The required value, x = 3 & y = 4.
We know, 1 = $\frac{Prt}{100}$, Where
I = Interest
p = Prinicipal
t = Time
He, according to the question for 1$^{st}$ & 2$^{st}$ part
$\frac{4000\times x\times 1}{100}+\frac{5000\times y\times 1}{100}$ = 320………………(i)
And,$\frac{5000\times x\times 1}{100}+\frac{4000\times y\times 1}{100}$= 310 ………….(ii)
Now, from (i) we have,
$\frac{4000x}{100}+\frac{5000y}{100}$ = 320
=> $\frac{4000x+5000y}{100}$ = 320
=> 4,000x + 5,000y = 320 x 100
=> 1,000(4x + 5y) = 32 X 1,000
=> 4x + 5y = 32
4x + 5y = 32………… (iii)
Again, from (ii) we have,
$\frac{4000x}{100}+{5000y}{100}$=310
=> $\frac{4000x+5000y}{100}$ = 310
=> 5,000x+ 4,000y = 310 x 100
=> 1,000(5x+ 4y) = 31 x 1,000
=> 5x + 4y= 31
5x + 4y= 31…………… (iv)
Now, we have
(iii) $\times$ =>16x+ 20y= 128
(iv) $\times$ =>25x+20y = 155
-9x = -27 [ বিয়োগ করা হল]
=> x = $\frac{-27}{-9}$
x=3
Now, putting x = 3 in (iii)
(4 x 3)+ 5y= 32
=> 12+5y=32
=> 5y=32- 12
=> 5y=20
=> y = $\frac{20}{5}$
y=4
17$\frac{1}{2}$ hours
(p + R + T) take 5 hours to fill up 1 tank
So, (P + R + T ) take 1 hour to fill up $\frac{1}{5}$ part of tank
Again. (P + R) take 7 hours to fill up 1 tank
SO, (P + R) take I hour to fill up $\frac{1}{7}$ part of tank
SO, ln 1 hour, T can fill up =($\frac{1}{5}-\frac{1}{7}$ ) = $\frac{7-5}{35}$ = $\frac{2}{35}$ part of tank
Now, $\frac{2}{35}$ pant of tank is filled by Tin 1 hour
so, 1 or complete part of tank is filled by T in = $\frac{35×1}{2}$ = 17$\frac{1}{2}$ hours
$\frac{41}{44}$
Given that. number of green ball = 5
so,Number of yellow ball = 4 and
so,Number of white ball = 3
so,Total ball=5+4+3= 12
so,Probability that the balls are same colour is
P(3 balls are same colours) = P(3 green) + P(3 yellow) + P (3 white)
= $\frac{5c_{3}}{12c_{3}}+\frac{4c_{3}}{12c_{3}}+\frac{3c_{3}}{12c_{3}}$
= $\frac{10}{220}+\frac{1}{220}+\frac{1}{220}$
= $\frac{10+4+1}{220}$ = $\frac{15}{220}$ = $\frac{3}{44}$
so, P (3 balls are same NOT colours) = 1-$\frac{3}{44}$ = $\frac{44-3}{44}$
= $\frac{41}{44}$
60%
let, total number of cars = x
The number of black cars = $\frac{x}{2}$
The remaining number of cars = x- $\frac{x}{2}$ = $\frac{x}{2}$
As. remaining cars were equally blue, white and red,
so, we will multiply the remaining number of cars us $\frac{x}{2}$ by $\frac{1}{3}$
so,The number of blue car = $\frac{x}{2}x\frac{1}{3}$ = $\frac{x}{6}$
so,The number of white ear = $\frac{x}{2}x\frac{1}{3}$ = $\frac{x}{6}$
so, The number of red ear =$\frac{x}{2}x\frac{1}{3}$ = $\frac{x}{6}$
Now, the number of black cars that are sold = $\frac{x}{2}$ of 70%
so,The number of blue cars that are sold = $\frac{x}{6}$ of 80%
so, The number of white ears that are sold = $\frac{x}{6}$ of 30%
so, The number of red cars that are sold = $\frac{x}{6}$ of 40%
The total number of cars that are sold = $\frac{7x}{20}+\frac{2x}{15}+\frac{x}{20}+\frac{x}{15}=\frac{36x}{60}=\frac{3x}{5}$
so, Percentage of cars that are sold = $(\frac{\frac{3x}{5}}{x}\times 100)\%=(\frac{3x}{5}\times \frac{1}{x}\times 100)\%$ = 60%
Time taken by the two train to cross each other is 3.42 seconds.
Let, length of each train be X meter
Now, we know that time taken by the train to pass a standing man is the same time to pass the length of its own.
Speed of the 1$^{st}$ train, v$_{1}$= $\frac{x}{3}ms^{-1}$
As the two trains are running in opposite direction. so we add their speed and distance
This two tram pass a distance of (x + x) meter = 2x meter with speed of $\frac{x}{3}+\frac{x}{4} ms^{-1}$
Now. we know required time = $\frac{Distance}{Speed}$
=> Time = $\frac{2x}{\frac{x}{3}+\frac{x}{4}}$ =3.42 seconds.
16$\frac{50}{3}$%
Let, the hourly income = 100 Tk.
After salary increase by 20%. his present hourly salary = 100 +20 = 120Tk.
Now,
If present salary is 120 TK then previous salary was = 100 TK
So, If present salary is 100 TK then previous salary was = $\frac{100\times 100}{120}$ = $\frac{250}{3}$ TK
so, He will reduce his working hour by = $(100-\frac{250}{3})$% = $(100-\frac{300-250}{3})$% = $\frac{50}{3}$ =16$\frac{50}{3}$%
