Number System - Questions & Discussions

MBA CAT 2022
161
is the ans 24?
My approach...
for the sum of coeficients to be prime the roots shud be even:)
let the quadratic eqn be (x-a)(x-b) = 0
also given that for sum integer the value is -55 = -5 * 11
from the options let the roots be 20 n 4. we can see tht if we subsitute 9 we get (9-4)(9-20) = 5*-11 = -55
hence all conditions are satisfied here...
so the sum of the roots must be 20+4 = 24.
PS-Did a bit of hit n trial before taking the values:)


dude can u please explain the bolden logic above ?
162

you are right...
actually i have not understud the question properly now getting it a little bit but very confusing language so probably its not correct
do this two

17).If the integers
m and n are chosen at random from integers from integers
1 to 100 with replacement, then the probability that a no. of the form
7^m + 7^n is divisible by 5 equals?
a)1
2 b)1
4 c)1
8 d) 1
16 e) none of these

1

abcd is a 4-digit number in base of 7 such that 2(abcd) = bcda (a, b not equal to 0).
Then a is
a. 1 b. 2 c. 3 d. Data insufficient

163
randeepsingh Says
dude can u please explain the bolden logic above ?

its not a set rule...as we knw p shud be in the form of 6k+1 or 6k-1 but doesnt mean all 6k+1 or 6k-1 are prime. same applies here. whn u are multiplying such tht(x-even)(x-even) = x^2 -2even + even^2 hence the sum of coeficitents turns out to be odd(not neccessarily prime but as all prime except 2 are odd so we can conclude tht the roots shud be even for the sum of coeficients to be prime.
If i m goin wrong sumwhere pls correct me...
thnxx
164
shanks4mba Says
koool...in tht case even 20,4 is a case...:)

in that case the sum of co efficients will be 57 i believe
165
its not a set rule...as we knw p shud be in the form of 6k+1 or 6k-1 but doesnt mean all 6k+1 or 6k-1 are prime. same applies here. whn u are multiplying such tht(x-even)(x-even) = x^2 -2even + even^2 hence the sum of coeficitents turns out to be odd(not neccessarily prime but as all prime except 2 are odd so we can conclude tht the roots shud be even for the sum of coeficients to be prime.
If i m goin wrong sumwhere pls correct me...
thnxx

that is what has happend it was not prime
anyways do the next two:-P
166
you are right...
actually i have not understud the question properly now getting it a little bit but very confusing language so probably its not correct
do this two
17).If the integers m and n are chosen at random from integers from integers
1 to 100 with replacement, then the probability that a no. of the form
7m + 7n is divisible by 5 equals?
a)1
2 b)1
4 c)1
8 d) 1
16 e) none of these

1abcd is a 4-digit number in base of 7 such that 2(abcd) = bcda (a, b 6= 0).
Then a is
a. 1 b. 2 c. 3 d. Data insufficient

total no. of ways of choosing two nos. = 100^2
for 7m + 7n to be divisable by 5 if m+n = 5,10,15....200, so total values will be
for m+n=5 -> 4
m+n = 10 ->9 similarly we have total favourable cases = 4+9+14+...199 = 40(4+199)/2= 4060
hence prob =4060/100^2 ?
167
that is what has happend it was not prime
anyways do the next two:-P

DAMN!! i took 57 as prime:banghead:...dunno wat i m doin...thnxx for tellin me:)
168
abcd is a 4-digit number in base of 7 such that 2(abcd) = bcda (a, b 6= 0).


Then a is

a. 1 b. 2 c. 3 d. Data insufficient


MY TAKE : a = 2

Solution

Converting given equation to base 10,

2 ( 343 a + 49 b + 7 c + d ) = 343 b+ 49 c + 7d + a ;

=> 685 a = 245 b + 35 c + 5 d

Values of a,b,c,d can vary from 0 to 6 only because they are in base 7.

