4/5. So the remainder is of the form 4x, so the answer is 20 and 40, right?
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If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 27 May 2018, 00:46
push12345 wrote:
menonrit wrote:
Vyshak wrote:
a/4b = 127/20 --> a/b = 127/5

4a/2b = 508/10 --> 50 + (8/10) --> Remainder will be in the form of 8x.

Only 40 matches the above form.

Answer: C



Hi Just wanted to know how you got 127 for the value of a


a/4b=6.35

a/4b=635/100

Reduce 635/100 into small fraction by dividing by 5 in both numerator and denominator

a/4b=127/20
a/b= 127*4/20

a/b=127/5

Give kudos if it helps

Posted from my mobile device


Bunuel
hello , can someone please explain , why cant we reduce the remainder from 8/10 to 4/5 == further to the above steps , i calculated the rest as 4*127/2*5 (4a/2b), which resulted in 254/5 . further simplifying, the expression gave me a remainder of 4 i.e 50 +4/5 --- Hence shouldn't both 20 and 40 be the possible remainders .
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If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 27 May 2018, 03:20
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saswata4s wrote:
If a and b are positive integers such that a/(4b) = 6.35, which of the following could be the remainder when 4a is divided by 2b?

I. 10
II. 20
III. 40


A) I only
B) II only
C) III only
D) I and II only
E) I, II and III


The key point to remember for this question is ratios may be similar but without reducing you will get different remainders.

i.e. \(\frac{10}{6}=\frac{20}{12}=\frac{40}{24}\)
If you reduce them you will get \(\frac{5}{3}\)

But note that the remainder of \(\frac{5}{3}\) is 2
remainder of \(\frac{10}{6}\) without reducing is 4
remainder of \(\frac{20}{12}\) without reducing is 8
remainder of \(\frac{40}{24}\) without reducing is 16
and so on...

Hence \(\frac{a}{(4b)}\) = 6.35 (multiply both sides by 100 to remove decimal point) implies \(\frac{a}{b}\) = \(\frac{635}{25}\)
Kindly note that the remainder is 10 in this case.

As per question stem the remainder when 4a is divided by 2b is:-
\(\frac{(635*4)}{(25*2)}\) = \(\frac{1540}{50}\) will leave a remainder of 40.

Hence option C is the correct answer!
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Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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New post 13 Jul 2018, 18:03
\(\frac{a}{4b} = 6.35\)

Multiply both sides by 8.

\(\frac{8a}{4b}=50.8\)

\(\frac{4a}{2b}=50+\frac{4}{5}\)

\(\frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}\)

So, remainder must be a multiple of 4 and divisor must be even. Let's evaluate the options.

I. 10 is not a multiple of 4, so this is out.
II. To have 20 as the remainder, the divisor has to be 25 (\(\frac{remainder}{divisor}=\frac{4}{5}=\frac{20}{25}\)) , which is not even. This is out too.
III. To have 40 as the remainder, the divisor has to be 50. The remainder is a multiple of 4 and the divisor is even. This one fits.

Answer: C
Re: If a and b are positive integers such that a/4b=6.35, which of the &nbs [#permalink] 13 Jul 2018, 18:03
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