20%
Given, total allotted time = 3 hours = 3×60 = 180 minutes
so, Remaining time = 180-30 = 150 minutes
so, Remaining papers = 35-5 = 30 minutes
per hour speed of grading first 5 papers = $\frac{60\times 5}{30}$ = 10 papers
& to complete the work in lime. required per hour speed of grading 30 papers = $\frac{30\times 60}{150}$ = 12
So. she should enhance the work Speed by = $\frac{12-10}{10}$x100 = 20%
24
Let. Tenth digit is x and unit digit is y
The number = (1Ox + y)
According to first condition, 4(x + y) = 10x + y ………(i)
And second condition, (10x + y) + 18 = 10y + x………………… (ii)
From equation (ii), 10x+4+y+18 = 10y +x
=> 9x-9y = -18 . .
=> 9y – 9x = 18
=> 9(y-x) = 18
= (y – x) = $\frac{18}{9}$ = 2
=> y – x = 2
so, y = x + 2 ……….(iii)
From equation (i) we get 4(x + y) = 10x + y
=> 4x + 4y = 10x + y
=> 3y =6x
=> y = 2
so, y = x+2………….(III)
Rrom eduatine (i), we get 4(x+y) = 10x+y
=> 4x+4y = 10x+y
=> 3y = 6x
=> y = 2y
=> x+2 = 2x [ putting the value of y]
so, x=2
Now, putting the valu of x in equation (iii) we get
y= x+2
=> y = 2+2 =4
so, y=4
The number is |(10x+y) = (10×2)+4 =20+4 =24
Question 3:
(x,y) = (1,-6) or $[\frac{7}{\sqrt{2}},\frac{5}{\sqrt{2}}]$
Given, x$^{2}$-yx = 7………(I)
=> xy = x$^{2}$-7
=> y = $\frac{x^{2}-7}{x}$
=> y = x-$\frac{7}{x}$……….(ii)
Another equation is y$^{2}$ +xy = 30 ………..(iii)
Adding equation (i) and (iii) we get
x$^{2}$-xy = 7
x$^{2}$-xy = 30
x$^{2}$+y$^{2}$ = 37………………(iv)
Now, putting the valu of y in equation (iv) we get
x$^{2}+(x-\frac{7}{x})^{2}$ = 37
=> $x^{2}+x^{2}-2\times x\times\frac{7}{x}+(\frac{7}{x})^{2}$ = 37
=> $2x^{2}-14+\frac{49}{x^{2}} = 37$
=> $2x^{2}+\frac{49}{x^{2}}$ = 51
=> $\frac{2x^{4}+49}{x^{2}}$ = 51
=> 2x$^{4}$-51x$^{2}$+49= 0
=> 2$x^{4}-2x^{2}-49x^{2}+49$ = 0
=> $2x^{2}(x^{2}-1)-49(x^{2}-1)$ = 0
=> $(x^{2}-1)(2x^{2}-49)$ = 0
Here $x^{2}-1$= 0
=> x$^{2}$ = 1
=> x = $\sqrt{2}$ = 1
=> x=1
and 2x$\sqrt{2}$-49 = 0
2x$^{2}$= 49
=> x$^{2}$ = $\frac{49}{2}$
so, x= $\sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}$
Thus, we get x = 1,$\frac{7}{\sqrt{2}}$
When x = 1,then y = 1-$\frac{7}{1}$ = 1-7 = -6
and when x =$\frac{7}{\sqrt{2}}-\frac{7}{\frac{7}{\sqrt{2}}}=\frac{7}{1}\times \frac{\sqrt{2}}{7}$= $\frac{7}{\sqrt{2}}-\sqrt{2}=\frac{7-(\sqrt{2})^{2}}{\sqrt{2}}=\frac{7-2}{\sqrt{2}}= \frac{5}{\sqrt{2}}$
so, (x,y) = (1,-6) or $[\frac{7}{\sqrt{2}},\frac{5}{\sqrt{2}}]$
