maths- solved sums

7/30/2019 Maths- Solved Sums

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Maths - Measures of Central Tendency (Solved)

Question 1. The marks of 20 students in a test were as follows:5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20.

Solution: Data is already arranged in ascending order.

(i)Mean = (5 + 6 + 8 + 9 + 10 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20)/20= 260/20 = 13. [Ans.]

(ii) n = 20 [even number]

Median = [(n/2) th + {(n/2) + 1}]/2 observation

= [(20/2) th observation + {(20/2) + 1}th observation ]/2

=(10 th observation + 11 th observation)/2

= (13 + 14)/2 = 13.5. [Ans.]

(iii) Mode = 15, as it occurs 3 times (The most often.). [Ans.]

Question 2. Given below are the weekly wages of 200 workers in a factory :

Weekly wages inrupees

Number of workers

80 100 20

100 120 30

120 140 20

140 160 40

160 180 90

Calculate the mean weekly wages of the workers.

Solution :

Method I

Weekly wages in Rs.No. of workers f Mid-values (x) fx

80 100 20 90 1,800

100 120 30 110 3,300

120 140 20 130 2,600

140 160 40 150 6,000


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160 180 90 170 15,300

Total f = 200 fx = 29,000

Mean = fx/f = 29,000/200 = Rs.145. [Ans.]

Method II

Weekly wages in Rs,No. of workers fi Mid-value xi di = xi A difi

80 100 20 90 40 0 800

100 120 30 110 20 0 600

120 140 20 130 = A 0 0

140 160 40 150 20 800

160 180 90 170 40 3600

Total fi = 200 difi = 3000

Mean = A + difi/fi = 130 + 3000/200 = 130 + 15 = 145. [Ans.]

Method III

Weekly wages in Rs. No. of workers fi Mid-values xiui = (xi A )/h =(xi

130)/20uifi

80 100 20 90 2 40

100 120 30 110 1 30

120 140 20 130 = A 0 0

140 160 40 150 1 40

160 180 90 170 2 180

Total fi = 200 uifi = 150

Mean = A + (uifi/fi)h = 130 + ( 150/200)20

= 130 + 15 = 145. [Ans.]


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Question 3. If the mean of 5, 4, 6, 3, 2x + 4, 3x 6, 5, 6, and 7 is 10, find the value of x.Solution :

By definition, mean = x/n.

Here, x = 5 + 4 + 6 + 3 +(2x + 4) + (3x 6) + 5 + 6 + 7 = 5x + 34,

mean = 10 and n = 9,

Hence, 10 = (5x + 34)/9

Or, 5x + 34 = 90Or, 5x = 90 34 = 56

Or, x = 56/5 = 11.2. [Ans.]

Question 4. The marks obtained by a set of students in an examination are given below:

Marks5 10 15 20 25 30

No. ofstudents

6 4 6 12 x 4

Given that the mean marks of the set is 18, calculate the value of x.

Solution :

Marks , (x)No. of students, (f) fx

5 6 30

10 4 40

15 6 90

20 12 240

25 x 25x

30 4 120

Total f = 32 + x

fx = 520

+ 25x

As, Mean = fx/f

18 = (520 + 25x)/(32 + x)

Or, 18(32 + x) = 520 + 25x

Or, 576 + 18x = 520 + 25x

Or, 25x 18x = 576 520

Or, 7x = 56


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Or, x = 56/7 = 8. [Ans.]

Question 5. If the median of 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 is 24, find the value of x.Solution :

Here we suppose that the data is arranged in ascending order.

Number of observation = 10.

Median = 1/2(5th observation + 6th observation)= 1/2(x + 2 + x + 4)

Or,24 = 1/2(2x + 6) = x + 3

Or,x = 24 3 = 21. [Ans.]

Maths - Trigonometrical Identities (Solved)

Question 1. Prove the following identities :

(i) (secA 1)/(secA + 1) = (1 cosA)(1 + cosA)

(ii) 1/(sin + cos) + 1/(sin cos) = 2sin/(1 2cos2)

(iii)cot2A cos2A = cot2A cos2A

(iv)(1 + sin)/(1 sin) = sec + tan

(v) (1 sin)/(1 + sin) = (sec - tan)2

(vi) sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA

(vii) cosA/(1 tanA) + sinA/(1 cotA)

(viii)1 cos2/(1 + sin) = sin

(ix) (1 cosA)/(1 + cosA) = sinA/(1 + cosA)(x) sin tan /(1 cos) = 1 + sec

(xi) (1 + tanA)2 + (1 tanA)2 = 2 sec2A

Solution :

(i) LHS = (secA 1)/(secA + 1) = (1/cosA 1)/(1/cos + 1)

= [(1 cosA)/cosA]/[(1 + cosA)/cosA]

= (1 cosA)/(1 + cosA) = RHS.

(ii) LHS = 1/(sin + cos) + 1/(sin cos)

= {(sin cos) + (sin + cos)}/(sin + cos)(sin cos)

= 2sin/(sin2 cos2)

= 2sin/[(1 cos2) cos2]

= 2sin/(1 2cos2) = RHS.

(iii) LHS = cot2A cos2A = cos2A/sin2A cos2A


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= cos2A(1/sin2A 1)

= cos2A(cosec2A 1)

= cos2A cot2A = RHS

(iv) LHS = (1 + sin)/(1 sin) = (1 + sin)/(1 sin)(1 + sin)/(1 + sin)

= (1 + sin)2/(1 sin2)

= (1 + sin)2/(cos2)

= (1 + sin)/cos = 1/cos + sin/cos

= sec + tan = RHS

(v) RHS = (sec tan)2 = (1/cos sin/cos)2 = {(1 sin)/cos}2

= (1 sin)2/(cos)2 = (1 sin)2/(1 sin2)

= (1 sin)2/{(1 + sin)(1 sin)}

= (1 sin)/(1 + sin) = LHS

(vi) LHS = sinA/(1 + cosA) + (1 + cosA)/sinA

= {sin2A + (1 + cosA)2}/{(1 + cosA) sinA}

= {sin2A + 1 + 2cosA + cos2A}/{(1 + cosA) sinA}

= {(sin2A + cos2A) + 1 + 2cosA}/{(1 + cosA) sinA}

= {1 + 1 + 2cosA}/{(1 + cosA) sinA}

= {2(1 + cosA)}/{(1 + cosA) sinA}

= 2/sinA = 2 cosecA = RHS

(vii) LHS = cosA/(1 tanA) + sinA/(1 cotA)

= cosA/(1 sinA/cosA) + sinA/(1 cosA/sinA)

= cos2A/(cosA sinA) + sin2A/(sinA cosA)

= cos2A/(cosA sinA) sin2A/(cosA sinA)

= (cos2A sin2A)/(cosA sinA)

= (cosA + sinA)(cosA sinA)/(cosA sinA)

= cosA + sinA = RHS

(viii) LHS = 1 cos2/(1 + sin) = (1 + sin cos2)/(1 + sin)

= (sin2 + sin)/(1 + sin) = sin (1 + sin)/(1 + sin)

= sin = RHS

(ix) LHS = (1cosA)/(1+cosA) = (1cosA)/(1+cosA)(1+ cosA)/(1+cosA)

= (1 cos2A)/(1 + cosA)2= (sin2A)/(1 + cosA)2

= sinA /(1 + cosA) = RHS


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(x) LHS = sin tan/(1 cos) = {sin (sin/cos)}/(1 cos)

= sin2/cos (1 cos) = (1 cos2)/cos (1 cos)

= {(1 + cos)(1 cos)}/cos (1 cos) = (1 + cos)/cos

= 1/cos + cos/cos = sec + 1 = RHS

(xi) LHS = (1 + tanA)2 + (1 tanA)2 = 1+ 2tanA + tan2A + 1 2tanA + tan2A

= 2 + 2 tan2A = 2 (1 + tan2A) = 2 sec2A = RHSQuestion 2.Evaluate, without the use of trigonometric tables :

(i) sin80/cos10 + sin59sec31

(ii) 3sin72/cos18 sec32/cosec58

(iii) 3cos80cosec10 + 2cos59cosec31

(iv) cos75/sin15 + sin12/cos78 cos18/sin72

(v) 2tan53/cot37 cot80/tan10

Solution :

(i) sin80/cos10 + sin59 sec31 = sin(90 10 ) / cos10 + sin(90 31)1/cos31

= cos10/cos10 + cos31/cos31 [sin(90 ) = cos]

= 1 + 1 = 2. [Ans.]

(ii)3sin72/cos18 sec32/cosec58

= 3sin(90 18)/cos18 sec(90 58)/cosec58

= 3 cos18/cos18 cose58/cosec58

= 31 1 = 2. [Ans.]

