Official Quant thread for CAT 2013

MBA CAT 2024
9161
@chillfactor said:
C(16, 6) or C(16, 10) is correct for the marks question (as explained by@Angadbir and @scrabbler )Let S be the set of proper three digit positive integers less than 1000. What is the minimum number of integers that must be chosen from S to ensure that there are at least two integers which have at least one digit in common?
10 numers ..with any frm 100 to 199,200-299,.............,aage tak


9162
@chillfactor said:
C(16, 6) or C(16, 10) is correct for the marks question (as explained by@Angadbir and @scrabbler )Let S be the set of proper three digit positive integers less than 1000. What is the minimum number of integers that must be chosen from S to ensure that there are at least two integers which have at least one digit in common?

is it 10 ?

111, 222, 333, 444, 555, 666, 777, 888, 999 next number will have common digits



9163
@sujamait said:
The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?OPTIONS1) 66 2) 67 3) 68 4) None
Take out 6 multiples of 6.Now,left integers are 30..There are 31 spaces.Place 6 integers in these 31 spaces
30!*31C6

67
9164
@chillfactor said:
C(16, 6) or C(16, 10) is correct for the marks question (as explained by@Angadbir and @scrabbler )Let S be the set of proper three digit positive integers less than 1000. What is the minimum number of integers that must be chosen from S to ensure that there are at least two integers which have at least one digit in common?
total 3 digit numbers with all different digits=9 —9 —8=648,now to get atlest 2 numbers with common digits,648+2=650 numbers shud be selected.
not sure though ,chill sir ka questn hai
9165
@sonamaries7 said:
The HM of 2 nos to their GM is 24:25. FInd the ratio of the nos.2/3,3/24/3,3/44/9,9/416/9, 9/16
2ab/(a+b)sqrt(ab) = 24/25
a/b=k
2k(sqrt(k)(k+1))=24/25
k=16/9 or 9/16
9166


@sonamaries7 said:
The HM of 2 nos to their GM is 24:25. FInd the ratio of the nos.2/3,3/24/3,3/44/9,9/416/9, 9/16
root(ab)/a+b=12/25
clearly,a=16,b=9
9167
@chillfactor said:
10 is correct Find all integers a, b, c such thata² + b² + c² = 2(bc + 1) and a + b + c = 2002
Getting 4 sets....shayad kuch galat kiya...2 with a = 1 and 2 with a = -1

(Goes to recheck)

Edit:
1, 1001, 1000
1, 1000, 1001
-1, 1002, 1001
-1, 1001, 1002
Not sure if all satisfy all conditions...
@chillfactor sirji OA?

regards
scrabbler
9168
@sonamaries7 said:
The HM of 2 nos to their GM is 24:25. FInd the ratio of the nos.2/3,3/24/3,3/44/9,9/416/9, 9/16
Numbers are a and b
HM = 2ab/(a + b)
GM = √(ab)

√(ab)/(a + b) = 12/25
625ab = 144(a + b)²
144a² + 144b² - 337ab = 0
144a² - 256ab - 81ab + 144b² = 0
(16a - 9b)(9a - 16b) = 0

So, ratio is 9/16 or 16/9
9169
@chillfactor said:
Find all integers a, b, c such thata² + b² + c² = 2(bc + 1) and a + b + c = 2002
2 pairs,i.e.
(1,1000,1001) and (1,1001,1000)

9170
@akansh_1 said:
2 pairs,i.e.(1,1000,1001) and (1,1001,1000)
appraoch ?

9171
@sonamaries7 said:
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
175=total units
A and B efficiency=7units perday
C efficiency=5 units per day

Let efficiency of A be x units per day
So,B efficiency be 7-x units per day

175/x=175/(7-x+5)
x=12-x
2x=12
x=6

B time =175/1

175 days
9172
@chillfactor said:
10 is correct Find all integers a, b, c such thata² + b² + c² = 2(bc + 1) and a + b + c = 2002


a + b + c = 2002
(a + b + c)^2=a² + b² + c² + 2(ab+bc+ca)
2002^2 = 2bc + 2 + 2ab+2bc+2ca
here i stopped and started another way...
kuch ni ho sakta mera..

a² + b² + c² = 2(bc + 1)
a^2 + (b-c)^2 = 2
as a,b,c are integrs
either
a=+/-1
b-c=+/-1

so,
if i take,
a=1,b=c+1 then we get 1 soln
a=1,b=c-1 .......
..
...

4 solutions ?

@chillfactor nice Q bydway..
9173
@chillfactor said:
10 is correct Find all integers a, b, c such thata² + b² + c² = 2(bc + 1) and a + b + c = 2002
a^2 + (b-c)^2 = 2

Both terms in LHS is non-negative, and none can be 0 (since 2 is not a perfect square)
So both terms = 1

So integer solns are of the nature:
a = +/-1
b = c +/-1

Since a + b + c = 2002, taking a = +1 will yeild 2 solns for b and c
Similarly, 2 solns for b,c when a = -1

So total 4 solns


9174
@Brooklyn : Missed the other two cases.Already explained by @AngadbirďťżAngadbirďťż and @sujamait.
9175
Let N be the smallest positive number such that N^3 ends in 888. Find the sum of digits of N.
OPTIONS

1) 12
2) 17
3) 8
4) 19
5) None of these

Does any one has got his/her hands on PDF unlocker software ?
9176

Yup, 4 solutions :thumbsup:


a(n) denotes a n-digit positive number having all of its digits as 1. Find the greatest common divisor of a(45) and a(140).
9177
@chillfactor said:
Yup, 4 solutionsa(n) denotes a n-digit positive number having all of its digits as 1. Find the greatest common divisor of a(45) and a(140).
11111? Both will be divisible by this...

regards
scrabbler
9178
@chillfactor said:
Yup, 4 solutionsa(n) denotes a n-digit positive number having all of its digits as 1. Find the greatest common divisor of a(45) and a(140).


HCF (11...45time , 1111..140 times)

11111 ?

9179
@sujamait said:
Let N be the smallest positive number such that N^3 ends in 888. Find the sum of digits of N.OPTIONS1) 12 2) 17 3) 8 4) 19 5) None of these Does any one has got his/her hands on PDF unlocker ?
Unit digit has to be 2
say, a2 is the number
(10a + 2)^3 = ...888
1000a^3 + 600a^2 + 120a = ....880
100a^3 + 60a^2 + 12a = ....88

Unit digit of 2a is 8
=> Unit digit of a is 4 or 9

When unit digit of a is 4
b42
(100b + 42)^3 = ...888
Unit digit of 3*42*42*b + 740 is 8
Unit digit of 3*42*42*b is 8
Unit digit of b is 4 or 9

When unit digit of a is 9
b92
(100b + 92)^3 = ...888
Unit digit of 3*92*92*b + 7786 is 8
Unit digit of 3*92*92*b is 2
Unit digit of b is 1 or 6

So, least number will be 192

9180
@sujamait said:
Let N be the smallest positive number such that N^3 ends in 888. Find the sum of digits of N.OPTIONS1) 12 2) 17 3) 8 4) 19 5) None of these Does any one has got his/her hands on PDF unlocker ?
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