Official Quant thread for CAT 2013

@Estallar12 said:
So much spamming on Quant Thread ? Move to SB. Question:A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?P.S. Spam Posts delete kar dena!
5 pm??

5 pm...

@Estallar12 said:
So much spamming on Quant Thread ? Move to SB. Question:A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?P.S. Spam Posts delete kar dena!
I : 4:30PM
II : 5:37:30PM
III : 4:25:00 PM
IV : 5:53:33 PM
@Estallar12 said:
So much spamming on Quant Thread ? Move to SB. Question:A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?P.S. Spam Posts delete kar dena!
clock 1 moves 45 mins in 1 hour
clock 2 moves 75 mins when clock 1 hour
60 mins clock 1 =75 mins clock 2
45 mins clock 1=5*45/4=225/4 mins in hour of actual time
60 mins clock 2=40 mins clock 3
225/4 mins clock 2 = 2*225/3*4 = 225/6=75/2 mins in one hour
60 mins clock 3 = 80 mins clock 4
225/6 mins clock 3= 4*225/3*6 = 900/18=50 mins


in 6 hours clock 1 moves 6*45/60 = 4.5 hours=4:30:00
clock 2 = 6*225/4*60 = 225/40 hours = 45/8 hours=5 hours and 37.5 mins=5:37:30

clock 3 = 75*6/2*60 = 75/20=15/4=3.75 hours = 3:45:00

clock 4=50*6/60 = 5 hours=5:00:00

@amresh_maverick said:
Given that w, x, y and z are natural numbers such that 4w = xyz, 4x = wyz and 4z = wxy. Which of the following is a possible value of (w + x + y €“ z)^2? 25 9 36 1
25...
y = 4
z = w=x = 1
@amresh_maverick said:
OA :9 :20 :18 Let S be the sum of an arithmetic series. The arithmetic mean of every two consecutive terms and every three consecutive terms of S form the consecutive terms of series S1 and S2 respectively. If the sum of all the terms in series S1 and in series S2 are 1375 and 690 respectively, then find the sum of all the terms in series S.
OA and solution please..
@Estallar12 said:
So much spamming on Quant Thread ? Move to SB. Question:A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?P.S. Spam Posts delete kar dena!
5pm
@saurav205 said:
OA and solution please..
1375-690 = 685
1375+685 = 2060 (OA)
@amresh_maverick said:
A triangle ABC has 2 points marked on side BC, 5 points marked on side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have at least one vertex common with the triangle ABC?
@Logrhythm

iska sol batao na plzzz
@Dexian said:
@Logrhythmiska sol batao na plzzz
bhai mere 127 aa rahe hai...answer kya hai??
@Logrhythm said:
bhai mere 127 aa rahe hai...answer kya hai??
answer yahi hai...........
solution batao
@Dexian said:
answer yahi hai...........solution batao
bhai meine aese kara...

total possible triangles = 13c3 - 7c3 - 5c3 - 4c3 = 237..

now take two from AB and one from BC or AC = 3c2(2c1+5c1) = 21 triangles...
take two from BC and one from AB or AC = 2c2(C(3,1)+C(5,1)) = 8 triangles..
take two from AC and one from AB or BC = 5c2(3c1+2c1) = 50 triangles...
take one from AB, one from AC and one from BC = 3c1*5c1*2c1 = 30 triangles...
so, total with no common vertex to the triangle = 21+8+50+30 = 109

hence, triangles with a common vertex to the original triangle = 237-109 = 128..

but in this 128 we have also counted trngl ABC, and the question asks for all the "new triangles", so 128-1 = 127 such triangles...
What a slow start to the sunday...post karo yaar questions..

Q: Amit has 2n marbles numbered 1 through 2n. He removes n marbles that are numbered consecutively. The sum of the numbers on the remaining marbles is 1615. Find all possible values of n.

Find the area of the traingle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a square of side 'A' ?

@shinoda said:
Find the area of the traingle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a square of side 'A' ?
A^2/8 (for right isosceles triangle)
3rt(3)/32 (for equilateral triangle)
@shinoda whats the OA?
@iLoveTorres said:
A^2/8?
nope
@shinoda said:
nope
see my edited post
@iLoveTorres said:
@shinoda whats the OA?
(3root3 * a^2)32
@iLoveTorres said:
A^2/8 (for right isosceles triangle)3rt(3)/32 (for equilateral triangle)
share ur approach