Q.1. What sum of money will amount to Rs3704.40 in 3 years at 5% compound interest.
Solution :
We have, A = Rs3704.40, n = 3, r = 5%.
Using A = P(1 + r/100)n , we get
3704.40 = P(1 + 5/100)3 => 370440/100 = P ×21/20×21/20×21/20
Or, P = 370440/100×20/21×20/21×20/21= 3200
Hence, required sum of money = Rs3200. [Ans.]
Q.2. What sum of money will amount to Rs3630 in two years at 10% per annum compound interest?
Solution :
Do yourself [Ans. = Rs3,000]
Q.3. Calculate the compound interest for the second year on Rs8,000 invested for 3 years at 10% p.a.
Solution :
Do yourself. [Ans. = Rs880.]
Q.4. At what rate percent p.a. compound interest would Rs80000 amount to Rs88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.
Solution :
Principal = Rs80000, Amount = Rs88200, Time = 2 years, Rate = ? n = 2.
Using, A = P (1 + r/100)n, we get
88200 = 80000 (1 + r/100)2
Or, (1 + r/100)2 = 88200/80000 = 441/400
Or, (1 + r/100) = 21/20
Or, r/100 = 21/20 – 1 = 1/20
Or, r = 1/20×100 = 5%
Second part, P = Rs80000, Time = 3 years, n = 3, r = 5%. [Ans.]
A = 80000(1 + 5/100)3
= 80000(21/20)(21/20)(21/20)
= Rs92610. [Ans.]
Q.5. 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 ndyear = (14080 × 10 × 1) / 100 = Rs. 1408.
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.]
Q.6. A man borrows Rs.5000 at 12% compound interest per annum, interest payable after six months. 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.]
Q.7. A man invests Rs.5000 for three years at a certain rate of interest, compounded annually. At the 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 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.]
Q.8. 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.]
Q.1. Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, factorize the above polynomial completely.
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.]
Q.2. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence factorize the given expression completely using factor theorem.
Solution :
If 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14 , then putting 2x + 7 = 0
Or x = – 7/2 we get, f (– 7/2) = 2 (– 7/2)3 + 5 (– 7/2)2 – 11 (– 7/2) – 14
= – 343/4 + 245/4 + 77/4 – 14 = – 399/4 + 245/4 + 154/4
= (399 – 399)/4 = 0.
Hence, 2x + 7 is a factor.
Dividing 2x3+ 5x2 – 11x – 14 by 2x + 7 we get
2x3 + 5x2 – 11x – 14 = (2x + 7)(x2 – x – 2).
Now, x2 – x – 2 = x(x – 2) + (x – 2) = (x + 1)(x – 2).
Hence, 2x3 + 5x2 – 11x – 14 = (2x + 7)(x + 1)(x – 2). [Ans.]
Q.3. 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.]
Q.4. Find the remainder when 2x3 – 3x2 + 7x – 8 is divided by x – 1.
Solution :
Do yourself. [Ans. = – 2.]
Q.5. 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.]
Q.6. Find the value of a, if x – a is a factor of x3 – a2x + x + 2.
Solution :
Do yourself [Ans. a = – 2 ]
Q.7. 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.
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,
9×3 + 3q – 3 = 0
Or, 3q – 30 = 0
Or, q = 10. [Ans.]
Q.8. Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x3 + ax2 + bx – 12.
Solution :
Do yourself. [Ans. a = 3, b = – 4.]
Q.1. Find the equation of a line passing through (2, –3) and inclined at an angle of 135º with positive direction of x-axis.
Solution :
The point is (x1,y1) = (2, –3), slope = m = tan135º = –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 y – (–3) = –1(x - 2)
Or, y + 3 = –x + 2
Or, x + y + 1 = 0. [Ans.]
Q.2. Find the equation of the line parallel to 3x + 2y = 8 and passing through the point (0, 1).
Solution :
The equation of line parallel to 3x + 2y = 8 may be written as,
3x + 2y = k -------- (i) we have to find the value of k.
Line (i) passes through (0, 1), therefore, 3(0) + 2(1) = k
Or, k = 2,
Hence, equation of required line is 3x + 2y = 2. [Ans.]
Q.3. Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.
Solution :
Do yourself. [Ans. 3x + 5y – 20 = 0.]
Q.4. The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A. Determine the 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 of line we get,
4x – 3×0 + 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 m1×m2 = –1
Or, (4/3) ×m2 = –1 => 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.]
Q.5. Write down the equation of the line whose gradient is 3/2 and which passes through P, where P 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 = {2×3 + 3×(–2)}/(2 + 3) = 0, y = {2×(–4) + 3×6}(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.]
Q.6. ABCD is a square. The co-ordinates of A and C are (3,6) and (–1,2) respectively. Write down the 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.
As, diagonals of a square bisect each other at right angle, let slope of BD be m2, then m1×m2 = –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 is
0 + y – 4 = (–1)(x – 1) or, x + y – 5 = 0 . [Ans.]
Q.7. If 3y – 2x – 4 = 0 and 4y – ax – 2 = 0 are perpendicular to each other, find the value of a.
Solution :
The 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, m1×m2 = –1, or, (2/3)(a/4) = –1 or, a = –6. [Ans.]
Q.8. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution :
The given equation of lines are : y = 3x + 7 -------------- (i) and 2y + px = 3 -------------------- (ii)
Slope of (i) , m1 = 3;
(ii) can be written as y = (– p/2)x + 3/2
Slope of (ii), m2 = – p/2;
As, (i) and (ii) are perpendicular to each other,
Therefore, m1×m2 = – 1
Or, 3×(– p/2) = – 1 => p = 2/3. [Ans.]
Q.1. Construct triangle BCP, where CB = 5 cm, BP = 4 cm, LPBC = 45º. Complete the rectangle ABCD such that i. P is equidistant from AB and BC; and ii. P is equidistant from C and D.
Measure and write down the length of AB.
Solution :
CB = 5 cm is drawn. CR and BQ perpendiculars to CB are drawn. L CBP = 45º is drawn. BP = 4 cm is taken and with P as centre and radius PD = BP an arc is drawn to meet CR at D. With B as centre, radius equal to CD, an arc is drawn to meet BQ at A. AD is joined. Length of AB = 5.6 cm approximately.

Q.2. i. Construct a Δ ABC, in which BC = 6 cm, AB = 9 cm and ےABC = 60º. ii. Construct the locus of all points inside Δ ABC, which are equidistant from B and C. iii. Construct the locus of the vertices of the triangle with BC as base, which are equal in area to triangle ABC. iv. Mark the point Q in your construction, which would make Δ QBC equal in area to Δ ABC, and isosceles. v. Measure and record the length of CQ.
Solution : i. A line XY is drawn and BC equal to 6 cm is taken on it. ےABC = 60º is drawn with arm AB = 9 cm. A and C is joined to get the required Δ ABC. ii. AD perpendicular to BC is drawn. iii. A line X’Y’ is drawn perpendicular to AD (i.e. parallel to XY and passing through A). X’Y’ is the required locus. iv. PQ the right bisector of BC is drawn meeting X’Y’ at Q. Q is the required point such that area(ΔQBC) = area(ΔABC). v. CQ = 8.4 cm.

Q.3. Use graph paper for this question. Take2 cm = 1 unit on both axes. i. Plot the points A(1,1), B(5,3) and C(2,7); ii. Construct the locus of points equidistant from A and B. iii. Construct the locus of points equidistant from AB and AC; iv. Locate the point P such that PA = PB and P is equidistant from AB and AC; v. measure and record the length Pa in cm.
Solution :

i. Points A(1,1), B(5,3) and C(2,7) are plotted as shown above. ii. AB is joined and right bisector of l of AB is drawn, which is the locus of points equidistant from A and B. iii. AC is joined and angle bisector m of ےBAC is drawn, which is locus of points equidistant from AB and AC. iv. The point of intersection of l and m is P, which is equidistant from AB and AC. v. PA = 2.5 cm.
Q.4. If the bisector of angles A and B of a quadrilateral ABCD intersect each other at the point P, prove that P is equidistant from AD and BC.
Solution :

