Kudos !! yeah dis is d OA !! cool.. bt I was kinda unsatisfied with the logic given in d source.. nw clear.. Tx dude :)
you dont need bt (if at all it helps) still d other way to word what u hav written could be: 35 seats.. 7 students.. so 28 would be vacant.. 28 vacant seats would give 29 slots which will be the only available seats for 7 students.. so 29P7
well ..u have amazing source of questions !! Really scratchy on this one !! Tried various combinations from particular 2 not being together to nobody being together ...
Ok ...for nobody to be together ...this is what i cud think : there need to be atleast 6 empty spaces between 7 ppl when they seat alternate ...OR degree of freedom is shortened by n-1 for n people
So we have 35 seats and 7 ppl ....so degree of freedon reduces (35-6)P7
i.e 29P7 = 29*28*27*26*25*24*23 ways of seating ... Not sure on this ...if you are having OA pls do post !
I generalised by taking 2 egs .. 3 ppl and 7 chairs ---> 5*4*3 ways 3 ppl and 6 chairs ---> 4*3*2 ways ...
So, i generalised that for n people and m chairs total ways = Pn ways ..
If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25 a. p^2 b. q^2 c. pq d. p^2q^2 e. p^3q
If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25 a. p^2 b. q^2 c. pq d. p^2q^2 e. p^3q
I have two questions ...please provide a solution to the same .
A certain list of 100 data has an average(arithmetic mean ) of 6 and standard deviation of d where d is positive .which of the following pairs of data when added to the list of 102 with standard deviation less than d ?
Ann $ 450,000 Bob $ 360,000 Cal $ 190,000 Dot $ 210,000 Ed $ 680,000
The table shows that the total sales recorded in July for the 5 salespeople at Acme Truck Sales. I was discovered that one of the Cal's sales was incorrectly recorded as one of the Ann's Sales .After this error was corrected , Ann's total sales were still higher then Cal's total sales and the median of the 5 sales is 330,000 .what was the incorrectly recorded sale .
Ann $ 450,000 Bob $ 360,000 Cal $ 190,000 Dot $ 210,000 Ed $ 680,000
The table shows that the total sales recorded in July for the 5 salespeople at Acme Truck Sales. I was discovered that one of the Cal's sales was incorrectly recorded as one of the Ann's Sales .After this error was corrected , Ann's total sales were still higher then Cal's total sales and the median of the 5 sales is 330,000 .what was the incorrectly recorded sale .
write all the sales in increasing order 190000 210000 360000 450000 680000 there numbers 190000 and 450000 would be changed to 190000+x and 450000-x but 450000-x >190000+x order could be 210000 190000+x 450000-x 360000 680000 becoz median is the middle number and it is 330 so only possibility is 450-x is median if 450000-x = 330000 x=120000
I have two questions ...please provide a solution to the same .
A certain list of 100 data has an average(arithmetic mean ) of 6 and standard deviation of d where d is positive .which of the following pairs of data when added to the list of 102 with standard deviation less than d ?
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT a. x = w b. x > w c. x/y is an integer d. w/z is an integer e. x/z is an integer
Just in case if anyone is looking for the explanation of this answer:
For consecutive integers: If the number of integers is odd: Then the sum is a multiple of that odd number. If the number of integers is even: Then the sum is never a multiple of that number.
ex: Sum of any 3 consecutive integers, will be a multiple of 3. Ex: 10,11,12. Sum=33 div by 3 or 100,101,102. Sum=303 div by 3.
ex: Sum of any 4 consecutive integers, will never be a multiple of 4. Ex: 10,11,12,13. Sum =46 not div by 4 or 100,101,102,103. Sum: 406, not div by 4.
Now here, we know that y is even for all z, as y=2z. Therefore, sum of its integers (which is x) cannot be divisible by y. So C can never happen.
