Ans .
160
\(A \times\frac{80}{100} = B \times\frac{50}{100}\) \( ∴ B = \frac{A\times80}{100}= 1.6A\) \( ∴ B = 160% of A\) \( ∴ x = 160\)
Ans .
125% of x
\( y = \frac{100 \times 100}{80}\) of x∴ y = 125% of x
Ans .
10% of y
\(\frac{8x}{100}=\frac{4y}{100}\)∴y = 2x∴20% of x = 10% of y
Ans .
64%
Let x be the multiplicand. ∴Error = \(\frac{5}{3} x - \frac{3}{5}x\) \(=\frac{25x-9x}{15}=\frac{16x}{15}\) ∴Percentage error = \(\frac{\frac{16x}{15}}{\frac{5}{3}x}\times100\) =\(\frac{16x}{15}\times\frac{3}{5x}\times100=64%\)
Ans .
60
p% of p = 36 ∴\(\frac{p}{100}\times p=36\) ∴\(p^2=3600\) ∴p = 60
Ans .
4
Let 2 be x% of 50 x% of 50 = 2 \(\frac{x}{100}\times50=2\) \(\frac{x}{2}=2\) ∴ x = 4
Ans .
200%
Let x % of \(\frac{1}{3}\) = \(\frac{2}{3}\) x% = \(\frac{2\times3}{3}\) = 2 x = 200%
Ans .
₹5
0.15% of 33\(\frac{1}{3}%\) of ₹ 10000
Ans .
240
30% of x = 72 ∴ x = \(\frac{72\times100}{30}=240\)
Ans .
400%
15% of (A + B)
= 25% of (A – B)
=>\(\frac{15}{100}\)(A+B) = \(\frac{25}{100}\)(A-B)
=> 15 (A + B) = 25 (A – B)
=> 15 A + 15 B = 25A – 25 B
=> 10 A = 40 B
=> A = 4 B
Now, let x% of B is equal to A
∴\(\frac{x}{100}\times B\) = A =>\(\frac{x}{100}\times B\) = 4B
∴ x = 400%
Ans .
15
20% of 25% of 300
=\(\frac{20}{100}\times \frac{25}{100} \times 300\)
=\(\frac{1}{5}\times \frac{1}{4} \times 300\) = 15
Ans .
1200
x% of \(\frac{25}{2}\) = 150
=>\(\frac{x}{100} \times \frac{25}{2}\) = 150
=>\(\frac{x}{8}\) = 150
=> x = 150 × 8 = 1200
Ans .
25%
50% of (x – y)
= 30% of (x + y)
=>\(\frac{1}{2}\)(x-y)=\(\frac{3}{10}\)(x+y)
=>\(\frac{x}{2}\) - \(\frac{3x}{10}\)=\(\frac{3y}{10}\) + \(\frac{y}{2}\)
=>\(\frac{5x-3x}{10}\)=\(\frac{3y+5y}{10}\)
=>\(\frac{x}{5}\)=\(\frac{4y}{5}\)
∴ x = 4y
=>y =\(\frac{x}{4}\) or \(\frac{x}{4}\times 100\) % = 25x
Ans .
50
P\(\times \frac{50}{100} \) = Q\(\times \frac{25}{100} \)
=> P × 50 = Q × 25
=> P = \(\frac{Q \times 25}{50}\) => P = \(\frac{Q}{2}\)
P = Q x %
∴ \(\frac{Q \times x}{100}\) = \(\frac{Q}{2}\)
=> x = \(\frac{100}{2}\) = 50
Ans .
40%
20% of A = 50% of B
2A = 5B => A = \(\frac{5B}{2}\)
Let B is x % of A.
\(\frac{5B}{2} \times \frac{x}{100} = B\)
=> x = \frac{200}{5} = 40%
Ans .
8%
Since 18% of the students
neither play football nor cricket.
It means 82% of the students either
play football or cricket or
both.
Using set theory
∴n(A\({\displaystyle \cup }\)B) = n(A)+n(B)-n(A\({\displaystyle \cap }\)B)
=> 82 = 40 + 50 - n(A\({\displaystyle \cap }\)B)
∴n(A\({\displaystyle \cap }\)B) = 90 – 82 = 8
∴ 8% students play both games.
Ans .
