Let maximum marks of the examination $=x$
Marks that Arun got $=30%$ of $x$ $=\dfrac{30x}{100}$
Given that Arun failed by 10 marks.
=> pass mark = $\dfrac{30x}{100}+10~~\cdots(1)$
Marks that Sujith got $=40%$ of $x$ $=\dfrac{40x}{100}$
Given that Sujith got 15 marks more than the passing marks.
=> pass mark $=\dfrac{40x}{100}-15~~\cdots(2)$
From $(1)$ and $(2)$,
$\dfrac{30x}{100}+10=\dfrac{40x}{100}-15\dfrac{10x}{100}=25\dfrac{x}{10}=25x=10×25=250$
pass mark
$=\dfrac{30x}{100}+10$
=$\dfrac{30×250}{100}+10$
=75+10=85
Equal number of candidates appeared in each state.
In state A, 6% candidates got selected.
In state B, 7% candidates got selected.
Given that 80 more candidates got selected in state B than A.
Therefore, 1% of candidates appeared in each state =80
=> 100% of candidates appeared in each state =80×100=8000
i.e., number of candidates appeared from each state = 8000
Let total number of men = 100
Then,
20 men play football.
80 men are less than or equal to 50 years old.
Remaining 20 men are above 50 years old.
Number of football players above 50 years old
=20×$\dfrac{20}{100}$=4
Number of football players less than or equal to 50 years old
$=20-4=16$
Required percentage$=\dfrac{16}{20}×100=80%$Price of the car = Rs.3,25,000
Car was insured to 85% of its price
Insured price $=325000×\dfrac{85}{100}$
Insurance company paid 90% of the insurance.
Amount paid by insurance company
=325000×$\dfrac{85}{100}$×$\dfrac{90}{100}$=325×85×9=248625
Difference between the price of the car and the amount received
=$325000-248625$
=$ Rs.76375$
p>5% of A + 4% of B =$ \dfrac{2}{3} $ (6% of A + 8% of B)
$\Rightarrow \dfrac{5}{100} $ A +$ \dfrac{4}{100} $ B=$ \dfrac{2}{3} \left(\dfrac{6}{100} A +\dfrac{8}{100} B\right) $
$\Rightarrow \dfrac{1}{20} $ A +$ \dfrac{1}{25} $ B=$ \dfrac{1}{25} $ A +$ \dfrac{4}{75} $ B
$\Rightarrow \left(\dfrac{1}{20} -\dfrac{1}{25} \right) $ A = $ \left(\dfrac{4}{75} -\dfrac{1}{25} \right) $ B
$\Rightarrow \dfrac{1}{100} $ A =$ \dfrac{1}{75} $ B
$ \dfrac{A}{B} $=$ \dfrac{100}{75} $=$ \dfrac{4}{3} $.
$\therefore$ Required ratio = 4 : 3
Given that $\dfrac{Q}{P}=\dfrac{P}{P+Q}\cdots(1)$
Since Q can be written as a certain percentage of P, we can assume that $Q=kP$
Hence $(1)$ becomes
$\dfrac{kP}{P}=\dfrac{P}{P + kP}$
$\Rightarrow k=\dfrac{1}{1+k}\\\Rightarrow k(k+1)=1~~\cdots(2)$
Q as a percentage of P
$=\dfrac{Q}{P}×100\\=\dfrac{kP}{P}×100=100k\%~~\cdots(3)$
From here we have two approaches.
Approach 1 - trial and error method
Here we use the values given in the choices to find out the answer.
Take 50% from the given choice.
If 50% is the answer,
$100k=50 \Rightarrow k=\dfrac{50}{100}=\dfrac{1}{2}$
But if we substitute $k=\dfrac{1}{2}$ in $(2),$
$k(k+1)=\dfrac{1}{2}\left(\dfrac{1}{2}+ 1\right)$ $=\dfrac{1}{2}×\dfrac{3}{2}=\dfrac{3}{4} \neq 1$
Now Take another choice, say 62%
If 62% is the answer,
$100k=62 \Rightarrow k=\dfrac{62}{100}$
If we substitute $k=\dfrac{62}{100}$ in $(2),$
$k(k+1)=\dfrac{62}{100}\left(\dfrac{62}{100}+ 1\right)$ $= \dfrac{62}{100}×\dfrac{162}{100}=\dfrac{10044}{10000}\approx 1$
Hence 62% is the answer.
Approach 2
lets solve the quadratic equation $(2)$
$k(k+1)=1\\k^2+k-1=0\\k=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\dfrac{-1 \pm \sqrt{1^2-\left[4×1×(-1)\right]}}{2×1}=\dfrac{-1 \pm \sqrt{5}}{2}\\= \dfrac{-1 \pm 2.24}{2}\\ = \dfrac{1.24}{2}\text{ or }\dfrac{-3.24}{2}\\= 0.62$
avoiding -ve value
From $(3)$, Q as a percentage of P
$=100k\%=(100×.62)\%=62\%$
Assume that initial price of 1 litre petrol = Rs.100,
Benson spends Rs.100 for petrol such that he buys 1 litre of petrol.
After the increase by 25%, price of 1 litre petrol
$=100+25=125$
Since Benson spends additional 15% on petrol, amount spent by him
$=100+15=115$
Hence quantity of petrol that he can purchase
$=\dfrac{115}{125} litre$
Quantity of petrol reduced
$=\left(1-\dfrac{115}{125}\right)$=$\dfrac{10}{125}$ litre
Required reduction percent
=$\dfrac{\left(\dfrac{10}{125}\right)}{1}$×100=$\dfrac{10}{125}$×100=$\dfrac{10}{5}$×4=2×4=8%Let their marks be $\left( x + 9\right)$ and $ x $.
Then, $ x $ + 9 =$ \dfrac{56}{100}$ $\left( x + 9 + x \right)$
$\Rightarrow$ 25$\left( x + 9\right)$ = 14$\left(2 x + 9\right)$
$\Rightarrow$ 3$ x $ = 99
$\Rightarrow x $ = 33
So, their marks are 42 and 33.
with an increase of 10 percent. B weight is increased by what percent.
Weight of A and B are 3x and 5x.
Initial weight before increase = $\dfrac{132 \times 100)}{110}$ = 120
8x = 120. X = 15
Initial weight of A and B are 45 and 75 kg respectively.
New weight of A = 54 so weight of B = 132 – 54 = 78.
So % increase = $[\dfrac{78-75}{75}] \times 100$ = 4 %
neither English nor Mathematics?
n(M or E) = n(M) + n(E) – n(M and E)
n(M or E) = 40+50-10 = 80
so % of the students who like neither English
nor Mathematics = 100 – 80 = 20%