CAT Profit and Loss Questions

Let the cost price of the item be C
We are given that Bina sells this at 19% loss or at (1 - 0.19)C = 0.81C at 4860
This gives us the value of C at Rs. 6000

If Bina had sold this at 17% profit, the selling price would have been $$1.17\times\ 6000\ =\ 7020$$

So Shyam bought the product at 4860 and sold it to Hari at 7020
Giving the profit made by Shyam to be $$7020-4860\ =\ 2160$$

Therefore, 2160 is the correct answer.

Let us say the cost price of an item is X
It is said that it is marked to make a profit of 20%. 
That means it is marked at 1.2X

Ravi gets a 10% discount on the marked price, 
$$0.9\left(1.2X\right)=1.08X$$

Saves 15 rupees, so 1.2X-1.08X
0.12X=15
X=125

Profit made by Gopi is 0.08(125)=10 rupees. 

Let us fix the Cost Price of the product to be X, and the Selling Price of the product to be 1.4X, since it is given that it is fixed to have a profit of 40%. 
If the CP has been 40% less, making the CP 0.6X, 
And the selling price is 5 rupees less, making it 1.4X-5
Profit will be 50%, 
So, $$1.5\left(0.6X\right)=1.4X-5$$
$$0.9X=1.4X-5$$
$$0.5X=5$$
$$X=10$$

Original selling price will be 14. 

We are told that there was two successive increments in the salary, with the second increment percentage twice the first one. Total Increment was 187.5%. 

Drawing up the equation
$$\left(1+z\right)\left(1+2z\right)=1.875$$
$$1+3z+2z^2=1.875$$
$$2z^2+3z-0.875$$
$$z=\frac{\left(-3\pm\sqrt{9+8\left(0.875\right)}\right)}{4}$$
$$z=\frac{\left(-3\pm\ 4\right)}{4}$$
$$z=0.25$$

Answer is 25%. 

It is given that a merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him.

Hence, the cost price of 105 cm clothes is 100 rupees. 

It is also known that he marked the price of 100 cm clothes as 110 rupees, and gave a 5% discount, and he cheated his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers.

Hence, the selling price of 95 cm clothes is 110*(19/20) rupees.

Therefore, the selling price of 105 cm clothes is 115.5 rupees.

Hence, the profit is 15.5%

The correct option is C

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A is $$\dfrac{5}{6}p$$

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of B is $$\dfrac{10}{9}p$$

Now, she sold them at a price such that a 10% profit is made on B

=> Selling price = s = 11/10 * 10/9 p => $$\dfrac{11}{9}p$$

=> Profit % on A = $$\dfrac{\left(\dfrac{11}{9}-\dfrac{5}{6}\right)}{\left(\dfrac{5}{6}\right)}\times\ 100$$ = 46.66% = nearly 47%

The cost price of the sunglass for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the selling price of the sunglass is 1200 rupees, which Kanu purchased. Hence, the profit made by Meenu is (1200-1000) = 200 rupees.

Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.

The cost price of the same sunglass for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.

Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.

Hence, the profit made by Minu in the second transaction is (300/960)*100% = 31.25%

The correct option is C

Let the number of white shirts be m, and the number of blue shirts be n. Hence, the total cost of the shirts = (1000m+1125n), and the number of shirts is (m+n)

The average price of the shirts is $$\ \frac{\ 1000m+1125n}{m+n}$$.

It is given that he set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10%.

Hence, the average selling price of the shirts = $$\left(\ \frac{\ 1000m+1125n}{m+n}\right)\times\ \frac{5}{4}\times\ \frac{9}{10}=\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)$$

The average profit of the shirts = $$\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)-\frac{\ 1000m+1125n}{m+n}=\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)$$

The total profit of the shirts = $$\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)\times\ \left(m+n\right)\ =\ \frac{1}{8}\left(1000m+1125n\right)$$

Now, $$=>\frac{1}{8}\left(1000m+1125n\right)=51000$$

$$=>1000m+1125n=51000\times\ 8=408000$$

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero. Therefore, m has to be maximum.

$$m\ =\ \ \frac{\ 408000-1125n}{1000}$$

The maximum value of m such that m, and both are integers is m = 399, and n = 8 (by inspection)

Hence, the maximum number of shirts = m+n = 399+8 = 407

Total syrup - 110 kg

Total juice - 120 kg

It is given, cost price of syrup is 20% less than the cost price of juice.