From P, perpendiculars PL on AD, PM on AB and PN on BC respectively are drawn.
P lies on the bisector of angle A, hence, PL = PM -------------------- (i)
Again, P lies on bisector of angle B, PM = PN ------------------------ (ii)
From (i) and (ii) we get, PL = PN i.e. P is equidistant from AD and BC. [Proved.]
Q.1. Prove the following identities : i. (secA – 1)/(secA + 1) = (1 – cosA)(1 + cosA) ii. 1/(sinθ + cosθ) + 1/(sinθ – cosθ) = 2sinθ/(1 – 2cos2θ) iii. cot2A – cos2A = cot2A cos2A iv. √(1 + sinθ)/√(1 – sinθ) = secθ + tanθ v. (1 – sinθ)/(1 + sinθ) = (secθ – tanθ)2 vi. sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA vii. cosA/(1 – tanA) + sinA/(1 – cotA) viii. 1 – cos2θ/(1 + sinθ) = sinθ ix. √(1 – cosA)/√(1 + cosA) = sinA/(1 + cosA) x. sinθ tanθ /(1 – cosθ) = 1 + secθ 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/(sinθ + cosθ) + 1/(sinθ – cosθ)
= {(sinθ – cosθ) + (sinθ + cosθ)}/(sinθ + cosθ)(sinθ – cosθ)
= 2sinθ/(sin2θ – cos2θ)
= 2sinθ/[(1 – cos2θ) – cos2θ]
= 2sinθ/(1 – 2cos2θ) = RHS. iii. LHS = cot2A – cos2A = cos2A/sin2A – cos2A
= cos2A(1/sin2A – 1)
= cos2A(cosec2A – 1)
= cos2A cot2A = RHS iv. LHS = √(1 + sinθ)/√(1 – sinθ) = √(1 + sinθ)/√(1 – sinθ)×√(1 + sinθ)/√(1 + sinθ)
= √(1 + sinθ)2/√(1 – sin2θ)
= √(1 + sinθ)2/√(cos2θ)
= (1 + sinθ)/cosθ = 1/cosθ + sinθ/cosθ
= secθ + tanθ = RHS v. RHS = (secθ – tanθ)2 = (1/cosθ – sinθ/cosθ)2 = {(1 – sinθ)/cosθ}2
= (1 – sinθ)2/(cosθ)2 = (1 – sinθ)2/(1 – sin2θ)
= (1 – sinθ)2/{(1 + sinθ)(1 – sinθ)}
= (1 – sinθ)/(1 + sinθ) = 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 – cos2θ/(1 + sinθ) = (1 + sinθ – cos2θ)/(1 + sinθ)
= (sin2θ + sinθ)/(1 + sinθ) = sinθ (1 + sinθ)/(1 + sinθ)
= sinθ = RHS ix. LHS = √(1–cosA)/√(1+cosA) = √(1–cosA)/√(1+cosA)×√(1+ cosA)/√(1+cosA)
= √(1 – cos2A)/√(1 + cosA)2 = √(sin2A)/√(1 + cosA)2
= sinA /(1 + cosA) = RHS x. LHS = sinθ tanθ/(1 – cosθ) = {sinθ (sinθ/cosθ)}/(1 – cosθ)
= sin2θ/cosθ (1 – cosθ) = (1 – cos2θ)/cosθ (1 – cosθ)
= {(1 + cosθ)(1 – cosθ)}/cosθ (1 – cosθ) = (1 + cosθ)/cosθ
= 1/cosθ + cosθ/cosθ = secθ + 1 = RHS xi. LHS = (1 + tanA)2 + (1 – tanA)2 = 1+ 2tanA + tan2A + 1 – 2tanA + tan2A
= 2 + 2 tan2A = 2 (1 + tan2A) = 2 sec2A = RHS
Q.2.Evaluate, without the use of trigonometric tables : i. sin80º/cos10º + sin59ºsec31º ii. 3sin72º/cos18º – sec32º/cosec58º iii. 3cos80ºcosec10º + 2cos59ºcosec31º iv. cos75º/sin15º + sin12º/cos78º – cos18º/sin72º v. 2tan53º/cot37º – cot80º/tan10º
Solution : i. sin80º/cos10º + sin59º sec31º = sin(90º – 10º ) / cos10º + sin(90º – 31º)×1/cos31º
= cos10º/cos10º + cos31º/cos31º [sin(90º – θ) = cosθ]
= 1 + 1 = 2. [Ans.] ii. 3sin72º/cos18º – sec32º/cosec58º
= 3sin(90º – 18º)/cos18º– sec(90º – 58º)/cosec58º
= 3 cos18º/cos18º – cose58º/cosec58º
= 3×1 – 1 = 2. [Ans.] iii. 3cos80ºcosec10º + 2cos59ºcosec31º
= 3cos80ºcosec(90º – 80º) + 2cos59ºcosec(90º – 59º)
= 3cos80ºsec80º + 2cos59ºsec59º
= 3cos80º×1/cos80º + 2cos59º×1/cos59º
= 3 + 2 = 5. [Ans.] iv. cos75º/sin15º + sin12º/cos78º – cos18º/sin72º
= cos(90º – 15º)/sin15º + sin(90º – 78º)/cos78º – cos(90º – 72º)/sin72º
= sin15º/sin15º + cos78º/cos78º – sin72º/sin72º
= 1 + 1 – 1 = 1. [Ans.] v. 2tan53º/cot37º – cos80º/tan10º = 2tan(90º – 37º)/cot37º – cot(90º – 10º)/tan10º
= 2cot37º/cot37º – tan10º/tan10º
= 2 – 1 = 1. [Ans.]
Q.3. Without using mathematical tables, find the value of x if cosx = cos 60º cos 30º + sin60º sin 30º.
Solution : cos x = cos 60º cos 30º + sin 60º sin 30º = cos (60º – 30º) = cos 30º
Therefore, x = 30º. [Ans.]
Alternatively cos x = 1/2×√3/2 + √3/2×1/2 = √3/4 + √3/4 = √3/2 = cos 30º.
Therefore, x = 30º. [Ans.]
Q.4. Without using table, find the value of : 14 sin 30º + 6 cos 60º – 5 tan 45º.
Solution :
14 sin 30º + 6 cos 60º – 5 tan45º = 14×1/2 + 6×1/2 – 5×1
= 7 + 3 – 5 = 5. [Ans.]
Q.5. If sin x = 3/5 and cos y = 12/13; evaluate : (a) Secant2 x. (b) tan x + tan y.
Solution : sin x = 3/5, cos x = √(1 – sin2 x) = √(1 – 9/25) = √(16/25) = 4/5. tan x = sin x/cos x = (3/5)/(4/5) = 3/4. cos y = 12/13, sin y = √(1 – cos2 y) = √(1 – 144/169) = √(25/169) = 5/13. tan y = sin y/cos y = (5/13)/(12/13) = 5/12.
(a) sec2 x = 1/cos2 x = 1/(4/5)2 = 1/(16/25) = 25/16. [Ans.]
(b) tan x + tan y = 3/4 + 5/12 = (9 + 5)/12 = 14/12 = 7/6 [Ans.]
Q.6. If 2 sin A – 1 = 0, show that : sin 3A = 3 sin A – 4 sin3 A.
Solution :
We have, 2 sin A – 1 = 0
Or, 2 sin A = 1
Or, sin A = 1/2 = sin 30º
Therefore, A = 30º.
L.H.S. = sin 3A = sin 3×30º = sin 90º = 1.
R.H.S. = 3 sin A – 4 sin3A = 3 sin 30º – 4 sin3 30º = 3×1/2 – 4×(1/2)3 = 3/2 – 1/2 = 1.
Therefore, L.H.S. = R.H.S. [Proved.]
Q.1. From the top of a hill, the angle of depression of two consecutive kilometer stones, due east are found to be 30º and 45º respectively. Find the distance of the two stones from the foot of the hill. [Use √3 = 1.732]
Solution :

Fig.
Let AB be the hill whose foot is B and D and C are two kilometer stones.
Therefore DC = 1 km = 1000 m, Let AB = h and BC = x.
In right angled Δ ABC, tan 45º = AB/BC => 1 = h/x => x = h ------------- (1)
In right angled Δ ABC, tan 30º = AB/BD => 1/√3 = h/(x + 1000)
=> x + 1000 = h√3 ----------------(2) using (1) and (2) we get x + 1000 = x√3 => x(√3 – 1) = 1000
Or, x = [1000/(√3 – 1)] × (√3 + 1)/(√3 + 1)
Or, x = 1000 (√3 + 1)/2 = 500(√3 + 1) = 500 × 2.732
= 1366 m = 1.366 km.
Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill. [Ans.]
Q.2. A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60º. When he moves 50 m away from the bank, he finds the angle of elevation to be 30º. Calculate : i. the width of the river and ii. the height of the tree.
Solution :

Fig.
Let AD be the tree of height h, In ΔADC, tan60º = h/CD
Or, √3 = h/CD
Or, CD = h/√3
In ΔADB, tan30º = h/BD
Or, 1/√3 = h/BD
Or, BD = h√3
BD – CD = 50 h√3 – h/√3 = 50
(3h – h)/√3 = 50
2h = 50√3 i. height of the tree, h = 50√3/2 = 25√3 25×1.732 = 43.3 m. [Ans.] ii. Width of the river = CD = h/√3 = 25√3/√3 = 25 m. [Ans.]
Q.3. In the figure (not drawn to scale) TF is a tower. The elevation of T, from A is xº where tan xº = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m is yº. Calculate : i. the height of the tower TF ii. the angle ’y’ correct to nearest degree.

