A man bought a certain number of toys for Rs.180, he kept one for his own use and sold the rest for

Abhishek Kumar   11 years AGO

let no. of toys=n
cost of 1 toy=180/n
now, (n-1)*(180/n + 1)=180+10; as 1 toy cost nothing,
it gives (n-20)(n+9)=0;
which gives n=20 toys

Like? Yes (16) | No (3)  

Sourangsu Bandyopadhyay   11 years AGO

The question is a bit Tricky...But let's explain it..
The solution can be derived as
Let the number of toys the man brought be 'x'
Then, Cost Price of each toy=180/x
(Trick: The cost price is just the average cost price used here.As per our question it might be confusing from the language of the question that each of the toy was brought for a different cost price.)
Since he paid Rs.1 extra for each of the toy he brought and sold(except the one he kept for himself!!)
He made a profit of Rs.10 on the whole bargain.
Thus,k we could claim that on an he gained Rs.1 on his average cost price.(Completely rejecting the fact that he might have brought each of the toy for different cost price)
Total selling price=Rs.(180+10)=Rs.190
Then,our equation becomes quite simple:
(190/(x-1))-(180/x)=1
Solving, we get x=20 or -9.
Since quantity cannot be negative.
Thus, x=20 [Answer]

Like? Yes (10) | No (5)  

prakash kumar   11 years AGO

let a man bought x toys
buying cost of each toy=180/x
a/q,
((180/x)+1)(x-1)-180=10
(180+x)(x-1)-180x=10x
180x-180+x^2-x-180x=10x
-180+x^2-x=10x
x^2-11x-180=0
x^2-20x+9x-180=0
x(x-20)+9(x-20)=0
(x-20)(x+9)=0
x cant be -ve
so,x=20

Like? Yes (7) | No (2)  

Raisul Alam   11 years AGO

Suppose the no of toys=x
So the cost price for each toys is=180/x
He kept 1 toy for his own.
So now (x-1) no of toys are remaining.
And the selling price of (x-1) no of toys are (x-1)*(180/x+1)(given in the qstn)
now (x-1)*(180/x+1)-180=10
=>x^-11x-180=0
by middle term factor we get (x-20)(x+9)=0
So the value of x will be either 20 or -9.But the no of toys can't be negative . So the no of toys are=20

Like? Yes (5) | No (1)  

HARI om kashyap   11 years AGO

let X be price of toys and Y be number of toys
then acc to question
XY=180
and (X+1)(Y-1)=190 since 10 rs is profit
solving both eq we get y-x=11
and XY=180.
then X=9
Y=20

Like? Yes (4) | No (1)  

ishita gorai   11 years AGO

let X be price of toys and Y be number of toys
then acc to question
XY=180
and (X+1)(Y-1)=190 since 10 rs is profit
solving both eq we get y-x=11
and XY=180.
then X=9
Y=20

Like? Yes (6) | No (3)  

ChinnaChowdaiah   11 years AGO

Let man bought x toys
Then, cost of one toy = Rs 180/x
He kept one toy for his own,
According to question

After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs (180/x)+1

so
(x-1)(180/x +1)=180+10
(x-1)(180/x)+(x-1)=190
180- 180/x +x -1=190
x- 180/x -11=0
x^2-11x-180=0
x^2-20x+9x-180=0
x(x-20)+9(x-20)=0
x=20 (or)-9
so the no of toys are 20

Like? Yes (3) | No (1)  

Ashik Kuppili   11 years AGO

Let man bought x toys
Then, cost of one toy = Rs
He kept one toy for his own,
According to question
After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs
so

Like? Yes (3) | No (2)  

Kumar Ankit   11 years AGO

Let man bought x toys

Then, cost of one toy = Rs

He kept one toy for his own,

According to question

After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs

so

since the number of toys can never be negative

Hence the man has bought 20 toys

Like? Yes (4) | No (3)  

Chandni Maurya   11 years AGO

ans=20
let n be the no of toys and x be the cost,
then,
n*x=180----(1)
(n-1)(x+1)=180+10=190---(2)
put n*x=180 in 2,
n-x=11----(3),
180=2*2*3*3*5= 20*9(satisfies the condition 3)
=>n=20

Like? Yes (2) | No (1)  

ravi singh   11 years AGO

answer is 20

let toys is x then cost of one toy is 180/x

now according to the question
(x-1)*( 180/x +1) = 190

=>x^2 - 11x -180 = 0
by solving this we get,
x=20, -9
since no. of toys is not negative hence answer is 20

Like? Yes (1) | No   

Md.Sultan Khichi   11 years AGO

Let the no. of toys are x..
So price of single toy is 180/x.
New price of one toy is (180/x +1)..
So according to the ques
(x-1)*(180/x+1)=190.
after solving we get x=20,-9.
So the no. of toys are 20.
Ans:3.

