Solution of Profit & Loss Related Complicated Problems

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Results Home » Profit And Loss-High School  School   » Solution of Profit & Loss related complicated Problems

Solution of Profit & Loss related complicated Problems Question 01:  01:  A man sells a car to his friend at 10% loss. If the friend sells it for Tk. 54000 and gains 20%, the original cost price of the car was:

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Solution: S.P. =Tk. 54000, Gain earned=20%.

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So, C.P. =Tk. (100*54000/120) =Tk. 45000. Now, S.P. =Tk. 45000 and loss=10%. Then, C.P. =Tk. (100*45000/90) = Tk. 50000. Question 02:  02:  A dealer sells a radio at a gain of 10%. If he had bought it at 10% less and sold it for Tk. 132 less, he would have still gained 10%. The cost price of the radio is:

BCS MATH Percentage Related Math Solutions for Any Written Aptitude Test Question 01: The value of a machine depreciates 10% annually. If its present value is

Solution olution::

Tk.4000, its value 2 years hence will be:

Let C.P C.P.. =Tk. =Tk. x.

Soluti...

Then, S.P. =Tk. (100*x/100) =Tk. 11x/10. New, C.P. = (90*x/100) = 9x/10 Tk. So, 11x/10 – 99x/100=132 Or, x=1200 So, C.P. =Tk. 1200.

Arithmetic (Average) math problems for BCS, BBA, MBA Exam Arithmetic Average part mathematics solution for BCS, MBA Bank Job an d Other Examin ation Question 01: Find the average of first 25 ...

Question 03:  03:  When the price of fans was reduced by 20%, the number of 

Solution of Percentage Related Problems for Any

fans sold increased by 40%, what was the effect on the sales in Taka?

Aptitude Tests Question 01: After deducting a commission of 

Solution: Consider, original cost of each is Tk. 100 and number originally sold is 100.

5% of T.V. set costs Tk. 9595. Its gross value is: Solution: 95% of x = 9595 Or, x= (...

Original sale proceed=Tk. (100*100) =Tk. 10000;

Complicated Percentage Related Problem

New sale proceed=Tk. (80*140) =Tk. 11200.

Solutions

Increase % = (1200*100/10000) %=12%.

Question 01: Mr. Hassan Arif’s basic pay for a 40 hours week is Tk.20. Overtime is paid for at

Question 04:  04:  When the price of a toy was increased by 30%, the number of toys sold fell by 30%. What is the effect on the sale of the shop?

25% above the basic rate. In a certain wee... Solutions of Percentage (Profit & Loss) related Math Problems

Solution:

Question 01: 6% more is gained by selling a

Let original cost of each toy is Tk. 100 and number sold is 100.

radio for Tk. 475 than by selling for Tk. 451. The

Then, sale proceed=Tk. (100*100) =Tk. 10000;

cost price of the radio is: Solution:...

New sale proceed = Tk. (130*70) =Tk. 9100 Decrease %=( 900*100/10000) %=9% Question 05:  05:  By selling 100 bananas, a fruit-seller gains the selling price of 20 bananas. His gain percent is:

http://bcsm athschool .bl ogspot.com /2013/07/sol uti on- of- pr ofi t- l oss- r el ated.htm l#.V3- cZzV97IU

Percentage Related Questions of MBA Admission and Bank Exam Question 01: In a survey of the people in Chittagong, it was found that 65% of the people surveyed watched the news on television, 40%

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Sol uti on of Pr ofi t & Loss r el ated com pl icated Pr obl em s - BC S M ath School re...

Solution: Gain= (S.P. of 100 bananas)-(C.P. of 100 bananas) (S.P. of 20) = (S.P. of 100)-(C.P. of 100); So, S.P. of 80=C.P. of 100. Let C.P. of each=Tk.1.

Profit Loss related MBA/E-MBA admission test questions Question 01: A trader normally makes a profit of  40% on items he sells. If he reduces the price by 10%, his sales increase by 40%. Let x ...

Then, C.P. of 80=Tk. 80; S.P. of 80=Tk. 100.

Profit & Loss Related Solutions of BCS, MBA, and

Gain %=( 20*100/80) %=25%

BBA Exam Math Questions Question 01: A man sold an article for Tk. 161,

Question 06: By selling 36 oranges, a vendor of Shibgonj market loses the

gaining 1/6 of his outlay; Find the cost price of  the article. Solution: Let C.P. ...

selling price of 4 oranges. Determine his loss percent. Profit & Loss related Problem's Solutions

Solution:

Question 01: If books bought at prices ranging

Loss= (C.P. of 36 oranges)-(S.P. of 36 oranges);

from Tk. 200 to tk. 250 are sold at prices

(S.P. of 4) = (C.P. of36)-(S.P. of 36)

ranging from Tk. 300 to Tk. 425, what is the

(S.P. of 40) = (C.P.of 36; Let C.P. of each =TK.1 C.P. of 40=Tk. 40,

g... Solution of simple interest related easy problems Question 01: A shopkeeper borrowed Tk. 25000

S.P. of 40=Tk. 36

from two money lenders. For one loan he paid

Loss= (4*100/40) %=10%

12% per annum and for the other 14% per annum. ...

