XAT Profit and Loss Questions

Given that the plate costs 120, they are selling it for 160, and the number of buyers is 300; the current profit is 40 per customer.

It is crucial to note that any increase of price by 10 will decrease the number of customers by 10, and conversely, any decrease of price will increase the number of buyers by 10.

Let's say the maximum profit is gained by increasing the profit by x times 10. In that case, the number of customers will decrease by x times 10.

The profit will be (40+10x)(300-10x); this has to be maximum.

Let's expand this expression.

(40+10x)(300-10x)=$$12000+3000x-400x-100x^2$$

$$12000+2600x-100x^2$$

This has to be maximum.

Let's multiply this expression by -1 for having positive coefficient of $$x^2$$

$$100x^2-12000-2600x=100x^2-2*130*10x+130^2= (10x-130)^2-12000-130^2$$

$$(10x-130)^2-28900$$

Before, we multiplied the expression by -1 to revert that again, multiply by -1.

The new expression is $$28900 - (10x-130)^2$$. The maximum value of this expression is 28900.

Let the amount invested by Jack and Sristi be $$'x'$$.

Jack's amount after four months will become $$\frac{x}{2^4}$$ and after six months it will become $$\frac{x}{2^6}$$.

Sristi's amount after four months will become $$x-\left(4\times\ 15000\right)$$ and after six months it will become $$x-\left(6\times\ 15000\right)$$

From the given information,

$$\ \frac{\ \ \frac{\ x}{2^4}}{x-\left(4\times\ 15000\right)}=\ \frac{\ \ \frac{\ x}{2^6}}{x-\left(6\times\ 15000\right)}$$

$$\ \ \left(\ x-\left(4\times\ 15000\right)\right)=\ 2^2\times\ \left(x-\left(6\times\ 15000\right)\right)$$

$$3x\ =\ 300000$$

$$x\ =\ 100000$$.

The amount invested by Jack and Sristi is 100000.

Option 'A' is correct.

Let printed price be 100.

Rajnish bought it at 25% discount, C.P of Rajnish = 75% of 100 = 75.

He sold it at 10% discount on printed price, S.P of Rajnish = 90% of 100 = 90

Hence, Rajnish, profit% = $$\ \frac{\ 90-75}{75}\times\ 100=20%$$.

The answer is option D.

The two types of apples sold A and B are bought in the ratio of 5: 8.

Considering the weights to be 5x and 8x.

The cost price of A is 20 percent higher than that of B.

Considering the cost price of B = y, A = 6y/5.

The total cost price of A = $$\left(5x\right)\cdot\left(\frac{6y}{5}\right)$$

The total cost price of B = $$\left(8x\right)\cdot\left(y\right)$$

THe total cost price = 8xy + 6xy = 14xy

14xy = 2800.

xy = 200.

THe cost price of A = 1200.

THe cost price of B = 1600.

A is sold a profit of 15 percent. 15 percent of 1200 = 180.

B is sold at a profit of 10 percent. 10 percent of 1600 = 160.

The total profit is 180 + 160

= 340

Let the part of the amount divided between Y and Z be 5k => Y gets 2k and Z gets 3k.

The overall profit is Rs 200000.

Hence the remaining profit is Rs 200000 - 5k. =

Left over profit of 2-5k is divided in the ratio 2.5:3.5:4

The final profit distribution among X, Y and Z.

=> Finally, X gets $$\frac{2.5}{10}\left(200000-5k\right)\ $$, Y gets $$2k+\frac{3.5}{10}\left(200000-5k\right)\ $$ and Z gets $$3k+\frac{4}{10}\left(200000-5k\right)\ $$.

Given the ratio of profit distribution of X and Z is 1 : 4

Given, $$3k+\frac{4}{10}\left(200000-5k\right)\ $$ = 4($$\frac{2.5}{10}\left(200000-5k\right)\ $$) => 3k=$$\frac{6}{10}\left(200000-5k\right)\ $$ =>10k=400000-10k => 20k=400000 => k=20000.