Let the cost price of juice per kg be 10CP

Cost price of syrup per kg is 8CP

10kg syrup -> cost price = 80CP

It is given, 10kg syrup is sold at 10% profit. This implies selling price = 1.1*80CP = 88CP

20kg juice -> cost price = 200CP

It is given, 20kg juice is sold at 20% profit. This implies selling price = 1.2*200CP = 240CP

It is given, Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg

Selling price of the remaining mixture = 308.32*200 = Rs 61664

Total S.P = 61664 + 328CP

Total C.P = 880CP + 1200CP = 2080CP

Overall profit = 64%

$$61664+328CP=\frac{164}{100}\left(2080CP\right)$$

Solving, we get CP = 20

Cost price for syrup per kg = 8CP = 8*20 = Rs 160

Raj invested Rs 10000 in the first year. Assuming the loss he faced was x%.

The amount after 1 year is 10,000*(1 - x/100). = 10000 - 100*x.

Given the balance was greater than Rs 5000 and hence x < 50 percent.

When Raj invested this amount in the second year he earned a profit which is five times that of the first-year percentage.

Hence the amount after the second year is : (10000 - 100x)(1+$$\frac{\left(5\cdot x\right)}{100}$$).

Raj gained a total of 35 percent over the period of two years and hence the 35 percent is Rs 3500.

Hence the final amount is Rs 13,500.

(10000 - 100x)(1+$$\frac{\left(5\cdot x\right)}{100}$$) = 13,500

$$\left(100+5\cdot x\right)\cdot\left(100\ -\ x\right)\ =\ 13500$$

10000 - 100*x +500*x - 5*$$x^2$$ = 13500.

$$5x^2-400x+3500\ =\ 0$$

Solving the equation the roots are :

x = 10, x = 70.

Since x < 50, x = 10 percent.

Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n+W, where W refers to the wage.

Then SP in the first case = $$12\times\ 100+11\times\ \left(n-100\right)$$

Given profit is 300 in this case: 1200+11n-1100-8n-W=300 =>3n-W = 200

In second case: 1200+9n-900-8n-W=-300 (Loss). => W-n = 600.

Adding the two equations: 2n = 800

n = 400.

Thus W = 600 + 400 = 1000

Let the amount invested by Anil Bobby and Chintu be x, y, and z.

Considering x+y+z = 100*p.

Given Anil's share was 70 percent = 70*p.

As per the information provided :

His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%.

Since the profits are distributed in the ratio of their investments :

With a 3% decrease in the profits the value of profit earned by A decreased by Rs 420 which was 70 percent of the total invested.

Hence for all three of them would be combinedly losing $$\left(420\right)\cdot\left(\frac{10}{7}\right)\ =\ 600$$

Hence 3 percent profit was equivalent to  Rs 600.

The initial investment is equivalent to Rs 20000.

This is the total amount invested.

Chintu's profit share increased by Rs 80 when the profit percentage increased by 2 %. A 2 percent increase in profit is equivalent to Rs 20000*2/100 = Rs 400.

Of which Rs 80 is earned by Chintu which is 20% of the total Rs 400.

Hence he invested 20% of the total amount.

Bobby invested the other 10 percent.

