Percentage questions

Here x% increases and x % decreases , then change in the percentage =\( \frac{x^{2}}{100} \)% decreases.

Here x = 30%

Then , \( \frac{30^{2}}{100} \) =\( \frac{900}{100} \) = 9% decreases.

4 hours \(=4 \times 60 \times 60=14,400\) seconds

Required fraction \(=\frac{18}{14400}=\frac{1}{800}\)

\(50\%\) of \((p - q) =30\%\) of \((p+q)\)

\(\displaystyle \Rightarrow 5 (p – q)= 3(p + q)\)

\(\displaystyle \Rightarrow 5p – 5q= 3p + 3q\)

\(\displaystyle \Rightarrow p = 4q\)

\(\displaystyle \Rightarrow (p : q)=(4 : 1)\)

Hence, the required answer is \((4 : 1)\).

Let Number be 100.
10% increase = \(\frac{110}{100}\times100 \) = 110
10% decrease= \( \frac{90}{110}\times100 \) = 99
Net change= \(\frac{99-100}{100}\times100 \) =-1.That is 1% decrease.

\(20\)% of \(5\) litre of alcohol \(\frac{20}{100} \times5 =1 \) litre of alcohol

Now add one liter of water now the mixture become \(6\) litres ,that contains \(1\) litre of alcohol

Strength of alcohol in new solution is equal to \(\frac{1 \:litre}{ 6 \:litre} \times100\) %

which is equal to \(16 \frac{ 2}{3}\) %

Let \(x\) be the number.
33% of \(33\frac{1}{3} = \frac{33}{100}\times \frac{100}{3}\) [\(\because 33\frac{1}{3} = \frac{100}{3}\)]
\(= 11\)
50% of \(x \)= 11
\(\Rightarrow \frac{50}{100}\times x = 11\)
\( \Rightarrow x = 11\times 2 = 22\)

\( x \) = 80 % of \( y \)

=\( x =(\frac{80}{100} )y\)

\(=\frac{ y}{x} = \frac{5}{4}\)

Required percentage = \(\frac{y}{2x}\times 100 = (\frac{5}{8}\times 100) =62.5\% \)

Let \(B=100\)

Then \(A=165\)

Now B’s income is 65 less than A’s

Hence it is,

\( \begin{array}{l} =\frac{65}{165} \times 100 \\ =\frac{13}{33} \times 100=\frac{1300}{33}=39 \frac{13}{33} \end{array} \)

\( \frac{x}{100} \times y = \frac{7}{8} \times 90 \)
\( \frac{xy}{100} = \frac{630}{8} \)
xy = \( \frac{630}{8} \times 100 \)
xy = \( \frac{63000}{8} \)
xy = 7875.

= \(\frac{\frac{1}{2}}{\frac{1}{3}}\)\(\times\) 100

= \(\frac{3}{2}\)\(\times\) 100

= 150 %

\( 1000 \times \frac{0.15}{100} \times \frac{100}{300} \)

\( = \frac{0.15 \times 100 \times 10000}{100 \times 300} \)

\( = 5 \)

sol: 80% of A =50% of B

\( A\times \frac{80}{100}= B\times \frac{50}{100} \)

\( = 8A = 5B \)

\( A = \frac{5}{8} B \)

Put value of A in given equation

\( B = x \)% of \( A \)

\( B = \frac{x \times \frac{5}{8} B}{100 } \)

\( x = \frac{100\times8}{5} \)

\(x = 160\)

Given that

80 % of A = 20 % of B

\( \frac{80\times A}{100} \)\( = \frac{20\times B}{100} \)

\( \frac{80\times A}{100} \)\( = \frac{1\times B}{5} \)

B =\( \frac{400 \times A}{100} \)

B= 400 % of B

X= 400

\(90 \%\) of \(A=30 \%\) of \(B\) and \(B=2x \%\) of \(A\)

\( A \times\frac{90}{100} = B \times\frac{30}{100} \)

