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bijan kumar patra

· started a discussion

· 1 Months ago

question no 93 (Question: 607200.0 )

please check the answer again

Question:
A man calculates his loss% as 16 \(\cfrac{2}{3}\)% on S.P. 

Quantity I - What is the actual loss percent?

Quantity II - if cost price had been 1400 and selling price would have been 1200 then find his actual 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:

Quantity I: Loss% = 16 \(\cfrac{2}{3}\)% = \(\cfrac{50}{3 × 100}\) = \(\cfrac{1}{6}\) →L

S.P. = 6

LOSS = 1

then C.P. = 7 (as C.P. = S.P. + L)

= \(\cfrac{1}{7}\) × 100 = 14 \(\cfrac{2}{7}\)%

Quantity II:  C P=Rs 1400

                      S P=Rs 1200


Loss= Rs 200


Actual loss %=(200*100)/1400

                      =(100/7)% 


Knowledge Expert

· commented

· 1 Months ago

1=2
The answer is correct.
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