(iii) 3cos80cosec10 + 2cos59cosec31= 3cos80cosec(90 80) + 2cos59cosec(90 59)

= 3cos80sec80 + 2cos59sec59

= 3cos801/cos80 + 2cos591/cos59

= 3 + 2 = 5. [Ans.]

(iv) cos75/sin15 + sin12/cos78 cos18/sin72

= cos(90 15)/sin15 + sin(90 78)/cos78 cos(90 72)/sin72

= sin15/sin15 + cos78/cos78 sin72/sin72

= 1 + 1 1 = 1. [Ans.](v) 2tan53/cot37 cos80/tan10 = 2tan(90 37)/cot37 cot(90 10)/tan10

= 2cot37/cot37 tan10/tan10

= 2 1 = 1. [Ans.]

Maths - Construction (Solved)Question 1. Use ruler and compass only in this question.

(i) Draw a circle, with centre O and radius 4 cm.


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(ii) Mark a point P such that OP = 7 cm. Construct the two tangents to the circle from P. Measure andrecord the length of one of the tangents.

Solution :

A circle with centre O and radius 4 cm is drawn.

A point P is taken 7 cm from O.

O is joined with P.

OP is bisected at C.Taking C as centre ad OC as radius a circle is drawn which intersect the previous circle at T and T.

PT and PT are joined which is the required tangents. PT = PT = 5.7 cm (aprox.)

Question 2. Construct an angle PQR = 45. Mark a point S on QR such that QS = 4.5 cm.Construct acircle to touch PQ at Q and also to pass through S.Solution :

L PQR = 45 is drawn.

QX is drawn perpendicular to PQ at Q.

QS = 4.5 cm is taken from QR.

Perpendicular bisector of QS is drawn which intersects QX at O.

Taking O as centre and OQ as radius a circle is drawn which pass through Q and S.

Question 3. Construct a ABC, in which AC = 5 cm, NC = 7 cm and AB = 6 cm.

(i) Mark D, the midpoint of AB.

(ii) Construct the circle which touches BC at C and passes through D.

Solution :

AB = 6 cm is drawn. From B an arc of radius 7 cm and from A an arc of radius 5 cm are drawn. They intersect


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at C. Ac and BC are joined. The point D, the mid-point of AB is drawn. CD is joined. At C perpendicular to ACis drawn. Perpendicular bisector of CD is drawn which intersect OC at O. Taking O as centre and OD as radius,a circle is drawn. This is the required circle to pass through C and D.

Question 4. Using ruler and compass only, construct a ABC such that AB = 5 cm,L

ABC = 75 and theradius of the circum-circle of triangle ABC 3.5 cm. On the same figure, construct a circle touching AB atits middle point, and also touching the side AC.Solution :

AB = 5 cm is drawn. Perpendicular bisector PQ of AB is drawn. Taking A as centre and radius of 3.5 cm, an arcis drawn to intersect PQ at O. With O as centre and OA as radius, a circle is drawn. At B an angle ABR = 75 isdrawn, here BR intersect the circle at C. AC is joined. ABC is the required triangle. Angle bisector of L BAC isdrawn which intersect PQ at D. Taking D as centre and DM as radius, another circle is drawn. It is the anotherrequired circle.

Question 5. Only ruler and compass may be used in this question,

(i) Construct ABC, such that AB = AC = 7 cm, BC = 5 cm.

(ii) Draw AX, the perpendicular bisector of side BC.

(iii) Draw a circle with centre A and radius 3 cm cutting AX at Y.

(iv) Construct another circle to touch the circle with centre A externally at Y and passing through B andC.


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Solution :

(i) BC = 5 cm is drawn. Taking B and C as centre two arcs are drawn to intersect at A. AB and AC are joined.ABC is the required triangle.

(ii) Perpendicular bisector AD of BC is drawn.

(iii) With A as centre a circle of radius 3 cm is drawn which intersect AD at P.

(iv) PB and PC are joined. Perpendicular bisector of PC is drawn which intersect AD at O. With O as centre and

OP as radius, a circle is drawn. This is the required circle through B and C.

Question 6. Ruler and compass only may be used in this question. All construction lines and arcs must beclearly shown, and be of sufficient length and clarity to permit assessment.

(i) Construct ABC, in which AB = 9 cm, BC = 10 cm and L ABC = 45.

(ii) Draw a circle, with centre A and radius 2.5 cm. Let it meet AB at D.

(iii) Construct a circle to touch the circle with centre A externally at D and also to touch the line BC.

Solution :AB = 9 cm is drawn. At B, L ABC = 45 is drawn. BC = 10 cm is taken. A circle with A as centre and radius 2.5cm is drawn which intersect AB at D. A perpendicular on AB at D is drawn which intersect BC at E. Anglebisector ofL BED is drawn which intersect DB at F. Taking F as centre and FD as radius a circle is drawn whichtouch previous circle at D and the line BC at ..

Question 7. Using ruler and compass only, construct

(i) A triangle ABC in which AB = 9 cm, BC = 10 cm and L ABC = 45.


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(ii) construct a circle of radius 2 cm to touch the arms ofL ACB in (i) above.

Solution :

AB = 9 cm is drawn. At B L ABC is drawn 45. BC = 10 cm ia taken. AC is joined and we get ABC. Anglebisector CD ofL ACB is drawn. A line parallel to BC and at a distance of 2 cm from BC is drawn whichintersect CD at D. With D as centre and radius of 2 cm a circle is drawn which touch arms of L ACB.

Maths - Circle (Solved)Question 1. In the figure given below, there are two concentric circles and AD is a chord of larger circle.

Prove that AB = CD.

Solution :

OM perpendicular to AD is drawn. We know that perpendicular from centre to the chord of a circle bisect thechord.

Therefore, AM = MD (for bigger circle) ------------ (i)

and BM = MC (for smaller circle) ------------ (ii)

On subtracting (ii) from (i) we get

AM BM = MD MC , Or, AB = CD. [Proved.]

Question 2. In the figure given below, AOE is a diameter of a circle, write down the measure of sum ofangles ABC and CDE. Give reasons of your answer.


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Solution :

AD is joined.

ADE is a right angle, being angle in a semi-circle.

L ABC + L ADC = 180 [opp. angles of a cyclic quad. ]L ADE = 90 [angle in the semi-circle]

Hence, L ABC + L ADE + L ADC = 270

Or, L ABC + L CDE = 270. [Proved.]

Question 3. In the figure given below, AC is a diameter of a circle with centre O. Chord BD isperpendicular to AC. Write down the angles p, q, r in terms of x .

Solution :

LAOB = 2LADB, Or, LADB = x/2.Now, LADB + LDAC + 90 = 180, Or, p = 90 x/2.

q = LADB = x/2.

r = LCAB = 1/2LCOB = 1/2(180 x ) = 90 x/2.

Question 4. In the given circle below, find the value of x.


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Solution :

LABD = LACD [angles in the same segment are equal]= 30 [LACD = 30 given]

AB is a diameter of the circle, LADB = 90 [angle in a semi-circle is 90]

In ABD, LDAB + LABD + LADB = 180

[sum of all the three angles of a triangle is 180]

Or, x + 30 + 90 = 180 Or, x = 180 90 30 = 60. [Ans.]

Question 5. A circle with centre O, diameter AB and a chord AD is drawn. Another circle is drawn withAO as diameter to cut AD at C. Prove that BD = 2OC.

Solution :

The following figure is drawn according to the question :

C to O and D to B are joined.

L ADB = 90 [angle in a semi-circle is 90, AB is diameter]

L ACO = 90 [same reason, OA is diameter]

In s ACO and ADB, L ACO = L ADB = 90

L CAO = L DAB [common]

Hence, ACO ~ ADB [A A similarity rule]

Therefore, OA/AB = OC/BD

Or, OA/2OA = OC/BD [AB = 2OC, given]

Or, BD = 2OC. [Proved.]


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Solution :

PRT is tangent at R and RQ is a chord from the contact point,

Therefore, L PRQ = L QSR = y [Angle in the alternate segment]

and L QRS = 90 [angle in semi-circle is 90 as, QS is a diameter.]

In PRS, L SPR +L PRS + LRSP = 180 [sum of angles of a = 180]

Or, x + y + 90 + y = 180

Or, x + 2y = 180 90 = 90. [Ans].

Question 7. In the figure given below, O is the centre of the circle and L AOC = 160. Prove that 3Ly 2Lx = 140.

Solution :

By arc property of the circle, Lx = 1/2(LAOC) = 1/2(160) = 80The vertices of the quadrilateral lie on the circle, so it is a cyclic quadrilateral.

