p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

Cant be determined,
Check the case of 8^2, it both a square as well as a cube(of 4)
Also check 729 which is 9^3, its both cube and square(of 27)

Waise to 6 hona chahiye par agar p 1 hua to q kuch bhi ho skta hai..
so wl go with cbd

check with two three examples.. 64 , 81 , 729

64 can be 2^6 or 4 ^3 or 8^2

81 can be 3^4 or 9^2

729 can be 9^3 or 27^2

so.. cannot be determined..

my take 2

p^q = 2^12 or 4^6 or 8^4 or 16^3 or 64^2

hence q can take values 12,6,4,3,2.

hence option a

it should be 6

question says it must be a multiple of .... means it should satisfy all values of p and q

so q = 2k' as well as 3k"

so q is multiple of both 2 and 3, means 6

2 does not satisfy all values. for ex 3^2 is square but not cube similarily 3^3 is cube but not square

so 6 should be ans (as it satisfies all values of p and q)

OA given: CBD
bt i think it shud be 6,,,

8. How many two-digit numbers have exactly four factors?
(a) 30 (b) 31 (c) 32 (d) 28

Ans is d) 28

Q5Find the two natural nos whose diff is 66 & LCM is 360
a)120;54 b)90;24 c)180;114 d)130;64 e)none
any formula to solve the problem?

take 2 numbers a*x and a*y (a and y are coprime) ax > ay

ax - ay = 66
a(x-y) = 66
x-y = 66/a

LCM = a*x*y = 360
x*y = 360/a

a should be a factor of both 66 and 360
common factor of 66 ans 360 = 1 and 6

if a = 1
then x-y = 66 and x*y = 360
x(x-66) = 360
x^2-66x-360 = 0
no integral value of x

now take a = 6
x-y = 11 and x*y = 60
=> x(x-11) = 60
=> x^2-11x-60 = 0
x = 15 (only positive value)
then y = 60/15 = 4

so 2 numbers will be 6*15 and 6*4 = 90 and 24

how many kilograms of sugar worth 3.6/kg should be mixed with sugar worth 4.2/kg such that by selling the mixture at 4.40/kg there may be profit of 10%

1) 6kg 2) 3kg 3)2kg

please explain the process.

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

Cant be determined,
Check the case of 8^2, it both a square as well as a cube(of 4)
Also check 729 which is 9^3, its both cube and square(of 27)

Waise to 6 hona chahiye par agar p 1 hua to q kuch bhi ho skta hai..
so wl go with cbd

check with two three examples.. 64 , 81 , 729

64 can be 2^6 or 4 ^3 or 8^2

81 can be 3^4 or 9^2

729 can be 9^3 or 27^2

so.. cannot be determined..

my take 2

p^q = 2^12 or 4^6 or 8^4 or 16^3 or 64^2

hence q can take values 12,6,4,3,2.

hence option a

it should be 6

question says it must be a multiple of .... means it should satisfy all values of p and q

so q = 2k' as well as 3k"

so q is multiple of both 2 and 3, means 6

2 does not satisfy all values. for ex 3^2 is square but not cube similarily 3^3 is cube but not square

so 6 should be ans (as it satisfies all values of p and q)

OA given: CBD
bt i think it shud be 6,,,

In the month of January, what is the ratio of probability of someones birthday falling on a date that is a prime
number to the probability of the birthday falling on a date that is perfect square of a prime number? (It is known
that the birthday falls in the month of January.)

bhai is it 11/2
bday falling on prime number prob= 11/31
prob of bday falling on square of prime (3,5) =2/31

In the month of January, what is the ratio of probability of someones birthday falling on a date that is a prime
number to the probability of the birthday falling on a date that is perfect square of a prime number? (It is known
that the birthday falls in the month of January.)

prime numbers : 2 3 5 7 11 13 17 19 23 29 31 = 11
square of primes 4 9 25 = 3

ratio = 11/3

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

it should be 6

question says it must be a multiple of .... means it should satisfy all values of p and q

so q = 2k' as well as 3k"

so q is multiple of both 2 and 3, means 6

2 does not satisfy all values. for ex 3^2 is square but not cube similarily 3^3 is cube but not square

so 6 should be ans (as it satisfies all values of p and q)

Q5Find the two natural nos whose diff is 66 & LCM is 360
a)120;54 b)90;24 c)180;114 d)130;64 e)none
any formula to solve the problem?

