Official Quant thread for CAT 2013

MBA CAT 2024
21766
@ScareCrow28 said:
a+c = 12 nai aya...but there are 12 cases where a+c = even!Put a=0-3 and find satisfying values of c..you will find there are 12 cases
bhai need one example i am still not getting it how 12 cases give an example suppose a=2 then c =?
21767
@raopradeep said:
bhai need one example i am still not getting it how 12 cases give an example suppose a=2 then c =?
a + c = even
0___0
0___2
0___4
1___1
1___3
1___5
2___0
2___2
2___4
3___1
3___3
3___5
Total = 12 cases
21768

One sitter from my side:


J is 11/8 times as fast as K. If J gives K a start of 150 m, how far must be the winning post so that the race ends in a dead heat?

(1) 100 m (2) 440 m (3) 550 m (4) 200 m

21769
@ScareCrow28 said:
a + c = even0___00___20___41___11___31___52___02___22___43___13___33___5 Total = 12 cases
thanks a lot man :)
21770
@raopradeep said:
5c2 wala case borrowed only quants but not di hai na
@vijay_chandola said:
Bhai is there any relation between the cards and books ? I didn't get it.
sorry sir, correct nahi hai mera approach...will solve and post it again......getting confused
sir cards ka meaning to shayad yeh hai ki...i think she has three chances to borrow these books
21772
@vijay_chandola said:
One sitter from my side:J is 11/8 times as fast as K. If J gives K a start of 150 m, how far must be the winning post so that the race ends in a dead heat?(1) 100 m (2) 440 m (3) 550 m (4) 200 m
550 m?

d/11k/8 = d - 150/k
d = 550 m
21773
@mailtoankit said:
sorry sir, correct nahi hai mera approach...will solve and post it again....sir cards ka meaning to shayad yeh hai ki...i think she has three chances to borrow these books
yaar dont call sir
@vijay_chandola
i think when both are not included then 5c3
when quants is there and di is not there 5c2
when both are there 5c1
5c3+5c2+5c1 =25
21774
@vijay_chandola said:
One sitter from my side:J is 11/8 times as fast as K. If J gives K a start of 150 m, how far must be the winning post so that the race ends in a dead heat?(1) 100 m (2) 440 m (3) 550 m (4) 200 m
D/[(11/8)*s] = (D-150)/S
D = 550
21775
@vijay_chandola said:
One sitter from my side:J is 11/8 times as fast as K. If J gives K a start of 150 m, how far must be the winning post so that the race ends in a dead heat?(1) 100 m (2) 440 m (3) 550 m (4) 200 m
550
21776
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?

(a) 24% (b) 31.58% (c) 35% (d) 53.75%
21777
@vijay_chandola said:
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?(a) 24% (b) 31.58% (c) 35% (d) 53.75%
31.58? 24/76 aa raha hai mera...

regards
scrabbler
21778
@vijay_chandola said:
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?(a) 24% (b) 31.58% (c) 35% (d) 53.75%
total price of trader = list price of 19*4 = lp of 76...
sp for him = lp of 5*20 = lp of 100...
p% = (100-76)/76 = .3158 or 31.58%...
21779
@vijay_chandola said:
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?(a) 24% (b) 31.58% (c) 35% (d) 53.75%
say 1 chocolate = Re1.
4 will be of rs 4.
but kadbury is giving 5 chocolates for 4. which means Rs.0.8 per chocolate.

now.. he buys 100(19 packets + 1 packet free) chocolates for rs 95*0.8=76
he sells them for re 1 each.. 100chocolates for 100. or

profit = 100-76=24

profit percentage = profit/cp = 24/76=31.578%
21780
@maverick.ashu said:
is oa 31.58
21781

find the number of 6 digit numbers that can be found using 1,2,3,4,5,6 once such that the 6 digit number is divisible by its unit digit (the unit digit is not 1)

620
456
520
372

21782
@vijay_chandola said:
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?(a) 24% (b) 31.58% (c) 35% (d) 53.75%
31.58 % ?
let list price of 1 chocolate be re 1
cp = 19*5*4/5 = 19*4 = 76
sp = 20*5*1 = 100
profit = 100 - 76/76 = 31.58%
21783
@vijay_chandola said:
Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?(a) 24% (b) 31.58% (c) 35% (d) 53.75%
OA: B
21784
@raopradeep said:
find the number of 6 digit numbers that can be found using 1,2,3,4,5,6 once such that the 6 digit number is divisible by its unit digit (the unit digit is not 1)620456520372
Sure of options?

regards
scrabbler
21785
@raopradeep said:
find the number of 6 digit numbers that can be found using 1,2,3,4,5,6 once such that the 6 digit number is divisible by its unit digit (the unit digit is not 1)620456520372
for unit digit 2, 3, 5 = 5! combinations each
for 4 and 6 = 2*4! each

total = 5!+5!+2*4!+5!+2*4! = 456
21786
@ChirpiBird said:
A.

divide the line into 5 parts (since their speeds are 2:3 )and... move them to and fro accordingly.

P(A)3rd______.________.1st/5th_________._______2nd/4th._______Q(B)

so 1st meet at 0.2D from P, next will be 0.8D from P , next will be P. and 4th will be 0.8 D.
Fifth meet ... 0.2D from P.

correct me if i am wrong.
actually u r doing a little mistake
total time taken
d/(p+q) + 6d/(p+q) = 7d/(p+q)
distance travelled by a from p = 7dp/(p+q)
now p:q = 2:3
putting the ratio it gives distance travelled as 2.8d
so 0.8d from p