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15 Jul 2025, 22:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
85% (01:34) correct
15%
(01:34)
wrong
based on 66
sessions
History
Date
Time
Result
Not Attempted Yet
In a certain box of of widgets there are only brass widgets and iron widgets. If 42 of the widgets are brass and 14 are iron, what is the total weight, in kilograms of the box?
(1) The weight of seven brass widgets is three times the weight of an iron widget.
(2) The total weight of a brass widget and five iron widgets is 38 kilograms.
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
(1) The weight of seven brass widgets is three times the weight of an iron widget.
(2) The total weight of a brass widget and five iron widgets is 38 kilograms.
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
16 Jul 2025, 01:39
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No of Brass widgets, \(B=42\)
No of Iron widgets, \(I=14\)
weight of brass widget, \(b\) & weight of iron widget, \(i\)
\(Bb+Ii=?\)
Statement 1: The weight of seven brass widgets is three times the weight of an iron widget.
\(7b=3i\)
\(b=\frac{3}{7}i\)
Not Sufficient
Statement 2: The total weight of a brass widget and five iron widgets is 38 kilograms.
\(b+5i=38\)
There is only one equation.
Not Sufficient
Statements Combined:
\(b=\frac{3}{7}i\) & \(b+5i=38\)
\(\frac{3}{7}i+5i=38\)
\(i=7\) & \(b=3\)
\(Bb+Ii=(42*3)+(14*7)=126+98=224\)
Sufficient
Answer: C
No of Iron widgets, \(I=14\)
weight of brass widget, \(b\) & weight of iron widget, \(i\)
\(Bb+Ii=?\)
Statement 1: The weight of seven brass widgets is three times the weight of an iron widget.
\(7b=3i\)
\(b=\frac{3}{7}i\)
Not Sufficient
Statement 2: The total weight of a brass widget and five iron widgets is 38 kilograms.
\(b+5i=38\)
There is only one equation.
Not Sufficient
Statements Combined:
\(b=\frac{3}{7}i\) & \(b+5i=38\)
\(\frac{3}{7}i+5i=38\)
\(i=7\) & \(b=3\)
\(Bb+Ii=(42*3)+(14*7)=126+98=224\)
Sufficient
Answer: C
16 Jul 2025, 03:37
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Bunuel
Box = 42 Brass (B) + 14 iron (F)
We need to find the weight of the box .
Statement 1:
(1) The weight of seven brass widgets is three times the weight of an iron widget.
7 B = 3 F
Hence, insufficient
Statement 2:
(2) The total weight of a brass widget and five iron widgets is 38 kilograms.
B + 5F = 38
Hence, Insufficient
Combining statements 1 and 2, we get
B = (3/7)*F
(3/7)*F + 5F = 38
38F = 7* 38
F = 7 , then B = 3
Thus, 42 B + 14 F = 42*3 + 14*7 = 224 kg.
Hence, Sufficient
Option C