7 : 3
\(\frac{20(P+Q)}{100}\) = \(\frac{50}{100}(P-Q)\)
\(\frac{P+Q}{P-Q}\) = \(\frac{5}{2}\)
\(\frac{2P}{2Q}\) = \(\frac{5+2}{5-2}_{\;\;\;\;\;\;\ By \; componendo \; and \; dividendo}\)
\(\frac{P}{Q}\) = \(\frac{7}{3}\) or 7 : 3
Ans .
10
Let x % × 0.1 = 0.01
\(\frac{x}{100}\times 0.1\) = 0.01
x = \(\frac{0.01 \times 100}{0.1}\) = 10
Ans .
\(\frac{13}{4}\)
Required percentage
= \(\frac{65}{2000}\times 100\) = \(\frac{13}{4}\)
[∵ 2kg = 2000g]
Ans .
0.005
1% = \(\frac{1}{100}\)
∴ \(\frac{1}{100}\times \frac{1}{2}\) = \(\frac{1}{200}\) = 0.005
Ans .
7.292 %
1 hour 45 minutes
= 1\(\frac{3}{4} hours = \frac{7}{4} hours\)
1 day = 24 hours
∴ Required per cent
= \(\frac{\frac{7}{4}}{24} \times 100\)
= \(\frac{7}{4 \times 24} \times 100\)
= 7.292%
Ans .
60%
Required percentage = \(\frac{1.14}{1.91} \times 100 = 60%\)
Ans .
40%
Required percentage = \(\frac{32}{40} \times 100 = 40%\)
Ans .
300
\(A \times \frac{90}{100} = \frac{B \times 30}{100}\)
\(→\) A × 3 = B
\(→\) A × x% = A × 3
\(→\) \(\frac{x}{100} = 3 → x = 300 \)
Ans .
400
\(\frac{A \times 90}{100} = \frac{B \times 30}{100}\)
\(→\) 3A = B
\(→\) \(3A = \frac{A \times 2x}{100}\)
\(→\) 300 = 2x \(→\) x = 150
Ans .
75%
\(A \times \frac{30}{100} + \frac{B \times 40}{100} = \frac{B \times 80}{100}\)
\(→\) A × 30 = B × 40
\(→\) \(\frac{A}{B} = \frac{40}{30} = \frac{4}{3}\)
\(→\) \(\frac{B}{A} = \frac{3}{4}\)
\(→\) \(\frac{B}{A} \times 100 = \frac{3}{4} \times 100 = 75%\)
Ans .
\(\frac{7}{6}\)
\((A+B) \times \frac{40}{100}\)
\((A-B) \times \frac{60}{100}\)
\(→\) 2 (A + B) = 3 (A – B)
\(→\) 2A + 2B = 3A – 3B
\(→\) A = 5B
∴\(\frac{2A-3B}{A+B} = \frac{10B-3B}{5B+B}\)
= \(\frac{7B}{6B} = \frac{7}{6}\)
Ans .
0.1%
\( 0.1% = \frac{0.1}{100} = 0.001 \)
Ans .
2%
Required percentage = \( \frac{70}{3.5 \times 1000} \times 100 = 2\% \)
Ans .
300%
\(\frac{1}{3}\) of 1206 = 1206 \(\times \frac{1}{3} \) = 402
∴ Required percent
= \(\frac{402}{134} \times 100 = 300\% \)
Ans .
5
\(a \times \frac{120}{100} = b \times \frac{80}{100} \)
\(→\) \(\frac{b}{a}\) = \( \frac{120}{80} \) = \( \frac{3}{2} \)
∴ \(\frac{b+a}{b-a}\) = \(\frac{\frac{b}{a}+1}{\frac{b}{a}-1}\) = \(\frac{\frac{3}{2}+1}{\frac{3}{2}-1}\) = \(\frac{\frac{5}{2}}{\frac{1}{2}}\) = 5
Ans .
\(\frac{1}{2}\)
(A + B) \(\times \frac{20}{100}\) = B \(\times \frac{50}{100}\)
\(\frac{A + B}{5}\) = \(\frac{B}{2}\)
\(2A + 2B = 5B\)
\(2A = 3B\)
\(\frac{2A}{B}\) = 3 or 2A = 3B
∴ \(\frac{2A - B}{2A + B}\) = \(\frac{2\frac{A}{B} - 1}{2\frac{A}{B} + 1}\) = \(\frac{3 - 1}{3 + 1}\)
= \(\frac{2}{4}\) = \(\frac{1}{2}\) = \(\frac{3B - B}{3B + B}\) = \(\frac{2B}{4B}\)
Ans .
