Bracelets cost p dollars each at a discount store. At a neighboring re

Question

Bracelets cost p dollars each at a discount store. At a neighboring retail chain store, the same bracelets cost $1 more each, which means that x dollars will buy 10 more bracelets at the discount store than at the retail chain store. What is the value of x in terms of p?

A. 10(p + 1)
B. 10(p - 1)
C. 10(p^2 + p)
D. 10(p^2 - p)
E. 10(p^2 + p + 1)

A. 10(p + 1) 
B. 10(p - 1)
C. 10( p^2  + p) 
D. 10( p^2  - p)
E. 10( p^2  + p + 1)

BrushMyQuant · Accepted Answer

Discount Store  
Price per piece: $p
Total Money: $x
Total Pieces which we can buy = $x / $p = x/p

Retail Chain Store  
Price per piece: $(p+1)
Total Money: $x
Total Pieces which we can buy = $x / $(p+1) = x/(p+1)

Given:   At discount store we can buy 10 bracelets more as compared to retail store in $x money
=> x/p = x/(p+1) + 10
multiply both sides with p and (p+1) we get
x(p+1) = xp + 10p*(p+1)
x(p+1-p) = 10p(p+1)
x = 10(p^2 + p)

Answer will be C 
Hope it helps!

LeenaSai · Answer

Can you please solve this using plugging in

Posted from my mobile device

BrushMyQuant · Answer

LeenaSai 
Once I have the equation x/p = x/(p+1) + 10 ...(1)
I can start substituting the answer choices. Now, before i blindly substitute I will analyze the equation i have above. I see that x is divided by p in LHS and (p+1) in RHS and i have an integer(10) on the right hand side. So, i will try to find a term in which x is a multiple of both p and p+1
=> x = something*p*(p+1) = something*(p^2+p) and luckily C is the only choice which has that.
If I was getting 2-3 terms which have p^2 + p then I would have substituted each one of them (one by one) in (1) and checked.
Hope it helps!

ScottTargetTestPrep · Answer

We see that we can buy x/p bracelets at the discount store and x/(p + 1) bracelets at the retail chain store. We can create the equation:

x/p = x/(p + 1) +10

Multiplying by p(p + 1), we have:

xp + x = xp + 10p^2 + 10p

x = 10p^2 + 10p

x = 10(p^2 + p)

Answer: C

AVMachine · Answer

Item Rate 1 = P $/Item
Item Rate 2 = (P+1) $/Item
Total Amount = X $

Given: X / P = 10 + X / (P+1).

Solve this by taking X on one side:

steffe440 · Answer

10 more bracelets for x dollars on the cheap store, hence we get p(q+10)=x and we get the second equation to be (p+1)(q)=x

Solving for this we get: 10p = q

Hence we can plug it in

(p+1)(10p)

Hence we get 10p^2 + 10p which is equal to C, 10(p^2+p)

dianyang · Answer

I have a right answer by giving example, for instance, price is 2$ then another is 3$ → LCM(2,3)=6 → So I can take 60$ as a total money that I can buy 30 and 20 (satisfy the question's requirement) and I looking for the correct expression in the answer section, only C satisfies.

Bracelets cost p dollars each at a discount store. At a neighboring re

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