When a certain tree was first planted,it was 4 ft tall. each year the height increased by a constant amount. at the end of the 6th year, the tree was 1/5 taller than it was at the end of 4th year. by how many feet did the height of the tree increase each year? 1) 3/10 2) 2/5 3) 1/2 4) 2/3 5) 6/5
assuming k as the constant increase..doesnt it lead to 4+5k = 1/5(4+3k) + (4+3k) ?
For this i would rather suggest that we can use the options and come to the answer.So by using 2/3 as the height of tree increase each year we get: Height after 1 year : 14/3 After 2 year: 16/3 After 3 year: 18/3 After 4 year: 20/3 After 5 year: 22/3 After 6 year: 24/3 = 8. Now the height after 6year is 6/5th of height after 4 year i.e. 20/3 * 6/5. So the answer is 2/3.
Can somebody help me out with this problem: Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451
Can somebody help me out with this problem: Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451
Sumit,
Ans - 50 Its tricky but not lengthy theres one 25 n one 2that product will make the last 2 digits 50. now 50 x any odd no. will always leave 50 as last 2 digitsn all the other nos. are odd.
Ans - 50 Its tricky but not lengthy theres one 25 n one 2that product will make the last 2 digits 50. now 50 x any odd no. will always leave 50 as last 2 digitsn all the other nos. are odd.
Regds, Preeti
hi preeti
can you pls be more clear on your explanation. i am unable to understand your logic
its simple there is only number which has 25 as the last two digits and only one number whose last digit is 2 so last two digits when you multiply those two is 50. since rest are all odd numbers which when multiplied give an odd number. ANd any odd number multiplied with a number with last two digits as 50 will have the last two digits as 50.
its simple there is only number which has 25 as the last two digits and only one number whose last digit is 2 so last two digits when you multiply those two is 50. since rest are all odd numbers which when multiplied give an odd number. ANd any odd number multiplied with a number with last two digits as 50 will have the last two digits as 50.
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.
why are we considering only 25..why not others..pls be more clear. and what you mean by " and only one number whose last digit is 2"
Hii, Look we are considering 25 bcoz in the given series 25 is the only number which gives us the clue(or rather have sum cyclic order).So lets see its cyclicity for last 2 digit: 25 * 01 = 25 25 * 02 = 50 25 * 03 = 75 25 * 04 = 00 25 * 05 = 25
And by " and only one number whose last digit is 2", it means in the given question there is only one number which ends in 2.
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.
(Answer in factorials)
It is best to find the number of ways with John and Susan sitting together and subtract this from the total without the restriction.
Let us bracket J and S and consider it to be a single entity. So a possible arrangement is (B=boy, G=girl)
(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252
(SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252
So total with JS together is 504
Now unrestricted arrangements are B G B G B G B G = 4! x 4! = 576 or G B G B G B G B = 4! x 4! = 576
and this total is 1152.
To satisfy condition that J and S are NOT together we must subtract 504
Required number of ways = 1152 - 504 = 648.
Its a bit tricky question.So if its still not clear let me know.I will explain it in complete detail.
Hii, Look we are considering 25 bcoz in the given series 25 is the only number which gives us the clue(or rather have sum cyclic order).So lets see its cyclicity for last 2 digit: 25 * 01 = 25 25 * 02 = 50 25 * 03 = 75 25 * 04 = 00 25 * 05 = 25
And by " and only one number whose last digit is 2", it means in the given question there is only one number which ends in 2.
I suppose all your doubts are clear now.
Regds, Preeti
hi preeti,
thanks for your clarification. But i would like to know how are you coming to the conclusion of considering "25" & "2", wherein there is every possibility to consider other digits as well. pls throw more light on this.
cyclicity of the tenth digit exists for 7 and 5, here i could have used that but didnt know before.
