## Farmer Thomas

Farmer Thomas sold a pair of cows for $210. On one he made 10%, and on the other he lost 10%- earning just 5% on his transaction. What did the cows originally cost him?

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bobg| Profile April 2nd, 2008 - 12:40 am1st cow sold for 45 (worth 50)

2nd cow sold for 165 (worth 150)

210 is the total

200 is the value of the cows

10$ is 5 percent profit total.

How I did it:

I subtracted 5% profit to find the price of the cows.

Each cow now costs 100$

But this doesn’t allow for profit if +/-10%

So:

to gain a net profit of 1/2 the price, the cows must be 100% apart in price. the most reasonable first guess is 50% and 150% of the original 100$

Given that +10% – 10% = 5% total it meant that the more expensive cow must be sold at a profit

150+15

50-5

15-5=10 or 5% of 200!

Skwert78| Profile April 2nd, 2008 - 12:52 am$50.00 and $150.00

thus selling them for $45.00 and $165.00 for a total of $210.00

falwan| Profile April 2nd, 2008 - 1:25 amoriginal coast of cows = $199.50

Thank You.

da Moose| Profile April 2nd, 2008 - 5:00 amFarmer Thomas originally paid £150 for Bossie and £50 for Bessie.

When it came time to sell the cows, he had to price them taking into consideration how well they’d performed, dairy-wise. You see, when he sold a bit of butter to Betty Botter, she’d bought butter from Bessie. But she found the butter bitter. If she put it in her batter, it would make the batter bitter. So ’twas better Betty Botter bought a bit of Bossie butter that would make her batter better, yet. :o)

vikram.mankal| Profile April 2nd, 2008 - 5:33 amOne cow was bought for 150

Other one for 50

10% gain on 1st = 15

10% loss on 2nd = -5

15 – 5 = 10

Total cost = 200

Gain = 210 – 200 = 10

Gain % = (gain / cost) * 100

= (10/200) * 100

= 5%

gunturi| Profile April 2nd, 2008 - 5:48 amFirst cow =150

Second cow=50

joe| Profile April 2nd, 2008 - 7:14 amThe cost prices of the cow are 152.25 and 47.25 (makes 10% on 152.25 which he sells for 167.475, and loses 10% in the 47.25 which he sells for 42.525 to give a total sale of 210). Why? well assume cost of one cow is x and the other y, then if he makes 5% on the sale then he must have bought them for 5% less than 210 = 199.50, so x + y = 199.50. The selling totals can be represented as 1.1x and 0.9y (1.1 is like 10pct gain and 0.9 a loss of 10%), which we are told equals 210. So we can solve the simultaneous equations:

x + y = 199.5

1.1x + 0.9y = 210

and it gives x as 152.25 and y as 47.25.

QED as we used to say…

T| Profile April 2nd, 2008 - 7:43 amI can’t find my equations for this… so I’ll just give my answer

Cow with 10% profit = $150 (sold for $165)

Cow with 10% loss = $50 (sold for $45)

Together the cows cost him $200 and he sold for $210 (a 5% profit)

SolarFlare| Profile April 2nd, 2008 - 10:33 amOriginal price of the two cows: $50 and $150. Explanation is as follows.

Let original price of the two cows be x and y, then from percentages gained and lost for the individual cows:

1.1x + 0.9y = 210 (Eqn1)

And from the overall percentage gain on the transaction:

1.05(x + y) = 210

1.05x + 1.05y = 210 (Eqn2)

We have simultaneous equations. From here the strategy is very straightforward. The simplest technique is to use eqn 1 to express y in terms of x:

y = (210 – 1.1x) / 0.9 (Eqn 3)

Then substitute this for y in eqn 2, rearrange and do the arithmetic. This gives x = 150. Substitute this value for x in the expression for y found earlier (Eqn3) & do the arithmetic. This gives y = 50. (Optionally, these values of x and y can then be checked by using them in the original two equations.)

pedstar| Profile April 2nd, 2008 - 11:00 amLet’s call the two cows Yvette and Xander…ette, hmm?

So, a 10% profit on the cost of Xanderette, and a 10% loss on the cost of Yvette meant 5% profit on the original cost of the cows for Farmer Thomas.

1.1x + 0.9y = 1.05z

But we know that Farmer Thomas sold his cows for $210.

