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Hovkial
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A man purchases five tables and seven chairs for $5850. 29 Sep 2019, 10:01
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55% (hard)

Question Stats:

71% (03:07) correct 29% (03:11) wrong based on 213 sessions

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A man purchases five tables and seven chairs for $5850. He sells the tables at a profit of 10% and the chairs at a profit of 16%. His overall gain is $711. What total price did he pay for one table and one chair?

(A) 300

(B) 750

(C) 900

(D) 1050

(E) 1500
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D
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A man purchases five tables and seven chairs for $5850. 29 Sep 2019, 10:46
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{5×1.1x+7×116y/100=6561
{5x+7y=5850
(812-770)y=656100-643500=12600
42y=12600
y=300$ , x=750$
750+300=1050
Option D

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GMAT 2: 760 Q51 V40
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A man purchases five tables and seven chairs for $5850. 13 Apr 2020, 05:05
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Hovkial
A man purchases five tables and seven chairs for $5850. He sells the tables at a profit of 10% and the chairs at a profit of 16%. His overall gain is $711. What total price did he pay for one table and one chair?

(A) 300

(B) 750

(C) 900

(D) 1050

(E) 1500
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Solution:


Let a and b be the price of one table and chair, respectively.
    • CP of Tables and chairs =\( 5*a + 7*b\)
    • \(5*a + 7*b = 5850\)…(i)
      o The selling price of Tables = \(5a*\frac{110}{100}\)
      o The selling price of chairs = \(7*b*\frac{116}{100}\)
    • Profit = SP – CP = \(5a*\frac{110}{100} + 7*b*\frac{116}{100} – 5a – 7b\)
    • \(\frac{550*a -500*a + 812*b-700*b}{100} = 711\)
    • \(50*a + 112*b = 71100\)…(ii)
On multiplying equation (i) by 10 and subtracting it from equation (ii), we get
    • \(50*a + 112*b – 50*a -70*b = 71100-58500\)
    • \(42*b = 12600\)
    • \(b = 300\)
      o \(5*a = 5850 – 7*300\)
      o \(a = \frac{5850-2100}{5} = 750\)
    • \(a + b = 750 + 300=1050\)
Hence, the correct answer is Option D.
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A man purchases five tables and seven chairs for $5850. 13 Apr 2020, 07:01
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5t+7c=5850
0.5t+7(0.16t) =711----gain multiply with 10 and subtact equations
C=300
So t=750
Total 1050(D)
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A man purchases five tables and seven chairs for $5850. 14 Apr 2020, 05:38
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I am sorry it's typo it should be 0.16c in second equation.

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A man purchases five tables and seven chairs for $5850. 27 Jan 2024, 02:45
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Hovkial
A man purchases five tables and seven chairs for $5850. He sells the tables at a profit of 10% and the chairs at a profit of 16%. His overall gain is $711. What total price did he pay for one table and one chair?

(A) 300

(B) 750

(C) 900

(D) 1050

(E) 1500
Show more
\(5T + 7C = 5850\)

\(5T*0.10 + 7C*0.16 = 711\)

Now, \(0.50T + 1.12C = 711\)

Multiply \(0.50T + 1.12C = 711\) by 10 -

\(5T + 11.20C = 7110\)

Now, use substitution -

\(5850 - 7C + 11.20C = 7110\)

Or,\(4.2C = 1260\)

Or, \(C = 300\)

Now again substitute \(C = 300\) in \(5T + 7C = 5850\)

Thus, \(5T + 7*300 = 5850\)

Or, \(5T = 3750\)

So, \(T = 750\)

Quote:
What total price did he pay for one table and one chair?
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\(T + C = 750 + 300 => 1050\), Answer will be (D) 1050
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A man purchases five tables and seven chairs for $5850. 02 Mar 2026, 05:51
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Deconstructing the Question

Let:

T = price of one table
C = price of one chair

Total cost:

\(5T + 7C = 5850\)

Profit:
Tables → 10%
Chairs → 16%

Step-by-step

Profit from tables:

\(0.10(5T) = 0.5T\)

Profit from chairs:

\(0.16(7C) = 1.12C\)

Total profit:

\(0.5T + 1.12C = 711\)

Multiply second equation by 10:

\(5T + 11.2C = 7110\)

Subtract first equation:

\((5T + 11.2C) - (5T + 7C) = 7110 - 5850\)

\(4.2C = 1260\)

\(C = 300\)

Substitute into first equation:

\(5T + 2100 = 5850\)

\(5T = 3750\)

\(T = 750\)

Total price of one table and one chair:

\(750 + 300 = 1050\)

Answer: D
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