This section deals with Crompton Greaves Aptitude questions. Here in this session, we have provided Question and answers with detailed explanation. We suggest the candidates can go through this page for secure good marks.

1. What are the points in the stress strain curve for steel?

A. elastic limit or yield point

B. ultimate stress

C. stress at failure

D. all of these

Ans. D

2. The ratio of inertial force and viscous force. It is a dimensionless number. It determines the type of fluid flow.

A. Reynolds number.

B. viscous force

C. torque

D. none of these

Ans. A

3. How many joules is 1 BTU?

A. 1055.056 joules.

B. 1054 joules

C. 950 joules

D. none of these

Ans. A

4. It is the temperature below which the tendency of a material to fracture increases rather than forming. Below this temperature, the material loses its ductility

A. annealing

B. Nil Ductility

C. Ductile- brittle transition

D. Both B&C

Ans. D

5. For a perfectly incompressible material, the Poisson’s ratio would be

A. greater than 0.5

B. less than 0.0

C. equal to 0.1

D. none of these

Ans. D

Solution:

Exactly 0.5. Most practical engineering materials have ? between 0.0 and 0.5. Cork is close to 0.0, most steels are around 0.3, and rubber is almost 0.5

6. A man sells an article with a 20% discount and gain a profit of 20%.What would be the profit percentage if he sells it with 10% discount?

A. 25%

B. 50% C. 35% D. none of these

Ans. C

Solution:

Let initial selling price be 100 Rs Cost price after 20% discount be 80* (100/120) = 200/3 Selling price after 10% discount is 90 Rs Profit = (90 – (200/3)) / (200/3) = 7/20 % profit = (7/20)* 100 = 35%

7. a radioactive element disintegrates by 20 th part every hour and find the probability that no matter is left out in duration of 45 min?

A. e ^ (-15) B. 15 C. e ^ (-20) D. none of these

Ans. A

Solution:

Since it deals with radioactive element ‘n’ is usually large. so poisons distribution is to be applieD. .. 1hr=60min———-20 45min———-? (45*20)/(60)=15.. Hence lambda=15 P(x=0) = (e-15)*(150)/ (0!) = (e-15)

8. If side of the square is x+2 and side of equilateral triangle is 2x and the perimeters of both square and equilateral triangle are equal .Then find the value of x ?

A. 4 B. 6 C. 8 D. none of these

Ans. A

Solution:

4*(x+2) = (3*2x) i.e.; x=4

9. If side of the square of increased by 5 and change in area were 165, then find the value of side of the square ?

A. 16 B. 14 C. 18 D. none of these

Ans. B

Solution: x+5)2-x2=165 10x+25=165 10x=140 x=14

10. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

A. 45 B. 60 C. 75 D. 90

Ans. D

Solution:

Let number of notes of each denomination be x. Then x + 5x + 10x = 480 16x = 480 x = 30. Hence, total number of notes = 3x = 3 x 30 = 90.

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