All the nos. can be xpressed in the form :
9x,9x+1,9x+2,9x+3,9x+4,9x+5.....9x+9
Since the set contains max. no. of elements, the combination of nos. must be taken in such a way that :
1) It does nt adds up to be a multiple of 9
2) Priority must be given to smaller nos., so that greater no. of elements can be considered. Eg : 9x and 9x +9 add up to giv a multiple of 9, so we neglect 9x+9 instead of 9x.
In 9x+1 and 9x+8, neglect 9x+8
In 9x+2 and 9x+7, neglect 9x+7
In 9x+3 and 9x+6, neglect 9x+6
In 9x+4 and 9x+5, neglect 9x+5
Type of nos. to be considered : 9x , 9x+1, 9x+2 , 9x+3, 9x+4
No. of the form : 9x = 333, 9x+1 = 333, 9x+2 = 333, 9x+3 = 332 , 9x+4 = 332
Total = 1663
If the correct ans. is 1331, I guess we'll hv to neglect nos. of the form 9x+4 so that the total comes out to be 1331...BUT Y?? :/:-(
Hi Puys ,
My Query,
A watch gains by 2% per hour when the temperature is in the range of 40 -50 C and it loses at the same rate when the temperature is in the range of 20 - 30 C . However the watch owner is fortunate since it runs on time in all other temperature ranges . On a sunny day the temperature started soaring up from 8 am in hte morning at the uniform rate of 2 C per hour and sometime during the afternoon it started coming doen at the same rate . Find what time will it be by the watch at 7pm if at 8 am the temperature was 32 C and at 4 pm it was 40C....
a) 6:55 pm
b) 6:55:12 pm
c) 6:55:24 pm
d) None of these
f(x)=f(x-2)-f(x-1), x is a natural number
f(1)=0,
f(2)=1
Find the value of f(
.
Please help me understand. The naswer is 13 as per the book.
Six white and six black balls of the same size are distributed among 10urns so that there is atleast 1 ball in each urn. What are the no. of different distributions of the balls?
a) 25000
b) 26250
c) 28250
d) 13125
(The correct ans option according to the book is b )
Plz explain the approach.
f(x)=f(x-2)-f(x-1), x is a natural number
f(1)=0,
f(2)=1
Find the value of f(.
Please help me understand. The naswer is 13 as per the book.
f(1) = 0
f(2) = 1
f(3) = f(1) - f(2) = -1
f(4) = 2
f(5) = -3
f(6) = 5
f(7) = -8
f(8 ) = 13
So that's your answer.
Hi all,
I have a couple of questions which are relatively simple but am not able to solve. Its a question from the alligations chapter in Arun Sharma's Quantitative Aptitude book. Here goes:
Q1) 400 students appeared for a test. 80% of the girls who appeared for the test passed, while 60% of the boys who appeared for the test cleared. Overall 65% of the students who appeared cleared the test cleared the test. How many girls appeared for the test?
Q2) A barrel contains 125 gallons of a liquid which has 80% Wine and 20% Water. How many gallons of water is to be added to the liquid to make it a 25% water solution?
Thanks in advance.
Please note that I would like to know the solution using the alligation method only if possible..Thanks
Hi all,
I have a couple of questions which are relatively simple but am not able to solve. Its a question from the alligations chapter in Arun Sharma's Quantitative Aptitude book. Here goes:
Q1) 400 students appeared for a test. 80% of the girls who appeared for the test passed, while 60% of the boys who appeared for the test cleared. Overall 65% of the students who appeared cleared the test cleared the test. How many girls appeared for the test?
Q2) A barrel contains 125 gallons of a liquid which has 80% Wine and 20% Water. How many gallons of water is to be added to the liquid to make it a 25% water solution?
Thanks in advance.
Any reasons you want these to be solved by the alligation method?? Personally, I find it to be confusing and prefer the simple method and anyway, it only matters that the problems are solved. If you want I can explain using the normal method.
Hey!
This is a question on Numbers which I'm unable to solve, plz tell me how to go about it:
N=202x20002x200000002x20000000000000002x200...2(31zeros)
The sum of digits in this multiplication will be-
(options)
a)112
b)160
c)144
d)Cannot be determined
Hi all,
I have a couple of questions which are relatively simple but am not able to solve. Its a question from the alligations chapter in Arun Sharma's Quantitative Aptitude book. Here goes:
Q1) 400 students appeared for a test. 80% of the girls who appeared for the test passed, while 60% of the boys who appeared for the test cleared. Overall 65% of the students who appeared cleared the test cleared the test. How many girls appeared for the test?
Q2) A barrel contains 125 gallons of a liquid which has 80% Wine and 20% Water. How many gallons of water is to be added to the liquid to make it a 25% water solution?
Thanks in advance.
for 1st question
A1- 60, A2-80, Aw-65
so u will get n1/n2 = 3:1
n1+n2 = 400 (given)
so n1= 300
n2=100 (which is a answer).
