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Bunuel
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Esther bought a computer at a price of p dollars, and she paid a sales 21 Apr 2024, 08:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

48% (02:15) correct 52% (02:14) wrong based on 486 sessions

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­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
 ­

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E
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gmatophobia
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Esther bought a computer at a price of p dollars, and she paid a sales 21 Apr 2024, 09:11
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Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
Show more
Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000

\(­p[1+\frac{t}{100}] > 1000 - r\)­

Constraint: \(0 \leq r \leq 1000\)

Question: Is \(p > $800\)

Statement 1

(1) r = 200

\(­p[1+\frac{t}{100}] > 1000 - r\)­

\(­p[1+\frac{t}{100}] > 1000 - 200\)­

\(­p[1+\frac{t}{100}] > 800\)­

We can infer that the cost of the computer, inclusive of the sales tax, is greater than $800. However, we cannot infer whether the cost of the computer exclusive of the sales tax is greater than $800.

Statement 1, alone, is not sufficient to answer the question. Hence, we can eliminate A and D. 

Statement 2

(2) t = 6

\(­p[1+\frac{6}{100}] > 1000 - r\)­

\(­p[1.06] > 1000 - r\)­

Dividing by 1.06 on both sides we get

\(­p > 943.39 - 0.943r\)­

As the value of r is not known, we cannot comment on whether \(p > 800\). For some value of ­\(r\) the value of \(p\) can be greater than \(800\), for some other value of \(r\), the cost can be equal to \(800\) (or less)

We can eliminate B.

Combined

\(­p[1+\frac{6}{100}] > 1000 - 200\)­

\(­p > \frac{800}{1.06}\)­

\(p > $754.71\)

Using the information from Statement 1 and Statement 2, we know that \(p > $754.71\), however we do not know whether\(p > 800\).

Ex:
If p = $760  → Is \(p > $800\) ⇒ The answer is No
If p = $810  → Is \(p > $800\) ⇒ The answer is Yes

As we are getting two different answers, the statements combined don't help.

Option E­
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Esther bought a computer at a price of p dollars, and she paid a sales 01 Nov 2024, 04:56
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gmatophobia

Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000

\(­p[1+\frac{t}{100}] > 1000 - r\)­

Constraint: \(0 \leq r \leq 1000\)

Question: Is \(p > $800\)

Statement 1

(1) r = 200

\(­p[1+\frac{t}{100}] > 1000 - r\)­

\(­p[1+\frac{t}{100}] > 1000 - 200\)­

\(­p[1+\frac{t}{100}] > 800\)­

We can infer that the cost of the computer, inclusive of the sales tax, is greater than $800. However, we cannot infer whether the cost of the computer exclusive of the sales tax is greater than $800.

Statement 1, alone, is not sufficient to answer the question. Hence, we can eliminate A and D.

Statement 2

(2) t = 6

\(­p[1+\frac{6}{100}] > 1000 - r\)­

\(­p[1.06] > 1000 - r\)­

Dividing by 1.06 on both sides we get

\(­p > 943.39 - 0.943r\)­

As the value of r is not known, we cannot comment on whether \(p > 800\). For some value of ­\(r\) the value of \(p\) can be greater than \(800\), for some other value of \(r\), the cost can be equal to \(800\) (or less)

We can eliminate B.

Combined

\(­p[1+\frac{6}{100}] > 1000 - 200\)­

\(­p > \frac{800}{1.06}\)­

\(p > $754.71\)

Using the information from Statement 1 and Statement 2, we know that \(p > $754.71\), however we do not know whether\(p > 800\).

Ex:
If p = $760 → Is \(p > $800\) ⇒ The answer is No
If p = $810 → Is \(p > $800\) ⇒ The answer is Yes

As we are getting two different answers, the statements combined don't help.

