In these exercises the problems of moderate to hard sums are solved from all the chapters
R1 When 25% of the is a number is added to the number, then the resulting number is half of the number.
what is the original number?
Option:
A)200
B)150
C)100
D)300
Solution
Let the number be x
25% of the number=25x/100
25% 0f the number added to 25 is = 25x/100 +25
the resulting number is half of the number that is = x/2
Then 25x/100 +25 =x/2
multiply by 100 on both side of the above equation
then we will get
(25x/100) X 100 +(25×100)=(x/2)x100
25x100x/100 +2500=100x/2
25x+2500=50x
Re arrange we will get 50x -25x = 2500
simplify 25x=2500
divide 25 on both side
25x/25=2500/25
that is x=100
The Correct ANS=C
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R2 What Percentage 60 represent in relation to 300?
Option:
A)10
B)35
C)30
D)20
Solution:
Percentage = part/whole X100
part =60 and whole =300
therefore percentage=60/300 x100
simplify then we will get 20%
The correct ANS=D
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R3 In a right angled triangle the sides are 6√7,9√7 and √90 .
Find the are of the triangle.
Option:
A)180
B)185
C)189
D)165
Solution:
Area of the triangle =base x height/2 (formula for area of the Triangle)
Here base=9√7 and height =9√7
Area of the triangle=9√7×6√7/2=6x9x7/2=189 sq .units
Correct ANS:C
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R4 If 17/x=20 , What is the ratio of x/17?
Option :
A)1.05
B)0.05
0.06
D) 0.04
.solution:
it is given 17/x=25multiply by x on both side
17/x Xx=25Xx17=25x divide by 25
17/25=25x/25 17/25=x divide by 171/25=x/17
x/17=1/25=0.04
Correct ANS:D
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Q5 In a circle With center “O” and Angle P0Q measure 240.
What is degree measure of Are PQ ,that corresponds to the Angle POQ
Option,
A)120
B)220
c)320
D)420
Solution:
the angle measure POQ= 240 is the angle subtended by the Arc PQ at the center
The are PQ substance angle 240 at the center
measure of angle Arc PQ is half the angle sustained at the center ( circle and angles theorem)
therefore The Measure of angle PQ is = 1/2 x240=120
Correct ANS:A
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R6)The velocity of a vehicle is180 kilometers per hour. Convert in the velocity in meters per second.
We can write 180 kilometers per hour as 180kilometer/hour
Options:
A)50meterspersecond
B)40meterspersecond
C)60 meterpersecond
D)55 meterpersecond
Solution:
1 hour =3600 second and one kilometer=1000 meter
We can express 180 kilometer per hour as 180 X 1000/3600
Simplify we will get=50 meter per second.
CORRECT ANS:A
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R7) Area of a rectangle is 1500 squire units .the sides of the rectangles are in the ratio 5:3.
Find the diameter of the rectangle.
OPTIONS:
A)10√34 units
B)20√34 units
C)5√34 units
D)34 units
Solution :
Area of the Rectangle is =Length X breath =lXb
Here l and b are in the ratio 5:3
Let the length = 5x
And breath = 3x
Therefore Area of the Rectangles = 5xX3x=15x2 =1500 (given)
Divide both side by 15
That is 15/15 x2 =1500/15
x2=100
X=√100=10
Length of the rectangle= 5x=5X10=50
Breath of the rectangle =3x=3X10=30
Diagonal of the rectangle=√(502)+302=√3400=10√34 units
correct ANS:A
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R8)If the circle lies on XY-plan. And the center C(5 ,0) and end point of a radius is (9,-3).
Find the equation of the given circle.
solution:
We know thar equation of a circle with radius r and point of tne center ( a,b) is
(x-a)2+(y-b)2=r2
It is given the center point of the circle is (5.0)
Radius =r
Here it is given the center point (5,0) and end point of the radius (9,-3)
By applying two point equation for the length of a line=radius =√{(5-9)2+(0+3)2
=√16+9=√25=5
The equation of the circle with radius 5 and center point(5,0)
(x-5)2+(y-0)2=52
The equation is(x-5) 2+y2=25
ANS:(x-5) 2+y2=25
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R9) For the given -9x+my=22, Find the value of K If it has no solution? where m is content.
Solution:
the given equation is -9x+my=22
it is also given that ,the equation has nonsolution
so the equation should be worng by considering a value for the unknown constant
so we can assume that the value of m=9
by assuming m=9 the right hand side of the equation is = 0
therefore the equation become 0=22 that is not correct and it has no solution
Correct ANS: M=9
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R10)How many solution has the following quadratic equation y2=-49
Option:
A)one
2)None
3)many
D)two
Solution:
consider the given equation
y2=-49
so y=√-49/p>
it is complex number , so the given equation has no solution
Correct ANS:B
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R11)At what value of x does the function f(x)=2x2-36x-19 reaches its minimum point.