\( A \times\frac{90 \times 100}{100\times 30} = B \)

\( A \times\frac{300}{100} = B \)

B = 300 % A

We know B=2x%

So x= 150

• easy method

• x - y -\( \frac{xy}{100}\)

  10 - 15 - \(\frac{10\times 15}{100}\) = -6.5 % , decrease of 6.5%

Simple trick

• Let the Orginal price = 100

  Increased by 10% = 100+10 = 110

  {110 \(\times \frac{15}{100 }=16.5\)}

  Decreased by \( 15% = 110 - 16.5 = 93.5 \)

  Total % decrease = 100% -93.5 = 6.5%

\(60\% of A = 30\% of B\)
\(60 \times A = 30 \times B\)
\(B = 2A\)
\(A: B = 1: 2\)
\(B = 40\% of C\)
\(10 B = 4 C \)
\(2 C = 5 B\)
\(B: C = 2: 5\)
\(A: B: C = 2:4:10 = 1:2:5\)
Given C is x% of A
From the above equation we get
\(x = \frac{5}{1} \times 100 = 500\% \)

Required percentage

\(=\frac{24}{40} \times 100=60 \%\)

4% of x = \(\frac{4x}{100}\)
2% of y=\(\frac{2y}{100}\)
\(\frac{4x}{100}\)=\(\frac{2y}{100}\)
4x=2y
y=2x
multiplying both sides by 10% we get
20 % of x = 10% of y

Required percentage = \(\frac{50}{100 - 50}\times 100\) = 100%

Hence , the answer is option B .

40% of x = \(\frac {40}{100}x=y\)---1 <br> <br>y% of 40 = \(\frac {\frac {40}{100}x}{100}\times40\)---2 <br> <br>Equating 1 and 2, \(y\) % of\( 40 \) =\(\frac {16}{100}x\) = 16%

\(( A + B ) \times \frac{40}{100}\)

\(=(A-B) \times \frac{60}{100}\)

\(\Rightarrow 2(A+B)=3(A-B)\)

\(\Rightarrow 2 A+2 B=3 A-3 B\)

\(\Rightarrow A=5 B\)

\(\therefore \frac{2 A-3 B}{A+B}=\frac{10 B-3 B}{5 B+B}\)

\(=\frac{7 B}{6 B}=\frac{7}{6}\)

\(\frac{x}{100}\times y=\frac{4}{5}\times 80\) , xy=\(64\times 100=6400\)

Required percentage

\(=\frac{1.14}{1.9} \times 100=60 \%\)

Given: \({(\frac{70}{100})}\times{A}\) =\({(\frac{50}{100})}\times{B}\)

B=\({(\frac{70}{100})}\times{A}\times 2\)

B = 140% of A

\( X = 140 \)

60% of (x-y) = 20% of (x+y)

\( 6(x-y) = 2 (x+y) \)

\( 6x-6y = 2x+2y \)

\( 6x-2x = 6y+2y \)

\( \frac{x}{y} = \frac{8}{4} = 2 \)

Required Percentage is \( =\frac{ y}{x }\times 100 \)

\( = \frac{1}{2} \times 100 \)

= 50 %

Let the income of B = 100,
then income of A = 140% of 100 \(=100 \times \frac{140}{100}=140\)

Thus, the income of B is less than the income of A = 140 – 100 = 40

\(\therefore \ Less \ \%=\frac{40}{140} \times 100 \%\\ =28 \frac{4}{7} \%\)

x is 50% of y<br> <br>X= \( Y\times\frac{{50}}{100} \) <br> <br>y= \( X\times\frac{{100\times2}}{50 \times2 } \) <br>

Y = 200 % of x

Given p% of p = 81\( \Rightarrow \frac{p}{100}\times p = 81 \)

\( \Rightarrow p^{2} = \)8100

\( \therefore p = 90 \)

\(\frac{20}{100} \times A = B\)

\(\frac{1}{5}\times A = B\)