He will invest 4.000 Tk. in company ‘M’ and (9,000 – 4.000) = 5.000 Tk. in company ‘N
Let. He invested Tk. x in company ‘M’ at 10%
In ‘N’ Company, he invested tk. (9.000 -x) at 9%
We kaow, I = $\frac{P\times r\times n}{100}$
The amount of interest at 10% of Tk. x =$\frac{x\times 10\times 1}{10}$=$\frac{x}{10}$ tk
And the amount of interest at 9% of Tk. (9.000 -x)=$\frac{9(9000-x)}{100}$
According to question. $\frac{x}{10}$+$\frac{9(9000-x)}{100}$= 850
=> $\frac{10x+81000-9x}{100}$ = 850
=> x+81,000 = 85,000
=> x = 85,000-81,000
so, x = 4,000
He will invest 4.000 Tk. in company ‘M’ and (9,000 – 4.000) = 5.000 Tk. in company ‘N
| 6 male members | 5 female members |
| 1 | 4 |
| 2 | 3 |
| 3 | 2 |
| 4 | 1 |
$6_{c_{1}}\times 5_{c_{4}}=6\times 5$ = 30
$6_{c_{2}}\times 5_{c_{3}}=15\times10$ = 150
$6_{c_{3}}\times 5_{c_{2}}=20\times10$= 200
$6_{c_{4}}\times 5_{c_{1}}=15\times 75$ = 30
Total number of ways to form the committee is = 30+150+200+75=455
$(24-16\sqrt{2})\pi $ square cm
Depict the following figure according to question
Area of ABC triangle = $\frac{1}{2}\times base\times height$
= $\frac{1}{2}\times4\times4$ = $8cm^{2}$
now, $AC^{2}$= $(AB)^{2}+(BC)^{2}$ = $4^{2}+4^{2}$ =16+16 = 32
AC = $\sqrt{32}$
The radius of a circle inscribed in a triangle =$\frac{2\times \text{area of triangle}}{\text{perimeter of triangle}}$ = $\frac{2\times 8}{(4+4)+\sqrt{32}}=\frac{16}{8+(2\times 4^{2})}$
=$\frac{16}{8+4\sqrt{2}}=\frac{16}{4(2+\sqrt{2})}=\frac{4}{2+\sqrt{2}}=\frac{2\times \sqrt{2\times \sqrt{2}}}{\sqrt{2}(\sqrt{2}+1)}=\frac{2\sqrt{2}}{\sqrt{2}+1}$
= $\frac{2\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2-1})(\sqrt{2}-1)}=\frac{2\sqrt{2}(\sqrt{2}-1)}{2-1}=2\sqrt{2(\sqrt{2}-1)}$
so radus = $2\sqrt{2}(\sqrt{2}-1)$cm
so, area of circle = $\pi R^{2 }= {[2\sqrt{2}(\sqrt{2}-1)]}^{2}$
= $\pi 4\times 2(\sqrt{2}-1)^{2}cm^{2}$
= $8\pi(\sqrt{2}-1)^{2}cm^{2}=(24-16\sqrt{2})\pi \text{square cm}$
6 men
2 men and 5 women require 3 days to complete $\frac{1}{4}$ Job
They complete $\frac{1}{4\times 3}$ =$\frac{1}{12}$ job in 1 day
3 men + 5 women require 2 days to complete $\frac{1}{12}$ job
They complete $\frac{1}{4\times 3}$ = $\frac{1}{8}$ in 1 days
3men + 5 women = $\frac{1}{8}$
2 men + 5 women = $\frac{1}{12}$
(-) 1 man = $\frac{1}{24}$
$\frac{1}{24}$ job is done in 1 day by 1 men
1 job is done in 1 day by 24 men
1 job is done in 4 day by $\frac{1}{24}$ = 6 men.
