C
Since, numbers are consecutive
So, for the case of calculation,
Let, a = 1
b = 2
c = 3
d = 4
e = 5
Sum of this five numbers= l +2+3 +4+5= 15
After deleting one of the five numbers sum of the remaining number = 80% of 15=15$\times \frac{80}{100}$=12
The sum decreased = 15 – 12 = 3. Which is equal to C.
C was deleted from a. b. c, d & e.
Alternative Solution:
Let, a = 1
b = 2
c = 3
d = 4
e = 5
Sum = a + b + c + d+ e = 5x +10 = 5(x +2) ———— (i)
Since, deleting one number of the above the sum in decreased by 20%
So, the value of the number will be the 20% of the sum of number.
Deleted number = 20% of sum
=$\frac{1}{5}\times 5$(x+2) [20%=$\frac{1}{5}$ and formula ….(i) ]
=x+2
=C
ans: C [ we let, c = x+2]
TK. 9$\frac{1}{11}$
Let, the family use Tk. 100 for sugar.
After increasing 10%, the present price of sugar = (100 + 100 x 10%) = Tk.100
So, they have to reduce = (110 — 100) = Tk. 10 for no change in the expenditure.
When Tk. 100 they reduce = Tk. 10
When Tk. 100 they reduce =$\frac{10\times 100}{110}$= Tk. 9$\frac{1}{11}$
850.
Let, the cost price is x Tk.
At 20% profit, the selling price = x + 20% of x = x+$\frac{x}{5}$ = $\frac{6x}{5}$
When cost price is increased by Tk. 50, then new cost price will be = (x + 50) Tk.
Again, if the selling price is also increased by Tk. 30, new selling price will be = ( $\frac{6x}{5}$+30) Tk
According to question,
( $\frac{6x}{5}$+30) – (x + 50) = (20% – $\frac{10}{3}$ %)(x + 50)
x = 850
Cost price is Tk. 850
Alternative:
Let, the cost price be = x
According to the question,
=>($\frac{x\times \frac{120}{100}+30)-(x+50)}{5(x+50)}$=20%- $\frac{10}{3}$%
=>$\frac{6x+150-5x-250}{5(x+50)}$=$\frac{60-10}{300}$
=>$\frac{x-100}{5(x+50)}=\frac{50}{300}$
=> 6x-600=5x+250
x=850.
