﻿ CBSE Class 10 Mathematics Solved Question Paper 2017 | Previous Year Papers | Zigya

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# CBSE Class 10 Mathematics Solved Question Paper 2017

#### Short Answer Type

1.

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.

3719 Views

2.

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 18184 Views

3.

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. 5037 Views

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, 7314 Views

5.

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 Hence, the diameter of the hemisphere is equal to 9 units.

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6.

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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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.

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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9.

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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10.

Find the roots of the quadratic equation  3563 Views