10 percent of Rs 20000 = Rs 2000

Let the total marks be 100x

Marks obtained by Bishnu = 52x

Marks obtained by Asha = 64x

Marks obtained by Ramesh = 52x+23

Marks obtained by Ramesh = 64x-34

=> 52x+23 = 64x-34

=> x = $$\frac{19}{4}$$

Marks obtained by Geeta =84x = 84*19/4 = 399

Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000...(i)

1.2x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

Let the cost price of 1kg of sugar = Rs 100

The total cost price of 35 kg = Rs3500

Marked up price per kg = Rs 120

GIven, the final profit is 15% => Final SP of 35 kg = 3500 *1.15 = Rs 4025

First 5 kg's are sold at 20% marked up price => $$SP_1=5\cdot100\cdot1.2$$ = Rs 600

Next 15 kgs are sold after giving 10% discount => $$SP_2=15\cdot100\cdot1.2\cdot0.9\ =\ 1620$$

3kgs of sugar got wasted

=> 23 kg of sugar was sold at Rs (600 +1620) = Rs  2220

Remaining 12kg should be sold at Rs 4025 - 2220 = Rs1805

=> SP of 1kg = 1805/12 $$\simeq\ 150$$

Hence, the seller should further mark up by $$\frac{\left(150-120\right)}{120}\cdot100\ =\ 25\%$$

Let the CP of the each toy be "x". CP of 12 toys will be "12x". Now the shopkeeper made a 10% profit on CP. This means that 

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be "m". Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m  Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

Let 'x' be the strength of group G. Based on the information, $$0.65x$$ constitutes of literate people {the rest $$0.35x$$ = illiterate}
Of this $$0.65x$$, 75% are old people =(0.75)0.65x old literates. The total number of old people in group G is $$0.72x$$ {72% of the total}. Thus, the total number of old people who are illiterate = $$0.72x-0.4875x\ =\ 0.2325x$$. This is $$\frac{0.2325x}{0.35x}\times\ 100\ \approx\ \ 66\%$$ of the total number of illiterates. Hence, Option D is the correct answer.

Assuming the income of Bimla = 100a, then the income of Amala will be 120a.

And the income of Kamala will be 120a*100/80=150a

If Kamala's income goes down by 4%, then new income of Kamala = 150a-150a(4/100) = 150a-6a=144a

If Bimla's income goes up by 10 percent, her new income will be 100a+100a(10/100)=110a

=> Hence the Kamala income will exceed Bimla income by (144a-110a)*100/110a=31

Assuming the maximum marks =100a, then Meena got 40a

After increasing her score by 50%, she will get 40a(1+50/100)=60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42   =>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

Assuming the cost price of pen = 100p and the cost price of book = 100b

So, on selling a pen at 5% loss and a book at 15% gain, net gain =  -5p+15b = 7 ....1

On selling the pen at 5% gain and the book at 10% gain, net gain = 5p+10b = 13  .....2

Adding 1 and 2 we get, 25b=20

Hence 100b= 20*4=80, 

C is the answer. 

CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

(x-y)/p = (1.56p-0.56p)/p = p/p=1

Let the cost of each bicycle= 100b

CP of 10 bicycles = 1000b

It is given that he sold six of these at a profit of 25% and the remaining four at a loss of 25%

SP of 10 bicycles = 125b*6+75b*4 

=1050b

Profit = 1050b-1000b =50b

50b=2000

CP = 100b = 4000

Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000

Let the retail price be M and cost price be C.

Given,

0.85 M = 1.02 C

M = 1.2 C

If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.

Let the manufacturing price of the table = $$x$$
Hence the price at which the wholesaler bought from the manufacturer = $$1.1 \times x$$
The price at which the retailer bought from the wholesaler = $$1.3 \times 1.1 \times x$$
The price at which the customer bought from the retailer  = $$1.5 \times 1.3 \times 1.1 \times x$$
$$1.5 \times 1.3 \times 1.1 \times x = 4290$$
=> x = 2000