Therefore, Lx + Ly = 180 Or, 80 + Ly = 180 Or, Ly = 100

Hence, 3Ly 2Lx = 3100 280 = 140. [Proved.]

Question 8. A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given thatL ATC = 36 and that the L ACT = 48. Calculate the angle subtended by AB at the centre of the circle.

Solution :

Let O be the centre of the circle. OA is joined.

In TAC, L TAC = 180 (48 + 36) = 96,

L CAB = 48 + 36 = 84,

Now OC is perpendicular to TC, [C being point of contact]


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L OAC = L OCA = 90 48 = 42 [OA = OC, being radii]

L OAB = L CAB L OAC = 84 42 = 42,

Again AOB is isosceles, L OBA = L OAB = 42

Therefore, L AOB = 180 (42 + 42) = 180 84 = 96. [Ans.]

Question 9. In the given figure below, find TP if AT = 16 cm and AB = 12 cm.

Solution :

In s PAT and PBT, L TPB = L PAT [angles in the alternate segment are equal]

L ATP is common

Therefore, PAT ~ PBT [AA similarity rule]

Hence, PT/AT = BT/PT Or, PT2 = ATBT = AT(AT - AB) = 16(16 12) = 64

Or, PT = 64 = 8 cm. [Ans.]

Maths - Symmetry (Solved)Question 1. Using ruler and compass only, construct a rectangle ABCD with AB = 5 cm and AD = 3 cmand construct its lines of symmetry. How many lines of symmetry are there ? Solution :A line AB = 5 cm is taken. At A an angle BAD equal to 90 is drawn. Taking A as centre, an arc of 3 cm isdrawn which intersect AD at D. From D an arc of 5 cm is drawn and from B an arc of 3 cm is drawn. These twoarcs intersect at C. BC and CD are joined. ABCD is the required rectangle. Lines joining mid-points of AD and


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BC is one line of symmetry and other is line joining mid-points of AB and CD as shown in the figure.

Question 2. Using ruler and compass only, construct a regular hexagon of side 2.5 cm. Draw all its lines ofsymmetry.

Solution :

A line AB of length 2.5 cm is drawn. At A and B angles BAF and ABC equal to 120 are drawn. BC and AFequal to 2.5 cm is taken and at C and F angles BCD and AFE equal to 120 are drawn. CD and FE equal to 2.5

cm is taken. DE is joined. ABCDEF is the required hexagon. The six lines of symmetry has been drawn in thefig.

Question 3. Construct a ABC in which AB = AC = 3 cm and BC = 2 cm. Using a ruler and compass onlydraw the reflection ABC of ABC, in BC. Draw the lines of symmetry of the figure ABAC.Solution :

BC equal to 2 cm is drawn. Taking B and C as centre, arcs BA and CA of length 3 cm are drawn which intersectat A. AB and AC are joined. ABC is the required triangle. Again taking B and C as centre, arcs equal to 3 cmradius are drawn which intersect at A. A is the reflection of A in BC. BC and AA are the two lines of symmetryas shown in the figure.


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Question 4. Use graph paper for this question. Plot the points A(8,2) and B(6,4). These two points are thevertices of a figure which is symmetrical about x = 6 and y = 2. Complete the figure on the graph. Writedown the geometrical name of the figure.Solution :

Co-ordinate axes are selected as shown below. A(8,2) and B(6,4) are plotted . The lines of symmetry x = 6 and y= 2 are drawn.As, the figure is symmetrical about x = 6 and y = 2, the image of A is C and the image of B is D as shown in thefigure below.

Maths - Equation of a Straight Line (Solved)

Question 1. Find the equation of a line passing through (2, 3) and inclined at an angle of 135 withpositive direction of x-axis.Solution :

The point is (x1,y1) = (2, 3), slope = m = tan135 = 1,

The equation of a line passing through (x1,y1) with slope m is :

y y1 = m(x x1)

Hence, equation of line passing through (2, 3) with slope 1 is given by


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y (3) = 1(x - 2)

Or, y + 3 = x + 2

Or, x + y + 1 = 0. [Ans.]

Question 2. Find the equation of the line parallel to 3x + 2y = 8 and passing through the point (0, 1).

Question 3. The line 4x 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A. Determinethe equation of the line passing through A and perpendicular to 4x 3y + 12 = 0.Solution :

At x-axis, y-co-ordinate is zero. The line 4x 3y + 12 = 0, meets x-axis. Hence putting y = 0 in the equation ofline we get,

4x 30 + 12 = 0 Or, 4x = 12 or, x = 3.

Therefore, line 4x 3y + 12 = 0, meets x-axis at A(3,0).[Ans.]

Slope m1 of the line 4x 3y + 12 = 0, = (coefficient of x /coefficient of y)

= (4/3) = 4/3

Let the slope of the required line be m2.

Then m1m2 = 1

Or, (4/3) m2 = 1 Or, m2 = 3/4

Therefore, line through A(3,0 ) with slope 3/4 is given by

y 0 = (3/4){x (3)}

Or, 4y = 3(x + 3) = 3x 9

Or, 3x + 4y + 9 = 0. [Ans.]

Question 4. Write down the equation of the line whose gradient is 3/2 and which passes through P, whereP divides the line segment joining A(2,6) and B(3, 4) in the ratio

2 : 3.Solution :

Let the co-ordinates of P be (x,y).

x = {23 + 3(2)}/(2 + 3) = 0, y = {2(4) + 36}(2 + 3) = 2,

Hence, P is (0,2), m = 3/2, Using the formula, y y1= m (x x1),

The equation of the line is given by, y 2 = 3/2 (x 0 )

Or, 2y 4 = 3x Or, 3x 2y + 4 = 0. [Ans.]

Question 5. ABCD is a square. The co-ordinates of A and C are (3,6) and (1,2) respectively. Write downthe equation of BD.Solution :

Co-ordinate of A and C are (3,6) and (1,2).

Slope of AC = m1 = (y2 y1)/(x2 x1) = (2 6)/(1 3) = 1.


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As, diagonals of a square bisect each other at right angle,

let slope of BD be m2, then

m1m2 = 1 or, m2 = 1/1 = 1

Co-ordinates of mid point of AC are {(x1 + x2)/2,(y1 + y2)/2}

= {(3 1 )/2, (6 + 2)/2} = (1,4).

Using the formula y y1 = m(x x1) , the equation of BD is0 + y 4 = (1)(x 1)

or, x + y 5 = 0 . [Ans.]

Question 6. If 3y 2x 4 = 0 and 4y ax 2 = 0 are perpendicular to each other, find the value of a.Solution :

he given equation of lines are , 3y 2x 4 = 0 --------- (i)

and 4y ax 4 = 0 ---------- (ii)

m1 = slope of (i) = coefficient of x/coefficient of y = (2)/3 = 2/3,m2 = slope of (ii) = coefficient of x/coefficient of y = ( a)/4 = a/4,

lines (i) and (ii) are perpendicular to each other, therefore, m1m2 = 1,

or, (2/3)(a/4) = 1 or, a = 6. [Ans.]

Maths - Distance and Section Formulae (Solved)Question 1. The mid point of the line segment joining (2a, 4) and ( 2, 2b) is

(1, 2a + 1). Find the value of a and b.

Solution :Mid point of (2a, 4) and ( 2, 2b) is (1, 2a + 1)

i.e. x = (x1 + x2)/2 Or, 1 = (2a 2)/2 => a = 2;

and y = (y1 + y2)/2 Or, 2a + 1 = (4 + 2b)/2 => 2a + 1 = 2 + b => b = 3;

Hence, a = 2, b = 3. [Ans.]

Question 2. If the points (2,1) and (1, 2) are equidistant from the point (x,y), show that x + 3y = 0.Solution :

Let A(2,1) and B(1, 2) is equidistant from P(x,y), then PA = PB

Or, [(x 2)2 + (y 1)2] = [(x 1)2 + (y +2)2] [by distance formula]

Or, x2 4x + 4 + y2 2y + 1 = x2 2x + 1 + y2 + 4y + 4

Or, 4x 2y + 2x 4y = 0

Or, 2x 6y = 0 Or, x + 3y = 0 [Proved.]

Question 3. For what value of a, the point (a,1), (1, 1) and (11,4) are collinear.

Solution :

Let line joining the points (a,1) and (11,4) is divided by the point (1, 1) in the ratio K : 1, then


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1 = (1a + K11)/(K+1) ------------------ (i)

and 1 = (11+K4)/(K+1) ------------------ (ii)

From (ii) we get 4K + 1 = K 1

Or, 5K = 2 Or, K = 2/5

Putting the value of K in (i) we get,

11K + a = K + 1 or, 10K + a = 1

or, 10 ( 2 /5) + a = 1 or, a = 1 + 4 = 5. [Ans.]

