The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
59
59
Official Quant Thread for CAT 2012
Quantitative
Add more users
Please continue here for all the Quant queries and discussions.
The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
The links to the previous threads are:
Part 1
http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a.html
Part 2
http://www.pagalguy.com/forum/quantitative-questions-and...
total work = 6*(3w+2m)
now,
{6(3w + 2m) / 9w} - {6(3w+2m)/3m} =5
= 1/3 +2m/9w - w/m - 2/3 =5/6
let m/w=p
=> 2p/9 - 1/p =7/6
=>4p^2 -21p - 18 =0
=> (4p+3)(p-6) =0 => p=6 =m/w => exceed by 5 times
now,
{6(3w + 2m) / 9w} - {6(3w+2m)/3m} =5
= 1/3 +2m/9w - w/m - 2/3 =5/6
let m/w=p
=> 2p/9 - 1/p =7/6
=>4p^2 -21p - 18 =0
=> (4p+3)(p-6) =0 => p=6 =m/w => exceed by 5 times
1
output man=6*output woman
exceed by 5 times
time taken by 3 men=t-5time taken by 9 women=t
efficiency of 1 man=1/3(t-5)efficiency of 1 woman=1/9t
2/3(t-5) + 3/9t=1/6
t=10 days
efficiency of man/woman=90:15 or 6:1
exceed by 5 times
time taken by 3 men=t-5time taken by 9 women=t
efficiency of 1 man=1/3(t-5)efficiency of 1 woman=1/9t
2/3(t-5) + 3/9t=1/6
t=10 days
efficiency of man/woman=90:15 or 6:1
In a triangle ABC, let E be a point on AB such that AE : EB = 1 : 3, D is a point on B C such that BD : DC = 4 : 1, and F is a point on ED joined such that EF : FD = 5 : 1. Let G be a point on AC. Show that
AG : GC = 4 : 1 and BF : FG = 17 : 7.
AG : GC = 4 : 1 and BF : FG = 17 : 7.
take efficiency of 1st man a 2nd man b and of woman c, total work 1
1/b - 1/(b+c) = 3 (from 1st condition)
a = b+c (from 2nd condition)
=> 1/b - 1/a = 3
1/a = 2*1/b - 8 (from 3rd condition)2/b - 1/a = 8
solving above 2 equations
1/b = 5b = 1/5a = 1/2c = 3/10
so all 3 workin...
1/b - 1/(b+c) = 3 (from 1st condition)
a = b+c (from 2nd condition)
=> 1/b - 1/a = 3
1/a = 2*1/b - 8 (from 3rd condition)2/b - 1/a = 8
solving above 2 equations
1/b = 5b = 1/5a = 1/2c = 3/10
so all 3 workin...
2
Q.3 It takes six days for three woman and two men working together to
complete a work. Three men would do the same work five days sooner
than nine woman. How many times does the output of a man exceed that
of a woman?
a> 3 b>4 c>5 d>6 e>7
complete a work. Three men would do the same work five days sooner
than nine woman. How many times does the output of a man exceed that
of a woman?
a> 3 b>4 c>5 d>6 e>7
It has to be zero solutions.
n^2 + 3n + 5 = n^2 + 3n - 28 + 33 = (n - 4)(n+7) + 33
Now (n + 7) - (n - 4) = 11. It implies that either of them is a multiple of 11 or not a multiple of 11.
In the former case (n-4)*(n+7) is a multiple of 121 but 33 is not a multiple of 121. So no such case...
n^2 + 3n + 5 = n^2 + 3n - 28 + 33 = (n - 4)(n+7) + 33
Now (n + 7) - (n - 4) = 11. It implies that either of them is a multiple of 11 or not a multiple of 11.
In the former case (n-4)*(n+7) is a multiple of 121 but 33 is not a multiple of 121. So no such case...
1
the answers are 1) 642) 1
@[251735:chillfactor] sir,, use cntrl nd den click on quote,,dey will keep adding in text box,,, :)
1
n^2+3n+5=121k
n^2+3n+5-121k=0
D=9-20+4*121k
D=121*4k-11
D=11*(44k-1)
for n to be integer D must be a perfect squarewhich is not possible as (44k-1) will never be divisible by 11 so there will only be one 11 in D.
so no solutions
PS: But still I feel there must be a better me...
n^2+3n+5-121k=0
D=9-20+4*121k
D=121*4k-11
D=11*(44k-1)
for n to be integer D must be a perfect squarewhich is not possible as (44k-1) will never be divisible by 11 so there will only be one 11 in D.
so no solutions
PS: But still I feel there must be a better me...
2
yes.. but plz share the approach..
0 hoga pehle bhi kiya hai lag rha hai........ :)
(a^13+b^13+c^13+d^13) mod (a+b+c+d) is not correct, it will be correct only if a, b, c, d are in AP
(a + b + c + d)^13 - (a + b + c + d) is always divisible by 13 ............(1)
x^13 - x is always divisible by 13So, (a^13 + b^13 + c^13 + d^13) - (a + b + c + d) is divisible by 13 ........
(a + b + c + d)^13 - (a + b + c + d) is always divisible by 13 ............(1)
x^13 - x is always divisible by 13So, (a^13 + b^13 + c^13 + d^13) - (a + b + c + d) is divisible by 13 ........
4
option A zero solutions hai kya
1
bhai,,i just did,,,mein bhi option B hi daalne waala tha,,bt den saw,, @budokai 's comment,,
confused,,mujhe bhi multiple lag rhe hain,,,
confused,,mujhe bhi multiple lag rhe hain,,,
1
yes bhai... formula for total distance travelled together when both are moving in same direction = 2nd, where n=no of rounds, d is distance between PQ.
Pranaam sood bhai, aizen bhai..
Pranaam sood bhai, aizen bhai..
2
How many numbers between 1 and 1000 are there such that n^2 + 3n + 5 is divisible by 121 ??
A. 0B. 5C. 7D. 10
A. 0B. 5C. 7D. 10
But yar why (a+b+c+d) is not possible???
EDITED: Dhanyawad @[251735:chillfactor] sir dhyan me hi nai rha ..... Agey kabhi galti nai hogi
(a^13+b^13+c^13+d^13) mod (a+b+c+d) =0 [Only when a,b,c,d are in AP]
:banghead:
Also (a+b+c+d)^3 mod (a+b+c+d) = 0
@[577366:chevy] bhai ye...
EDITED: Dhanyawad @[251735:chillfactor] sir dhyan me hi nai rha ..... Agey kabhi galti nai hogi
(a^13+b^13+c^13+d^13) mod (a+b+c+d) =0 [Only when a,b,c,d are in AP]
:banghead:
Also (a+b+c+d)^3 mod (a+b+c+d) = 0
@[577366:chevy] bhai ye...
4
Agey kabhi galti nai hogi
@[555978:yashasvi5]
Let the total units of work be W units. Let x units/hour be the rate of work of the second person and let y units/hour be the rate of work of the woman. Let a units/hour be the rate of the first person.
Now, as per the given condition,
(W/x)=3+[(W)/(x+y)] ....(i)<...
Let the total units of work be W units. Let x units/hour be the rate of work of the second person and let y units/hour be the rate of work of the woman. Let a units/hour be the rate of the first person.
Now, as per the given condition,
(W/x)=3+[(W)/(x+y)] ....(i)<...
1
log5^2=log5+logy2log5=log5+logylog5=logyy=5.
iska answer i guess will be Cannot be determined,,
523abc
for divisibility wid 9,,sum of numbers shud be multiple of 9
10+a+b+c=18
a+b+c=8
a=1,b=5,c=2
523152 is divisible by all 7,8,9
a*b*c=10
similarly wen a+b+c+10=27
a+b+c=17
6,5,6
523656 is also divisible by 7...
523abc
for divisibility wid 9,,sum of numbers shud be multiple of 9
10+a+b+c=18
a+b+c=8
a=1,b=5,c=2
523152 is divisible by all 7,8,9
a*b*c=10
similarly wen a+b+c+10=27
a+b+c=17
6,5,6
523656 is also divisible by 7...
3
Aizen bhai,OA is d)26 .... you have the first term as well na ?
i proceeded like...
(a+b)^2 -(a^2+b^2)= 2ab --->divisble by 2
(a+b)^3-(a^3+b^3)=3(a^2b+aB^2) --. divisible by 3
and so extending this(a+b+c+d)^13 - (a^13+b^13+c^13+d^13) will be divisble by 13
From options 26 is the...
i proceeded like...
