Practice MCQ Questions and Answer on Profit and Loss

Solution: $$\eqalign{ & 355 \times \frac{{100}}{{71}} = {\text{Cost price}} \cr & {\text{Sale price}} = 355 \times \frac{{100}}{{71}} \times \frac{{121}}{{100}} = 605 \cr} $$

Solution: Let the C.P. be Rs. 100 Then, Profit = 20% S.P. = Rs. 120 New C.P. = 20% less = Rs. 80 If he gain is 25% then S.P. $$\eqalign{ & = {\text{Rs}}\left( {\frac{{125}}{{100}} \times 80} \right) \cr & = {\text{Rs}}.100 \cr} $$ Difference in S.P. = Rs. (120 - 100) =Rs. 20 If difference in S.P. is Rs. 20, then C.P. = Rs. 100 If difference in S.P. is Rs. 18, then $$\eqalign{ & {\text{C}}{\text{.P}}{\text{.}} = {\text{Rs}}.\left( {\frac{{100}}{{20}} \times 18} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}90 \cr} $$

Solution: $$\eqalign{ & {\text{Marked price of shirt}} \cr & = {\text{Rs}}{\text{.1080}} \cr & {\text{Selling price of shirt}} \cr & = {\text{Rs}}{\text{.1080}} - \frac{{1080 \times 12}}{{100}} \cr & = {\text{Rs}}{\text{.}}\left( {1080 - 129.60} \right) \cr & = {\text{Rs}}.950.40 \cr & {\text{Let CP of shirt is Rs}}{\text{. }}x \cr & \therefore x + \frac{{x \times 8}}{{100}} = 950.40 \cr & \Rightarrow \frac{{108x}}{{100}} = 950.40 \cr & \Rightarrow x = 950.40 \times \frac{{100}}{{108}} \cr & \Rightarrow x = {\text{Rs}}{\text{. }}880 \cr} $$

Solution: Let the marked price = 100 units According to the question, \[{\text{100(MP)}}\xrightarrow{{30\% {\text{ discount}}}}{\text{70(SP)}}\]      \[ \to \] CP of retailer Cost price of retailer = 70 Retailer sold at Marked price = 100 Profit = Marked price - Cost price = 100 - 70 = 30 units profit $$\eqalign{ & \therefore {\text{Profit }}\% = \frac{{30}}{{70}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 42\frac{6}{7}\% \cr} $$

Solution: Total profit = 7.76 In Ist case, (Anand) instead of 1 kg, he gave 950 gm So, profit = 50 gm ⇒ 1000 gm = Rs. 22 1 gm = $$\frac{{22}}{{1000}}$$ 50 gm = $$\frac{{1100}}{{1000}}$$  = 1.1 In Ist case, total profit = 3 + 1.1 = Rs. 4.1 So, left profit = 7.76 - 4.1 = Rs. 3.66 In IInd case, (Sakshi) profit is Rs. 3 as he sold (in Rs. 25) and left Rs. 0.66 is because he sold less of instead of 1000 gm So, Rs. 22 → 1000 gm Rs. 1 → $$\frac{{1000}}{{22}}$$  × 0.66 = 30 gm So he sold 970 gm instead of 1000 gm Answer

Solution: Advertised price of the item = N ∴ Commission on advertised price of the article = $$\frac{{N \times 23}}{{100}}$$ Price after commission = $$N - \frac{{23N}}{{100}} = \frac{{77N}}{{100}}$$ $$\eqalign{ & \therefore {\text{CP}}\,{\text{of}}\,{\text{the}}\,{\text{item}} \cr & \frac{{\frac{{77N}}{{100}} \times 100}}{{100 + 10}} \cr & = \frac{{77N}}{{110}} \cr & = \frac{{7N}}{{10}} \cr & {\text{Profit}} = \frac{{77N}}{{110}} - \frac{{7N}}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7N}}{{110}} \cr & \Rightarrow 56 = \frac{{7N}}{{110}} \cr & \therefore N = \frac{{56 \times 100}}{7} = 800 \cr} $$

Solution: $$\eqalign{ & {\text{C}}{\text{.P}}{\text{.}}\,{\text{of}}\,{1^{st}}\,{\text{transistor}} \cr & = Rs.\,\left( {\frac{{100}}{{120}} \times 840} \right) = Rs.\,700 \cr & {\text{C}}{\text{.P}}{\text{.}}\,{\text{of}}\,{2^{nd}}\,{\text{transistor}} \cr & = Rs.\,\left( {\frac{{100}}{{96}} \times 960} \right) = Rs.\,1000 \cr & {\text{So,}}\,{\text{total}}\,{\text{C}}{\text{.P}}{\text{.}}\, = Rs.\,\left( {700 + 1000} \right) = Rs.\,1700 \cr & {\text{Total}}\,{\text{S}}{\text{.P}}{\text{.}}\, = \,Rs.\,\left( {840 + 960} \right) = Rs.\,1800 \cr & \therefore {\text{Gain}}\,\% = \left( {\frac{{100}}{{1700}} \times 100} \right)\% = 5\frac{{15}}{{17}}\% \cr} $$

Solution: Let first radio CP is x second radio CP is 1920 - x As is given, $$x \times \frac{{\left( {100 + 20} \right)}}{{100}} = \left( {1920 - x} \right) \times $$      $$\frac{{\left( {100 - \frac{{20}}{3}} \right)}}{{100}}$$ $$\eqalign{ & 120 \times x = \left( {1920 - x} \right) \times \left( {\frac{{280}}{3}} \right) \cr & 360 \times x = \left( {1920 - x} \right) \times 280 \cr & 640 \times x = 1920 \times 28 \cr & \therefore x = 840 \cr} $$ Hence, second radio cp is 1920 - x = 1920 - 840 = 1080 CP of both Radio = 840 and 1080

Solution: Let the cost of product A be Rs. x and B be Rs. y According to first sell     0.80x + 1.3y = x + y ⇒ 0.2x = 0.3y ⇒ x = $$\frac{3\text{y}}{2}$$ . . . . . . (i) According to second sell     0.85x + 1.15y = (x + y) - 6 ⇒ 0.15y = 0.15x - 6 (Multiply by 100) ⇒ 15y = 15x - 600 (Putting the value of x) ⇒ 15y = 15 × $$\frac{3\text{y}}{2}$$ - 600 ⇒ 15y = $$\frac{45\text{y}}{2}$$ - 600 (Multiply by 2 both side) ⇒ 30y = 45y - 1200 ⇒ 15y = 1200 ⇒ y = 80 million

Solution: Let the total value be Rs. x $$\eqalign{ & {\text{value of }}\frac{2}{3}{\text{rd}} = {\text{Rs}}{\text{.}}\frac{{2x}}{3} \cr & {\text{value of}}\frac{1}{3}{\text{rd}} = {\text{Rs}}.\frac{x}{3} \cr & {\text{Total S}}{\text{.P}}{\text{.}} \cr & = {\text{Rs}}.\left[ {\left( {106\% {\text{ of }}\frac{{2x}}{3}} \right) + \left( {97\% {\text{ of }}\frac{x}{3}} \right)} \right] \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{53x}}{{75}} + \frac{{97x}}{{300}}} \right) \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{309x}}{{300}}} \right) \cr & \therefore \frac{{309x}}{{300}} - x = 540 \cr & \Rightarrow \frac{{9x}}{{300}} = 540 \cr & \Rightarrow x = \left( {\frac{{540 \times 300}}{9}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 18000 \cr} $$