Question 4. Find the value of k if the triangle formed by A(8, 10), B(7, 3) and C(0,k) is right angledat B.

Solution :

As B is a right angle, so, AC must be the hypotenuse and hence by

Pythagoras Theorem, AC2 = AB2 + BC2

Or, (0 8)2 + (k + 10)2 = (7 8)2+ ( 3 + 10)2 + (0 7)2 + (k +3)2

Or, 64 + k2 + 20k + 100 = 1 + 49 + 49 + k2 + 6k + 9Or, 14k = 108 164 = 56 Or, k = 56 /14 = 4. [Ans.]

Question 5.The three vertices of a parallelogram are (3,4), (3,8) and (9,8). Find the fourth vertex.

Solution :

Let the vertices of the given parallelogram be A(3,4), B(3,8), C(9,8) and D(x,y). We have to find x and y.

As, we know diagonals of a parallelogram bisects each other, so, the mid-point of AC and BD must be thesame.

Mid-point of AC are {(9+3)/2, (8+4)/2}= (6,6)

And mid-point of BD are {(x+3)/2, (y+8)/2}Therefore, (x+3)/2 = 6 and (y+8)/2 = 6, which gives x = 9 and y = 4.[Ans.]

Question 6.(2, 2), (x,8), (6,y) are three cyclic points whose centre is (2,5). Find the values of x and y.Solution :

Let P(2, 2), Q(x,8) and R(6,y) be cyclic points whose centre is O(2,5). Then OP = OQ = OR = r, radius of thecircle.

OP2 = (2 + 2)2 + (5 2)2 = 16 + 9 = 25 , Or, OP = 5

OQ2 = (x 2)2 + (8 5)2, Or, x2 4x + 4 + 9 = 25 [OP = OQ]Or, x2 4x 12 = 0 , Or, x2 6x + 2x 12 = 0

Or, (x+2)(x 6) = 0 which gives x = 6, 2. [Ans.]

OR2 = (2 6)2 + (5 y)2, Or, y2 10y + 25 + 16 = 25 [OP=OR]

Or, y2 10y + 16 = 0, Or, y2 8y 2y + 16 = 0

Or, (y 2)(y 8) = 0 which gives y = 2,8. [Ans.]


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Question 7. In what ratio does the y-axis divide the line AB, where A(4, 1) and B(17.10) ?

Solution :

Let P(0,y) be a point on the y-axis which divides the given line AB in the ratio AP : PB = K : 1.

Hence, [1(4) + K17]/[K +1] = 0

Or, 17K 4 = 0 Or, K = 4/17 so the ratio is 4 : 17. [Ans.]

Question 8. Two vertices of a triangle are (3, 5) and (7,4). If its centroid is (2, 1), find the third vertex.

Solution :

Let the third vertex be (x,y). Then, using formula of centroid,

2 = (3 7 + x)/3 Or, x 4 = 6 Or, x = 10.

Similarly, 1 = (4 5 + y)/3 Or, y 1 = 3 Or, y = 2.

Hence, third vertex is (10, 2). [Ans.]

Maths - Marices (Solved)

Question 1. Find x and y if

Solution :

or,6x 10 = 8..(i)

and 2 x +14 = 4y(ii)

From(i), 6x 10 = 8

or, 6x = 18

or, x = 3 [Ans.]

From(ii), 2x +14 = 4y


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or, 4y = 2 3 + 14 = 8

or, y = 2 [ Ans.]

Question 2. Evaluate without using tables :

Solution :

Question 3.

find the matrix X such that A + X = 2 B + C.

Solution :

A + X = 2 B + C

Or, X = 2 B + C A


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Maths - Factor Theorem (Solved)

Question 1. Show that (x 1) is a factor of x3 7x2 + 14x 8. Hence, factorize the above polynomialcompletely.

Solution :

If (x 1) is a factor of f(x) = x3 7x2 + 14x 8, then f(1) = 0.

Now, f(1) = 13 7(1)2 + 14(1) 8 = 1 7 + 14 8 = 0.

Hence, (x 1) is a factor of f(x).

To find other factor, f(x) = x3 7x2 + 14x 8 = x2(x 1) 6(x 1) + 8(x 1)

=(x 1) (x2 6x +8) = (x 1){x(x 4) 2(x 4)}

= (x 1)(x 2)(x 4).[Ans.]

Question 2. Find the remainder when f(x) = x3 6x2 + 9x + 7 is divided by g(x) = x 1.

Solution :

When f(x) is divided by g(x) = x 1, then remainder, R = f(1), by remainder theorem.

Hence, R = f(1) = (1)3 6 (1)2+9(1) + 7

= 1 6 + 9 + 7 = 11. [Ans.]

Question 3. For what value of a , the polynomial g(x) = x a is a factor of f(x) = x3 ax2 +x + 2.

Solution :

As, x a is a factor of f(x), therefore, f(a) = 0

i.e. a3 a a2 + a + 2 = 0

or, a3 a3 + a + 2 = 0

or, a + 2 = 0

or, a = 2. [Ans.]

Question 4. Find the value of p and q, if (x + 3) and (x 4) are the factors of x3 px2 qx + 24.

Solution :

Le f(x) = x3 px2 qx + 24.


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As, x + 3 is a factor of f(x), f( 3) = 0

i.e. ( 3)3 p( 3)2 q( 3) + 24 = 0

or, 27 9p + 3q + 24 = 0

or, 9p + 3q 3 = 0 --------------------------- (i)

Also x 4 is a factor of f(x), then f(4) = 0

i.e. (4)3 p(4)2 q(4) + 24 = 0

or, 64 16p 4q + 24 = 0

or, 16p 4q + 88 = 0 ---------------------- (ii)

Multiplying eqn.(i) by 4 and eqn.(ii) by 3 and then adding we get

84p + 252 = 0 or, p = 3. [Ans.]

Putting value of p in eqn.(i) we get,

93 + 3q 3 = 0

Or, 3q 30 = 0

Or, q = 10. [Ans.]

Question 5. Find the value of q if the polynomial f(x) = x3 + 2x2 13x + q is divisible by g(x) = x 2. Hencefind all the factors.

Solution :

As, f(x) = x3 + 2x2 13x + q is divisible by g(x) = x 2,

Therefore, f(2) = 0

Or, (2)3 + 2(2)2 13(2) + q = 0

Or, 8 + 8 26 + q = 0

Or, q = 10. [Ans.]

Now, f(x) = x3 + 2x2 13x + 10

= x3 2x2 + 4x2 8x 5x + 10

= x2(x 2) + 4x(x 2) 5(x 2)

= (x 2)(x2 + 4x 5)


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= (x 2)(x2 + 5x x 5)

= (x 2){x(x + 5) 1(x + 5)}

= (x 2)(x 1)(x + 5). [Ans.]

Maths - Ratio and Proportion (Solved)

Question 1. If A:B = 4:5, B:C = 6:7and C:D = 14:15, find A:D.

Solution :

Here we have, A:B = 4:5, B:C = 6:7 and C:D = 14:15

Or, A/B = 4/5, B/C = 6/7 and C/D = 14/15

Or, A/BB/CC/D = 4/56/714/15 [Multiplying both sides]

Or, A/D = 16/25 Or, A:D = 16:25. [Ans.]

Question 2. If x : y = 9:10, find the value of (5x + 3y) : (5x 3y).

Solution :

x : y = 9:10 Or, x/y = 9/10

(5x + 3y) / (5x 3y) = (5x/y + 3) / (5x/y 3)

= (59/10 +3)/(59/10 3)

= (9/2 + 3)/ (9/2 3)

= [(9 + 6)/3]/[(9 6)/3]

= (15/3)/(3/3) = 15/3 = 5/1 = 5. [Ans.]

Question 3. What must be subtracted from each term of 5 :7, so that it is equal to 3 : 4 ?

Solution :

Let x be subtracted from each term of 5 : 7, so that it become 3 : 4.

i.e. (5 x)/(7 x) = 3/4

Or, 4(5 x) = 3(7 x)

Or, 20 4x = 21 3x

Or, 20 21 = 4x 3x


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Or, 1 = x Or, x = 1. [Ans.]

Question 4. What must be added to each of 7, 16, 43 and 79 so that they become proportion ? Find thenumber, which are in proportion.

