hi,
For the problem on parallelogram:
Conisder a parallelogram ABCD. Drop a perpendicular AF from vertex A to side CD.
In parallelogram ABCD,
let height = AF = x
area = base * height = 6x = 18 (given)
which implies, x = 3 = AF
Now in triangle ADF,
AD = 6 and AF = 3
sin(angle ADF) = AF/AD = 1/2
we know that sin(30) = 1/2, so we can conclude that angle ADF = 30 degrees.
Therefore, one angle of the parallelogram is 30 degrees
Thanks!
To Dopa,
hey I have found out why sqrt(x) can't be -ve. The definition of square root has it. I got it from wikipedia. Here it goes:
"The principal square root function f(x) = x (usually just referred to as the "square root function") is a function which maps the set of non-negative real numbers R+ {0} onto itself, and, like all functions, always returns a unique value."
This means that the set( f(x) ) is only 0 to +ve infinity. Therefore sqrt(25) can't be -5. It can only be +5.
I hope this will answer the long-lasting answer for the sqrt(-xx) quest. The answer is surely and only A i.e. -x.
ashishjha100 SaysA small square is inscribed in a large square.What is the perimeter of larger square?(measurements given are3 & 4):sarcasm:
@ Ashish,
Its not clear 3 and 4 are sides of which two sides.cud u elaborate further?
thanks
What is the greatest possible triangular area with one vertex at the centre of the circle of radius 1 and the other two vertices on the circle?
I got sqrt(3)/4 as the ans
But GMAC says its 0.5
:idea:
The greatest possible triangle area in the case stated below will occur when the triangle is a right triangle.
The triangle will also be an isosceles right triangle. (as we know the radius of the circle = 1)
Therefore, the greatest possible area = 1/2 * 1 * 1 = 1/2 = 0.5
Hope this helps!.. Plz correct me if i'm wrong.
What is the greatest possible triangular area with one vertex at the centre of the circle of radius 1 and the other two vertices on the circle?
I got sqrt(3)/4 as the ans
But GMAC says its 0.5
:idea:
The greatest possible triangle area in the case stated below will occur when the triangle is a right triangle.
(To understand why, draw a square inscribed within a circle and let the diagonal of the square pass through the center of the circle. The diagonal will partition the square into two triangles each of which will have the greatest area)
The triangle will also be an isosceles right triangle. (as we know the radius of the circle = 1)
Therefore, the greatest possible area = 1/2 * 1 * 1 = 1/2 = 0.5
Hope this helps!.. Plz correct me if i'm wrong.
Arch,
If we inscribe a square in a circle and cut it through circle's center, the 3 vertices of so formed 2 right triangles will be on the circle.......Ques: is one vertex on center of circle.....
Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent stamps, what is the least number of 1-cent stamps she could have bought ?
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
Oh yes!.
Thanks a lot Ashish for pointing out my mistake.
The triangle area will be the largest when it is a right triangle. I was just elaborating on that.
Arch,
If we inscribe a square in a circle and cut it through circle's center, the 3 vertices of so formed 2 right triangles will be on the circle.......Ques: is one vertex on center of circle.....
The greatest possible triangle area in the case stated below will occur when the triangle is a right triangle.
(To understand why, draw a square inscribed within a circle and let the diagonal of the square pass through the center of the circle. The diagonal will partition the square into two triangles each of which will have the greatest area)
The triangle will also be an isosceles right triangle. (as we know the radius of the circle = 1)
Therefore, the greatest possible area = 1/2 * 1 * 1 = 1/2 = 0.5
Hope this helps!.. Plz correct me if i'm wrong.
Consider triangle AOB with O as centre of circle and A and B points of the chord of the circle.
In the max possible case, angle AOB = 90 deg
Hence area = 1/2*(1)*(1) = 0.5
In the diagram above, points A, B, C, D, and E represent the five teams in a certain league in which each team must play each of the other teams exactly once. The segments connecting pairs of points indicate that the two corresponding teams have already played their game. The arrows on the segments point to the teams that lost; the lack of an arrow on a segment indicates that the game ended in a tie. After all games have been played, which of the following could NOT be the percent of games played that ended in a tie?
(A) 10%
(B) 20%
(C) 30%
(D) 40%
(E) 50%
(A) 10%
(B) 20%
(C) 30%
(D) 40%
(E) 50%
Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent stamps, what is the least number of 1-cent stamps she could have bought ?
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
No of 5 cent stamps = x = no. of 25 cent stamps
no of 10 cent stamps = 2x
Let no. of 1 cent stamps = y
Given 0.05x+0.25x+0.1(2x)+0.01y = 2.65
=> 0.5x+0.01y = 2.65
At most, the value of 5 cent stamps can be 2.5$
hence leastno. of 1 cent stamps= 0.15/0.01 = 15
OA is C
In the diagram above, points A, B, C, D, and E represent the five teams in a certain league in which each team must play each of the other teams exactly once. The segments connecting pairs of points indicate that the two corresponding teams have already played their game. The arrows on the segments point to the teams that lost; the lack of an arrow on a segment indicates that the game ended in a tie. After all games have been played, which of the following could NOT be the percent of games played that ended in a tie?
(A) 10%
(B) 20%
(C) 30%
(D) 40%
(E) 50%
The answer should be A i.e 10%
total matches = 10
already tie = 2
minimum tied % = 2*100/10=20%
so it can not be less than 20% hence 10% of matches can never tie
hope m correct
The answer should be A i.e 10%
total matches = 10
already tie = 2
minimum tied % = 2*100/10=20%
so it can not be less than 20% hence 10% of matches can never tie
hope m correct
I can spot only 7 matches......could u pls help me......
ashishjha100 SaysI can spot only 7 matches......could u pls help me......
There are only 7 shown in the Diagram but there are three more to be played between B-D,B-E,C-E ( as the problem says that each team plays with each of the other team once)
hope u got it
There are only 7 shown in the Diagram but there are three more to be played between B-D,B-E,C-E ( as the problem says that each team plays with each of the other team once)
hope u got it
Thank u Inder.........now its clear
An RC aggarwal question -
In how many ways can 21 books on English and 19 books on Spanish be placed in a row on a shelf so that two books on Spanish are never together?
1. 3990
2. 1540
3. 1995
4. 3672
5. None of these.
OA - 2
I think
answer should be 5... what say ya
An RC aggarwal question -
In how many ways can 21 books on English and 19 books on Spanish be placed in a row on a shelf so that two books on Spanish are never together?
1. 3990
2. 1540
3. 1995
4. 3672
5. None of these.
OA - 2
I think
answer should be 5... what say ya
Yes - I too think it should be 5....
shouldn't the answer be (21!) * (22 C 19) * (19!)
happy to hear that !
A die is rolled 4 times. What is the probability of not getting two even numbers on any two consecutive rolls?
(A) 0.25
(B) 0.4
(C) 0.5
(D) 0.75
(E) 0.8
85 people at a certain college study atleast one of the three languages: english, german and swedish. 30 students study english and one more language. 21 students study german and one more language. 27 students study swedish and one more language. If only one person studies all three languages, how many students study both swedish and german but not english?
(A) 9
(B) 12
(C) 18
(D) 21
(E) 27