Given,

C1 : C2 : C3 = 9 : 10 : 8 ... i

C2 : C4 : C5 = 18 : 19 : 20 ... ii

Let's multiply i by 9 and ii by 5

C1 : C2 : C3 = 81 : 90 : 72

C2 : C4 : C5 = 90 : 95 : 100

Therefore, C1 : C2 : C3 : C4 : C5 = 81 : 90 : 72 : 95 : 100

Given,

100x - 81x = 19

x = 1

Hence, total profit = 100 + 95 + 72 + 90 + 81 = 438

Let the number of dozens of candies he bought of each variety be x
Hence total cost = 12x + 15x = 27x
Total selling price = 16.50*2x = 33x
Profit = 33x - 27x = 6x
Given 6x = 150 => x = 25
Hence he bought 50 dozens of candies in total

Let the cost of good mangoes be 2x per kg. The cost of medium mangoes be x per kg. 

CP of good mangoes = 160x

CP of medium mangoes = 40x

His selling price = 0.9*2x = 1.80x

Therefore, total revenue generated by selling all the mangoes = 120*1.8x = 216x

Hence, the profit % = $$\frac{16x}{200x} * 100 $$ = 8%

Let the cost price be C and the marked price be M.

Given,

0.6 M = 1.2 C

M = 2C

CP of 60 toys = 60C

Now only 50 are remaining.

Hence, 

M (1 - d) * 50 = 72C

1- d = 0.72

d = .28

Hence 28%

In total Raju bets 6000Rs and ends up with no profit - no loss. So there are 3 possibilities.
1) White comes 2nd, Black comes 4th and Red comes 5th.
2) Black comes 1st, White comes 3rd and Red comes 4th or 5th.
3) Black comes 2nd, Red comes 3rd and White comes 4th or 5th.

So there can never be 3 horses between white and red according to above to possibilities. Hence option D cannot be true.

There are total 3 cases which satisfies the condition "no profit and no loss."

Case 1: White comes 2nd.(remaining two horses(red/black) come 4th/5th)

Profit from white horse = Final Amount - Initial Amount = 2000*3 - 2000 = 4000

Loss from Red and Black horse = 3000+1000 = 4000

Net profit = 4000-4000 = 0

Case 2: Black, Red come second, third respectively.(remaining one horse(white) comes 4th/5th)

Profit from Black = 1000*3-1000 = 2000

Profit from Red = 3000 - 3000 = 0

Loss from white = 2000

Net profit = 2000-2000 = 0

Case 3: black, white come first, third respectively.(remaining one horse(red) comes 4th/5th)

Profit from Black = 1000*4-1000 = 3000

Profit from White = 2000 - 2000 = 0

Loss from Red = 3000

Net Profit = 3000-3000=0

And it is mentioned that grey case 4th. ==> case 1 is wrong.(because, in that case red, black should come 4th,5th)

So option C cannot be true.

Cost of 20 units = 240+20b+400c

Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c

=> 480 = 20b + 2800c => 120 = 5b + 700c

Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c

=> 240 = 2400c => c = 1/10 and b = 10

Let the number of items needed for max profit be k

CP = $$240+10k+k^2/10$$

SP = 30k

Profit = SP - CP = $$30k - 240 - 10k - k^2/10$$ = $$20k - 240 - k^2/10$$

or Profit = $$\frac{1}{10} (-k^2 + 200k - 2400)$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400 + 7600 - 7600))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 10000) + 7600)$$

or, Profit = $$\frac{1}{10} (-(k - 100)^2 + 7600)$$

To maximise the value of Profit, $$-(k - 100)^2$$ must be 0.

So, $$k$$ must be equal to 100.

Hence, option B is the correct answer.

