(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

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(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

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(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.

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(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)

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(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.

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(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

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(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)