[Aarav]

Originally Posted by dhruv.dingliwal

Hey Tati Bhai,
Hats off to this soln..
And thanx to Aarav for such a fundoo one.. In my opinion this one should be counted as one of the best of this series..

Cheers,
Dhruv

Vineet deserves the credit for this as he had set this problem. My only contribution was to solve this and put in the NL. Vineet will be back tomorrow, you tell him as to how much you liked this problem :-)

[5ter]

[quote=Aarav;796637]------------------------------------------------------------
Quantitative Question # 55
------------------------------------------------------------

Question:

The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the costprice of 6 kg oranges and 5 kg apples (is)

(a) can not be determined (b) Rs 318 (c) Rs 375 (d) Rs 384 (e) none of the foregoing

option (b)

20 org + 1 kg app --> cp = 60 , sp = 72
1kg org + 10 app --> cp = 50 , sp = 60

After struggling with arbit approaches was about to give up but then I realised in 2nd equation 50* 1.2 = 60 and 60 * 1.2 = 72.!! now i knew what was coming next...hehe

--> 1.2kg org + 12 app = cp = 60 , sp = 72
As profit% for each fruit are different

1.2kg org = 20 oranges and 1kg app = 12 app

sp of 5 org + 3 app = 18 (dividing 1st eq by 4)
cp of 6kg org + 5kg app = 300

sum = 318 option (b)

btw aarav that was a very good one..almost gave up until i realised the trick!

-5ter

[Obsessed_bout_mba]

[QUOTE=5ter;796757]

Originally Posted by Aarav

------------------------------------------------------------
Quantitative Question # 55
------------------------------------------------------------

Question:

The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the costprice of 6 kg oranges and 5 kg apples (is)

(a) can not be determined (b) Rs 318 (c) Rs 375 (d) Rs 384 (e) none of the foregoing

option (b)

20 org + 1 kg app --> cp = 60 , sp = 72
1kg org + 10 app --> cp = 50 , sp = 60

After struggling with arbit approaches was about to give up but then I realised in 2nd equation 50* 1.2 = 60 and 60 * 1.2 = 72.!! now i knew what was coming next...hehe

--> 1.2kg org + 12 app = cp = 60 , sp = 72
As profit% for each fruit are different

1.2kg org = 20 oranges and 1kg app = 12 app

sp of 5 org + 3 app = 18 (dividing 1st eq by 4)
cp of 6kg org + 5kg app = 300

sum = 318 option (b)

btw aarav that was a very good one..almost gave up until i realised the trick!

-5ter

offffffff..thanks buddy
i was strugglin a lot with this prob and that 1.2x didnt strike me..
anyway..intelli.ly solved...thanks again

[scarletyoke]

Really good one!

Originally Posted by Aarav

------------------------------------------------------------
Quantitative Question # 55
------------------------------------------------------------

Question:

The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the costprice of 6 kg oranges and 5 kg apples (is)

(a) can not be determined (b) Rs 318 (c) Rs 375 (d) Rs 384 (e) none of the foregoing

[JeetendraKhanna]

great solutions, i got the qty/kg for both... (12 apples per kg and 50/3 oranges per kg) but was trying to find out the individual profit percentages. that was not required, just using the 2 basic equations did not strike. anyways, alls well that ends well. the answer is 318 option (b)

[crappy_happy]

lovely question

[Ravitosh]

Oh Dear!!! Brilliant question...

[Fermions]

The moment I checked my mail n found the QAD 55 I realised its a REAL challenging type :

However, here's my solution with no assumptions.

Given :
the profit % of the 2 fruits r different.
20 orng + 1kg app = > cp=60, sp=72 => profit % =20
1kg orng + 10 app = > cp=50, sp=60 => profit % =20

Now, with d profit % of 2 diffent entities sold being diffent the profit % of the combined lot can still b same in 2 cases ONLY if the ratio of the 2 diffnt fruits in each set is the same.( USING ALLIGATION FUNDA )............(1)

let 1kg of orng =x no. of d same(may not b integer)
let 1kg of app = y no. of d same.............................................. ..(2)

using 1 & 2:
20/y = x/10 = k(say)
x=10k ; y= 20/k . ......................................(3)

let c.p. of orng & app be co & ca respectively :
using the given info :
20 co + 20/k ca =60 = > (1co + ca/k) =3 ...................(4)
10k co + 10 ca =50 = > ( k co +ca) = 5 ...................(5)
(4) /(5) gives k= 5/3 ( chk this out)

=> 1 kg of orng = x = 50/3 ;
= > 1 kg of app = y = 12 ; ...........................................(6)

using these values in C.P. info:
20 co + 12 ca =60
=>100 co + 60 ca = 300

similarly for S.P info given :
20 so + 12 sa = 72
5 so + 3 sa = 18.

WHAT IS WANTED ??

S.P of 5 orng + 2 app = 5 s0 + 3 sa =18
C.P of 6kg orng + 5 kg app = 100 co + 60 ca = 300

sum =318. option (b)

[5ter]

Fermion,

Here is my take on this cos I solved it by a different approach which I think is equivalent.

in your solution y = 20/k and k = 5/3 --> y = 12 and not 6. So what we get for 1kg orange and 1kg of apple is same 50/3 and 12 respectively.

if you forget for the k factor for the moment there are just 2 variables apples and oranges ( as kgs and numbers are dependent on each other) and equations are :

A1x + B1y = C1
A2x + B2y = C2

you say A1/B1 = A2/B2 (allegation you use 20/y = x/10 =k)
which is equivalent to A1/A2 = B1/B2

now we know that this ratio should also be equal to C1/C2 for the system of eqution to have a solution.

What I did was I made C1 and C2 equal by a constant factor in both equations and then A1=A2 and B1=B2 and obtained the same answer. I think as equations are not really independent it doesnt matter whether you take ratios of coefficients or equate them using a factor.

Any takers?

-5ter

[tatimatla]

Originally Posted by doomsayer

Nice soln tati...but why we need profit percenatge to be diff here?

See, if the profit %ages are equal (to 20%, in this case), the final profit %age comes to 20% irrespective of the ratio of apple : oranges.
Only when the profit %ages are different can we safely assume that there exists a unique ratio of appl : oranges. It is only at this ratio that the profit %age comes to 20% and at no other ratio.