Akash, Bhanu and Sam work on a project. After they complete 1/4th of the project, Sam takes a break. For 7 days only Akash and Bhanu work on the project. After that Sam relieves both of them. He completes the project in 5 days. Bhanu works 50% faster than Akash while Sam alone can finish the entire project in 20 days, How long would Bhanu take to finish the entire project?
Akash, Bhanu and Sam work on a project. After they complete 1/4th of the project, Sam takes a break. For 7 days only Akash and Bhanu work on the project. After that Sam relieves both of them. He completes the project in 5 days. Bhanu works 50% faster than Akash while Sam alone can finish the entire project in 20 days, How long would Bhanu take to finish the entire project?
ans 23 1/3 days
(A + B)*7 + 5*(S) = 3/4
Now, it is given in the question that S can complete the entire project in 20 days, so S complete 1/20 of the work in one day.
Using above, (A + B)*7 + 1/4 = 3/4
(A + B)*7 = 1/2
(A + 3A/2)*7 = 1/2 ( Using B = 3A/2)
On solving above, we get A = 1/35
Hence, B = 3/70
This means that B completes 3/70 of the work in one day. So, he will take 70/3 days to complete the entire project on his own.😃
Akash, Bhanu and Sam work on a project. After they complete 1/4th of the project, Sam takes a break. For 7 days only Akash and Bhanu work on the project. After that Sam relieves both of them. He completes the project in 5 days. Bhanu works 50% faster than Akash while Sam alone can finish the entire project in 20 days, How long would Bhanu take to finish the entire project?
ans 23 1/3 days
(a+b)7 + 5s = 3/4 (work that remained to be completed)
and s = 1/20 , b = 1.5a
so we get a=1/35
b = 3/70
so b will take 70/3 = 23 1/3 days to complete the task
first divide the coins in to three groups. 10-10-8 . weight 10-10 together and you l get to know whether the heavier coin is in the group of 10-10 or 8. so we are done with one weighing. Now lets say the heavier coin is in the group of 10. divide it into three groups of 4-4-3. weigh 4-4 together and now u l get to know whether the coin is in the group of 4 or in the group of 3. if in group of 4 then weigh 2 -2 and then once more and you get the coin in fourth weighing. similar approach can be followed if the coin was in the group of 8.
How many ordered sets of positive integer triples (a,b,c)are there such that a+b×c=100?
t
t
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We have to count all factors of 1 to 99 and then add the no of factors of each number. 1 is factor of all numbers, so should be counted 99 times 2 will be counted 49 times 3, 33 times 4 - 24 times 5 - 19 times 6 - 16 times 7 - 14 times 8 - 12 times 9 - 11 times 10 - 9 times 11 - 9 times 12 - 8 times 13 & 14 - 7 times 15 & 16 - 6 times 17, 18, 19 - 5 times 20, 21, .., 24 - 4 times 25, 26, .., 33 - 3 times 34, 35, ..., 49 - 2 times 50, 51, .., 99 - 1 time
1/10! ? 10 no.s gen by it can be arranged in 10! ways. That means in those many ways 10 no.s can be generated. Out of which only one way is there where 10 no.s are in ascending order. :)
take ths for example.. 472+559 = 1031 all conditions are statisified...
A mixture comprises two chemicals A and B. The price of A is Rs. 100/- per litre and that of B is Rs.200/- per litre. We can spend a maximum of Rs. 600/- for making the mixture. The densities of A andB are 10 kg/litre and 12 kg/litre respectively. The mixture must contain each of the chemicals to theextent of at least 25% by weight. The maximum weight of the mixture that can be made is closestto:a. 60 kg b. 51 kg c. 54 kg d. 48 kg
closest to 48 kg 100x+200y=600 we have to minimise liq b contents (too costly) 12y/10x=1/3 ==> x=3.6y ==>y=6/5.6 for our optimal solution and weight= 10X+12y=48y=48*6/5.6 which is closest to 48 kg
take ths for example.. 472+559 = 1031 all conditions are statisified...
A mixture comprises two chemicals A and B. The price of A is Rs. 100/- per litre and that of B is Rs.200/- per litre. We can spend a maximum of Rs. 600/- for making the mixture. The densities of A andB are 10 kg/litre and 12 kg/litre respectively. The mixture must contain each of the chemicals to theextent of at least 25% by weight. The maximum weight of the mixture that can be made is closestto:a. 60 kg b. 51 kg c. 54 kg d. 48 kg
closest to 48 kg 100x+200y=600 we have to minimise liq b contents (too costly) 12y/10x=1/3 ==> x=3.6y ==>y=6/5.6 for our optimal solution and weight= 10X+12y=48y=48*6/5.6 which is closest to 48 kg