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

[implex]

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Quantitative Question # 009
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A shokeeper sells 2 different brands of an article, one for Rs 10 and other for Rs 12 each. One day he left the shop in the hands of his management graduate son who confused the two brands and sold them at each other's price. Due to this, the shopkeeper lost Rs 40 which amounted to 10% fall in his expected profit. If shopkeeper's expected profit was one among 10%, 15%, 20%, or 25% then

(1) the number of cheaper brand sold at higher cost was 60
(2) the expected profit was 20%
(3) if the profit per article of expensive brand is 40 paise more than that of cheaper brand => the least number of articles of expensive brand he had to sell to recover his loss would be 34
(4) at least two of the foregoing
(5) none of the foregoing

[implex]

Originally Posted by implex

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Quantitative Question # 009
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A shokeeper sells 2 different brands of an article, one for Rs 10 and other for Rs 12 each. One day he left the shop in the hands of his management graduate son who confused the two brands and sold them at each other's price. Due to this, the shopkeeper lost Rs 40 which amounted to 10% fall in his expected profit. If shopkeeper's expected profit was one among 10%, 15%, 20%, or 25% then

(1) the number of cheaper brand sold at higher cost was 60
(2) the expected profit was 20%
(3) if the profit per article of expensive brand is 40 paise more than that of cheaper brand => the least number of articles of expensive brand he had to sell to recover his loss would be 34
(4) at least two of the foregoing
(5) none of the foregoing

Let the cost price of 12 rupee article be a and that of 10 rupees articles be b rupees

And the amount sold be x and y respectively
SO cost =ax+by
expected sales=12x+10y
actual sales=10x+12y

reduction in profit
2(x-y)=40
x-y=20...(1)

now the reduction in profit is 10%
10[ (10-a)x+(12-b)y]=9[(12-a)x+(10-b)y]
-ax-by+8x+30y=
(8-a)x +(30-b)y=0 (2)
x(38-a-b)=20(30-b)...(3)

now!!
suppose y=60 this means x=80, then
4(8-a)+3(30-b)=0

4a+3b=122
(12-a)=(10-b) +0.4
a-b=1.6
7a=126.8 this gives negative profit ( one of them is untrue)


take expected profit is 20%

5(12x+10y)=6(ax+by)
(30-3a)x +(25-3b)y=0 using y=x-20 we get
(55-3a-3b)x=20(25-3b)
(30-b)/(38-a-b)=(25-3b)/(55-3a-3b)
(30-b)/(36.4-2b)=(25-3b)/(50.2-6b)
-230.2b+1506=910-159.2 b
71b=594
b=594/71=8.4 approx
a =10


x(t+0.4)+yt- x(t-1.6)-y(t+2)=40

so none of the pairs !!

option 5)

[pavanpadekal]

I have a strange feeling that this is a crap solution , nevertheless
10% 40
100% 400
ExPECTED initial profit =400
10%
400-10%
4000 Combined CP
400 - 15%
40000/15=8000/3=2666.66 CCP
400 - 20%
40000/20=2000 CCP
400 - 25%
40000/25=8000/5=1600 CCP

4000 4400
12E+10C=4400(Expected)
10E+12C=4360(Actual)
E-C=20
22E=4600
E=2300/11

2666.66 3066.66
12E+10C=3066.66
10E+12C=3026.66
E-C=20
22E=3266.66
E=3266.66/22

2000 2400
12E+10C=2400
10E+12C=2360
E-C=20
22E=2600
E=1300/11


1600 2000
12E+10C=2000
10E+12C=1960
E-C=20
22E=2200
E=2200/22=100

profit per article of cheaper brand = x
profit per article of expensive brand = x+0.40
100(x+0.4)+80x=400
180x+40=400
180x=360
x=2
40=n(2.4)
n=40 /2.4 <20
if the profit per article of expensive brand is 40 paise more than that of cheaper brand => the least number of articles of expensive brand he had to sell to recover his loss would be <20

Number of cheaper brand sold at higher cost = 80
EXpected Profit Percentage = 400*100/1600=100/8=50/4=25/2=25%
Hence Ans(5) none of the foregoing

[implex]

alternate solution !
clealry we have found x-y=20

we can easily show that condition 3 is equivalent, so no extra info !