Solution : i. In Δ ATF, tan xº = TF/AF
Or, 2/5 = TF/200
Or, 5TF = 400
Or, TF = 400/5 = 80 m. [Ans.] ii. BF = AF – AB = 200 – 80 = 120 m
In Δ BTF, tan yº = TF/BF = 80/120 = 2/3
Or, tan yº = 0.6667
From the table we get y = 34º. [Ans.]
Q.4. The angle of elevation of the top of a tower from two points P and Q at a distance of ‘a’ and ‘b’ respectively, from the base and in the same straight line with it are complementary. Prove that height of the tower is √(ab).
Solution :

Let L SPR = θ then L SQR = 90º – θ and RS = h,
In Δ SPR, tan θ = h/a Or, h = a tan θ --------------------------------- (i)
In ΔSQR, tan (90º – θ) = h/b Or, h = b tan (90º – θ) = h cot θ ----- (ii)
Multiplying (i) and (ii) we get, h2 = a tan θ b cot θ = a b
Or, h = √(a b). [Proved.]
Q.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/B×B/C×C/D = 4/5×6/7×14/15 [Multiplying both sides]
Or, A/D = 16/25 Or, A:D = 16:25. [Ans.]
Q.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)
= (5×9/10 +3)/(5×9/10 – 3)
= (9/2 + 3)/ (9/2 – 3)
= [(9 + 6)/3]/[(9 – 6)/3]
= (15/3)/(3/3) = 15/3 = 5/1 = 5. [Ans.]
Q.3. If a : b = 5 : 3, find (5a + 8b) : (6a – 7b).
Solution :
Do yourself. [Ans. = 49 : 9]
Q.4. If (3x + 5y)/(3x – 5y) = 7/3, find x:y.
Solution :
We have, (3x + 5y)/(3x – 5y) = 7/3
Applying componendo and dividendo
[(3x + 5y) + (3x – 5y)]/[(3x + 5y) – (3x – 5y)] = (7 + 3)/(7 – 3)
Or, 6x/10y = 10/4
Or, x/y = (10×10)/(4×6) = 100/24 = 25/6
Hence, x:y = 25:6 [Ans.]
Q.5. Two numbers are in the ratio of 3 : 5, if 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.
Solution :
Let the numbers be 3x and 5x,
As per question, (3x + 8)/(5x + 8) = 2/3
Or, 2(5x + 8) = 3(3x + 8)
Or, 10x + 16 = 9x + 24
Or, 10x – 9x = 24 – 16
Or, x = 8
Therefore, numbers are 3x = 3×8 = 24 and 5x = 5×8 = 40. [Ans.]
Q.6. 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
Or, – 1 = x Or, x = – 1. [Ans.]
Q.7. What must be added to each of 7, 16, 43 and 79 so that they become proportion ? Find the number, 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.]
Q.8. What number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
Solution:
Do yourself. [Ans. = 3]
Q.9. What number should be subtracted from each of the following numbers 23, 30, 57 and 78, so that the remainders are in proportion?
Solution :
Do yourself. [Ans. = 6]
Q.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.
Calculate mean, median and mode.
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.]
Q.2. Calculate the mean, the median and the mode of the following numbers :
3, 1, 5, 6, 3, 4, 5, 3, 7, 2.
Solution :
Do yourself. [Ans. Mean = 3.9, Median = 3.5, Mode = 3.]
Q.3. Given below are the weekly wages of 200 workers in a factory : Weekly wages in rupees | 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 | 160 – 180 | 90 | 170 | 15,300 | Total | Σf = 200 | | Σfx = 29,000 |
Mean = Σfx/Σf = 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 | Σfi = 200 | | | Σdifi = 3000 |
Mean = A + Σdifi/Σfi = 130 + 3000/200 = 130 + 15 = 145. [Ans.]
Method III Weekly wages in Rs, | No. of workers fi | Mid-value xi | ui = (xi – A )/h =(xi – 130)/20 | uifi | 80 – 100 | 20 | 90 | – 2 | – 40 | 100 – 120 | 30 | 110 | 0 | 0 | 120 – 140 | 20 | 130 = A | 0 | 0 | 140 – 160 | 40 | 150 | 1 | 40 | 160 – 180 | 90 | 170 | 2 | 180 | Total | Σfi = 200 | | | Σuifi = 150 |
Mean = A + (Σuifi/Σfi)×h = 130 + ( 150/200)×20
= 130 + 15 = 145. [Ans.]
Q.1. Express the given relation as the set of ordered pairs (x,y), where :
R = {(x,y) : 2x + 3y ≤ 10, x,y ε N}.
Solution :
R = {(1, 1), (1, 2), (2, 1), (3, 1)}. [Ans.]
Q.2. Let A = {1, 2, 3, 4} and B = {5, 6, 7}. Which of the following are relations from A to B : i. {(1,5),(1,7),(2,6)} ii. {(2,5),(3,7),(1,6),(1,7),(4,7)} iii. {(5,1),(6,2),(7,3)} iv. {(4,5),(4,6),(3,7),(2,5),(1,6)} v. {(1,2),(5,6),(1,7)}.
Solution :
(i) is a relation as first elements belongs to A and second element to B. Similarly , (ii) and (iv) are relations.
Q.3. Decide whether the following relation is i. symmetric ii. transitive iii. reflexive. Justify by giving examples. R = “Is less than” on N.
Solution :
“Is less than” is i. not reflexive, as 3 < 3 i.e. aRa is not true. ii. not symmetric, as 3<5 but 5>3 i.e. aRb does not imply bRa. iii. transitive, as 3<5 and 5<7 then 3<7 i.e. aRb and bRc imply aRc.
Q.4. A function is defined by f(x) = (3x2 + 2x + 1)/(x +1) , x ε R, x ≠ – 1.
Find the value of i. f(2) ii. f(– 3) and iii. f(2)/f(– 3) + 1.
Solution : i. [3(2)2 + 2(2) + 1]/(2 + 1) = 17/3. [Ans.] ii. [3(– 3)2 + 2(– 3) + 1]/(– 3 +1) = 21/(– 2) = –21/2. [Ans.] iii. f(2)/f(– 3) + 1 = (17/3) / (–21/2) + 1 = 1 – 34/63 = 29/63 . [Ans.]
Q.5. Given f(x + 1) = 3x + 2 , evaluate : i. f (– 2) ii. f (2x).
Solution :
As, f (x + 1) = 3x +2 = 3(x + 1) – 1 ,
Replacing x + 1 by X we have, f(X) = 3X – 1, i. f(– 2) = 3(– 2) – 1 = – 6 – 1 = – 7 . [Ans.] ii. f(2x) = 3(2x) – 1 = 6x – 1 . [Ans.]
Q.6. The domain and range of a function f(x) = 3/x + 1 are subsets of A and B respectively, where A = {– 1/2, 0, 2/3, 6/7, 1} and B = {– 5, 0, 9/2, 5, 11/2}. List the elements of the function as ordered pairs.
Solution : f(– 1/2 ) = 3/(– 1/2) + 1 = – 6 + 1 = – 5 ε B, f(0) = 3/0 + 1 = not defined, f(2/3) = 3/(2/3) + 1 = 9/2 + 1 = 11/2 ε B, f(6/7) = 3/(6/7) + 1 = 21/6 + 1 = 7/2 + 1 = 9/2 ε B, f(1) = 3/1 + 1 = 4. f = {(– 1/2, – 5), (2/3, 11/2), (6/7, 9/2) }. [Ans.]
Q.7. A = {– 2 , – 1 , 1 , 2} and f = {(x, 1/x), x ε A} i. List the domain of f. ii. List the range of f. iii. Is f a function ? Give reasons for your answer.
Solution : i. List of domain of f = {– 2, – 1, 1, 2} ii. List of Range of f = (– 1/2, – 1, 1, 1/2) iii. Yes it is a function. Each domain has range and any element of domain is not repeated.
Q.8. If (a, b) ε R, name the kind of relation between a and b if aRb => bRa.
Is R = {(a, b) : a < b; a, b ε N}, also a relation of this kind? Explain.
Solution : i. We have, (a, b) ε R and aRb => bRa
Therefore, the relation is symmetric. [Ans.] ii. R = {(a, b) : a < b; a, b ε N}
* does not represent symmetric relation as relation between order pair (a, b) and (b, a) are not the same.
* a is not less than a.
* a < b, b < c => a < c ; for all a, b, c ε R
Therefore, relation is transitive. [Ans.]
Q.1. Use graph paper for this question.
The point A (2, 3), B (4, 5) and C (7, 2) are the vertices of a Δ ABC, i. Write down thw coordinates of A’, B’, C’ if Δ A’B’C’ is the image of Δ ABC, when reflected in the origin. ii. Write down the co-ordinates of A”, B”, C” if Δ A”B”C” is the image of Δ ABC, when reflected in the x-axis. iii. Mention the special name of the quadrilateral BCC”B” and find its area.
Solution :
The point A (2,3), B (4,5) and C (7,2) are taken in the graph paper as shown :

