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

[slam]

10 % of CP is 40

10% of the expected profit is Rs.40. (not 10% of the cost price.

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slam.

[schinnar]

The Answer is (1) the number of cheaper brand sold at higher cost was 60

Let the number of cheaper item sold be x
Let the number of costlier item sold be y

Originally x is sold for Rs 10 and y for 12

The total amount is 10x + 12y -(1)

Now the x is sold for 12 and y is sold for 10

The total amount is 12x + 10y (2)

Due to this loss is Rs 40 and this makes the profit to come down 10%

So the actual profit is Rs 400

Now (1) - (2)

10x+12y - 12x-10y = 40 => y-x=20

So the difference between the number of articles sold is 20

As choice (1)

the number of cheaper brand sold at higher cost was 60 = > x=60 and y=80

Hence total original Selling price is Rs 1560

his management graduate son is Rs 1520

This satisfy only for 25% profit. CP is = 1248 + 25% profit 312 = 1560


As we found the profit percentage is 25% for the choice 1 , even if some other number it gives 20% profit, it is not true for all the cases because it is false

Apply the choice (3) for the prevoius solution
Say x paise is the profit for cheaper brand then x+40 paise is for the higher brand

60(x)+80(x+40) = 31200(Rs 312) => x = 200(Rs 2)

profit for y = 2.40 orginal cost price is Rs 9.60 , but he is selling it for 10 so there is no loss. The choice 3 cannot be true.

[slam]

Originally Posted by Aarav

Didn't understand this -> If (1) and (2) are false then why the answer is (5)?

My mistake. I somehow did not consider the option of marking only (3), so crossed out (4) and went straight to (5).
Thank you for pointing this out Aarav Sir.
Continuing my solution,
80(Pa) + 100 (Pb) = 40 (expected profit)
put Pb = Pa + 0.40 (for option 3)
to get Pa = 2, and Pb = 2.40

=> to make up the lost profit of Rs 40, he needs to sell 40/2.4 = 16.66 => 17 items of the higher price, which is less than 34.
Now we can justify not marking (3) here.

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slam.

[fiction]

After solving, finally I got :

Profit = Rs 400
Expected profit = 25%
Cost Price = Rs 1600
Number of expensive item sold = 100
Number of cheaper item sold = 80

So none of the options from 1 to 4 match.

Hence answer is (5)

[pavanpadekal]

Originally Posted by mukulorama

Namaste,

I am getting a different answer


=> 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

You have not considered that 10% of the original profit is lost and that the expected profit percentage before transaction has to be either 10,15,20 or 25%

,i.e., 10% of original profit = 40 implying total profit expected is 400 and

in your trial and error case you need to check if 400*100/CP= either 10/15/20/25

Now 40000/CP=10/15/20/25

Evaluating for extreme values of CP only
CP must lie between 4000 and 1600.

Now CP+Profit = SP

Evaluating at the least value of 1600CP and Profit of 400 SP cant be lesser than 2000 while your solution clearly requires SP to be =<15xx
Hence the answer doesn't satisfy all the constraints in the question.
Hope it helps

[Aarav]

Originally Posted by slam

My mistake. I somehow did not consider the option of marking only (3), so crossed out (4) and went straight to (5).
Thank you for pointing this out Aarav Sir.
Continuing my solution,
80(Pa) + 100 (Pb) = 40 (expected profit)
put Pb = Pa + 0.40 (for option 3)
to get Pa = 2, and Pb = 2.40

=> to make up the lost profit of Rs 40, he needs to sell 40/2.4 = 16.66 => 17 items of the higher price, which is less than 34.
Now we can justify not marking (3) here.

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slam.

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

[dewan_iitr]

Brand A - SP = 10 Profit = a CP = 10-a Quantity sold = x
Brand B - SP = 12 Profit = b CP = 12-b Quantity sold = b
so
ax+by=400
(a+2)x + (b-2)y = 360
so y-x = 20 [Assumed son sold same quantiy as his father expected... don understand why buts without ques cannot be solved]

Profit %age = (ax+by)/[(10-a)x+(12-b)y]*100
using equations - Profit %age = 20000/(11x-80) one can now put options to eliminate 1 and 2

Also only 25% profit will yield x as an integer.which gives x=80 and y=100

option 3 says b-a = .4
80a + 100 b =400 so b= 256/90 and a= 44/18
Quanity of brand B required to be sold = 40/b = 40/ 256/90 so option 3 does hold good