\( A = 5B \)

\(\frac{40}{100} \times B = C\)

\(\frac{2B}{5} = C\)

\(\frac{5C}{2} = B\)

\( A+B= 5B + B = 6B \)

\( A +B = 6B = 6\times\frac{5}{2}\times C= 15 C\)

33\(\frac{1}{3}\)

\(\frac{100}{3}\)%

\(\frac{100}{3 \times 100}\)

\(\frac{100}{300}\) = \(\frac{1}{3}\)

\( \frac{3}{24} \times 100= \frac{100}{8}=12.5 \%\)

p% of p = 64
\({(\frac{p}{100}) }\times{p}\) =64
\(p^{2}\)= 6400
p=80

\(x = 80\)% of \(y=\frac{80y}{100}\)
\(\frac{5x}{4}=y\)
\(\frac{(5\times25)}{(4\times25)}\times x=y\)
\(y=\frac{125}{100} \times x\)
\(y=125\)% of \(x\)

total number of games = 30
game won= 24
Required percentage= \({(\frac{24}{30})}\times {100}\)
= \(\frac{2400}{30}\) = 80%

\(x = 40\)% of y , \(x=\frac{40y}{100}\)
\(\frac{5x}{2}=y\)
\(\frac{(5\times50)}{(2\times50)}x=y\)
\(y=\frac{250}{100}x\)
\(y=250\)% of \(x\)

Let x% of \(\frac{1}{3}\)=\(\frac{2}{3}\)
x%= \(\frac{({2}\times{3})}{3}\) =2
x=200%

\(x = 20\)% of y , \(x=\frac{20y}{100}\)
\(y= \frac{5x}{1}\)
\(y= \frac{(5\times100)}{(1\times 100)}x\)
\(y=\frac{500}{100}x\)
\(y=500\)% of \(x\)

\(\frac{p}{100}\times p =36\)

\(\Rightarrow p^{2} =3600 \)

\( p=60\)

60% of A = \(\frac{3}{4}\) of B

=> \(A \times\frac{60}{100}\ = B \times\ \frac{3}{4}\)

=> \(\frac{3}{5}\ \times A = \frac{3}{4}\ \times B\)

=> \(\frac{A}{B}\ = \frac{3}{4}\ × \frac{5}{3}\ = \frac{5}{4}\)

Let, the required number be \(x\).
\(x \times \frac35 \times 60 \% = 36\)
\(\Rightarrow x \times \frac35 \times \frac{60}{100} = 36\)
\(\Rightarrow \frac{18x}{50} = 36 \Rightarrow 18x = 50 \times 36\)
\(\Rightarrow x = \frac{50 \times 36}{18 }= 100\)

No of boys who didn't write the exam\(= \frac{50}{100}\times 1500=750\)
No of girls who didn't write the exam\(= \frac{60}{100}\times 800=480\)
Total number of student who didn't write exam\(=750+480=1230\)
The percentage of the candidates who didn't write the examination\(= \frac{1230}{2300}\times100=53.48\%\)

Let B = 100

Then A = 48% more than B = 100 + 48 = 148

A + B = 148 + 100 = 248

C is 60% less than the sum of A and B;

C = 40% of (A+B) = (40/100)×248 = 99.2

Now, the required Percentage = 49.2%

Quantity of alcohol = \((\frac{40}{100}\times 5 )=2 litres\)

New strength = \((\frac{2}{6}\times 100)\%=33\frac{1}{3}\%\)

\(\frac{8\frac{1}{2} }{ 11 \frac{1}{3} }\)

=\(\frac{17}{2}\) × \(\frac{3}{34}\)× 100

=3× 25

=75%

Let the income be X.

Then,

\(x\times\frac{80}{100}\times\frac{70}{100}\times\frac{90}{100}\) = 10,080

\( x \times 0.504 = 10080 \)

\( x = \frac{1,0080}{0.504} \)

\( = 20,000 \)