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Bunuel
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 27 Dec 2019, 01:49
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

75% (02:53) correct 25% (02:59) wrong based on 260 sessions

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If \((a + b) : (b + c) : (c + a)\) is \(3 : 4 : 5\) and \(a + b + c = 18\), find the value of \(a × b × c\).

A. 60
B. 120
C. 144
D. 162
E. 172
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D
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 27 Dec 2019, 02:27
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Bunuel
If \((a + b) : (b + c) : (c + a)\) is \(3 : 4 : 5\) and \(a + b + c = 18\), find the value of \(a × b × c\).

A. 60
B. 120
C. 144
D. 162
E. 172
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Given ,

(a + b) : ( b + c) : (c + a) = 3:4:5

we can formulate an equation:

a + b + b + c + c + a = 3x + 4x + 5x

2(a + b + c ) = 12x

12x = 2*18

12x = 36

x = 3.

Now,

a + b = 3*3 = 9

b + c = 4*3 = 12

deduct them,

b + c - a - b = 12 - 9 = c - a = 3

c - a = 3

a + c = 15

add them ,

a + c + c - a = 15 + 3 = 18

2c = 18

c = 9.

now we have substitute the values,

b + c = 12

b = 12 - 9 = 3

a + b = 9

a = 9 - 3 = 6

so,

a*b*c = 6*3*9 = 162.

D is the correct answer
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 27 Dec 2019, 04:59
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(a+b):(b+c):(c+a) = 3x:4x:5x

(a+b)+(b+c)+(c+a) = 3x+4x+5x
2(a+b+c) = 12x = 2*18
x=3.

So, (a+b),(b+c),(c+a) are 9, 12, 15, respectively.

(a+b)-(b+c)+(c+a)= +9-12+15=12
2a=12 --> a=6, and consequently, b=3 and c=9.

a*b*c=6*3*9=162

Final answer is (D)

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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 28 Dec 2019, 00:17
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Approach:

- Given (a+b):(b+c):(c+a) = 3:4:5 & a+b+c = 18

- Assume (a+b), (b+c), and (c+a) be 3x, 4x, and 5x

- On addition we get 2(a+b+c) = 12x => x = 3

- To get 'a' : (a+b+c)-(b+c) => 18 - 4(3) = 6

- To get 'b' : (a+b+c)-(c+a) => 18 - 5(3) = 3

- To get 'c' : 18-a-b => 18-6-3 = 9

Finally \(a×b×c= 6×3×9 = 162\)

Option D!
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 28 Dec 2019, 00:29
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Say a+b = 3k, b+c=4k, c+a=5k
So we can deduce from (a+b and b+c) => c-a=k
Solving the 2 equations, these are the values:
c=3k
a=2k
b=k

a+b+c = 18
So, k=3

a*b*c = 2*3*1*(3^3) = 162
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 01 Mar 2020, 02:38
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If \((a + b) : (b + c) : (c + a)\) is \(3 : 4 : 5\) and \(a + b + c = 18\), find the value of \(a × b × c\).

A. 60
B. 120
C. 144
D. 162 --> correct

E. 172

Solution:
\((a + b) : (b + c) : (c + a)\) is \(3 : 4 : 5\)
=> let's say, a+b = 3x, b+c = 4x & c+a = 5x;
so adding all three, we get 2(a+b+c)=12x=2*18=> x = 3;
so a+b = 3*3 --(i)
b+c = 4*3 ----(ii)
& c+a = 5*3 ----(iii)
so (ii)-(i)=> (b+c)-(a+b)=12-9=> c-a =3 --(iv)
-> so (iii) + (iv) => (c+a) + (c-a) = 15+3 => 2x=18=> c= 9,
so from (iii)=> a=15-9=6,
so from (i)=> b = 9-6=3,
finally \(a × b × c\) = 6*3*9=162
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If (a + b) : (b + c) : (c + a) is 3 : 4 : 5 and a + b + c = 18, find 05 Feb 2026, 09:59
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Deconstructing the Question
We are told:
\((a+b): (b+c): (c+a)=3:4:5\)
So set:
\(a+b=3k,\ b+c=4k,\ c+a=5k\)
Also \(a+b+c=18\).
Find \(a\cdot b\cdot c\).

Step-by-step
From the ratio:
\(a+b=3k\)
\(b+c=4k\)
\(c+a=5k\)

Add all three:
\((a+b)+(b+c)+(c+a)=3k+4k+5k=12k\)

Left side simplifies to:
\(2(a+b+c)=12k\)

Use \(a+b+c=18\):
\(2\cdot 18=12k \Rightarrow 36=12k \Rightarrow k=3\)

So:
\(a+b=9,\ b+c=12,\ c+a=15\)

Find \(b\):
\((a+b)+(b+c)=21 = (a+b+c)+b = 18+b\)
So:
\(b=3\)

Then:
\(a=9-3=6\)
\(c=12-3=9\)

Product:
\(a\cdot b\cdot c = 6\cdot 3\cdot 9 = 162\)

Answer: 162
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