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kiran120680
Moderator - Masters Forum
Joined: 18 Feb 2019
Last visit: 27 Jul 2022
Posts: 706
Given Kudos: 276
Location: India
Schools: AGSM '20 EDHEC Sept 2019 Intake Great Lakes '21 IE
GMAT 1: 460 Q42 V13
GPA: 3.6
14 Apr 2019, 07:25
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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)!
Difficulty:
95%
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Question Stats:
57% (03:09) correct
43%
(02:45)
wrong
based on 362
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If p:q=r:s and r:s=t:v , find the value of ((p^2∗k^z+r^2∗k^z+t^2∗k^z)/(q^2+s^2+v^2))^1/z
A. k
B. (p/s)^1/z∗k^2
C. (r/v)^z/2∗k
D. (r/s)^2/z∗k
E. (t/s)^2/z∗k
A. k
B. (p/s)^1/z∗k^2
C. (r/v)^z/2∗k
D. (r/s)^2/z∗k
E. (t/s)^2/z∗k
16 Apr 2019, 03:40
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kiran120680
Here is the solution..
\(( \frac{(p^2∗k^z+r^2∗k^z+t^2∗k^z)}{(q^2+s^2+v^2)} )^{1/z}\)
Multiply both denominator and numerator by r^2..
\(( \frac{r^2(p^2∗k^z+r^2∗k^z+t^2∗k^z)}{(r^2q^2+r^2s^2+r^2v^2)} )^{1/z}\)...
Now, p:q=r:s means ps=rq and similarly r:s=t:v means RV=at...
\(( \frac{r^2(p^2∗k^z+r^2∗k^z+t^2∗k^z)}{(p^2s^2+r^2s^2+s^2t^2)} )^{1/z}\)
Take out common terms s^2 in denominator and k^z in numerator..
\(( \frac{r^2k^z(p^2+r^2+t^2)}{s^2(p^2+r^2+t^2)} )^{1/z}\)
\((\frac{r}{s})^{2/z}*k^{z/z}\)
D
01 Apr 2022, 17:47
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If unable to see the Algebraic solution, you can substitute easy numbers and match the expression up with the answer choices
P/Q = R/S = T/V
Let
P = 1 and Q = 2
R = 3 and S = 6
T = 4 and V = 8
(1st) there is a common (k)^z in the numerator, so you can take the common factor
(K)^z * ( p^2 + r^2 + t^2)
________________________
(q^2 + s^2 + v^2)
This is all raised to the power of: ^(1/z)
Or
The “Z-Root” is taken on the expression
2nd: Exponent Rule: Power of a Product Rule
(K)^Z power ——-> taking the Z root will make this expression K
And then (1/Z) power is taken on the remaining fraction —- substituting numbers
K * [ (1^2 + 3^2 + 4^2) ]^(1/z)
_________________________
[ (2^2 + 6^2 + 8^2) ]^(1/z)
K * (26 / 104)^(1/z)
K * (1/4)^(1/z)
K * (1^2 / 2^2) ^(1/z)
Using the nested exponent rule backwards —- the numerator of the fractional exponent is the power the Base is taken to
K * (1/2) ^ (2/z)
The base of 1/2 is equal to any of the given ratio, so we just need to find an answer that fits
D: since (r/s) = (1/2)
k * (r/s)^ (2/z)
D is the correct answer (can be accomplished in around 2 minutes with accurate and effective arithmetic)
Posted from my mobile device
P/Q = R/S = T/V
Let
P = 1 and Q = 2
R = 3 and S = 6
T = 4 and V = 8
(1st) there is a common (k)^z in the numerator, so you can take the common factor
(K)^z * ( p^2 + r^2 + t^2)
________________________
(q^2 + s^2 + v^2)
This is all raised to the power of: ^(1/z)
Or
The “Z-Root” is taken on the expression
2nd: Exponent Rule: Power of a Product Rule
(K)^Z power ——-> taking the Z root will make this expression K
And then (1/Z) power is taken on the remaining fraction —- substituting numbers
K * [ (1^2 + 3^2 + 4^2) ]^(1/z)
_________________________
[ (2^2 + 6^2 + 8^2) ]^(1/z)
K * (26 / 104)^(1/z)
K * (1/4)^(1/z)
K * (1^2 / 2^2) ^(1/z)
Using the nested exponent rule backwards —- the numerator of the fractional exponent is the power the Base is taken to
K * (1/2) ^ (2/z)
The base of 1/2 is equal to any of the given ratio, so we just need to find an answer that fits
D: since (r/s) = (1/2)
k * (r/s)^ (2/z)
D is the correct answer (can be accomplished in around 2 minutes with accurate and effective arithmetic)
Posted from my mobile device