I am unable to use alligation method in second example, but solved by normal method .
Answer = 8.33 gal
Hi all,
I have a couple of questions which are relatively simple but am not able to solve. Its a question from the alligations chapter in Arun Sharma's Quantitative Aptitude book. Here goes:
Q1) 400 students appeared for a test. 80% of the girls who appeared for the test passed, while 60% of the boys who appeared for the test cleared. Overall 65% of the students who appeared cleared the test cleared the test. How many girls appeared for the test?
Q2) A barrel contains 125 gallons of a liquid which has 80% Wine and 20% Water. How many gallons of water is to be added to the liquid to make it a 25% water solution?
Thanks in advance.
see
boys gals
60 80
65
15 5
b/g=3/1
=> 1/4th of 400 = 100 gals
q2)
water pure water
20 100
25
75 5
15 1
i.e 15/1=125/x
x= 125/25....
hopefully m rite..did it in half sleep
Hey!
This is a question on Numbers which I'm unable to solve, plz tell me how to go about it:
N=202x20002x200000002x20000000000000002x200...2(31zeros)
The sum of digits in this multiplication will be-
(options)
a)112
b)160
c)144
d)Cannot be determined
ans: sum of teh digs is the no obtained on div by 9
so here we hav----4.4.4.4.4.4= 1024 sum is 7
check frm options only 160 gives sum of digs as 7 hence the ans
1. a number xy is multipied by another number ab and the result comes as pqr, where r=2y, q=2(x+y) and p=2x, where x,ya. 11
b.13
c.31
d.22
e. none of these
2. the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110
3. there is a natural number that becomes equal to the sqaure of a natural number when 100 is added to it, and to the square of another natural number when 169 is added to it. find the number?
4. find two three digit numbers whoes sum is multiple of 504 and the quotient is multiple of 6.
5. define a number K such that it is the sum of squares of the first M nartural numbers
6. M is a two digit number which has the property that: The product of factorials of it's digit> sum of factorials of its digits.
how mant value of M exists
a. 56
b. 64
c. 63
d. none of these
7. if you form a subset of integars chosen from between 1 to 3000, such that no two integers add up to multiple of nine, what can be the maximum number of elements in the subset
a.1668
b.1332
c.1333
d.1334
8. K is a three digit number such that the ration of the number to the sum of its digit is least.
how many values of K will the ratio be the highest
a. 9
b. 8
c. 7
d. none of these
9. a triangular number is defiened as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000 have the property that they are the difference of squares of two consecutive natural numbers
a. 20
b. 21
c. 22
d. 23
10. what is the highest power of 3 available on the expression 58!-38!
a.17
b.18
c.19
d.none of these
11. a set S is formed by including some of the first One thousand natural numbers. S contains the maximum numbers such that they satisfy the following conditions
1. no number of set S is prime
2. when the numbers of the set S are selected two at a time, we always see a co prime numbers
what is the number of elements in the set S
a. 11
b 12
c. 13
d. 7
Nice finally found quant link...............
1. how many integer value of x and y are such that 4x+7y=3, while xa. 144
b. 141
c. 143
d. 142
2. what is the reminder when 2(8!)-21(6!) divides 14(7!)+14(13!)
a. 1
b.7!
c. 8!
d. 9!
3. arun, bikas and chetakar have a total of 80 coins among them. arun triples the number of coins with the others by giving them some coins from his own collection. next bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at teh begning.
a.11
b.10
c.9
d.12
4. the super computer at Ram Mohan Roy seminar takes an input of a number N and a X where X is a factor of the number N. in a particular case N is equal to 83p796161q and X is equal to 11, where 0
1. how many integer value of x and y are such that 4x+7y=3, while xa. 144
b. 141
c. 143
d. 142
2. what is the reminder when 2(8!)-21(6!) divides 14(7!)+14(13!)
a. 1
b.7!
c. 8!
d. 9!
3. arun, bikas and chetakar have a total of 80 coins among them. arun triples the number of coins with the others by giving them some coins from his own collection. next bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at teh begning.
a.11
b.10
c.9
d.12
4. the super computer at Ram Mohan Roy seminar takes an input of a number N and a X where X is a factor of the number N. in a particular case N is equal to 83p796161q and X is equal to 11, where 0
1.ANS 143
y=(3-4x)/7
Case 1: when x is +ve then y is -ve
x=6 then y= -3
here x increment by 7 and Y increment by -4
thus x set = {6,13.....upto 496} i.e n(x)=71 note use AP to find out number of terms
Case2 : when x is -ve then y is +ve
x={-1,-8,.......upto -498 } i.e n(x)=72
hence total 71+72=143
2.Ans 7!
2(8!)-21(6!) = 16(7!) - 3(7!) = 13(7!)
