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CBSE X
All India (2)
MATHS PAPER 2012
Time allowed: 180 minutes; Maximum Marks: 90
General Instructions: | |
1) | All questions are compulsory. |
2) | The question paper consists of thirty questions divided into 4 sections A, B, C and D. Section A comprises of ten questions of 01 mark each, Section B comprises of five questions of 02 marks each, Section C comprises ten questions of 03 marks each and Section D comprises of five questions of 06 marks each. |
3) | All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. |
4) | There is no overall choice. However, internal choice has been provided in one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt only one of the alternatives in all such questions. |
5) | In question on construction, drawing should be near and exactly as per the given measurements. |
6) | Use of calculators is not permitted. |
Section A | Section B | Section C | Section D |
1. If 1 is a root of the equations a${y}^{2}$ + ay + 3 = 0 and${y}^{2}$+ y + b = 0 then ab equals:
Consider y as the variable of the quadratic equations.
1) Consider the first equation and substitute the value of y as 1.
2) Simplify the equation and find the value of a.
3) Consider the second equation and Substitute the value of y as 1.
4) Simplify the equation and find the value of b.
5) Find the product of a and b.
Given that the equations a${y}^{2}$ + ay + 3 = 0 and ${y}^{2}$ + y + b = 0 have 1 as the common root.
âˆ´Both the equations will be satisfied by y = 1.
Substituting y = 1 in the equation a${y}^{2}$+ ay + 3 = 0, we get
a${\left(1\right)}^{2}$+ (a × 1) + 3 = 0
âˆ´ a + a + 3 = 0
⇒ 2a + 3 = 0
⇒2a = - 3
⇒a = $\frac{\xe2\u02c6\u20193}{2}$ -------- (1)
Substituting y = 1 in the equation ${y}^{2}$+ y + b = 0, we get
${\left(1\right)}^{2}$+ 1 + b = 0
⇒1 + 1 + b = 0
⇒ 2 + b = 0
⇒ b = - 2 --------- (2)
From ( 1) and ( 2)
⇒ ab = $\frac{\xe2\u02c6\u20193}{2}$×(- 2) = 3
Therefore, the value of ab is 3.
2. The sum of first 20 odd natural numbers is:
Consider the formula to find the sum of the natural numbers involving the first term, number of terms and the common difference.
1) List the odd natural numbers.
2) The numbers form an A.P.
3) Find the common difference of the A.P. and note the first term.
4) Find the sum of the 20 terms of the A.P. by considering the value of n as 20.
5) Substitute the values and simplify to find the sum of the first 20 odd natural numbers.
Sequence of odd natural numbers is 1, 3, 5, 7, 9â€¦,.
The numbers form an A.P. where a = 1, d = 3âˆ’1 = 2.
Sum of the n natural numbers, ${S}_{n}$= $\frac{n}{2}$[2a + ( n âˆ’ 1)d ]
${S}_{n}$= $\frac{20}{2}$[ 2×1 + ( 20âˆ’1) ×2]
${S}_{n}$= 10[ 2 + 38 ]
= 10 × 40
= 400.
âˆ´The sum of first 20 odd natural numbers is 400.
3. In Fig. 1, the sides AB, BC and CA of a triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then the length of BC (in cm) is:
Subtract the length of AR from AC to find the length of CR.
1) Lengths of tangents drawn from the external points A, B and C are equal.
2) Equate the lengths of tangents from the points A, B and C.
3) Use this concept and find the lengths of BQ and CQ.
4) Find the sum of lengths of BQ and CQ to find the length of BC.
Given, AP = 4 cm, BP = 3 cm and AC = 11 cm.
The lengths of tangents drawn from an external point to a circle are equal.
AP = AR, BP = BQ, CQ = CR -------- (i)
AC = 11 cm
⇒ AR + RC = 11 cm
⇒ AP + CQ = 11 cm [From equation (i)]
⇒ 4 cm + CQ = 11 cm
⇒ CQ = (11 âˆ’ 4) cm
⇒ CQ = 7 cm
BP = BQ = 3 cm
BC = BQ + QC
⇒ BC = (3 + 7) cm
⇒ BC = 10 cm .
4. In Fig 2, a circle touches the side DF of Î”EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of Î”EDF (in cm) is:
DK, DH, FH and FM are the tangents from D and F.
1) Tangents from E, EK and EM are equal.
2) Tangents from D are DK and DH. Similarly, tangents from F are FH and FM.
3) Find the perimeter of the triangle EDF. Write EK and EM as two parts.
4) Use the concept of equal chords and find the perimeter of the triangle.
Given that EK = 9 cm
Also length of tangents drawn from an external point to the circle are equal.
âˆ´EK = EM = 9 cm
Also, DH = DK and FH = FM â€¦ (i)
EK = EM = 9 cm
⇒ ED + DK = 9 cm and EF + FM = 9 cm
⇒ ED + DH = 9 cm and EF + HF = 9 cm [From equation (i)] â€¦ (ii)
Perimeter of Î”EDF = ED + DF + EF
= ED + DH + HF + EF
= (9 + 9) cm [From equation (ii)]
= 18 cm.
5. If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is:
Write the ratio of the volumes as a fraction and simplify.
1) Assume the radius of the base and the height of the cylinder as r and h respectively.
2) Find the volume of the cylinder.
3) Consider the radius as $\frac{r}{2}$, and find the volume of the new cylinder.
4) Find the ratio of the volume of the new cylinder and the volume of the original cylinder by writing them as fractions.
Consider the radius and height of the right circular cylinder be r and h respectively.
âˆ´Volume of the original cylinder = Ï€${r}^{2}$h
According to the question, radius of the new cylinder is halved keeping the height same.
⇒ Radius of the new cylinder = $\frac{r}{2}$
Also, height of the new cylinder = h
âˆ´ Volume of the new cylinder = Ï€${\left(\frac{r}{2}\right)}^{2}h$ = $\frac{\mathrm{\xcf\u20ac}{r}^{2}h}{4}$
âˆ´$\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{cylinder}}{\mathrm{Volume}\mathrm{of}\mathrm{original}\mathrm{cylinder}}$= $\frac{\frac{\mathrm{\xcf\u20ac}{r}^{2}h}{4}}{\mathrm{\xcf\u20ac}{r}^{2}h}$= $\frac{1}{4}$= 1:4.
6. If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, then the diameter of the larger circle (in cm) is:
Divide both the sides of the equation of the sum of areas by Ï€.
1) Assume the radii of the smaller circles as ${r}_{1}$and ${r}_{2}$ and the radius of the larger circle as R.
2) Equate the area of the larger circle to the sum of the areas of the smaller circles.
3) Substitute the radii of the smaller circles and simplify to find the radius of the larger circle.
Consider${r}_{1}$ and${r}_{2}$ as the radii of the two given circles.
Given that, 2${r}_{1}$= 10 âˆ´${r}_{1}$ = 5 cm
Also, 2${r}_{2}$= 24 ⇒${r}_{2}$= 12 cm
Assume R as the radius of the larger circle.
