Practice questions here, for every subject and every exam. Unlimited questions for unlimited attempts, given with answers and explanations.
In an election $15 \%$ voters could not caste their votes $20 \%$ caste votes become invalid. If winner secured $40 \%$ of total votes and he had won by 9,600 votes. Find number of votes which did not caste:
Let
total number of votes $=100$%
According to Question
$\because 12 \%=9,600$ votes
Number of voters who did not participate in voting
$\therefore 15 \%=\frac{9,600}{12} \times 15=12,000$
If $x \%$ of 240 is equal to $60 \%$ of $(x+450)$, then $20 \%$ of $(x+50)$ is what percentage less than $40 \%$ of $x$ ?
$x \%$ of $240$ = 60 $\%$ of $(x+450)$
$\frac{x}{100} \times 240 = \frac{60}{100}(x+450)$
$2.4x = 0.6x+270$
$1.8x = 270$
$x = 150$
So $20\%\times(150+50) = \frac{1}{5}\times 200 = 40$
and $40\%\times 150 = 60$
So $40$ is $20$ less than $60$.
required $\% = \frac{20\times100}{60}\% = 33\frac{1}{3}\%$
Rakesh got $40 \%$ marks in an examination and failed by 30 marks. Aditya got $45\%$ marks and got 40 marks more than the marks required to pass. What is the percentage of marks required to pass?
ATQ,
$\Rightarrow 40 \%+30=45 \%-40$
$\Rightarrow 5 \%=70$
$1 \%=14$
Total marks of exam $=100 \times14=1400$
Passing marks $=40 \times 14+30=590$
$\therefore$ Percentage required to pass the exam = $\frac{590}{1400} \times 100$
$=42.15 \%$
In an election, 60% of voters cast their vote. 25% of the casted votes were invalid. 60% of the valid votes were secured by BJP and he won the Election by 18000 votes. Find the number of voters who could not cast their votes?
Let the number of Votes =100%
Cast vote=60%
Not cast vote=40%
In cast vote two types
First type, Valid=75% $\times{\frac{60}{100}}$=45
Second, Invalid=60 - 45 = 15
BJP votes in % = 60%$\times{\frac{45}{100}}$=27
Looser looses by = 45 - 27 = 18%
Difference = 27%-18%=9%
9%=18000
$40 \%=\frac{18000 \times 40}{9}=80000$
Population of Aligarh is increased by $30 \%$ in 2018. In the next year i.e., $2019,25 \%$ populatiion left from Aligarh and in $2020,20 \%$ of the population die due to Corona. If present population of Aligarh is $3,90,000$, then find initial population.
Let initial population of Aligarh $=x$
$\therefore x \times \frac{130}{100} \times \frac{75}{100} \times \frac{80}{100}=3,90,000$
$x=3,90,000 \times \frac{100}{130} \times \frac{100}{75} \times \frac{100}{80}$$ = 5,00,000$
So, the initial population of Aligarh is 5,00,000.
If 20% of 50% of x is twice the value of 50% of 25% of y, then x is what percent more or less than y ?
$20 \% \times 50 \% \times \mathrm{x}=2 \times \mathrm{y} \times 25 \% \times 50 \%$
$\frac{20}{100} \times \frac{50}{100} \times x=2 \times y \times \frac{25}{100} \times \frac{50}{100}$
$\frac{x}{10}=\frac{y}{4}$
$\frac{x}{y}=\frac{5}{2}$
$\%=\frac{5-2}{2} \times 100=150 \%$
A saves $35 \%$ of his income. If his income increases by $20.1 \%$ and his expenditure increases by $20 \%$, then by what percentage do his savings increase or decrease? (correct to one decimal place)
Let, income = 100, Saving = 35 and Expenditure = 65
New income = 100 + $100\times\frac{20.1}{100}$ = 120.1
New expenditure = 65 + $65\times\frac{20}{100}$ =78
New savings = 120.1 – 78 = 42.1
Increase in saving = 42.1 – 35 = 7.1
Percentage =$\frac{7.1}{35}\times100$= 20.3%
When the price of an item was reduced by $20 \%$, its sale increased by $x \%$. If there is an increase of $25 \%$ in receipt of the revenue, then the value of $x$ is:
Price × Sale = Revenue
10 10 = 100
8 (10+x) = 125
10 + x = $\frac{125}{8}$
10+x = 15.625
x = 5.625
Req % = $\frac{5.625}{10}\times 100$=56.25
40 liters of milk contains $10 \%$ water. How much water should be added to the mixture so that the water in the mixture becomes $20\%$?
$10 \%$ of 40 litres is 4 litres.
To make it $20 \%$ water.
you'll have to add ' $x$ ' litres of water to the mixture.
$\frac{x+4}{x+40}=\frac{20}{100}$
$ x+4=(x+40) 0.2 $
$ x+4=0.2 x+8 $
$ 0.8 x=4 $
$ x=\frac{4}{0.8} $
$x=5$
Rate of rice reduced by $20 \%$, then consumption got increased by $15 \mathrm{~kg}$ to make the expenditure constant. If the expenditure is Rs. 450 , find the initial rate?