.'. Share of Y = $$2k+\frac{3.5}{10}\left(200000-5k\right)\ $$  = 75000. 

There are 9 employees in the firm. Median of any set is the element which occurs in the middle of the set when the elements are positioned in increasing or decreasing order. The number of elements in this set = 5+3+1 = 9

Thus, the middle element will be in the position $$\frac {9+1}{2}$$ = 5th. Since we know that the clerks are paid the least so the clerk salaries would be in the beginning of the set if arranged in ascending order. Thus, the 5th element will be a clerk's salary as the number of clerks is 5. 

Thus, median of the set = Rs 15,000.

To find the mean salary, we first need to find the sum total salary of the people involved. 

Total salary = (Number of clerks*salary of 1 clerk) + (Number of assistants*salary of 1 assistant) + (Number of accountants*salary of 1 accountant) 

$$ \Rightarrow$$ Total salary = (5*15,000)+(3*40,000)+(1*66,000)

$$ \Rightarrow$$ Total salary = 75,000+1,20,000+66,000 

$$ \Rightarrow$$ Total salary = 2,61,000 

Mean salary = $$\frac{\textrm{Total Salary}}{\textrm{Total number of employees}} $$

$$ \Rightarrow $$ Mean salary = $$ \frac{261000}{9} $$

$$ \Rightarrow $$ Mean salary = Rs 29,000 /-

Thus, difference between mean and median salary = 29,000-15,000 = Rs 14,000

Let CP of object be a. 

It is given that SP = $$(1+\frac{4}{100} )$$ x CP 

$$ \Rightarrow $$ SP = 1.04 x CP

It is given that MP = $$(1+\frac{x}{100} )$$ x CP

It is also given that SP = $$(1-\frac{\frac {2x}{3}}{100} )$$ x MP 

$$ \Rightarrow $$ SP = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ x CP 

$$ \Rightarrow (1+\frac{4}{100})$$ x CP  = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ x CP 

$$ \Rightarrow (1+\frac{4}{100})$$   = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ 

$$ \Rightarrow \frac{104}{100}$$   = $$(1-\frac{2x}{300} )$$ x $$(1+\frac{x}{100} )$$ 

$$ \Rightarrow \frac{104}{100}$$   = $$(\frac{300-2x}{300} )$$ x $$(\frac{x+100}{100} )$$ 

$$ \Rightarrow \frac{104}{100} \times 300 \times 100$$   = $$(300-2x)$$ x $$(x+100)$$ 

$$ \Rightarrow 31200 $$   = $$(300-2x)$$ x $$(x+100)$$ 

Now, we look at the options. 

Since the question says that $$x\epsilon[25,30]$$ so 50% of x ie 0.5x cannot be less than 12.50 ie $$\frac {25}{2}$$ and cannot be more than 25 ie $$\frac {50}{2}$$

This eliminates option A.

Putting values of $$x$$ given in the remaining options in the final expression, [here we need to be careful to use the value of x and not the value of 50% of x as given in the expression]

Look carefully in the remaining options : 16,13,15 as 0.5x ie 32,26,30 as possible values of x. 

In the question since the discount rate offered is $$\frac{2x}{3}$$%, then a safer choice would be to check for the option of 30 in the beginning. It is a safer choice because the percentage of discount that we get in the other options are not whole numbers. 

NOTE : THIS IS JUST A SAFE CHOICE AND NEVER MARK AN ANSWER DIRECTLY ON THIS PRESUMPTION WITHOUT CHECKING IT.

Putting x=30 in the expression : 

(300 - (2x30))x(30+100) = (300-60)x(100+30) = 240x130 = 24x13x100= 312x100 = 31200= LHS of expression.

Thus the value of $$x$$ is 30.