[5] is correct option

[rajeev_hts]

let n are the cheaper articles and m are costly articles.

if person sold them at each other's rate then

diference = 10n+12m-12n-10n = 2m-2n

this disserence is 40
so m-n = 20 => m=n+20

projected Sell Price SP= 10n+12m = 22n+240

CP = 22n+240-400= 22n-160

profit% = 400/(22n-160)

for profit = 25%

n=80
m=100

Answer is (5)

[IIMA_2009]

[FONT='Verdana','sans-serif']Let the item sold for 10 rs be a and item sold for 12rs be b…[/font]
[FONT='Verdana','sans-serif']10a+12b=Total SP [/font]
[FONT='Verdana','sans-serif']10b+12a=Reduced SP.[/font]
[FONT='Verdana','sans-serif']Total SP = 40 + Reduced SP.= > b=20+a [/font]à[FONT='Verdana','sans-serif']1[/font]
[FONT='Verdana','sans-serif']40 is 10% of expected profit = > expected profit as 400 rs….[/font]
[FONT='Verdana','sans-serif']Option a)[/font]
[FONT='Verdana','sans-serif']a=60 and b=80[/font]
[FONT='Verdana','sans-serif']therefore [/font]
[FONT='Verdana','sans-serif']10*60+12*80=1560[/font]
[FONT='Verdana','sans-serif']Profit = 400[/font]
[FONT='Verdana','sans-serif']CP=1260[/font]
[FONT='Verdana','sans-serif']400/1260 is not equal to the given profit percentage…[/font]
[FONT='Verdana','sans-serif']Option b)[/font]
[FONT='Verdana','sans-serif']400 – 20% profit [/font]
[FONT='Verdana','sans-serif']Hence Total Selling price is 2000[/font]
[FONT='Verdana','sans-serif']10a+12b=2000[/font]
[FONT='Verdana','sans-serif']10b+12b=1960[/font]
[FONT='Verdana','sans-serif']a+b=180 [/font]à[FONT='Verdana','sans-serif']2[/font]

[FONT='Verdana','sans-serif']Solving 1 and 2 [/font]
[FONT='Verdana','sans-serif']a=80 and b=100[/font]
[FONT='Verdana','sans-serif']Now if 360/1960 the percentage decrease is not 10 in expected profit..[/font]

[FONT='Verdana','sans-serif']Option d is thus also ruled out..[/font]
[FONT='Verdana','sans-serif']Option c)[/font]
[FONT='Verdana','sans-serif']Since 10, 20 are ruled out = > Expected profit = 25%[/font]
[FONT='Verdana','sans-serif']Now solving we get that the items solved to cancel Profit should be 17 and not 34..[/font]

[FONT='Verdana','sans-serif']Hence 5 is the answer..[/font]

[getneonow]

Editing my post: found some silly mistakes..

Had to sit patiently to solve this qn but i guess i didnt fully complete it
anyways here's how i went abt it

Let a, b be the quantities of 10rs and 12rs articles.. The revenue be R, P be the profit in %

R = 10+12b -> profit here is P ..(1)

In the second case when articles are reversed
R-40 = 12a+10b -> profit here is 0.9P...(2)

From (1) and (2) we get b = a+20 => R = 22a+240..(3)
Note that a,b,R must be whole numbers, not some decimals..

Let C be the cost price for those a and b number of cheap and expensive articles

Then R = (1+ P/100)*C

and R-40 = (1+0.9P/100)*C

Eliminating C from above eqns and simplifying to get R

R = 20 + 20(2000/P +19)

Case1) P = 10

R = 4400; (3) => a = 2080/11 ; so not possible => P != 10

Case2) P =15

clearly not possible ; R comes out in decimals

Case3) P =20

R =4000; (3)=> a = 1880/11 ..Not Possible

Case4) P =25

R=2000; (3)=> a = 80 =>b = 100; Possible

so choice (1) and (2) are wrong; expected profit is 25% with a = 80(cheap article)

Tried to check (3) but got bored here and somehow after getting familiar with the problem, i feel this wudn't be true; a hunch

If choices (1) and (2) are wrong then choice(4) cannot be true.. So ans is (5) None

.. I wish i cud verify (3) perfectly.

Waiting for a more complete soln

Neo