@gnehagarg said:(2,33)(4,33)(8,33)(1,66)(1,132)(1,252)(3,88)(11,24)8 solutions
Ans is 4...
Read the question again
Read the question again
@pirateiim478 said:Two pipes A and B would fill a cistern in 37.5 minutes and 36 minutes respectively. Both pipes being opened, find when the 2nd pipe must be turned off, so that the cistern may be filled in half an hour?
Let the time by x
So 30/37.5 + x/36 = 1 => 4/5 + x/36 = 1 => x = 7.2minutes
So 30/37.5 + x/36 = 1 => 4/5 + x/36 = 1 => x = 7.2minutes
@pirateiim478 said:Two pipes A and B would fill a cistern in 37.5 minutes and 36 minutes respectively. Both pipes being opened, find when the 2nd pipe must be turned off, so that the cistern may be filled in half an hour?
Let total work = 900
A- work/min = 24
B - Work/min = 25
given condition -
49x + 24(30-x) = 900
x = 7.2 min
By giving one rubber free with 4 pencils, it means that a discount of 10% is given on the sale of pencils. Then by giving 1 pencil free with 6 rubbers, it means that a discount of x% is given on the sale of rubbers. Then x (approximately) equals (1) 57% (2) 61% (3) 64% (4) 73% (5) none of these
@The_Loser
@anytomdickandhary said:1080 = 135*8 = 5*27*8 = (2^3)*(3^3)*(5^1)hence total number of factors = (3+1)*(3+1)*(1+1) = 32for find the square factors we write 1080 as 4*9*(2*3*5) .now square factors can come only by using 4 and 9 hence number of square factors = (1+1)*(1+1) = 4hence required no. = 32 - 4 = 28.ATDH.
Thank you very much.
I had got the answer with a slightly different method.
1080 = 2^3*3^3*5^1
Total no.of factors = 4*4*2 = 32
For getting the square factors, we can only include even powers + zero powers
So, 1080 = 2^3*3^3*5^1
= (2^0+2^1+2^2+2^3)(3^0+3^1+3^2+3^3)(5^0+5^1)
becomes
= (2^0+2^2)(3^0+3^2)(5^0)
No.of square factors = 2*2*1 = 4
No.of non square factors = 32 - 4 = 28
I was just wondering if their is a direct way for solving this, in which we don't need subtraction at the end i.e. if we can directly calculate no.of non square factors as we directly calculated no.of square factors.
I had got the answer with a slightly different method.
1080 = 2^3*3^3*5^1
Total no.of factors = 4*4*2 = 32
For getting the square factors, we can only include even powers + zero powers
So, 1080 = 2^3*3^3*5^1
= (2^0+2^1+2^2+2^3)(3^0+3^1+3^2+3^3)(5^0+5^1)
becomes
= (2^0+2^2)(3^0+3^2)(5^0)
No.of square factors = 2*2*1 = 4
No.of non square factors = 32 - 4 = 28
I was just wondering if their is a direct way for solving this, in which we don't need subtraction at the end i.e. if we can directly calculate no.of non square factors as we directly calculated no.of square factors.
@gnehagarg said:(n+12)^2-4=k^2-13^2(n+14)*(n+10)=(k+13)*(k-13)n=odd k=evenn=even k=oddno value satisfies it
try again. Ans is 2........
@bs0409 said:By giving one rubber free with 4 pencils, it means that a discount of 10% is given on the sale of pencils. Then by giving 1 pencil free with 6 rubbers, it means that a discount of x% is given on the sale of rubbers. Then x (approximately) equals (1) 57% (2) 61% (3) 64% (4) 73% (5) none of these
We can say (4p+r)(90/100) = 4p = > p/r = 9/4
We need , (p+6r) (x/100) = 6r => x/100 = (6r) / (p+6r) = 24/33 (Say p=9 and r=4)
=>x = 72.7% = 73(appr)
We need , (p+6r) (x/100) = 6r => x/100 = (6r) / (p+6r) = 24/33 (Say p=9 and r=4)
=>x = 72.7% = 73(appr)
In how many ways can 11 distinct toys given to 4 children so that each child gets atleast 1 toy?
The numbers a,b,c are in Arithmetic Progression while a-k,b,c+k are in geometric progression, with common ratio r. Which of the following is true?
a) r b) -1 c) 0 d) 1
if a=b=c=1 Are they in AP???doubt??
a) r b) -1 c) 0 d) 1
if a=b=c=1 Are they in AP???doubt??
In an arithmetic progression, the sum of the first p terms is q and the sum to the first q terms is p. What is the sum to (p + q) terms?
@vbhvgupta said:In an arithmetic progression, the sum of the first p terms is q and the sum to the first q terms is p. What is the sum to (p + q) terms?
0
@vbhvgupta said:The numbers a,b,c are in Arithmetic Progression while a-k,b,c+k are in geometric progression, with common ratio r. Which of the following is true? a) r c) 0 if a=b=c=1 Are they in AP???doubt??
Yes 1,1, 1 can be considered as both AP and GP
@vbhvgupta said:In an arithmetic progression, the sum of the first p terms is q and the sum to the first q terms is p. What is the sum to (p + q) terms?
It will be -(p+q).
@pirateiim478 said:In how many ways can 11 distinct toys given to 4 children so that each child gets atleast 1 toy?
a+b+c+d= 7 as each min=1
7 distinct toys hence proceeding as below
0 0 0 7 ->4 ways
0 0 1 6 ->4!/2! ways
0 0 2 5 ->4!/2! ways
0 0 3 4 ->4!/2! ways
2 2 2 1 -> 4 ways
3 1 2 1 ->4!/2! ways
0 1 1 5 ->4!/2! ways
0 2 1 4 -> 4! ways
0 3 1 3 -> 4!/2! ways
Add all the numbers
Bet i wouldve missed something :splat:..
will edit this post in a while
How many natural number not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat.
@rkshtsurana said:can be yaar not sure...sum of digit is 30..so may be a five digit multiple of 3... but digit sum is 3..so cannot be
Look closely dost - sum of digits is 30 as you said....so it is divisible by 3 but not by 9. So cannot be a perfect square as any perfect square divisible by 3 will also be divisible by 9!
regards
scrabbler
regards
scrabbler
@vbhvgupta said:How many natural number not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat.
4 Digit nos= 229
1xxx -> 64 nos
2xxx -> 64
3xxx-> 64
41xx ->16
42xx ->16
431x ->4
4321 ->1
3 Digit Nos=64
2 Digit Nos = 16
1 Digit Nos = 4
229+64+16+4= 313