(a+b)^2 -(a^2+b^2)= 2ab --->divisble by 2
(a+b)^3-(a^3+b^3)=3(a^2b+aB^2) --. divisible by 3
and so extending this(a+b+c+d)^13 - (a^13+b^13+c^13+d^13) will be divisble by 13
From options 26 is the...
@[562114:Budokai001] .... my bad !!! left out one term
1
options are:
252141
252141
If log25 = log5y, then what is the appropriate value of y?
Please help
Please help
@[521225:Vinaysastra]
Bhai i had done it using culdip sir method ... check and revert if your doubt persists.Check aizen bhai solution too
Bhai i had done it using culdip sir method ... check and revert if your doubt persists.Check aizen bhai solution too
@[367896:sumeet1489] Thanks bhai,mera aajkal dimaag kaam karna band ho chuka hai.
Iska option B hoga...
a^m+b^n+c^s is divisible by (a+b+c) when (m,n,s) are odd.
p.s: @[577366:chevy] bhai , @[515110:culdip] sir __/\__
a^m+b^n+c^s is divisible by (a+b+c) when (m,n,s) are odd.
p.s: @[577366:chevy] bhai , @[515110:culdip] sir __/\__
1
find out root of 603000it is 776.xxx
so all numbers in the subset should be greater than 776 (as 776 is not member of set A, but 780 is and 773*780 < 603000)
so elements in subset will be 780 787 794 801 . . . 1004=> (1004 - 780)/7 +1 = 33 elements
so all numbers in the subset should be greater than 776 (as 776 is not member of set A, but 780 is and 773*780 < 603000)
so elements in subset will be 780 787 794 801 . . . 1004=> (1004 - 780)/7 +1 = 33 elements
4
yes correct answer......cud you please give d approach?
nothing yaar
total area=9*(6+12+.........24 terms/planks)that would give you total area in inches then divide by 12^2 to convert it in ft^2
total area=9*(6+12+.........24 terms/planks)that would give you total area in inches then divide by 12^2 to convert it in ft^2
1
@[555978:yashasvi5]
Q.2 Two men and a women are entrusted with a task. The second man needs three hours more to cope with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as the second man and the woman working togeth...
Q.2 Two men and a women are entrusted with a task. The second man needs three hours more to cope with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as the second man and the woman working togeth...
my take is 32...
sqrt of 603000 is approx equal to 780...
so no of elements in set S with value more than or equal to 780 is 32.. hence the answer.
Please let me know if my ans is wrong..
sqrt of 603000 is approx equal to 780...
so no of elements in set S with value more than or equal to 780 is 32.. hence the answer.
Please let me know if my ans is wrong..
@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30
@sumeet1489 said: option a hai kya iska
OA is A
(a+b+c+d)^13 - (a^13+b^13+c^13+d^13) is always divisble by
a)156
b)a+b+c+d
c)12
d)26
Season 1 : CAT'12 :( || XAT'13 :) || XL dinged :( |||| Season 2 : Will make it BIG!!
bhai maine padha tha,,,ki we have to reduce dem in simplest form,,before applying formula,,,,
1
@[367896:sumeet1489] Yes saar,method please.
Kindly help,
The number 523abc is divisible by 7,8 and 9. Then the value of a*b*c is
A. 10B. 60C. 180D. Cannot be determined
Kindly share the approach as well..
Thank you..
The number 523abc is divisible by 7,8 and 9. Then the value of a*b*c is
A. 10B. 60C. 180D. Cannot be determined
Kindly share the approach as well..
Thank you..
is it 112.5ft^2
1
@[500559:ImCATian] Oh yes, one should simplify and then apply that formula.
Iska a. 33 hai kya???
Elements of S is in AP.
So we are asked the product of any two elements...
1004 + (33-1)*-7 = 780
So when multiplied by next lowest number should be > 603000
So 780*787 = 613860 ....
p.s: @[419941:Enceladus] : sood sir __/\__ ; @[522923:allan89] bhai __/\__
Elements of S is in AP.
So we are asked the product of any two elements...