Solution :

Let x be added to each of 7, 16, 43 and 79 so that the resulting number are in proportion. i.e. (7 + x) : (16 + x) : :

(43 + x) : (79 + x)

Or, (7 + x)/(16 + x) = (43 + x)/(79 + x)

Or, (7 + x)(79 + x) = (16 + x)(43 + x)

Or, 553 + 7x + 79x + x2 = 688 + 16x + 43x + x2

Or, 553 + 86x + x2 = 688 + 59x + x2

Or, 86x 59x + x2 x2 = 688 553

Or, 27x = 135

Or, x = 135/27 = 5. [Ans.]

The numbers are : (7 + 5), (16 + 5), (43 + 5) and (79 + 5)

Or, 12, 21, 48 and 84. [Ans.]

Question 5. In a regiment, the ratio of number of officers to the number of soldiers was 3 : 31 before abattle. In the battle 6 officers and 22 soldiers were killed. The ratio between the numbers of officers and

the number of soldiers now is 1 : 13. Find the number of officers and soldiers in the regiment before thebattle.Solution :

Before battle,

Let the number of officers be 3x and that of soldiers 31x

After battle,

Number of officers = 3x 6 and that of soldiers = 31x 22

As per question,

(3x 6)/(31x 22) = 1/13

Or, 13(3x 6) = 31x 22

Or, 39x 78 = 31x 22

Or, 39x 31x = 78 22


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Or, 8x = 56

Or, x = 7

Hence, no. of officers = 3 7 = 21, and no. of soldiers = 31 7 = 217. [Ans.]

Question 6.Find the mean proportion of : 25, 64

Solution :

Let x be the mean proportional between 25 and 64,

Therefore, 25 : x : : x : 64

Or, 25/x = x/64

Or, x2 = 25 64 = 1600

Or, x = 40. [Ans.]

Question 7. If x : y : : y :z , show that x :z = x2 : y2.

Solution :

Let x/y = y/z = k,

Then, y = zk , x = yk = (zk)k = zk2,

L.H.S. = x/z = zk2/z = k2,

R.H.S. = x2/y2 = (zk2)2/(zk)2 = z2k4/z2k2 = k2.

Hence, L.H.S. = R.H.S. Proved.

Question 8. If (4a + 9b) : (4c + 9d) = (4a 9b) : (4c 9d) , show that a : b = c : d.

Solution :

Here we have, (4a + 9b)/(4c + 9d) = (4a 9b)/(4c 9d)

Or, (4a + 9b)/(4a 9b) = (4c + 9d)/(4c 9d) [By alternendo]

Or, [(4a + 9b) + (4a 9b)]/[(4a + 9b) (4a 9b)]

= [(4c +9d) + (4c 9d)]/[(4c + 9d) (4c 9d)]

[By componendo and dividendo]

Or, 4a/18b = 8c/18d Or, a/b = c/d Or, a : b = c: d. Proved.


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Question 9. If x = [(a + 3b) + (a 3b)]/[(a + 3b) (a 3b)],

prove that 3bx2 2ax + 3b = 0.

Solution :

We have, x/1 = [ (a + 3b) + (a 3b)]/[(a + 3b) (a 3b)]

Applying componendo and dividendo we get

(x + 1)/(x 1)

= [(a + 3b) + (a 3b) + (a + 3b) (a 3b )]/[(a + 3b) + (a 3b) (a + 3b) + (a 3b)]

Or, (x + 1)/(x 1) = 2(a + 3b)/2(a 3b) = (a + 3b)/(a 3b)

Squaring both sides we get,

(x2 + 2x + 1)/(x2 2x + 1) = (a + 3b)/(a 3b)

Again applying componendo and dividendo we get,

(x2 + 2x + 1 + x2 2x + 1)/(x2 + 2x + 1 x2 + 2x 1 )

= (a + 3b + a 3b )/(a + 3b a + 3b)

2(x2 + 1)/2(2x) = 2(a)/2(3b)

Or, 3bx2 + 3b = 2ax [By Cross Multiplication]

Or, 3bx2 2ax + 3b = 0. [Proved.]

Maths - Quadratic Equation (Solved)

Question 1. Solve the following quadratic equation for x and give your answer correct to two decimalplaces :

x2 3x 9 = 0

Solution :

We have x2 3x 9 = 0

Comparing with ax2 + bx + c = 0

We get, a = 1, b = 3, c = 9 and D = b2 4ac = ( 3)2 4 (1)( 9)

= 9 + 36 = 45.


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Hence, x = [ b D]/2a = [ ( 3) (45)] / 2

= [3 35]/2 = [3 6.708]/2

Or, x = 4.85 and 0 1.85. [Ans.]

Question 2. Solve (7x + 1)/(7x + 5) = (3x + 1)/(5x + 1).

Solution :

Here we have, (7x + 1)/(7x + 5) = (3x +1)/(5x +1)

Multiplying both sides by (7x +5)(5x +1) [LCM of the fraction], we get

(7x +1)(5x +1) = (3x +1)(7x +5)

or, 35x2 + 7x + 5x + 1 = 21x2 + 7x + 15x + 5

or, 35x2+ 12x + 1 = 21x2 + 22x + 5

or, 35x2 21x2 +12x 22x + 1 5 = 0

or, 14x2 10x 4 = 0

or, 7x2 5 x 2 = 0

or, 7x2 (7 2)x 2 = 0 [By splitting middle term]

or, 7x2 7x + 2x 2 = 0

or, 7x(x 1) + 2(x 1) = 0

or, (x 1)(7x +2) = 0 [By factorization]

either x 1 = 0 or, 7x + 2 = 0

either x = 1 or, x = 2/7

Hence, roots of the quadratic equation are 1, 2/7. [Ans.]

Question 3. Solve the quadratic equation :

21x2 8x 4 = 0

Solution :

Here we have, 21x2 8x 4 = 0

Or, 21x2 (14 6)x 4 = 0 [By splitting middle term]

Or, 21x2 14x + 6x 4 = 0


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Or, 7x(3x 2) + 2(3x 2) = 0

Or, (3x 2 )(7x + 2) = 0 [By factorization]

Either 3x 2 = 0 or, 7x + 2 = 0

Either 3x = 2 or, 7x = 2

Either x = 2/3 or, x = 2/7

Hence, roots of the quadratic equation are 2/3, 2/7. [Ans.]

Question 4. Solve the equation : (3x2 2) = 2x 1.

Solution :

Hers we have, (3x2 2) = 2x 1,

Squaring both sides we get, 3x2 2 = (2x 1)2

Or, 3x2 2 = 4x2 4x + 1

Or, 3x2 4x2 + 4x 2 1 = 0

Or, x 2 + 4x 3 = 0

Or, x2 4x + 3 = 0

Or, x2 (3 + 1)x + 3 = 0

Or, x2 3x x + 3 = 0

Or, x(x 3) (x 3) = 0

Or, (x 3)(x 1) = 0

Either , x 3 = 0 Or, x 1 = 0

Either , x = 3 Or, x = 1 . Hence roots are 1,3. [Ans.]

Question 5. Solve the quadratic equation :

3 x2 + 10x 83 = 0.

Solution :

Here we have, 3 x2 + 10x 83 = 0; where, a = 3, b = 10, c = 83.

Using, D = b2 4ac = (10)2 4 3 ( 83) = 196 >0

Hence, x = [ b D]/2a = [ 10 (196)]/23


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Or, x = [ 10 14]/23. Or, x = 4/23 , 0 24 /23

Therefore , x = (23)/3 , 43. [Ans.]

Maths - HCF and LCM of Polynomials by factorization (Solved)

Question 1. Find the LCM and HCF of the following polynomials :

f(x) = 2x2 5x 3, g(x) = 4x2 36.

Solution :

(x) = 2x2 5x 3 = (2x + 1)(x 3),

g(x) = 4x2 36 = 4(x2 9) = 22 (x + 3)(x 3).

LCM = 22 (2x + 1)(x + 3)(x 3) = 4(2x + 1)(x + 3)(x 3).[Ans.]

HCF = x 3.[Ans.]

Question 2. Find the HCF of the following polynomials :

30x2y2(x4 x3 2x2); 42(x6 8x3) and 24x6y6(x2 5x +6)

Solution :

30x2y2(x4 x3 2x2) = 30x2y2.x2(x2 x 2)

= 30x4y2(x2 2x + 1x 2)

= 30x4y2[x(x 2) + 1(x 2)]

= 235x4y2(x 2)(x + 1) ---------- (i)

42(x6 8x3) = 42x3(x3 8) = 42x3(x3 23)

= 237x3(x 2)(x2+ 2x + 4) ----------- (ii)

24x6y6(x2 5x + 6) = 24x6y6[x2 2x 3x + 6]

=24x6y6[x(x 2) 3(x 2)]

= 233x6y6(x 2)(x 3) ----------- (iii)

Hence, HCF = 23x3(x 2) = 6x3(x 2). [Ans.]


