CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

Aarav

This problem requires some patience and so far only one solution deserves 4 points. The rest are either incorrect or incomplete.

siddooba · 11:21 AM

Let x be the number of items sold of costlier brand with price 12
Let y be the other with price 10

Let E be the expected profit.

He lost 40 which amounted to 10%
E = 400

case (i) Sales = 4000 if E is 10% of Total Sales
case (ii) Sales = 8000/3 if E is 15% of Total Sales
case (iii) Sales = 2000 if E is 20% of Total Sales
case (iv) Sales = 1600 if E is 25% of Total Sales

We know (12x+10y)-(10x+12y)=40
x-y = 20

Case (i)
12x+10(x-20)=4000
22x=4200
x does not have an integral solution - hence rejected.

Case (ii)
22x = 8000/3 + 200
x does nnot have an integral solution - hence rejected

Case (iii)
22x=2200
x=100

case (iv)
22x=1800
x does nnot have an integral solution - hence rejected

Hence
x=100
y=80
Total Sales=2000
Profit = 20%

~~To prove or disprove (3)~~

He gets a profit of 400 .
Let the profit per piece of the costlier article be a
The profit per article of cheaper one is a - .4

100(a)+80(a-.4) = 400
180a-32=400
a=432/180
a=2.4

So he gets 2.4 Rs Profit on every costly article sold
So he has to sell at least 17 costly articles to recover his loss of 40 rs.
Hence (3) is wrong.

(2) is the answer

chidvi · 11:24 AM

My Option is 2 .

let X be the brand which costs 10 Rs and y be the brand which costs 12 Rs.

There will be loss only in case of higher brand seling for lower price.

this gives us y - x = 20;

and as loos is 10% of profits total profit expect is 400 RS

let p be the percent of profit

p/100 *10 *x + p/100*12 *y = 400;

only case where we get an integer value is for p =20%

ie y =100 ; x = 80 ;

If you consider option 3 we will be left with 3 equations and 4 unknows.

let`s say profit on lowe brand is r and higher brand is s . given

s-r = .4

s*y+r*x =400

y-x=20

Option 2 .

slam

Quote:

Originally Posted by Aarav

Now you have the numbers (CP/SP/Quantity) -> explain this with reverse calculation. Few are getting 25% as expected profit %.

I didn't really get what I should explain by reverse calculations using the values that I have, could you please explain?

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slam.

slam

Quote:

Originally Posted by siddooba

Let x be the number of items sold of costlier brand with price 12
Let y be the other with price 10

Let E be the expected profit.

He lost 40 which amounted to 10%
E = 400

case (i) Sales = 4000 if E is 10% of Total Sales
case (ii) Sales = 8000/3 if E is 15% of Total Sales
case (iii) Sales = 2000 if E is 20% of Total Sales
case (iv) Sales = 1600 if E is 25% of Total Sales

We know (12x+10y)-(10x+12y)=40
x-y = 20

Case (i)
12x+10(x-20)=4000
22x=4200
x does not have an integral solution - hence rejected.

Case (ii)
22x = 8000/3 + 200
x does nnot have an integral solution - hence rejected

Case (iii)
22x=2200
x=100

case (iv)
22x=1800
x does nnot have an integral solution - hence rejected

Hence
x=100
y=80
Total Sales=2000
Profit = 20%

~~To prove or disprove (3)~~

He gets a profit of 400 .
Let the profit per piece of the costlier article be a
The profit per article of cheaper one is a - .4

100(a)+80(a-.4) = 400
180a-32=400
a=432/180
a=2.4

So he gets 2.4 Rs Profit on every costly article sold
So he has to sell at least 17 costly articles to recover his loss of 40 rs.
Hence (3) is wrong.

(2) is the answer

(profit percentage = profit / cost price)

E would x % of total cost price and not Total Sales that you have used.

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slam.

Fuzon

My option too wd be option 2. Coz Its the only profit percentage giving an integral values for x and y

Also is you calculate the items sold by his son - they come out to be 12 * 80 + 10 *100 = 1960 which is 40 Rs Less than the expected profit of 400 which is a 10 % fall .

Also If profit on the 10rs article is x and that on 12 Rs article is x+ 0.4 then we have

80 x + 100 (x +0.4) = 400
180 x = 360
x = 2

Cheaper article profit = 2
Dearer article profit i = 2.4

Thus 40 / 2.4 ~ 17 Thus 3 is wrong

so the option that satisifes the condition is Option 2 !