Area of larger circle = Sum of areas of two circles of given radius
âˆ´ Ï€${R}^{2}$ = Ï€${{r}_{1}}^{2}$+ Ï€${{r}_{2}}^{2}$
⇒${R}^{2}$= ${\left(5\right)}^{2}$+ ${\left(12\right)}^{2}$
⇒${R}^{2}$= 25 + 144
⇒R = $\sqrt{169}$
⇒ R = 13 cm
âˆ´ The diameter of the larger circle = 2R = (2 × 13) = 26 cm.
7. The length of shadow of a tower on the plane ground is $\sqrt{3}$times the height of the tower. The angle of elevation of sun is:
Find tangent of the angle of elevation of the sun.
1) Draw a figure representing the data given in the figure.
2) BC, the length of the shadow of the tower is $\sqrt{3}$ times of the height AB.
3) Find the tangent of the angle of elevation of the sun at end of the shadow on the plane ground.
4) Substitute the values of the sides in the ratio and simplify to find the value.
5) Equate the value to the known trigonometric ratio and find the angle of elevation of the sun.
Consider AB is the tower and BC is the length of the shadow of the tower.
Let Î¸ be the angle of elevation of the sun.
Length of the shadow of the tower =$\sqrt{3}$ × Height of the tower
BC = $\sqrt{3}$AB .............. (1)
In a right angled triangle ABC,
⇒ tan Î¸ = $\frac{\mathrm{AB}}{\mathrm{BC}}$
⇒ tan Î¸ = $\frac{\mathrm{AB}}{\sqrt{3}\mathrm{AB}}$ [Using (1) ]
⇒ tan Î¸ = $\frac{1}{\sqrt{3}}$
⇒ tan Î¸ = tan 30Â°
âˆ´ Î¸ = 30Â°
Therefore, the angle of elevation of the sun is 30Â°.
8. If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (âˆ’2, 5), then the coordinates of the other end of the diameter are:
Equate the corresponding coordinates of midpoints to find the coordinates of the other end of the diameter.
1) Assume the coordinates of the other end point of the diameter as (x, y).
2) Find the coordinates of the midpoint of the diameter.
3) Equate the coordinates of the midpoint to the given coordinates.
4) Equate the corresponding coordinates and find the values of x and y.
Consider AB as the diameter and O as the centre of the circle.
Coordinates of one end point of circle and coordinates of its centre are given.
Coordinates of A are (2, 3) and centre O are (âˆ’2, 5).
Assume the coordinates of the point B as (x, y).
Centre of a circle is the midpoint of the diameter.
âˆ´ According to the midpoint formula,$(\frac{2+x}{2},\frac{3+y}{2})$ = (âˆ’2 , 5)
⇒$\frac{2+x}{2}$ = âˆ’2 and $\frac{3+y}{2}$ = 5
⇒ 2 + x = âˆ’4 and 3 + y = 10
⇒x = âˆ’6 and y = 7
Therefore, coordinates of the other end of the diameter are (âˆ’6, 7).
9. The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are:
Substitute the values of the coordinates according to the formula correctly.
1) Use the section formula to find the coordinates of the point P which divides the line segment AB in the ratio of 2:1.
2) Substitute the values and simplify to find the coordinates of the point P.
Assume the coordinates of the point P which divides line segment joining A (1, 3) and B (4, 6) in the ratio of 2 : 1 as (x, y).
Coordinates of a point (x, y) dividing the line segment joining the points (${x}_{1}$, ${y}_{1}$) and (${x}_{2}$, ${y}_{2}$) in the ratio of${m}_{1}$ :${m}_{2}$are given by
x = $\frac{{m}_{1}{x}_{2}+{m}_{2}{x}_{1}}{{m}_{1}+{m}_{2}}$ and y = $\frac{{m}_{1}{y}_{2}+{m}_{2}{y}_{1}}{{m}_{1}+{m}_{2}}$
Now, x = $\frac{2\left(4\right)+1\left(1\right)}{2+1}$ and y = $\frac{2\left(6\right)+1\left(3\right)}{2+1}$
x = $\frac{9}{3}$ and y = $\frac{15}{3}$
x = 3 and y = 5
Therefore, (3, 5) divides the line segment AB in the ratio 2 : 1.
10. Two dice are thrown together. The probability of getting the same number on both dice is:
Reduce the probability obtained into lowest terms.
1) List the possible outcomes when two dice are thrown together.
2) Write the number of possible outcomes.
3) List the favourable outcomes of getting the same number in the possible outcomes.
4) Write the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of possible outcomes gives the required probability.
In a random experiment when two dice are thrown. The number of possible outcomes are:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6).
âˆ´ Total number of possible outcomes = 36
Let E be the event of getting same number on both dice.
The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
⇒ Number of favourable outcomes = 6
âˆ´ P(E) =$\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{6}{36}$ = $\frac{1}{6}$
So, the required probability is $\frac{1}{6}$.
11. Find the value(s) of k so that the quadratic equation ${x}^{2}$âˆ’4kx + k = 0 has equal roots.
Substitute the values of the coordinates in the formula of the discriminant along with their signs.
1) Roots of the given quadratic equation are equal, so the discriminant of the equation is equal to zero.
2) Equate the discriminant of the quadratic equation to zero.
3) Substitute the values and simplify to find the value of k.
Given quadratic equation is${x}^{2}$âˆ’4kx + k = 0
Given equation has equal roots. So, the discriminant of the equation, D = 0.
Discriminant of the equation a${x}^{2}$+ bx + c = 0 is ${b}^{2}$-4ac.
⇒${b}^{2}$âˆ’4ac = 0
⇒${\left(\xe2\u02c6\u20194k\right)}^{2}$âˆ’4$\left(1\right)\left(k\right)$= 0
⇒16${k}^{2}$âˆ’4k = 0
⇒4k( 4kâˆ’1) = 0
⇒4k = 0 or 4kâˆ’1 = 0
⇒ k = 0 or k = $\frac{1}{4}$
Therefore, the values of k are 0 or$\frac{1}{4}$ for which the given equation will have equal roots.
12. Find the sum of all three digit natural numbers, which are multiples of 11.
n value of the last term gives the number of terms of the A.P.
1) List the three digit numbers which are divisible by 11.
2) Note the first and the last term of the series.
3) Equate the last term of the series to the ${n}^{\mathrm{th}}$term of an A.P. and find the value of n.
4) Find the sum of n terms of the A.P. using the values obtained.
The smallest and the largest three digit natural numbers, that are divisible by 11 are 110 and 990 respectively.
So, the three digit numbers which are divisible by 11 are 110, 121, 132, â€¦, 990.
The numbers form an A.P. with first term, a = 110 and the common difference, d = 11.
Assume that there are n terms in the sequence.
Therefore,${a}_{n}$ = 990
${a}_{n}$= a + ( n - 1) d = 990
⇒110 + ( n - 1)11 = 990
⇒110 + 11n - 11 = 990
⇒11n = 990 - 99
⇒11n = 891
⇒ n = 81.
