ans: The number of students in room A = 100.
Let, the number of students in rooms A and B be x and y respectively.
Then, x-10 = y+ 10
=> x-y= 20 …………….(i)
and x+20=2(y—20)
=> x- 2y = – 6O……………… (ii)
Solving (i) and (ii) we get: x = 100,y = 80.
ans: 8.
Let, Tenth position digit =X and Unit position digit = y
Number = 10x + y
According to the questions
(10x +y) – (10y + x) = 36
=> 10x+y-10y-x =36
=> 9x – 9y = 36
x-y= 4……………….. (i)
Again, Since the number is greater than the number obtained on reversing the digits, So the ten’s digit is greater than the unit’s digit.
y : x = 1 : 2
=> x : y = 2 : 1
=> $\frac{X}{Y}$ = 2
=> x = 2y
x – 2y = 0 …………….(ii)
Solving (i) and (ii) we get: x = 8, y = 4
Sum of the digit 8 + 4 = 12
Difference of the digit 8 – 4 = 4
Difference = 8
Alternative: Since the number is greater than the number obtained on reversing the digits, So the ten‘s digit is greater than the unit’s digit. Let ten‘s and unit’s digits be 2x and x respectively.
Then, according to the question
(10 x 2x + x)-(10x + 2x)=36
=> 9x = 36
x = 4.
Required difference = (2x + x) – (2x – x) = 2x = 8.
ans: 17
Let the cost price of the manufacturer x Tk
After 18% profit sells price = x + x ($\frac{18}{100}$) = $\frac{118x}{100}$
After adding 20% profit on it new sells price = $\frac{118x}{100}$ +$\frac{118x}{100}$+ $\frac{20}{100}$
= $\frac{118x}{100}$ (1+$\frac{1}{5}$
= $\frac{118x}{100}$($\frac{6}{5}$)
= 118x (g) = $\frac{708x}{500}$
There by earning profit 25% on It new sells price = $\frac{708x}{500}$(1+ $\frac{1}{4}$)
= $\frac{708x}{500}$($\frac{5}{4}$) = $\frac{177x}{100}$
Here according to the questions, $\frac{177x}{100}$ = 30.09
=> 177x = 3009
=> x = $\frac{3009}{177}$
x = 17
