BDBL Written Math solution

(1) The profit of a company is given in Taka by P = 3x²-35x+50, where x is the amount in Taka spent on advertising. For what values of x does the company make a profit?

3x² – 35x + 50 > 0
=> 3x² – 30x – 5x + 50 > 0
=> 3x(x-10) – 5(x-10) > 0
=> (x-10)(3x-5) > 0

Here, the Roots of the equation are: 10 and 5/3

Since, the in-equation is Greater than 0; i.e. positive
So, range of x < 5/3; Or x>10

Ans. x < 5/3; Or x>10



——————————–

(2) An amount of Tk. 7200 is spent to cover the floor of a room by carpet. An amount of Tk. 576 would be saved if the breadth were 3 meters less. What is the breadth of the room?

7200 – 576 = 6624

So, L*B / L*(B-3) = 7200 / 6624 = 25/23
=> 25B – 75 = 23B
=> 2B = 75
=> B = 37.5 metres

—————————–

(3) Find the three digit prime number whose sum of the digits is 11 and each digit representing a prime number. Justify your answer.

Sum of digit to be 11 of a 3-digit number and each digit being a prime number, the number is: 227 or 353

In case of 227, sum of the digits is 2+2+7 = 11.
And 2, 2, 7 all the digits are prime.

Similarly, In case of 353, sum of the digits is 3+5+3 = 11.
And 3, 5, 3 all the digits are prime.



—————————-

(4) If a/(q-r) = b/(r-p) = c/(p-q) then show that, a+b+c = pa+qb+rc
Let, a/(q-r) = b/(r-p) = c/(p-q) = k
So, a = k(q-r); b = k(r-p); and c = k(p-q)

Now, L.H.S. => a+b+c = k(q-r)+ k(r-p)+ k(p-q) = k(q-r+r-p+p-q) = k x 0 = 0
And, R.H.S. => pa+qb+rc = p*k(q-r)+ q*k(r-p)+ r*k(p-q) = kpq-kpr+kqr-kpq+kpr-kqr = 0
So, L.H.S. = R.H.S. (Showed)

——————————

(5) Prove that a cyclic parallelogram must be a rectangle.

Opposite angles in a parallelogram are congruent (equal), while opposite angles in a cyclic quadrilateral are supplementary (add up to 180°). Congruent supplementary angles are right angles, so opposite angles in a cyclic parallelogram are right angles. Thus all four angles are right angles, and it’s a rectangle.

Let ABCD be the cyclic parallelogram with angles A and C being opposite angles.
A = C
And, A+C = 180°
since A = C
So, A+A = 180°
2A = 180°
A = 90°
if any one angle of parallelogram is 90°, the parallelogram is a rectangle.

————————-

(6) After traveling 108 km, a cyclist observed that he would have required 3 hrs less if he could have traveled at a speed 3 km/hr more. At what speed did he travel?
S*T = 108
Again, (S+3)(T-3) = 108
=> ST – 3S + 3T – 9 = ST
=> 3T – 3S = 9
=> T – S = 3
=> T = S+3
Now, S*(S+3) = 108
=> S² + 3S – 108 = 0
=> S² + 12S – 9S – 108 = 0
=> S(S+12) – 9(S+12) = 0
=> (S+12)(S-9) = 0
S = 9 (Ans.)

SHORTCUT: 108 = 9*12 = 12*9
D = S*T
So, S = 9 and T = 12

—————————–

(7) Solve:
x/2 + 6/y = 9
x/3 + 2/y = 4

(ii) x 3 – (i)
=> x – x/2 = 3
=> x/2 = 3
=> x = 6
From (i) => 6/2 + 6/y = 9
=> 6/y = 6
Y = 1
Ans. (x, y) = (6, 1)

Be the first to comment

Leave a Reply

Your email address will not be published.


*