Subject

Mathematics

Class

CBSE Class 10

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

Draw a line segment of length 8 cm and divide it internally in the ratio 4: 5.
Steps of construction.


1) Draw a line segment AB = 8 cm.
2) Draw a ray AX making an acute angle ∠BAX=60° with AB.
3) Draw a ray BY parallel to AX by making an acute angle ∠ABY=∠BAX.
4) Mark four points A1, A2, A3, A4 on AX and five points B1,B2,B3,B4,B5 on BY in such a way that AA1=A1A2=A2A3=A3A4.
5) Join A4B5
6) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4:5.

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2.

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere? 


Let the radius of the hemisphere be x units.
Volume of a hemisphere = Surface area of the hemisphere
rightwards double arrow 2 over 3 πr cubed equals 3 πr squared
rightwards double arrow 2 over 3 straight r equals 3 space
rightwards double arrow straight r equals 92
rightwards double arrow straight d equals 9 space units
Hence, the diameter of the hemisphere is equal to 9 units.

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3.

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.


Given: PA and PB are the tangents to the circle.
PA = 12 cm
QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm
(The lengths of tangents drawn from an external point to a circle are equal)
Similarly, QC = AC = 3 cm
and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm
Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

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4.

If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?


Let the points X (4,k) and Y(1,0)
It is given that the distance XY is 5 units.
By using the distance formula,
XY space equals space square root of left parenthesis 4 minus 1 right parenthesis squared space plus left parenthesis straight k minus 0 right parenthesis squared end root
rightwards double arrow space 5 space equals space square root of 9 plus left parenthesis straight k right parenthesis squared end root
rightwards double arrow space 25 space equals space 9 space plus straight k squared
rightwards double arrow 16 space equals space straight k squared
rightwards double arrow plus-or-minus space 4 space equals space straight k

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5.

The ratio of the height of a tower and the length of its shadow on the ground is √3:1. What is the angle of elevation of the sun?



Let the height of the tower be x and y the length of the shadow on the ground be x:y.

The angle of elevation of the sun from the ground is θ.
We have, x:y =√3:1
Now, In ∆ABC
tan space straight theta space equals fraction numerator space Height space over denominator Base end fraction
tan space straight theta space equals space AB over BC
space tan space straight theta space equals fraction numerator square root of 3 over denominator 1 end fraction
therefore space tan space straight theta space space equals space 60 degree

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6.

Find the value of k for which the equation x2 + k(2x + k − 1) + 2 = 0 has real and equal roots. 


The given equation is x2+k(2x+k−1)+2=0.
⇒x2+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b2−4ac
⇒D=(2k)2−4×1×[k(k−1)+2]
⇒D=4k2−4[k2−k+2]
⇒D=4k2−4k2+4k−8
⇒D=4k−8=4(k−2)
For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.

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7.

Find how many integers between 200 and 500 are divisible by 8.


The first term between 200 and 500 divisible by 8 is 208, and the last term is 496.
So, first term (a) = 208
Common difference (d) = 8
an=a+(n−1)d=496
⇒208+(n−1)8=496
⇒(n−1)8=288
⇒n−1=36⇒n=37
Hence, there are 37 integers between 200 and 500 which are divisible by 8.

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8.

Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other.


Let AB be the diameter of a circle, with centre O. The tangents PQ and RS are drawn at points A and B, respectively.

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OA ⊥ PQ and OB ⊥ RS
⇒ ∠OBR = 90°
∠OBS = 90°
∠OAP = 90°
∠OAQ = 90°
We can observe the following:
∠OBR = ∠OAQ and ∠OBS = ∠OAP
Also, these are the pair of alternate interior angles.
Since alternate interior angles are equal, the lines PQ and RS are parallel to each other.
Hence, proved.

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9.

Find the roots of the quadratic equation square root of 2 straight x squared space plus 7 straight x space plus 5 square root of 2 space equals space 0


square root of 2 straight x squared space plus 7 straight x space plus 5 square root of 2 space equals space 0
or
rightwards double arrow square root of 2 straight x squared space plus 5 straight x space plus 2 straight x space plus 5 square root of 2 space equals space 0
rightwards double arrow straight x left parenthesis square root of 2 straight x space plus 5 right parenthesis space plus square root of 2 left parenthesis square root of 2 straight x end root plus 5 right parenthesis space equals space 0
rightwards double arrow space left parenthesis square root of 2 straight x end root space plus 5 right parenthesis left parenthesis straight x plus square root of 2 right parenthesis space equals space 0
rightwards double arrow left parenthesis square root of 2 straight x end root space plus 5 right parenthesis space left parenthesis straight x plus square root of 2 right parenthesis space equals space 0
rightwards double arrow space straight x space plus space square root of 2 space equals space 0
square root of 2 straight x end root space plus 5 space equals space 0
rightwards double arrow space straight x space equals negative square root of 2
straight x space equals negative space fraction numerator 5 over denominator square root of 2 end fraction space equals negative space fraction numerator 5 square root of 2 over denominator 2 end fraction space
Hence space roots space are space minus square root of 2 comma space minus fraction numerator 5 square root of 2 over denominator 2 end fraction space
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10.

A number is chosen at random from the number –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1? 


S = {−3, −2, −1, 0, 1, 2, 3}
Let E be the event of getting a number whose square is less than or equal to 1.
So, E = {−1, 1, 0}
P(E)=3/7.
Hence, the probability of getting a number whose square is less than or equal to is 3/7.

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