1004 + (33-1)*-7 = 780
So when multiplied by next lowest number should be > 603000
So 780*787 = 613860 ....
p.s: @[419941:Enceladus] : sood sir __/\__ ; @[522923:allan89] bhai __/\__
1
why cant we derive the solution without simplyfying..
I too got the answer after getting fractions simplified.
Thanks for attempt..
I too got the answer after getting fractions simplified.
Thanks for attempt..
1
@[567546:shraymishra] It's a good book.Toughness level is easy to moderate.But also solve Arun Sharma.
1
sorry ...plz check OAs..
its 10 & 1/15.
thanks for attempt.
its 10 & 1/15.
thanks for attempt.
haan yaar,,galti kar di maine,,
1)
4/6=2/3
6/9=2/3
5/4
LCM will be= LCM of numer/HCF of denomi = 10/1=10
2)
4/6=2/3
3/9=1/3
6/5
HCF will be=HCF on nume/LCM of denomi = 1/15
P.S.--- Aaj toh mela laga hua hai yahaan pe,,,lage raho bhai log
1)
4/6=2/3
6/9=2/3
5/4
LCM will be= LCM of numer/HCF of denomi = 10/1=10
2)
4/6=2/3
3/9=1/3
6/5
HCF will be=HCF on nume/LCM of denomi = 1/15
P.S.--- Aaj toh mela laga hua hai yahaan pe,,,lage raho bhai log
2
my answer is 1
10^3 + 9 ^3=1729
12^3=1728
in case of (mx+1)^n/m=1 always read somewherehere is the caseso rem shud be 1
10^3 + 9 ^3=1729
12^3=1728
in case of (mx+1)^n/m=1 always read somewherehere is the caseso rem shud be 1
how 2nd..??
ny formula..??
ny formula..??
LCM of fractions= (LCM of numerators)/(HCF of denominators) HCF of fractions=(HCF of numerators)/(LCM of denominators)Hence, the answers :1) 60/1. 2) 1/90
Hey Puys..Can someone plz tell me how is arihant's quantum CAT for QA..Am hoping someone will help me out..
Answers are
1) 10
2) 1/15
1) 10
2) 1/15
option a hai kya iska
1
In how many ways can (2^4)(3^6) be written as a product of two distinct numbers?
In how many ways can (5^6)(7^5) be written as a product of two distinct factors?
In how many ways can (2^7)(3^11) be written as a product of two co-primes?
In how many ways can (5^6)(7^5) be written as a product of two distinct factors?
In how many ways can (2^7)(3^11) be written as a product of two co-primes?
1
@ varun
How u got those two functions..??
Plz tell..
How u got those two functions..??
Plz tell..
1
One side of a staircase is to be closed in by rectangular planks from the floor to each step.The width of each plank is 9 inches and their height are successively 6 inches ,12 inches , 18 inches.... and so on.There are 24 planks required in total.Find the area in square feet.
total distance travelled together = 2nd, where n=5; d= 100
=>1000m.
A travels =2/5 *1000 =400m.
=> A and B will be at P during 5th meeting . Hence ans should be 0m
=>1000m.
A travels =2/5 *1000 =400m.
=> A and B will be at P during 5th meeting . Hence ans should be 0m
Plz solve this
1) Find LCM of 4/6,6/9,5/4.
2) Find HCF of 4/6,3/9,6/5.
1) Find LCM of 4/6,6/9,5/4.
2) Find HCF of 4/6,3/9,6/5.
PLEASE EXPLAIN PROCESS>>>
set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A
a)33 b)32c)31d)30
a)33 b)32c)31d)30
capacity u have assumed as 24 l randomly or on some obeservation?
It can be written as (100 + 1)(100 + 2)(100 + 3)(200 - 1)(200 - 2)(200 - 3). The last two digits of the product of the first two terms will be 06 (100n + 6) The last two digits of the product of the last two terms will be 94 (200n - 6) The product of 94 and 6 is 564. Hence the last two ...
@[590022:yudh]
The answer is 0m. They both will meet at point P for the fifth time...
The answer is 0m. They both will meet at point P for the fifth time...
1)we have to find out remainder when divide by 100
1*2*3*-1*-2*-3remainder=-36 or 64
so last two digits=64
1*2*3*-1*-2*-3remainder=-36 or 64
so last two digits=64
1