Hi .... Can you please solve this on a word doc or take a picture of the solved sum...I am unable to understand the logic and the syntax... thanks for your help! Jishnu
Hi .... Can you please solve this on a word doc or take a picture of the solved sum...I am unable to understand the logic and the syntax... thanks for your help! Jishnu
Hi,
I just expanded 7 binomially in terms of 10 and 3 and xxxx..6 denotes a number ending in 6. The logic mainly was to express the binomial expansion of (10-3)^202 as 100X+ 10Y+Z.
If you see fourth eqn from the top as we want the 10th digit, we focus on Y and Z.
Y was obtained as (a neg no. ending in 6 + 101) and Z as -1
so on adding up you get Y as a neg no. ending in 5,
It is best to find the number of ways with John and Susan sitting together and subtract this from the total without the restriction.
Let us bracket J and S and consider it to be a single entity. So a possible arrangement is (B=boy, G=girl)
(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252 (SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252 So total with JS together is 504
Now unrestricted arrangements are B G B G B G B G = 4! x 4! = 576 or G B G B G B G B = 4! x 4! = 576
and this total is 1152.
To satisfy condition that J and S are NOT together we must subtract 504
Required number of ways = 1152 - 504 = 648.
Its a bit tricky question.So if its still not clear let me know.I will explain it in complete detail.
Regds, Preeti.
Hi in the boldend arrangement the combination B (JS) G B G B G is also possible, where the boys and girls wont be alternate,similarily for (SJ), hence the no. of arrangements cant be 7x they would be 4x
Hi in the boldend arrangement the combination B (JS) G B G B G is also possible, where the boys and girls wont be alternate,similarily for (SJ), hence the no. of arrangements cant be 7x they would be 4x
considering the above, I got 864 as the answer.
Bang on Riva !
Its (2*4!*4!) - (2*4*3!*3!) = Whatever !!!
Initially I did the same mistake that Preeti had done.
Hi in the boldend arrangement the combination B (JS) G B G B G is also possible, where the boys and girls wont be alternate,similarily for (SJ), hence the no. of arrangements cant be 7x they would be 4x
considering the above, I got 864 as the answer.
Hi Riva, I have got your problem,but then you are subtracting those cases which are not added.I mean to say that if we had multiplied it with 2(one for JS and other for JS) then there would be a possibility of a B coming near to a boy or a girl coming near to a girl.But we have taken only one case for this particular combination. like if a boy comes next to this pair then the pair will sit like JS and if a girl comes next then the pair will sit like SJ. so in all there will be no possiblity of a Boy coming next to a boy or a girl coming next to a girl. So the example u have given will be like: B (SJ) G B G B G . and if the row start with a girl than it will be: G (JS) B G B G B .
Initially I did the same mistake that Preeti had done.
Let me add a little detail to the answer I gave for this question.
I start by bracketing John and Susan together, and denote this by (JS). I treat this bracket as a single entity, and it could occupy one of seven positions in the line-up. (There are six other people to be arranged in the line). Now the three remaining boys could be arranged in 3! (=6) ways, and the three girls also in 3! ways. So with the bracket (JS) plus six others, the total number of acceptable arrangements (i.e. with boys and girls alternating) is 7 x 3! x 3! = 252. We could of course have bracketed John and Susan in the order (SJ), and again this could give a total of 7 x 3! x 3! = 252 ways.
The total of possible arrangements with boys and girls alternating and with a boy at the start is 4! x 4! = 576. And the total with a girl at the start is also 4! x 4! = 576 giving a grand total of 1152.
Now to satisfy the condition that John and Susan are NOT together we subtract the number of ways when they ARE together (=504)
The ratio of new cars to old cars in a certain city subway system is 5:3. If 9 tenths of the new cars are considered "Graffiti free", which of the following is the largest possible ratio of "Graffiti free" cars to non-graffiti free cars in the entire system?
The ratio of new cars to old cars in a certain city subway system is 5:3. If 9 tenths of the new cars are considered "Graffiti free", which of the following is the largest possible ratio of "Graffiti free" cars to non-graffiti free cars in the entire system?