The Trade

$Let\quad the\quad no\quad of\quad toys\quad bought\quad be\quad equal\quad to\quad x.\\ \therefore \quad Cost\quad Price\quad of\quad each\quad toy\quad =\quad Rs\quad \frac { 900 }{ x } \\ He\quad increased\quad the\quad price\quad of\quad each\quad toy\quad by\quad Rs.\quad 2\quad \\ \therefore \quad Selling\quad Price\quad of\quad each\quad toy\quad is\quad =\quad Rs\quad \frac { 900 }{ x } +2\\ Now\quad he\quad threw\quad away\quad 5\quad toys\quad and\quad sold\quad x-5\quad toys.\\ \therefore \quad Total\quad money\quad earned\quad =Rs\quad (\frac { 900 }{ x } +2)\times (x-5)\\ \therefore \quad Profit\quad =\quad Total\quad money\quad earned\quad -\quad Total\quad money\quad spent\\ \quad \quad 80\quad =\quad \quad (\frac { 900 }{ x } +2)\times (x-5)\quad -\quad 900\\ Solving\quad the\quad above\quad equation\\ \Rightarrow \quad \boxed { x\quad =\quad 75 }$

Let x be the no. of article sold and y is the purchase price per article. Therefore, (x+5) y=900 AND x (y+2)=980. Solve for x and y. x=70 and y=12. So, number of articles purchased originally is 70+5=75.

let the number of toys is x let the initial cost of the toy be y so what we have is

xy=900

therefore the cost of a toy is

y=900/x

five toys are defective, so the final number of toys is x-5 and the cost is increased by two. so the cost is y+2. now the equation we have is:

(x-5)(y+2)=980 (is equal to 980 because he makes a profit of 80. so the total amount he got is original cost + gain) we have found y in terms of x which is 900/x

(x-5)*((900/x)+2)=980

900-(4800/x)+2x-10=980

-(4500/x)+2x=90

(2x^2-4500-90x)/x=0

this is a inverse function. the forbidden value is 0 because the denominator MUST not be 0. now take only the numerator and then equate it to 0

2x^2-4500-90x=0

there will be two answers as it is a quadratic equation. the answers are -30 and 75. as this problem is related to a real life experience, the answer cannot be in negative. SO THE ANSWER IS 75.

consider x as total number of toys bought Spent= Rs 900...........(1) On selling (x-5) toys at rate of (900/x +2), total income= (x-5)(900/x +2)........(2) Now calculate (2)-(1)=80 and solve quadratic in x, to get 75.

let the number of toys be x
(900/x+2)(x-5)=980
x=75

(900/y + 2) * (y-5) = 980 solving this we get, y=75 where y is the no of objects bought initially.

et the no of toys be x cost of each toy=900/x selling price of each toy =900/x+2 no of toys sold =x-5 selling price=980 hence (900/x +2 )(x-5)=980 x=75

75 toys for Rs 900. each toy costs Rs. 12. if 5 were damaged, only 70 toys were sold for Rs. 14, which earned Rs. 980.

LET NO. OF TOYS BE X AND COST OF EACH TOY=Y SO X Y=900 NOW 5 TOYS ARE DEFECTIVE SO (X-5) (Y+2)=980 SOLVING IT AND SUBSTITUTING VALUE OF X AS 900/Y WE GET Y=12 SO X=75

Let the number of kitty purchased be x and the cost of each kitty be y then according to the question xy=900 and (x-5)(y+2)=900+80 ...... as he made a profit of 80 rupee hence, on solving we get x=75 and y=12 .

let x be the hello kitty toys ,5 toys are not good so he sold x-5 hello kitty toys if x=75 5 toys are not good so,total toys are 70 =>75 12=900 2 rupees increased so 70 14=>980 so total toys=75