CMAT Profit, Loss, and Interest Questions

Compound Interest after 3 years = Amount - Principal

= $$P\left(1+\dfrac{r}{100}\right)^3-P$$

= $$1000\left(1+\dfrac{30}{100}\right)^3-1000$$

= $$1000\left(1.3\right)^3-1000$$

= ₹ $$1197$$

Simple Interest after 3 years = $$\dfrac{1000\times\ 30\times\ 3}{100}$$ = ₹ $$900$$

So, required difference =₹ $$\left(1197-900\right)=297$$

So, required percentage = $$\dfrac{297}{900}\times\ 100=33\%$$

Let say the merchant has 100 consignments and let say C.P. of one consignment be ₹ $$x$$.

So, total C.P. = ₹ $$100x$$

Now, 25% consignment was abducted forever by some thieves.

So, remaining consignments = $$100\left(1-\dfrac{25}{100}\right)=75$$

Now, selling price of these 75 consignments = $$75\times\ x\times\ \left(1+\dfrac{20}{100}\right)\left(1+\dfrac{25}{100}\right)=112.5x$$

So, total S.P. = $$112.5x$$

So, profit % = $$\dfrac{112.5x-100x}{100x}\times\ 100\%=12.5\%$$ profit.

A person sold an article for ₹ $$136$$ and incurred 15% loss.

Now, cost price of the article = ₹ $$\dfrac{136}{0.85}=160$$

Now, after selling it for ₹ $$x$$, he would have obtained a profit of 15%.

So, $$x=160\times\ 1.15$$ = ₹ $$184$$

So, option B is the correct answer.

Let the C.P of one mobile phone be $$₹x$$. 

S.P of first mobile phone = ₹18,750

S.P of second mobile phone = ₹14,250

Let the loss percent, incurred on the second mobile phone be $$p\ \%$$. 

Profit percentage, incurred on the first mobile phone = $$\left(5\times p\%\right)=5p\ \%$$

Now, we will use the profit-and-loss percentage formula. 

For the first mobile phone: 

Profit percentage = $$\dfrac{\left(\text{S.P}\ -\ \text{C.P}\right)}{\text{C.P}}\times100$$

$$5p=\dfrac{\left(₹18,750-₹x\right)}{₹x}\times100$$                                        $$\longrightarrow\ i$$

For the second mobile phone: 

Loss percentage = $$\dfrac{\left(\text{C.P}\ -\ \text{S.P}\right)}{\text{C.P}}\times100$$

$$p=\dfrac{\left(₹x-₹14,250\right)}{₹x}\times100$$                                        $$\longrightarrow\ ii$$

From equations $$i$$ and $$ii$$, we have to find the value of $$x$$.

Divide equation $$i$$ by equation $$ii$$,

$$\dfrac{5p}{p}=\dfrac{\dfrac{\left(₹18,750-₹x\right)}{₹x}\times100}{\dfrac{\left(₹x-₹14,250\right)}{₹x}\times100}$$

$$5=\dfrac{\left(18,750-x\right)}{\left(x-14,250\right)}$$

$$5\left(x-14,250\right)=\left(18,750-x\right)$$

$$5x-\left(5\times14,250\right)=18,750-x$$

$$5x+x=18,750+\left(5\times14,250\right)$$

$$6x=18,750+71,250=90,000$$

$$x=15,000$$

Hence, the C.P of one mobile phone is ₹15,000.

Total S.P = ₹18,750 + ₹14,250 = ₹33,000

Total C.P = $$₹2x$$ = $$₹\left(2\times15,000\right)=₹30,000$$

Overall Profit = Total S.P - Total C.P = ₹33,000 - ₹30,000 = ₹3,000

Overall Profit percentage = $$\dfrac{₹3,000}{₹30,000}\times100=10\%$$

Hence, the overall profit percentage of the transaction is 10%.

$$\therefore$$ The required answer is D.   

It is given that the vegetable vendor by means of his false balance defrauds to the extent of 10% in buying goods.

So, if the vegetable vendor is buying 100 gm goods then he is buying 10 gm (10% of 100 = 10) extra good.

So, multiplying factor for profit % in this case = $$\left(1+\dfrac{10}{100}\right)$$

Now while selling, he defrauds to 10% in selling

So, if the vegetable vendor is claiming to sell 100 gm goods then in reality he is selling 90 gm (since, he defrauds 10%)

So, the profit he made = (100-90) gm = 10 gm

So, multiplying factor for profit % in this case = $$\left(1+\dfrac{10}{90}\right)$$

So, overall gain % = $$\left(1+\dfrac{10}{100}\right)\left(1+\dfrac{10}{90}\right)-1=\dfrac{110}{100}\times\ \dfrac{100}{90}-1=\dfrac{11}{9}-1=\dfrac{2}{9}$$

So, gain% = $$\dfrac{2}{9}\times\ 100\%=22\dfrac{2}{9}\%$$

After the 2-year course gets completed, ₹ 6 lakh loan will amount to: ₹ $$6\left(1+\dfrac{8}{100}\right)^2$$

Now, he has repayed half of the amount.

So, amount outstanding after completion of course = $$\dfrac{6}{2}\left(1+\dfrac{8}{100}\right)^2=3\left(1.08\right)^2=3.4992$$ lakh

Now after another 2 years, when rate of interest will be 10% p.a., the amount will become = $$3.4992\left(1+\dfrac{10}{100}\right)^2=3.4992\times\ 1.1^2=4.234032$$ lakh

So, amount returned by Ravi = ₹ $$(3.4992+4.234032)=7.733232$$ lakh.

So, option A is the correct answer.

Let the cost price be Rs. 1000 per kg

So, Selling price = Rs. 900 per kg

But he sells only 70% of the quantity claiming it to be 100% quanitity.

So, the actual cost price of 700 grams = Rs. 700

Actual selling price = Rs. 900

Profit = Rs. 200

Profit % = (Profit/Cost Price)*100 = 200/700 = 200/7 % = $$28\dfrac{4}{7}$$%

Let A be his additional investement. The interest obtained on A will be 0.08A.

The interest obtained on Rs. 1,200 will be Rs. 60.

60+0.08A = 0.06(1200+A)

0.02A = 12

A = 600

Statement I :

Mixing the pulses in ratio 7 : 3, i.e. $$7x$$ kg and $$3x$$ kg

we get the total price of the mixture as $$7x\times15+3x\times20=165x$$

So, the price of 1 kg is $$\dfrac{165x}{10x}=16.5$$

Hence, Statement I is true.

Statement II :

If the quantity of rice sold at 16 % profit is 750 kg, then the quantity of rice sold at 8 % profit is 250 kg.

The overall profit is $$\dfrac{\left(16\times750+8\times250\right)}{1000}=\frac{14000}{1000}=14$$

Hence, Statement II is also true

From 3 years to 4 years, the interest earned is 73205-66550 = ₹ 6,655

Assuming Rate of Interest = r%

Principal = 66,550

For 1 year from 3rd to 4th year, the amount becomes = ₹ 73,205

So, Interest = PRT/100

6655 = 66500*r*1/100

r = 10%

Percentage profit = $$\dfrac{\Pr ofit}{Cost\ \Pr ice}\times100$$

(A) Cost price : ₹36, Profit: ₹17

Percentage profit = $$\dfrac{17}{36}\times100$$ = 47.22%

(B) Cost price : ₹50, Profit : ₹24

Percentage profit = $$\dfrac{24}{50}\times100$$ = 48%

(C) Cost price : ₹40, Profit : ₹19

Percentage profit = $$\dfrac{19}{40}\times100$$ = 47.5%

(D) Cost price : ₹60, Profit : ₹29

Percentage profit = $$\dfrac{29}{60}\times100$$ = 48.33%

Thus, percentage profits are in this sequence D > B > C > A