Cost of 20 units = 240+20b+400c
Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c
=> 480 = 20b + 2800c => 120 = 5b + 700c
Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c
=> 240 = 2400c => c = 1/10 and b = 10
Let the number of items needed for max profit be k
CP = $$240+10k+k^2/10$$
SP = 30k
Profit = SP - CP = $$30k - 240 - 10k - k^2/10$$ = $$20k - 240 - k^2/10$$
Maximum when 20 - k/5 = 0 or k = 100
Profit = 2000 - 240 - 1000 = 760

Let a, b and c be the percentages of amount invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b - 0.025c

Return attained if there is a fall in the stock market => 0.001a - 0.03b + 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b - 0.025c = 0.001a - 0.03b + 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let's put the values for a, b and c that satisfy the above equation.

b = 9, c = 16, a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed return is 0.2%

Let a, b and c be the percentages of amount invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b - 0.025c

Return attained if there is a fall in the stock market => 0.001a - 0.03b + 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b - 0.025c = 0.001a - 0.03b + 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let's put the values for a, b and c that satisfy the above equation.

b = 9, c = 16, a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed return is 0.2% and it is attained when 36% is invested in option B and 64% is invested in option C.

Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $$x^2/100 * P = 441$$
Let the new price be N => $$P - x^2/100 * P = N$$
Price after the second cycle = $$N - x^2/100 * N$$ = 1944.81
=> $$(P - x^2/100 * P)(1 - x^2/100) = 1944.81$$
=> $$(P - 441)(1 - 441/P) = 1944.81$$
=> $$P - 441 - 441 + 441^2/P = 1944.81$$
=> $$P^2 - (882 + 1944.81)P + 441^2 = 0$$
=> $$P^2 - 2826.81P + 441^2 = 0$$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25

After the first game, Ghosh Babu picked 8 of clubs => He gets Rs 8. Then the dealer picked Queen of clubs => Ghosh Babu pays Rs 16 => Ghosh Babu is at a loss of Rs 8 after 1st game.

After the second game, Ghosh Babu picked 10 of hearts => He gets Rs 10. Then the dealer picked 2 of spades => Ghosh Babu gets another Rs 10 => Ghosh Babu is now at a profit of Rs 12.

After the third game, Ghosh Babu picked six of diamonds => He gets Rs 6. Then the dealer picked ace of hearts => Ghosh Babu pays Rs 6 to dealer => Ghosh Babu is still at a profit of Rs 12.

In the fourth game, Ghosh Babu picks 8 of spades => He gets Rs 8. The the dealer picks jack of spades => Ghosh Babu pays Rs 16 to dealer => Ghosh Babu is at a profit of Rs 4.

Hence, the maximum profit earned is Rs 12.

After the first game, Ghosh Babu picked 8 of clubs => He gets Rs 8. Then the dealer picked Queen of clubs => Ghosh Babu pays Rs 16 => Ghosh Babu is at a loss of Rs 8 after 1st game.

After the second game, Ghosh Babu picked 10 of hearts => He gets Rs 10. Then the dealer picked 2 of spades => Ghosh Babu gets another Rs 10 => Ghosh Babu is now at a profit of Rs 12.

After the third game, Ghosh Babu picked six of diamonds => He gets Rs 6. Then the dealer picked ace of hearts => Ghosh Babu pays Rs 6 to dealer => Ghosh Babu is still at a profit of Rs 12.

In the fourth game, Ghosh Babu picks 8 of spades => He gets Rs 8. The the dealer picks jack of spades => Ghosh Babu pays Rs 16 to dealer => Ghosh Babu is at a profit of Rs 4.

Hence, the maximum profit earned is Rs 12.

The maximum loss that Ghosh Babu had was Rs 8.

He must have had at least Rs 8 so that he did not have to borrow any amount from others.

After the first game, Ghosh Babu picked 8 of clubs => He gets Rs 8. Then the dealer picked Queen of clubs => Ghosh Babu pays Rs 16 => Ghosh Babu is at a loss of Rs 8 after 1st game.

After the second game, Ghosh Babu picked 10 of hearts => He gets Rs 10. Then the dealer picked 2 of spades => Ghosh Babu gets another Rs 10 => Ghosh Babu is now at a profit of Rs 12.