\(\frac{zx}{y}\)% of a
\(\frac{ax}{100}\) = \(\frac{by}{100}\)
\(→\) b = \(\frac{ax}{y}\)
∴ z% of b = \(\frac{ax}{y}\times \frac{z}{100}\)
= \(\frac{xz}{y}\)% of a
Ans .
2
60 × 60 × \(\frac{y}{100}\)
= 1 minute 12 seconds
\(→\) 36y = 72 \(→\) y = 2
Ans .
2%
Required percentage = \(\frac{72}{3.6\times 1000} \times 100 = 2%\)
Ans .
30,000
Let the total number of employees
be x.
∴ x \(\times \frac{402}{134}\) = 20700
∴ x = \(\frac{20700 \times 100}{69}\) = 30000
Ans .
60%
Required percentage = \(\frac{24}{40} \times 100\) = 60%
Ans .
80
\(x \times \frac{125}{100}\) = 100
\(x = \frac{100 \times 100}{125}\) = 80
Ans .
600
\(x \times \frac{83}{100}\) = 498
\(x = \frac{498 \times 100}{83}\) = 600
Ans .
20%
Let C = 100
Then, A = 150
B = 125
∴ Required percentage
= \(\frac{150-125}{125} \times 100\) = 20%
Ans .
50000
If the number of trees in the garden be x, then
\(x \times \frac{60}{100} \times \frac{25}{100} \times \frac{20}{100}\) = 1500
\(→\)\(x \times \frac{3}{5} \times \frac{1}{4} \times \frac{1}{5}\) = 1500
\(→\)x = \(\frac{1500 \times 5 \times 4 \times 5}{3}\)
= 50000
Ans .
88%
Males = 25000 \(\times \frac{4}{5}\) = 20000
Females = 5000
Educated males = \( \frac{20000 \times 95}{100}\) = 19000
Educated Females = \( \frac{5000 \times 60}{100}\) = 3000
Total educated persons = 22000
∴ Required per cent = \( \frac{22000}{25000} \times 100\) = 88%
Ans .
36
Required number = \(\frac{240 \times 25}{100} - \frac{160 \times 15}{100}\)
= 60 – 24 = 36
Ans .
135
First part = ₹x and second part = ₹y
∴ \(\frac{x \times 80}{100}\) = \(\frac{y \times 60}{100}\) + 3
\(\frac{4x}{5}\) = \(\frac{3y}{5}\) + 3
4x – 3y = 15 ...(i)
Again,
\(\frac{4}{5}\) = \(\frac{9x}{10}\) + 3
8y = 9x + 60
8y – 9x = 60 ...(ii)
By equation (i) × 8 + (ii) × 3,
32x – 24y = 120
24y – 27x = 180
\(\frac{\hspace{80pt}}{\:}\)
5x = 300 \(→\) x = 60
From equation (i)
4 × 60– 3y = 15
\(→\) 3y = 240 – 15 = 225
\(→\) y = \(\frac{225}{3}\) = 75
∴ x + y = 60 + 75 = 135
Ans .
80
Group A = 40%
Group B = \(\frac{60 \times 75}{100}\) = 45%
Group C = 15%
If the total number of students be x, then
\( \frac{x\times 15}{100} \)= 12
x = \( \frac{12\times 100}{15} \) = 80
Ans .
33\(\frac{1}{3}\)%
After taking away respective balls,
Number of balls in the box= 75 + 25 + 50 = 150
∴ Percentage of black balls = \( \frac{50}{150} \times 100\) = \( \frac{100}{30}\) = 33\(\frac{1}{3}\)%
Ans .
4 pairs
∵ S.P. of a dozen pairs of socks
= \(\frac{180 \times 80}{100}\) = ₹144
S.P. of 1 pair of socks
∴ S.P. of 1 pair of socks = \(\frac{144}{12}\) = ₹12
No of pairs available for ₹48 = \(\frac{48}{12}\)= 4
Ans .
100
Let the number be x
\(\frac{3}{5} \times \frac{60}{100} \times x\) = 36
x = \(\frac{36 \times 5\times 5}{3\times 3}\) = 100
Ans .
