A bookseller says to a customer \"The prize of water bag is greater than 400 rupees and I shall giv

M4maths Previous Puzzles

29 Jan 2014 Solve

A bookseller says to a customer "The prize of water bag is greater than 400 rupees and I shall give you a scheme of 50%+50%+50%+50% off after giving you a discount of 100 rupees." If after then the cost of water bag is square root of its original value, then prize of water bag is…… (Leaving paisa if any).

OPtion

1) 400
2) 441
3) 432
4) 484
5) 510
6) 424
7) 525
8) 529
9) 584
10)None of these

Solution

First method:
(x-100)(1/2)(1/2)(1/2)(1/2) = √x (eq.1)
On solving we get the solution
apply shridharacharya rule (-b +_√(b^2-4ac))/2a

Option 3)

Correct Option: 3 Best Solution (10)

Chandni Maurya   10 years AGO

ans= (3)432 rs

let x be the price of water bag.

according to ques,
(x-100)*0.5*.5*.5*.5 = sq. rt (x)
(x-100)*0.0625 = sq. rt (x)
0.0625 x - sq. rt (x) - 6.25 = 0 --- (1)

solving equ. 1
sq. rt (x) = 20.806 rs & - 4.80 rs
= 20.806 (neglecting -ve value)

=> x = (20.806)^2
x = 432.889 rs
after neglecting paisa:
x = 432 => opt (3)

Like? Yes (12) | No (2)  

S.SHAFI   10 years AGO

Let the prize of water bag is x..
The prize of water bag is greater than 400rupees,
so x > 400
The scheme is 50%+50%+50%+50% off after giving a discount of 100,
so (1/2)*(1/2)*(1/2)*(1/2)*(x-100) is the amount left and it is equal to square root of its original value

so, (x-100)/16 = sqrt(x);
squaring on both sides, we will get

x^2 + 10^4 - 200x = 256x

x^2 - 456x + 10^4 = 0

x = {456 + sqrt(456^2 - 4*1*10^4)}/2

+ because x is greater than 400,

x = 432

Like? Yes (4) | No (3)  

Leelanand Kumar Yalla   10 years AGO

Let The Cost of water Bag is 400+x;
After Giving a Discount of 100 Rupees it is 300+x;
And again A scheme of Discount 50%+50%+50%+50%..
I.e., (300+x)*50/100*50/100*50/100*50/100= (300+x)/16;
The Original Price of Water Bag is 400+x;
square root(400+x)= (300+x)/16;
By solving This we will get x= 32.8999756;
so the actual Value of Water bag is 400+32.899= 432 Rupees by Removing Paisa...... This is the answer...

Like? Yes (3) | No (2)  

Shruti Joshi   10 years AGO

(n-100) = 16*(n)^1/2

This gives the equation : n^2 - 456n + 10000 = 0

Solve to get answer = 432 and 23
And as given the price is greater than 400, so discard 23

Like? Yes (1) | No   

Vinay Gupta   10 years AGO

Let list price = x^2
and cost price = x
and total off after discount of 100 rs. = 50%+50%+50%+50%=375/4%

So the equation becomes
(x^2-100)[{1-(375/4)}*100]=x
=>(x^2-100)[(400-375)/400]=x
=>(x^2-100)*(25/400)=x
=>(x^2-100)*(1/16)=x
=>x^2-100=16x
=>x^2-16x-100=0

So x=20.805(approx.)
and x^2=432(approx.)Ans

Like? Yes (2) | No (1)  

hena samanta   10 years AGO

let the original prize is x
(x-100)/16=square root of x
x=432 approx

Like? Yes (1) | No   

Priyanshi Agarwal   10 years AGO

initially take the prize of water bag as (400+x)
then apply discount of 100
=300+x
then apply 50% discount on (300+x)
we get 150+(x/2)
subtracting (300+x)-(150+(x/2))
=150+x/2
repeat the above process on the amount obtained after subtraction in each case three more times
after that,
sqrt(400+x)=(18.75+(x/16))
solve above eq. we get,
x=32
i.e original prize is 400+32= 432

Like? Yes (1) | No   

Keshav Kishlay   10 years AGO

given,
(x-100)*1/2*1/2*1/2*1/2=sqrt(x)
so,
x*x-456*x+10000=0
x=432

Like? Yes (2) | No (2)  

jatin patware   10 years AGO

the original value of water bag be x
discounted =x-100;
off amount=x-100/16;
atc
((x-100)/16)=x^1/2;
solve for x we get x is eual to the=432
so option 3 is correct

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

original value of water bag=x,then
[[[(x-100)0.5]0.5]0.5]0.5=Sqrt(x)
0.0625x-6.25=Sqrt(x)
x-100=16*Sqrt(x)
Squaring both sides
x^2-456x+10000=0
Solving x=432.8

Like? Yes (1) | No (2)