TK 200
1n 6% profit, the selling price = 100 + 6 = Tk. 106
& 4% less, the cost of the item = 100 — 4 = Tk. 96
Again, in 12.5% profit the selling price = (96 + $\frac{96 \cdot 12.5}{100}$)
= Tk. 108
Difference between selling price = 108 — 106 = Tk. 2
1n difference Tk. 2 the cost = Tk. 100
In difference Tk. 4 the cost = Tk. $\frac{100 \cdot 4}{2}$
= Tk. 200
18 Kgs Apple
7 kgs mango = 10 kgs apple + 1 kg orange ………..(i)
7 kgs orange = 1 kgs mango + 2 kgs apple ………….(ii)
Or, 1 kg mango = 7 kgs orange – 2 kgs apple ………….(iii)
Now from (i) x 7 — (iii) we get,
49 kgs mango = 70 kgs apple + 7 kgs orange
1 kg mango = 7 kgs orange — 2 kgs apple
48 kgs mango = 72 kgs apple
12 kgs mango = $\frac{72 x 12}{48}$ kgs apple
= 18 kgs apple
3 pens
Let, he sold x Number of pens at profit
(22 – x) Number of pens at loss
So, 1 unit makes profit = Tk. 35
x unit makes profit = Tk. 35 x X = 35x
1 unit makes loss = Tk. 10
(22 – x) unit makes loss = Tk. 10 (22 — x)
According to the question,
35x — 10 (22 — x) = 635
=> 35x —220 + 10x = 635
=> 45x = 855
So, x = 19
He sold at loss = (22 — 19) = 3 pens
TK 250
Arif takes 4 days for 1 (full) work
Arif takes 1 day for $\frac{1}{4}$(full) work
Afzal takes 5 days for 1 (full) work
Afzal takes 1 days for $\frac{1}{5}$(full) work
Arif & Afzal 1 days work = $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{9}{20}$
So, they can do $\frac{9}{20}$work in 1 day
They can do 1 (full) work in $\frac{20}{9}$ day
Arif takes 1 day for = $\frac{1}{4}$work
Arif takes $\frac{20}{9}$ day for = $\frac{20}{9 x 4}$ work = $\frac{5}{9}$ work
Afzal done = 1 – $\frac{5}{9}$= $\frac{4}{9}$work
So, Arif : Afjsl = $\frac{5}{9}$ : $\frac{4}{9}$
Sum of this two ratio = (5 + 4) = 9
Arif will get = 450 of $\frac{5}{9}$= Tk. 250
4 non-smokers
Total peOpIe = 40
Female = 24
Male= 40—24 =16
Total smokers = 40 + 2 = 20
Female smokers = 24 of $\frac{1}{3}$ = 8
Male smokers = 20 — 8 = 12
Number of male nonesmokers = (16 – 12) = 4
6 and 2
Speed of boat = x; Speed of current = y
While traveling against current, it can sail (x — y)km/hr & while sailing with. the current, it can
travel (x + y) km/hr.
8(x —y) = 32……….. (i)
4(x +y) = 32………… (ii)
From equation no(i) 8(x —y) = 32
=> x—y=4
From equation no(ii) 4(x +y) = 32
=> x+y=8
Adding and subtracting (i) & (ii)
x = 6
y = 2
The original price of per meter cloth is Tk. 50 and current price is Tk. 40
At 20% reduction, When past price Tk. 100 then present price = (100 – 20) = Tk. 80
When past price Tk. 800 then present price = Tk. $\frac{80 \cdot 800}{100}$
= Tk. 640
The present price of 4 meters cloth = 800 — 640 = Tk. 160
The current price of 4 meters cloth = Tk. 160
The current price of 1 meters cloth = Tk. $\frac{160}{4}$
= Tk. 40
If present price Tk. 80 then past price = Tk. 100
If present price Tk. 40 then past price = Tk. $\frac{100 \cdot 40}{80}$ = Tk. 50
Number of row is 70
Let, each row contains x No. of parking spaces.
The No. of row =$\frac{70}{x}$
After expansion, each row contains (x — 2) No. of parking Spaces.
Afier expansron, each No. of row = $\frac{70}{x-2}$
According to the question,
$\frac{70}{x}$ + 4 = $\frac{70}{x-2}$
=> 70x+4x² – 140 —8x=70x
=> 4x²-8x—140= 0
=> x² —2x — 35 = o [Dividing by 4]
=> x(x—7)+5(x—7)= o
=> (X+5)(x—7)=o
So, x=—5 or x=7
But x will never be negative, so x = 7