5(12x+10y)=6(ax+by)
(30-3a)x=y(3b-25)

(30-3a)x=(x-20)(3a-29.

20(3a-29.=x(6a-60)
x=10

so none of the options are true

option 5)

[slam]

Originally Posted by implex

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Quantitative Question # 009
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A shokeeper sells 2 different brands of an article, one for Rs 10 and other for Rs 12 each. One day he left the shop in the hands of his management graduate son who confused the two brands and sold them at each other's price. Due to this, the shopkeeper lost Rs 40 which amounted to 10% fall in his expected profit. If shopkeeper's expected profit was one among 10%, 15%, 20%, or 25% then

(1) the number of cheaper brand sold at higher cost was 60
(2) the expected profit was 20%
(3) if the profit per article of expensive brand is 40 paise more than that of cheaper brand => the least number of articles of expensive brand he had to sell to recover his loss would be 34
(4) at least two of the foregoing
(5) none of the foregoing

good morning people. Here's my take:
First article:
SP = 10, sold a items, profit per item = (Pa), sold at 12 rs per item
Second article
SP = 12, sold b items, profit per item = (Pb), sold at 10 rs per item

expected profit = a(Pa) + b(Pb)
actual profit = a(Pa +2) + b(Pb -2)

expected - actual = 40 => b = a+20
also, 40 is 10% of expected profit => expected profit = 400 rs.
now,
expected profit = total cost * profit percentage/100
total cost = 10a + 12b - 400 = 22a + 240 -400 = 22a -160
if profit% = 10, 22a = 4160
if profit% = 15, 22a = 400 * 100/15 + 160
if profit% = 20, 22a = 2160
if profit% = 25, 22a = 1760
(a has to be a positive integer since it is the number of items sold, so only profit = 25% would be possible).
expected profit = 25%
also,
a = 1760/22 = 80
=> b = 100

so options (A) and (B) are incorrect. so we can choose option (5) here.

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slam.

[shankey_595]

suppose expensive products are y and others are x
so 10x+12y-10y-12x=40 >> y-x=20

10% of cp =40 >> cost price =400
now values of x&y are only feasible when % profit = 15 >> sp=460
giving x=10 and y=30
by dis we eliminate otpion 1,2 and 4

for d third part
10p+30(p+2/5)=60 which gives p=6/5 and hence we require 25(2/5+6/5)=40 expensive products to cover d loss

so ans 5th is d right option

[Rockeeze]

hi all
kind of late to post my soln.
here is my take
let the quantity of rs 12 artical be X
quantity of Rs 10 article be Y
so, expected SP 12X+10Y
but actual SP, 10X+12Y

so clearly Y-X=20

nowSP/CP can be 1.1 or 1.15 or 1.2 or 1.25
now equating expected SP and above expression
we find CP to be 400, 460,480,500............for each case of profit

again we hv SP-CP=400
putting the soln for each case we find tht, profit of 25% to be valid
as X and Y have to be whole numbers

so we find X=30
Y=50

option A and B are negated
for option C.. we find its nt consistent, just as implex bro, explained

hence, i ll with OPTION 5

[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

[Aarav]

Originally Posted by slam

good morning people. Here's my take:
First article:
SP = 10, sold a items, profit per item = (Pa), sold at 12 rs per item
Second article
SP = 12, sold b items, profit per item = (Pb), sold at 10 rs per item

expected profit = a(Pa) + b(Pb)
actual profit = a(Pa +2) + b(Pb -2)

expected - actual = 40 => b = a+20
also, 40 is 10% of expected profit => expected profit = 400 rs.
now,
expected profit = total cost * profit percentage/100
total cost = 10a + 12b - 400 = 22a + 240 -400 = 22a -160
if profit% = 10, 22a = 4160
if profit% = 15, 22a = 400 * 100/15 + 160
if profit% = 20, 22a = 2160
if profit% = 25, 22a = 1760
(a has to be a positive integer since it is the number of items sold, so only profit = 25% would be possible).
expected profit = 25%
also,
a = 1760/22 = 80
=> b = 100

so options (A) and (B) are incorrect. so we can choose option (5) here.

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slam.

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

[nbangalorekar]

once u get the difference b'ween the quantities as 20 then use the options,.
answer is option '5'
CHEERS...