Quant by Arun Sharma

MBA CAT 2024
42
abhijit111 Says
Hey the Question is very simple. The answer is 3 . U can solve such questions using options...

So plz explain using options.Which will be the ans.
43
solve dis problem

Some birds settled on the branches of a tree . First, they sat one to a branch and there was one bird too many. Next they sat sat two to a branch and there was one branch too many. How many branches were there ? (A.S. Pg 13)
a)3 b)4 c)5 d)6

After solving, explain the answer



See, these types of probs can better be solved thru options. Say, no of branches = 4.
Then no of birds acc to the first part of the statement = 4*1+1 = 5.
Acc to the second part, no of birds = (4-1)*2=6.
So this is not the correct option. At the same time you can see from the above calc that our assumed no of brances was somewhat more. Actually it shud be somewhat less to match with the prev value of no of birds. So it becomes obvious at this pt of time that the ans will be somewhat less than 4. There's only one option remaining which is somewhat less than 4. So, without checking furthur with 3, you can tick it as the ans thus saving time for you.

:):)
44
mejogi Says
I m not able to understand ur logic.Plz explain.


bi - # of birds
br - # of branches
bi = br+1 ------------- I
bi = 2(br-1) ------------- II
br = 3


let bi is the number of birds and br be the number of branches
from the info in the question we can arrive at I and II
solve the two eqns to get br = 3
45

hi folks,
pls help me on these...

1)What will be the new equation of the straight line 3x+4y=6 if the origin gets shifted to (3,-4)
a) 3x+4y=5 (b) 4x-3y=4 (c) 3x+4y+1 = 0 (d) 3x+4y-13=0


2)What will be the value of "p" if the equation of the straight line 2x+5y=4 gets changed to 2x+5y=p after shifting the origin at (3,3)
(a) 16 (b) -17 (c)12 (d) 10


also pls tell me the method to find the co-ordinates of orthocentre of a triangle when the co-ordinates of the 3 vertices are given (or equations of sides being given)...are there methods, other than finding the co-ordinates by solving equations (i.e the equations of altitudes)

thanks

46
hi folks,
pls help me on these...

1)What will be the new equation of the straight line 3x+4y=6 if the origin gets shifted to (3,-4)
a) 3x+4y=5 (b) 4x-3y=4 (c) 3x+4y+1 = 0 (d) 3x+4y-13=0



Lets say new corordinates will be (X,Y)
Original origin is (0,0)
then,
X=x+3

&

Y=y-4

substitute it in 3x+4y=6
3(X-3)+4(Y+4)=6
3X-9+4Y+16=6
3X+4Y=-1

ans 3x+4y=-1
47
hi folks,
pls help me on these...


2)What will be the value of "p" if the equation of the straight line 2x+5y=4 gets changed to 2x+5y=p after shifting the origin at (3,3)
(a) 16 (b) -17 (c)12 (d) 10


X=x+3
Y=y+3

2(X-3)+5(Y-3)=4
2X+5Y-6-15=4
2X+5Y=25

thus p=25
48

thanks chetna for that....i can now apply the change co-ordinates concept...

but i found you use X=x+3....isnt it the other way round? i.e x=X+3 ??? then the answer would be -17....also 25 doesnt figure in the answer option...but this is the case when the co-ordinates being changed are of +ve values like (3,3) what happens when the co-ordinates are in the negative?
still confused about that....if i apply x=X+h concept to the first sum that you solved, then i get option (d) as the answer but (c) is correct...
also folks,any tips for the orthocentre question that i had??

thanks again for solving the prob.

49

I think u shud c it this way -->

suppose u have a eq -> 3x + 4y =0
then this line will b passing thru origin .Now , u make ur origin to be (3,3) so now the line shud pass thru (3,3) .The slope of the orignal line is -3/4 so the eq of the new line becomes -> y-3 = -3/4(x-3) => 3x + 4y - 21=0 .The same eq u can get by doing ->

3(x-3) + 4(y-3) = 0. Therefore when u shift a point towards the +ve side u wud subtract and when u shift it towards -ve side u wud add.

I hope this wud make things clear now.:)

for the orthocenter ,its a point of the intersection of altitudes so it wud b better to solve it by logic rather than use formulas. Although i think thr may b a formula but dont remem it now.

