p,q,r are non negative integers such that p + q + r = 10 the maximum value of pqr + pr + qr + pq isa- >=40 and b- >=50 and c- >=60 and d- >=70 and e- >=80 and dunno the oa. plz post ur approach.
3 3 4 will give u a value of 69 which shud be d max... my opinion...
in distribution of values wen their sum is fixed..........idea is to put them as close to as possible.... no concrete method though....
More to the point y cannot be 2 as the number 222 does not exist in base 2. Hence x =7 and y = 6 are only possible case, giving 79 as the answer. regards scrabbler
p,q,r are non negative integers such that p + q + r = 10 the maximum value of pqr + pr + qr + pq isa- >=40 and b- >=50 and c- >=60 and d- >=70 and e- >=80 and dunno the oa. plz post ur approach.
p = q = r = 10/3
max value = 10/3^3 + 10/3^2 + 10/3^2 + 10/3^2 = 70.3...ab ans option c hai ya d yeh samjh nahi aa raha..
p,q,r are non negative integers such that p + q + r = 10 the maximum value of pqr + pr + qr + pq isa- >=40 and b- >=50 and c- >=60 and d- >=70 and e- >=80 and dunno the oa. plz post ur approach.
A special kind of wine requires to be kept for 4 years in a special chamber as as to acquire the required taste.The cost of manufacturing one bottle is 250 and inventory cost for each bottle is rs 4 per year.The company makes 6000 bottles and keeps them in special chambers, but every year 2% of the bottles break.After 4 years at what price should they sell remaining bottles to make a profit of 10%
OA :9 :20 :18 Let S be the sum of an arithmetic series. The arithmetic mean of every two consecutive terms and every three consecutive terms of S form the consecutive terms of series S1 and S2 respectively. If the sum of all the terms in series S1 and in series S2 are 1375 and 690 respectively, then find the sum of all the terms in series S.
A special kind of wine requires to be kept for 4 years in a special chamber as as to acquire the required taste.The cost of manufacturing one bottle is 250 and inventory cost for each bottle is rs 4 per year.The company makes 6000 bottles and keeps them in special chambers, but every year 2% of the bottles break.After 4 years at what price should they sell remaining bottles to make a profit of 10%
manufacturing cost = 6000*250 = 1500000..
for first year -> 6000*4 = 24000
cost after first year = 1524000 --- (1)
2% bottles break after 1st year, hence bottles left -> 5880
cost of maintaining in chamber = 5880*4 = 23520 --- (2)
2% bottles break after 2nd year, hence bottles left -> 4704
cost of maintaining in chamber =4704*4 = 18816 --- (3)
2% bottles break after 3rd year, hence bottles left -> 4609
cost of maintaining in chamber = 4609*4 = 18436 --- (4)
2% bottles break after 4th year, hence bottles left -> 4516
cost of maintaining in chamber = 4516*4 = 18064 --- (4)
p,q,r are non negative integers such that p + q + r = 10 the maximum value of pqr + pr + qr + pq isa- >=40 and b- >=50 and c- >=60 and d- >=70 and e- >=80 and dunno the oa. plz post ur approach.
manufacturing cost = 6000*250 = 1500000..for first year -> 6000*4 = 24000cost after first year = 1524000 --- (1)2% bottles break after 1st year, hence bottles left -> 5880cost of maintaining in chamber = 5880*4 = 23520 --- (2)2% bottles break after 2nd year, hence bottles left -> 4704 cost of maintaining in chamber =4704*4 = 18816 --- (3)2% bottles break after 3rd year, hence bottles left -> 4609cost of maintaining in chamber = 4609*4 = 18436 --- (4)2% bottles break after 4th year, hence bottles left -> 4516cost of maintaining in chamber = 4516*4 = 18064 --- (4)hence, cost = 1524000+23520+18816+18436+18064 = 1602836so, selling price should be = 1602836*1.1 = 1763119.6 ??
A triangle ABC has 2 points marked on side BC, 5 points marked on side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have at least one vertex common with the triangle ABC?
Q: 123123............300 times div by 1001 find the remainder?