this is the solution for the sum where the min marks and max marks are given for subjects and the total percent is 60 . a+b+c+d=150(60 % of 250) the maximum value a,b,c can take is 50 and d can take a maximum value of 100. 153c3-(3*103c3)-53c3. this sum can be handled in a similar fashion as that of 4 dice whose sum is 10 .(sum iin the sense sum of the values in the dice).
prakash_miku Saysjust saw in my book (Arun sharma 3rd edition)..the answer given is 10320. they may have done the correction or if i have an old book then a typo later :D
Thanks for confirming...mine is 2nd edition
here u know there is always a cycle of 4............like 2,4,8,6 for 2............8,4,2,6 for 8 ..............
now 4 is a number which divides only those which last 2 digits are divisble..............
so,78^5562 has a 4n+2 cycle.........so last unit digit=4
now,56^256 ...............it always give unit digit 6 becoz 0,5,6 always give unit digit as it is.
so unit digit=6
now,97^1250 has a cycle of 4n+2 so unit digit=9
now................answer is---------->(6*9)*4=(54)*4=6 unit digit
"x" dealt some cards to "y" and himself from a pack of playing cards and laid the rest aside. "x" then said to "y" -If you give me a certain number of your cards, i will have four times as may cards as you will have. If i give you the same number of cards, I will have thrice as many cards as you will have . Of the given choices, which could represent the number of cards with "x" ?
a. 9
b. 31
c. 12
d. 35
Hi this question is from arun sharma pre assessment test for number systems.....how many 3-digit even numbers can be formed such that if one of the digits is 5 the following digit must be 7?
1)5
2)405
3)365
4)495
i am getting the answer as 405 ..but the correct one given in the book happens to be 365.
Hi this question is from arun sharma pre assessment test for number systems.....how many 3-digit even numbers can be formed such that if one of the digits is 5 the following digit must be 7?
1)5
2)405
3)365
4)495
i am getting the answer as 405 ..but the correct one given in the book happens to be 365.
My take to this is:
All the even numbers that do not have 5 in any of the digits = 8*9*5=360
and
If thr is a 5 in the 3 digit even number it can be only at the hundreds place. Hence the even number will look like :
57_ which has 5 such numbers.
Thus the total numbers like this will be = 365.
Hi this question is from arun sharma pre assessment test for number systems.....how many 3-digit even numbers can be formed such that if one of the digits is 5 the following digit must be 7?
1)5
2)405
3)365
4)495
i am getting the answer as 405 ..but the correct one given in the book happens to be 365.
lets first eliminate 500 to 599 in our calulation in the very first place..
100......498 has got 200 even numbers,and then 600.........998 has got 200 again..so 400 3 digit numbers till now..
though we have already eliminated all the digits containing 5 but there are few more to be eliminated from 400..they are:
150,152,154,156,158(total 5)
250,252,254,256,258..till 450 to 458 in the first series
and in second series..
650,652,654,656,658 til 950 to 958...therefore, there are 40 such numbers so now final number is 400-40..360 and now between 500 to 599 as per the given conditions there are only 5 such numbers viz. 570,572,574,576,578 so final answer is 365!!!!
My take to this is:
All the even numbers that do not have 5 in any of the digits = 8*9*5=360
and
If thr is a 5 in the 3 digit even number it can be only at the hundreds place. Hence the even number will look like :
57_ which has 5 such numbers.
Thus the total numbers like this will be = 365.
My take on this was :
The first digit could be any of the 8 digits 1,2,3,4,6,7,8,9 the second digit can be any of the 10 digits and the third digit any of the even numbers from 0 - 9 which is 5 therefore in all the ans would be 8*10*5 which is 400.
Now if you consider the case where the number starts with 5 and its followed by 7...then there are only 5 such cases. so 400 + 5 = 405
My take on this was :
The first digit could be any of the 8 digits 1,2,3,4,6,7,8,9 the second digit can be any of the 10 digits and the third digit any of the even numbers from 0 - 9 which is 5 therefore in all the ans would be 8*10*5 which is 400.
Now if you consider the case where the number starts with 5 and its followed by 7...then there are only 5 such cases. so 400 + 5 = 405 :splat:
For the 2nd digit u hv included 5 also hence u r getting 405 as ans.
"x" dealt some cards to "y" and himself from a pack of playing cards and laid the rest aside. "x" then said to "y" -If you give me a certain number of your cards, i will have four times as may cards as you will have. If i give you the same number of cards, I will have thrice as many cards as you will have . Of the given choices, which could represent the number of cards with "x" ?
a. 9
b. 31
c. 12
d. 35
It would be 31 and "y" in that case would have 9 cards. If "y" gives "x" 1 card then the count with both would be 8 and 32. When "x" gives 1 card to "y", the count would be 10 and 30.
UtsavGambhir SaysIt would be 31 and "y" in that case would have 9 cards. If "y" gives "x" 1 card then the count with both would be 8 and 32. When "x" gives 1 card to "y", the count would be 10 and 30.
explanation to this would be highly appreciated
There are 5 blue socks,4 red socks and 3 green socks in Debu's wardrobe .He has to select 4 socks from this set.In how many ways can he do so ?
(a)245 (b)120 (c)495 (d)60
There are 5 blue socks,4 red socks and 3 green socks in Debu's wardrobe .He has to select 4 socks from this set.In how many ways can he do so ?
(a)245 (b)120 (c)495 (d)60
it's 495...12C4...
then what is the difference between previous socks ques and the below ques:
no. of ways of selcting 5 letters out of 5A's,4 B's, 3C's. 2 D's and 1 E..
please reply
there is no difference until any condition is given
there is no condition given.i posted the ques as it is..in the solution of second ques actually, they considered conditions..like 5 different letters, 2 different 3 similar letters,3 different 2 similar and so on...am confused:S
puys pls solve this...and post your approach.
a manufacturer estimates that on inspection 12 % of the articles he produces will be rejected. he accepts an order to supply 22000 articles at rs 7.5 each..he estimates the profit on his outlay including the manufacturing of rejected articles to be 20%..find the cost of manufacturing each article..
a)Rs.6 b) Rs 5.50 c) Rs.5 d) Rs. 4.5 e) Rs. 6.5
answer is 5.5rs
12% of 22000 are ejected means 88% are accepted..88% of 22000=19360
total selling price =19360*7.5=145200
profit =20% so, 120/100*(total cost price for 22000 articles)=145200
total cost price for 22000 articles=121000
cost price of one article=121000/22000=5.5 rs
please let me know if the answer is correct
answer is 5.5rs
12% of 22000 are ejected means 88% are accepted..88% of 22000=19360
total selling price =19360*7.5=145200
profit =20% so, 120/100*(total cost price for 22000 articles)=145200
total cost price for 22000 articles=121000
cost price of one article=121000/22000=5.5 rs
please let me know if the answer is correct
the answer is correct!!
alekhyaalekhya Saysthere is no condition given.i posted the ques as it is..in the solution of second ques actually, they considered conditions..like 5 different letters, 2 different 3 similar letters,3 different 2 similar and so on...am confused:S
I can't get why they did by that method unless given.