Fig. i. Reflection in origin, Mo (x,y) = (– x, – y)
Mo [A (2, 3)] = A’ (– 2, – 3)
Mo [B (4, 5)] = B’ (– 4, – 5)
Mo [C (7, 2)] = C’ (– 7, – 2) ii. Now, A,B,C is reflected in x-axis. Reflection in x-axis Mx (x ,y) = (x, – y)
Mx [A (2, 3)] = A” (2, – 3)
Mx [B (4, 5)] = B” (4, – 5)
Mx [C (7, 2)] = C” (7, – 2) iii. BCC”B” is an isosceles trapezium.
Area of Trapezium = 1/2(CC” + BB”) × h = 1/2(4 + 10) × 3
= 1/2×14×3 = 21 sq. unit. [Ans.]
Q.2.
i. Point P(a,b) is reflected in the x-axis to P’(5, –2). Write down the values of a and b. ii. If P” is the image of P when reflected in the y-axis. Write down the co-ordinates of P”. iii. Name the single transformation that maps P’ to P”.
Solution :
Let Rx , Ry and Ro represents reflection in x-axis, y-axis and in origin respectively. i. Rx P(a, b) = P’(a, – b), so, P’(a, – b) = P’(5, – 2)
Hence, a = 5 and b = 2. [Ans.] ii. Ry P(a, b) = P”(– a, b), so, Ry (5, 2) = P”(– 5, 2). [Ans.] iii. RoP = P”. [Ans.]
Q.3.
i. Plot A(3, 2) and B(5, 4) on graph paper . Take 2 cm = 1 unit on both axes. ii. Reflect A and B in the x-axis to A’, B’. Plot these on the same graph paper. iii. Write down :
(a) the geometrical name of the figure ABB’A’.
(b) the axis of symmetry of ABB’A’.
(c) the measure of the angle ABB’.
(d) the image A” of A, when A is reflected in the origin.
(e) the single transformation that maps A’ to A”.
Solution :
The graph is given below :-

(iii) (a) Isosceles trapezium (b)x-axis (c) 45º (d) A”(– 3, – 2) (e) y-axis. [Ans.]
Q.4. Use graph paper for this question. The point P(2,– 4)is reflected in the x = 0 to get the image P’. i. Write down the co-ordinates of P’. ii. Point P’ is reflected in the line y = 0, to get the image P”. Write down the co-ordinates of P”. iii. Name the figure PP’P”. iv. Find the area of the figure PP’P”.
Solution :
The graph is given below :-

i. P’(– 2, – 4) ii. ”(– 2, 4) iii. Right angled triangle. iv. Area of Δ PP’P” = 1/2×P’P×P”P’ = 1/2×4 units × 8 units = 16 sq. units. [Ans.] v. Q.1. Using ruler and compass only, construct a rectangle ABCD with AB = 5 cm and AD = 3 cm and construct its lines of symmetry. How many lines of symmetry are there ? vi. Solution : vii. A line AB = 5 cm is taken. At A an angle BAD equal to 90º is drawn. Taking A as centre, an arc of 3 cm is drawn 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 two arcs intersect at C. BC and CD are joined. ABCD is the required rectangle. Lines joining mid-points of AD and BC is one line of symmetry and other is line joining mid-points of AB and CD as shown in the figure. viii. ix. Q.2. Using ruler and compass only, construct a regular hexagon of side 2.5 cm. Draw all its lines of symmetry. x. Solution : xi. A line AB of length 2.5 cm is drawn. At A and B angles BAF and ABC equal to 120º are drawn. BC and AF equal to 2.5 cm is taken and at C and F angles BCD and AFE equal to 120º 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 the fig. xii. xiii. Q.3. Construct a Δ ABC in which AB = AC = 3 cm and BC = 2 cm. Using a ruler and compass only, draw the reflection A’BC of Δ ABC, in BC. Draw the lines of symmetry of the figure ABA’C. xiv. Solution : xv. BC equal to 2 cm is drawn. Taking B and C as centre, arcs BA and CA of length 3 cm are drawn which intersect at A. AB and AC are joined. ABC is the required triangle. Again taking B and C as centre, arcs equal to 3 cm radius are drawn which intersect at A’. A’ is the reflection of A in BC. BC and AA’ are the two lines of symmetry as shown in the figure. xvi.
Q.4. Use graph paper for this question. Plot the points A(8,2) and B(6,4). These two points are the vertices of a figure which is symmetrical about x = 6 and y = 2. Complete the figure on the graph. Write down 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 the figure below.

Q.5. Use a ruler and compass only in this question. (i) Construct the quadrilateral ABCD in which AB = 5 cm, BC = 7 cm and LABC = 120º, given that AC is its only line of symmetry. (ii) Write down the geometrical name of the quadrilateral. (iii) Measure and record the length of BD in cm.
Solution :

Fig i. Steps of construction : –
1. BC = 7 cm is drawn. At B angle XBC = 120º is drawn.
2. BA = 5 cm is cut from BX.
3. At A with A as centre an arc of radius 5 cm is drawn. At C with C as centre an arc of radius 7 cm is drawn to cut previous arc at D. AD and CD are joined.
ABCD is the required quadrilateral. ii. It is a kite. iii. BD = 5.7 cm. [Ans.]
Q.6. Parts of a geometrical figure is given in each of the diagrams below. Complete the figures so that the line ‘m’, in each case, is the line of symmetry of the copmpleted figure. Recognizable free hand sketches would be awarded full marks.

Fig
Solution :

Fig
Q.1. Use ruler and compass only in this question. i. Draw a circle, with centre O and radius 4 cm. ii. Mark a point P such that OP = 7 cm.
Construct the two tangents to the circle from P. Measure and record 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.)

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

Q.3. Construct a Δ ABC, in which AC = 5 cm, BC = 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 at C. Ac and BC are joined. The point D, the mid-point of AB is drawn. CD is joined. At C perpendicular to AC is 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.

Q.4. Using ruler and compass only, construct a Δ ABC such that AB = 5 cm, LABC = 75º and the radius of the circum-circle of triangle ABC 3.5 cm. On the same figure, construct a circle touching AB at its 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 arc is drawn to intersect PQ at O. With O as centre and OA as radius, a circle is drawn. At B an angle ABR = 75º is drawn, here BR intersect the circle at C. AC is joined. ABC is the required triangle. Angle bisector of L BAC is drawn which intersect PQ at D. Taking D as centre and DM as radius, another circle is drawn. It is the another required circle.

Q.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 the chord.

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

Solution :

AD is joined.
ADE is a right angle, being angle in a semi-circle.
L ABC + L ADC = 180º [opp. angles of a cyclic quad. ]
L ADE = 90º [angle in the semi-circle]
Hence, L ABC + L ADE + L ADC = 270º
Or, L ABC + L CDE = 270º. [Proved.]
Q.3. In the figure given below, AC is a diameter of a circle with centre O. Chord BD is perpendicular to AC. Write down the angles p, q, r in terms of x .

Solution :
LAOB = 2×LADB, Or, LADB = x/2.
Now, LADB + LDAC + 90º = 180º, Or, p = 90º– x/2. [Ans.] q = LADB = x/2. [Ans.] r = LCAB = 1/2×LCOB = 1/2×(180º– x ) = 90º– x/2. [Ans.]
Q.4. In the given circle with diameter AB, find the value of x.

