Problem Solving

$$Let\ no\ of\ chicken=x$$
$$Let\ no\ of\ days=d$$
If the farmer sells 75 of his chicken, his stock of feed will last for 20 more days than planned
If he buys 100 more chicken, he will run out of feeds 15 days earlier than planned
$$Amount\ of\ feed\ =xd$$
$$xd=\left(x+100\right)\left(d-15\right)$$
$$where,\ xd=\left(x-75\right)\left(d+20\right)\ $$
$$\left(x-75\right)\left(d+20\right)\ =\left(x+100\right)\left(d-15\right)$$
$$\frac{\left(d+20\right)}{\left(d-15\right)}=\frac{\left(x+100\right)}{\left(x-75\right)}$$
$$\left(x=5d\right)$$
$$xd=\left(x-75\right)\left(d+20\right)$$
$$5d^2=\left(5d-75\right)\left(d+20\right)$$
$$d^2=\left(d-15\right)\left(d+20\right)$$
$$d=60$$
$$x=5d$$
$$x=5\cdot60=300$$
$$answer\ is\ OptionE$$

BTGmoderatorDC wrote:If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300
Source: Manhattan Prep

$$\left( {\# {\rm{chicken}}} \right)\left( {\# days} \right) = {\rm{constant}}$$
$$? = x = \# {\rm{chicken}}$$
$$\left\{ \matrix{
\left( {x - 75} \right)\left( {d + 20} \right) = \left( {x + 100} \right)\left( {d - 15} \right)\,\,\,\,\left( {\rm{I}} \right) \hfill \cr
\left( {x - 75} \right)\left( {d + 20} \right) = xd\,\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.$$
$$\left( {\rm{I}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd - 15x + 100d - 15 \cdot 100$$
$$ \Rightarrow \,\,\,35x - 175d = 20\left( {75 - 5 \cdot 15} \right) = 0\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,35} \,\,\,\,\,x = 5d\,\,\,\left( {{\rm{III}}} \right)$$
$$\left( {{\rm{II}}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{III}}} \right)} \,\,\,\,\,20x - {{75 \cdot x} \over 5} = 75 \cdot 20$$
$$ \Rightarrow \,\,\,5x = 75 \cdot 20\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 15 \cdot 20 = 300\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

BTGmoderatorDC wrote:If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

Total amount of feed = (number of chickens)(number of days).
The number of chickens is INVERSELY PROPORTIONAL to the number of days.
If the number of chickens DOUBLES, then the feed will last for 1/2 the number of days.
If the number of chickens TRIPLES, then the feed will last for 1/3 the number of days.

Let f = feed, c = chickens, and d = days.
Thus:
f = cd.

If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned.

The colored portions and the answer choices indicate that the problem is constrained to POSITIVE INTEGERS.
The portions in blue imply that the value of c -- a value that must divide evenly into f -- is probably a multiple of 75 and 100.
The LCM of 75 and 100 is 300.
Thus, it is almost certain that c = a multiple of 300.

The portions in red imply that the value of d -- a value that also must divide evenly into f -- is probably a multiple of 20 and 15.
The LCM of 20 and 15 is 60.
Thus, it is almost certain that d = multiple of 60.

Of the five answer choices, only E is a multiple of 300.
Test E, letting c = 300 and d = 60, implying that f = 300*60 = 18,000.
When the correct answer is plugged in, the value of f will remain constant.

E: 300
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned.
(c-75)(d+20) = (300-75)(60+20) = (225)(80)= 18,000.
if he buys 100 more chickens, he will run out of feed 15 days earlier than planned.
(c+100)(d-15) = (300+100)(60-15) = (400)(45) = 18,000.
Success!
In each case, the value of f remains constant at 18,000.

The correct answer is E.