8. How many two-digit numbers have exactly four factors?
(a) 30 (b) 31 (c) 32 (d) 28

A test has 50 qstns....1 mark for correct answer , -1/3 for wrong answer and -1/6 for every unattempted qstn.A student scored 32,Atleast how many qstns can he mark wrong?
a)9 b)12 c)6 d)3

Conceptually samjha do yaar and how to solve these kind of qstns algebraically???

let x be correct ques
y be incorrect and z be unattempted ques
x+y+z = 50
x-y/3-z/6=32

solving these two eq u will get 7x = y+242
now use options u will get 3 as answer

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

check with two three examples.. 64 , 81 , 729

64 can be 2^6 or 4 ^3 or 8^2

81 can be 3^4 or 9^2

729 can be 9^3 or 27^2

so.. cannot be determined..

In the month of January, what is the ratio of probability of someone's birthday falling on a date that is a prime
number to the probability of the birthday falling on a date that is perfect square of a prime number? (It is known
that the birthday falls in the month of January.)

bro,
pls clarify colored line
lower limit of x-y=lower of x-lower limit of y
=2-1=1
so ratio is 7:1
am i right?

as y is -ve so for lower limit of x-y we need to take min of x and max of y, so 2-3 = -1

and if lower limit is coming -1 then 1 can not be correct as -1 < 1

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

my take 2

p^q = 2^12 or 4^6 or 8^4 or 16^3 or 64^2

hence q can take values 12,6,4,3,2.

hence option a

A test has 50 qstns....1 mark for correct answer , -1/3 for wrong answer and -1/6 for every unattempted qstn.A student scored 32,Atleast how many qstns can he mark wrong?
a)9 b)12 c)6 d)3

Conceptually samjha do yaar and how to solve these kind of qstns algebraically???

x correct, y wrong, z unattempted

x -y/3 - z/6 = 32
6x - 2y - z = 192 ...(1)

and x+y+z = 50
or 6x+6y+6z = 300 ..(2)

subtract 1 from 2

it becomes 8y + 7z = 108
we need to minimize y

y = (108-7z)/8

z can not be more than 15
so check by taking values of z
at z = 12, y = 3 (that is least value of y)

so ans is 3

in 2 < x < 4
upper limit of x it 4 and lower limit is 2

come to question

upper limit of x+y = upper limit of x + upper limit of y = 4+3 = 7

lower limit of x-y = lower limit of x - upper limit of y = 2-3 = -1

so ratio will be 7 : -1 = 7/-1 = -7

bro,
pls clarify colored line
lower limit of x-y=lower of x-lower limit of y
=2-1=1
so ratio is 7:1
am i right?

shashi khawani Says

minimum value of x-y = -1 and not 1... Please correct me if worng..

Absolutely, its my mistake. Thus none of these option? Is that how its done? We get answer as -7

Hi

How did u find the remainder of 13!-1! mod 13?

R(13!-1!)/13

In this case 13! has 13 hence its perfectly divisible by 13, no remainder, now we move on to the next term, -1, it leaves a remainder, ie 13-1=12. not sure how much of this u understood, i had the same doubt

Check Arun Sharma quant book for more clarity ATB

shashi khawani Says

minimum value of x-y = -1 and not 1... Please correct me if worng..

well ,as per arun sharma,the answer is e)none of above

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

Waise to 6 hona chahiye par agar p 1 hua to q kuch bhi ho skta hai..
so wl go with cbd

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

Cant be determined,
Check the case of 8^2, it both a square as well as a cube(of 4)
Also check 729 which is 9^3, its both cube and square(of 27)

p^q is a perfect square as well as a perfect cube, (p&q; are natural nos.), then q must b a multiple of
a.2 b.3 c.6 d.9 e.can't be determined

Wonder if it simply means, max value eof x+y and min value of x-y, if so then,

max value of x+y putting x=4&y;=3 is 7 & min value is for x=2&y;=1, ratio 7/1=7

Is that the answer?

minimum value of x-y = -1 and not 1... Please correct me if worng..