----------------
Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us.

50
thanks chetna for that....i can now apply the change co-ordinates concept...

but i found you use X=x+3....isnt it the other way round? i.e x=X+3 ??? then the answer would be -17....also 25 doesnt figure in the answer option...but this is the case when the co-ordinates being changed are of +ve values like (3,3) what happens when the co-ordinates are in the negative?
still confused about that....if i apply x=X+h concept to the first sum that you solved, then i get option (d) as the answer but (c) is correct...
also folks,any tips for the orthocentre question that i had??

thanks again for solving the prob.


For the first sum , C) is the ans and i think i have got the same .... u check again ... i hav already solved it

for the second problem,as far as i know ..... the previous origin is (0,0) and new one is 3,3)

if i write

X=x+3
and put X=3 and x=0 it makes sense

but if i write

x=X+3
and put x=0 n X=3.....it doesnt work ..... 0!=6 right??
So i will still go with that ..... i m not bothered about options ..... ur concept shud be right

but still then ,u check if the new origin is (3,3) and not something else
51

Qn no. 6 : LOD 2..
There are three circular garbage cans each of diameter 2m. the three are touching each other extarnally at one point only.find the circumferance of the rope encompassing the three cans.


Ans : 1. 2pi + 6
2. 3pi +4
3. 4pi + 6
4. 6pi + 6

52

Grt initiative guys....its grt for ppl who r in office...wld get my doubts tomo onwards....Cheers

53

Im not able to give u the dig for this ?. Sorry for tat
The perimeter will ->
P=3[ (120/360)2*pi*r ] + 2r+2r+2r
=2*pi*r + 6*r
=2*pi + 6

54

ok so now ive attatched the dig. also ..........hope tat helps and answers ur ? .


------------------
Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us.

55
Qn no. 6 : LOD 2..
There are three circular garbage cans each of diameter 2m. the three are touching each other extarnally at one point only.find the circumferance of the rope encompassing the three cans.


Ans : 1. 2pi + 6
2. 3pi +4
3. 4pi + 6
4. 6pi + 6


3(2+(120/360)*2*pi*1)

=6+2pi

1) option
56

Here is a question ppl:

Find no. of zeros in 1^1.2^2.3^3.4^4.5^5........98^98.99^99.100^100

Happy solving....

57

i dont think there shud be power sign in between the term coz then the no. of zeros will be 0,as anything raised to the power of 1 will always result in 1.
plz clarify if these are power sign or multiplication signs.

58
Im not able to give u the dig for this ?. Sorry for tat
The perimeter will ->
P=3[ (120/360)2*pi*r ] + 2r+2r+2r
=2*pi*r + 6*r
=2*pi + 6



can u plz tell me how did u assume that angles as 120 deg(the circular part)and 90 degs(for the straight part):(
59
Here is a question ppl:

Find no. of zeros in 1^1.2^2.3^3.4^4.5^5........98^98.99^99.100^100

Happy solving....

Hi
U can solve this by finding no. os 5's
5(1+2+3+...+20)=5*20*21/2=1050
25(1+2+3+4)=250
Answer 1300 zeroes...Is it correct?
60
i dont think there shud be power sign in between the term coz then the no. of zeros will be 0,as anything raised to the power of 1 will always result in 1.
plz clarify if these are power sign or multiplication signs.



1^1.2^2.3^3.4^4.5^5........98^98.99^99.100^100

by '.' he mean '*' multiplication

no of zeroes dues to terms like (10,20....100) will be =10+20+30+40+50+60+70+80+90+200=650

no of 5's =(5+15+25+35+45...........+95)+25+75
=10/2(100)+100=600

as no of 5's will be less than 2's

total no of zeroes=600+650=1250
61
1^1.2^2.3^3.4^4.5^5........98^98.99^99.100^100

by '.' he mean '*' multiplication

no of zeroes dues to terms like (10,20....100) will be =10+20+30+40+50+60+70+80+90+200=650

no of 5's =(5+15+25+35+45...........+95)+25+75
=10/2(100)+100=600

as no of 5's will be less than 2's

total no of zeroes=600+650=1250

Hi
can u check whats wrong with my answer? i got answer as 1300