**genius_yogi** · 12:16 PM

Here I go

Lets Make it bit simpler by avoiding much calculations

Let items be X & Y

......................X(x no.s sold).......... Y(y no.s sold)........
Actual Price ............10 ...........................12................
Error state ..............12............................10.... ............
profit per article .......a..............................b........... .....

now let the actual selling price of combined items be K,
10x + 12y = K -(1)
But in error state, acc. to question
12x + 10y = K-40 -(2)

solving both the above Equations we get y = x +20 -(3)
Putting this value of y in equation (1), we get

22x = K - 240 -(4)

Now from the Question that profit is Rs. 400

...................................CP............. ........... SP (K)....................
case 1 10% profit ........4000 ......................4400 ...................
case 2 15% profit........ 8000/3 ..................9200/3....................
case 3 20% profit ........2000 ......................2400..................
case 4 25% profit ........1600 ......................2000...................

using equation (4) in all the 4 cases, we will find that in cases 1, 2 & 3 value of x will not be an integer thus this cases woill be dropped and going with case 4

22x - 240 = 2000
x = 80
y = x +20
y = 100

therefore, profit % is 25%
which cqancels option 1, 2 & 4

Now according to option 3

b = a + .40

80a + 100(a + .40) = 400
solving this a comes out to be 2

Loss is Rs 40, if it has to be covered by article Y (@2.40)
which will require around 17 articles of Y
thus option 3 is also wrong

answer 5 - none of the foregoing

siddooba

Quote:

Originally Posted by slam

(profit percentage = profit / cost price)

E would x % of total cost price and not Total Sales that you have used.

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slam.

Oh . Thanks - Changing the solution to reflect this

Let x be the number of items sold of costlier brand with price 12
Let y be the other with price 10

Let E be the expected profit.

He lost 40 which amounted to 10%
E = 400

case (i)
Cost = 4000 if E is 10% of Cost
Sales = Cost + Profit = 4400

case (ii)
Cost = 8000/3 if E is 15% of Cost
Sales = Cost + Profit = 8000/3 + 400

case (iii)
Cost = 2000 if E is 20% of Cost
Sales = Cost + Profit = 2400

case (iv)
Cost = 1600 if E is 25% of Cost
Sales = Cost + Profit = 2000

We know (12x+10y)-(10x+12y)=40
x-y = 20

Case (i)
12x+10(x-20)=4400
22x=4600
x does not have an integral solution - hence rejected.

Case (ii)
22x = 8000/3 + 600
x does nnot have an integral solution - hence rejected

Case (iii)
22x=2600
x does nnot have an integral solution - hence rejected

case (iv)
22x=2200
x=100

Hence
x=100
y=80
Total Sales=2000
Profit = 25%

~~To prove or disprove (3)~~

He gets a profit of 400 .
Let the profit per piece of the costlier article be a
The profit per article of cheaper one is a - .4

100(a)+80(a-.4) = 400
180a-32=400
a=432/180
a=2.4

So he gets 2.4 Rs Profit on every costly article sold
So he has to sell at least 17 costly articles to recover his loss of 40 rs.
Hence (3) is wrong.

(5) is the answer

man_on_mission · 12:28 PM

Let the number of Expensive articles sold be x and
numbrer of Cheaper articles sold be y.

So, (12x + 10y) - (12y+10x) = 40
=> x - y = 20..................................(1)

Now, due to interchange in selling Price of two articles loss = Rs 40 which is 10% of expected profit so expected profit = 400.

Now, If expected profit is 10%,

Cost price = 4000
Selling price = 4000 + 400
So 12x + 10y =4400
240 + 22y = 4160 (from 1)
y = 4160/22, which is not an integer...

Therefore profit cant b 10%

if the expected profit is 15%

Cost Price = 2666.66
Selling Price = 2666.66 + 400
240 + 22y = 3066.66
y = 2826.66/22
again y is not an integer

Therefore profit cant b 15%

if the expected profit is 20%

Cost Price = 2000
Selling Price = 2000 + 400
240 + 22y = 2400
y = 2160/22 which is not an integer...

Therefore profit cant b 20%

if the expected profit is 25%

Cost Price = 1600
Selling Price = 1600 + 400
240 + 22y = 2000
y = 1760/22,
y = 80

Therefore profit is 25%

And y = 80
from (1), x = 100
So number of Expensive articles sold = 100
number of cheaper articles sold = 80
Thus option (1) is incorrect.
Also, Expected profit = 25%, Option (2) is incorrect

For option (3)
Let prfit per article for cheaper brand = p
so profit per article forexpensive brand = p + 0.40

So, 100(p+.4) + 80p = 400
p = 2
so profit on expensive item = 2.4
so to recover Rs 40 number of expensive articles to be sold = 40/2.4 = 100/6 = 16.66

So option 3 is also incorrect.

Therefore the correct answer is option (5) none of the foregoing

**genius_yogi**

Quote:

Originally Posted by mukulorama

Namaste,

I am getting a different answer

Lets assume the guy sold M articles originally worth Rs 12 at Rs 10 and N articles originally worth Rs 10 at Rs 12

then his total sale price is

M(10) + N(12) = some number say X

If he had stuck to the original price, then the total sale value of the items would have been

M(12) + N(10) = Y

now since shopkeeper suffered loss of Rs 40, we see that Y-X should be 40

=> M-N = 20

looking at statement 1) the number of cheaper brand sold at higher cost was 60

=> N = 60 and M = 80

The total price for which the son sold is 80*10 + 60*12 = 1520

Actual price for which the son should have sold is 80*12+60*10 = 1560

therefore loss of 40

Which means first statement is correct.

Can anyone please clarify

Look at the question completey which say profit% can be 10, 15, 20 or 25 which is not satisfied here