The required sum is ${S}_{n}=\frac{n}{2}$[2a + (n -1)d]
= $\frac{81}{2}$[ 2$\left(10\right)$+ ( 81 -1)11]
= $\frac{81}{2}$[ 220 + 880 ]
= $\frac{81}{2}$[ 1100 ]
= 44550.
âˆ´The sum of all three digit numbers which are multiples of 11 is 44550.
13. Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then find the length of BP.
OP is the hypotenuse in both the triangles OAP and OBP.
1) Draw the figure representing the data given in the question.
2) Join OA, OB and OP.
3) Radius of a circle is perpendicular to the tangent at the point of contact.
4) So, the triangles OAP and OBP are the right angled triangles.
5) Apply Pythagoras theorem in triangle OAP and find the length of OP.
6) Apply Pythagoras theorem in triangle OBP and find the length of OB.
Given : OA = 8 cm, OB = 5 cm and AP = 15 cm
To find: BP
Construction: Join O and P i.e OP.
Now OA âŠ¥AP and OB âŠ¥BP [ Tangent to a circle is perpendicular to the radius through the point of contact ]
⇒ âˆ OAP = âˆ OBP = 90Â°
On applying Pythagoras theorem in Î”OAP, we get
${\left(\mathrm{OP}\right)}^{2}$= ${\left(\mathrm{OA}\right)}^{2}$+ ${\left(\mathrm{AP}\right)}^{2}$
⇒ ${\left(\mathrm{OP}\right)}^{2}$ = ${\left(8\right)}^{2}$+ ${\left(15\right)}^{2}$
⇒ ${\left(\mathrm{OP}\right)}^{2}$= 64 + 225
⇒${\left(\mathrm{OP}\right)}^{2}$ = 289
⇒ OP = 17
âˆ´The length of OP is 17 cm.
Now applying Pythagoras theorem in Î”OBP, we get
${\mathrm{OP}}^{2}$= ${\mathrm{OB}}^{2}$+ ${\mathrm{BP}}^{2}$
⇒ ${\left(17\right)}^{2}$= ${\left(5\right)}^{2}$+ ${\left(\mathrm{BP}\right)}^{2}$
⇒ 289 = 25 + ${\left(\mathrm{BP}\right)}^{2}$
⇒ ${\left(\mathrm{BP}\right)}^{2}$= 289 âˆ’ 25
⇒ ${\left(\mathrm{BP}\right)}^{2}$= 264
⇒ BP = 16.25 cm (approximately)
âˆ´The length of BP is 16.25 cm.
14. In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
OR
In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
If BP = PC then, P is the midpoint of BC.
OR
If OC bisects AB then, AC = BC.
1) Tangents drawn from an external point to a circle are equal. So, equate the tangents drawn from the vertices of the triangle.
2) Equate AB and AC of the triangle.
3) Write the parts of AB and AC.
4) Replace the equal tangents of the circle and simplify to prove that BP and CP are equal and hence P bisects BC.
OR
1) Join OC which is a radius at the point of contact of the tangent to the inner circle.
2) The radius at the point of contact of a tangent at any point on the circle is perpendicular to the tangent.
3) Tangent of the inner circle is also the chord of the outer circle.
4) Perpendicular from the centre to the chord bisects the chord.
Given: In isosceles triangle ABC with AB = AC, circumscribing a circle.
To prove: P bisects BC
Proof: AR and AQ are the tangents drawn to the circle from an external point A.
âˆ´ AR = AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR = BP and CP = CQ.
Given that in Î”ABC, AB = AC.
⇒ AR + RB = AQ + QC
⇒ BR = QC (Since AR = AQ)
⇒ BP = CP (Since BR = BP and CP = CQ)
⇒ P bisects BC
Hence, it is proved.
OR
Given: Two concentric circles${C}_{1}$ and${C}_{2}$ with centre O, and AB is the chord of${C}_{1}$ touching${C}_{2}$ at C.
To prove: AC = CB
Construction: Join OC.
Proof: AB is the chord of${C}_{1}$ touching${C}_{2}$ at C, then AB is the tangent to the circle${C}_{2}$ at C with OC as radius.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
âˆ´ OC âŠ¥ AB
But AB as the chord of the circle ${C}_{1}$. So, OC âŠ¥ AB.
âˆ´ OC is the bisector of the chord AB.
Therefore, AC = CB (Perpendicular from the centre to the chord bisects the chord).
15. The volume of a hemisphere is 2425$\frac{1}{2}{\mathrm{cm}}^{3}$. Find its curved surface area. [ Use Ï€ = $\frac{22}{7}$]
Consider the volume of the hemisphere and radius in the form of a fraction.
1) Convert the volume of the hemisphere into an improper fraction.
2) Assume the radius of the hemisphere as 'r'.
3) Find the volume of the hemisphere and equate it to the given value.
4) Simplify and find the radius of the hemisphere.
5) Find the curved surface area of the hemisphere using the radius obtained.
Given that volume of hemisphere = 2425$\frac{1}{2}{\mathrm{cm}}^{3}$= $\frac{4851}{2}{\mathrm{cm}}^{3}$
Consider the radius of the hemisphere as â€˜râ€™ cm.
Volume of a hemisphere = $\frac{2}{3}$Ï€${r}^{3}$
⇒$\frac{2}{3}$Ï€${r}^{3}$= $\frac{4851}{2}$
⇒$\frac{2}{3}$ × $\frac{22}{7}\times {r}^{3}$= $\frac{4851}{2}$
⇒${r}^{3}$= $\frac{4851\times 3\times 7}{2\times 2\times 22}$
⇒${r}^{3}$= $\frac{441\times 21}{2\times 2\times 2}$
⇒${r}^{3}$= $\frac{21\times 21\times 21}{2\times 2\times 2}$
⇒r = $\sqrt[3]{\frac{21\times 21\times 21}{2\times 2\times 2}}$
⇒r = $\frac{21}{2}$ cm ----------- (1)
âˆ´Curved surface area of a hemisphere = 2Ï€${r}^{2}$
⇒2 × $\frac{22}{7}$ × ${\left(\frac{21}{2}\right)}^{2}$
= 2 × $\frac{22}{7}$ × $\frac{21\times 21}{2\times 2}$
= 693 ${\mathrm{cm}}^{2}$.
16. In Fig. 6, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. [ Use Ï€ = $\frac{22}{7}$]
Radius of the quadrant of the circle is equal to the side of the square.
1) Find the area of the square using the given length of the side.
2) Radius of the quadrant of the circle is equal to the side of the square.
3) Find the area of the quadrant of the circle.
4) Subtract the area of the quadrant of the circle from the area of the square to find the area of the shaded region of the square.
Given that OABC is a square of side 7 cm.
âˆ´ Area of square OABC = ${\left(7\right)}^{2}$= 49${\mathrm{cm}}^{2}$
Given that OAPC is a quadrant of circle with centre O.
âˆ´Radius of the quadrant of the circle = OA = 7 cm
âˆ´ Area of quadrant of the circle = $\frac{1}{4}$Ï€${r}^{2}$
= $\frac{1}{4}$ × $\frac{22}{7}$ × ${7}^{2}$
= $\frac{1}{2}$ × 11 × 7
= $\frac{77}{2}{\mathrm{cm}}^{2}$
Area of the shaded region = Area of the square âˆ’ Area of the quadrant of the circle.
= ( 49 - $\frac{77}{2}$)
=$\frac{(98\xe2\u02c6\u201977)}{2}$
= $\frac{21}{2}$
= 10.5${\mathrm{cm}}^{2}$
âˆ´ Area of the shaded region is 10.5${\mathrm{cm}}^{2}$
17. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.
Find the distance between the points by substituting the coordinates in the distance formula in correct order.
1) Point A is equidistant from B and C. So, equate the distances AB and AC.
2) Substitute the coordinates of the points in the distance formula.
3) Simplify and find the value of p.
Coordinates of the given points are A (0, 2), B, (3, p) and C (p, 5).
Given that A is equidistant from the points B and C.
âˆ´ AB = AC
⇒$\sqrt{{\left(3\xe2\u02c6\u20190\right)}^{2}+{\left(p\xe2\u02c6\u20192\right)}^{2}}$= $\sqrt{{\left(p\xe2\u02c6\u20190\right)}^{2}+{\left(5\xe2\u02c6\u20192\right)}^{2}}$
⇒$\sqrt{{\left(3\right)}^{2}+{\left(p\xe2\u02c6\u20192\right)}^{2}}$= $\sqrt{{p}^{2}+{3}^{2}}$
Squaring on both sides, we get
⇒${p}^{2}$âˆ’4p + 13 =${p}^{2}$+ 9
⇒ âˆ’4p = âˆ’ 4
⇒ p = 1.
18. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Number selected up to 50 must be a multiple of both 3 and 4 i.e. a multiple of 12.
1) List the natural numbers from 1 to 50.
2) Note the number of possible outcomes.
3) List the numbers which are divisible by 3 and 4 between 1 and 50.
4) Note the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of possible outcomes gives the required probability.
The first fifty natural numbers are 1, 2, 3,..........50.
Total number of possible outcomes = 50
Multiples of 3 and 4 which are less than or equal to 50 are:
12, 24, 36, 48
Number of favourable outcomes = 4
Let E be the favourable outcome.
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$
= $\frac{4}{50}$
= $\frac{2}{25}$
Thus, the probability of the number selected is a multiple of 3 and 4 is $\frac{2}{25}$.
19. Solve for x: 4${x}^{2}$ âˆ’ 4ax + (${a}^{2}$âˆ’${b}^{2}$) = 0
OR
Solve for x: 3${x}^{2}$- 2$\sqrt{6}$x + 2 = 0
Split the middle term using the constant term and the coefficient of ${x}^{2}$.
OR
Consider the coefficients of the terms along with their sign.
1) Split the middle term of the quadratic equation using the factors of the constant term and the coefficient of ${x}^{2}$.
2) Factorise the equation by taking common in the first two and the last two terms.
3) Equate each of the factors to zero and find the value of x.
OR
1) Equate the given quadratic equation to the standard form of the quadratic equation and find the values of a, b and c.
2) Find the value of the discriminant by using the coefficients.
3) Find the roots of the quadratic equation by substituting the given values in the formula to find the roots of the quadratic equation.
The given quadratic equation is 4${x}^{2}$ âˆ’ 4ax + (${a}^{2}$âˆ’${b}^{2}$) = 0
⇒ 4${x}^{2}$ âˆ’ 4ax + ( a - b)( a + b) = 0
⇒ 4${x}^{2}$ âˆ’( a - b)( a + b) = 0
⇒ 2x [ 2x - ( a - b) ] - ( a + b)[ 2x - ( a - b) ] = 0
⇒ [2x - ( a - b)] [ 2x - ( a + b) ] = 0
⇒ 2x = ( a - b) or 2x = ( a + b)
⇒ x = $\frac{a\xe2\u02c6\u2019b}{2}$ or x = $\frac{a+b}{2}$
âˆ´ The solution of the given quadratic equation is given by x = $\frac{a\xe2\u02c6\u2019b}{2}$ or x = $\frac{a+b}{2}$
OR
Given quadratic equation is 3${x}^{2}$-2$\sqrt{6}$x + 2 = 0.
Comparing coefficients with the quadratic equation a${x}^{2}$+ bx + c = 0, we have a = 3 , b = - 2$\sqrt{6}$ and c = 2.
Discriminant of the given quadratic equation,
D = ${b}^{2}$-4ac = ${\left(2\sqrt{6}\right)}^{2}$-4$\left(3\right)\left(2\right)=24\xe2\u02c6\u201924=0\text{}$
Roots of a quadratic equation a${x}^{2}$+ bx + c = 0 are $\frac{-b\xc2\pm \sqrt{D}}{2a}$.
x = $\frac{\xe2\u02c6\u2019\left(\xe2\u02c6\u20192\sqrt{6}\right)\xc2\pm \sqrt{0}}{2\times 3}$
x = $\frac{2\sqrt{6}}{6}$
x = $\frac{\sqrt{6}}{3}$
Hence, the solution of the given quadratic equation is x = $\frac{\sqrt{6}}{3}$.
20. Prove that the parallelogram circumscribing a circle is a rhombus.
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
1) Lengths of tangents drawn from an external point to the circle are equal.2) Group the tangents according to the sides of the parallelogram.
OR
Prove that the corresponding sides of the triangles are equal to prove that the triangle are congruent.
1) In a parallelogram ABCD , the circle touch the sides AB, BC, CD and AD at P, Q, R and S respectively.
2) Since, ABCD is a parallelogram then AB = CD and BC = AD.
3) Lengths of tangents drawn from an external point to the circle are equal.
4) Adding all these and grouping the lengths of these tangents according to the sides of the parallelogram.
5) Therefore, ABCD is a parallelogram that has sides of equal length.
6) Hence, the parallelogram ABCD is also a rhombus.
OR
1) Draw a quadrilateral ABCD circumscribing a circle touching the circle at P, Q, R and S.
2) Draw lines joining the vertices of the quadrilateral to the centre of the circle.
3) Draw perpendiculars OP, OQ, OR and OS on the sides AB, BC, CD and AD of the quadrilateral.
4) Prove that the triangles OAP and OAS are congruent and hence the angles âˆ POA and âˆ AOS are equal.
5) Similarly prove that the angles in the pairs of adjacent right angled triangles are equal.
6) Add the equations of angles and regroup the angles.
7) Simplify and prove that the angles subtended by the opposite sides of the quadrilateral are supplementary.
Given that, ABCD is a parallelogram.
AB = CD â€¦ (1)
BC = AD â€¦ (2)
We know that
DR = DS (Tangents to the circle from point D)
CR = CQ (Tangents to the circle from point C)
BP = BQ (Tangents to the circle from point B)
AP = AS (Tangents to the circle from point A)
Adding all these equations, we get
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On substituting the values of equations (1) and (2) in this equation, we get
2AB = 2BC
AB = BC â€¦ (3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
âˆ´ABCD is a rhombus.