Solution :
LABD = LACD [angles in the same segment are equal]
= 30º [LACD = 30º given]
AB is a diameter of the circle, LADB = 90º [angle in a semi-circle is 90º]
In Δ ABD, LDAB + LABD + LADB = 180º
[sum of all the three angles of a triangle is 180º]
Or, x + 30º + 90º = 180º Or, x = 180º – 90º – 30º = 60º. [Ans.]
Q.1. A train covers a distance of 600 km at x km/hr. Had the speed been (x + 20 ) km/hr, the time taken to cover the distance would have been reduced by 5 hours. Write down an equation in x and solve it to evaluate x.
Solution :
Time required to cover 600 km at x km/hr = 600/x hours
Time required to cover 600 km at (x + 20) km/hr = 600/(x + 20) hours
As per problem, 600/x – 600/(x + 20) = 5
Or, [600(x +20) – 600x]/x(x +20) = 5
Or, [600x + 12000 – 600x]/(x2 + 20x) = 5
Or, 12000/(x2 + 20x) = 5
Or, x2 + 20x = 2400
Or, x2 + 20x – 2400 = 0 [Ans.]
Or, x2 + (60 – 40)x – 2400 = 0
Or, x2 + 60x – 40x – 2400 = 0
Or, x(x + 60) – 40(x + 60) = 0
Or, (x + 60)(x – 40) = 0
Either, x + 60 = 0 Or, x – 40 = 0
Either x = – 60 Or, x = 40
Neglecting x = – 60, as negative speed is never possible, x = 40 km/hr. [Ans.]
Q.2. In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find : i. The number of rows in the original arrangement. ii. The number of seats in the auditorium after re-arrangement.
Solution :
Let the number of rows in original arrangement be x
Then the number of seats in original arrangement is also x
Total number of seats = x×x = x2 .
As per problem, 2x(x – 10) = x2 + 300
Or, 2x2 – 20 = x2 + 300
Or, x2– 20x – 300 = 0
Or, x2 – (30 – 10)x – 300 = 0
Or, x2 – 30x + 10x – 300 = 0
Or, x(x – 30) + 10(x – 30) = 0
Or, (x – 30)(x +10) = 0
Either, x – 30 = 0 Or, x +10 = 0
Either, x = 30 Or, x = – 10
Neglecting negative value, as rows cannot be negative, x = 30. i. No. of rows in the original arrangement = x = 30. [Ans.] ii. No. of seats after rearrangement = x2 + 300 = (30)2 + 300
= 900 + 300 = 1200. [Ans.]
Q.3. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, The digits interchange their places. Find the number.
Solution :
Let digit at unit’s place b x, then the digit at ten’s place will be 18/x.
Therefore, the original number = 1x + 10(18/x) = x + 180/x ---------- (i)
The reversed number = 10x + 1(18/x) = 10x + 18/x ----------- (ii)
As per question, x + 180/x – 63 = 10x + 18/x
Or, x – 10x + 180/x – 18/x – 63 = 0
Or, – 9 x + 162/x – 63 = 0
Or, 9x – 162/x + 63 = 0
Or, (9x2 – 162 + 63x)/x = 0
Or, 9x2 + 63x – 162 = 0
Or, x2 + 7x – 18 = 0
Or, x2 + (9 – 2)x – 18 = 0
Or, x2 + 9x – 2x – 18 =0
Or, x(x + 9) – 2(x + 9) = 0
Or, (x + 9)(x – 2) = 0
Either, x + 9 = 0 Or, x – 2 = 0
Either, x = – 9 Or, x = 2
As digits cannot be negative, therefore, x = 2.
Thus, the original number = x + 180/x = 2 + 180/2 = 2 + 90 = 92. [Ans.]
Q.4. Five years ago, a woman’s age was the square of her son’s age. Ten years later her age will be twice that of her son’s age. Find : i. The age of the son five years ago. ii. The present age of the woman.
Solution :
Let five years ago age of son = x and mother’s age = x2.
After 10 years, age of son = x + 15 and mother’s age = x2 + 15.
As per question, x2 + 15 = 2(x + 15)
Or, x2 + 15 = 2x + 30
Or, x2 – 2x – 15 = 0
Or, (x – 5)(x +3) = 0
Or, x = 5 Or, x = – 3
As age is never negative, x = 5. i. Five years ago age of son = 5 years. [Ans.] ii. Present age of woman = 52 + 5 = 25 + 5 = 30 years. [Ans.] iii. Q.5. A shopkeeper buys a certain number of books for Rs720. If the cost per book was Rs5 less the number of books that could be brought for Rs720 would be 2 more. Taking the original cost of each book to be Rsx, write an equation in x and solve it. iv. Solution : v. Number of books = Total Cost/Original Cost per book = 720/x.
Number of books when price decreased by Rs5 = 720/(x – 5).
As per question, 720/(x – 5) – 720/x = 2.
Or, 720[x – (x – 5)]/[(x – 5)x] = 2
Or, 720[x – x + 5] = 2x2 – 10x
Or, 2x2 – 10x – 3600 = 0
Or, x2 – 5x – 1800 = 0
Or, x2 + 40x – 45x – 1800 = 0
Or, x(x + 40) – 45(x + 40) = 0
Or, (x + 40)(x – 45) = 0
Or, x = 45 or x = – 40 (not possible)
Therefore, x = 45. [Ans.] vi. Q.6. By increasing the speed of a car by 10 km/hr, the time of journey for a distance of 72 km is reduced by 36 minutes. Find the original speed of the car. vii. Solution : viii. Let original speed of the car be x km/hr.
Therefore, time = 72/x hr.
Increased speed = x + 10 km/hr
Therefore, new time = 72/(x + 10) hr.
As per question, 72/x – 72/(x + 10) = 36/60
Or, (72 x + 720 – 72 x)/x(x + 10) = 3/5
Or, 3600 = 3(x2 + 10x)
Or, 1200 = x2 + 10x
Or, x2 + 10x – 1200 = 0
Or, x2 + 40x – 30x – 1200 = 0
Or, x(x + 40) – 30(x + 40) = 0
Or, (x – 30)(x + 40) = 0
Therefore, x = 30
As, x = – 40 is not possible.
Therefore, original speed = 30 km/hr. [Ans.]
Q.7. An aeroplane traveled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for : (i) The onward journey. (ii) The return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Solution :
() Time taken for onward journey, t = 400/x ----------------- (1)
() Time taken for return journey, t = 400/(x + 40) ------------------ (2) from (1) and (2) we get, 400/x = 400/(x + 40) + 1/2
Or, 400/x = (800 + x + 400)/{2(x + 40)}
Or, 800x + x2 + 40x = 800x + 32000
Or, x2 + 40x – 32000 = 0
Or, x2 + 200x – 160x – 32000 = 0
Or, x(x + 200) – 160(x + 200) = 0
Or, (x + 200)(x – 160) = 0
Either x + 200 = 0 or x – 160 = 0 => x = – 200 or x = 160
But x = – 200 is not possible. Hence, x = 160 km/hr. [Ans.]
Q.7. The hotel bill for a number of people for overnight stay is Rs4,800. If there were 4 more, the bill each person had to pay would have reduced by Rs200. Find the number of people staying overnight.
Solution :
Let the number of people stayed be x.
In the first case amount paid by each person = Rs4,800/x,
In the second case amount paid by each person = Rs4,800/(x + 4)
As per question, Rs4,800/x – Rs4,800/(x + 4) = Rs200
Or, [4,800(x + 4) – 4,800x]/x(x + 4) = 200
Or, [4,800x + 19,200 – 4,800x]/x(x + 4) = 200
Or, 19,200/x(x + 4) = 200
Or, x (x + 4) = 19,200/200 = 96
Or, x2 + 4x – 96 = 0
Or, x2 + 12x – 8x – 96 = 0
Or, x(x + 12) – 8 (x + 12) = 0
Or, (x + 12)(x – 8) = 0
Or, x = 8, – 12.
Rejecting negative value, we get x = 8.
Hence, number of person staying overnight = 8. [Ans.]
Q.8. A trader buys x articles for a total cost of Rs600. i. Write down the cost of one article in terms of x. If the cost per article were Rs5 more, the number of articles that can be bought for Rs600 would be four less. ii. Write down the equation in x for the above situation and solve it to find x.
Solution :
Do yourself. [(i) 600/x, (ii) x2 – 4x -480 = 0, x = 20.]
Q.9. The distance travel by a car between two towns A and B is 216 km and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car.
Calculate : i. The time taken by the car to reach town B from A in terms of x; ii. The time taken by the train to reach town B from A in terms of x; iii. If the train takes 2 hours less than the car to reach town B, obtain the equation in x and solve it. iv. Hence find the speed of the train.
Solution : i. Time taken by the car to reach the town B = 216/x hrs. ii. Time taken by the train to reach the town B = 208/(x + 16) hrs. iii. The difference in time is 2 hours, hence,
216/x – 208/(x + 16) = 2
Or, 216(x + 16) – 208x = 2x(x + 16)
Or, 216x + 3456 – 208x = 2x2 + 32x
Or, 2x2 + 24x – 3456 = 0
Or, x2 + 12x – 1724 = 0
Or, (x + 48)(x – 36) = 0
Or, x = 36, – 48.
Rejecting x = – 48, we get x = 36 iv. Speed of car = 36 km/hr [Ans.] and speed of train = 36km/hr + 16 km/hr = 52 km/hr. [Ans. v. Q.1. Solve the equation : 2x – 1/x = 7. Write your answer correct to two decimal places. vi. Solution : vii. We have, 2x – 1/x = 7
Or, 2x2 – 1 = 7x
Or, 2x2 – 7x – 1 = 0 ,
Here a = 2, b = – 7, c = – 1, D = b2 – 4ac = (–7)2 – 4×2×–1 = 57.
Using formula , x = [– b ± √D]/2a, we get x = [– (–7) ± √57] /2×2
= [7 ± √57]/4
= [7 ± 7.550]/4
Therefore, x = [7 + 7.550]/4 or, x = [7 – 7.550]/4
Either, x = 3.64 or, x = – 0.14 [Ans.] viii. Q.2. Solve the following quadratic equation for x and give your answer correct to two decimal places : x2 – 3x – 9 = 0 ix. Solution : x. 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.
Hence, x = [– b ± √D]/2a = [– (– 3) ± √ (45)] / 2
= [3 ± 3√5]/2 = [3 ± 6.708]/2
Or, x = 4.85 and 0 – 1.85. [Ans.] xi. Q.3. Solve the following equation and give your answer up to two decimal places : x2 – 5x – 10 = 0. xii. Solution : xiii. We have, x2 – 5x – 10 = 0
Here a = 1, b = – 5, c = – 10, D = b2 – 4ac = (– 5)2 – 4×1×(– 10) = 25 + 40 = 65.
As, x = [– b ± √D]/2a = [–(– 5) ± √65]/2
= [5 ± 8.06]/2.
Therefore, x = [5 + 8.06]/2 = 13.02/2 = 6.53 or [5 – 8.06]/2 = – 3.06/2 = – 1.53.
Hence, x = 6.53 ; – 1.53 [Ans.] xiv. Q.4. Solve the equation 3x2 – x – 7 = 0 and give your answer correct to two decimal places. xv. Solution : xvi. Do yourself [Ans. = 1.703, – 1.37] xvii. Q.5. Solve (7x + 1)/(7x + 5) = (3x + 1)/(5x + 1). xviii. Solution : xix. 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.] xx. Q.6. Solve the quadratic equation : 21x2 – 8x – 4 = 0 xxi. Solution : xxii. Here we have, 21x2 – 8x – 4 = 0
Or, 21x2 – (14 – 6)x – 4 = 0 [By splitting middle term]
Or, 21x2 – 14x + 6x – 4 = 0
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.]
Q.1. Ajay owns 560 shares of a company. The face value of each share is Rs. 25. The company declares 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.]
Q.2. A man invests Rs.20020 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 = (20020×5)/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.]
Q.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 income is (Re.0.80 × Rs.120 × 13.50)/Rs.13.50 = Rs.96.00
Therefore, second investment is better. [Ans.]
Q.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.]
Q.5. A man invested Rs.45000 in 15% Rs.100 shares quoted at Rs.125. When the market value of these 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.]
Q.6. A man wants to buy 62 shares available at Rs132(par value of Rs100). i. How much should he invest? ii. If the dividend is 7.5%, what will be his annual income? iii. If he wants to increase his annual income by Rs150, how many extra shares should he buy?
Solution : i. Market value of 62 shares = Rs(132×62) = Rs8184.
Therefore, he should invest Rs8184. [Ans.] ii. Dividend per share = 7.5% of Rs100 = Rs7.5.
Therefore, dividend for 62 shares = Rs7.5×62 = Rs465.
Hence, annual income = Rs465. [Ans.] iii. Income on one share = Rs7.5
Therefore, for income of Rs150, number of shares = 150/7.5 = 20.
Hence, to increase his income by Rs150, number of extra shares should be 20. [Ans.]
Q.7. A dividend of 9% was declared on Rs100 shares selling at a certain price. If the rate of return is 7 ½ %, calculate : i. the market value of the share. ii. the amount to be invested to obtain an annual dividend of Rs630.
Solution :
Dividend per share = 9% of Rs100 = Rs9. i. Let market value of one share be x.
Therefore, profit on 1 share = 7 ½ % of Rsx = Rs(15/2×1/100×x) = Rs3x/40.
As per question, 3x/40 = 9
Or, x = 9(40/3) = Rs120.
Therefore, market value of each share = Rs120. [Ans.] ii. Annual dividend = Rs630.
Therefore, number of shares = 630/9 = 70.
The amount to be invested = Rs120×70 = Rs8400.
Hence, amount to be invested = Rs8400. [Ans.]
Q.8. A man invest Rs8800 on buying shares of face value of rupees hundred each at a premium of 10%. If he earns Rs1200 at the end of year as dividend, find : i. the number of shares he has in the company. ii. the dividend percentage per share.
Solution :
Investment = Rs8800, Market value of 1 share = Rs100 + 10% of Rs100
= Rs110. i. No of shares = Rs8800/Rs110 = 80. [Ans.] ii. Face value of 80 shares = Rs100×80 = Rs8000.
Dividend = Rs1200 ,
Percentage dividend = (Dividend/face value of shares)×100
= (1200/8000)×100
= 15%. [Ans.] iii. Q.1. Draw a histogram to represent the following data : Pocket money in Rs. | No. of students | 150 – 200 | 10 | 200 – 250 | 5 | 250 – 300 | 7 | 300 – 350 | 4 | 350 – 400 | 3 | iv. Solution : v.
Fig.
vi. Q.2. For the following frequency distribution draw a histogram. Hence calculate the mode. Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | Frequency | 2 | 7 | 18 | 10 | 8 | 5 | vii. Solution : viii.
Fig.
ix. The perpendicular on x-axis from the point of intersection of AB and CD give the mode.
Here mode = 12.5 or 12.6. [Ans.]
Q.3. Draw a histogram and hence estimate the mode for the following frequency distribution : Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | Frequency | 2 | 8 | 10 | 5 | 4 | 3 |
Solution :