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Question 3. Find the LCM of the following polynomials :

2x3 128 ; x2 9x + 20 ; x2 16.

Solution :

2x3 128 = 2(x3 64) = 2(x3 43) = 2(x 4)(x2 + 4x + 16) --- (i)

x2 9x + 20 = x2 4x 5x + 20 = x(x 4) 5(x 4)

= (x 4)(x 5) ----------- (ii)

x2 16 = x2 42 = (x + 4)(x 4) ------------------------------- (iii)

Hence, LCM = 2(x 4)(x + 4)(x 5)(x2 + 4x +16). [Ans.]

Question 4. The HCF of two polynomials is x + 3 and their LCM is x3 7x + 6. If one of the polynomials is

x2 + 2x 3, find the other polynomial.

Solution :

Let P(x) = x2 + 2x 3 = x2 + 3x x 3 = x(x + 3) 1(x + 3)

= (x + 3)(x 1)

LCM = x3 7x + 6 = x2(x 1) + x(x 1) 6(x 1) = (x 1)(x2 + x 6)

Let other polynomial be Q(x),

P(x)Q(x) = (HCF)(LCM)

Hence, Q(x) = [(HCF)(LCM)]/[P(x)]

= [(x + 3)(x 1)(x2 + x 6)]/[(x + 3)(x 1)]

= (x2 + x 6) [Ans.]

Question 5. If (x 3)(x + 2) is the GCD (HCF) of the polynomials

P(x) = (x2 2x 3)(2x2 + ax 2) and Q(x) = (x2 + x 2)(3x2 + bx 3), find the value of a and b.

Solution :

P(x) = (x2 2x 3)(2x2 + ax 2) = (x2 3x + x 3)(2x2 + ax 2)

= [x(x 3) + 1(x 3)] (2x2 + ax 2) = (x + 1)(x 3)(2x2 + ax 2)

As, GCD = (x 3)(x + 2),


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Hence, P(2) = 0 , Or, x = 2

Or, ( 2 +1)( 2 3)[24 + a( 2) 2 ] = 0

Or, ( 1)( 5)(8 2a 2) = 0

Or, 5(6 2a) = 0 , as 5 0 , then 6 2a = 0 or, a = 3 ------------ (i)

Q(x) = (x2+ x 2)(3x2 + bx 3) = (x2 + 2x x 2)(3x2 + bx 3)

= [x(x +2) 1(x + 2)](3x2 + bx 3) = (x 1)(x + 2)(3x2 + bx 3)

As, GCD = (x 3)(x + 2),

Hence, Q(3) = 0 , Or, x = 3

Or, (3 1)(3 + 2).(3.9 + 3b 3) = 0

Or, (2)(5)(27 + 3b 3) = 0

Or, 10(24 + 3b) = 0 , as 10 0 , then 24 + 3b = 0 or, b = 8 ------------ (ii)

Therefore, a = 3, b = 8. [Ans.]

Question 6. If x is the HCF of polynomials

p(x) = x3 7x + 6 and

q(x) = x3 x2 + x 1, find the value of .

Solution :

p(x) = x3 7x + 6,

p() = 3 7 + 6,

When = 1, p(1) = (1)3 7 (1) + 6 = 0

Therefore, ( 1) is a factor of p()

Therefore, p() = 2( 1) + ( 1) 6( 1)

= ( 1)(2 + 6)

= ( 1)(2 + 3 2 6)

= ( 1)[( + 3) 2( + 3)]

= ( 1)( 2)( + 3)

Since, (x ) is the HCF of p(x) and q(x), then p() = 0


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Or, ( 1)( 2)( + 3) = 0

Therefore, ( 1) = 0 or, ( 2) = 0 or, ( + 3) = 0

Hence, = 1, 2, 3 ------------------------------------------- (i)

q(x) = x3 x2 + x 1

q() = 3 2 + 1 = 2( 1) + 1( 1)

= ( 1)(2 + 1)

Since, (x ) is HCF of p(x) and q(x), then q() = 0

Or, ( 1)(2+ 1) = 0

Therefore, ( 1) = 0 or, (2 + 1) = 0

Hence, = 1, 2 = 1 --------------------------------------------- (ii)

Thus, = 1 is the only common solution. [Ans.]

Maths - Shares and Dividends (Solved)

Question 1. Ajay owns 560 shares of a company. The face value of each share is Rs. 25. The company de-clares a dividend of 9%. Calculate :

(i) The dividend that Ajay will get.

(ii) The rate of interest on his investment if Ajay had paid Rs. 30 for each share.

Solution :No. of shares = 560,

Nominal Value of one share = Rs. 25,

Rate of dividend = 9%.

(i) Dividend per share = 9% of Rs. 25 = 9/100 Rs.25 = Rs. 9/4.

Dividend for 560 shares = 560 Rs. 9/4 = Rs. 1260.

(ii) Investment = No. of shares Market value of one share

= 560 Rs. 30 = Rs. 16800.

Rate of interest on investment = (Dividend / Investment) 100

= (1260/16800) 100 = 7.5%.[Ans.]


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Question 2. Aman invests Rs.29929 in shares of par value Rs.26 at 10% premium. The dividend is 15%per annum. Calculate :

(i) the number of shares

(ii) the dividend received by him annually

(iii) the rate of interest he gets on his money.

Solution : (i) Here, Par value = Rs.26, Premium = 10% , Let the no. of shares be x

Market value of one share = Rs.26 + 10% of Rs.26 = Rs.26(1 + 10/100)

= Rs.26 (11/10) = Rs.143/5.

As per question, (143/5)x = 20020 or, x = (200205)/143 = 700. [Ans.]

(ii) Dividend per share = 15% of Rs.26 = (15/100) Rs.26 = Rs.3.90

Hence, dividend on 700 shares = Rs.3.90 700 = Rs.2730. [Ans.]

(iii) On Rs.20020 his income is Rs.2730

On Rs.100 his income is (Rs.2730/Rs.20020) 100 = 150/11 = 13(7/11)

Hence, his income = 13(7/11)%. [Ans.]

Question 3. Which is better investment :

7% Rs.100 shares at Rs.120 or 8% Rs.10 shares at Rs.13.50?

Solution : Let the investment be Rs.120 13.50

In the first case :

As, investment is Rs.120 income is Rs.7

Hence, when investment is Rs.120 13.50 income is (Rs.7 Rs120 13.50)/Rs.120

= Rs.94.50

In the second case :

As, investment is Rs.13.50 income is 8% of Rs.10 = Re.0.80 Hence, when investment is Rs.120 13.50 incomeis

(Re.0.80 Rs.120 13.50)/Rs.13.50

= Rs.96.00


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Therefore, second investment is better. [Ans.]

Question 4. A company with 10000 shares of Rs.100 each, declares an annual dividend of 5%.

(i) What is the total amount of dividend paid by the company ?

(ii) What would be the annual income of a man, who has 72 shares, in the company ?

(iii) If he received only 4% on his investment, find the price he paid for each share.

Solution : (i) Dividend per share = 5% of Rs.100 = Rs.5.

Dividend paid by the company for 10000 shares = 10000 Rs.5

=Rs.50000. [Ans.]

(ii) Annual income of a man for 72 shares = 72 Rs.5 = Rs.360. [Ans.]

(iii) Let his investment per share be x.

As per question, 4% of x = Rs.5

Or, (4/100) x = Rs.5

Or, x = Rs.5 (100/4) = Rs.125. [Ans.]

Question 5. A man invested Rs.45000 in 15% Rs.100 shares quoted at Rs.125. When the market value ofthese shares rose to Rs.140, he sold some shares, just enough to raise Rs.8400. Calculate :

(i) the number of shares he still holds.

(ii) the dividend due to him on these shares.

Solution :Number of shares which he purchased = Rs.45000/Rs.125 = 360.

(i) Let the number of shares he sold be x ,

As per question, Rs.140 x = Rs.8400

Or, x = Rs.8400 /Rs.140 = 60.

Hence number of shares still with him = 360 60 = 300. [Ans.]

(ii) Dividend per share = 15% of Rs.100 = Rs.15.

Hence dividend for 300 shares = Rs.15 300 = Rs.4500. [Ans.]]


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Maths - Banking (Solved)

Question 1. Shyam deposited Rs. 150 per month in his bank for eight months under the RecurringDeposit Scheme. Find the maturity value of his deposit, if the rate of interest is 8% per annum and the

interest is calculated at the end of every month ?

Solution : Sum deposited in 8 months @ Rs. 150 per month = Rs. 150 8 = Rs. 1200.

No of months of interest = {8 (8 + 1)}/2 = 36.

