Profit And Loss - Aptitude MCQ Questions and Solutions with Explanations

Let the cost price of the watch is = Rs. x
According to the question,
⇒ Profit = Selling price - Cost price
⇒ 4x = 14 × 450 - 14x
⇒ 18x = 14 × 450
$$ \Rightarrow x = \frac{{14 \times 450}}{{18}} = 350$$
Cost Price of a watch = Rs. 350

$$20\% = \frac{{1 \to {\text{Profit}}}}{{5 \to {\text{CP}}}}$$
Selling price = 6 units → 390
$$\eqalign{ & {\text{5 units }} \to \frac{{390 \times 5}}{6} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 325 \cr} $$

In such type of pattern based question adopt option approach,
1^st → Check largest value of cost price
2^nd → Check smallest value of cost price
mark the answer which is greatest.
$$\eqalign{ & {\text{Option D}} \cr & {{\text{1}}^{{\text{st}}}} \to \frac{{29}}{{60}} \times 100 \cr & \,\,\,\,\,\,\,\, = 48.33 \cr & {\text{Option A}} \cr & {{\text{2}}^{{\text{nd}}}} \to \frac{{17}}{{36}} \times 100 \cr & \,\,\,\,\,\,\,\,\, = 47.22\left( {{\text{wrong}}} \right) \cr} $$

let the M.P. of an article be Rs. 100. We may calculate the final price of this article under each of the three offers:
i. Final Price = 85% of 75% of Rs. 100
$$\eqalign{ & {\text{ = Rs}}{\text{.}}\left( {\frac{{85}}{{100}} \times \frac{{75}}{{100}} \times 100} \right) \cr & = {\text{Rs}}{\text{. }}63.75 \cr} $$

ii. Final price = 90% of 70% of Rs. 100
$$\eqalign{ & = {\text{ Rs}}{\text{.}}\left( {\frac{{90}}{{100}} \times \frac{{70}}{{100}} \times 100} \right) \cr & = {\text{Rs}}{\text{. }}63 \cr} $$

iii. Final price = 95% of 65% of Rs. 100
$$\eqalign{ & {\text{ = Rs}}{\text{.}}\left( {\frac{{95}}{{100}} \times \frac{{65}}{{100}} \times 100} \right) \cr & = {\text{Rs}}{\text{. }}61.75 \cr} $$
The final price is lowest in (iii). So, it is the best offer.

Final price in 1^st case
= 70% of 90% of 90% of Rs. 10000
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{70}}{{100}} \times \frac{{90}}{{100}} \times \frac{{90}}{{100}} \times 10000} \right) \cr & = {\text{Rs}}{\text{. 5670}}{\text{}} \cr} $$
Final price in 2^nd case
= 95% of 95% of 60% of Rs. 10000
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{95}}{{100}} \times \frac{{95}}{{100}} \times \frac{{60}}{{100}} \times 10000} \right) \cr & = {\text{Rs}}{\text{. 5415}}{\text{}} \cr} $$
∴ Money saved by choosing the better offer
= Rs. (5670 - 5415)
= Rs. 255

Let the second discount rate be x%
Then,
= (100 - x)% of 90% of 450 = 344.25
$$\eqalign{ & \Rightarrow \frac{{100 - x}}{{100}} \times \frac{{90}}{{100}} \times 450 = 344.25 \cr & \Rightarrow \left( {100 - x} \right) = \left( {\frac{{34425}}{{9 \times 45}}} \right) = 85 \cr & \Rightarrow x = 15 \cr} $$
∴ Second discount rate = 15%

According to the question,
Cost price = 1200
Repair = 200
Total Cost price = Rs. 1400
Selling price = Rs. 1680
Profit = 280
$$\eqalign{ & {\text{Profit }}\% = \frac{{280}}{{1400}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\% \cr} $$

Lets assume the original price of the jewel = Rs. P
and profit earned by the third seller= x%
Then, ( 100 + x )% of 125% of 120% of P = 165% of p
$$\left[ {\frac{{\left( {100 + x} \right)}}{{100}} \times \frac{{125}}{{100}} \times \frac{{120}}{{100}} \times P} \right] = $$       $$\left[ {\frac{{165}}{{100}} \times P} \right]$$
$$\eqalign{ & \Rightarrow \left( {100 + x} \right) = \frac{{165 \times 100 \times 100}}{{125 \times 120}} \cr & \Rightarrow x = 10\% \cr} $$

$$\eqalign{ & {\text{CP}} = 345600 \times \frac{{100}}{{120}} \times \frac{{100}}{{120}} \times \frac{{100}}{{120}} \cr & {\text{CP}} = \frac{{345600000000}}{{1728000}} \cr & {\text{CP}} = {\text{Rs}}{\text{.}}\,200000 \cr} $$

Let the second discount rate be x%
Then,
(100 - x)% of 80% of 820 = 570.72
$$\eqalign{ & \Rightarrow \frac{{\left( {100 - x} \right)}}{{100}} \times \frac{{80}}{{100}} \times 820 = 570.72 \cr & \Rightarrow \left( {100 - x} \right) = \left( {\frac{{57072}}{{8 \times 82}}} \right) = 87 \cr & \Rightarrow x = 13 \cr} $$
∴ Second discount rate = 13%