1.05z=210. So, z, the original cost of the cows, is $200.

x + y = 200

1.1x + 0.9y = 210

Solving simultaneously,

x = 150, and y = 50. Thus, Xanderette cost $150, and Yvette cost $50?

foger1979| Profile April 2nd, 2008 - 11:40 amThe cow that he made 10% he bought originally for $150

Thus adding 15 dollars for a total of $165.

The second cow he bought was originally $50.

Thus losing $5 for a total of $45.

Sales of $165 + $45 = $210 ($10 profit)

$10 is a 5% earning over the original cost of $200.

zxo| Profile April 2nd, 2008 - 12:25 pmThere are a few ways to interpret this one, but the simplest answer seems to be $50 and $150.

aaronlau| Profile April 2nd, 2008 - 1:01 pmthink it is too simple if its just $210/105% = $200

Unless its each cow?

Then $150 cow sold for $165, profit $15.

Then $50 cow sold for $45, loss of $5.

suineg| PUZZLE MASTER | Profile April 2nd, 2008 - 9:07 pmtotal price = total cost * 1.05–> 210/1.05=200

C1+C2= total cost–> C1+C2=200–> C1= 200-C2

P1+P2= total price–> P1+P2=210–> P1= 210- P2

P1= C1+0.1C1–>1=1.1C1

P2= C2-0.1C2–>2=0.9C2

—————Solving equations————————-

1.1C1+0.9C2 = 210->1.1(200-C2)+0.9C2=210-> 220-0.2C2=210

->-0.2C2=-10–>C2=50->C1=200-50=150

cost of cow with 10% earnings= 150

cost of cow with 10% loss= 50

proof= 15-5= 10 total earning–> 200*0.05= 10

I think

LadyInsomnia| Profile April 2nd, 2008 - 9:24 pmThe cows originally cost him $200.

The first cow cost him $150, and he sold it for $165. The second cow cost him $50, and he sold it for $45.

jasc| Profile April 2nd, 2008 - 9:40 pm$150 and $50

Cost1 + Cost2 = $210/1.05

= $200

Cost2 = 200 – Cost1

1.1Cost1 + .9(200-Cost1) = 210

Cost1 = 150

paprika60| Profile April 2nd, 2008 - 10:22 pmThe original cost of 1cow was 110 the other was 90. So total the original cost was 200.

tad| Profile April 3rd, 2008 - 12:23 pmHe paid $200 total for the cows.

$50 for the cow he lost 10%

$150 for the cow he made 10%

hex| PUZZLE MASTER | Profile April 3rd, 2008 - 1:19 pmCosts: C1 & C2

Selling: S1 & S2

Given:

S1+S2=210

S1=1.1 x C1

S2= 0.9 x C2

S1+S2=1.05 x (C1+C2)

So:

Solving for C1 and C2:

C1=150

C2=45

Shawn| PUZZLE GRANDMASTER | Profile April 3rd, 2008 - 3:59 pmHe made money on the cow that originally cost $150.

He lost money on the cow that originally cost $50.

xpress84| Profile April 5th, 2008 - 12:23 am1st cow $150

2nd cow $50

sold the first for 10% profit

second for 10% loss

RK| Founder | Profile April 5th, 2008 - 12:40 pmThe cost of the 2 cows:

$150

$50

It may be worthwhile checking out the good explanations given by Suineg, SolarFlare, Pedstar, Bobg, Jasc, Foger1979, & Vikram.

hex| PUZZLE MASTER | Profile April 6th, 2008 - 2:59 amSince the comments are now visible, I see that I must have deleted a part of my post without noticing.

Solving for C1 and C2:

C1=150

C2=50

and hence S1 and S2:

S1=165

S2=45

seanRhoades| Profile April 24th, 2008 - 6:02 pmI get the same answer as Joe, the original cost is cow1 = 152.25; cow2 = 47.25, I assume our algebra is the same but I suppose I should show some work, total profit = 10.50, so

cow1/10 – cow2/10 = 10.5 and

cow1 + cow2 =199.5, two variables two equations, so use substitution, solve for cow1 here and substitute we get

(199.5 – cow2)/10 – cow2/10 = 10.5

cow2 = 47.25

cow1 = 152.25