ans: 60 percent of the total caps in the shop is blue.
Given the number of blue cap = 42
The number of red cap = 42 x $\frac{1}{3}$ = 14
& the number ofgreen cap =14 x $\frac{1}{2}$ = 7
As, the green and yellow caps are same of number, so the number of yellow cap = 7.
Total number of cap = 42 +14 + 7 + 7 = 70
The percentage of blue cap =
As, the green and yellow caps are same of number, so the number of yellow cap = 7.
Total number ofcap = 42 +14 + 7 + 7 = 70
The percentage of blue cap = $\frac{42}{70}$ x 100% = 60%
ans: Man’s and Woman’s income are 10,000 Tic & 6,000 Tk respectively.
Let, the amount of income and expenditure of man = 5x and 3x
& the amount of income and expenditure of woman = 3y and y
The man’s savings = 5x – 3x = 2x and
The woman’s savings = 3y – y = 2y
Here, 2x = 2y = 4,000
=> 2x = 4,000
=> 9x = 2,000
=> 2y = 4,000
=> y = 2,000
The man’s income = 5x = (5 x 2,000) Tk. = 10,000 Tk.
The woman’s income = 3y = (3 x 2,000) Tk. = 6,000 Tk.
ans: His overall 15.294% of loss.
At 20% loss, if cost price is Tk. 100, then selling price = Tk. (100 – 20) = Tk. 80
Now, while selling price is Tk. 80, then cost price = Tk. 100
While selling price is Tk. 1, then cost price = $\frac{100}{80}$ Tk.
While selling price is Tk. 1,040, then cost price = $\frac{100×1040}{80}$ = Tk. 1,300
Again, at 10% less, if cost price is Tk. 100, then selling price = Tk. (100 – 10) = Tk. 90
While selling price is Tk. 90, then cost price = Tk. 100
While selling price is Tk. 1, then cost price = Tk. $\frac{100}{90}$
While selling price is Tk. 1040, then cost price = TK. $\frac{100×1040}{90}$ =TK. $\frac{10400}{9}$
Total selling price of two articles = 2 x 1,040 = Tk. 2,080
And Total cost price of two artIcles = Tk. (1300 +$\frac{10400}{9}$)
= Tk. ($\frac{11700+10400}{9}$) = TK. $\frac{22100}{9}$
Overall loss percentage =( $\frac{\frac{22100}{9}-2080)}{\frac{22100}{9}}$ x 100%
= $\frac{\frac{22100-18720}{9}}{\frac{22100}{9}}\times 100$%
= ($\frac{3380}{9}$x$\frac{9}{22100}$)x100%
=15.294%
Question 4:
ans: The Largest integer $\frac{s+10}{5}$
Let, 5 consecutive integers are x, x + l, x + 2, x + 3 and x + 4
According to the question
x+(x+1)+(x+2)+(x+3)+(x+4)=S
=> 5x +10 = S
=> 5x = S -10
=> x = $\frac{s-10}{5}$
The Largest integer =x + 4
= $\frac{s-10}{5}$+4
= $\frac{s-10+20}{5}$
= $\frac{s+10}{5}$
ans: Larger number is 9.
Let, the larger number be x
And the small one is y
According to the question
x – y = 5 …………….(i)
x$^{2}$- y$^{2}$ = 65 ……………..(ii)
Now, from the equation (ii), we have
x$^{2}$- y$^{2}$ = 65
=> (x +y) (x—y) = 65
=> (x +y) (x—y) = 65
=> (x + y) x 5 = 65
x + y = 13……….. (iii)
Adding equation (iii) & (i)
Now, we get
x +y =13
x -y = 5
(+) 2x = 18
x = 9
Let, Robi’s speed was = x mile/hour
According to questions
Distance = 100 mile
Speed = $\frac{Distance}{Time}$
Time = $\frac{Distance}{Speed}$
$T_{1}$ = $\frac{100}{X}$
Again, if he drives 8 miles per hour faster then, new speed = (x + 8) mile/hour
And distance = 100 miles
New time, $T_{1}$ = $\frac{100}{x+8}$
Now, from the question
$T_{1}$ = $\frac{5}{6}$x$T_{1}$
=>$\frac{100}{x+8}$ = $\frac{5}{6}(\frac{100}{x})$
=> $\frac{5}{6x}$ = $\frac{1}{x+8}$
=> 5x + 40 = 6x
=> x = 40
Time taken by the trip = $\frac{100}{40}$ hr$
= $\frac{10}{4} hr$
= 2.5 hr
ans , $\frac{8}{5}$
$\frac{\sqrt{2+x+\sqrt{2-x}}}{\sqrt{2+x-\sqrt{2-x}}}$ = 2
=> $\frac{\sqrt{2+x+\sqrt{2-x+\sqrt{2+x-\sqrt{2-x}}}}}{\sqrt{2+x+\sqrt{2-x-\sqrt{2+x+\sqrt{2-x}}}}}$ = $\frac{2+1}{2-1}$
=> $\frac{2\sqrt{2+x}}{2\sqrt{2-x}} =\frac{3}{1}$
=> $(\frac{\sqrt{2+x}}{\sqrt{2-x}})^{2}=3^{2}$
=> $\frac{2+x}{2-x}=9$
=> 2+x=18-9x
=> 10x=l6
=> x=$\frac{16}{10}$
x=$\frac{8}{5}$
Let, the 2nd number = x
1st number will be $\frac{x}{2}$
And 3rd number will be $\frac{x}{3}$
According to question
=> $\frac{\frac{x}{2}+x+\frac{x}{3}}{3}$ = 44
=> $\frac{x}{2}+x+\frac{x}{3}$ = 44$\times$3
=> $\frac{3x+6x+2x}{6}$ = 132
=> 11x = 132$\times$6
=> x = $\frac{132\times 6}{11}$
x=72