Ann $ 450,000 Bob $ 360,000 Cal $ 190,000 Dot $ 210,000 Ed $ 680,000
The table shows that the total sales recorded in July for the 5 salespeople at Acme Truck Sales. I was discovered that one of the Cal's sales was incorrectly recorded as one of the Ann's Sales .After this error was corrected , Ann's total sales were still higher then Cal's total sales and the median of the 5 sales is 330,000 .what was the incorrectly recorded sale .
Just in case if anyone is looking for the explanation of this answer:
For consecutive integers: If the number of integers is odd: Then the sum is a multiple of that odd number. If the number of integers is even: Then the sum is never a multiple of that number.
ex: Sum of any 3 consecutive integers, will be a multiple of 3. Ex: 10,11,12. Sum=33 div by 3 or 100,101,102. Sum=303 div by 3.
ex: Sum of any 4 consecutive integers, will never be a multiple of 4. Ex: 10,11,12,13. Sum =46 not div by 4 or 100,101,102,103. Sum: 406, not div by 4.
Now here, we know that y is even for all z, as y=2z. Therefore, sum of its integers (which is x) cannot be divisible by y. So C can never happen.
I think we handle this problem w/o any calculation. Standard deviation a an average deviation of elements from the mean. the farther the element from the mean, the more will be its standard deviation. Now lets look at our options:
1. -6 & 0 ==> one element is 12 units and another 6 units far from the mean. This is increase the SD 2. 0 & 0 ==> both elemnts are 6 units far. Similar pattern for options C & D
In E, we see no deviation from the mean and in addition two elements are added which will bring down the avg deviation.
P.S: At first I tried to solve this problem and didnt take long to realize i cant get the ans in 2 mins :grin:
I have two questions ...please provide a solution to the same .
A certain list of 100 data has an average(arithmetic mean ) of 6 and standard deviation of d where d is positive .which of the following pairs of data when added to the list of 102 with standard deviation less than d ?
How many trailing zeros will be there after the rightmost non-zero digit in the value of 25!? A. 25 B. 8 C. 6 D. 5 E. 2
Ans in red ..
explanation: for one 0 u need one 10 , for one 10 u need one 5
no of 5 in 25 ! = 25/5 + 5/5 = 5 + 1 = 6
p.s : 25! = 15,511,210,043,330,985,984,000,000
Alternative approach: Not for the faint hearted 25 x 24 = 600 From here on start multiplying the units digits, 600x23 = yy800 x 22 = yy600x21 = yy600 x 20 = yy12000.....yy2000 x 15 = yy10000 .....yy40000 x 10 = yy400000.......yy600000 x 5 = TT000,000
It might look tedious, but once you start crunching, its not. All you will have to eye for is what multiplication in the units, will yield a 10, 20, 30 etc, so that a zero can be added.
A cube of side 5cm is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted? A. 9 B. 61 C. 98 D. 54 E. 64
A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year? A. 11236 B. 11025 C. 14400 D. 12696 E. Cannot be determined
A cube of side 5cm is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted? A. 9 B. 61 C. 98 D. 54 E. 64
IMO : D
For each square surface of the original cube, there are 25 1x1 squares. Out of these squares, eliminate the bordering squares, as they will touch the adjacent side of the original cube, and will have more than 1 surface painted. So 25 - 16 = 9 are the non-bordering middle squares, for each of the 6 square sides of the cube.
Therefore overall 6 x9 = 54 squares. with only one side painted.
A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year? A. 11236 B. 11025 C. 14400 D. 12696 E. Cannot be determined
This one is nice and tricky.
Let T be the side of the square field this year. So T^2 is the area of the field this year. Analogously, let L^2 be the area of the field last year.
Since each cabbage takes 1 sqft, T^2 - L^2 = 211 x 1(sft)
Up until here, things are clear, but here is when I had to think for a little bit. Was tempted to choose E, but then...it dawned....
211 is a prime number, so its factors are 211 and 1.
