The numbers are 7, 10 & l3.
Let. second term = a
Common difference = d
First term will be = a -d
And third term will be = a + d
According to question, a – d + a + a + d = 30
=> 3a = 30
so, a = $\frac{30}{3}$ = 10
So, second term = 10
so, First term = ( l0 – d )
And third = (10+d)
According to question, (10 – d) $^{2}$ + (10) $^{2}$+ (10 + d)$^{2}$ = 318
=> (10)$^{2}$ – 2 x10 x d +d$^{2}$+100+100+20d+d$^{2}$ = 318
=> 2d$^{2}$+300 = 318
=> 2d$^{2}$ = 318-300
=> 2d$^{2}$ = 18
=> d $^{2}$ = $\frac{18}{2}$ =9
=> d = $\sqrt{9}$
so, d = 3
so,1$^{st}$ term = 10-3=7;2$^{nd}$ term = 10 and 3$^{rd}$ term = 10+3=13
Only 15 people speak in Bangla.
Here, the necessary rule is
Total = English + Bangla – Both + None
=> 50 = 35 + Bangla – 25 + 0
so, Bangla = 40
So total 40 people speak Bangla
Speak only Bangla = 40 – 25 = 15
22$\sqrt{2}$
Let, x be the perpendicular drawn from the vertex of triangle
We know, Sin 45 ° = $\frac{1}{\sqrt{2}}$
=> $\frac{1}{\sqrt{2}}$ = $\frac{x}{10}$
=> x = $\frac{10}{\sqrt{2}}$
Area of triangle = $\frac{1}{2}$ x Base x Height
= $\frac{1}{2}$ x 10 x $\frac{10}{\sqrt{2}}$ = 25$\sqrt{2}$
Question 4:
a=16 & b=27.
a এবং b এর মান কত হলে (64x$^{3}$-9ax$^{3}$+108$^{3}$-b) একটি পুর্ণ ঘন হবে?
We know, (p-q)$^{3}$=p$^{3}$-3p$^{2}$q+3pq$^{2}$-q$^{3}$……..(i)
Now to make the term 64x$^{3}$-9ax$^{3}$+108$^{3}$-b into a perfect cube, we will equate all its term with the equation (i)
so, 1$^{st}$ term, p$^{3}$ = 64x$^{3}$=(4x)$^{3}$ => p=4x
so,3$^{nd}$ term, 3pq$^{2}$ = 108x
=> $3\times 4\times q^{2}\times =3\times 4x\times 3^{2}$ => q$^{2}$= 3$^{3}$ =>q=3
so, 2$^{nd}$ term , 3p$^{2}$q = 9ax$^{2}$
=> $9\times (4x)^{2}\times 3=9ax^{2}$
=> $9\times 16\times x^{2}=9ax^{2}$
so, a= 16
4$^{th}$ trem , q$^{3}$= b => 3$^{3}$= b so, b =27
The price of 1 table is 500 Tk. and l chair is 100 tk
Let. the price of one table is x Tk.
and the price of one chair is y Tk.
According to the question
3x + 5y = 2,000 ………………………………………(i)
5x + 7y = 3,200…………………………………………(ii)
Now, multiplying equation (i) by 5 & (ii) by 3 & subtracting equation (ii) from equation (i)
15x + 25y = 10,000 ………….(iii)
5x + 7y = 9,600 …………………….(iv) [ বিয়োগ করা হল ]
4y = 400
=> y = $\frac{400}{4}$ = 100
Putting the value of y in equation (i) we get
3x + 5y = 2,000
=> 3x + (5 x 100) = 2000
=> 3x + 500 = 2,000
=> 3x = 2,000 – 500
=> 3x = 1,500
so, x = $\frac{1500}{3}$
220 ways.
Here, 3 members is to be included in
Available men = 7 & Available women = 5
so, the combination can be
(i) $^{7}C_{3}x ^{5}C_{0} = \frac{7!}{3!x4!} x \frac{5!}{5!x0!}$ = 35
(ii) $^{7}C_{2}x ^{5}C_{1} = \frac{7!}{2!x5!} x \frac{5!}{1!x4!}$ = 105
(iii) $^{7}C_{1}x ^{5}C_{2} = \frac{7!}{1!x6!} x \frac{5!}{2!x3!}$ = 70
(iv) $^{7}C_{0}x ^{5}C_{3} = \frac{7!}{7!x0!} x \frac{5!}{3!x2!}$ = 10
so, total combination = 35 + 105 + 70 + 10 = 220 ways.
The share of profit of A = 500 Tk. & B = 480 Tk.
Total business time = (8 + 7) months = 15 months.
A’s time weighted investment = =[(3,000 x 8) + (3,000 + 2,500) x 7 ] TK.
B’s time weighted investment = 4,000 X 15 = = 60,000 TK.
So, investment ratio = 62,500 : 60,000 = 25 : 24 [Dividing by 2,500]
New, sum of their investment = 25 + 24 = 49
So A’s profit = (980 $\times \frac{24}{49}$) Tk. = 500 Tk &
B’s profit = (980 $\times \frac{25}{49}$) = 480 Tk
The share of profit of A = 500 Tk. & B = 480 Tk.
$(a+\frac{1}{a})(a+\frac{1}{a}-2)$
Given , a$^{2}$+$\frac{1}{a^{2}}$+2-2a-$\frac{2}{a^{a}}$
= a$^{2}$+$(\frac{1}{a})^{2}+2-2a\frac{2}{a}$
= $= (a+\frac{1}{a})^{2}-2\times a\times \frac{1}{a}+2-2a-\frac{2}{a}$
= $(a+\frac{1}{a})^{2}$-2+2-2a-$\frac{2}{a}$
= $(a+\frac{1}{a})^{2}$-2a-$\frac{2}{a}$
= $(a+\frac{1}{a})$($(a+\frac{1}{a})$-($(a+\frac{1}{a})$
= $(a+\frac{1}{a})(a+\frac{1}{a}-2)$