Therefore value of 50% of $$x$$ = 0.5x30 = 15 

Let the cost price of the article be Rs.100.
The shopkeeper raises the price by x% and then decreases it by y%.
As a result, he reaches the cost price of the article. 
Also, it has been given that y is the difference between y% and x%.
$$y = x-y$$
$$2y = x$$
We know that $$(1+2y)(1-y)*100 = 100$$
$$(1+2y)(1-y) = 1$$
$$1-y+2y-2y^2 = 1$$
$$2y^2-y=0$$
$$2y = 1$$
$$y = 1/2$$ or $$0.5$$
$$x = 2y$$ 
=> $$x = 1$$ or $$x = 100$$%
Therefore, option D is the right answer. 

Patel bought 20 shirts. He would have gotten every fourth shirt at Rs. 100. He would have paid the marked price for 15 shirts and would have gotten 5 shirts at Rs. 100 each. 

Let the marked price of one shirt be 'x'.
15x + 5*100 = 20000
15x = 19500
x = 19500/15 = Rs. 1300.
The marked price of a shirt is Rs. 1300. Therefore, option B is the right answer. 

Let number of apples = $$x$$ and oranges = $$y$$

=> $$23x + 10y = 653$$   $$(x > y)$$

Since, 653 has last digit 3, which is possible when 23 is multiplies by 1,11,21,31 and so on.

Also, $$x > y$$ => $$x = 21$$ and $$y = 17$$

=> C.P. of 1 apple = $$\frac{100}{115} \times 23 = 20$$

C.P. of 1 orange = $$\frac{100}{125} \times 10 = 8$$

=> Total C.P. = $$(21 \times 20) + (17 \times 8) = 420 + 136 = 556$$

$$\therefore$$ Profit % = $$\frac{653 - 556}{556} \times 100 = 17.4 \%$$

Let Manufacturing Cost of the product = $$Rs. 100$$

=> Maximum Retail Price(MRP) = $$100 + \frac{55}{100} \times 100 = Rs. 155$$

Retailer gives 10% discount on MRP

=> Retailer's selling price = $$155 - \frac{10}{100} \times 155 = Rs. 139.5$$

It is given that the retailer earned 23% profit on his purchase price, say $$Rs. x$$

=> $$\frac{123 x}{100} = 139.5$$

=> $$x = \frac{13950}{123} = 113.41$$

Now, the purchase price of retailer = $$x$$ = selling price of Manufacturer

$$\therefore$$ Profit earned by Manufacturer = $$113.41 - 100 = 13.41$$

$$\approx 13 \%$$

Let Nikhil buy $$x, y$$ and $$z$$ pieces of kajubarfi, gulabjamun and sandesh respectively.     $$(x,y,z \geq 1)$$

=> $$x + y + z = 100$$ --------------Eqn(I)

Also, $$10x + 3y + \frac{1}{2} z = 100$$

=> $$20x + 6y + z = 200$$ ----------Eqn(II)

Subtracting eqn(I) from (II), we get :

=> $$19x + 5y = 100$$

=> $$y = \frac{100 - 19x}{5}$$

If $$x = 1$$, $$y$$ will not be natural. The only value of $$x$$ for natural $$y$$ is $$x = 5$$

=> $$y = \frac{100 - 95}{5} = 1$$

$$\therefore$$ Nikhil must buy 1 gulabjamun.

Let 'n' be the number of potato chips bought by the brothers. Also let 'x' be the cost price of a banana chips. ( x < 10) 

Total number of chips purchases = (n + 1). It is given that each brother has equal number of chips packets i.e. (n + 1) is an even number or we can say that 'n' is odd. 

Total amount spend by the brother on these chips packets = 10n + x. It is given that the amount received for each pot was same as the number of pots sold.

Hence, we can say that 10n + x is a perfect square. We can see that the tens place digit is an odd number. 

Perfect squares ending with an odd digit in the tens place = 16, 36, 196, 256 and so on {All (10a $$\pm$$ 4)$$^2$$ type numbers}

We can see that unit place is 6 in all cases and that will be the same as cost price of a banana chips packet. 

The difference between the amount with the two friends = Cost price of 1 potato chips packet - Cost price of 1 banana chips packet = 10 - 6 = 4

Hence,we can say that the brother, who has only chips packets with him, should given Rs.2 to the other brother so that they have the same amount with them.

