**Directions(1-5):** Find the values of x and y and then compare and choose a correct option.

- (i) 𝑥^2 − 20𝑥 + 96 = 0

(ii) 𝑦^2 = 64

y > xy >= xx > y𝑥 >= yNo relation can be established between x and y.Option D

(i) 𝑥^2 − 20𝑥 + 96 = 0

𝑥^2 − 12𝑥 − 8𝑥 + 96 = 0

𝑥(𝑥 − 12) − 8(𝑥 − 12) = 0

(𝑥 − 12)(𝑥 − 8) = 0

𝑥 = 12,8

(ii) 𝑦^2 = 64 𝑦 = ±8

𝑥 ≥ y - (i) x^2 − 11x + 24 = 0

(ii) y^2 − 12y + 27 = 0

y > x𝑥 >= yx > yy >= xNo relation can be established between x and y.Option E

(i) 𝑥^2 − 11𝑥 + 24 = 0

𝑥^2 − 8𝑥 − 3𝑥 + 24 = 0

𝑥(𝑥 − 8) − 3(𝑥 − 8) = 0

(𝑥 − 3)(𝑥 − 8) = 0

𝑥 = 3 , 8

(ii) 𝑦^2 − 12y + 27 = 0

𝑦^2 – 9𝑦 − 3𝑦 + 27 = 0

𝑦(𝑦 − 9) − 3(𝑦 − 9) = 0

(𝑦 − 9)(𝑦 − 3) = 0

𝑦 = 9 , 3

No relation can be established between x and y. - (i) x^2 + 12x + 35 =0

(ii) 5y^2 + 33y + 40 =0

x > yy > xy >= x𝑥 >= yNo relation can be established between x and y.Option C

(i) 𝑥^2 + 12𝑥 + 35 = 0

𝑥^2 + 7𝑥 + 5𝑥 + 35 = 0

𝑥(𝑥 + 7) + 5(𝑥 + 7) = 0

(𝑥 + 7)(𝑥 + 5) = 0

𝑥 = −7 , −5

(ii) 5𝑦^2 + 33y + 40 = 0

5𝑦^2 + 25𝑦 + 8𝑦 + 40 = 0

5𝑦(𝑦 + 5) + 8(𝑦 + 5) = 0

(𝑦 + 5)(5𝑦 + 8) = 0

𝑦 = − 8/5 , −5

𝑦 ≥ 𝑥 - (i) 4x^2 + 9x + 5 =0

(ii) 3y^2 + 5y + 2 =0

x > y𝑥 >= yy > xy >= xNo relation can be established between x and y.Option D

(i) 4𝑥^2 + 9𝑥 + 5 = 0

4𝑥^2 + 4𝑥 + 5𝑥 + 5 = 0

4𝑥(𝑥 + 1) + 5(𝑥 + 1) = 0

(4𝑥 + 5)(𝑥 + 1) = 0

𝑥 = −1 , − 5/4

(ii) 3𝑦^2 + 5y + 2 = 0

3𝑦^2 + 3y + 2y + 2 = 0

3𝑦(𝑦 + 1) + 2(𝑦 + 1) = 0

(3𝑦 + 2)(𝑦 + 1) = 0

𝑦 = − 2/3 , −1

𝑦 ≥ x - (i) 4𝑥^2 − 21𝑥 + 20 = 0

(ii) 3y^2 − 19y + 30 = 0

y > xx > yy >= x𝑥 >= yNo relation can be established between x and y.Option E

(i) 4𝑥^2 − 21𝑥 + 20 = 0

4𝑥^2 − 16𝑥 − 5𝑥 + 20 = 0

4𝑥(𝑥 − 4) − 5(𝑥 − 4) = 0

(4𝑥 − 5)(𝑥 − 4) = 0

𝑥 = 5/4 , 4

(ii) 3𝑦^2 − 19𝑦 + 30 = 0

3𝑦^2 – 9𝑦 − 10𝑦 + 30 = 0

3𝑦(𝑦 − 3) − 10(𝑦 − 3) = 0

(3𝑦 − 10)(𝑦 − 3) = 0

𝑦 = 10/3 , 3

No relation can be established between x and y. - In a vessel, there are two types of liquids A and B in the ratio of 5 : 9. 28 lit of the mixture is taken out and 2 lit of type B liquid is poured into it, the new ratio(A:B) thus formed is 1 : 2. Find the initial quantity of mixture in the vessel?
40 litres48 litres44 litres56 litres50 litresOption D

Let the initial quantity of mixture in vessel be x litres.

[𝑥× 5/14 −10]/[𝑥× 9/14 −18+2] = 1/2

⇒ [5𝑥−140]/[9𝑥−224] = 1/2

⇒ 10x – 280 = 9x – 224

⇒ x = 56 litres - The average weight of 5 students in a class is 25.8 kg. When a new student joined them, the average weight is increased by 3.9 kg. Then find the approximate weight of the new student.
40 kg45 kg38 kg30 kg49 kgOption E

Weight of new student = 6 × (25.8 + 3.9) – 5 × 25.8 = 49 kg - The difference between downstream speed and upstream speed of boat is 6 km/hr and boat travels 72 km from P to Q (downstream) in 4 hours. Then find the speed of boat in still water?
13 km/hr.17 km/hr.15 km/hr.12 km/hr.10 km/hr.Option C

Let the speed of boat in still water be x km/hr and that of stream be y km/hr.

(x + y) – (x – y) = 6

⇒ 2y = 6

⇒ y = 3 km/hr.

Downstream stream = (x + y) = 72 4 = 18 km/hr.

⇒ x = 15 km/hr. - A shopkeeper marked his article 50% above the cost price and gives a discount of 20% on it. If he had marked his article 75% above the cost price and gives a discount of 20% on it then find the earlier profit is what percent of the profit earned latter?
40%50%30%10%20%Option B

Let the CP be Rs. 100x.

Then, MP = Rs. 150x

SP = 150x × 80/100 = Rs. 120x

Profits = Rs. 20x

New MP = Rs. 175x

New SP = 175x × 80/100 = Rs. 140x

New Profit = Rs. 40x

Required % = 20𝑥/40𝑥 × 100 = 50% - A train of some length passes the platform of length 524 m in 55 seconds. Find the length of train if the speed of train is 72 km/hr.
522 m550 m576 m510 m576 mOption E

Speed of train in m/s. = 72 × 5/18 = 20 m/s

Let length of train be x m.

(524 + 𝑥)/55 = 20

x = 1100 – 524 = 576 m