This set of Aptitude Questions and Answers (MCQs) focuses on “Simplification – Set 3”.
1. If \(\frac{a}{b}=\frac{1}{2},\frac{b}{c}=3,\frac{c}{d}=\frac{1}{3},\frac{d}{e}\)=4 and \(\frac{e}{f}\)=2, then find the value of \(\frac{abc}{def}\).
a) 16/3
b) 8/3
c) 16/6
d) 8/5
View Answer
Explanation: \(\frac{a}{b}=\frac{1}{2}\)==>b=2a.
\(\frac{b}{c}\)=3==>\(\frac{2a}{c}\)=3==>c=\(\frac{2a}{3}\).
\(\frac{c}{d}=\frac{1}{3}\)==>\(\frac{2a}{3d}=\frac{1}{3}\)==>d=2a.
\(\frac{d}{e}\)=4==>\(\frac{2a}{e}\)=4==>e=\(\frac{a}{2}\).
\(\frac{e}{f}\)=2==>\(\frac{a}{2f}\)=2==>f=\(\frac{a}{4}\).
Therefore, \(\frac{abc}{def}=\frac{a*2a*\frac{2a}{3}}{2a*\frac{a}{2}*\frac{a}{4}}=\frac{16}{3}\).
2. If x=\(\frac{a}{a+b}\) and y=\(\frac{b}{a-b}\), then find the value of \(\frac{ab}{a+b}\).
a) \(\frac{a+b}{a^2+b^2}\)
b) \(\frac{ab}{a^2+b^2}\)
c) \(\frac{ab}{a^2-b^2}\)
d) \(\frac{2ab}{a^2+b^2}\)
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Explanation: \(\frac{ab}{a+b}=\frac{\frac{a}{a+b}*\frac{b}{a-b}}{\frac{a}{a+b}+\frac{b}{a-b}}=\frac{\frac{a}{a+b}*\frac{b}{a-b}}{\frac{a^2-ab+ab+b^2}{(a+b)(a-b)}}=\frac{ab}{a^2+b^2}\).
3. If x=1-2q and y=2q+5, then for what value of q, x is equal to y?
a) 0
b) 1
c) -2
d) -1
View Answer
Explanation: Given, x and y are equal.
1 – 2q = 2q + 5
4q = -4
q = -1.
4. If \(\frac{a}{b}=\frac{2}{3}\) and \(\frac{b}{c}=\frac{15}{8}\), then find \(\frac{c^2-a^2}{c^2+a^2}\).
a) -9/39
b) 9/31
c) -9/41
d) 7/41
View Answer
Explanation: \(\frac{a}{b}=\frac{2}{3}\)==>b=\(\frac{3a}{2}\).
\(\frac{b}{c}=\frac{15}{8}\)==>\(\frac{3a}{2c}=\frac{15}{8}\)==>\(\frac{a}{c}=\frac{5}{4}==>\frac{a^2}{c^2}=\frac{25}{16}\).
\(\frac{c^2-a^2}{c^2+a^2}=\frac{1-\frac{a^2}{c^2}}{1+\frac{a^2}{c^2}}=\frac{1-\frac{25}{16}}{1+\frac{25}{16}}=\frac{16-25}{16+25}=-\frac{9}{41}\).
5. If \(\frac{a}{b}=\frac{7}{6}\), then find the value of \(\Big(\frac{6}{7}-\frac{6a-b}{6a+b}\Big)\).
a) 3/28
b) 5/28
c) 6/28
d) 3/56
View Answer
Explanation: \(\Big(\frac{6}{7}-\frac{6a-b}{6a+b}\Big)=\bigg(\frac{6}{7}-\frac{b(\frac{6a}{b}-1)}{b(\frac{6a}{b}+1)}\bigg)=\Big(\frac{6}{7}-\frac{7-1}{7+1}\Big)=\Big(\frac{6}{7}-\frac{6}{8}\Big)=\frac{6}{58}=\frac{3}{28}\).
6. If \(\frac{p}{q}=\frac{1}{4}\), then find the value of \(\frac{p^2-q^2}{p^2+q^2}\).
a) -15/16
b) -15/17
c) -16/15
d) -17/15
View Answer
Explanation: \(\frac{p}{q}=\frac{1}{4}\)==>\(\frac{p^2}{q^2}=\frac{1}{16}\).
\(\frac{p^2-q^2}{p^2+q^2}=\frac{\frac{1}{16}-1}{\frac{1}{16}+1}=\frac{1-16}{1+16}=-\frac{15}{17}\).
7. If \(\frac{x}{2y}=\frac{3}{5}\), then find \(\frac{x+y}{x-y}+\frac{15}{16}\).
a) 11\(\frac{1}{16}\)
b) 11\(\frac{13}{16}\)
c) 11\(\frac{15}{16}\)
d) 11\(\frac{15}{14}\)
View Answer
Explanation: \(\frac{x}{2y}=\frac{3}{5}\)==>\(\frac{x}{y}=\frac{6}{5}\).
\(\frac{x+y}{x-y}+\frac{15}{16}=\frac{y(\frac{x}{y}+1)}{y(\frac{x}{y}-1)}+\frac{15}{16}=\frac{\frac{6}{5}+1}{\frac{6}{5}-1}+\frac{15}{16}\)=11+\(\frac{15}{16}\)=11\(\frac{15}{16}\).
8. Find the value of \(\Big(1+\frac{2}{x+2}\Big)\Big(1+\frac{2}{x+4}\Big)\Big(1+\frac{2}{x+6}\Big)\Big(1+\frac{2}{x+8}\Big)\).
a) \(\frac{x+10}{x-2} \)
b) \(\frac{x+8}{x+2} \)
c) \(\frac{x-10}{x+2} \)
d) \(\frac{x+10}{x+2} \)
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Explanation: \(\Big(1+\frac{2}{x+2}\Big)\Big(1+\frac{2}{x+4}\Big)\Big(1+\frac{2}{x+6}\Big)\Big(1+\frac{2}{x+8}\Big)=\frac{x+4}{x+2}*\frac{x+6}{x+4}*\frac{x+8}{x+6}*\frac{x+10}{x+8}=\frac{x+10}{x+2}\).
9. If (a-b) is 8 more than (c+d) and (a+b) is 5 less than (c-d), then find (a-c).
a) 1.5
b) 1
c) 0
d) -1
View Answer
Explanation: (a-b)-(c+d) = 8 → a-b-c-d = 8 …… (i)
(c-d)-(a+b) = 5 → a+b-c+d = -5 …… (ii)
On, adding (i)and (ii), we get,
2a -2c = 3
a-c = 3/2 = 1.5.
10. Which of the following is the correct value of (y+a)(y-b)?
a) y2+aby-ab
b) y2+ay-by+ab
c) y2+(a-b)y-ab
d) y2+(a-b)y+ab
View Answer
Explanation: (y+a)(y-b) = y2+ay-by-ab = y2+(a-b)y-ab.
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