option:
A)1.5
B)-1.5
C)-3/4
D)3/4
Solution:
consider the given equation f(x)=2x2-6x+15
It is a quadratic equation
for a general quadratic equation f(x)=ax2+bx+c
minimum value of a quadratic equation occurs at x= -b/2a ( fomula) in the given equation b=-6 and a=2
substitute the values of a and b in the formula x=-b/2a
we will get x=-(-6)/2X2
that is x=6/4=3/2 =1.5
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R12)Find the vale of 3x(2x-1),the value of x from the given equation
5(x-1)+5x=30
Option:
A)24
B)-24
C)14
D)-14
Solution:
consider the given expression 5(x-1)+5x=30
take common out for the left side of the Expression
now we will get 5x(5-1+1)=30
That is 5x(1/5+1)=30
now 5x(6/5)=30 (simplify)
5x=30X5/6
5x=5X5 (simplify)
5x=52
compare the powers ,because the the basis are the same by exponential lawwe will get the value of x=2
now substitute the value of x=2 in the given equation
that is 3x(2x-1)
now we will get 3X2(2X2-1)
That is 3X23
3X8=24
Correct ANS:24
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RQ12 Find the angle value of angle NMB from The given diagram.
The angle B is right angle
solution:
Consider the ∆ ABC
Sum of three angles ∠CAB+∠ABC+∠ACB=180 ( property of triangle sum of three angles)
∠CAB+90+55=180
∠CAB+145=180∠CAB=180-145=35
Consider the ∆ XAN .
∠XNA=∠CNM (AB and XZ intersects at N, Opposite angle is equal)
∠XNA=25 ( ∠ CNM=25 9given) in the ∆XNA ,∠NAB=∠ANX+∠NXA ( exterior angle is equal to sum of opposite angles
that is 35=25+∠NXA
35 – 25 =∠NXA
MBX=180now consider the ∆XMB
∠MXB+∠XMB+∠MBX=180 (properties of triangle)
10+90+∠XMB=180
∠XMB+100=180
simplify then we will get ∠XMB=180-100
Therefore ∠XMB=80
Correct ANS: ∠XMB=80
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RQ13 Find the value of angle∠OAX= y in the dia gram
in the given diagram O is the center of the circle
AB is the chord, X is a point on the circle
it is also given ∠OXB=55 degree
Solution:
Consider the ∆ AOC
∠OAB=55° (given)
∠XBO= ∠ OXC (angle opposite equal sides of the triangle )
Now consider ∆XOB
∠ OXB+ ∠XOB+ ∠OBX=180 (Sum of three angles in a triangle)
55+ ∠XOB+55 =180
110+ ∠XOB=180
add -180 on both side of the equation
110-110+ ∠XOB=180-110
∠XOB=70
Now consider the ∆XAO, ∠ XOB = ∠AXO+ ∠ XAO
Here ∠ XOB is the exterior angle of the triangle XAO
∠AXO+ ∠ XAO =70
But ∠AXO and ∠ XAO are equal (angle opposite to equal sides
∠ XAO =70/2=35
Correct ANMS= 35°
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RQ 14 In the given diagram O is the center of the circle. and AB is a diameter of the circle
AT is a tangent at point A on the circle. if the Angle AOM is 68° .Find the value angle ATM
Option:
A)44°
B)24°
C)56°
D)45°
Solution
In The diagram it is given ∠AOM= 68°
that is The arc MB of the given circle subtends ∠MOA at the center of the circle
∠ ABM is the angle subtends by the Arc AM on the circumference of the circle
therefore ∠ABM = 1/2∠AOM=68/2=34Now consider the ( by circle theorem)
Now consider the ∆ABT, AB is the diameter and AT is the tangent through the point A
Therefore ∠ TAB=90° (Properties of tangents and circle)
We know the ∠ ABT= 34°
Sum three Bangles in the ∆ABT is 180°
That is ∠ABT+∠TAM+∠ATB=180°
34°+90°+∠ATM=180°
124° + ∠ ATM=180°
Add -124 on both side of the equation
124°-124°+∠ATM =180°-124
therefore ∠ATM=56°
Correct ANS: C
RQ15) Find the degree measure of angle x in the given diagram
solution
Consider the ∆CDB in the above diagram
∠ ADC=∠ DBC+∠DCB=42+∠DCB (Exterior angle is equal to sum of inferior opposite angles in a triangle)
∠FAD=∠ DCF (angle subtended by the arc DF at A and C properties of circle)M
therefore ∠DCF= x
Therefore exterior ∠ADC=x+42
Now consider ∆AED
x+∠AED+∠ADE=180 (sum of three angles in a Triangle)
But ∠ADE=x°+42 (proved)
Now consider the Stright line DC
∠AEC+ ∠AED=180 (Propeties of Stright line)
But ∠AEC= 94 (given)
therefore 94+∠AED=189
add -94 on both side
94-94+∠AED=180-94
∠AED=86
substitute the value of ∠AED=86 in sum of angle in the Triangle AED
and substitute the value of ∠ADE=x+42 (proved )
therefore sum of three angles in the Triangle ADE
becomes x+x+42+86=180
That is 2x+128=180 (simplified)
Add -128 on both side, then we will get
2×128-128=180-128
2x=52 (simplyfied)
divide by 2 on both side, Then we will get
2x/2= 52/2
x=26>
Correct ANS:26°
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RQ16) Find the area of the Trapezium with parallel sides 120 and 90 withstanding side 25