**There are 6 consecutive odd numbers. The difference between the square of the average of the first three numbers and the square of the average of the last three numbers is 288. What is the last odd number?**- 31
- 27
- 29
- 25
- 33

Answer : Option C.

Let the 6 consecutive odd no.’s are:

X, X+2, X+4, X+6, X+8, X+10

Avg. of 1^{st}three no’s is X+2.

Avg. of Last three no’s is X+8.

Given that (X+8)^{2}-(X+2)^{2}=288

X=19

Last Odd no. is X+10= 29.

Answer is 29.

**In a bag there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red?**- 29/35
- 7/15
- 23/35
- 2/5
- 19/35

Answer : Option C.

Probability of atleast one of the balls drawn is red= 1- (9/15) x (8/14)=23/35.

Answer is 23/35.**A started a business with an investment of Rs. 28,000. After 5 months from the start of the business, B and C joined with Rs. 24,000and Rs. 32,000 respectively and withdrew Rs. 8000 from the business. If the difference between A’s share and B’s share in the annual profit is Rs. 2,400, what was the annual profit received?**- Rs. 15,600
- Rs. 14,400
- Rs. 14,040
- Rs. 15,360
- Rs. 13,440

Answer : Option B.

Equivalent Contribution of A= 28000 x5+20000 x 7= 280000

Equivalent Contribution of B= 24000x 7= 168000

Equivalent Contribution of C= 32000 x 7= 224000

Let total profit be X.

Given that,

280000X/672000 – 168000X/672000=2400

112000/672000 x X=2400

or X=2400 x 672/112

X=14400**At present, Ami’s age is twice Dio’s age and Cami is two years older than Ami. Two years ago, the respective ratio between Dio’s age at that time and Cami’s age at that time was 4 : 9. What will be Ami’s age four years hence?**- 40 years
- 30 years
- 42 years
- 36 years
- 48 years

Answer : Option A.

D; A = 2D; C = A+2 = 2D+2

Given, D-2/(2D+2)-2 = 4/9

Solving, D = 18 Years and A = 36+4 = 40 years.

- Relationship between X and Y cannot be established
- X < Y
- X > Y
- X ≤ Y
- X ≥ Y

**I. 9x**^{2}- 37x + 30 = 0

II. 3y^{2}– 19y + 30 = 0

Answer : Option D.

X =< Y

I. 9x2 – 37x + 30= 0 ; II. 3y2 – 19y +30 = 0

Solving these equations we get, y= 10/3, 3 and x = 3, 10/9

So X =< Y

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**I. 2x**^{2}– 17x + 36 = 0

II. 5y^{2}– 33y + 40 = 0

Answer : Option 1.

Relationship can’t be determined

I. 2x^{2}– 17x + 36 = 0

II. 5y^{2}– 33y + 40 = 0

Solving these equations we get, y= 5, 8/5 and x = 4, 9/2**I. 12x**^{2}- 23x + 11 = 0

II. 21y^{2}– 20y + 4 = 0

Answer : Option A

Relationship can’t be determined

I. 12x^{2}- 23x + 11 = 0

II. 21y^{2}– 20y + 4 = 0

Solving these equations we get, y= 2/7, 2/3 and x = 1, 11/12**I. x**^{2}+ 12x + 35 = 0

II. 7y^{2}+ 32y + 16 = 0

Answer : Option B

Y>X

I. x^{2}+ 12x + 35 = 0

II. 7y^{2}+ 32y + 16 = 0

Solving these equations we get, y= -4, -4/7 and x = -5, -7**I. 25x**^{2}+ 20x + 3 = 0

II. 4y^{2}+ 11y + 6 = 0

Answer : Option C.

X>Y

25x^{2}+ 20x + 3 = 0&

II. 4y^{2}+ 11y + 6 = 0

Solving these equations we get, y= -2, -3/4 and x = -1/5, -3/5**A and B can complete a piece of work in 80 days and 120 days respectively. They started working together but A left after 20 days. After another 12 days C joined B and they completed the work in 28 more days. In how many days can C alone complete the work?**- 110 days
- 112 days
- 114 days
- 120 days
- None of these

Answer : Option B.

So efficiency of A and B are 3 units and 2 units respectively.

As they worked for 20 days together after that A left so total unit contribution in 20 days= (3+2) units * 20 days = 100 units

Remaining units = 240 – 100 = 140 units

After another 12 days C joined B and they completed the work in 28

more days, so total units contribution of B in 40 days = 2*40 = 80 units

Remaining units i:e = 140-80 = 60 units

Now 60 units is done by C in 28 days

So to do 240 units C require = 28/60 *240 = 112 days