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Nullbyte
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Let the total price is. P
The amount paid by kamrul :

P-13%of P =315
It gives P=362

For the max diff , oishi will get 40 % discount (30 +10)
Which is 60%of P
So Oishi will pay 217 and the difference is 315-217=98
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No it is not net discount that 30+10 is 40 otherwise if some shop gives 90+10 % discount then the product is free

Either calculate using formula otherwise do two operation like nullbyte has done above

Hope you understand
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I have a doubt

Is he getting the additional discount on the marked price or on the previous discount?
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Things here is like suppose a product is of Rs 100 and it is sold 50%+50% discount, so the first time the price is rs 50 and then the second time price will be 25. Since they have not written here explicitly about MP here, first time price you can think as MP and then after all discount the sale of item you can say SP
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Hey team is there a better way to approach this question? I do understand the solution given below and online but find it too lengthy, is a there a quicker way by any chance?
y is a multiple of 5. y can’t be 0 because in that case 3x=200 which gives us no integral solution.
We take y=5k which makes our equation 3x=200-4*5k
=> 3x = 200-20k = 20(10-k) = 2*10*(10-k)

So it must be a multiple of 10
Correct option E
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BY WHEN WILL THE OFFICIAL GUIDE QUESTIONS 2023/2024 ADDED TO THE FORUM QUIZ
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Data Sufficiency Butler: February 2024
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I have a doubt in this questionhttps://gmatclub.com/forum/if-k ... 07071.html

why the first option cannot be solved algebraically to give k=1 If K is a non-zero integer, does K = 1?

(1) K^K = K
(2) K^K = 1


M36-132

in the first option?
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For K=-1 ; (-1)^(-1) = -1 ;

+1 and -1 are two possible cases for St-1. Whereas, for st-2 K=1 is the only solution
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why is solving for k in option 1 is wrong? --- k^k-k=0 leads to k(k^(k-1) -1)=0 which gives k=1
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why is solving for k in option 1 is wrong? --- k^k-k=0 leads to k(k^(k-1) -1)=0 which gives k=1
Are you trying to solve k^(k-1)=1?
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yes
Put k = - 1 in the equation
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but if i take 1 as k^0 on right side then it comes to k=1
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Yes so there’s 2 solutions to the equation
First is k=1 which gives 1^(1-1) =1
Second is (-1)^(-2)=1

So you can’t say if k=1 or not
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i do not need to put k=-1 because K^(k-1)=K^0 which implies k=1 or 0 but as k is non zero integer k=1 is the only answer
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