OR
Consider ABCD a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R and S.
Let us join the vertices of the quadrilateral ABCD to the centre of the circle.
In Î”OAP and Î”OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the same circle)
OA = OA (Common side)
Î”OAP â‰… Î”OAS (SSS congruence criterion)
Therefore, A â†” A, P â†” S, O â†” O
And thus, âˆ POA = âˆ AOS
âˆ 1 = âˆ 8
Similarly,
âˆ 2 = âˆ 3
âˆ 4 = âˆ 5
âˆ 6 = âˆ 7
âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 + âˆ 5 + âˆ 6 + âˆ 7 + âˆ 8 = 360Âº
(âˆ 1 + âˆ 8) + (âˆ 2 + âˆ 3) + (âˆ 4 + âˆ 5) + (âˆ 6 + âˆ 7) = 360Âº
2âˆ 1 + 2âˆ 2 + 2âˆ 5 + 2âˆ 6 = 360Âº
2(âˆ 1 + âˆ 2) + 2(âˆ 5 + âˆ 6) = 360Âº
(âˆ 1 + âˆ 2) + (âˆ 5 + âˆ 6) = 180Âº
âˆ AOB + âˆ COD = 180Âº
Similarly, we can prove that âˆ BOC + âˆ DOA = 180Âº
âˆ´ Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
21. Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm and 8 cm. Then construct another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of the given triangle.
Locate five points on BX such that they are equidistant.
1) Construct a triangle ABC using the measurements of three sides.
2) Draw a ray BX making an acute angle with BC.
3) Mark 5 points on BX and join the fifth point to C.
4) Draw a line ${B}_{3}$C' parallel to ${B}_{5}$C from ${B}_{3}$.
5) Draw a line C'A' parallel to CA.
6) A'BC' is the required triangle.
Given that
BC = 6 cm, AC = 8 cm
The triangle to be formed is to be right angled triangle.
Steps of construction:
Step 1: Draw a line segment, BC = 6 cm.
Step 2: Draw a ray CN at C making an angle of 90Â°.
Step 3: From C as centre, 8 cm as the radius draw an arc at CN intersecting it at A and Join AB.
Step 4: ABC is the triangle whose similar triangle is to be drawn.
Step 5: Draw any ray BX making an acute angle with BC on the opposite side of the vertex A.
Step 6: Locate 5 (Greater of 3 and 5 in $\frac{3}{5}$) points ${B}_{1}$, ${B}_{2}$,${B}_{3}$,${B}_{4}$and${B}_{5}$on BX such that B${B}_{1}$= ${B}_{1}{B}_{2}$= ${B}_{2}{B}_{3}$=${B}_{3}{B}_{4}$= ${B}_{4}{B}_{5}$.
Step 7: Join ${B}_{5}$C and draw a line through${B}_{3}$ (Smaller of 3 and 5 in $\frac{3}{5}$) parallel to ${B}_{5}$C to intersect BC at Câ€™.
Step 8: Draw a line through Câ€™parallel to CA to intersect BA at Aâ€™.
Step 9: Aâ€™BCâ€™ is the required similar triangle whose sides are $\frac{3}{5}$ times the corresponding sides of Î”ABC.
22. In Fig. 7, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm and centre O. If âˆ POQ = 30Â°, then find the area of the shaded region. [ Use Ï€ = $\frac{22}{7}$]
Angle subtended by both the arcs at the centre is 30Â°.
1) Shaded region lies between the sectors formed by the arcs.
2) Difference between the areas of the sectors formed by the arcs of the concentric circles gives the area of the shaded region.
3) Angle subtended by both the arcs at the centre is 30Â°.
4) substitute the values and simplify to find the area of the shaded region.
Let PQ and AB are the arcs of two concentric circles of radii 7 cm and 3.5 cm respectively.
Assume ${r}_{1}$and${r}_{2}$as the radii of the outer and the inner circle respectively.
Suppose the angle subtended by the arcs at the centre O as Î¸.
Then${r}_{1}$= 7 cm, ${r}_{2}$= 3.5 cm and Î¸ = 30Â°
Area of the shaded region = Area of sector OPQ âˆ’ Area of sector OAB
= $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}$× Ï€${{r}_{1}}^{2}$âˆ’$\frac{\mathrm{\xce\xb8}}{360}$ × Ï€${{r}_{2}}^{2}$
= $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}$Ï€(${{r}_{1}}^{2}$âˆ’${{r}_{2}}^{2}$)
= $\frac{30\xc2\xb0}{360\xc2\xb0}$ × $\frac{22}{7}$ × (${7}^{2}$âˆ’${\left(3.5\right)}^{2}$)
= $\frac{1}{12}$ × $\frac{22}{7}$ × 36.75
= 9.265 ${\mathrm{cm}}^{2}$
âˆ´ The area of the shaded region is 9.625 ${\mathrm{cm}}^{2}$.
23. From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. [ Ï€ = $\frac{22}{7}$]
OR
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then find the radius and slant height of the heap.
Consider one base area of the cylinder to find the total surface area of the remaining solid.
OR
Equate the volumes of the cylinder and the conical heap to find the radius of the conical heap.
1) Find the radius of the cylindrical part of the solid cylinder.
2) Find the slant height of the conical part using the radius and the height of the cylinder.
3) Total surface area of the remaining solid is the sum of the curved surface areas of the cylindrical part and the conical part, and the base area of the cylindrical part.
4) Substitute the values and simplify to find the total surface area of the remaining solid.
OR
1) Assume the radii of the cylindrical bucket and the conical heap as ${r}_{1}$and ${r}_{2}$.
2) Equate the volumes of the sand in the form of a cylinder and the conical heap.
3) Substitute the values and simplify to find the radius of the conical heap.
4) Find the slant height of the conical heap of sand using the vertical height and the radius.
Given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm
Diameter of the cylindrical part = 12 cm
âˆ´Radius (r) of the cylindrical part =$\frac{12}{2}$= 6 cm
âˆ´ Radius of conical part = 6 cm
Slant height (l) of conical part = $\sqrt{{r}^{2}+{h}^{2}}$
= $\sqrt{{6}^{2}+{7}^{2}}$
= $\sqrt{36+49}$
= $\sqrt{85}$
= 9.22 cm ( approximately)
Total surface area of the remaining solid
= CSA of cylindrical part + CSA of conical part + Base area of the circular part
= 2Ï€rh + Ï€rl + Ï€${r}^{2}$
= (2 × $\frac{22}{2}$ × 6 × 7) + ($\frac{22}{7}\times $6 × 9.22) + ($\frac{22}{7}$ × 6 × 6)
= (264 + 173.86 + 113.14) ${\mathrm{cm}}^{2}$
= 551 ${\mathrm{cm}}^{2}$
OR
Height (${h}_{1}$) of the cylindrical bucket = 32 cm
Radius (${r}_{1}$) of circular end of the bucket = 18 cm
Height (${h}_{2}$) of the conical heap = 24 cm
Let the radius of the circular end of conical heap be ${r}_{2}$.
