Flextronics – Aptitude Questions & Answers for Placement Tests
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Q.1 I. x2 – 16x + 63 = 0 II. y2 – 25y + 156 = 0
Check Solution
Ans: A
I. x2 – 16x + 63 = 0 => (x – 7)(x – 9) = 0 => x = 7, 9 II. y2 – 25y + 156 = 0 => (y – 12)(y – 13) = 0 => y = 12, 13 Since x can be 7 or 9, and y can be 12 or 13, x < y
Q.2 I. x2 + 10x + 21 = 0 II. y2 + 2y – 15 = 0
Check Solution
Ans: D
I. (x+7)(x+3) = 0 => x = -7, -3 II. (y+5)(y-3) = 0 => y = -5, 3 Comparing the values: -7 < -5, -7 < 3, -3 > -5, -3 < 3. Thus, the relationship cannot be established.
Q.3 3 5 14 41 122 365
Check Solution
Ans: D
The pattern is: (Number * 3) – 1. (3*3)-1=8, (8*3)-1=23, (23*3)-1=68, (68*3)-1=203
Q.4 7 11 19 33 51 83
Check Solution
Ans: D
The pattern is adding consecutive even numbers, starting from 4, to the previous number: 7+4=11, 11+8=19, 19+14=33, 33+18=51, 51+32=83. The number 51 does not fit the pattern, it should have been 33+16=49
Q.5 The figure given below shows a rectangle ABCD with AB = 8 cm and BC = 6 cm. E and F are points on AB and CD respectively such that AE = CF = 2 cm. What is the area of the shaded region (the two triangles) if the shape is folded along EF?
Check Solution
Ans: A
When folded, the shaded region forms a rectangle with length equal to the width of the rectangle and width equal to AE = CF. The area of each triangle is 0.5 * base * height. The height of the rectangle is BC which is 6. Each base = 2. Sum of area is equal to 2(0.5*2*6)= 12
Q.6 What is the area of a rectangular garden? I. The length of the garden is twice its breadth. II. The cost of fencing the garden at Rs. 10 per meter is Rs. 180.
Check Solution
Ans: D
Statement II allows us to calculate the perimeter. Combining the information from Statement I and II allows us to calculate both length and breadth.
Q.7 A cyclist travels a certain distance downstream in 4 hours and returns the same distance upstream in 6 hours. If the speed of the stream is 2 kmph, what is the speed of the cyclist in still water? (in kmph)
Check Solution
Ans: A
Let the speed of the cyclist in still water be ‘x’ kmph. Downstream speed = x + 2 kmph Upstream speed = x – 2 kmph Distance = Speed x Time Distance downstream = (x + 2) * 4 Distance upstream = (x – 2) * 6 Since the distance is the same: (x + 2) * 4 = (x – 2) * 6 4x + 8 = 6x – 12 2x = 20 x = 10
Q.8 The respective ratio of the speeds of a car and a train is 15:29. The train covers 580 km in 4 hours. Find the distance covered by the car in 6 hours.
Check Solution
Ans: A
Train’s speed = 580 km / 4 hours = 145 km/hr. Car’s speed = (15/29) * 145 km/hr = 75 km/hr. Distance covered by car in 6 hours = 75 km/hr * 6 hours = 450 km.
Q.9 A vessel contains 80 liters of a solution of acid and water in the ratio 3:5. If 16 liters of the solution are removed and replaced with 16 liters of water, what is the ratio of acid to water in the resulting solution?
Check Solution
Ans: C
Initially, Acid = (3/8) * 80 = 30 liters and Water = (5/8) * 80 = 50 liters. After removing 16 liters, Acid removed = (3/8) * 16 = 6 liters and Water removed = (5/8) * 16 = 10 liters. Remaining Acid = 30 – 6 = 24 liters and Remaining Water = 50 – 10 = 40 liters. After adding 16 liters of water, new water = 40 + 16 = 56 liters. Hence ratio of Acid : Water = 24:56 = 3:7
Q.10 A contractor estimated he could build a wall in 60 days with 15 workers. After working for 20 days, he found that only 1/3rd of the wall was built. How many additional workers does he need to hire to complete the wall on time?
Check Solution
Ans: D
Remaining work = 2/3. Remaining time = 60 – 20 = 40 days. 15 workers completed 1/3 work in 20 days. So, work done by 15 workers in 1 day = 1/(3*20) = 1/60. For 2/3 work, workers needed in 1 day = (2/3) / (1/60) = 40. Therefore, workers needed = 40 * 1 = 40. Additional workers needed = 40 – 15 = 25.
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