Official Quant thread for CAT 2013

MBA CAT 2024
10201
@amresh_maverick said:
A function F(n) is defined as F(n – 1) = 1/ [2-F(n)] for all natural numbers 'n'. If F(1) = 2, then whatis the value of [F(1)] + [F(2)] +…………+ [F(50)](Here, [x] is equal to the greatest integer less than or equal to 'x')
F(n) = 2 - 1/F(n-1)

F(1) = 2, given
F(2) = 2 - (1/2) = 3/2

F(3) = 2 - (2/3) = 4/3

=> F(n) = (n + 1)/n

-> Value of that thing = 2 + 49*(1) = 51 ?

10202
@amresh_maverick said:
A function F(n) is defined as F(n – 1) = 1/ [2-F(n)] for all natural numbers 'n'. If F(1) = 2, then whatis the value of [F(1)] + [F(2)] +…………+ [F(50)](Here, [x] is equal to the greatest integer less than or equal to 'x')
51? first term 2 and then all the other terms are 1.
10203
@vijay_chandola said:
Q: Isosceles triangle ABC has the property that, if D is a point on AC such that BD bisects angle ABC, then triangle ABC and BCD are similar. If BC has length of one unit, then what is the length of AB?
1/(2*cos(72)) :splat:

ABC similar to BCD => Angle A = (Angle B)/2 , and Angle B = Angle C => AB = AC

A + B + C = 180 => 5A = 180 => Angle A = 36 degrees

B = 2A = 72 degrees

BC/2AB = cosB => AB = (1/(2*cos(72)) ?
10204

koi solve karwa do yeh :@vijay_chandola ,@YouMadFellow
If the algebraic sum of deviations of 20 observations measured from 23 is 70, mean of these observations would be
24
25
26
none of the above

10205
@YouMadFellow said:
1/(2*sin(72)) ABC similar to BCD => Angle A = (Angle B)/2 , and Angle B = Angle C => AB = AC A + B + C = 180 => 5A = 180 => Angle A = 36 degreesB = 2A = 72 degreesBC/2AB = sinB => AB = (1/(2*sin(72)) ?
BC/2AB = sinB ye kya kiya? :neutral:

I used cos formula,
Cos 72=(x^2+1-x^2)/2*x*1
=> x=1/(2*cos 72).
isme kya galat h?
10206
@sonamaries7 said:
koi solve karwa do yeh :@vijay_chandola ,@YouMadFellowIf the algebraic sum of deviations of 20 observations measured from 23 is 70, mean of these observations would be242526none of the above
Numbers are of the form 23 + a1, 23 + a2 and so on

Sum = 23*20 + ( a1 + a2 + a3 + ... a20) = 23*20 + 70 = 530

Mean = Sum/20 = 530/20 =26.5 😞

None of these ? :splat: ..
10207
@vijay_chandola said:
BC/2AB = sinB ye kya kiya? I used cos formula,Cos 72=(x^2+1-x^2)/2*x*1=> x=1/(2*cos 72).isme kya galat h?
Yeah, I edited the post actually, Stupid Calculation mistake :splat: .. Exchanged sin and cos :|
10208
@sonamaries7 said:
koi solve karwa do yeh :@vijay_chandola ,@YouMadFellowIf the algebraic sum of deviations of 20 observations measured from 23 is 70, mean of these observations would be242526none of the above
26.5 I'm getting
galati kahan ho ri
10209
@vijay_chandola said:
26.5 I'm gettinggalati kahan ho ri
arre sahi hai...koi galati nahi hai...mujhe qs hi samajh me nahi aarha tha
ab aagya..@YouMadFellow : thanks!
10210
Vertices A, B and C of a parallelogram ABCD lie on a circle and D lies inside the circle such that line BD intersects the circle at point P. Given that ÐAPC = 75° and ÐPAD = 19°, what is the measure of ÐPCD?
10211
@vijay_chandola said:
Vertices A, B and C of a parallelogram ABCD lie on a circle and D lies inside the circle such that line BD intersects the circle at point P. Given that ÐAPC = 75° and ÐPAD = 19°, what is the measure of ÐPCD?
11 degrees ?

Angle ABC = 105
Angle C = 75

PCB + PAB = 180 -> x + 75 + 75 + 19 = 180 => x = 11 ?
10212

Only when a right triangle is given,i know which one is hypotenuse and which one is base.But in other cases,i can't identify.I know this may be a preposterous question to many,but nevertheless i don't know the logic.So how do we identify which side is hypotenuse or base if it is not right angled traingle?

10213
@nole By definition of hypotenuse, it is the longest side of a right-angled triangle.

When it is not a right angle triangle, how come there is a hypotenuse ? :splat:
10214
@vijay_chandola k i thought we can find the unknown side using sin,cos for which we might need p,b or h.So is there any way we can find the unkown side in a traingle(not right angled) by using sin,cos then?
10215
@nole said:
@vijay_chandola k i thought we can find the unknown side using sin,cos for which we might need p,b or h.So is there any way we can find the unkown side in a traingle(not right angled) by using sin,cos then?
Yes, we can definitely find the sides using cos or sin Formulae.
Some frequently used formulae are:
[1] b^2+c^2-a^2=2*b*c*Cos A
[2] Area=1/2*b*c*Sin A= 1/2*c*a*Sin B=1/2*a*b*Sin C

10216
@joyjitpal scroll up and u will find complete solution by vijay_chandola,scroll down,there u will find explanation by chillfactor. (scroll up and down on the page,where u are asking chillfactor the answer to the question by saying "then what ?") i.e post no 10350 and 10355.
10217
S is the set of nonnegative integers. It can be divided in 2012 distinct subsets ,each containing equal number of elements and the elements of each subsets add up to the same Number. What is the minimum possible integer which is the largest element in S?
10218
what will be the next number in the series:-
1. 32,14.23,7,5.2,12.03,?
2. 340,680,1428,3146.6,?
10219
Mr praveen has to build a wall 1000 meters long in 50 days. he employs 56 men but at the end of 27 days finds only 448 meters are built. How many more men must be employed so that the work may be finished in time?
10220
@sumit99 said:
Mr praveen has to build a wall 1000 meters long in 50 days. he employs 56 men but at the end of 27 days finds only 448 meters are built. How many more men must be employed so that the work may be finished in time?
56 men built 448 meters in 27 days
=> 1 man builds 8 meters in 27 days
=> 1 man builds (8/27) meters in 1 day

Now, remaining = 1000 - 448 = 552

So, 552/[(8/27)*23] = Total men required after that = 81

So, extra men required = 81 - 56 = 25 ?