Sunday, October 2, 2011
Saturday, August 27, 2011
Multiple Choice Questions...Ch-01,02,03,04...!!!
TOT MKS: 15(1 X 1)
·
1. Who gave the formula to find the roots of a quadratic
equation?
A. Arya Bhatta
B. Bhaskaracharya
C. Pythagoras
D. Shridhar Acharya
2. If
,
then m=______.
,
then m=______.
A. 3
B. -3
C. 9
D. -9
3. Find the H.C.F. of (3x3)3, 3x2
X 2x3 and 27x6 / 3x2.
A. 3x2
B. 3x4
C. 3
D. x4
4. The solution set of x+y-1=0 and 2x+2y-2=0 is ______.
A. {(1,0)}
B. {(0,1)}
C. An empty set
D. An infinite set.
5. If 4 is one of the roots of x2+6x+k=0, find
the value of k.
A. -10
B. 40
C. -40
D. 20
PAGE DOWN….
6. The reciprocal rational expression of the opposite of 
is ______.
is ______.
A. 
B.
C. 
D. 
7. Find the L.C.M. of (2x2)2, 12x3 / 4x2 and
.
.
A. 2x2
B. 3x2
C. 12x4
D. 6x4
8. Five years ago, the sum of ages of four persons was 60
years. Five years hence, the sum of ages of same four persons will be ______
years.
A. 100
B. 70
C. 65
D. 80
9. x4-5x2+3x-1=0 is ______.
A. A rational expression
B. A polynomial
C. A quadratic equation
D. Not a quadratic equation.
10. Simplify: 
+ 
+
A.
1
B. 0
C. -1
D. 2
11. Find the H.C.F. of 
,
and (2x)2.
,
and (2x)2.
A. x
B. 1
C. x2
D. x3
PAGE
DOWN….
12. If 3x+2y=7 and 2x+3y=8, find the value of x+y.
A. 1
B. 2
C. 3
D. 4
13. If he discriminant of the equation 2x2-5x-k=0 is 81,
find the value of k.
A. 2
B. 5
C. 7
D. 9
14. Which rational expression should be added to 
to get
?
to get ?
A. x+1
B. 
C. x-1
D. 
15. If 7x-9y=25 and 9x-7y=23, find the value of x-y.
A. 3
B. 4
C. 2
D. 1
MORE TO COME...!!!
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ANSWERS:::
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SOME QUESTIONS ARE NOT UNDERSTOOD>>>WILL SEE TO IT LATER...!!!1
Wednesday, July 27, 2011
Theorems for TRIANGLES...
THEOREM:1(FUNDAMENTAL THEOREM OF PROPORTIONALITY)
STATEMENT: A line drawn parallel to a side of a triangle intersects the other two sides in two distinct points so that each of these sides becomes union of two line segments.The lengths of line segments of each side lying in the same semi-plane of that line are proprtional to the lengths of the corresponding sides.
DATA: In Triangle ABC, l is parallel to line segment BC. l intersects line segment AB and line segment AC in D and E respectively.
TO PROVE: (1) AD/AB=AE/AC
(2)BD/AB=EC/AC
PROOF:Construct Line Segment BE and Line Segment CD . Also Let Line segment EF be Perpendicular to Line AB and Line Segment DN be Perpendicular to Line AC. F belongs to Line AB and N belongs to Line AC.
Area of Triangle ADE=1/2 * base * Altitude.
=1/2 * AD * EF ..........................................................................eq.1
ARea of Triangle DBE = 1/2 * base * Altitude.
=1/2 * BD * EF ..........................................................................eq.2
More theorems...7,8,9,10,11.(COMING SOON!!!!)
STATEMENT: A line drawn parallel to a side of a triangle intersects the other two sides in two distinct points so that each of these sides becomes union of two line segments.The lengths of line segments of each side lying in the same semi-plane of that line are proprtional to the lengths of the corresponding sides.DATA: In Triangle ABC, l is parallel to line segment BC. l intersects line segment AB and line segment AC in D and E respectively.
TO PROVE: (1) AD/AB=AE/AC
(2)BD/AB=EC/AC
PROOF:Construct Line Segment BE and Line Segment CD . Also Let Line segment EF be Perpendicular to Line AB and Line Segment DN be Perpendicular to Line AC. F belongs to Line AB and N belongs to Line AC.
Area of Triangle ADE=1/2 * base * Altitude.
=1/2 * AD * EF ..........................................................................eq.1
ARea of Triangle DBE = 1/2 * base * Altitude.
=1/2 * BD * EF ..........................................................................eq.2
More theorems...7,8,9,10,11.(COMING SOON!!!!)
Tuesday, July 26, 2011
CHAPTER:1-IMP SUMS
Q. If A gives 1/6th part of the amount with him to B, the Amounts with both becomes equal. If B gives Rs.16 to A, then A's Amount becomes two and a half that with B. Find the amount with each.
ANS. Let the amount with A be Rs.x and the amount with B be Rs.y.
According to the First Condition,
x-x/6 = y+x/6
Therefore, (6x-x)/6 = (6y+x)/6
Therefore, 6x-x = 6y+x
Therefore, 6x-x-6y-x = 0
Therefore, 6x-6y-2x = 0
Therefore, 4x-6y = 0 ...........................eq.1
According to the Second Condition,
x+16 = 5/2 (y-16)
Therefore, 2x+32 = 5y-80
Therefore, 2x-5y = -80-32
Therefore, 2x-5y = -112......................eq.2
Multiplying eq.1 by 2, and then eliminating(subtarction),
We Get...4y = 224
Therefore, y = 224/4
Therefore, y = 56
Putting y=56 in eq.1, we get...
4x-6y=0
Therefore, 4x-6(56)=0
Therefore, 4x-336=0
Therefore, 4x=336
Therefore, x=336/4
Therefore, x=84
ANS: The Amount with A is Rs.84 and The Amount with B is Rs.56
ANS. Let the amount with A be Rs.x and the amount with B be Rs.y.
According to the First Condition,
x-x/6 = y+x/6
Therefore, (6x-x)/6 = (6y+x)/6
Therefore, 6x-x = 6y+x
Therefore, 6x-x-6y-x = 0
Therefore, 6x-6y-2x = 0
Therefore, 4x-6y = 0 ...........................eq.1
According to the Second Condition,
x+16 = 5/2 (y-16)
Therefore, 2x+32 = 5y-80
Therefore, 2x-5y = -80-32
Therefore, 2x-5y = -112......................eq.2
Multiplying eq.1 by 2, and then eliminating(subtarction),
We Get...4y = 224
Therefore, y = 224/4
Therefore, y = 56
Putting y=56 in eq.1, we get...
4x-6y=0
Therefore, 4x-6(56)=0
Therefore, 4x-336=0
Therefore, 4x=336
Therefore, x=336/4
Therefore, x=84
ANS: The Amount with A is Rs.84 and The Amount with B is Rs.56
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