# Sample Aptitude Questions of ABB

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1. The total cost of 8 buckets and 5 mugs is Rs. 92 and the total cost of 5 buckets and 8 mugs is Rs. 77. Find the cost of 2 mugs and 3 buckets.
1. Rs. 35
2. Rs. 70
3. Rs. 30
4. Rs. 38
CP of 1 bucket = Rs. X
CP of 1 mug = Rs. Y
∴ 8x + 5y = 92………….. (i)
5x + 8y = 77 …………….(ii)
By equation (i) × 5 – equation (ii) × 8.
40x + 25y – 40x – 64y
= 460 – 616 ⇒ − 39y = - 156⇒ y = 4
From equation (i),
8x + 20 = 92 ⇒8x = 92 – 20 = 72 ⇒ x = 9
∴ CP of 2 mugs and 3 buckets
= 2 × 4 + 3 × 9 = 8 + 27 = Rs. 35
1. If a/(1-a) + b/(1-b) + c/(1-c) = 1 thenthe value of 1/(1-a) + 1/(1-b) + 1/(1-c) is
1. 1
2. 3
3. 4
4. 0
2. If (x – 3)2 + (y – 5)2 + (z – 4)2 = 0, then the value of x2/9 + y2/25 + z2/16 is
1. 12
2. 9
3. 3
4. 1
(x – 3)2 + (y – 5)2 + (z – 4)2 = 0
⇒ x – 3 = 0 ⇒ x= 3
Y – 5 = 0 ⇒ y = 5
Z – 4 = 0 ⇒ z = 4
3. If 4x/3 + 2P = 12 for what value of P, x = 6?
1. 6
2. 4
3. 2
4. 1
When x = 6, (4 * 6)/3 + 2P = 12
⇒ 8 + 2P = 12
⇒ 2P = 12 – 8 = 4
⇒ P = 2
4. The value of (4+3√3)/(7+4√3) is
1. 5√3 - 8
2. 5√3 + 8
3. 8√3 + 5
4. 8√3 - 5
Expression = (4+3√3)/(7+4√3)
Rationalizing the denominator
1. If x(3 - 2/x) = 3/x, then the value of x2 + 1/x2 is
1. 219
2. 249
3. 319
4. 349
2. What number must be added to the expression 16a2 – 12a to make it a perfect square?
1. 9/4
2. 11/2
3. 13/2
4. 16
a2 - 2ab + b2 = (a-b)2
∴ 16a2 – 12a = (4a)2 - 2*4a*3/2
Hence, on adding (3/2)2 = 9/4, expression will be a perfect square.
3. The straight line 2x + 3y = 12 passes through:
1. 1st, 2nd and 3rd quadrant
2. 1st, 2nd and 4th quadrant
3. 2nd, 3rd and 4th quadrant
4. 1st, 3rd and 4th quadrant
The usual way to solve these type of questions is to put x = 0 once and find y coordinate. This would represent the point where the line cuts the Y axis.
Similarly put y = 0 once and find x coordinate. This would represent the point where the line cuts the X axis. Then join these points and you will get the graph of the line.
So when we put x = 0 we get y = 4.
When we put y = 0 we get x = 6.
So when we join these points we see that we get a line in 1st quadrant, which when extended both sides would go to 4th and 2nd quadrants. So option B.
4. The sum of three altitudes of a triangle is
1. equal to the sum of three sides
2. Less than the sum of sides
3. greater than the sum of the sides
4. Twice the sum of sides

AP ∠ AB
BQ ∠ BC
CR ∠ AC
∴ AP + BQ + CR ∠AB + BC + AC
5. In ΔABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.
1. 40°
2. 25°
3. 35°
4. 20°