Model questions for TCS Placement Papers - Questions Profit And Loss 1

                                                   Profit and Loss

Sample questions for Placement Papers – Solved Problems 1

3 mangoes and 4 apples costs Rs 85. 5 apples and 6 peaches costs Rs. 122. 6 mangoes and 2 peaches cost Rs.114. what is the combined price of 1 apple,1 peach and 1 mango?

 (a)Rs 37                 (b)Rs 39                  (c) Rs 35                 (d)Rs 36        e)None

Answer : A 

Explanation :

3 mangoes and 4 apples costs Rs 85  => 3m+4a=85...(i)

5 apples and 6 peaches costs Rs. 122 =>  5a+6p=122....(ii)

6 mangoes and 2 peaches cost Rs.114.=> 6m+2p=114....(iii)

On  solving (i) & (iii),we get the equation in terms of a and p => -8a+2p=-56 ..(iv)

m=15,a=10,p=12

Now solving (ii) and (iv) , we get  58p=696 => p=12

Substituting p=12 in equation (iii)  , we get m=15

Substituting m=15 in equation (i), we get a=10

m=15,a=10 and p=12

The combined price of 1 apple , 1 peach and 1 mango = m +a+ p= 15 +10 +12=Rs37

Sample questions for Placement Papers – Solved Problems 2

A person buys a brand new Honda bicycle for Rs.800. He uses it for a year and then sells it. He makes a profit of 10% in this transaction. Then, again, he buys a new cycle but finds that the price has gone up by 20%. He uses the new cycle for a year and then sells it for a profit of 30%. Find the average annual profit of the person through these transactions.

a)160                      b)184                      c)168                      d)172

Answer: B

Explanation: He bought for 800 at first.

He Sold at a profit of 10% => Profit on first cycle = 800 $ \times\frac{10}{100}$=80

He again bought a new one at a price hike of 20% =-> New price= 800 $ \times\frac{120}{100}$ =960

He sold the second cycle at 30% profit = > Profit on the second cycle = 960 $ \times\frac{30}{100}$ =288

Total profit in 2 transaction is=80+288=368;

Average annual profit i = $\frac{368}{2}$=184rs

Sample questions for Placement Papers – Latest Pattern Solved Problems 3

A man purchased cow and horse at rs 200000 and sold a cow for 20% profit and horse at 20% a loss, if the overall gain is 4000 find the selling price of cow and horse?

          a)    Rs 132000,Rs 72000   b)Rs 140000,Rs64000   

c)Rs 114000,Rs 90000     d) Rs 124000,Rs 80000

Answer A

Explanation:

Let the price of a cow is 'C' and the price of the horse is 'H' 
The cost price of cow and horse => C + H =200000 ....(I)

He sold the cow for 20% profit and horse at 20% loss. In overall he gained Rs 4000

   Selling price of cow and horse => 1.2 C + 0.8 H  =204000....(II)

Solving (I) and (II) => From (i) C = 200000 -.H

Substituting this in (II) , we get H = 90000 and C= 110000

Cost price of horse = 90000 and cost prie of cow =110000

Selling price of Cow = 110000 $\times\frac{120}{100}=132000$

Selling price of Horse = 90000 $\times\frac{90}{100}$=72000

Sample questions for Placement Papers – Latest Pattern Solved Problems 4

If Suraj's sum of the salaries of 2003, 2004, 2005 is $36400 and the salary gets incremented 20% every year then find the salary drawn by Suraj in the year 2005?

a)     14400             b) 15200     c) 16500      d)18000

Answer : A

The salary of Suraj is increased by 20% every year.

When a salary is increased by 20%, its value after increase is salary $ \times\frac{120}{100}$

                                                                                           = salary $ \times\frac{6}{5}$

In 2004 , his salary = S $ \times\frac{6}{5}$

In  2005 , his salary = S $ \times\frac{6}{5} \times\frac{6}{5}$

Sum of his salaries in 2003,2004 and 2005 =>$ S \;+\,\frac{6}{5}S\,+ \;\frac{36}{25}S\;$= $36400  

                         On solving, we get S=10000

Salary in 2005 = 10000 $\times\frac{120}{100}\times\frac{120}{100}$   = $14400

                                               

Sample questions for Placement Papers – Solved Problems 5

A man sold 12 candies at 10 dollars had a loss of b% and again sold 12 candies at 12 dollars had a profit of b%. Find the value of b?

a)10%          b)9.09%                 c)8%            d)11.11%

Answer: B

Explanation:

Here the profit and loss in both transactions is the same when they are sold at two different prices, their Cost price will be the average of two selling prices.

