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# Maths Questions for Competitive Entrance exams – Profit and Loss with Solutions

Maths Questions for Competitive Entrance Exams – Profit and Loss with Solutions are created here for the aspirants who are going to appear in competitive entrance exams like CPO, Railways, Bank, Defense services etc. The Maths Questions for Competitive Entrance exams – Profit and Loss with Solutions will give you an idea about the questions asked in the Competitive Entrance Exams and the ways of solving them.The continuous practice of Maths Questions for Competitive Entrance Exams is needed in qualifying competitive entrance exams.Therefore you are required to study whole of blog post completely for the success.

## Maths Questions for Competitive Entrance exams – Profit and Loss with Solutions

Profit and Loss : If selling price(SP) > Cost price (CP) then it is a profit

Profit = SP – CP

If cost price(CP) > Selling price (SP) then it is a loss

Loss = CP – SP

Profit % = (Profit/CP)×100

Loss % = (Loss/CP)×100

## Maths Questions for Competitive Entrance exams – Profit and Loss with Solutions

Q1. Naina purchased an article in Rs 450 and sold it in Rs 500, find his loss or profit percentage.

a.(12 and 2/3)%     b.50 %      c.25 %   d. (11 and 1/9 )%

Ans. Cost price(CP) of the article is Rs 450

The selling price(SP) of the article is Rs 500

Since SP > CP,therefore Profit of Naina is

SP -CP =Rs (500 -450 ) =Rs 50

Profit % =[profit/CP]×100 =(50/450)×100 =100/9=(11 and 1/9 )%

Q2.Naresh bought 4 dozen pencils at Rs 10.80 a dozen and sold them for 80 paise each. Find his gain or loss percent.

a.9 %      b.8.11 %      c.9.11 %      d.11.11%

Ans.The cost of 4 dozen pencils Rs( 10.80 ×4) =Rs 43.20

CP of 4 dozen pencils = Rs 43.20

4 dozen pencils = 12 ×4 = 48 pencils

80 painse =Rs 0.80

Selling price(SP) of 4 dozen pencils = Rs(48× 0.80)=Rs 38.4

SP < CP

Therefore loss is

CP – SP =Rs (43.20 – 38.4)=Rs 4.80

Loss % = (Loss/CP)×100 =(4.80/43.20)×100 = 11.11 %

Q3.Ramesh bought two boxes for Rs 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.

a.Rs 550,Rs 750      b.Rs 450,Rs 850     c.Rs 350,Rs 950    d.Rs 650,Rs 650

Ans.Let cost price of one box is x then cost price of other box is 1300 – x

SP of one box is = CP + Profit% CP =Rs(x +20% of x) = Rs[x+20x/100]=Rs(x+x/5)=Rs 6x/5

SP of other box =CP – Loss% CP =Rs(1300-x-12% of (1300-x)

= Rs[1300-x-(12/100)×1300+12x/100]

=Rs[1300-x-156+3x/25]

SP of both boxes is same

6x/5 =1300-x-156+3x/25

6x/5 +x -3x/25 =1300 -156 =1144

(30x +25x-3x)/25 =1144

52x/25 =1144

x = 1144×25/52

x =22×25=550

Hence cost price of one box is Rs 550 and of other box is Rs(1300 -550) =Rs750

Q4.A vendor buys oranges at Rs 26 per dozen and sells them at 5 for Rs 13. Find his gain percent.

a.10 %      b.8 %      c.15 %      d.20%

Ans.CP of one dozen(12 oranges) orange is =26

SP of 5 oranges =Rs 13

SP of 12 oranges =Rs (13/5)×12 =Rs 31.20

SP>CP

Profit =Rs (31.20 -26) =Rs 5.20

Profit % =(5.20/26)×100 =20 %

Q5.The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.

a.9 %      b.(13 and 1/9)%    c.(12 and 1/9)%     d.(11 and 1/9)%

Ans.Let the cost one article is Rs 1

Cost price of 9 articles = Rs 9

Selling price of 9 articles =Cost price of 10 articles= Rs 10

SP > CP

Profit = SP -CP = 10 – 9 = 1

Profit % =(Profit /CP)×100

=(1/9)×100

=100/9 =(11 and 1/9)%

Q6.By selling a book for Rs 258, a bookseller gains 20%. For how much should he sell it to gain 30%?

Ans. Profit % = (Profit/CP)×100

20 = (Profit/CP)×100

100×Profit = 20 CP

100(SP -CP) = 20 CP

100 SP -100 CP = 20 CP

120 CP = 100SP (SP =Rs 258)

120CP = 100 ×258 = 25800

CP = Rs 215

If gain is 30 %

Profit % = (Profit/CP)×100

30 = (Profit/215)×100

Profit = (215×30)/100 = Rs 64.5

SP = CP +Profit = Rs(215+64.5) = Rs 279.5

Therefore he should sell the book at Rs 279.5 for the gain of 30%

Q7.A defective briefcase costing Rs 800 is being sold at a loss of 8%. If the price is further reduced by 5%, find its selling price.

Ans. Loss % = (Loss/CP)×100

8 = (Loss/CP)×100

100×Loss = 8 CP

100Loss = 8 ×800

Loss = Rs 64

SP-CP = 64

SP = 800 -64 =Rs 736

If price(SP) is further reduced by 5 %,then SP is

=SP- 5 % of SP = 736 – 5%of 736 =736 – 36.80=Rs 699.2

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## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions of class 9 science

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### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals