A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is consonant.
Answer :
There are 26 letters in English alphabets.
∴ Total number of outcomes = 26
We know that there are 5 vowels and 21 consonants in English alphabets.
∴ Total number of favourable outcomes = 21
∴ Probability that the chosen letter is a consonant = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{21}{26}$
In Fig. 1, PA and PB are tangents to the circle with centre O
such that ∠APB = 50°. Write the measure of ∠OAB.
Figure 1 
Answer :
PA and PB are tangents drawn from an external point P to the circle.
∴ PA = PB (Length of tangents drawn from an external point to the circle are equal.)
In ΔPAB,
PA = PB
⇒ $\angle $PBA = $\angle $PAB .....(1) (Angles opposite to equal sides are equal.)
Now,
$\angle $APB + $\angle $PBA + $\angle $PAB = 180°
⇒ 50º + $\angle $PAB + $\angle $PAB = 180° [Using (1)]
⇒ 2$\angle $PAB = 130°
⇒ $\angle $PAB = $\frac{130\xb0}{2}$ = 65°
We know that radius is perpendicular to the tangent at the point of contact.
∴ $\angle $OAP = 90° (OA ⊥ PA)
⇒ $\angle $PAB + $\angle $OAB = 90°
⇒ 65° + $\angle $OAB = 90°
⇒$\angle $OAB = 90° − 65° = 25°
Hence, the measure of $\angle $OAB is 25°.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x, y.
Answer :
Let AB and CD be the two towers of heights x and y, respectively.
Suppose E is the centre of the line joining the feet of the two towers i.e. BD.
Now, in ΔABE,
$\frac{\mathrm{AB}}{\mathrm{BE}}=\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{\mathrm{BE}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}=\sqrt{3}x.....\left(1\right)$
Also,
In ΔCDE,
$\frac{\mathrm{CD}}{\mathrm{DE}}=\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{y}{\mathrm{DE}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DE}=\frac{y}{\sqrt{3}}.....\left(2\right)$
Now, BE = DE .....(3) (E is midpoint of BD.)
So, from (1), (2) and (3), we get
$\sqrt{3}x=\frac{y}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}=\frac{1}{3}$
Hence, the ratio of x and y is 1 : 3.
If $x=\frac{1}{2},$ is a solution of the quadratic equation $3{x}^{2}+2kx3=0,$ find the value of k.
Answer :
If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
Answer :
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.
Answer :
In Fig. 2, AB is the diameter of a circle with centre O and AT is
a tangent. If ∠AOQ = 58°, find ∠ATQ.

Figure 2 
Answer :
Solve the following quadratic equation for x :
$4{x}^{2}4{a}^{2}x+\left({a}^{4}{b}^{4}\right)=0.$
Answer :
Find the ratio in which the point $\mathrm{P}\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points $\mathrm{A}\left(\frac{1}{2},\frac{3}{2}\right)$ and B(2, −5).
Answer :
Find the middle term of the A.P. 213, 205, 197, , 37.
Answer :
In Fig. 3, APB and AQO are semicircles, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Figure 3 
Answer :
A solid wooden toy is in the form of a hemisphere surrounded by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is $166\frac{5}{6}{\mathrm{cm}}^{3}$. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of Rs 10 per cm^{2}. $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Answer :
Find the area of the triangle ABC with A(1, −4) and midpoints of sides through A being (2, −1) and (0, −1).
Answer :
In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. $\left(\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\mathrm{and}\sqrt{5}=2.236\right)$
Figure 4 
Answer :
In Fig. 5, from a cuboidal solid metallic block, of dimensions
15cm âœ• 10cm âœ• 5cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Figure 5 
Answer :
In Fig. 6, find the area of the shaded region [Use π =
3.14]
Figure 6 
Answer :
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 30 m high, find the height of the building.
Answer :
If the sum of the first n terms of an A.P. is $\frac{1}{2}\left(3{n}^{2}+7n\right)$, then find its n^{th} term. Hence write its 20^{th} term.
Answer :
Three distinct coins are tossed together. Find the probability of
getting
(i) at least 2 heads
(ii) at most 2 heads
Answer :
Find that value of p for which the quadratic equation (p + 1)x^{2} − 6(p + 1)x + 3(p + 9) = 0, p ≠ − 1 has equal roots. Hence find the roots of the equation.
Answer :
In Fig. 7, tangents PQ and PR are drawn from an external point P
to a circle with centre O, such that ∠RPQ = 30°. A chord
RS is drawn parallel to the tangent PQ. Find ∠RQS.
Figure 7 
Answer :
From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length of the flag staff is 5 m, find the height of the tower.
Answer :
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
Answer :
A box contains 20 cards numbered from 1 to 20. A card is drawn at
random from the box. Find the probability that the number on the
drawn card is
(i) divisible by 2 or 3
(ii) a prime number
Answer :
Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
Answer :
A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment.
Answer :
Solve for x :
$\frac{2}{x+1}+\frac{3}{2\left(x2\right)}=\frac{23}{5x},x\ne 0,1,2$
Answer :
To fill a swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.
Answer :
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Answer :
Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are $\frac{3}{4}$times the corresponding sides of the isosceles triangle.
Answer :
If P(–5, –3), Q(–4, –6), R(2, –3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.
Answer :
10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2013 will be available online in PDF book soon. The solutions are absolutely Free. Soon you will be able to download the solutions.