25
\(\frac{P-Q}{2}\) = \((P+Q) \times \frac{30}{100}\)
\(→\) 5(P – Q) = (P + Q) × 3
\(→\) 5P – 3P = 5Q + 3Q
\(→\) 2P = 8Q
\(→\) P = 4Q = \(4 \times \frac{P \times x}{100})\
Ans .
90
Let greater number be x.
∴ Smaller number = 150 – x
According to the question,
\(\frac{40 \times x}{100}\) = \(\frac{60 \times (150-x)}{100}\)
\(→\) 2x = 3 × 150 – 3x
\(→\) 5x = 3 × 150
\(→\) x = 90
Ans .
400
Let the number be x. According to the question
80% of x + 80 = x
\(→\)\(\frac{80x}{100}\) + 80 = x
\(→\)\(\frac{4x}{5}\) + 80 = x
\(→\)\(\frac{x}{5}\) = 80 \(→\) x = 400
Ans .
720
Suppose number be x 20% of x = 120
\(x \times \frac{20}{100}\) = 120
\(x = \frac{120 \times 100}{20}\) = 600
600 \(\times\) 120% = 600 \(\times\) \(\frac{120}{100}\) = 720
Ans .
150
Let the number be x. Then x – 60% of x = 60
\(→\) x – 0.60x= 60
\(→\) 0.4x = 60
\(→\) x = \(\frac{60}{0.4}\)
\(→\) x = \(\frac{600}{4}\)
x = 150
∴ The number is 150
Ans .
300
Let number be x.
∴ According to question,
75% of x + 75 = x
\(\frac{3x}{4}\) + 75 = x
x - \(\frac{3x}{4}\) = 75
x = 75 × 4 = 300
Ans .
40%
Let the third number be x,
According to the question;
First number = \(\frac{20}{100} \times x\) = \(\frac{x}{5}\)
Second number = \(\frac{50}{100} \times x\) = \(\frac{x}{2}\)
∴ Required percentage
= \(\frac{\frac{x}{5} \times 100}{\frac{x}{3}}\) = \(\frac{x}{5} \times \frac{2}{x} \times 100\) = 40%
Ans .
93.75%
Rule : If two numbers are
respectively x% and y% less
than the third number, first
number as a percentage of
second is
\(\frac{100-x}{100-y} \times 100%\)
∴ Required percentage = \(\frac{100-25}{100-20} \times 100%\)
= \(\frac{75}{80} \times 100%\) = 93 75%
Ans .
60
According to question
x + \(\frac{x \times 150}{100}\) = 150
\(→\) x + \(\frac{x \times 3}{2}\) = 150
\(→\) 2x + 3x = 2 × 150 = 300
\(→\) 5x = 300 \(→\) x = 60
Ans .
50
Let the number be x.
According to the question,
\(x \times \frac{18}{100}\) = \(75 \times \frac{12}{100}\)
\(→\)x = \(\frac{75 \times 12}{18}\) = 50
Ans .
5395
Let the numbers be x and y and x > y.
According to the question,
\(6\frac{1}{2}%\) of x = \(8\frac{1}{2}%\) of y
or \(\frac{13}{2}%\) of x = \(\frac{17}{2}%\) of y
or 13x = 17y
or x = \(\frac{17}{13}y\)
∴ \(\frac{17}{13}y - y = 1660\)
or \(\frac{17y - 13y}{13} = 1660\)
or 4y = 1660 × 13
y = \(\frac{1660 \times 13}{4} = 5395\)
Ans .
120
If the number be x, then
\(x \times \frac{75}{100}\) + 75 = x
\(→\)\(\frac{3x}{4}\) + 75 = x
\(→\)x - \(\frac{3x}{4}\) = 75
\(→\) \(\frac{x}{4}\) = 75
\(→\) x = 4 × 75 = 300
∴ 40% of 300
= \(\frac{300 \times 40}{100}\) = 120
Ans .
37
Number to be added = x (let)
∴ \(\frac{320 \times 10}{100}\) + x = \(\frac{230 \times 30}{100}\)
\(→\) 32 + x = 69
\(→\) x = 69 – 32 = 37
Ans .
\(\frac{1}{5}\), -4
X is 20% less than Y.
If Y = 100, X = 80
\(\frac{Y-X}{Y}\) = \(\frac{100 - 80}{100}\)
= \(\frac{20}{100}\) = \(\frac{1}{5}\)
\(\frac{X}{X-Y}\) = \(\frac{80}{80-100}\)
= \(\frac{80}{-20}\) = -4
Ans .