Equivalent principal for 1 month = Rs. 150 36 = Rs. 5400.

Interest on Rs. 5400 for 1 month @ 8% p.a.

= Rs. (5400 1/12 8/100) = Rs. 36.

The maturity value = Rs. 1200 + Rs. 36 = Rs. 1236. [Ans.]

Question 2. Mr. Ajay Kumar has a saving account in a bank. His passbook has the following entries :

Date Year2007

ParticularsWithdrawals

Rs.PDeposits

Rs. PBalanceRs.

P

January01

B/F 1276.38

January09 By cheque 2307.25 3583.63

March 07 To self 2000.00 1583.63

March 25 By cash 6200.00 7783.63

June 10 To cheque 4500.00 3283.63

July 16 By clearing 2628.70 5912.33

October20

To cheque 524.50 5387.83

October25

To self 2700.00 2687.83

November5

By cash 1500.00 4187.83

December3

To cheque 1000.00 3187.83

December25

By transfer 2927.50 6115.33


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Calculate the interest due to him for the year 2007 at 4.5% per annum if the interest is paid once in ayear at the end of December. Also, find the total amount he will receive on 11th January, 2008, if he closeshis account.

Solution : As per entries we have the following information :

Month Balance of the month Minimum balance between 10thto the last day of the month

Qualifying balance

January Rs.1276.38, Rs.3583.63 Rs.3583.63 Rs.3580

February Rs.3583.63 Rs.3583.63 Rs.3580

MarchRs.3583.63, Rs.1583.63,Rs.7783.63

Rs.1583.63 Rs.1580

April Rs.7783.63 Rs.7783.63 Rs.7780

May Rs.7783.63 Rs.7783.63 Rs.7780

June Rs.7783.63, Rs.3283.63, Rs.3283.63 Rs.3280

July Rs.3283.63, Rs.5912.33 Rs.3283.63 Rs.3280

August Rs.5912.33 Rs.5912.33 Rs.5910

September Rs.5912.33 Rs.5912.33 Rs.5910

OctoberRs.5912.33, Rs.5387.83,Rs.2687.83

Rs.2687.83 Rs.2690

November Rs.2687.83, Rs.4187.83 Rs.4187.83 Rs.4190

DecemberRs.4187.83, Rs.3187.83,

Rs.6115.33Rs.3187.83 Rs.3190

Total Rs.52750

Here, principal for 1 month = Rs.52750, rate of interest = 4.5%, time = 1/12 year

Interest = Rs.(52750 4.5 1/12)/100 = Rs.197.81. [Ans.]

Amount received by Ajay on 11th January 2008 = last balance + interest

= Rs.6115.33 + Rs.197.81

= Rs.6313.14. [Ans.]

Question 3.The entries in a Saving Bank Passbook are as given below :

Date Particulars Withdrawal Deposit Balance

01-01-07

B/F Rs.14000


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01-02-07

By cash Rs.11500 Rs.25500

12-02-07

To cheque Rs.5000 Rs.20500

05-04-07


15-04-07

To cheque Rs.4250 Rs.20000

09-05-07


04-06-07


Calculate the interest for six months (January to June) at 4% per annum on the minimum balance on orafter the tenth day of each month.

Solution : As per passbook entries, we have the following information :

Month Balance of the monthMinimum balance between 10th

to the last day of the monthQualifying Balance

January Rs14000, Rs.14000 Rs.14000

February Rs.25500, Rs.20500 Rs.20500 Rs.20500

March Rs.20500 Rs.20500 Rs.20500

April Rs.20500, Rs.24250, Rs.20000 Rs.20000 Rs.20000

May Rs.20000, Rs.21500 Rs.21500 Rs.21500

June Rs.21500, Rs.23000 Rs.23000 Rs.23000

Total Rs.119500

Here, principal for 1 month = Rs.119500, rate of interest = 4%, time = 1/12 year

Interest = Rs.(119500 4 1/12)/100 = Rs.398.33. [Ans.]


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Question 4. Bharti has a recurring deposit account in a bank for 5 years at 9% per annum simpleinterest. If she gets Rs.51607.50 at the time of maturity, find the monthly instalment.

Solution : Let the monthly instalment be x,

Total money deposited by Bharti in 5 years = Rs.60 x = Rs.60x.

Principal for 1 month = Rs.x (60 + 59 + 58 + + 3 + 2 + 1)

=Rs.x 60 (60 + 1)/2 [Using, Sn = n (n + 1)/2]

= Rs.1830x.

Interest of Rs.1830x for 1 month at the rate of 9% = Rs.(1830x 9 1/12)/100

= Rs.(549/400x

The maturity amount = Rs.60x + Rs.(549/40)x = Rs.(2949/40)x.

As per question, (2949/40)x = 51607.50

Or, x = Rs.(51607.50 40)/2949 = Rs.700. [Ans.]

Question 5. Mira Kumar has an account with a bank. The following entries are from her pass-book :

Date ParticularsWithdrawals

Rs.p.DepositsRs. p.

Balance Rs.p.

08

02 08

B / F 8500.00

18 02 08

To self 4000.00

12 04 08

By cash 2238.00

15

06 08 To self 5000.00

08 07 08

By cash 6000.00

Compute the above page of her pass book and calculate the interest for the six months. February to July2008, at 4.5% per annum.

Solution : Mira Kumars Pass-Book.


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Month Balance of the monthMinimum balance between 10 th

to the last day of the month.Qualifying Balance.

February Rs. 8500.00, Rs. 4500.00. Rs. 4500.00 Rs. 4500.00

March Rs. 4500.00. Rs. 4500.00 Rs. 4500.00

April Rs. 4500.00, Rs. 6738.00. Rs. 4500.00 Rs. 4500.00

May Rs. 6738.00. Rs. 6738.00 Rs. 6740.00June Rs. 6738.00, Rs. 1738. Rs. 1738.00 Rs. 1740.00

July Rs. 1738, Rs. 7738.00. Rs. 7738.00 Rs. 7740.00

Total Rs.29720.00

Total principal for 1 month = Rs. 29720.00 = P, T = 1/12, R = 4.5%.

Interest = (PRT)/100 = Rs. (297200 4.5 1/12)/100 = Rs. 111.45.

Maths - Sales Tax and Value added Tax (Solved)Question 1. Dinesh bought an article for Rs. 374, which included a discount of 15% on the marked priceand a sales tax of 10% on the reduced price. Find the marked price of the article.

Solution: Let marked price be Rs. x, Discount = 15%, Rate of Sales Tax = 10%.

Discount on Rs. x = 15% of x = (15/100) x = 15x/100.

Price after discount = x 15x/100 = 85x/100.

Sales Tax = 10% of 85x/100 = (10/100)(85x/100) = 85x/1000.

Price including sales tax = 85x/100 + 85x/1000 = 935x/1000.

As per question, 935x/1000 = Rs. 374

Or,

x = Rs. 374(1000/935) = Rs. 400.

Hence, marked price = Rs. 400. [Ans.]

Question 2. Ms. Chawla goes to a shop to buy a leather coat which costs Rs.735. The rate of sales tax is5%. She tells the shopkeeper to reduce the price to such an extent that she has to pay Rs.735, inclusive ofsales tax. Find the reduction needed in the price of the coat.

Solution: Let the reduced price of the coat be Rs.x

Sales tax = 5% of x = (5/100) Rs.x = Rs.x/20

Amount paid by Ms Chawla = Rs.(x + x/20) = Rs.21x/20


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As per problem, 21x/20 = 735

Or, x = (20/21) 735 = 700

Reduced price of the coat = Rs.700

Hence, reduction needed = Rs.735 Rs.700 = Rs.35. [Ans.]

Question 3. a shopkeeper buys an article at a rebate of 30% on the printed price. He spends Rs.40 ontransportation of the article. After charging sales tax at the rate of 7% on the printed price, he sells thearticle for Rs.856.Find his profit percentage.

Solution: Let the printed price of the article be Rs.x.

Sales tax = 7% of Rs.x = (7/100) Rs.x = Rs.7x/100.

Selling price = Rs.x + Rs.7x/100 = Rs.107x/100.

As per question, 107x/100 = 856 or x = (856 100)/107 = 800.

Hence, printed price = Rs.800.

Rebate = 30% of Rs.800 = (30/100) Rs.800 = Rs.240.

Cost price of the article = Rs.800 Rs.240 = Rs.560.

Overhead = cost of transportation = Rs.40.

Actual cost price = Rs.560 + Rs.40 = Rs.600.

Profit = Printed price Actual cost price = Rs.800 Rs.600 = Rs.200.