CP=$\frac{12+10}{2}$

 Therefore the cost price is $11

Now

11 $ \times\frac{100+b}{100}$ =12

 =9.09%

Alternate Method :

Let the cost price be CP

 One sold at b% profit and other sold at b% loss

  Loss $ =\frac{CP-SP}{CP}\times100\implies\;\frac{CP-10}{CP}\times100% =b\% $ ----------->1

 Profit $=\frac{SP-CP}{CP}\times100\implies\;\frac{12-CP}{CP}\times100 % =b\% $ ------->2

 Here both profit and loss are equal

    $\frac{CP-10}{CP}\times100=\;\frac{12-CP}{CP}\times100 $

     On solving , we get CP=11

Now substituting the value of CP in equation =>   $\frac{11-10}{11} \times100  $ %= b%

= > b%= 9.09%

Sample questions for Placement Papers – Solved Problems 6

A dealer buys a product at Rs.1920. he sells at a discount of 20% still he gets the profit of 20%. what is the selling price?

a)Rs. 2304     b) Rs. 1536    c) Rs. 2200    d) it is not possible

Answer : A

Explanation :

Cost price of the product is Rs 1920.

Discount always given on marked price and profit is always obtained on the cost price

  $ CP\, \times\frac{120}{100} =\, MP\, \times \frac{80}{100}$ (20% profit got even after giving 20% discount on MP)

$1920\, \times\frac{120}{100} =\, MP\, \times \frac{80}{100} $

              MP=2880

Selling price = $MP\, \times \frac{80}{100}\, =\, 2880\times\frac{80}{100} $=  Rs 2304

Sample questions for Placement Papers –Latest Pattern Solved Problems 7

A Person buys a horse for 15 pounds, after one year he sells it for 20 pounds. After one year, again he buys the same horse at 30 pounds and sells it for 40 pounds. What is the profit for that person?
option 
(a) 20 pounds          (b) 15 pounds          (c) 25 pounds          (d) 10 pounds
Answer : b

Explanation :   Profit in first transaction = 20- 15 = 5 pounds

                          Profit in 2^nd transaction = 40-30 = 10 pounds

Total profit for that person = 5 + 10 = 15 pounds

Sample questions for Placement Papers – Solved Problems 8

A man sold 2 pens. The initial cost of each pen was Rs. 24. If he sells it together one at 25% profit and another 20% loss. Find the amount of loss or gain, if he sells them separately?
Answer : 
Explanation : 

Given CP of each pen=Rs 24

One pen is sold at 25% gain and another at 20% loss

The SP of the pen sold at profit 25%  = 24 $ \times\frac{120}{100}$ = Rs 30

 SP of the pen sold at 20% loss = 24 x $ \times\frac{80}{100}$= Rs 19.2

 SP of 2 pens = 30 + 19.2 = 49.2  and Total Cost price of 2 pens = 24 + 24 =48

Total SP=30+19.2=49.2 Rs

 If they sell them separately SP =24+24=48 Rs

If they sell them separately, they will get (48 -49.6)= Rs 1.6 loss

Sample questions for Placement Papers – Solved Problems 9

The marked price of a coat was 40% less than the suggested retail price.  Eesha purchased the coat for half the marked price at the fiftieth-anniversary sale.  What percentage less than the suggested retail price did Eesha pay?

a) 60                      b) 20                       c) 70                      d) 30

Answer : C

Explanation :

Let the retail price is Rs.100.

Then market price is  40% less than retail price.

Therefore marked price is (100-40) % of 100 = 60.  

Eesha purchased the coat for half the marked price.  = ½  x 60 =30 which is 70 less than the retail price. 

Sample questions for Placement Papers – Latest Pattern Solved Problems 10

A dealer originally bought 100 identical batteries at a total cost of q rupees. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many rupees was each battery sold?

a)150q          b)q/150                  c)3q/200                 d)2q/300

Answer: C

Explanation:

Cost of 100 batteries = q rupees
Cost of 1 battery =Rs  $\frac{q}{100}$ 
It is given that the selling price of 1 battery is 50 percent above the original cost per battery.

So the selling price of 1 battery = $ \frac{q}{100}+\left[\frac{q}{100}\times\frac{50}{100}\right] $

                                              = $\frac{q}{100}+ \frac{q}{200} \, = \:\frac{3q}{200}$

                                            

Sample questions for Placement Papers – New Pattern Question 11

Raj goes to the market to buy oranges. If he can bargain and reduce the price per orange by Rs 2 he can buy 30 oranges instead of 20 oranges with the money he has. How much money does he have?
a)Rs 100                 b) Rs 250                c) Rs 175                d) Rs 120

Answer: C

Explanation: Let the price of one orange is x

If he can reduce the price by Rs 2 he can buy 30 oranges...

=>Total money= 30(x -2)

Similarly, if the price is not reduced, then he can buy 20 oranges

      =>Total money = 20x

  20x = 30(x-2) => 30x -20x =60 => x =6

So total money he has => 20 x => 20 x 6 = 120