Q55)If 2<4 and 1<3,then find ratio of the upper limit for x+y and lower limit of x-y?
a)6 b)7 c)8 d)4 e)none
What is meaning of upper & lower limit?

in 2 < x < 4
upper limit of x it 4 and lower limit is 2

come to question

upper limit of x+y = upper limit of x + upper limit of y = 4+3 = 7

lower limit of x-y = lower limit of x - upper limit of y = 2-3 = -1

so ratio will be 7 : -1 = 7/-1 = -7

1*1!+2*2!+....+12*12! = 1!*(2-1) + 2!(3-1) + 3!(4-1) + ... + 12!(13-1)

= 2! - 1! + 3*2! - 2! + 4*3! - 3! + ... + 13*12! - 12!

Every term will be cancelled except 13! - 1!

=> 13! - 1! mod 13 = -1 or 12

Hi

How did u find the remainder of 13!-1! mod 13?

Q55)If 2<4 and 1<3,then find ratio of the upper limit for x+y and lower limit of x-y?
a)6 b)7 c)8 d)4 e)none
What is meaning of upper & lower limit?

Upper limit is max value
lower limit is min value
so m getting ans as 7...

Q55)If 2<4 and 1<3,then find ratio of the upper limit for x+y and lower limit of x-y?
a)6 b)7 c)8 d)4 e)none
What is meaning of upper & lower limit?

Wonder if it simply means, max value eof x+y and min value of x-y, if so then,

max value of x+y putting x=4&y;=3 is 7 & min value is for x=2&y;=1, ratio 7/1=7

Is that the answer?

Q55)If 2<4 and 1<3,then find ratio of the upper limit for x+y and lower limit of x-y?
a)6 b)7 c)8 d)4 e)none
What is meaning of upper & lower limit?

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

a+b = 15
or a = 15-b

AM = 15/2 = 7.5
GM = 7.5*80/100 = 6

GM = rt(ab) = 6
ab = 36

putting a = 15-b

b(15-b) = 36
15b-b^2 = 36
b^2-15b+36 = 0
(b-12)(b-3) = 0
b = 12, 3

so b = 12, a = 3 or vice versa

so 2 numbers are 3 and 12

why couplw is aa and bb?? if we assume m1 w1 as one couple and m2 w2 as 2nd couple then u r nt taking sm arrangement lyk
m1 w1 m2 w2 x..... 1 arrngement
w1 m1 m2 w2 x... 2 arrangement
bt u r tking both arrangement as 1 .. why so?????

Bhai it means the same... If i take them different... Then total arrangements = 12 * 21 * 2! = 48..

And total = 5! = 120..

Probability = 2/5... I had taken them same to had divided 5! by 2!*2!.

Hope you got it:grin:

1)Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)
(log 2 = 0.3010)
(a) 46 (b) 47 (c) 48 (d) 49

question-take log 20^-37=-37log20=-37(log2+log10)=-37(1.3010)=-48.137

concept - The characteristic (in the logarithm of a number) is one more than the number of zeroes to the right of the decimal in a positive number less than 1

now the decimal should be positive therefore -49 + .693

so the no of zeroes will be 1 less than characteristic no..so ans is 48..

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

The nos are 12&3 because the AM of 2 nos whose sum is 15 is 7.5 and 20% lower than 7.5 is 6. Is it not sufficient to just check the answers and see which one gives 6?

couples = aa and bb...

Possibilitites in our favour = abcab, abcba, bacba, bacab, acbab, bcaba, abacb, babca, cbaba, cabab, babac,ababc

Total = 5!/2!2! = 30...
Pobability = 12/30 = 2/5

Bhailog __________________//\\_________________

why couplw is aa and bb?? if we assume m1 w1 as one couple and m2 w2 as 2nd couple then u r nt taking sm arrangement lyk
m1 w1 m2 w2 x..... 1 arrngement
w1 m1 m2 w2 x... 2 arrangement
bt u r tking both arrangement as 1 .. why so?????

Q32)Four bells ring at regular intervals of 6,8,12,18 sec.They start ringing together at 12clock.How many times they will ring together during next 12 min?(including 12 min marks)
a)9 b)10 c)11 d)12 e)none of above

LCM of 6 8 12 and 18 is 72
total time duration is 12 mins = 720 seconds
thus number of times they ll ring together including 12: 00 and 12:!2
= 720/72 + 1 = 11 times

LCM = 72 sec...

Total time they ring together = 12*60 / 72 + 1 (for the 1st time)= 10 + 1 = 11

acc to soln key,correct ans is b

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

Sum = 15
so AM = 7.5
GM is 20% lower than AM
So GM = 6
check with options where a and b when multiplied gives 36 and hence a GM of 6

else solve it as
a+b = 15
ab = 36
a = 12 and b = 3

Datta Bhai, I think you didnt get the question... I may be wrong...

I think you took it that the 2 coupens should not be together.. But the question is that the people of a couple should not be together...

if aa is a couple, then a and a should be be besides each other...

you took aa =1.. Measn in all your cases, the pair is always together..

Hope you got it.:grin:

Eisa???? I totally mis-interpreted

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

This is easier with options

Answer = 12,3...

Bs bhai, Varun, Chill sir and other bhailog.. Iska koi method hai kya?

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

answer is b...just check for which value ab is a perfect square u will get your answer.

Q32)Four bells ring at regular intervals of 6,8,12,18 sec.They start ringing together at 12clock.How many times they will ring together during next 12 min?(including 12 min marks)
a)9 b)10 c)11 d)12 e)none of above

LCM = 72 sec...

Total time they ring together = 12*60 / 72 + 1 (for the 1st time)= 10 + 1 = 11

Q32)Four bells ring at regular intervals of 6,8,12,18 sec.They start ringing together at 12clock.How many times they will ring together during next 12 min?(including 12 min marks)
a)9 b)10 c)11 d)12 e)none of above

1)Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)
(log 2 = 0.3010)
(a) 46 (b) 47 (c) 48 (d) 49

log x = -37 * 1.3010 = -48.137 = -49 + 0.863...

So there are 48 zero?

Q4)The sum of two numbers is 15 and their geometric mean is 20% lower than their A.M.Find the number
a)11,4 b)12,3 c)13,2 d)10,5 e)9,6

dont u think it should be...

couples aa-1 bb-2 single c-3

total ways
123 132 213 231 321 312
favourable when 3 in middle
so 2/6....just asking :)

bcoz condition is just couples shoudnt sit together....Are we concerned with how the couples arrange themselves when they sit????

Datta Bhai, I think you didnt get the question... I may be wrong...

I think you took it that the 2 coupens should not be together.. But the question is that the people of a couple should not be together...

if aa is a couple, then a and a should be be besides each other...

you took aa =1.. Measn in all your cases, the pair is always together..

Hope you got it.:grin:

1)Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)
(log 2 = 0.3010)
(a) 46 (b) 47 (c) 48 (d) 49

couples = aa and bb...

Possibilitites in our favour = abcab, abcba, bacba, bacab, acbab, bcaba, abacb, babca, cbaba, cabab, babac,ababc

Total = 5!/2!2! = 30...
Pobability = 12/30 = 2/5

Bhailog __________________//\\_________________

dont u think it should be...

couples aa-1 bb-2 single c-3

total ways
123 132 213 231 321 312
favourable when 3 in middle
so 2/6....just asking :)

bcoz condition is just couples shoudnt sit together....Are we concerned with how the couples arrange themselves when they sit????

Foster87 Says

what are the last two digits of (1/5^903)??(approach please)

is it 05?????

2 couples and a single person is to be seated in 5 chairs.What is the probability that no couple is to be seated together?
Answer 2/5 kaise hai...

1 way is when single person sits in the middle chair??? what is the 2nd arrangement....

couples = aa and bb...

Possibilitites in our favour = abcab, abcba, bacba, bacab, acbab, bcaba, abacb, babca, cbaba, cabab, babac,ababc

Total = 5!/2!2! = 30...
Pobability = 12/30 = 2/5

Bhailog __________________//\\_________________

E(153) = 96

thus, 128^960/153 = 1

we r left with 128^40/153

splitting denominator to find the remainder now ,

153 = 9 x 17

where, 9 gives us a remainder of 7 where 17 gives 1

hence, 9k+7 = 17p +1

therefore, remainder is 52

========================================

bhai logo chinese remainder theoram kaise lagatey hain ??

2 couples and a single person is to be seated in 5 chairs.What is the probability that no couple is to be seated together?
Answer 2/5 kaise hai...

1 way is when single person sits in the middle chair??? what is the 2nd arrangement....

Anand and Bala started from towns P and Q simultaneously towards Q and P respectively. After 6 hours, they crossed each other at town R. After that, Anand took 5 hours more to reach Q than Bala took to reach P. Find the sum of the times taken by them to reach their destinations from R ( in hours).
(A) 15 (B) 11 (C) 13 (D) 17

Total Distance --> 6a+6b
Time taken by B to travel 6a is t
=> 6a/b = t
Time taken by A to travel 6b is (t+5)
=> 6b/a = t+5

So add --> 13

if a four digit number N has 15 factors?
then N^2 has how many factors?

a- 29
b- 30
c- 45
d- none
plz help........!!

N = p^14 or (p^4 * p^2)

Now p^14 > 4 digit number hence not possible

So N = (p^4 * p^2) = 9*5 = 45

So option (c)

if a four digit number N has 15 factors?
then N^2 has how many factors?

a- 29
b- 30
c- 45
d- none
plz help........!!

A no. "N" with 15 factors can be of the form,
(2^14)
(2^4)*(3^2)

But (2^14) is a a no. more than 4 digits.
Thus,we take (2^4)*(3^2)
Now,N^2= {(2^4)*(3^2)}^2
(2^8 )*(3^4)

Thus,no. of factors=(8+1)*(4+1)=45(option c)

Foster87 Says

what are the last two digits of (1/5^903)??(approach please)

getting ans as 08
rewrite.it as 2^903/10^903
now 2^903= (1024)^90*8
so last 2 digits wl b.. 76*8 = 08

if a four digit number N has 15 factors?
then N^2 has how many factors?

a- 29
b- 30
c- 45
d- none
plz help........!!

I got 45


15 can be written as 5X3
There the powers will be 5-1,3-1 =4,2
Squaring powers will become 8,4
The no of factors will be 9X5=45

If the number has only one prime number then
Power will be 15-1=14
Squaring 14X2=28
No of factors=28+1=29

Due to 4 digit restriction will be 45 only.

Anand and Bala started from towns P and Q simultaneously towards Q and P respectively. After 6 hours, they crossed each other at town R. After that, Anand took 5 hours more to reach Q than Bala took to reach P. Find the sum of the times taken by them to reach their destinations from R ( in hours).
(A) 15 (B) 11 (C) 13 (D) 17

Bheja fry ho gaya yeh karte karte...

Jara method bhi bata dena...

form 2 eqn..
6a = bx .... 1
6b = a(x+5) ... 2
so put value of a in terms of b from eq 1
u get.. x^2+5x-36 =0
so x is 4 and x+5 is 9
sum = 13

if a four digit number N has 15 factors?
then N^2 has how many factors?

a- 29
b- 30
c- 45
d- none
plz help........!!

If a+b+c=24,then the maximum value of (a)^2*(b)^3*(c) is

A) (2^3)*(3^10)
B) (6^3*2^10)
C) (6^3*2^8 )
D) (2^14*3^3)

here the expression is a^2*b^3*c....
hence we will write a+b+c=24 as
a/2+a/2+b/3+b/3+b/3+c=24.

to have maximum value,each f the no's should be 4...i.e..a/2=4=>a=8,b/3=4=>b=12,c=4
hence the maxm value is 2^14*3^3:)