14(7!)+14(13!) = 13(7!) + 7! + 14!
rem(13(7!)/13(7!) + 7!/13(7!) + 14!/13(7!)) = 0+7! +0 =7!
3.My ans is 20
let suppose arun is having x coins,bikas y and chetakar z
x+y+z =80
Case1 : After arun gives some a coins
(a-x) + 3b + 3c = 80 ----------------1
Case 2: After bikas gives back b coins
3(x-a) + (3y -b) + 3z = 80
3(x-a) + 20 + 9z = 80 : 3y-b =20 given
3(x-a) + 9z= 60
=> (x-a) + 3z = 20 -----------2
eq 1 and 2
y=20
4. 9
(3+7+6+6+q)-(8+p+9+1+1)=q-p+3
rem(q-p+3)/11 =0
now given that q>p
therefore
q-p+3>3
Now the max value that could occur is 11
For this q=9 and p=1
rem(N/p+q) = rem(N/10) =9
rem(N/p) = rem(N/1) = 0
rem = 9+0 =9
5. Ans 3
1 to 9 i.e number of digits is 9
10 t0 99 number of digits 90 *2 =180
100 t0 999 -----> 900 *3 = 2700
1000 t0 9999 ---> 9000 * 4 =36000
9 + 180 + 2700 = 2889
28383 - 2889 = 25494 is the remaining terms and these terms lies in 4 digits unit place number system
hence 25494/4 = (q=6373 and rem(2))
hence 6373 terms is = 1000 + (6373-1) *1 = 7372
and next term is 7373
1234.......7372 7373
hence ans is 3
1. a number xy is multipied by another number ab and the result comes as pqr, where r=2y, q=2(x+y) and p=2x, where x,ya. 11
b.13
c.31
d.22
e. none of these
2. the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110
3. there is a natural number that becomes equal to the sqaure of a natural number when 100 is added to it, and to the square of another natural number when 169 is added to it. find the number?
4. find two three digit numbers whoes sum is multiple of 504 and the quotient is multiple of 6.
5. define a number K such that it is the sum of squares of the first M nartural numbers
6. M is a two digit number which has the property that: The product of factorials of it's digit> sum of factorials of its digits.
how mant value of M exists
a. 56
b. 64
c. 63
d. none of these
7. if you form a subset of integars chosen from between 1 to 3000, such that no two integers add up to multiple of nine, what can be the maximum number of elements in the subset
a.1668
b.1332
c.1333
d.1334
8. K is a three digit number such that the ration of the number to the sum of its digit is least.
how many values of K will the ratio be the highest
a. 9
b. 8
c. 7
d. none of these
9. a triangular number is defiened as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000 have the property that they are the difference of squares of two consecutive natural numbers
a. 20
b. 21
c. 22
d. 23
10. what is the highest power of 3 available on the expression 58!-38!
a.17
b.18
c.19
d.none of these
11. a set S is formed by including some of the first One thousand natural numbers. S contains the maximum numbers such that they satisfy the following conditions
1. no number of set S is prime
2. when the numbers of the set S are selected two at a time, we always see a co prime numbers
what is the number of elements in the set S
a. 11
b 12
c. 13
d. 7
the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110
11. a set S is formed by including some of the first One thousand natural numbers. S contains the maximum numbers such that they satisfy the following conditions
1. no number of set S is prime
2. when the numbers of the set S are selected two at a time, we always see a co prime numbers
what is the number of elements in the set S
a. 11
b 12
c. 13
d. 7
***************
2.1090=9*11*10
11 is highest so we have 1090/11 + 1090/121=99+9=108(take quotitent)
11.
*I am getting 10 as ans*
see in s u can have only nos tht are coprime nd no prime which means tht if in s we put 25 so no other multiple of5
then we put 36 so no multiple of 2,3,6
similarly 49 then we can put squares of all prime no s less than 1000 as they would be coprime nd not prime so in that way we will see that we have 10 nos
Please explain how to solve:
1.The remainder when no 123456789101112..........484950 is divided by 16 is-
a.3 b.4c.5d.6
ans-d
2. 333^555 + 555^333 is divisible by
a.2 b.3 c.37 d.11 e.all of these
ans-d
3. a triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000 have the property that they are the difference of squares of two consecutive natural numbers
a. 20
b. 21
c. 22
d. 23
ans=c
4.If 2
ans=d
Please explain how to solve:
1.The remainder when no 123456789101112..........484950 is divided by 16 is-
a.3 b.4c.5d.6
ans-d
2. 333^555 + 555^333 is divisible by
a.2 b.3 c.37 d.11 e.all of these
ans-d
ans=d
1)divisibility test for 16 is that the last 4 number must be divisible by 16..
4950/16=6 as reminder
2)well this question has been asked many a times. The problem should be not divisible by becoz its divisible by all except 11
(333^555 + 555^333)/11
(3^555+5^333)/11
3^(5x111)+5^(3x111) by 11
=9