Official Quant thread for CAT 2013

@amresh_maverick said:
OA = 25Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C? 10 :23 :15 7 :15 :16 6 :13 :13 9 :20 :18 PS: sab phodu log padhar chuke hain
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.

@Sybar said:
From 4 gentleman and 4 ladies a committee of 5 is to be formed which consist of a president,vice-president and three secretaries.What will be the number of ways of selecting the commitee with atleast 3 women such that atleast one woman hols the post of president or vice-president?420610256512I'm having trouble with this one.
Case 1: 2M + 3F
Ways if 1 woman holds VP/P = 4C3*3C1*4C2*2*2 = 288
If 2 women hold P/VP position = 4C3*3C1*4C2*2 = 144

Case 2: 1M + 4F
Ways if 1 woman holds VP/P = 4C4*4C1*2*4 = 32
If 2 women hold P/VP position = 4C4*4C1*3*4 = 48

Total = 512 Ways
A went on a tour. He visited a total of 8 cities. In each city he spent Rs.2 less than half the amount he had with him. He spent Rs.100 in the last city he visited. Find the amount he had initially (in Rs.)?
@Dexian said:
512??
The answer is 512
@jain4444 said:
A went on a tour. He visited a total of 8 cities. In each city he spent Rs.2 less than half the amount he had with him. He spent Rs.100 in the last city he visited. Find the amount he had initially (in Rs.)?
25604??
@ScareCrow28 said:
Case 1: 2M + 3FWays if 1 woman holds VP/P = 4C3*3C1*4C2*2*2 = 288If 2 women hold P/VP position = 4C3*3C1*4C2*2 = 144Case 2: 1M + 4FWays if 1 woman holds VP/P = 4C4*4C1*2*4 = 32If 2 women hold P/VP position = 4C4*4C1*3*4 = 48Total = 512 Ways
Could you explain why the 2s and the 4s are multiplied?


@jain4444 said:
A went on a tour. He visited a total of 8 cities. In each city he spent Rs.2 less than half the amount he had with him. He spent Rs.100 in the last city he visited. Find the amount he had initially (in Rs.)?
26620?

@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.
1760?
@amresh_maverick said:
OA = 25

Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C. Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C?


10 :23 :15
7 :15 :16
6 :13 :13
9 :20 :18


PS: sab phodu log padhar chuke hain
i think b can never happen
v1/6 - 7v1/38 = 7v2/38 - v2/4
here v1/v2 comes out negative so not possible
@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.
2060???
@jain4444 said:
A went on a tour. He visited a total of 8 cities. In each city he spent Rs.2 less than half the amount he had with him. He spent Rs.100 in the last city he visited. Find the amount he had initially (in Rs.)?
26620
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?
@ChirpiBird said:
26620?
isnt it 26110?
@amresh_maverick said:
26620
iska koi aasan sol hai kya??
i mean i know the tedious one.............
@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?
114?
@mailtoankit said:
1 ?1 + 2 + ................9 = 9*10/2 = 4510 + 11 + 12 +.....19 = 10*11/2 = 5520 + 21 + ........... 29 = 11*12/2 - 1 = 6530 + 31 +.............40 = 12*13/2 - 1 - 2 + 4 = 75(45 + 55 + 65 + 75) mod 9 = 244 mod 9 = 1


correct answer..

but we can also take sum in one shot that is.

(40*41)/2 Mod 9
=(20*41) mod 9
=(2*5) mod 9 = 1


@jain4444 said:
A went on a tour. He visited a total of 8 cities. In each city he spent Rs.2 less than half the amount he had with him. He spent Rs.100 in the last city he visited. Find the amount he had initially (in Rs.)?
jain bhai ans toh bta do ???
@Vipul24 said:
correct answer..but we can also take sum in one shot that is.(40*41)/2 Mod 9=(20*41) mod 9=(2*5) mod 9 = 1
yup...i recognized that later..
@Sybar said:
Could you explain why the 2s and the 4s are multiplied?
Ways if 1 woman holds VP/P = 4C3*3C1*4C2*2*2
4C3 --- Taking 3 females of 4
3C1 --- Taking 1 lady of those 3 and assigning some position ( P/VP)
4C2 --- Taking 2 Males
2 ------ Ways of assigning a post to a female
2 ------ Ways of assigning a post to two males

Similarly others have been realised.
Hope you can make out others too. Sorry for late reply.
@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?
127??