Like? Yes (1) | No   

Dinesh thatti   11 years AGO

Let man bought x toys
Then, cost of one toy = Rs
He kept one toy for his own,
According to question
After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs
so


since the number of toys can never be negative
Hence the man has bought 20 toys

Like? Yes (1) | No   

Nigam Agrawal   11 years AGO

let x is the no. of toys...
then /toys price=180/x
we get one toys and sold rest of them
so (180/x+1)(x-1)=190
for solving this
we get x=20,-9 so 3 option is correct

Like? Yes (1) | No   

Jagadeesh   11 years AGO

toys bought for 20*9=180.
after using one toy and price increased by 1 rupee 19*10=190.
Profit achieved 190-180=10.
Therefore number of toys is 20.

Like? Yes (1) | No   

Paranthamakannan   11 years AGO

XY=180,
(X-1)(Y+1)=190....
solve tihs ...
X=20

Like? Yes (1) | No (1)  

VIKASH KUMAR SHARMA   11 years AGO

let the no. of toys are x.
if he sell the rest toys beside his toy, he is getting 10 rs profit, thrfor it is clear that after taking out his toy he is selling his toys 1 rs more as of the cost price , thrfor if the no. of toys are 20 thn after taking out a single toy, the cost of 20 toy is 180 rs, again for 19 toys the sellipng price = 19*10=190
thrfor the total no. of toys are 20 (ans)

Like? Yes (2) | No (2)  

Devendra Marghade   11 years AGO

If number of toys=x
cost of each toy=180/x
Given (x-1)[(180/x)+1]=190
x^2-11x-180=0
(x-20)(x+9)=0
x=20

Like? Yes (2) | No (2)  

moumita majee   11 years AGO

let X be price of toys and Y be number of toys
then acc to question
XY=180
and (X+1)(Y-1)=190 since 10 rs is profit
solving both eq we get y-x=11
and XY=180.
then X=9
Y=20

Like? Yes (2) | No (2)  

Rituparna Bhattacharyya   11 years AGO

ans. 3)20

let no. of toys be x and price of toys be y
&
AT LAST he is left with one of his toy which he didn't sell and Rs. 10

therefore
x*y = 180
&
(x-1)*(y+1) = 190
on solving for x we'll get x=20 and y =9.

Like? Yes (1) | No (1)  

Md Shahnawaz Shahab   11 years AGO

let no. of toys be x and price of toys be y
therefore x*y = 180
and (x-1)*(y+1) = 190
on solving for x we'll get x=20 and y =9.

Like? Yes (1) | No (1)  

deepak kumar   11 years AGO

Let man bought x toys

Then, cost of one toy = Rs

He kept one toy for his own,

According to question

After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs

so

(x-1)(180/x + 1) = 180+10
(x-1)*180/x + (x-1) = 190
180-180/x + x-1 = 190
x^2 -11x - 180=0
(x-20)(x+9)=0
x=20

Like? Yes (1) | No (1)  

vishal chandra pathak   11 years AGO

simple....1 toy cost 180/x....and profit loss equation...thats all
(180/x +1)(x-1)-180=10
=>(180+x)(x-1)-180x=10x
=>x^2-11x-180=0
so x=20.
thanks

Like? Yes (1) | No (1)  

Bholanath    11 years AGO

let total no. of toys =x
so, cost of one toy=180/x
as kept one toy for own use,
so, remaining toys =x-1
and the man increase the price of each toys by
1 Rs.
so total sold price=(x-1)(180/x +1)
and this must be equal to 190 because profit =10
so,(x-1)(180/x +1)=190
on solving, x=20

Like? Yes (2) | No (2)  

Harjit kaur   11 years AGO

Let no of toys =x
Cp of x toys=180
Cp of 1toy=180/x
SP of 1toy= 180/x+1
SP of x-1 toys=(180/x+1)(x-1)
Now SP-CP=10
(180/x+1)(x-1)-180=10
by solving we get
X^2-11x-180=0
solving this quadratic equation
X=20
thus no of toys =20
Opt is 3

Like? Yes (1) | No (1)  

chandu   11 years AGO

consider we have 90 toys each toy consist 2rps so he sold only 179 toys for 1 rupee he got profit of 10rps of toy that he had so 2*10=20

Like? Yes (1) | No (1)  