Fig.
From the graph, mode = 23. [Ans.]
Q.4. The marks obtained by 200 students in an examination are given below : Marks | No. of students | 0 – 10 | 05 | 10 – 20 | 10 | 20 – 30 | 11 | 30 – 40 | 20 | 40 – 50 | 27 | 50 – 60 | 38 | 60 – 70 | 40 | 70 – 80 | 29 | 80 – 90 | 14 | 90 – 100 | 06 |
Using a graph paper, draw an Ogive for the above distribution. Use your Ogive to estimate : i. The median; ii. The lower quartile; iii. The number of students who obtained more than 80% marks in the examination and iv. The number of students who did not pass, if the pass percentage was 35.
Solution :
The table is prepared Marks | No. of students | Marks less than | C . F. | | 0 – 10 | 05 | 10 | 5 | (10.5) | 10 – 20 | 10 | 20 | 15 | (20,15) | 20 – 30 | 11 | 30 | 26 | (30,26) | 30 – 40 | 20 | 40 | 46 | (40,46) | 40 – 50 | 27 | 50 | 73 | (50,73) | 50 – 60 | 38 | 60 | 111 | (60,111) | 60 – 70 | 40 | 70 | 151 | (70,151) | 70 – 80 | 29 | 80 | 180 | (80,180) | 80 – 90 | 14 | 90 | 194 | (90,194) | 90 – 100 | 06 | 100 | 200 | (100,200) |

Fig. i. Here, n = 200, n/2 = 100, from the point (0, 100) on the y-axis, a line parallel to x-axis is drawn to intersect the curve. From this point a line parallel to y-axis is drawn to intersect x – axis at (57, 0). Thus median is 57. [Ans.] ii. n/4 = 200/4 = 50, from the point (0, 50) on the y-axis, a line parallel to x-axis is drawn to intersect the curve. From this point a line parallel to y-axis is drawn to intersect x-axis at (38, 0). Thus lower quartile is 38. [Ans.] iii. From the table, the number of students obtaining more than 80% marks
= 200 – 180 = 20. [Ans.]
(iv) 38 students did not pass. [Ans.]
Q.1. Solve the following inequation and graph the solution on the number line.
– 8/3 ≤ x + 1/3 < 10/3; x ε R.
Solution :
The given inequation has two parts :
– 8/3 ≤ x + 1/3 and x + 1/3 < 10/3
Or, – 8/3 – 1/3 ≤ x and x < 10/3 – 1/3
Or, – 9/3 ≤ x and x < 9/3
Or, – 3 ≤ x and x < 3
Or, – 3 ≤ < 3. [Ans.]
The required graph is :

Fig.
Q.2. Find the range of values of x, which satisfy the inequation
– 1/5 ≤ 3x/10 + 1 < 2/5, x ε R. Graph the solution set on the number line.
Solution :
Here we have, - 1/5 ≤ 3x/10 + 1 < 2/5 .
Multiplying throughout by LCM of 5, 10, 5 i.e. 10 we get ,
– 2 ≤ 3x + 10 < 4
Adding – 10 to both sides , we get 1. 2 – 10 ≤ 3x + 10 + (– 10) < 4 + (– 10) 2. 12 ≤ 3x < – 6 3. 4 ≤ x < – 2 [dividing throughout by 3]
Hence, the solution set is {x: x ε R, – 4 ≤ x < – 2 }.[Ans.]

