Official Quant thread for CAT 2013

MBA CAT 2024
23667
@ravi.theja said:
sir can u please explain me how to apply cosine rule..am not able to proceed .tanx in advance sir
See angle ADC + angle ADB = 180
So, Cos(ADC) + Cos(ADB) = 0

Cos(ADC) = (AD^2 + CD^2 - AC^2)/(2*AD*CD)
Cos(ADB) = (AD^2 + BD^2 - AB^2)/(2*AD*BD)

Form the quadratic equation and get BD
23668
@anantn said:
@chillfactorI think i have misinterpreted the question, please clarify if this is what the question is asking:sum of natural numbers from 1-8= 36 . so take the negative component to x and positive one to be 36-x. hence the alternating sum will be of the form (36-x)-x = 36 -2x, and since 36-2x>0 because of largest number being placed first x can range from 1-18.so the sum will be 36*9 - 2(18*19) = 306. I haven't filtered out the invalidate cases and such yet, just want to confirm if my interpretation of the question is correct?
It says that for any subset {a, b, c, d, e, f}, if a > b > c > d > e > f, then
Alternating sum will be a - b + c - d + e - f

For subset {a, b, c}, if a > b > c, alternating sum = a - b + c
23669
@chillfactor but then there will be only one alternating sum possible....whats the catch...?
23670
Came up with a foolproof approach but too tired to solve the final equation, maybe tomorrow:


1. take angle apb =x, angle apc= y for convinience, so angle pbc will be 360-(x+y)
2. sum of the area of the triangles apb,apc and pbc will add up to the area of ABC so we can so 1/2*(12*sinx + 15*siny+20*sin(360-(x+y)) = root3/4*(side of triangle)^2
3. now sinx= root(1-cosx*^2), use cosine formula to get value of cos, then repeat the same for sin y, and substitute both in equation.
4. for sin (360-(x+y) use sin(a-b) and then simplify by using trignometric formulates and substitute in the equation.
5. substitute final values in equation of step 2, and take (side of triangle)^2 to be a single variable since it gets repeated say T, and then you get a single variable equation, using which you can find the value of the side of equilateral triangle and subsequently the area.

This method will net you the answer, but is probably the most time consuming approach consuming upwards of 15 mins to solve, not practical for any competitive exams =/.

Will think up a simpler solution once im re-energized tomorrow.

@bodhi_vriksha

EDIT: this refers to that equilateral triangle abc with a point b at distance 3,4,5 from abc respectively wala question!
23671
@ravi.theja said:
In triangle ABC, AB=AC and D is any point on BC. Find BD, if AB=65cm AD=63cm and CD=8cma) 8cmb) 24cmc) 32cmd) 16cm
23672
@ChirpiBird said:
you could have used stewart's theorem and solved it.. 2 line answer..
23673

even after a discount of q% on MP a trader gains by P%. what is the markup percentage over the cost price.

23674

A person sold an watch at 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells at twice the percentage profit of its previous percentage profit then the new SP will be

23675
@vbhvgupta said:
A person sold an watch at 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells at twice the percentage profit of its previous percentage profit then the new SP will be
126 is answer???
23676
@vbhvgupta said:
A person sold an watch at 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells at twice the percentage profit of its previous percentage profit then the new SP will be
132?
23677
@saurabhlumarrai said:
126 is answer???
oa 132
23678
@vbhvgupta said:
oa 132
calculation mistake:
60+72=132

arey uska answer kya hai wo P and q wale ka???
23679
@vbhvgupta said:
A person sold an watch at 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells at twice the percentage profit of its previous percentage profit then the new SP will be
sp=96
x/100 = (96-x)/x
x^2+100x-9600 = 0
x= 60 ... cp, gain

profit =120%
cp=60
1.2*60 + 60 = sp

132?

23680
@iLoveTorres said:
you could have used stewart's theorem and solved it.. 2 line answer..
what is that? :P
itte saare theorems h geometry mein ...i cant remember. 😛
Pythagoras works for me. 😃

that equilateral triangle wale mein bhi (right angle triangle area + equilateral triangle area) se hi kia h... ditto what chillfactor has done.

@vbhvgupta said:
even after a discount of q% on MP a trader gains by P%. what is the markup percentage over the cost price.
mp=100
discount =Q%
sp = 100 - Q

p/100 = (100-Q - CP)/CP
p*CP/100 + CP= 100-Q
CP=100*(100-Q)/(p+100)

(100 - CP)/CP

= (P+Q)/100-Q ??
23681
@saurabhlumarrai said:
calculation mistake:60+72=132arey uska answer kya hai wo P and q wale ka???
(P+Q)/100-Q * 100
23682
Ten boxes each contain 9 balls. The balls in one box each weigh 0.9kg ;the rest all weigh 1 kg. in how many least number of weighing you can determine the box with the light balls.
#easy
23683
@ChirpiBird said:
Ten boxes each contain 9 balls. The balls in one box each weigh 0.9kg ;the rest all weigh 1 kg. in how many least number of weighing you can determine the box with the light balls.#easy
3??
23684
@ChirpiBird said:
Ten boxes each contain 9 balls. The balls in one box each weigh 0.9kg ;the rest all weigh 1 kg. in how many least number of weighing you can determine the box with the light balls.#easy
minimum could be 2.??


23685
@ChirpiBird said:
Ten boxes each contain 9 balls. The balls in one box each weigh 0.9kg ;the rest all weigh 1 kg. in how many least number of weighing you can determine the box with the light balls.#easy
1 way

1st box = 1 ball
2nd box = 2 balls
.
.
9th box = 9 balls

now total sum should be = 45 kg
if sum would be = 44.80 then faulty box is box 2 and so on
23686
@vbhvgupta said:
A person sold an watch at 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells at twice the percentage profit of its previous percentage profit then the new SP will be
sp - cp/cp = cp/100
96 - cp/cp = cp/100
cp^2 + 100cp -9600 = 0
cp = 60
new sp = 220/100*60 = 132 ?