Problem Solving

BTGmoderatorDC wrote:A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?

i. 1 : 2 : 3
ii. 2 : 3 : 4
iii. 4 : 7 : 10

A. I only
B. II only
C. III only
D. I and III only
E. II and III only

Source: Manhattan Prep

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

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

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

$$\left( {\text{I}} \right)\,\,\,p:c:a = 1:2:3\,\,\,\, \Rightarrow \,\,\,{\text{possible}}\,\,\left( {k = 1,\,\,{\text{take}}\,\,1\,\,{\text{pecan}}\,\,{\text{nut}}\,\,{\text{out}}\,\,\,\, \Rightarrow \,\,\,\,\left( {p,c,a} \right) = \left( {5,10,15} \right)} \right)$$
$$\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( {\text{B}} \right),\left( {\text{C}} \right),\left( {\text{E}} \right)$$

$$\left( {{\text{III}}} \right)\,\,p:c:a = 4:7:10\,\,\,\,\mathop \Rightarrow \limits^{\left( {\text{below}} \right)} \,\,\,{\text{impossible:}}\,\,\,$$
$${\rm{some}}\,\,p\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k - {\rm{some}},10k,15k} \right) \hfill \cr
\,{2 \over 3} = {{10k} \over {15k}} = {c \over a} \ne {7 \over {10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$
$${\rm{some}}\,\,c\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k - {\rm{some}},15k} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k - {\rm{some}}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,40k - 4 \cdot {\rm{some}} = 42k\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$
$${\rm{some}}\,\,a\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k,15k - {\rm{some}}} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k}} = {3 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$


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.

I don't know the standard approach for this sum, but I just tried number plugin method
6:10:15
Which means 6x,10x,15x
So probably 6,10,15
12,20,30
18,30,45
24,40,60
Now we try to fit with answer 6,10, 15 to be fitted with
1:2:3 ( 2 & 3 direct fits when we multiply 5), reduce 6 by one number 5
5, 10 15 (1:2:3)
So A works

BTGmoderatorDC wrote:A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?

i. 1 : 2 : 3

ii. 2 : 3 : 4

iii. 4 : 7 : 10


A. I only

B. II only

C. III only

D. I and III only

E. II and III only

Original ratio values:
P.................................................C...............................................A
6x<---distance of 4x--->10x<---distance of 5x--->15x

Options for the new ratio:
I) 1 : 2 : 3
I) 2 : 3 : 4
III) 4 : 7 : 10

The ratio values in each option are EVENLY SPACED.
Implication:
By removing some of one type of nut, we must yield a ratio composed of EVENLY SPACED VALUES.
Two cases are possible:

Case 1: The value for P decreases by x units (from 5x to 4x)
P.................................................C...............................................A
5x<---distance of 5x--->10x<---distance of 5x--->15x
Yielded ratio:
5:10:15 = 1:2:3

Case 2: The value for A decreases by x units (from 15x to 14x)
P.................................................C...............................................A
6x<---distance of 4x--->10x<---distance of 4x--->14x
Resulting ratio:
6:10:14 = 3:5:7

Of options I, II, and III, only option I is possible.

The correct answer is A.