0.025
1% of 1% of 25% of 1000
\( = 1000 \times \frac{25}{100} \times \frac{1}{100} \times \frac{1}{100}\)
= 0.025
Ans .
\(\frac{2}{7}\)
\(\frac{120 \times 25}{100}\) + \(\frac{380 \times 40}{100}\)
= 637 × ?
\(→\) 30 + 152 = 637 × ?
\(→\) 182 = 637 × ?
\(→\) ? = \(\frac{182}{637}\) = \(\frac{2}{7}\)
Ans .
4620
Population of the illiterate in the village
= (100 – 30)% of 6600
= \(\frac{6600 \times 70}{100}\) = 4620
Ans .
10% of y
8% of x = 4% of y
\(→\) \(\frac{x \times 8}{100}\) = \(\frac{y \times 4}{100}\)
\(→\) x = \(\frac{4 \times y}{8}\) = \(\frac{y}{2}\)
∴ 20% of x = \(\frac{20}{100}\) of \(\frac{y}{2}\)
= \(\frac{10}{100}\) of y
= 10% of y
Ans .
2
Let the number be x.
∴ \(x \times \frac{3}{4} \times \frac{4}{5} \times \frac{40}{100}\) = 48
\(→\) \(x \frac{3}{5} \times \frac{2}{5}\) = 48
\(→\) \(x = \frac{48 \times 5 \times 5}{3 \times 2}\) = 200
∴ 1% of 200
= \(200 \times \frac{1}{100} = 2\)
Ans .
4.114
Required sum
= \(\frac{24.2 \times 16}{100}\) + \(\frac{2.42 \times 10}{100}\)
= 3.872 + 0.242
= 4.114
Ans .
\(\frac{1}{30}\)%
x% of 15 hours = 18 seconds
\(→\) x% of 15 × 60 × 60 seconds
= 18 seconds
\(→\) \(\frac{15 \times 60 \times 60 \times x}{100}\) = 18
\(→\) x = \(\frac{18}{15 \times 6 \times 6 }\)% = \(\frac{1}{30}\)%
Ans .
28125
\(80 \times \frac{y}{100}\times \frac{x}{100}\)
= \(\frac{900 \times 25}{100}\)
\(→\) \(\frac{xy \times 80}{10000} = 9 \times 25\)
\(→\) xy = \(\frac{9 \times 25 \times 10000}{80}\) = 28125
Ans .
25 days
Required time = \(\frac{35 \times 100}{140}\) = 25 days
Ans .
500
According to the question,
\(\frac{60A}{100}\) = \(\frac{30B}{100}\)
\(→\) \(\frac{3A}{5}\) = \(\frac{3B}{10}\) = \(\frac{3}{10} \times \frac{40C}{100}\)
\(→\) \(\frac{3A}{5}\) = \(\frac{3C}{25}\) = \(\frac{3A}{25} \times \frac{x}{100}\)
\(→\) \(\frac{3}{5}\) = \(\frac{3x}{2500}\)
\(→\) 5x = 2500
\(→\) x = \(\frac{2500}{5}\) = 500
Ans .
42
Total staff strength in the office = 100 (let)
Females = 40
Males = 60
Married females = \(\frac{40 \times 70}{100}\) = 28
Unmarried females = 40 – 28 = 12
Unmarried males = 30
∴ Unmarried staff
= 30 + 12 = 42
i.e. 42%
Ans .
100
Let the number be x.
According to the question,
\(\frac{x \times 50}{100} + 50\) = x
\(→\)\(\frac{x}{2} + 50\) = x
\(→\)\(x - \frac{x}{2}\) = 50
\(→\)\(\frac{x}{2}\) = 50
\(→\) x = 100
Ans .
125
Let the required amount be Rs. x.
According to the question,
\(90 \times 83\frac{1}{3}% = x \times 60%\)
\(→\)\(90 \times \frac{250}{3}% = x \times 60%\)
\(→\) x = \(\frac{30 \times 250}{60}\) = ₹ 125
Ans .
350
Let the whole number be x.
According to the question,
51% of x = 714
\(\frac{x \times 51}{100}\) = 714
x = \(\frac{714 \times 100}{51}\) = 1400
∴ 25% of 1400
= \(\frac{1400 \times 25}{100}\) = 350
Ans .
Rs. 27
Initial price of eggs = Rs. x per dozen (let).