$\begin{array}{ll}P & Q \\ \frac{5}{4} & \frac{4}{5}\end{array}$
$5 - 4 = 15 kg$
$1 = 15 kg $
$4= 60 kg$
Initial rate$ = \frac{450}{60}$
$= Rs 7.5 per kg$
In an election A received $11 \%$ of the votes and $B$ received 145680 votes. $69 \%$ of the candidates didn't cast their vote to any of the party. Find the total percentage of votes that $B$ received.
$69 \%$ of the candidates didn't cast their votes
$\%$ of candidates cast their votes $=31 \%$
Out of $31 \%$, A received $11 \%$ votes, then $B$ receives $=20 \%$ votes
Value of $20 \%$ is equal to 145680
$20 \%=145680$
$\Rightarrow 1 \%=7284$
$\Rightarrow 100 \%=728400$
Hence, total % of votes that B received $=\frac{145680}{728400}=0.2 \times 100=20 \%$
$10 \%$ of a number $\mathrm{A}$ is equal to $30 \%$ of a number $\mathrm{B}$. The number $\mathrm{B}$ is equal to $25 \%$ of the third number $\mathrm{C}$. If the value of $\mathrm{C}$ is 6000, then what is the sum of $60 \%$ of $\mathrm{A}$ and $20 \%$ of $\mathrm{B}$?
$A$ of $\frac{10}{100}=B$ of $\frac{30}{100}$
$\frac{A}{B}=\frac{3}{1}$
$C$ of $\frac{25}{100}=B$
$\frac{B}{C}=\frac{1}{4}$
$C=4$
$4= 6000$
$1=1500$
$B=1500$
$A=3\times1500=4500$
Hence,
$\mathrm{A}$ of $60 \%$ + $\mathrm{B}$ of $20 \%$
$\Rightarrow A \times \frac{60}{100}+B \times \frac{20}{100}$
$\Rightarrow 4500\times \frac{60}{100}+1500\times \frac{20}{100}$
$=2700+300$
$=3000$
Aman spends 75% of his income. If his expenditure increases by 40% and savings increase by 30%, then by what percentage will his income increase?
Let Income of Aman is 100.
Income increase by = 37.5%
The reduction of 25% in the price of salt enables a person to buy 5 kg more for ₹ 275 . The reduced price of salt per kg (in ₹) is:
$
25 \%=\frac{1}{4}
$
So, $4\text { unit }= 5 \times 4 =20 \mathrm{~kg}$
Reduced price $=\frac{275}{20}=\frac{55}{4}=$₹ 13.75
Quantity of wheat of A quality $=\frac{40}{100} \times 60=24 \mathrm{~kg}$
Quantity of wheat of B quality $=60-24=36 \mathrm{~kg}$
Let a $\mathrm{kg}$ of wheat of A quality is added.
According to question
$
\begin{aligned}
&(24+36+a) \times \frac{40}{100}=36 \\
&(60+a)=\frac{36 \times 10}{4}=90 \\
&a=90-60=30 \mathrm{~kg}
\end{aligned}
$
As per the question,
$11 \%=3190$
So , $100 \%=\frac{3190}{11} \times 100=29000$
29000 people went to see the matches in 2014.
In an election between two candidates, 8 % of the votes were invalid. The winning candidate got 60 % of the total valid votes and won the election by 5888 votes. How many voters were registered?
Let total no. of votes be 1000 .
Votes which are valid $=1000-1000 \times \frac{8}{100}=920$
Votes got by winning candidates $=920 \times \frac{60}{100}=552$
Votes got by loosing candidates $=920-552=368$
Given, (552 - 368)unit $\rightarrow 5888$
184 unit $\rightarrow 5888$
1 unit $\rightarrow 32$
So, total votes registered $=1000 \times 32=32000$
Class X has 5 sections which have 40,50,50,60 and 60 students, respectively. The pass percent of these sections are 80%, 76%, 64%, 65% and 70%, respectively. The pass per cent of the entire class X (correct to two decimal places) is:
Passed students in section $1=80 \%$ of $40=32$
Passed students in section 2 $=76 \%$ of $50=38$
Passed students in section $3=64 \%$ of $50=32$
Passed students in section $4=65 \%$ of $60=39$
Passed students in section $5=70 \%$ of $60=42$
Total passed students $=183$
Total students $=260$
Pass students in percentage $=\frac{183}{260} \times 100 \%$
$=70.38 \%$
The pass percentage in an examination 35% of the total marks. A student scored 390 marks and was declared a failure with 65 marks. The total marks of the examination must be
$\begin{aligned} \text { Total marks }=& 100 \% \\ \text { Marks scored }=& 390 \\ \text { Passing marks }=35 \% &=390+65 \\ 35 \% &=455 \\ 100 \% &=455 \times \frac{100}{35} \end{aligned}$
Total marks of the examination $=1300$
The boys and girls in a school are in the ratio 7:5. If 60% of the boys and 40% of the girls are aged 14 years and below the age of 14 years and the number of students above the age of 14 years is 870 , then the total number of students in the school is:
Boys: Girls $=7: 5$
Let boys are $=70 {x}$
And girls are $=50 {x}$
$60 \%$ of boys are below 14 years
Above 14 years boys $=40 \%$ of $70 x$$=28 {x}$
Above 14 years girls $=60 \%$ of $50 x$$=30 {x}$
Total students above 14 years $=870$
$28 x+30 x=870$
$58 x=870$
$x=15$
$\begin{aligned} \text { total number of students in the school } &=120 x \\ &=120 \times 15 \\ &=1800 \end{aligned}$