|
|
|
|
|
|
INFOSYS WRITTEN TEST ON 16th JULY, 2006 AT TAGORE INTERNATIONAL, DELHI |
Q1. A dice has faces 1 against 6, 3 against 5 and 2 against 4 always. How many such combinations of faces are possible in a dice?
Ans :( Very controversial question on this site indeed. You will find various answers for this. I have also one another answer ,but as per me it is correct ,please do check with you peers)
It can be considered as a problem where we have three pair of numbers and three pairs of opposite faces.
For first pair say (1 & 6) we need to choose one among three opposite faces pairs. i.e.3C1
Now after this we can arrange it in two ways (1&6) or (6 &1).so total no of ways for this becomes (3C1 * 2)
For the second pair we are left with two opposite pairs faces .Hence for this we have (2C1*2)
Going in this way for the last pair ,we have(1C1*2)
So the total no. of ways become ((3C1 * 2)* (2C1 * 2)* (1C1 * 2))=48
Q2. ). A software engineer starts from home at 3 pm for evening walk. He walks at a speed of 4 kmph on level ground and then at a speed of 3 kmph on the uphill and then down the hill at a speed of 6 kmph to the level ground and then at a speed of 4 kmph to the home at 9 pm. What is the distance on one way?
Answer:
Lets say the level ground is of distance x and hill of y on one way
Then while going the time taken is (x/4+y/3)
And while returning the time taken =(x/4+y/6)
Total time taken =(x/2+y/2)=6 hours(9pm-3pm)
Hence x+y= 12 hours
Q3. There were 2 systems A n B.14 degrees in A is equivalent to 36 in system B.and 133 in A is equivalent to 87 in B.Now what is the temperature when both r equal?
Ans:
Here we have to assume that the equality between two temperatures is based on linear equation, otherwise it cannot be solved
Let the relation between A & B be A=kB+c
Putting values given in the equation we would have two equations in two variables.then we find value of k as 7/3 and that of c as –70.
i.e A=(7/3)B-70
now putting B as A we find answer as:52.5 units.
Q4. 9. We are given 100 pieces of a puzzle. If fixing two components together is counted as 1 move ( a component can be one piece or an already fixed set of pieces), how many moves do we need to fix the entire puzzle.
Ans: 99
Say I have two dots then these can be joined by one line .Similarly if I have three then by two lines .Going on like this for 100 we need 99..
Q5. There is a peculiar island where a man always tells truth and a women never says two 2 consequtive truth or false statements that is if she says truth statement then she says false statement next and vice versa. a boy and girl also goes in the same way. one day i asked a child " what r u a boy or a girl" howver the child replied in their language that i dint understand
but the parents knew my language and one parent replied that "child said ‘ I am a boy’" the other one said that "no kibi is a girl, kibi lied"(8 marks)
a: is kibi a boy or a girl
b: who ansered first mother or father?
Ans:Please confirm for this.It was an easy question .(I don’t know the exact question)
Q6. Find the next term of the series:
a) 1,2,3,5,16,__
b) 1,2,3,8,__,224
Ans: a)There are two ways for this.I don’t know what answer they want
231 (2^2 - 1^2 = 3, 3^2 - 2^2 =5, ......16^2 - 5^2 = 231)
or
79(product of two last terms and then +1 or –1 alternatevly)
b)27 (2*1+1=3, 3*2+2=8, 8*3+3=27)
Q7.(I am wriiting very similar but not the exact question)
There are some card indexes .One guy comes and start picking the card indexes.
He goes like this.He firstly picks one and then skips one,then he picks one and skips two ,then picks one and skips three and so on.Finally when the card which he had picked were counted these turned out to be be 5% of the total card indexes.Find the no of card indexes he picked ?
Answer:The series for picked card indexes goes like this 1,1,1,1,1, and so on
While for the skipped card indexes series goes like this 1,2,3,4,5 and so on
So if he picks n indexes then he skips (n(n+1))/2 indexes.
So the total no of indexes are [n+(n(n+1))/2 ]
As per given situation
n=5% of [n+(n(n+1))/2 ]
here n comes to be 37.(Please check with others also for this question)
Q8. A horse travels half the distance with no load at a speed of 12 miles/hr. The other hal mile the horse is loaded and it travels with a speed of 4 miles\hr. What is the average speed of the horse?
Ans: 6 miles\hr(simple average question)
Q9.One day Harry and I set our watches together. None of us was aware that my watch was getting faster by 2 min per hour and Harry’s watch was getting slower by 1 min per hour. After sometime, we discovered that my watch was 1 hr ahead of Harry’s watch. Can you find out after how long we noticed this ? (3 Marks)
Ans: 20 hrs
Let it has been X hours since they have set their clocks
Time gaine be one =(2/60)(X),Hence in X actual hours the hours passed by this watch=X+(2/60)(X)
Time lost by other= (1/60)(X), Hence in X actual hours the hours passed by this watch=X-(1/60)(X)
So if we compare the time
(X+(2/60)(X))-(X-(1/60)(X))=1
X=20 Hours
Q 10. . A jeweller makes a window display. he has 7 gems of which he needs to display 6. three on the left side of the pane on three on the right. with following conditions.
let the gems be Armanet, diamond, emerald , sapphire , garnet, opal
and ruby.Armanet should be always displayed on left side while diamond always on the right.
Ruby should neither be kept with diamond or garnet. emerald and sapphire should always be on the same side.
(Here the question was very similar to this)
1]what are the possible combinations on the possible for left pane.
a] armanet,garnet , opal
b] armanet, ruby, garnet
c] diamond, emerald , sapphire
d] armanet , emerald, garnet
2]what are the possible combinations on the possible for right pane.
a]diamond, garnet , opal
b]diamond, ruby, opal
c]armanet,emerald, sapphire
d]diamond,garnet,ruby
3]If garnet,diamond,opal are on the right side what is the possible option for left side
a]Ruby, armanent,sapphire
b] armanet,emerald, sapphire
c] Armanent,emerald, ruby
d],ruby ,emerald, sapphire
3]If Armanent,, ruby,opal are on the left side what is the possible option for right side
a] diamond, garnet , emerald
b] diamond, garnet , sapphire
c] emerald , garnet , sapphire
d]emerald,sapphire,opal
e] diamond, sapphire , emerald
Answer:
Easy combination ,deduction and reduction problem
1]a;2]a;3]b;4]e
|
|
|
|
|
|
|
|
|
|
|