Maximum value of RHS of above equation is : 1470 + 210 + 30 = 1710
=> a cannot be greater than or equal to 3

Also , for a = 1 , no possible values of b ,c, d

for a = 2, atleast one set is possible ( a = 2 , b = 5, c = 4 , d = 1)

Hence , a = 2
169
you are right...
actually i have not understud the question properly now getting it a little bit but very confusing language so probably its not correct
do this two
17).If the integers m and n are chosen at random from integers from integers
1 to 100 with replacement, then the probability that a no. of the form
7m + 7n is divisible by 5 equals?
a)1
2 b)1
4 c)1
8 d) 1
16 e) none of these

1abcd is a 4-digit number in base of 7 such that 2(abcd) = bcda (a, b 6= 0).
Then a is
a. 1 b. 2 c. 3 d. Data insufficient


didn't comprehend the second que..can u post it clearly?
170
you are right...
actually i have not understud the question properly now getting it a little bit but very confusing language so probably its not correct
do this two
17).If the integers
m and n are chosen at random from integers from integers
1 to 100 with replacement, then the probability that a no. of the form
7m + 7n is divisible by 5 equals?
a)1
2 b)1
4 c)1
8 d) 1

16 e) none of these

if question is this:-
7^m + 7^n

m = 4k+1......16
n = 4k+3......16

m = 4k+2......16
n = 4k........17

P = (17*16 + 16*16)/10000 = 16*33/10000 = 33/625
171
shanks4mba Says
DAMN!! i took 57 as prime:banghead:...dunno wat i m doin...thnxx for tellin me:)

hey i am really sorry it was 7^m+7^n
i just copy pasted from a pdf file so it was not clear
172
kinscool Says
didn't comprehend the second que..can u post it clearly?

hey it is actually (a,b not equal to 0)
hope now u get it
i am editing the original post
173
if question is this:-
7^m + 7^n

m = 4k+1......16
n = 4k+3......16

m = 4k+2......16
n = 4k........17

P = (17*16 + 16*16)/10000 = 16*33/10000 = 33/625


It is 1/4.
174
MY TAKE : a = 2



Solution

Converting given equation to base 10,

2 ( 343 a + 49 b + 7 c + d ) = 343 b+ 49 c + 7d + a ;

=> 685 a = 245 b + 35 c + 5 d

Values of a,b,c,d can vary from 0 to 6 only because they are in base 7.

Maximum value of RHS of above equation is : 1470 + 210 + 30 = 1710
=> a cannot be greater than or equal to 3

Also , for a = 1 , no possible values of b ,c, d

for a = 2, atleast one set is possible ( a = 2 , b = 5, c = 4 , d = 1)

Hence , a = 2

hey why are you saying that there is no possible value for a=1
i mean if a=1 then we can have b=2,c=5,d=4(245x2+35x5+4x5=490+175+20=685)
in that case
the first number is 1254
that is 1x343+2x49+5x7+4x1=343+98+35+4=480
and
the 2nd number is
2541
i.e.
2x343+5x49+4x7+1x1=686+245+28+1=960
what is wrong with this solution please tell me
175

maskedmenace please explain it
even my answer was 1/4 but the answer in the material is different
so please come out with the solution

176
maskedmenace please explain it
even my answer was 1/4 but the answer in the material is different
so please come out with the solution


7^m + 7^n

m = 1,3,7,9

n = 1,3,7,9

The probability that it is divisible by 5 is 1/4.
177

I want little help in AP : can common difference of an AP be zero?
Because ia m stuck in question which goes lyk this.

x,y,z are real positive numbers which are in arithmetic progression, if xyz=4 find the minimum value of y.

answer is: 2^2/3

I am getting this answer wen x, y,z are equal i.e. common difference is 0,
correct me if i am wrong.

178

help me in solving the following sum:

a natural number N when divided by 10 leaves a remainder of R and quotient of Q. the number N is equal to the sum of the product QR and the sum(Q+R). If N is a multiple of 9 and is less than 10000, ho0w many values can N assume?

a)110
b)112
c)999
d)1000

179
I want little help in AP : can common difference of an AP be zero?
Because ia m stuck in question which goes lyk this.

x,y,z are real positive numbers which are in arithmetic progression, if xyz=4 find the minimum value of y.
answer is: 2^2/3

I am getting this answer wen x, y,z are equal i.e. common difference is 0,
correct me if i am wrong.


Yes CD of AP canbe zero...

Solution xyz = 4

Consider each x=y=z=4^1/3.. We will get the answer
180
help me in solving the following sum: