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Sanskrithi First solution has 1L milk in 5L solution, hence 0.2L milk/ liter of sol. 2nd sol has 0.33L milk/liter of sol. Then he take 3L of first sol and 2L of second sol. so, the final mixture has 0.2*3+0.33*2=1.26L of milk in 5L of Mixture. Let he sold first sol in Re1 with 20% gain, means 1L milk with Re0.2 gain. In the final mix he sold 0.26L free wich have a profit of (0.2/1)*0.26=0.052=5.2% loss can anyone please explain this solution? Log in to Reply
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Sonam Choubey Do correction on “”Tips & Tricks & Shortcuts to solve type of Profit & Loss “” In Type 3 problem ,You should write Rs60 in place of Rs.48 Log in to Reply
Hemanth A reduction of 20% in the price of salt enables a lady to obtain 10kgs more for Rs.100, find the original price per kg? Explanation (20/100-20)=25% 25% is equivalent to 10 kg 10*4=40 100/40=2.5 the original price is 2.5 kg explain this Log in to Reply
HelpPrepInsta Hi Hemanth, Here is your answer … Let price of salt = x Before reduction in price , x * y kg = 100 rs. i.e. xy = 100 or y = 100/x eq.1 After reduction of 20% in price , (80x/100) * (y +10) kg = 100 rs. i.e. (4x/5) * (y+10) = 100 eq. 2 put the value of y (from eq. 1) to eq. 2 we get x = 2.5 Hence, the original price is 2.5 . We Hope you get it , All The Best. Log in to Reply
Hemanth problem 5 please explain these steps profit 40%of 4x/5=140x/100*4x/5=56x/50 56x/50-9x/10=55rs Log in to Reply
HelpPrepInsta Hello avinash, we hope you are doing great.. In question 7 , eq 2 is formed by multiplying 12 to c+h = 200000 (written on the 2 line of the answer). Log in to Reply
ashima question no 2 explaination not justified for Thus, after giving 10% discount it becomes 0.90x Selling price = 0.90x ……(ii) From (1) and (2), 0.90x will gives = (25 * 0.90x)/ (0.75x) = 30 % profit. Log in to Reply
Ashish Mishra Hii Ashima, The solution is right as 25/0.75x is the profit on 0.75 selling price so on 0.90 selling price it will be (0.90x * 25)/0.75x Thank You. Log in to Reply