Here, – 4 is included and – 2 is not included.
Q.3. Solve the following inequation, and graph the solution on the number line :
2x – 5 ≤ 5x + 4 < 11 , x ε R.
Solution :
Here we have, 2x – 5 ≤ 5x + 4 < 11
i.e. 2x – 5 ≤ 5x + 4 and 5x + 4 < 11 or, 2x – 5 + (– 5x + 5) ≤ 5x + (– 5x + 5) and 5x + 4 + (– 4 ) < 11 + (– 4 ) or, 0 – 3x ≤ 9 and 5x < 7 or, x ≥ – 3 and x < 7/5 or, 0 – 3 ≤ x and x < 7/5
Hence, – 3 ≤ x < 7/5, x ε R.
Therefore, solution set is {x : x ε R, – 3 ≤ x < 7/5}. [Ans.]

Q.4. Solve the following inequation and graph the solution set on the number line :
2x – 3 < x + 2 ≤ 3x + 5, x ε R.
Solution :
Here we have, 2x – 3 < x + 2 ≤ 3x + 5, x ε R.
2x – 3 < x + 2 and x + 2 ≤ 3x + 5 or, 2x – 3 + (–x + 3) < x + 2 + (–x + 3) and x + 2 + (– 3x – 2 ) ≤ 3x + 5 + (– 3x – 2) or, x < 5 and – 2x ≤ 3 or, x < 5 and x ≥ – 3/2 or, x < 5 and – 3/2 ≤ x or, – 3/2 ≤ x and x < 5 or, – 3/2 ≤ x < 5 but x ε R. therefore , the solution set = {x : x ε R,– 3/2 ≤ x < 5}. [Ans.]

Q.5. Given that x ε R, solve the following inequality and graph the solution on the number line : 1 ≤ 3 + 4x ≤ 23.
Solution :
The equation has two parts
– 1 ≤ 3 + 4x and 3 + 4x ≤ 23
Or, – 1 – 3 ≤ 4x and 4x ≤ 23 – 3
Or, – 4 ≤ 4x and 4x ≤ 20
Or, – 1 ≤ x and x ≤ 5
Therefore, solution set = { x : – 1 ≤ x ≤ 5, x ε R } [Ans.]

Fig.
Q.6. Given that x ε I, solve the inequation and graph the solution on the number line:
3 ≥ (x – 4)/2 + x/3 ≥ 2.
Solution :
3 ≥ (x – 4)/2 + x/3 ≥ 2
Or, 3 ≥ (3x – 12 + 2x)/6 ≥ 2
Or, 3 ≥ (5x – 12)/6 ≥ 2
Or, 18 ≥ 5x – 12 ≥ 12
Or, 30 ≥ 5x ≥ 24
Or, 6 ≥ x ≥ 24/5
Or, 6 ≥ x ≥ 4.8 [Ans.]

Fig.
Q.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.]
Q.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.]
Q.3. Find the ratio in which the point (1 ,– 1) divides the line joining the points (a, 1) and (11, 4). Also find the value of a.
Solution :
Let line joining the points (a,1) and (11,4) is divided by the point (1, – 1) in the ratio K : 1, then
1 = (1×a + K×11)/(K+1) ------------------ (i) and –1 = (1×1+K×4)/(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.]
Q.4. Find the value of ‘k’ if the triangle formed by A(8, – 10), B(7, – 3) and C(0,k) is right angled at 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 + 9
Or, 14k = 108 – 164 = – 56 Or, k = – 56 /14 = – 4. [Ans.]

Q.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 the same.
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.]
Q.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 the circle.
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.]
Q.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) + K×17]/[K +1] = 0
Or, 17K – 4 = 0 Or, K = 4/17 so the ratio is 4 : 17. [Ans.]
Q.8. Calculate the ration in which the line joining A(6, 5) and B(4, – 3) is divided by the line y = 2.
Solution :
Do yourself. [Ans. = 3 : 5]
Q.1. Area of two similar triangles ABC and PQR are 25 cm2 and 49 cm2 respectively. If QR = 9.8 cm, find BC.
Solution :
Given :
Δ ABC ~ Δ PQR Therefore,{ar(Δ ABC)}/{ar(Δ PQR)}= BC2/QR2 or, 25/49 = BC2/(9.8)2 , [Taking square root of both sides]
5/7 = BC/9.8 or, 7BC = 49 or, BC = 7 cm. [Ans.]
Q.2. Two isosceles triangles have equal vertical angles and their areas are in the ratio 4 : 9. Find the ratio of their corresponding heights.
Solution :
Let ABC and PQR be two isosceles triangles, where, AB = AC, PQ = PR and vertical angles A and P are equal.

Fig
Also, (area of Δ ABC)/(area of Δ PQR) = 4/9, AD is perpendicular to BC and PM is perpendicular to QR. As, AB =AC and PQ = PR then AB/PQ = AC/PR ; and vertical angles A and P are equal ;
Therefore, Δ ABC ~ Δ PQR [S.A.S. similarity axiom]
(area of Δ ABC )/(area of Δ PQR) = AD2/PM2 or, 4/9 = AD2/PM2 or, AD/PM = 2/3. [Ans.]
Q.3. P and Q are points on the side AB and AC respectively of a triangle ABC such that PQ‌ ‌is parallel to BC and divides triangle ABC into two parts, equal in area. Find PB : AB.
Solution :

Fig
Area of Δ APQ = Area of trapezium PBCQ
= Area of Δ ABC – area of Δ APQ Or, 2×Area of Δ APQ = Area of Δ ABC ------------------------ (i)
In Δs APQ and ABC,
L APQ = L ABC [PQ is parallel to BC, corresponding angles are equal]
L A is common , Hence, Δ APQ ~ Δ ABC [A.A. axiom of similarity]
Therefore, area of Δ APQ / area of Δ ABC = AP2/AB2 ‌‌‌‌
Or, area of Δ APQ / 2×area of Δ APQ = AP2/AB2
Or, 2AP2 = AB2 Or, √2 AP = AB
Or, √2(AB – PB) = AB Or, (√2 – 1). AB = √2.PB
Hence, PB / AB = (√2 – 1)/√2 Or, PB : AB = (√2 – 1) : √2. [Ans.]
Q.4. In the figure given below, DE is parallel to BC, AD = 4 cm, BD = 2 cm and the area of Δ ABC = 12 cm2. Calculate : i. Area of Δ ADE ii. (Area of Δ ADE)/(Area of trapezium DBCE).

Fig
Solution : i. As, DE is parallel to BC, therefore, Δ ADE ~ Δ ABC
Therefore, (Area Δ ADE)/(Area of Δ ABC) = AD2/AB2
Or, (Area of Δ ADE)/12 = 42/(4 + 2)2 = 42/62 = 16/36 = 4/9 --- (1)
Or, Area of Δ ADE = 12×(4/9) = 16/3 = 5 1/3 cm2. [Ans.] ii. From part (i) we have, (Area of Δ ADE)/(Area of Δ ABC) = 4/9
Or, (Area of Δ ADE)/(Area of Δ ABC – Area of Δ ADE) = 4/(9 – 4)
Or, (Area of Δ ADE)/(Area of trapezium DBCE) = 4/5. [Ans.]
Q.5. In the given figure, ABC is a triangle, DE is parallel to BC and AD/DB = 3/2.

