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Prasanjeet Ghosh

· started a discussion

· 1 Months ago

not clear (Question:191410.0)

If there is a profit, then

3x – 3xy/100= 2x + 6xy/100 x = 12xy/100

∴ y = 8 13

Question:

DIRECTIONS: In the following question, two statements numbered I and II are given, on solving them, we get quantities I and II respectively. Compare both the quantities and choose the correct option.

   

Quantity I-A trader sells goods to a customer at a profit of k% over the cost price; besides it he cheats his customer by giving 880 g only instead of 1 kg. Thus his overall profit percentage is 25%. Find the value of k?


Quantity II- The marked price and the cost price of an article are in the ratio 3 2. The discount percentage offered before selling it and the profit/loss percentage made on it are in the ratio 2 : 3. Find the profit/loss percentage.

Options:
A) Quantity I > Quantity II 
B) Quantity I < Quantity II
C) Quantity I ≥ Quantity II
D) Quantity I ≤ Quantity II
E) Quantity I = Quantity II or the relation can’t be determined  
Solution:

From Quantity I, 

Profit % = \(\cfrac{25}{100}=\cfrac{120+k}{880}⇒ k = 100\)
Therefore, net profit % = \(\cfrac{100}{1000}× 100 = 10\%\)


From Quantity II, 

Let the marked price and the cost price be Rs.3x and Rs.2x respectively. Let the discount percentage and the profit / loss percentage be 2y% and 3y% respectively

S.P. = M.P – D = C.P. ± P/L

∴ 3x (\(\cfrac{1 – 2y}{100}\)) = 2x (\(\cfrac{1 ± 3y}{100}\))  ….(1)

If there is a profit, then

3x – \(\cfrac{3xy}{100}\)= 2x + \(\cfrac{6xy}{100 }\)  x = \(\cfrac{12xy}{100 }\)   

∴ y = 8 \(\cfrac{1}{3}\)

If there is a loss L.H.S of (1) > R.H.S of (1), which is contradicting (1)

There is no loss

Hence is Profit percentage = 3y % = 25% 

Knowledge Expert

· commented

· 1 Months ago

Read the solution stepwise
If there is a loss L.H.S of (1) > R.H.S of (1), which is contradicting (1)

There is no loss

Hence is Profit percentage = 3y % = 25%
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