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Volume of sand in the cylindrical bucket = Volume of sand in conical heap
⇒Ï€${{r}_{1}}^{2}{h}_{1}$= $\frac{1}{3}\mathrm{\xcf\u20ac}{{r}_{2}}^{2}{h}_{2}$
⇒Ï€$\times {18}^{2}\times $32 = $\frac{1}{3}$Ï€ ×${{r}_{2}}^{2}$× 24
⇒${{r}_{2}}^{2}$= $\frac{3\times {18}^{2}\times 32}{24}$= ${18}^{2}$× 4
⇒${r}_{2}$ = 18 ×2 = 36 cm
Slant height (l) = $\sqrt{{36}^{2}+{24}^{2}}$ = 12$\sqrt{13}$cm
Hence, the radius and slant height of the conical heap are 36 cm and 12$\sqrt{13}$cm respectively
24. The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45Â° and 30Â°. If the ships are 200 m apart, find the height of the light house
Apply appropriate trigonometric ratio to the angles of depression of two ships from the top of the light house.
1) Draw a figure representing the data given in the question.
2) Assume 'h' as the height of the light house and 'x' as the distance between the foot of the light house and the first ship.
3) Mark the alternate angles of the angles of depression.
4) Find the tangent of the alternate angle of the angle of depression of the first ship and find the value of 'x' in terms of 'h'.
5) Find the tangent of the alternate angle of the angle of depression of the second ship and find the value of h.
6) Substitute the value of 'x' in the terms of h in the trigonometric ratio and simplify to find the height of the light house.
The given solution can be represented as,
Consider AB as the light house and the ships are at the points C and D.
Let h be the height of the light house and BC = x.
In a right angled Î”ABC:
tan 45Â° = $\frac{\mathrm{AB}}{\mathrm{AC}}$
⇒1 = $\frac{h}{x}$
⇒x = h
In a right angled Î”ABD,
tan 30Â° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
⇒ tan 30Â° = $\frac{h}{x+200}$
⇒ tan 30Â° = $\frac{h}{h+200}$
⇒$\frac{1}{\sqrt{3}}$ = $\frac{h}{h+200}$
⇒$\sqrt{3}h$ - h = 200
⇒h($\sqrt{3}$ - 1) = 200
⇒h = $\frac{200}{\sqrt{3}\xe2\u02c6\u20191}$
⇒h = $\frac{200}{\sqrt{3}\xe2\u02c6\u20191}$× $\frac{\sqrt{3}+1}{\sqrt{3}+1}$
⇒h = $\frac{200}{2}(\sqrt{3}+1)$
⇒h = 100 ($\sqrt{3}$+ 1 )
âˆ´The height of the light house is 100 ($\sqrt{3}$+ 1) m.
25. A point P divides the line segment joining the points A (3, âˆ’5) and B (âˆ’4, 8) such that $\frac{\mathrm{AP}}{\mathrm{PB}}$= $\frac{k}{1}$. If P lies on the line x + y = 0, then find the value of k.
Substitute the values of the coordinates according to the section formula in order.
1) The point P divides the line segment AB in the ratio of k:1.
2) Use the section formula and find the coordinates of the point P.
3) The point P lies on the line x + y = 0, so it should satisfy the equation.
4) Substitute the coordinates of the point P for x and y in the given equation.
5) Simplify to find the value of k.
The given points are A (3, âˆ’5) and B (âˆ’ 4, 8).
Consider ${x}_{1}$= 3, ${y}_{1}$= âˆ’5, ${x}_{2}$= âˆ’4 and${y}_{2}$= 8.
Since $\frac{\mathrm{AP}}{\mathrm{PB}}$ = $\frac{k}{1}$, the point P divides the line segment joining the points A and B in the ratio of k : 1. The coordinates of P can be found using the section formula .
$(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n})$
Put, m = k and n = 1
Coordinates of P = $(\frac{k\times \xe2\u02c6\u20194+1\times 3}{k+1},\frac{k\times 8+1\times \xe2\u02c6\u20195}{k+1})$
= $(\frac{\xe2\u02c6\u20194k+3}{k+1},\frac{8k\xe2\u02c6\u20195}{k+1})$
Given that, P lies on the line x + y = 0.
$\frac{\xe2\u02c6\u20194k+3}{k+1}$+ $\frac{8k\xe2\u02c6\u20195}{k+1}$= 0
$\frac{\xe2\u02c6\u20194k+3+8k\xe2\u02c6\u20195}{k+1}$= 0
4k - 2 = 0
k = $\frac{1}{2}$
âˆ´ The required value of k is $\frac{1}{2}$.
26. If the vertices of a triangle are (1, âˆ’3), (4, p) and (âˆ’9, 7) and its area is 15 sq. units, find the value(s) of p.
Find two values of p, by equating the area to both positive and negative values of the given area.
1) Find the area of the triangle formed by the given points using the formula to find the area of the triangle.
2) Equate the expression obtained to both positive and negative values of the given area.
3) Simplify and find two values of p.
Given that vertices of a triangle are (1, âˆ’3), (4, p) and (âˆ’9, 7).
${x}_{1}$= 1, ${y}_{1}$= -3 , ${x}_{2}$= 4 , ${y}_{2}$= p , ${x}_{3}$= -9 and ${y}_{3}$= 7.
Area of given triangle = $\frac{1}{2}$[${x}_{1}$(${y}_{2}$- ${y}_{3}$) + ${x}_{2}$(${y}_{3}$- ${y}_{1}$) + ${x}_{3}$(${y}_{1}$- ${y}_{2}$)
= $\frac{1}{2}$[1$\left(p\xe2\u02c6\u20197\right)$+ 4$\left(7+3\right)$+ $\left(\xe2\u02c6\u20199\right)\left(\xe2\u02c6\u20193\xe2\u02c6\u2019p\right)$]
= $\frac{1}{2}$[p - 7 + 40 + 27 + 9p ]
= $\frac{1}{2}$[ 10p + 60 ]
= 5( p + 6)
The expression may be positive or negative.
⇒5(p + 6) = 15 or 5(p + 6) = âˆ’ 15
⇒ p + 6 = 3 or p + 6 = âˆ’ 3
âˆ´ p = âˆ’ 3 or p = âˆ’ 9.
27. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card.
1) Cards other than yellow includes red and blue cards.
2) Cards which are neither yellow nor blue are red.
1) Find the total number of cards of all the three colours.
2) Ratio of number of blue cards and the total number of cards gives the probability that the card drawn is a blue card.
3) Ratio of the sum of the number of red and blue cards, and the total number of cards gives the probability that the card drawn is not a yellow card.
4) Ratio of number of red cards and the total number of cards gives the probability that the card drawn is neither a yellow nor a blue card.
Number of red cards = 100
Number of yellow cards = 200
Number of blue cards = 50
Total number of cards = 100 + 200 + 50 = 350
(i)
P(A blue card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{blue}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$
= $\frac{50}{350}$
= $\frac{1}{7}$.
