well i think we should have also taken one more constraint into consideration that is n>=100-n2n>=100n>=50n/100-n100-n=xn=100-xsince n>=50x100-x/x100/x -1factors of 100=99*2=18remove factors >50which is only n=100and we are already not taking it into consideration since n=0 we are taking and n=100 we are not taking(as mentioned in my earlier post)so still it would be the same ans
but bhai among the 18 values, there are ones in which n is -ve also , so i think it's not necessary that n>=50
If the ratio of sines of angles of a triangle is 1 : 1 : sqrt2 , then the ratio of square of the greatest side to sum of the squares of other two sides is
If the ratio of sines of angles of a triangle is 1 : 1 : sqrt2 , then the ratio of square of the greatest side to sum of the squares of other two sides is
If the ratio of sines of angles of a triangle is 1 : 1 : sqrt2 , then the ratio of square of the greatest side to sum of the squares of other two sides is
Sin A : sin B : Sin C = 1: 1: root 2 ==> 1/root2 : 1/root2 : 1 ==> A:B:C = 45:45:90 ;
If the ratio of sines of angles of a triangle is 1 : 1 : sqrt2 , then the ratio of square of the greatest side to sum of the squares of other two sides is
quite simply it is a 45-45-90 triangle so it would be [rt(2)]^2/(1)^2+(1)^2=1?
a and b are two alloys of argentum and brass prepared by mixing metals in proportion 7:2 and 7:11 respectively. If equal quantities of two alloys are melted to form a third alloy C, the proportion of argentum and brass in C will be?
a and b are two alloys of argentum and brass prepared by mixing metals in proportion 7:2 and 7:11 respectively. If equal quantities of two alloys are melted to form a third alloy C, the proportion of argentum and brass in C will be?
You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.
Puys...I couldnt remember the funda in this question. ..but i remmeber that in problems like this we have to take powers of 3 or powers of 2 in some...can some body refresh the funda behind this type o problems...any links would help
You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.Puys...I couldnt remember the funda in this question. ..but i remmeber that in problems like this we have to take powers of 3 or powers of 2 in some...can some body refresh the funda behind this type o problems...any links would help
@techgeek2050take,abcd + abc + abd + acd + bcd + ab + ac + ad + bc + bd + cd + a + b + c + d=> Abcd + abc + abd + acd + bcd + ab + ac + ad + bc + bd + cd + a + b + c + d +1 -1=> abc(d+1) + abd + acd + bcd + ab + ac + ad + bc + bd + cd + a + b + c + d + 1 -1=> abc(d+1) + ab( d+1) + acd + bcd + ac + ad + bc + bd + cd + a + b + c + d + 1 -1=> abc(d+1) + ab( d+1) + ac(d+1) + bcd + ad + bc + bd + cd + a + b + c + d + 1 -1=>abc(d+1) + ab( d+1) + ac(d+1) + bc (d+1) + ad + bd + cd + a + b + c + d + 1 -1=> abc(d+1) + ab( d+1) + ac(d+1) + bc (d+1) + a( d+1) + b (d+1)+ c(d+1) + ( d + 1) -1=> (d+1) [ ... ] -1if we continue we will end up with (a+1)(b+1)(c+1)(d+1) -1 2010 = (a+1)(b+1)(c+1)(d+1) rest is easy.
..but i remmeber that in problems like this we have to take powers of 3 or powers of 2 in some...can some body refresh the funda behind this type o problems...any links would help 