New price = Rs \(\frac{3x}{4}\)per dozen
According to the question,
\(\frac{162}{\frac{3x}{4}}\) - \(\frac{162}{x}\) = 2
\(→\)\(\frac{216}{x}\) - \(\frac{162}{x}\) = 2
\(→\)\(\frac{54}{x}\) = 2
\(→\) 2x = 54
\(→\) x = Rs. 27 per dozen
Ans .
2.083
Required percent = \(\frac{30}{24 \times 60} \times 100\) = 2.083
Ans .
75%
Initial number of mangoes = 300
Number of remaining mangoes = 300 – 75 = 225
Required percent = \(\frac{225}{300} \times 100\) = 75%
Ans .
46\(\frac{2}{3}\)%
Required percent = \(\frac{3.5}{7.5}\times 100\)
= \(\frac{3500}{75}\) = \(\frac{140}{3}\)
= 46\(\frac{2}{3}\)%
Ans .
20%
Discount percent = \(\frac{1}{5} \times 100 = 20%\)
Ans .
20%
B’s income = Rs. 100
∴ A’s income = Rs. 125
∴ Required percent
= \(\frac{125-100}{125} \times 100\)
= \(\frac{2500}{125}\) = 20%
Ans .
2.5%
1 day = 24 hours = (24 × 60) minutes
∴ Required percent = \(\frac{36}{24 \times 60} \times 100\) = 2.5%
Ans .
16
Let the larger number be x.
∴ Smaller number = \(\frac{25x}{100}\) = \(\frac{x}{4}\)
According to the question,
\(x-\frac{x}{4} = 12\)
\(→\) \(\frac{3x}{4}\) = 12
\(→\) 3x = 12 × 4
x = \(\frac{12 \times 4}{3}\) = 16
Ans .
55
Initial number of students in the class = x
According to the question,
\(x \times \frac{120}{100}\) = 66
\(→\) x = \(\frac{66 \times 100}{120}\) = 55
Ans .
16\(\frac{2}{3}\)%
Required percent =
\(\frac{20}{100+20} \times 100\)
= \(\frac{2000}{140}\)= \(\frac{50}{3}\) = 16\(\frac{2}{3}\)%
Ans .
10000
Number of goats before flood = x (let)
According to the question,
\(x \times \frac{88}{100}\times \frac{95}{100}\) = 8360
\(→\) x =\(\frac{8360 \times 100 \times 100}{88 \times 95}\) = 10000
Ans .
5
Let, C = 100
∴ B = \(100 \times \frac{25}{100}\) = 25
∴ A = \( \frac{20}{100} \times 25\) = 5
∴ x % of C = 5
\(→\) \( \frac{x}{100} \times 100\) = 5
\(→\) x = 5
Ans .
Rs. 720900
Number of boys in the school
= \(\frac{1500 \times 56}{100}\) = 840
Number of girls = (1500 – 840) = 660
Monthly fee of each boy = Rs. 540
Monthly fee of each girl = Rs \(\frac{540 \times 75}{100}\) = Rs. 405
∴ Total monthly fee of boys and girls
= Rs. (840 × 540 + 660 × 405)
= Rs. (453600 + 267300)
= Rs. 720900
Ans .
756
Percentage of children
= (100 – 54 – 32)%
= 14%
According to the question,
∵ 14% = 196
∵ 1% = \(\frac{196}{14}\) = 14
∴ 54% = 54 × 14 = 756 men
Ans .
200
Expression
= \(\frac{25}{4}\)% of 1600 + \(\frac{25}{2}\)% of 800
= \(\frac{1600 \times 25}{400}\) + \(\frac{800 \times 25}{200}\)
= 100 + 100 = 200
Ans .
25
Required percent = \(\frac{25}{100} \times 100\) = 25%
Ans .
50
Required percent = \(\frac{40}{80} \times 100\) = 50%
Ans .
\(\frac{200}{11}\)%
Correct answer = 1 – \((\frac{1}{4}+\frac{1}{5})\)
= 1 – \(\frac{9}{20}\) = \(\frac{11}{20}\)
Incorrect answer = 0.45 = \(\frac{45}{100}\) = \(\frac{9}{20}\)
Error = \(\frac{11}{20}\) - \(\frac{9}{20}\)= \(\frac{1}{10}\)
Percentage error = \(\frac{\frac{1}{10}}{\frac{11}{20}} \times 100\) = \(\frac{200}{11}\)%