Quant Test for IBPS PO Prelims Exam set – 5

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

1. (i) 𝑥^2 − 20𝑥 + 96 = 0
   (ii) 𝑦^2 = 64
   
   y > x
   y >= x
   x > y
   𝑥 >= y
   No 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
   
   
    



2. (i) x^2 − 11x + 24 = 0
   (ii) y^2 − 12y + 27 = 0
   
   y > x
   𝑥 >= y
   x > y
   y >= x
   No 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.
   
   
    



3. (i) x^2 + 12x + 35 =0
   (ii) 5y^2 + 33y + 40 =0
   
   x > y
   y > x
   y >= x
   𝑥 >= y
   No 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
   
   
    



4. (i) 4x^2 + 9x + 5 =0
   (ii) 3y^2 + 5y + 2 =0
   
   x > y
   𝑥 >= y
   y > x
   y >= x
   No 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
   
   
    



5. (i) 4𝑥^2 − 21𝑥 + 20 = 0
   (ii) 3y^2 − 19y + 30 = 0
   
   y > x
   x > y
   y >= x
   𝑥 >= y
   No 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.
   
   
    



6. 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 litres
   48 litres
   44 litres
   56 litres
   50 litres
   Option 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
   
   
    



7. 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 kg
   45 kg
   38 kg
   30 kg
   49 kg
   Option E
   Weight of new student = 6 × (25.8 + 3.9) – 5 × 25.8 = 49 kg
   
    



8. 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.
   
   
    



9. 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%
   
   
    



10. 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 m
    550 m
    576 m
    510 m
    576 m
    Option 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