Option E­
Show more
Statement 1 can be derived upto ­Is p[1+t/100]>800? The answer would always be no right? since it can be re-written as Is p>800/(1+t/100)
The argument says she paid some sales tax. Hence, t has to be greater than 0. So, the equation will be Is p>800/(some number greater than 1). which is false. is this reasoning correct? GMATNinja KarishmaB @Bunuel
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Esther bought a computer at a price of p dollars, and she paid a sales 19 Apr 2025, 21:58
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say that t=300%; implying p> 200; falls apart.
Vasavan
gmatophobia

Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000

\(­p[1+\frac{t}{100}] > 1000 - r\)­

Constraint: \(0 \leq r \leq 1000\)

Question: Is \(p > $800\)

Statement 1

(1) r = 200

\(­p[1+\frac{t}{100}] > 1000 - r\)­

\(­p[1+\frac{t}{100}] > 1000 - 200\)­

\(­p[1+\frac{t}{100}] > 800\)­

We can infer that the cost of the computer, inclusive of the sales tax, is greater than $800. However, we cannot infer whether the cost of the computer exclusive of the sales tax is greater than $800.

Statement 1, alone, is not sufficient to answer the question. Hence, we can eliminate A and D.

Statement 2

(2) t = 6

\(­p[1+\frac{6}{100}] > 1000 - r\)­

\(­p[1.06] > 1000 - r\)­

Dividing by 1.06 on both sides we get

\(­p > 943.39 - 0.943r\)­

As the value of r is not known, we cannot comment on whether \(p > 800\). For some value of ­\(r\) the value of \(p\) can be greater than \(800\), for some other value of \(r\), the cost can be equal to \(800\) (or less)

We can eliminate B.

Combined

\(­p[1+\frac{6}{100}] > 1000 - 200\)­

\(­p > \frac{800}{1.06}\)­

\(p > $754.71\)

Using the information from Statement 1 and Statement 2, we know that \(p > $754.71\), however we do not know whether\(p > 800\).

Ex:
If p = $760 → Is \(p > $800\) ⇒ The answer is No
If p = $810 → Is \(p > $800\) ⇒ The answer is Yes

As we are getting two different answers, the statements combined don't help.

Option E­
Statement 1 can be derived upto ­Is p[1+t/100]>800? The answer would always be no right? since it can be re-written as Is p>800/(1+t/100)
The argument says she paid some sales tax. Hence, t has to be greater than 0. So, the equation will be Is p>800/(some number greater than 1). which is false. is this reasoning correct? GMATNinja KarishmaB @Bunuel
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Esther bought a computer at a price of p dollars, and she paid a sales 24 Apr 2025, 23:32
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Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
­

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Show more

Given: 1000 - p - (t% of p) < r

Question: Is p > 800?

(1) r = 200

This tells us that whatever is leftover is less than 200. Leftover could be 0 or 199.
So depending on the tax rate t, price of computer could be say 900 with say 10% tax or say 500 with say 90% tax.
p may or may not be more than 800.
Not sufficient.


(2) t = 6

Without a value of r, just the tax rate is insufficient. Leftover amount could be $500 (so p < 800 for sure) or leftover amount could be $1 (in which case p > 800).
Not sufficient.

Using both,
Leftover could be 0 or 199. So 1.06p could be anything from 801 to 1000. So p could be 799 or 900 or 950 etc.
Still not sufficient.

Answer (E)

Check out some more Math based tricky DS questions here:
https://youtu.be/wFor-e2sYHg
https://youtu.be/3zQ20VEdSa8
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Esther bought a computer at a price of p dollars, and she paid a sales 30 Apr 2025, 21:08
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Let marked price of computer be "x" .
" p " is price paid including tax .
so, p =x + (t/100)x .

Budget for computer = 1000
Actual money spend on buying computer = p
Money left after buying computer = (1000 -p )

so , as per given info , " Esther had less than r dollars left "
(1000 - p) < r - equation 1

Now to find , " was the price of the computer, excluding the sales tax, greater than $800 "
i.e. is x >800 ?

from equation 1
1000 - p < r
putting value of p
1000 - [ x + (t/100)x] < r

x(1+t/100) > 1000 -r
x > (1000 -r ) / (1 +t/100)

Check statement 1 ; r=200
x > 800 / (1 +t/100) - not sufficient

Check statement 2 ; t = 6
x > (1000 -r)/1.06 - not sufficent

check statement 3 ; r=200 ,t=6
x> 800/1.06
x> 754

even after putting both values of r and t , x > 754 , hence x can be 780 also and 820 also ( for example )
so no definite answer to the question , is x >800 ?

so , even both statements are not sufficient to find the definite answer to the question ( is x> 800 ? ).
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Esther bought a computer at a price of p dollars, and she paid a sales 16 May 2025, 03:07
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Hey Karsihma,
how're you getting 1000 - p - (t% of p) < r

from question we get p + (t% of p) = 1000 - r, that's it right?
I don't understand how you're getting that greater than or lesser than part.
pls explain.

Thanks in advance.
KarishmaB
Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
­

Show Spoiler

Given: 1000 - p - (t% of p) < r

Question: Is p > 800?

(1) r = 200

This tells us that whatever is leftover is less than 200. Leftover could be 0 or 199.
So depending on the tax rate t, price of computer could be say 900 with say 10% tax or say 500 with say 90% tax.
p may or may not be more than 800.
Not sufficient.


(2) t = 6

Without a value of r, just the tax rate is insufficient. Leftover amount could be $500 (so p < 800 for sure) or leftover amount could be $1 (in which case p > 800).
Not sufficient.

Using both,
Leftover could be 0 or 199. So 1.06p could be anything from 801 to 1000. So p could be 799 or 900 or 950 etc.
Still not sufficient.

Answer (E)

Check out some more Math based tricky DS questions here:
https://youtu.be/wFor-e2sYHg
https://youtu.be/3zQ20VEdSa8
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Esther bought a computer at a price of p dollars, and she paid a sales 27 May 2025, 21:00
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Hi Sir,

When we plug any values of T in statement 1 then value of P is coming lower than 800 then why statement 1 is not sufficient


Vasavan
gmatophobia

Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000

\(­p[1+\frac{t}{100}] > 1000 - r\)­

Constraint: \(0 \leq r \leq 1000\)

Question: Is \(p > $800\)

Statement 1

(1) r = 200

\(­p[1+\frac{t}{100}] > 1000 - r\)­

\(­p[1+\frac{t}{100}] > 1000 - 200\)­

\(­p[1+\frac{t}{100}] > 800\)­

We can infer that the cost of the computer, inclusive of the sales tax, is greater than $800. However, we cannot infer whether the cost of the computer exclusive of the sales tax is greater than $800.

Statement 1, alone, is not sufficient to answer the question. Hence, we can eliminate A and D.

Statement 2

(2) t = 6

\(­p[1+\frac{6}{100}] > 1000 - r\)­

\(­p[1.06] > 1000 - r\)­

Dividing by 1.06 on both sides we get

\(­p > 943.39 - 0.943r\)­

As the value of r is not known, we cannot comment on whether \(p > 800\). For some value of ­\(r\) the value of \(p\) can be greater than \(800\), for some other value of \(r\), the cost can be equal to \(800\) (or less)

We can eliminate B.

Combined

\(­p[1+\frac{6}{100}] > 1000 - 200\)­

\(­p > \frac{800}{1.06}\)­

\(p > $754.71\)

Using the information from Statement 1 and Statement 2, we know that \(p > $754.71\), however we do not know whether\(p > 800\).

Ex:
If p = $760 → Is \(p > $800\) ⇒ The answer is No
If p = $810 → Is \(p > $800\) ⇒ The answer is Yes

As we are getting two different answers, the statements combined don't help.

Option E­
Statement 1 can be derived upto ­Is p[1+t/100]>800? The answer would always be no right? since it can be re-written as Is p>800/(1+t/100)
The argument says she paid some sales tax. Hence, t has to be greater than 0. So, the equation will be Is p>800/(some number greater than 1). which is false. is this reasoning correct? GMATNinja KarishmaB @Bunuel
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Esther bought a computer at a price of p dollars, and she paid a sales 23 Sep 2025, 08:28
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Lets try...

1000 - (p + pt/100) < r...Is p>800?

r=200

800 < (p + pt/100)
Greater than 800...p = 900..t = 10%...990...Possible...
less than 800...p = 700 t = 20%....840..Possible....

So NS ...


Statement 2....
t = 6%

So
1000 - 1.06p < r
P - 799 x 1.06 ---No Case 846 (use calculator)
P- 801 x 1.06 ----849...Yes Case.....NS .....

Combine ....
1000 - 1.06p < 200
Both the above cases can be used..hence...E








Bunuel
­Esther bought a computer at a price of p dollars, and she paid a sales tax of t percent. If Esther had less than r dollars left of the $1,000 that she budgeted for the computer, was the price of the computer, excluding the sales tax, greater than $800?

(1) r = 200
(2) t = 6
­

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