A circle is drawn inside a trapezium such that it touches all the four sides of the trapezium. The linejoining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in theratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm)of the longer parallel side of the trapezium?(a) 8 (b) 10 (c) 12 (d) Cannot be determined
Should be 12....the total of the two parallel sides should be 16 and the ratio should be 2 : 6 regards scrabbler
A circle is drawn inside a trapezium such that it touches all the four sides of the trapezium. The linejoining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in theratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm)of the longer parallel side of the trapezium?(a) 8 (b) 10 (c) 12 (d) Cannot be determined
It is (c) 12.
The parallel sides (smaller, middle, longer) are in the ratio 1 : 2 : 3
Try to draw the diagram and extend non-parallel sides to complete the triangle. I am just giving this hint. It'll be better for you to apply some brain on it and get the result.
Basically, the number of ways in which N can be written as a sum of two or more consecutive natural numbers is the same as number of odd factors of N minus 1.
1000=2^3 * 5^3
Now odd factors will be contributed only by the powers of 5, ranging fom 5^0 to 5^3. These are 4 in number.
Basically, the number of ways in which N can be written as a sum of two or more consecutive natural numbers is the same as number of odd factors of N minus 1.1000=2^3 * 5^3Now odd factors will be contributed only by the powers of 5, ranging fom 5^0 to 5^3. These are 4 in number.Hence the required number of ways = 4-1=3
Any logic behind this? Why is it Number of odd factors - 1?
Basically, the number of ways in which N can be written as a sum of two or more consecutive natural numbers is the same as number of odd factors of N minus 1.1000=2^3 * 5^3Now odd factors will be contributed only by the powers of 5, ranging fom 5^0 to 5^3. These are 4 in number.Hence the required number of ways = 4-1=3
A circle is drawn inside a trapezium such that it touches all the four sides of the trapezium. The line joining the midpoints of the non-parallel sides divides the trapezium in two parts with areas in the ratio 3 : 5. If the lengths of the non-parallel sides are 6 cm and 10 cm, then what is the length (in cm) of the longer parallel side of the trapezium? (a) 8 (b) 10 (c) 12 (d) Cannot be determined
Did not see your thoughts on my query yet and the posts seem to have stopped. So, consider the following two cases:
Case1 (odd number of terms): Suppose the sequence has 2x+1 terms, and the middle term is y => N=(2x+1)y, where 2x+1 is the odd factor.
Case 2 (even number of terms): Suppose there are 2y terms, and middle terms are x and x+1 =>N=2y{(x+x+1)/2}=y(2x+1), where 2x+1 is the odd factor.
Clearly for each set of consecutive numbers that adds up to N, there is an odd factor of N and every odd factor greater that 1 leads to a distinct set.
Therefore the number of such sets are the same as the number of odd factors more than 1.
Another query for you to ponder on: Which natural numbers cannot be expressed as the sum of two or more consecutive integers?
can u pls tell how u r getting that 62.5???rest is ok....
If there are even number of terms in a sequence, there are two middle terms, whose average gives the overall middle term for the entire sequence.
In the case at hand the two middle terms for the sequence (from 55 to 70) are 62 and 63, whose average is 62.5.
Another way to view the same thing and which is a more generic one (also holds for odd number of terms) - since this is a set of consecutive numbers, the middle term will be the average of the first and the last terms -> (55+70) /2 = 62.5
can u pls tell how u r getting that 62.5???rest is ok....
If there are even number of terms in a sequence, there are two middle terms, whose average gives the overall middle term for the entire sequence.
In the case at hand the two middle terms for the sequence (from 55 to 70) are 62 and 63, whose average is 62.5.
Another way to view the same thing and which is a more generic one (also holds for odd number of terms) - since this is a set of consecutive numbers, the middle term will be the average of the first and the last terms -> (55+70) /2 = 62.5
If there are even number of terms in a sequence, there are two middle terms, whose average gives the overall middle term for the entire sequence.In the case at hand the two middle terms for the sequence (from 55 to 70) are 62 and 63, whose average is 62.5.Another way to view the same thing and which is a more generic one (also holds for odd number of terms) - since this is a set of consecutive numbers, the middle term will be the average of the first and the last terms -> (55+70) /2 = 62.5Hope your query is at rest now
yeah thats ok.....but
see... the odd factors of 1000 are 5, 25,125,1
Taking n=5,we are getting avg=200,hence 198+ 199+......+202
Taking n=25, we are getting avg=40,hence 28+29+30+.....+52
But for avg =62.5,we are getting n=16 on what logic are we taking n=16 or vice versa avg=62.5? is it by trial and error??? Hope my query is clear now.......
