Hi,
we have ratios of three items as 6:10:15..

restriction

we can remove nuts only from one bowl..

inference of restriction

the ratio between the other two will remain the same

solution

lets see the choices..

i. 1 : 2 : 3
the ratio 10:15 is same as 2:3..
so the nuts are removed from 1st bowl..
1:2:3 will mean 5:10:15.. so 1 nut is removed from 1st bowl...
POSSIBLE

ii. 2 : 3 : 4
the corresponding ratios are 6:10:15..
no two ratios are equal in earlier and present..
so not possible..
\(2:3\neq{6:10}\)..
\(2:4\neq{6:15}\)..
\(3:4\neq{10:15}\)..
so NOT POSSIBLE.

iii. 4 : 7 : 10
same reason as ii..
NOT POSSIBLE..

ANS A
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IMO : A

I don’t know the standard approach for this sum, but I just tried number plug in method
6:10:15
Which means 6x,10x,15x
So probably 6,10,15
12,20,30
18,30,45
24,40,60
No we try to fit with answer 6,10 ,15 to be fitted with
1:2:3 ( 2 & 3 direct fit when we multiply 5) , reduce 6 by one number 5
5, 10 15 (1:2:3)
So A works
Remaining I tried it doesn’t get fit

As always, get it from answer,

Given:
P:C:A = 6x:10x:15x
Some nuts removed from one of three types
Which of the following be the ratio?

i. 1 : 2 : 3

ii. 2 : 3 : 4

iii. 4 : 7 : 10

From the given ratio we need to get the above three answer,
NOTE: all the number should be divisible by common number(dividend)
Let's say some quantity removed from pecans, then we need to figure out the divisor divisible by common dividend

i) if removed 1 from 6 changes to 5 =>5:10:15 = 1:2:3
SATISFIED
ii) if anything removed from cashew, we not going to get divisor that is divisible by common dividend
NOT SATISFIED
iii) if removed a number we can able to solve,
OPTION 1:- 6:10:14 = 3:5:7
whatever may the number we get reduced still we gonna get same amount of peacans and cashew.
NOT SATISFIED

Only satisfied option is 1

Answer will be 'A'

For case #3 4:7:10 -- why does that not work? If each is multiplied by 1.5 and we remove from the middle, then it would match

\(?\,\,\,:\,\,\,p:c:a\,\,{\rm{possible}}\,\,\left( {{\rm{when}}\,\,{\rm{some}}\,\,{\rm{nuts}}\,\,{\rm{of}}\,\,{\rm{one}}\,\,{\rm{type}}\,\,{\rm{removed}}} \right)\)


\(p:c:a = 6:10:15\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,p = 6k \hfill \cr
\,c = 10k \hfill \cr
\,a = 15k \hfill \cr} \right.\,\,\,\,\,\,\left( {k > 0\,\,{\mathop{\rm int}} \left( * \right)} \right)\)

\(\left( * \right)\,\,\left\{ \matrix{
\,{\mathop{\rm int}} - {\mathop{\rm int}} = a - c = 15k - 10k = 5k\,\,{\mathop{\rm int}} \hfill \cr
\,{\mathop{\rm int}} - {\mathop{\rm int}} = p - 5k = 6k - 5k = k\,\,\,{\mathop{\rm int}} \hfill \cr} \right.\)


\(\left( {\rm{I}} \right)\,\,\,p:c:a = 1:2:3\,\,\,\, \Rightarrow \,\,\,{\rm{possible}}\,\,\left( {k = 1,\,\,{\rm{take}}\,\,1\,\,{\rm{pecan}}\,\,{\rm{nut}}\,\,{\rm{out}}\,\,\,\, \Rightarrow \,\,\,\,\left( {p,c,a} \right) = \left( {5,10,15} \right)} \right)\)

\(\,\,\, \Rightarrow \,\,\,\,\,{\rm{refute}}\,\,\left( {\rm{B}} \right),\left( {\rm{C}} \right),\left( {\rm{E}} \right)\)


\(\left( {{\rm{III}}} \right)\,\,p:c:a = 4:7:10\,\,\,\,\mathop \Rightarrow \limits^{\left( {\text{below}} \right)} \,\,\,{\rm{impossible:}}\,\,\,\)

\({\rm{some}}\,\,p\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k - {\rm{some}},10k,15k} \right) \hfill \cr
\,{2 \over 3} = {{10k} \over {15k}} = {c \over a} \ne {7 \over {10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\)

\({\rm{some}}\,\,c\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k - {\rm{some}},15k} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k - {\rm{some}}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,40k - 4 \cdot {\rm{some}} = 42k\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\)

\({\rm{some}}\,\,a\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k,15k - {\rm{some}}} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k}} = {3 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\)


The correct answer is (A).

(Note that (II) does not need to be evaluated!)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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