Therefore, option B is the correct answer. 

Ratio of profits earned by Ram : Shyam = 4000 : 6000

= 2 : 3

If profit < 2,00,000

% of profit earned by Shyam = $$\frac{3}{5} \times$$ 100 = 60%

If 2,00,000 < profit < 4,00,000, he gets 20 % and 60 % of the remaining profit.

% of profit earned by Shyam = 20% + .80 $$\times$$ 60% = 68%

If profit > 4,00,000

% of profit earned by Shyam = 35 % + .65 $$\times$$ 60% = 74%

Now, for first 2,00,000 profit earned by Shyam = $$\frac{60}{100} \times$$ 2,00,000 = Rs. 1,20,000

For second 2,00,000 profit earned by Shyam = $$\frac{68}{100} \times$$ 2,00,000 = Rs. 1,36,000

Let total profit earned by them = Rs. (4,00,000 + $$x$$)

=> From $$Rs. x$$ profit, Shyam received = 3,67,000 - 1,20,000 - 1,36,000 = Rs. 1,11,000

=> $$\frac{74}{100} \times x$$ = 1,11,000

=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000

$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000

Let the number of units of A and B produced be $$x$$ and $$y$$ respectively.

For product A, time taken for grinding = $$2x$$ and polishing = $$3x$$

For product B, time taken for grinding = $$3y$$ and polishing = $$2y$$

Total number of hours of grinding done per day = $$10 \times 12 = 120$$ hrs

Total number of hours of polishing done per day = $$15 \times 10 = 150$$ hrs

=> $$2x + 3y = 120$$ ---------Eqn(I)

and $$3x+ 2y = 150$$ ----------Eqn(II)

Applying the operation : 3* Eqn(I) - 2* Eqn(II), we get :

=> $$(6x - 6x) + (9y - 4y) = 360 - 300$$

=> $$y = \frac{60}{5} = 12$$

=> $$x = 42$$

$$\therefore$$ Profit made by the manufacturer = $$(42 \times 5) + (12 \times 7)$$

= $$210 + 84 = Rs. 294$$

Let total cost price of Books and More = $$RS. 100x$$

It is given that Books and More have earned 20% profit overall

=> Total S.P. = $$\frac{120}{100} \times 100x = 120x$$

Thus, S.P. of music CDs = $$\frac{35}{100} \times 120x = 42x$$

S.P. of books = $$42x + \frac{50}{100} \times 42x = 63x$$

=> S.P. of DVDs = $$120x - 42x - 63x = 15x$$

40% profit is earned in music CDs and 25% profit in books.

=> C.P. of music CDs = $$\frac{100}{140} \times 42x = 30x$$

and C.P. of books = $$\frac{100}{125} \times 63x = 50.4x$$

=> C.P. of DVDs = $$100x - 30x - 50.4x = 19.6x$$

=> Loss made on DVDs = $$19.6x - 15x = 4.6x$$

$$\therefore$$ Loss % on DVDs = $$\frac{4.6 x}{19.6 x} \times 100 = 23.46 \%$$

Let total selling price of books and more = $$Rs. 100x$$

Thus, S.P. of music CDs = $$\frac{35}{100} \times 100x = 35x$$

S.P. of books = $$35x + \frac{50}{100} \times 35x = 52.5x$$

=> S.P. of DVDs = $$100x - 35x - 52.5x = 12.5x$$

40% profit is earned in music CDs and 25% profit in books.

=> C.P. of music CDs = $$\frac{100}{140} \times 35x = 25x$$

and C.P. of books = $$\frac{100}{125} \times 52.5x = 42x$$

It is given that Books and More made 50% loss in film DVDs

=> C.P. of DVDs = $$\frac{100}{50} \times 12.5x = 25x$$

Thus, total C.P. of books and more = $$25x + 42x + 25x = 92x$$

$$\therefore$$ Profit made by books and more = $$\frac{100x - 92x}{92x} \times 100$$

= $$\frac{8}{92} \times 100 = 8.69 \% \approx 8.7 \%$$