Question 07:  By selling an article for Tk. 144, a man loses 1/7 of his outlay. By selling it for Tk. 168, his gain or loss percent is:

RECENT UPDATES Solution: Let C.P. =Tk. x. Then, loss=Tk. (x/7); S.P. = (C.P.)-(Loss) =(x-x/7) =Tk. 6x/7. So, 6x/7=144 X= (144*7/6) =168; Thus, C.P= Tk. 168. If S.P. is Tk. 168, then gain percent is 0% Question 08:  When the price of a toy was increased by 20%, the number of toys sold was decreased by 15%. Determine the real effect on the sales of the shop? Solution: Suppose the original cost of each toy=Tk. 100 and number originally sold=100. So, Original sale proceed= 100*100= Tk. 10000; New sale proceed=Tk. (120*85) =Tk.10200; So, Increased% = (200*100/10000) %=2% Question 09:  By selling 45 oranges for Tk. 40, Mr. Rasel khan loses 20%. How many should he sell for Tk. 24 so as to gain 20% in the transaction? Solution: S.P. of orange=Tk. (40/45) =Tk. 8/9; 80% of C.P. =8/9 Or, C.P. = (8/9 * 100/80) =Tk. 10/9; S.P. = (120% of Tk. 10/9) =Tk. 4/3. For Tk. 4/3, he sells 1 orange. Then, for Tk. 24, He would sell (3*24/4) =18 oranges. Question 10:  A owns a house worth Tk. 10000. He sells it to B at a profit of  10% based on the worth of his house. B sells the house back to A at a loss of 10%. In this transaction A gets: Solution: C.P. of B=110% of Tk. 10000=Tk. 11000;

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Sol uti on of Pr ofi t & Loss r el ated com pl icated Pr obl em s - BC S M ath School

Loss of B=10%; S.P. of B=90% of Tk. 11000=Tk. 9900. Thus, C.P. of A=Tk. 9900; So, A gets [(10% of Tk. 10000) + (10000-9900)] =TK. 1100 Question 11: A grocer of Dharmaghor Bazar sells rice at a profit of 10% and uses weights which are 20% less than the market weight. What is the total gain earned by the grocer? Solution: Let us consider a packet of rice marked 1 kg. Then, its actual weight= (80% of 1Kg) =0.8 kg. Let C.P. of 1 kg be Tk. x. Then, C.P. of 0.8 kg=Tk. 0.8x Now, S.P. =110% of C.P. of 1 kg= (110*x/100) = Tk. 1.1x; Gain %=( 0.3x * 100/0.8x)%=37.5% Question 12:  Due to an increase of 30% in the price of eggs of a local market of Shonirakra, 3 fewer eggs are available for Tk. 7.80. Determine the present rate of eggs per dozen. Solution: Let the original rate be x Taka per egg. Number of eggs bought for Tk. 7.80=780/x New rate = (130% of x) Taka per egg=13x/10 Taka per egg. Number of eggs bought for Tk. 7.80=(780*10/13x)=600/x. So, 780/x – 600/x=3 Or, 3x =180 Or, x=60 So, percent rate= (13*60/10) Taka per egg=78 Taka per egg= Tk. 9.36 per dozen. Question 13: A shopkeeper sells ¾ the of its articles at a gain of 20% and the remaining at cost price. His real gain in the transaction is: Solution: Let total C.P. of all the articles=Tk. 100; C.P. of 3/4th part= Tk. (3*100/4) =Tk. 75. S.P. of 3/4Th part=TK. (120*75/100) =Tk. 90; S.P. of ¼ th part=Tk. (1*100/4) =Tk. 25. Total S.P. = Tk. (90+25) =Tk. 115 Gain=15% Question 14:  A retailer of Shibrampur market buys a sewing machine at a discount of 15% and sells it for Tk. 1955. Thus, he makes a profit of 15%. Now, determine the amount of discount. Solution: Let us consider, the marked price be Tk. x. Discount availed by the retailer=15% of Tk. x. C.P. of the machine by the retailer=(x-15% of x) =TK. 17x/20 So, 15% of 17x/20 = 1955 – 17x/20 So, 51x/400 + 17x/20=1955 Or, x=2000; Discount received by retailer= (15% of Tk. 2000) =Tk. 300. Question 15:  A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Tk.55 more, he would have had a profit of 40%. The cost price of the article is: Solution: Let us consider, C.P. =Tk. x.

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Sol uti on of Pr ofi t & Loss r el ated com pl icated Pr obl em s - BC S M ath School

Then, S.P. = (90*x/100) = Tk. (9x/10); New C.P. = (80*x/100) =Tk. (4x/5) Now, gain=40%. So, New S.P. = (140/100 * 4x/5) = Tk. (28x/25) Then, 28x/25 – 9x/10 = 55 Or, x=250. C.P. =Tk. 250. Question 16:  At what price must Hanif sell a mixture of 80 kg sugar at Tk. 6.75 per kg with 120 kg at Tk. 8 per kg to gain 20%? Solution: Total C.P. of 200 kg of sugar = (80 *6.75 + 120*8) = Tk. 1500. C.P. of 1 kg= Tk. (1500/200) =TK. 7.50. Gain required=20% So, S.P. of 1 kg= (120% of Tk. 7.50) =Tk. (120*7.50/100) = Tk. 9 per kg. Kindergarten Primary School Junior School   High School   Exam Questions

Category: Profit And Loss, Profit And Loss-High School

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