(ii)
P(Not a yellow card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{other}\mathrm{than}\mathrm{yellow}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$
= $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{card}s+\mathrm{Number}\mathrm{of}\mathrm{blue}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$
= $\frac{100+50}{350}$
= $\frac{150}{350}$
= $\frac{3}{7}$.
(iii)
P(Neither yellow nor a blue card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{that}\mathrm{are}\mathrm{neither}\mathrm{yellow}\mathrm{nor}\mathrm{blue}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$
= $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$
= $\frac{100}{350}$
= $\frac{2}{7}$.
28. The${17}^{\mathrm{th}}$term of an A.P. is 5 more than twice its${8}^{\mathrm{th}}$term. If the ${11}^{\mathrm{th}}$term of the A.P. is 43, then find its${n}^{\mathrm{th}}$term.
Find the value of a in terms of d.
1) Write the equation representing the given condition.
2) Substitute the formulas to find the ${17}^{\mathrm{th}}$term and ${8}^{\mathrm{th}}$term in the given condition.
3) Simplify the equation and find the value of a in terms of d.
4) Find the ${11}^{\mathrm{th}}$term of the A.P. and equate it to the given value.
5) Substitute the value of a in terms of d.
6) Simplify to find the value of d.
7) Find the value of a using the value of d.
8) Find the ${n}^{\mathrm{th}}$term of the A.P. by substituting the values of a and d in the formula to find the ${n}^{\mathrm{th}}$term of an A.P.
Let a be the first term and d be the common difference of the given A.P.
According to the given condition,
${17}^{\mathrm{th}}$term = (2 ×${8}^{\mathrm{th}}$ term) + 5
i.e.,${a}_{17}$ = 2${a}_{8}$ + 5
⇒a + (17âˆ’1)d = 2[a + (8âˆ’ 1)d] + 5
⇒a + 16d = 2[a + 7d] + 5
⇒a + 16d = 2a + 14d + 5
⇒a = 2d âˆ’ 5 ----------- (1)
But${11}^{\mathrm{th}}$term,${a}_{11}$= 43
⇒a + ( 11 âˆ’ 1)d = 43
⇒a + 10d = 43
⇒2d âˆ’ 5 + 10 d = 43 [From (i)]
⇒12d = 48
⇒ d = 4
âˆ´a = 2dâˆ’5 = (2 × 4) âˆ’ 5 = 3
But ${n}^{\mathrm{th}}$term of the A.P., ${a}_{n}$= a + (n âˆ’ 1)d
Substituting the values of a and d, we get
${a}_{n}$= 3 + (n âˆ’ 1) 4 = 3 + 4n âˆ’ 4 = 4n - 1
âˆ´${n}^{\mathrm{th}}$ term of the given A.P. is 4n âˆ’ 1.
29. A shopkeeper buys some books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Find the number of books he bought.
OR
The sum of two numbers is 9 and the sum of their reciprocals is $\frac{1}{2}$. Find the numbers.
Subtract the fractions by taking LCM.
1) Assume the number of purchased by the shopkeeper as x.
2) Find the original cost price of the book in terms of x.
3) Find the new cost price for the increased number of books.
4) Difference between the cost prices is Rs 1.
5) Write the equation representing the difference between the cost prices as Rs 1.
6) Simplify and solve the quadratic equation obtained to find the number of books.
OR
1) Assume the first number as x.
2) Find the second number as (9 - x)
3) Obtain a quadratic equation by applying the given condition to the numbers.
4) Solve the quadratic equation and find the assumed number.
5) Find the other number by using the sum of the numbers.
Consider the number of books purchased by the shopkeeper be x.
Cost price of x books = Rs 80
âˆ´Original cost price of one book = Rs.$\frac{80}{x}$
If the shopkeeper had purchased 4 more books, then the number of books purchased by him would be (x + 4).
âˆ´New cost price of one book = Rs $\frac{80}{(x+4)}$
Given that, original cost price of one book âˆ’ New cost price of one book = Rs 1
⇒$\frac{80}{x}$ - $\frac{80}{(x+4)}$= 1
⇒$\frac{80(x+4)\xe2\u02c6\u201980x}{x(x+4)}$= 1
⇒80x + 320 - 80x = x(x + 4)
⇒${x}^{2}$+ 4x = 320
⇒${x}^{2}$+ 4x - 320 = 0
⇒${x}^{2}$+ 20x - 16x - 320 = 0
⇒x(x + 20) - 16( x + 20) = 0
⇒x - 16 = 0 or x + 20 = 0
⇒x = 16 or x = -20 ( Number of books cannot be negative)
âˆ´ x = 16
âˆ´ The number of books purchased by the shopkeeper is 16.
OR
Assume the first number as x.
Given that, the sum of the first number and the second number is 9.
âˆ´ (x + Second number) = 9
⇒ Second number = 9 âˆ’ x
Sum of their reciprocals is $\frac{1}{2}$.
âˆ´ $\frac{1}{x}$ + $\frac{1}{9\xe2\u02c6\u2019x}$= $\frac{1}{2}$
⇒$\frac{9\xe2\u02c6\u2019x+x}{x(9\xe2\u02c6\u2019x)}$= $\frac{1}{2}$
⇒ 9 × 2 = 9x âˆ’${x}^{2}$ ⇒
⇒${x}^{2}$âˆ’9x + 18 = 0 ⇒
⇒ ${x}^{2}$âˆ’6x âˆ’ 3x + 18 = 0
⇒ x(x âˆ’ 6) âˆ’ 3(x âˆ’ 6) = 0 ⇒
⇒ (x âˆ’ 3) (x âˆ’ 6) = 0 ⇒
⇒ x âˆ’ 3 = 0 or x âˆ’ 6 = 0
⇒ x = 3 or x = 6
When x = 3, we have 9 âˆ’ x = 9 âˆ’ 3 = 6
When x = 6, we have 9 âˆ’ x = 9 âˆ’ 6 = 3
âˆ´ The two numbers are 3 and 6.
30. Sum of the first 14 terms of an A.P. is 1505, and its first term is 10. Find its ${25}^{\mathrm{th}}$term.
Use the formula to find the sum of the n terms of the A.P. involving the first term and the common difference.
1) Assume the common difference of the A.P. as d.
2) Find the sum of the 14 terms of the A.P. by substituting the values of a and n and equate it to the given value.
3) Simplify the equation to find the value of d.
4) Find the ${25}^{\mathrm{th}}$term of the A.P. by using the values of a, n and d.
Given that first term, a = 10 and sum of first 14 terms is 1505.
Let the common difference of the A.P. be d.
The sum of first n terms of an A.P., ${S}_{n}$= $\frac{n}{2}$[2a + ( nâˆ’1)d]
The sum of first 14 terms of the A.P.:
${S}_{14}$ = $\frac{14}{2}$[2×10 + (14 âˆ’1)d] = 1505
⇒7[20 + 13d] = 1505
⇒20 + 13d = $\frac{1505}{7}$
⇒20 + 13d = 215
⇒13d = 215 âˆ’ 20
⇒13d = 195
⇒ d = 15.
