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

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Joined: 27 May 2014
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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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02 May 2018, 03:08
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95% (hard)

Question Stats:

55% (02:45) correct 45% (02:33) wrong based on 155 sessions

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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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02 May 2018, 09:30
1
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.

Intern
Joined: 04 Apr 2017
Posts: 23
If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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15 May 2018, 06:32
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.

Vyshak
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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15 May 2018, 07:28
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.

Vyshak
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 )

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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15 May 2018, 10:36
2
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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Joined: 07 May 2018
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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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17 May 2018, 02:43
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.

Hi Just wanted to know how you got 127 for the value of a
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Joined: 07 Dec 2014
Posts: 1152
If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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17 May 2018, 18:55
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
C
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Joined: 02 Oct 2017
Posts: 735
Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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19 May 2018, 09:45
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.

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

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Joined: 22 Jan 2015
Posts: 23
Re: If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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24 May 2018, 04:44
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.

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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Joined: 28 Mar 2018
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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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26 May 2018, 23:46
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.

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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Joined: 11 Feb 2015
Posts: 678
If a and b are positive integers such that a/4b=6.35, which of the  [#permalink]

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27 May 2018, 02:20
1
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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13 Jul 2018, 17:03
1
$$\frac{a}{4b} = 6.35$$

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.

Re: If a and b are positive integers such that a/4b=6.35, which of the &nbs [#permalink] 13 Jul 2018, 17:03
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