Prakash kumar   11 years AGO

let x be no of toys of 180 toys.
so cost of 1 toy = 180/x.
so x-1 toys sold at 180/x + 1

therefore a/c
(x-1) * [(180/x) + 1] = 190
x^2 - 11x - 180 = 0

x =20 (neglect x = -9)

Like? Yes (1) | No (1)  

prabhat kumar   11 years AGO

let no. of toys = x
each toys price = y
according to question,
180/x=y & xy=180-------------(1)
180+10/x+1=y-1
190/x+1=y-1
190=xy-x+y-1
putting euation value in above
190=180-x+y-1
y-x=11
y=10+x---------------(2)
now from first equation
180/x=11+x
x^2-11x-180=0
x=20

Like? Yes (3) | No (3)  

IFFAT IRAM   11 years AGO

let the number of toys be x
c.p. of x toys =180
no. of toys sold=x-1
s.p. of 1 toy=180/x + 1
= (180+x)/x
s.p. of x toys=(180+x)(x-1)/x
(180+x)(x-1)/x - 180 = 10
x^2-11x-180=0
x=20

Like? Yes (1) | No (1)  

Rituraj Singh   11 years AGO

sp-cp=profit

let no. of toys be x
[(x-1)(180/x +1)]-180=10
==>x^2 - 11x - 180 =0
on solving we get

x= -9 and x=20
therefore x=20

Like? Yes (1) | No (1)  

Srinivas    11 years AGO

number of toys =n,cp=180/n
sold n-1,sp=(180/n)+1
sp-cp=profit=10
gives n=20

Like? Yes (1) | No (1)  

susmit   11 years AGO

Let x be number of toys.
Let y be cost per toy.
So, x*y = 180...........(1)
Given :
He sold x-1 toys at the price of y+1.
So, S.P. = (x-1)*(y+1) = xy+x-y-1...........(2)
But x*y = 180 & S.P. = 180+10 = 190.
So 180+x-y-1 = 190
So x-y = 11.................................(3)
Solving (1) & (3) x = 20 & y = 9.
So, number of toys = 20.
Hence option no. (3)

Like? Yes (1) | No (2)  

Dharmendra Singh   11 years AGO

abs 3) 20
equation: x= no of toys;
(180/x +1)(x - 1)=10 put the value and get the result otherwise solve you will get x=20,-9

Like? Yes (1) | No (2)  

anil   11 years AGO

Let man bought x toys
Then, cost of one toy = Rs
He kept one toy for his own,
According to question
After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs
so


since the number of toys can never be negative
Hence the man has bought 20 toys

Like? Yes (1) | No (2)  

achyuth   11 years AGO

Let man bought x toys

Then, cost of one toy = Rs 180/x

He kept one toy for his own,

According to question

After selling (x – 1) toys, he made a profit of Rs 10 by selling each toy for Rs 180/x +1

so
(x-1)(180/x +1)=180+10
x.180/x +x -180/x -1=190
180-180/x +x-1=190
x^2 -11x -180=0
(x-20)(x+9)=0
x=20 or -9

Like? Yes (1) | No (2)  

anurag   11 years AGO

let x toys=rs.180
=> 1 toy cost=rs.180/x
sold x-1 toys each rs.1 extra
=>(x-1)((180/x)+1)=180+10=190
=> x^2-11x-180=0
=> x=20 =>ans=20 toys

Like? Yes (1) | No (2)  

PINKY   11 years AGO

let no of toys be x and their cost be y
then, x*y= Rs 180 (equation 1)
again A/q
(x-1)(y+1)= Rs 190
xy+x-y-1=190
180+x-y=191
x-y=11
x-180/x=11 (from above eqn 1)
by solving above quadratic eqn we will have
x=-9
x=20
so, the no. of toys are 20.

Like? Yes (1) | No (2)  

chaitanya   11 years AGO

Let the number toys be x and cost price of each toy is y,
so xy = 180
and after removing one item, and increasing cost price by 1 rupee
Total cost is (x-1)(y+1) = 180 + 10 ( this 10 is profit)
xy-y+x-1=190
180-y+x=191
x-y=11 [since xy=180, y= 180/x)
x-180/x=11
x^2-180=11x
x^2-11x-180=0
x^2-20x+9x-180=0
x(x-20)+9(x-20)=0
(x-20)(x+9)=0

so x=20, because the number of toys will never be negative.

Like? Yes (1) | No (2)  

lakshay bansal   11 years AGO

let he buys x toys
so cp of 1 toy=180/x
sp of 1 toy = (180/x)+1
sp of x-1 toys = ((180/x)+1)(x-1)=190
so x=20,-9
but -9 not possible so ans is 20

Like? Yes (1) | No (2)