Profit percent = (profit/cost price) 100% = (200/800) 100

= 100/3% = 33(1/3)%. [Ans.]

Question 4. The catalogue price of a colour TV is Rs.2400. The shopkeeper gives a discount of 8% on thelisted price. He gives a further off-season discount of 5% on the balance. But sales tax at the rate of 10%is charged on the remaining amount. Find :

(i) the sales tax amount a customer has to pay.

(ii) the final price he has to pay for the colour TV.

Solution: Catalogue price = Rs.24000.

Discount = 8% of Rs.24000 = (8/100) Rs.24000 = Rs.1920.

Price after discount = Rs.24000 Rs.1920 = Rs.22080.

Off-season discount = 5% of Rs.22080 = (5/100) Rs.22080 = Rs.1104.


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Price after off-season discount = Rs.22080 Rs.1104 = Rs.20976.

(i) Sales tax = 10% of Rs.20976 = (10/100) Rs.20976 = Rs.2097.60. [Ans.]

(ii) The final price a customer has to pay = Rs.20976 + Rs.2097.60

= Rs.23073.60. [Ans.]

Question 5. [VAT] A manufacturer produces a good which cost him Rs.500. He sells it to a wholesaler at aprice of Rs.500 and wholesaler sells it to retailer at a price of Rs.600. The retailer sells it to the customerat a price of Rs.800. If the sales tax charged is 5%. Find the tax charged under VAT by : (i) manufacturer,(ii) wholesaler and (iii) retailer. Find the tax paid by the customer.

Solution: Cost of production of manufacturer = Cost of wholesaler = Rs.500.

Tax charged by manufacturer from wholesaler = 5% of Rs.500 = Rs.25.

Cost charged by wholesaler from retailer = Rs.600

Tax charged by the wholesaler = 5% of Rs.600 = Rs.30.

Cost charged by retailer from customer = Rs.800.

Tax charged by retailer from customer = 5% of Rs.800 = Rs.40.

Tax paid to Govt. Tax paid to Govt. bymanufacturer

Tax paid to Govt. bywholesaler

Tax paid to Govt. byretailer

Tax paid toGovt. by cus-tomer

Rs.40 Rs.25 Rs.5 Rs.10 Rs.40

The Government collects the same tax at different stages equal to Rs.40. [Ans.]

Maths - Compound Interest (Solved)

Question 1. Ramesh invests Rs. 12800 for three years at the rate of 10% per annum compound interest.Find :

(i) The sum due to Ramesh at the end of the first year.(ii) The interest he earns for the second year.(iii) The total amount due to him at the end of the third year.

Solution: Principal = Rs. 12800, Rate = 10%.

Interest for one year = (PRT)/100 = (12800 10 1) / 100 = Rs. 1280.

(i) Sum due after one year = 12800 + 1280 = Rs. 14080.[Ans.]

(ii) Interest for 2 nd year = (14080 10 1) / 100 = Rs. 1408.


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Amount after 2 nd year = Rs. 14080 + Rs. 1408 = Rs. 15488.

Interest for 3 rd year = (15488 10 1) / 100 = 1548.80. [Ans.]

(iii) The amount due after 3 year = 15488 + 1548.80 = Rs. 17036.80. [Ans.]

Question 2. A man borrows Rs.5000 at 12% compound interest per annum, interest payable after sixmonths. He pays back Rs.1800 at the end of every six months. Calculate the third payment he has to

make at the end of 18 months in order to clear the entire loan.

Solution: As 12% is the interest p.a. so half-yearly interest will be 6%.

Principal for the first six months = Rs.5000.

Interest for the first six months = Rs.(5000 6 1)/100 = Rs.300.

Amount after six month = Rs.5000 + Rs.300 = Rs.5300.

Money refunded after six months = Rs.1800.

Principal for the second six months = Rs.5300 Rs.1800 = Rs.3500.

Interest for the second six months = Rs.(3500 6 1)/100 = Rs.210.

Amount after second six months = Rs.3500 + Rs.210 = Rs.3710.

Money refunded after second six months = Rs.1800.

Principal for the third six months = Rs.3710 Rs.1800 = Rs.1910.

Interest for the third six months = Rs.(1910 6 1)/100 = Rs.114.60.

Hence payment he has to make after 18 months to clear the entire loan = Rs.1910 + Rs.114.60 = Rs.2024.60[Ans]

Question 3. A man invests Rs.5000 for three years at a certain rate of interest, compounded annually. Atthe end of one year it amounts to Rs.5600. Calculate :

(i) the rate of interest per annum.(ii) the interest occurred in the second year.(iii) the amount at the end of the third year.

Solution: Here, P = Rs.5000, A = Rs.5600, n = 1, r = ?

(i) Using A = P (1 + r/100)n,

Rs.5600 = Rs.5000 (1 + r/100)1

or, 5600/5000 = 1 + r/100

or, 56/50 = 1 + r/100


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or, r/100 = 56/50 1 = 6/50

or, r = (6 100)/50 = 12 % p.a. [Ans.]

(ii) Interest occurred in the second year, I = PRT/100 = (Rs.5600 12 1)/100 = Rs. 672.[Ans.]

(iii) Amount at the end of three years, Using A = P (1 + r/100)n, where, P= Rs.5000, r = 12%, n = 3.

A = Rs.5000 (1 + 12/100)3

= Rs.(5000 112 112 112)/1000000

= Rs.7024.64. [Ans.]

Question 4. The compound interest on a certain sum of money at 5% per annum for two years is Rs.246.Calculate the simple interest on the same sum for three years at 6% per annum.

Solution: Using C.I. = P [(1 + r/100)n 1], we get

Rs.246 = P [(1 + 5/100)2 1]

Or, Rs.246 = P [(21/20)(21/20) 1] = P (41/400)

Or, P = Rs.(246 400)/41 = Rs.2400.

Now, P = Rs.2400, r = 6% p.a., t = 3 years

Using S.I = Prt/100,

S.I. = Rs.(2400 6 3)/100 = Rs.432. [Ans.]

Question 5. On a certain sum of money, the difference between the compound interest for a year, payablehalf-yearly, and the simple interest for a year is Rs.180. Find the sum lent out, if the rate of interest inboth the cases is 10%.

Solution: Let the principal, P = Rs.x, r = 10%, t = 1 year

Using S.I. = Prt/100,

S.I. = Rs.(x 10 1)/100 = Rs.x/10

No of conversion period, n = 2 1 = 2,

r = 10%/2 = 5% per conversion period, P = x,

Using C.I. = P [(1 + r/100)n 1],

C.I. = Rs.x [(1 + 5/100)2 1]

= Rs.x(21/20 21/20 1)


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= Rs.x(441/400 1)

= Rs.(41/400)x.

As per problem, C.I. S.I. = Rs.180

Or, Rs.(41/400)x Rs.x/10 = 180

Or, Rs.(41/400 1/10)x = 180

Or, Rs.x/400 = 180

Or, x = Rs.180 400 = Rs.7200. [Ans.]

Question 6. A certain sum amounts to Rs.5292 in 2 years and to Rs.5556.60 in 3 years at compoundinterest. Find the rate and the sum.

Solution: Here, Rs.5556.60 Rs.5292 = Rs.264.60 = Interest for 1 year on Rs.5292.

i.e. P = 5292, t = 1 year, S.I. = Rs.264.60, r = ?

Using, S.I. = prt/100

264.60 = (5292 r 1)/100

or, r = (264.60 100)/5292 = 5%

2nd part : Let P = x, A = 5292, t = 2 years, n = 2, r = 5% p.a.

Using, A = P(1 + r/100)n

5292 = x (1 + 5/100)2

or, 5292 = x (105/100)2 = x (2121)/(2020)

or, 5292 = x (441/400)

or, x = (5292400)/441 = 4800

Hence, rate = 5% and the sum = Rs.4800. [Ans.]

Question 7. The cost of car, purchased 2 years ago, depreciates at the rate of 20% every year. If itspresent worth is Rs.315600, find :

(i) its purchase price(ii) its value after 4 years.

Solution: (i) Here, V = present value = Rs315600, t = 2 years, r = 20%,

V0 = purchase price = ?


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Using V = V0(1 r /100)n,

315600 = V0 (1 20/100)2

or, 315600 = V0 (80/100)2 = V0 (44) (55) =V0(16/25)

or, V0 = 31560025/16 = Rs.493125. [Ans.]

(ii) Again, V0 = Rs.493125, V = value after 4 years, t = 4years, n = 4, r = 20%.

Using, V = V0(1 r /100)n

V = 493125 (1 20/100)4

= 493125 (80/100)4

= 493125 (256/625)

= Rs.201884. [Ans.]

maths- solved sums

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