The number of students who were travelling is 60.
r of students be = x
Then the per student cost was = $\frac{2400}{X}$ Tk.
After joining 10 students, total number of students (x + 10) then the per student cost decreased to $\frac{2400}{x+10}$
According to question,
=> $\frac{300}{x}$-$\frac{300}{x+300}$ =1 Tk.
=> $\frac{300(x+10)-300x}{x(x+10)}$ =1 Tk.
=> 300x + 3,000 — 300x =x$^{2}$+10x
=< x^{2} +10x-3,000 = 0
=> x^{2}2 + 60x-50x-3,000= 0
=> x(x + 60) – 50(x + 60) = 0
=> (x+ 60) (x—50)=0
=> (x — 50) = 0
x = 50 or x + 60 = 0
x = – 60 Which is not acceptable.
The number of students who were travelling is (x + 10) = 50 + 10 = 60.
The ratio of number of boys to the number of girls in the clas is 4:5
Let, the number of boys is = x and the number of girls is = y
According to question,
16.41x + 15.4y= 15.8(x + y)
=> 16.4x+15.4y=15.8x+15.8y I mm: 7115
=> 16.4x -15.8x = 15.8y—15.4y
=> 0.6x = 0.4y
=> 6x = 4y
=> $\frac{x}{y}=\frac{4}{5}$
ans: 322.
x+($\frac{1}{x}$)=3
Now, (x^{6}+$\frac{1}{x^{6}}$= $(x^{2})^{3}+(\frac{1}{x^{2}})^{3}$
= $(x^{2}+\frac{1}{x^{2}})^{3}$-3$\times x^{2}\times \frac{1}{x^{2}}(x^{2}+\frac{1}{x^{2}})$
= $(x^{2}+\frac{1}{x^{2}})^{3}-3(x^{2}+\frac{1}{x^{2}})$
Now, $x^{2}+\frac{1}{x^{2}} =(x+\frac{1}{x})^{2}-2x\frac{1}{x}=3^{2}-2=7$
So, putting the values,
= 7$^{3}$-3×7
=> 343-21 = 322
2,000 Tk.
Let, the cost price be Tk. x
Profit = Selling price – Cost price.
& Loss = Cost price -Selling price.
Now, According to question
=> 1,920 -x = x- 1,280
=> x + x = 1,280 + 1,920
=> 2x= 3,200
=> x = $\frac{3200}{2}$
=> x = 1,600
The cost price of the article is Tk. 1,600.
Now at 25% profit, the selling price will be = (1600 + 1600 x $\frac{25}{100}$ ) Tk.
= (1,600 + 400) Tk. = 2,000 Tk.
A’s share 2,250 Tk.; B’s share 1,500 Tk.; C’s share 750 Tk.
(A + B), s 5 days work= 5($\frac{1}{10}+\frac{1}{15}$)
=$\frac{5}{6}$
Remaining work = 1-$\frac{5}{6}$= $\frac{1}{6}$
C’s 2 day work = $\frac{1}{6}$
18 বর্গএকক ।
চিত্রে যেহেতু, $\angle$AOB = এক সমকোণ
তাই ABO একটি সমকোণী ত্রিভুজ যার AB অতিভুজ ।
পিথাগোরাসের সৃত্রানু
AB$^{2}$= AO$^{2}$+BO$^{2}$
=> AB$^{2}$ = 3$^{2}$+3$^{2}$
=> AB$^{2}$ = 9+9
=> AB$^{2}$ = 18
=> AB = $\sqrt{2\times 9}$
=> AB = 3$\sqrt{2}$
অর্থাৎ বর্গক্ষেত্রের একবাছ্ পাওয়া গেল 3$\sqrt{2}$
অতএব বর্গক্ষেত্রের ক্ষেত্রফল = (বাহু)² বর্গএকক
= $(3\sqrt{2}) ^{2}$বর্গএকক
=18 বর্গএকক
