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Prachi jain

· started a discussion

· 1 Months ago

maths (Question: 100189.0 )

in this question isn't it last step should be
2x+x=70

Question:

A trader buys two types of pulses. In which, price of first type of pulse is twice to second type of pulse. He mixes both types of pulse in the ratio of 2:3 and sells the mixture at \(\unicode{x20B9} \)17.50 / kg and gets a profit of 25%, then find the price of first type of pulse. 

Options:
A) \(\unicode{x20B9} \)10/kg
B) \(\unicode{x20B9} \)20/ kg
C) \(\unicode{x20B9} \)30/ kg
D) \(\unicode{x20B9} \) 40/ kg 
E) None of these
Solution:

Let, quantity of first type of pulse = 2 kg

And quantity of second type of pulse = 3 kg 

Cost price of 1 kg pulse = (S.P.× 100 )/(100+profit)

= (17.50×100 )/(100+25 )=  1750/125=14 Rs 

∴ Cost price of 5 kg pulse = 14 × 5 = 70

Again let price of second type of pulse = Rs x kg 

∴ Cost of 1st type of pulse = \(\unicode{x20B9} \) 2x /kg 

Now, (x × 2) + 3x = 70   

Hence, cost of first type of pulse = 1x/kg  = \(\unicode{x20B9} \)20 /kg  

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