Simplification Questions and Answers – Set 6

This set of Aptitude Questions and Answers (MCQs) focuses on “Simplification – Set 6”.

1. If (2p+3q)(2r-3s)=(2p-3q)(2r+3s), then which of the following is true?
a) \(\frac{p}{q}=\frac{r}{s} \)
b) \(\frac{p}{q}=\frac{s}{r} \)
c) \(\frac{p}{s}=\frac{r}{q} \)
d) \(\frac{q}{p}=\frac{r}{s} \)
View Answer

Answer: a
Explanation: (2p+3q)(2r-3s)=(2p-3q)(2r+3s).
\(\frac{(2p+3q)}{(2p-3q)}=\frac{(2r+3s)}{(2r-3s)}\)==>\(\frac{2 \frac{p}{q}+3}{2 \frac{p}{q}-3}=\frac{2\frac{r}{s}+3}{2\frac{r}{s}-3}\)==>\(\frac{p}{q}=\frac{r}{s}\).

2. If (p+q+2r+3s)(p-q-2r+3s)=(p-q+2r-3s)(p+q-2r-3s), then find the value of 2qr.
a) 3/2
b) 3p/2s
c) 3ps
d) (ps)2
View Answer

Answer: c
Explanation: (p+q+2r+3s)(p-q-2r+3s)=(p-q+2r-3s)(p+q-2r-3s)
[(p+q)+(2r+3s)][(p-q)-(2r-3s)]=[(p-q)+(2r-3s)][(p+q)-(2r+3s)].
(p+q)(p-q)-(p+q)(2r-3s)+(2r+3s)(p-q)-(2r+3s)(2r-3s)=(p+q)(p-q)-(p-q)(2r+3s)+(2r-3s)(p+q)-(2r+3s)(2r-3s).
-(p+q)(2r-3s)+(2r+3s)(p-q)=-(p-q)(2r+3s)+(2r-3s)(p+q).
2(2r+3s)(p-q)=2(2r-3s)(p+q).
(2r+3s)(p-q)=(2r-3s)(p+q).
2pr-2qr+3ps-3qs=2pr+2qr-3ps-3qs.
-2qr+3ps=2qr-3ps.
6ps=4qr==>3ps=2qr.

3. Find the value of the expression \(\frac{a+b}{a-b}÷\frac{(a+b)^2}{a^2-b^2}\).
a) -1
b) a+b
c) 1
d) a-b
View Answer

Answer: c
Explanation: \(\frac{a+b}{a-b}÷\frac{(a+b)^2}{a^2-b^2}=\frac{a+b}{a-b}*\frac{a^2-b^2}{(a+b)^2}=\frac{a+b}{a-b}*\frac{(a+b)(a-b)}{(a+b)^2}\)=1.
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4. \(\frac{x^2-y^2-z^2-2yz}{x^2+y^2-z^2+2xy}\) is equivalent to which of the following?
a) \(\frac{x-y+z}{x+y+z}\)
b) \(\frac{x-y-z}{x-y+z}\)
c) \(\frac{x-y-z}{x+y-z}\)
d) \(\frac{x+y+z}{x-y+z}\)
View Answer

Answer: c
Explanation: \(\frac{x^2-y^2-z^2-2yz}{x^2+y^2-z^2+2xy}=\frac{x^2-(y^2∓z^2+2yz)}{(x^2+y^2+2xy)-z^2}=\frac{x^2-(y+z)^2}{(x+y)^2-z^2}=\frac{(x+y+z)(x-y-z)}{(x+y+z)(x+y-z)}=\frac{x-y-z}{x+y-z}\).

5. If a+b+c=0, then a2+ab+b2 is equal to which of the following?
a) c2+bc+b2
b) c2-bc+b2
c) a2-ac+c2
d) a2+ac+c2
View Answer

Answer: d
Explanation: a+b+c=0 🡪 b = -(a+c).
a2+ab+b2=a2-a(a+c)+(a+c)2=a2-a2-ac+a2+c2+2ac=a2+ac+c2.

6. If x+y+z = 2t, then find the value of (t-x)2+(t-y)2+(t-z)2+t2.
a) s2-x2-y2-z2
b) s2+x2+y2+z2
c) x2+y2+z2
d) 4s2-x2-y2-z2
View Answer

Answer: c
Explanation: =(t-x)2+(t-y)2+(t-z)2+t2.
=t2+x2-2tx+t2+y2-2ty+t2+z2-2tz+t2.
=4t2+(x2+y2+z2)-2t(x+y+z).
=(2t)2+(x2+y2+z2)-2t*2t.
=(x2+y2+z2).

7. If (a+b)2-c2=4,(b+c)2-a2=9 and (c+a)2-b2=36, then what is/are value(s) of (a+b+c)?
a) ±3
b) ±9
c) ±1
d) ±7
View Answer

Answer: d
Explanation: Given (a+b)2-c2=4,(b+c)2-a2=9 and (c+a)2-b2=36.
On adding all three equations, we get,
(a+b)2-c2+(b+c)2-a2+(c+a)2-b2=4+9+36.
(a+b+c)(a+b-c)+(a+b+c)(b+c-a)+(a+b+c)(a+c-b)=49.
(a+b+c)[a+b-c+b+c-a+a+c-b]=49.
(a+b+c)2=49==>(a+b+c)=±7.
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8. If x+y = 2z, then find the value of \(\frac{x}{x-z}+\frac{y}{y-z}\).
a) 1/2
b) 1
c) 2
d) 3
View Answer

Answer: c
Explanation: x+y = 2z 🡪 x = 2z-y.
\(\frac{x}{x-z}+\frac{y}{y-z}=\frac{2z-y}{2z-y-z}+\frac{y}{y-z}=\frac{2z-y}{z-y}+\frac{y}{y-z}=-\frac{2z-y}{y-z}+\frac{y}{y-z}=\frac{-2z+y+y}{y-z}=\frac{2(y-z)}{y-z}\)=2.

9. If xy+yz+xz=0, then what is the value of \(\frac{1}{x^2-yz}+\frac{1}{y^2-xz}+\frac{1}{z^2-xy}\)?
a) 0
b) 1
c) 3
d) x+y+z
View Answer

Answer: a
Explanation: xy+yz+xz=0==>xy=-yz-xz or yz=-xy-xz or xz=-xy-yz.
=\(\frac{1}{x^2-yz}+\frac{1}{y^2-xz}+\frac{1}{z^2-xy}\).
=\(\frac{1}{x^2+xy+xz}+\frac{1}{y^2+xy+yz}+\frac{1}{z^2+yz+xz}\).
=\(\frac{1}{x(x+y+z)}+\frac{1}{y(x+y+z)}+\frac{1}{z(x+y+z)}\).
=\(\frac{yz+xz+xy}{xyz(x+y+z)}\)=0.
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10. If x*y=\(\frac{xy}{x+y}\), then find the value of 5*(4*-2).
a) 5
b) -20
c) 16
d) -40
View Answer

Answer: b
Explanation: 5*(4*-2)=5*\(\frac{4(-2)}{4-2}\)=5*-4=\(\frac{5(-4)}{5-4}\)=-20.

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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