ans: The required value, x = 3 & y = 4.
We know, I = $\frac{prt}{100}$, Where
I = Interest
p = Prinicipal
t = Time
Here, according to the question for 1$^{st}$ & 2$^{nd}$ part
$\frac{4000x \times 1}{100}+\frac{5000y \times 1}{100}=320$——–(1)
$\frac{5000x \times 1}{100}+\frac{4000y \times 1}{100}=310$——–(1)
From (1) 4000x+5000y = 32000
=> 1,000(4x + 5y) = 32 x 1,000
=> 4x + 5y = 32
4x + 5y = 32………………………………………. (iii)
Again, from (ii) we have,
=> $\frac{5000x}{100}$$\frac{4000y}{100}$= 310
=> $\frac{5000x+4000x}{100}$= 310
=> 5,000+ 4,000y = 310 x 100
=> 1,000(5x + 4y) = 31 x 1,000
=> 5x + 4y = 31
5x + 4y = 31……………………….. (iv)
Now, we have
(iii)) x4 => 16x+20y=128
(iv) x5 => 25x+20y=155
-9x = – 27
=> x = $\frac{-27}{-9}$
x=3
Now, puttingx= 3 in (iii)
(4x 3)+5y=32
=> 12+5y=32
=> 5y=32—12
=> 5y=20
=> y =$\frac{20}{5}$
y=4
The distance between Mr. Zaman’s office from his home is 6km.
Let, the distance between Mr. Zarnan’s office from his home is x km.
We know, 60 minutes = 1 hour
Here, 7 min = $\frac{7}{60}$
And, 5 min = $\frac{5}{60}$
We know, speed = $\frac{Distance}{Time}$
Time = $\frac{Distance}{speed}$
At 5 miles per hour, we have real/actual/required time from office to home = $\frac{x}{5}$-$\frac{7}{60}$hour
And, At 6 miles per hour, we have real/actual/required time from office to home = $\frac{x}{6}$-$\frac{5}{60}$hour
According to the question
$\frac{x}{5}-\frac{7}{60} = \frac{x}{6}+\frac{x5}{60}$
=> $\frac{x}{5}-\frac{7}{6}=\frac{5}{60}+\frac{7}{60}$
=> $\frac{6x-5x}{30}=\frac{ 5+7}{60}$
=> $\frac{x}{30}$=$\frac{12}{60}$
=> x =$\frac{12}{60}$x30
x=6
The required value, x$^{4}$+ $\frac{1}{x^{4}}$= 322.
Given that, x = $\sqrt{5}-\sqrt{4}$……………………….. (i)
Now, Reversing Eqution …………………………..(i)
$\frac{1}{x}$= $\frac{1}{\sqrt{5-\sqrt{4}}}$
=> $\frac{1}{x}$ = $\frac{1(\sqrt{5+\sqrt{4}})}{(\sqrt{5-\sqrt{4}})(\sqrt{5+\sqrt{4}})}$
=> $\frac{1}{x}$ = $\frac{\sqrt{5+\sqrt{4}}}{(\sqrt{5})^{2}-(\sqrt{4})^{2}}$
=> $\frac{1}{x}$ = $\frac{\sqrt{5+\sqrt{4}}}{5-4}$
=> $\frac{1}{x}$ = $\sqrt{5}+\sqrt{4}$…………………..(ii)
Now, {(i) + (17)} we have,
$x+\frac{1}{x}=(\sqrt{5}-\sqrt{4})+(\sqrt{5}+\sqrt{4})$
=> $x+\frac{1}{x} = \sqrt{5}-\sqrt{4}+\sqrt{5}+\sqrt{4}$
=> $x+\frac{1}{x} = 2\sqrt{5}$……….(iii)
Now, $x^{4}+\frac{1}{x^{4}}= (x^{2}+\frac{1}{x^{2}})^{2}-2x^{2}\frac{1}{x^{2}}$
= $((x+\frac{1}{x})^{2}-2x\frac{1}{x})^{2}-2$
=$((2\sqrt{5})^{2}-2)^{2}-2$
=322
The required solution, x = -$\frac{3}{5}$
$\frac{3}{x+1}+\frac{6}{2x+1}=\frac{18}{3x+1}$
=>$\frac{3}{x+1}+\frac{6}{2x+1}=\frac{9+9}{3x+1}$
=>$\frac{3}{x+1}+\frac{6}{2x+1}=\frac{9}{3x+1}+\frac{9}{3x+1}$
=>$\frac{3}{x+1}-\frac{9}{3x+1}=\frac{9}{3x+1}-\frac{6}{2x+1}$
=>$\frac{9x+3-9x-9}{(x+1)(3x+1)}=\frac{18x+9-18x-6}{(3x+1)(2x+1)}$
=>$\frac{-6}{(x+1)(3x+1)}=\frac{3}{(3x+1)(2x+1)}$
=>$\frac{2}{(x+1)(3x+1)}=\frac{-1}{(3x+1)(2x+1)}$
=>$\frac{2}{(x+1)(3x+1)}+\frac{1}{(3x+1)(2x+1)}=0$
=>$(\frac{1}{3x+1})[\frac{2}{(x+1)}+\frac{1}{(2x+1)}]=0$
Here, we cant the value of x using $\frac{1}{3x+1}=0$
So, =>$\frac{4x+2+x+1}{(x+1)(2x+1)}=0$
=>$\frac{5x+3}{(x+1)(2x+1)}=0$
=> 5x+3=0
=> x= $\frac{-3}{5}$
