Let Johns initial salary = Rs.100
After decreasing by 50%, Johns salary = Rs. 50 [because it will become half]
After subsequently increasing by 50%, Johns salary
=50×$\dfrac{100+50}{100}$=50×$\dfrac{150}{100}$= Rs.$75$
Loss = $100-75=$ Rs.$25$
Loss percent =$\dfrac{25}{100}$×100=25%
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
$\therefore$ Required percentage =$ \left(\dfrac{11628}{20400} \times 100\right) $% = 57%.
Number of valid votes = 80% of 7500 = 6000.
$\therefore$ Valid votes polled by other candidate = 45% of 6000
=$ \left(\dfrac{45}{100} \times 6000\right) $= 2700.
Number of runs made by running = 110 - $\left(3 \times 4 + 8 \times 6\right)$
= 110 - (60)
= 50.
$\therefore$ Required percentage =$ \left(\dfrac{50}{110} \times 100\right) $% = 45$ \dfrac{5}{11} $%
height in the previous year
$=\dfrac{159 \times 100}{106}$=150 cm
Rebate = 6% of Rs. 6650 = Rs.$ \left(\dfrac{6}{100} \times 6650\right) $= Rs. 399.
Sales tax = 10% of Rs. (6650 - 399) = Rs.$ \left(\dfrac{10}{100} \times 6251\right) $= Rs. 625.10
$\therefore$ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
Let the number of students be $ x $. Then,
Number of students above 8 years of age = (100 - 20)% of $ x $ = 80% of $ x $.
$\therefore$ 80% of $ x $ = 48 +$ \dfrac{2}{3} $of 48
$\Rightarrow \dfrac{80}{100} x $ = 80
$\Rightarrow x $ = 100.
consisting of 40 lines and 32 character oer line then the required no of pages will be how much percentage more than the
previous no of pages ?
E)None of these
45×30×60 = X×40×32
Page =X = $\dfrac{45 \times 30 \times 60}{40 \times 32}$ = 63.28 = 63
63 – 45 = 18
(18/45)×100 = 40%
liquid and 5 parts of the second liquid.The water percentage in the new mixture is
E)None of these
$\dfrac{(20×8)+(35×5)}{(8+5)} = \dfrac{335}{13}$ = 25.76%
the expenditure is
Explanation :
Formula : x*100/x+100
Reduction % = $\dfrac{15 \times 100}{100+15} = \dfrac{1500}{115}$ = 13%