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Assume a =3= 1+2=1+1+1
b=1; c=2; d=1; e=1; f=1

[3(3+1+1)]+[1(2+1)]+[(1+1)*1][2(2+1)]+ 1= 27=3^3

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\(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)

= \(a(ad+be+df) + be(c+f) + ce(c+f) + (d+e) f^2 + f*f^2\)

= \(a(ad+be+df) + e(c+f)(b+c) + (d+e+f) f^2\)

= \(a( ad+be +df + ec+ef +f^2)\)

= \(a[ ad + e (b+ c) + f(d+e+f)]\)

= \(a[ad+ae+af]\)

=\(a^2(d+e+f)\)

= \(a^3\)


MathRevolution
[GMAT math practice question]

\(a = b + c = d + e + f.\) What is the expression of \(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)?

A. \(a\)

B. \(a^2\)

C. \(a^3\)

D. \(a^4\)

E. \(a^5\)


any shortcuts ? this is taking a lot longer than usual GMAT questions

Posted from my mobile device
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[GMAT math practice question]

\(a = b + c = d + e + f.\) What is the expression of \(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)?

A. \(a\)

B. \(a^2\)

C. \(a^3\)

D. \(a^4\)

E. \(a^5\)

Given: a = b + c = d + e + f.

Asked: What is the expression of a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3?

a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3
a^2d + abe + adf + bec + bef + df^2 + ef^2 +c^2e +cef + f^3
f^2(d + e + f) + a^2d + abe + adf + bec + bef + c^2e +cef
af^2 + a^2d + abe + adf + ce (b + c) + ef(b+c)
a [f^2 + ad + be + df] + ace + aef
a [f^2 + ad + be + df + ce + ef]
a [f(d + e + f) + ad + e(b+c)]
a [af + ad +ae]
a^2 [d+e+f]
a^3

IMO C
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=>

\(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)

\(= a(ad + be + df) + be(c + f) + (d + e + f)f^2 + ce(c + f)\) (combining the terms (\(d + 3)f^2\) and \(f^3\))

\(= a(ad + be + df) + (b + c)e(c + f) + (d + e + f)f^2\) (taking out a common factor of \(e(c + f)\) from be\((c + f)\) and \(ce(c + f))\)

\(= a(ad + be + df) + ae(c + f) + af^2\) (since \(a = b + c = d + e + f\))

\(= a(ad + be + df] + a(ce + ef + f^2)\)

\(= a[ad + be + df + ce + ef + f^2]\)

\(= a[ad + (b + c)e + (d + e + f)f]\)

\(= a[ad + ae + af]\), since \(a = b + c = d + e + f\)

\(= a^2(d + e + f)\)

\(= a^3,\) since \(a = d + e + f\)

Therefore, C is the answer.
Answer: C
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jackfr2
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\(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)

= \(a(ad+be+df) + be(c+f) + ce(c+f) + (d+e) f^2 + f*f^2\)

= \(a(ad+be+df) + e(c+f)(b+c) + (d+e+f) f^2\)

= \(a( ad+be +df + ec+ef +f^2)\)

= \(a[ ad + e (b+ c) + f(d+e+f)]\)

= \(a[ad+ae+af]\)

=\(a^2(d+e+f)\)

= \(a^3\)


MathRevolution
[GMAT math practice question]

\(a = b + c = d + e + f.\) What is the expression of \(a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f^3\)?

A. \(a\)

B. \(a^2\)

C. \(a^3\)

D. \(a^4\)

E. \(a^5\)


any shortcuts ? this is taking a lot longer than usual GMAT questions


Yes - You can observe, since all of them are addition, hence no change in degree.
Also, maximum degree in the expression is 3 - f^3

So ans will be a^3
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Solving it algebraically will be too time-consuming.

Let's assume values for the given variables. We can assume 1 = c,d,e,f and b= 2 and a = 3

This will give you the correct answer, but I think choosing unique values would be a much safer option.

d=5
e=3
f=2
b=4
c=6
a=10
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Able to solve the question but taking very long time ,any way to reduce time?
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