After the third game, Ghosh Babu picked six of diamonds => He gets Rs 6. Then the dealer picked ace of hearts => Ghosh Babu pays Rs 6 to dealer => Ghosh Babu is still at a profit of Rs 12.

In the fourth game, Ghosh Babu picks 8 of spades => He gets Rs 8. The the dealer picks jack of spades => Ghosh Babu pays Rs 16 to dealer => Ghosh Babu is at a profit of Rs 4.

As Ghosh Babu earned a profit of Rs 4 and now he has Rs 100, he initially would have had 100 - 4 = Rs 96.

Let's say price of turban is x.

So total price for 12 months will be = $$90+x$$

total price for 9 months = $$\frac{(90+x) \times 9}{12} = (65+x) $$

By solving above equation, we will get value of x= 10.

Let the total production cost be 100.
Hence, selling price is 120.
Price of German component A is 30 and the price of the US component B is 50

After change in exchange rate, price of German component is 30*1.3 = 39
and price of US component is 50*1.22=61

Total increase equals 39+61-30-50 = 20
Hence, the minimum production cost is 100+20=120
The maximum possible selling price is 120*110% = 132.

So, maximum possible gain is (132-120)/120 = 10%

Let the total production cost be 100.
Hence, selling price is 120.
Price of German component A is 30 and the price of the US component B is 50

After change in exchange rate, price of German component is 30*1.2 = 36
and price of US component is 50*0.88=44

Total increase equals (36+44)-(30+50) = 0
Hence, the total production cost did not change.
As the selling price also did not change, the gain percentage equals 20%

Let's say cost price is 100
gain = 14.28
selling price = 114.28
Marked price = x(say)
So $$x- \frac{11.11x}{100} = \frac{8x}{9} = 114.28$$
Or $$x = 128.52$$
So marked price is 28.52% more than cost price.
 

Let's say he buy fruits of weights 3 kg., 4kg., 5 kg.
Total kilograms of dry fruits $$=3+4+5=12$$

Overall cost price $$=3\cdot100+4\cdot80+5\cdot60=300+320+300=920$$

So cost price per kg. $$=\dfrac{300+320+300}{12} = \dfrac{920}{12}$$

Selling price = $$\dfrac{920}{12} \times \dfrac{3}{2}$$ = 115 per kg (Since Profit is 50%)

Hence answer will be D.

He earns x% on first 2000 and y% on rest of his income.

So on 4000 rs. , he will earn as follows:

$$2000 \frac{x}{100} + 2000 \frac{y}{100}$$ = 700

Or $$x+y = 35$$

Similarly on 5000 rs. ,he will earn 900 as follows:

$$2000 \frac{x}{100} + 3000 \frac{y}{100} = 900$$

Or $$20x + 30y = 900$$

On solving above equations, we will get value of x = 15

Let's say the cost of the cloth is x rs per metre. Because of the faulty meter, he is paying x for 120 cms when buying. 

So cost of 100 cms = 100x/120.

He is selling 80 cms for x, so selling price of 100cms of cloth is 100x/80. 

discount = 20% 

so the effective selling price is .8*100x/80= x

profit = SP-CP= x - 100x/120 = x/6 

Profit % = x/6 divided by 100x/120 = 20%
 
 

Let the original weight of the diamond be equal to $$10k$$. So, after breaking into 4 pieces, the parts of the diamond weight $$k, 2k, 3k,4k$$

The price of the diamond varies directly in proportion to the weight. Let us assume, the $$P=C*W^2$$ where $$C$$ is a constant and $$W$$ is the weight of the diamond.

Therefore, the original price is $$C*10k*10k = 100k^2*C$$
The new weight is $$Ck^2 + C(2k)^2 + C(3k)^2 + C(4k)^2 = 30k^2C$$

The decrease in the price equals 70,000. So, $$100k^2C-30k^2C = 70000$$

Or, $$k^2C = 1000$$

Therefore the original price = $$100k^2C = 100000$$

As shopkeeper gave 3 one-rupee change for 20 rs. change, Buyer must have ordered for a total of 17 rs. stamps.
Now buyer ordered for at least more than 1 stamp for each type
Hence the minimum he bought was:

2 stamp for 5 rupees = 10 rs.
2 stamp for 2 rupees = 4 rs.
2 stamp for 1 rupee = 2 rs.

