Periyar University 2008 B.E Electronics

Three parts x, y, z

x+y+z=12

x=y+z

2x=12

x=6

So the largest part is six hrs


Ans: six hrs





55) (1- 1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
Sol: Leaving the 1st numerator and last denominator, all the numerator and denominator will cancelled out 1 a different.

Ans: 5/ (n+5)





56) Ten boxes are there. every ball weighs 100 gms. 1 ball is weighing 90 gms.

i) If there are three balls (n=3) in every box, how many times will it take to obtain 90 gms ball? ii) identical ques. with n=10

iii) identical ques. with n=9

Sol: The chances are

When n=3

(i) nC1= 3C1 =3 for 10 boxes.. 10*3=30


(ii) nC1=10C1=10 for 10 boxes ....10*10=100


(iii) nC1=9C1=9 for 10 boxes.....10*9=90





57) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels?

Sol: 4/5 full tank= 12 mile

1 full tank= 12/ (4/5)

1/3 full tank= 12/ (4/5)*(1/3) = five miles


Ans: five miles





58) Wind flows 160 miles in 330min.for 80 miles how much time needed

160 miles?

Sol: one mile = 330/160

80 miles= (330*80)/160=165 min.


Ans: 165 min.





59) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the identical speed limit by twice the identical if the 2nd person was traveling at a speed of 35 mph. obtain the speed limit

Sol :( x+10) =(x+35)/2

Solving the eqn we get x=15


Ans: 15




60) A sales person multiplied a number and get the ans is three instead of that number divided by 3.what is the ans he truly has to get.

Sol: presume 1

1* three = 3

1*1/3=1/3

So he has to got 1/3

Ans: 1/3





61) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent?

Earning:   Approval pending.