âˆ´${25}^{\mathrm{th}}$term of the A.P., ${a}_{25}$= 10 + (25 âˆ’ 1) 15 = 10 + (24 × 15) = 370. [${a}_{n}$= a + (n âˆ’ 1)d]
âˆ´ ${25}^{\mathrm{th}}$term of the A.P is 370.
31. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
OR
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
The shortest distance between two points is the perpendicular distance.
OR
Arrange the tangents of the circle according to the sides of the quadrilateral.
1) Draw a figure representing the data given in the question.
2) Mark any point B on the tangent l and draw a line segment joining the centre O and the point B.
3) Let the line segment intersect the circle at C.
4) OB is written as the sum of two parts OC and CB.
5) OB is greater than OC. So, OB is greater than OA since OA and OC are radii of the circle.
6) OA is shorter than OB.
7) OA is shorter than any other segment joining the centre of the circle and a point on the tangent.
8) This proves that OA, the radius at the point of contact of the tangent is perpendicular to the tangent.
OR
1) Draw a figure representing the data given in the question.
2) Write each side of the quadrilateral as sum of two parts.
3) Write the tangents from each vertex of the quadrilateral and they are equal.
4) Add one pair of opposite sides of the quadrilateral and equate to the sum of their parts.
5) Replace the equivalent tangents of the parts of the opposite sides considered.
6) Rearrange the parts of the sides and prove that the sums of the opposite sides of a quadrilateral are equal.
Given: A circle of centre O and a tangent l at point A.
To prove: OA âŠ¥ l
Construction: Consider any point B other than A on the tangent l. Join O and B.
Suppose OB intersects the circle at C.
Proof: Among all line segments joining the centre O to any point on l, the perpendicular is the shortest line.
So, in order to prove OA âŠ¥ l, we need to prove that OA is shorter than OB.
OA = OC (Radius of same circle)
Now, OB = OC + BC
⇒ OB > OC
⇒ OB > OA
⇒ OA < OB
B is an arbitrary point on the tangent l.
Thus, OA is shorter than any other line segment joining O to any point on l.
âˆ´ OA âŠ¥ l.
OR
Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure.
Length of the tangents drawn from an external point to the circle are equal.
Therefore,
AP = AS ---------(1)
BP = BQ ---------(2)
CQ = CR -----------(3)
DR = DS ------------(4)
âˆ´ AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [Using (1) ,(2),(3) and (4)]
= (AS + DS) + (BQ + CQ)
= AD + BC
âˆ´ AB + CD = AD + BC
Hence, it is proved.
32. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. [ Use Ï€ = $\frac{22}{7}$]
Radius and the height of the hemisphere are equal.
1) Find the height of the cone by subtracting the radius i.e. height of the hemisphere from the total height of the solid.
2) Find the volume of the hemisphere and the volume of the cone using the given dimensions.
3) Sum of the volumes of the hemisphere and the cone gives the volume of the solid.
Given that radius of cone = Radius of hemisphere = 3.5 cm
Total height of solid (OC) = 9.5 cm
⇒ OD + DC = 9.5
⇒ OD + 3.5 = 9.5 [DC (radius of hemisphere) = 3.5 cm]
⇒ OD = 6 cm
âˆ´ Height of cone = OD = 6 cm
Volume of solid = Volume of cone + Volume of hemisphere
= ($\frac{1}{3}$Ï€${r}^{2}$h) + ($\frac{2}{3}$Ï€${r}^{3}$)
= $\frac{\mathrm{\xcf\u20ac}{r}^{2}}{3}$[ h + 2r]
= $\frac{22}{7}\times \frac{3.5\times 3.5}{3}$× [ 6 + (2 × 3.5)]
= $\frac{22\times 0.5\times 3.5\times 13}{3}$
= $\frac{500.5}{3}$
= 166.83 $$
âˆ´ The volume of the solid is 166.83 ${\mathrm{cm}}^{3}$.
33. A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket.[ Use Ï€ = $\frac{22}{7}$]
Convert the volume of the water into${\mathrm{cm}}^{3}$.
1) Assume the height of the bucket in the form of frustum of a cone as h.
2) Find the volume of the bucket using the formula to find the volume of the frustum of the cone.
3) Convert the given volume of the water into${\mathrm{cm}}^{3}$.
4) Equate the volumes and simplify to find the value of h which gives the height of the bucket.
Consider the height of the bucket be h cm.
Let${r}_{1}$and${r}_{2}$ be the radii of the circular ends of the bucket.
Given that,${r}_{1}$ = 28 cm and${r}_{2}$= 21 cm
Capacity of bucket = 28.49 litres
âˆ´Volume of the bucket = 28.49 × 1000${\mathrm{cm}}^{3}$ [1 litre = 1000 ${\mathrm{cm}}^{3}$]
⇒$\frac{1}{3}\mathrm{\xcf\u20ach}({{r}_{1}}^{2}+{{r}_{2}}^{2}+{r}_{1}{r}_{2})=28.49\times 1000$
⇒$\frac{1}{3}$ × $\frac{22}{7}$ × h × [ ${28}^{2}$+ ${21}^{2}$+ (28 × 21)] = 28490 $$
⇒$\frac{22}{21}$ × h × 1813 = 28490
⇒ h = $\frac{28940\times 21}{22\times 1813}$cm
⇒ h = 15 cm
âˆ´ Height of the bucket is 15 cm.
34. The angle of elevation of the top of a hill at the foot of a tower is 60Â° and the angle of depression from the top of the tower to the foot of the hill is 30Â°. If the tower is 50 m high, find the height of the hill.
Apply appropriate trigonometric ratio of the angle of elevation and the angle of depression.
1) Draw a figure representing the data given in the question.
2) Form right angled triangles involving the angle of elevation and the angle of depression.
3) Mark the alternate angle of the angle of depression.
4) Find the tangent of the alternate angle of the angle of depression and find the distance between the tower and the hill.
5) Find the tangent of the angle of elevation and find the height of the hill.
Consider AB is the hill and CD is the tower.
Angle of elevation of the hill at the foot of the tower is 60Â°, i.e., âˆ ADB = 60Â°
The angle of depression of the foot of hill from the top of the tower is 30Â°, i.e., âˆ CBD = 30Â°.
In a right angled triangle Î”CBD:
⇒ tan 30Â° = $\frac{\mathrm{CD}}{\mathrm{BD}}$
⇒ BD = $\frac{\mathrm{CD}}{\mathrm{tan}30\xc2\xb0}$
⇒ BD = $\frac{50}{\frac{1}{\sqrt{3}}}$ = 50$\sqrt{3}$
In a right angled triangle Î”ABD:
⇒tan 60Â° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
⇒ AB = BD × tan 60Â°
⇒ AB = 50$\sqrt{3}$ × $\sqrt{3}$
⇒ AB = 150 m
âˆ´ The height of the hill is 150 m.
Section A | Section B | Section C | Section D |
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