For the total to be seventeen, the buyer must have purchased 3 one rupee stamps.  
And 3 one rupee stamps were also there as changes given by shopkeeper.
So total number of stamps = 2+2+(3+3) = 10
 

let's say price of maruti car is x rs.
Sales = y
revenue = xy
Changed price = 1.3x
changed value of sales = 0.8y
new revenue = 1.04 xy
Percentage change in revenue = 4%

Selling price of first watch = 300
Profit = 10%
cost price = $$\frac{300}{1.1}$$
Selling price of second watch = 300
Loss = 10%
cost price = $$\frac{300}{0.9}$$

Total selling price of transaction= 600
Total cost price of transaction = $$300(\frac{10}{11} + \frac{10}{9}) = 600 (\frac{100}{99})$$
Loss = $$600 (\frac{100}{99} - 1)$$
%loss = $$(600 (\frac{100}{99} - 1)) \div (600(\frac{100}{99})) \times 100 = 1$$

Cost price per watch = 150
Cost price for 1500 watches = $$1500 \times 150$$ = 225000
Total expense = 225000 + 30000 = 255000
Selling price for season = $$1200 \times 250$$ = 300000
For out of season = $$300 \times 100$$ = 30000
Total selling = 300000 + 30000 = 330000
Profit = 330000 - 255000 = 75000

Break even implies that cost price is equal to selling price
Hence let's say in season x watches were sold
Cost price will be = $$1500 \times 150 + 30000 = 255000$$
So total selling price = $$250x + (1500 -x) 100 = 255000 $$
Or $$x = 700$$

 

Let's say total money was $$x$$ rs.
So cost price of 40 mango will be = $$x$$ ;
Hence cost price of 20 mangoes will be = $$\frac{x}{2}$$
Taxi fare = $$\frac{10x}{100}$$
Total expense = $$\frac{x}{2}$$ +  $$\frac{10x}{100}$$ = $$\frac{6x}{10}$$
Remaining money =$$ \frac{4x}{10}$$
Cost price of 1 orange will be = $$\frac{x}{50}$$
Hence in $$\frac{4x}{10}$$ rs. 20 oranges can be purchased.

The total property consists of cash, gold coins and silver bars.
And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.
As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.
So total property will be =$$\frac{ 25000 \times 100}{5}$$ = 500000 
Hence property amounting only gold coins and silver bars will be = Total property - Total Cash
i.e. = 500000 - (25000 + 50000 + 75000 + 75000)
= 2,75,000 

The total property consists of cash, gold coins and silver bars.
And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.
As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.
So total property will be =$$\frac{ 25000 \times 100}{5}$$ = 500000 

Let the number of gold bars be 7a and the number of silver bars be 27a.
The total value of the gold and silver bars is 500000 - 225000 = 275000
Therefore, 4000*7a + 1000*27a = 275000
Or, 55000a = 275000
Or, a = 5

Therefore the number of silver bars with Ghosh Babu is 27a = 135

It is given that the maximum cost is 12 lakhs
$$1200000 <= T \times 10000 + V \times 15000$$ (Where T is number of TV's and V is number of VCR's)
and as we know $$T+V = 100$$
By solving the above two equation, we will get v<=40

We have to keep the no. of VCRs as maximum, as they fetch higher profit than TV, so

$$T = 60$$
$$V = 40$$​​​​​

As we know for maximizing profit, Total Revenue = Marginal Cost
i.e. $$T \times 10000 + V \times 15000 = 1200000$$
and $$T+V = 120$$
By solving above two equations we will get
$$T=120$$
$$V=0$$
Hence ratio will be 0.​​​​​​