Fig i. Determine the ratio AD/AB, DE/BC. ii. Prove that ΔDEF is similar to Δ CBF.
Hence, find EF/FB. iii. What is the ratio of the areas of Δ DEF and Δ BFC ?
Solution : i. Given – DE || BC and AD/DB = 3/2.
In ΔS ADE and ABC, LA = LA, (Common Angle)
LD = LB, (Corresponding Angles)
Therefore, Δ ADE ~ Δ ABC [By A A similarity]
Hence, AD/AB = AE/AC = DE/BC
Now, AD/AB = AD/(AD + DB) = 3/(3 + 2) = 3/5 = DE/BC. [Ans.] ii. In ΔS DEF and CBF,
LFDE = LFCB (Alternate Angle)
LDFE = LBFC (Vertically Opposite Angle)
Therefore, Δ DEF ~ Δ CBF [By A A Similarity]
Hence, EF/FB = DE/BC = 3/5 => EF/FB = 3/5 [Ans.] iii. Area of Δ DEF / Area of Δ CBF = EF2/FB2 = 32/52 = 9/25 [Ans.] iv. Q.6. In the figure given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ΔPOB = 120 cm2, find the area of ΔQOA.

Fig v. Solution : vi. We have, PO = 6 cm, QO = 9 cm, Area of Δ POB = 120 cm2.
In Δs POB and QOB, LB = LA [Each 90º]
LPOB = LQOA [Vertically opposite angle]
Therefore, ΔPOB ~ ΔQOA [AA – criteria]
Hence, Area ofΔQOA/Area of Δ POB = OQ2/OP2
Or, Area of ΔQOA/120 = 92/62
Or, Area of Δ QOA = (81×120)/36 = 270 cm2. vii. Q.7.In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD. viii. Solution : ix.
Fig
x. In Δs ABC and DEC,
LB = LE = 90º and LC = LC
Therefore, Δ ABC ~ Δ DEC [A.A. similarity rule]
Hence, AB/DE = AC/DC
Or, 9/3 = 24/DC
Or, DC = (24×3)/9 = 8 cm.
Therefore, AD = AC – DC = 24 – 8 = 16 cm. [Ans.] xi. Q.1. Dinesh bought an article for Rs. 374, which included a discount of 15% on the marked price and a sales tax of 10% on the reduced price. Find the marked price of the article. xii. Solution : xiii. 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.] xiv. Q.2. The price of a T.V. set inclusive of sales tax of 9% is Rs13407. Find its marked price. If the sales tax is increased to 13%, how much more does the customer pay for the T.V.? xv. Solution : xvi. Let marked price of T.V. be x, sales tax = 9%.
Therefore, sales tax = 9% of x = Rs9x/100.
Total cost of T.V. = x + 9x/100 = 109x/100.
As per question, 109x/100 = 13407
Or, x = (13407×100)/109 = 12300.
Thus marked price of T.V. = Rs12300.
New tax rate = 13%
Sales tax = 13% 0f Rs12300 = Rs(13/100)×12300 = Rs1599.
Total amount paid by the customer = Rs12300 + Rs1599 = Rs13899.
Extra amount paid by the customer = Rs13899 – Rs13407 = Rs492. [Ans.] xvii. Q.3. Ms. Chawla goes to a shop to buy a leather coat which costs Rs.735. The rate of sales tax is 5%. She tells the shopkeeper to reduce the price to such an extent that she has to pay Rs.735, inclusive of sales tax. Find the reduction needed in the price of the coat. xviii. Solution : xix. 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
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.]
Sales Tax and Value added Tax

Q.4. A shopkeeper buys an article at a rebate of 30% on the printed price. He spends Rs.40 on transportation of the article. After charging sales tax at the rate of 7% on the printed price, he sells the article 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.]
Q.5. The price of a washing machine, inclusive of sales tax, is Rs13530. If the sales tax is 10%, find its basic price.
Solution :
Let basic price of the washing machine be x, sales tax = 10%.
Therefore, sales tax = 10% of x = (10/100)x = x/10.
Total price of washing machine = x + x/10 = 11x/10.
As per question, 11x/10 = 13530
Or, x = 13530(10/11) = 12300.
Therefore, basic price = Rs12300. [Ans.]
Q.6. A colour T.V. is marked for sale for Rs17,600 which includes sales tax at 10%. Calculate the sales tax in rupees.
Solution :
Do yourself. [Ans. = Rs1,600.]
Q.7. The catalogue price of a colour TV is Rs.2400. The shopkeeper gives a discount of 8% on the listed 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.
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.] iii. Q.1. In an equilateral Δ ABC of side 14 cm, side BC is the diameter of a semi-circle as shown in the figure below. Find the area of the shaded region. [Take π = 22/7 and √3 = 1.732] iv.
Fig.
v. Solution : vi. Area of the shaded portion = Area of equilateral Δ ABC + Area of semi-circle.
= √3/4 a2 + 1/2πr2,
Where, a = 14 cm and r = 14/2 = 7 cm.
Therefore area of shaded portion = √3/4×142+ 1/2×22/7×72
= 1.732×196/4 + 11×7
= 84.868 + 77 = 161.868 sq. cm. [Ans.] vii. Q.2. In the given figure, the area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.(π = 22/7) viii. ix. Solution : x. Let R = radius of the outer circle = 21 cm, r = radius of the smaller circle = x,
As per question, π R2 – π r2 = 770
Or, 22/7 (212 – x2) = 770
Or, 22/7 (441) – (22/7) x2 = 770
Or, 22 × 63 – 770 = (22/7) x2
Or, (22/7) x2 = 1386 – 770 = 616
Or, x2 = 616 (7/22) = 28 × 7
Or, x = 14 cm. i.e. radius of smaller circle = 14 cm. [Ans.]
Q.3. In the given figure, the shape of the top of a table in a restaurant is that of a segment of a circle with centre O and LBOD = 90º . BO = OD = 60 cm. Find : i. The area of the top of the table. ii. The perimeter of the table. [Take π = 3.14]

Solution : i. Area of the top of the table = (πr2θ)/360º = (3.14×60×60×270º)/360º
= (3.14 ×360×270º)/360º
= 3.14 ×2700 = 8478 cm2.[Ans.] ii. Perimeter = (3/4) ×Circumference + OB + OD = (3/4) ×2π r + r + r
= (3/4)×2×3.14×60 + 60 + 60 = 282.6 + 120 = 402.6 cm. [Ans.]
Q.4. A sheet is 11 cm long and 2 cm wide. Circular pieces of 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
Solution :
No of discs = (Area of sheet )/(area of discs) = (l×b)/(πr2)
= (11×2)/{(22/7)×(0.25)2} = (22 ×7)/(22×0.0625)
= 70000/625 = 112. [Ans.]
Q.5. In the given figure, AC and BD are two perpendicular diameters of a circle ABCD. Given that the area of the shaded portion is 308 cm2, calculate : i. the length of AC and ii. the circumference of the circle. [Take π = 22/7]

Solution :
The shaded portion is the quadrant of the circle
The area of two quadrants = 308 cm2
Or, 2× {(1/4)π r2} = 308
Or, 2×{(1/4)(22/7)r2} = 308
Or, r2 = 308×(7/11) = 28×7
Or, r = 14 cm. i. AC = diameter of the circle = 2 ×14 = 28 cm. [Ans.] ii. Circumference = 2πr = 2(22/7)(14) = 88 cm.[Ans.] iii. Q.1. Shyam deposited Rs. 150 per month in his bank for eight months under the Recurring Deposit 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 ? iv. Solution : v. 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.] vi. Q.2. Mr. Ajay Kumar has a saving account in a bank. His passbook has the following entries : Date Year 2007 | Particulars | Withdrawals Rs. P | Deposits Rs. P | Balance Rs. P | January 01 | B/F | | | 1276.38 | January 09 | By cheque | | 2307.25 | 3583.63 | March 07 | To self | 2000.00 | 6200.00 | 1583.63 | March 25 | By cash | | | 7783.63 | June 10 | To cheque | 4500.00 | 2628.70 | 3283.63 | July 16 | By clearing | | | 5912.33 | October 20 | To cheque | 524.50 | | 5387.83 | October 25 | To self | 2700.00 | | 2687.83 | November 5 | By cash | | 1500.00 | 4187.83 | December 3 | To cheque | 1000.00 | | 3187.83 | December 25 | By transfer | 2927.50 | | 6115.33 | vii. Calculate the interest due to him for the year 2007 at 4.5% per annum if the interest is paid once in a year at the end of December. Also, find the total amount he will receive on 11th January, 2008, if he closes his account. viii. Solution : ix. As per entries we have the following information : Month | Balance of the month | Minimum balance between 10th to 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 | March | Rs.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 | October | Rs.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 | December | Rs.4187.83, Rs.3187.83, Rs.6115.33 | Rs.3187.83 | Rs.3190 | Total | | | Rs.52750 | x. 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.] xi. Q.3. The entries in a Saving Bank Passbook are as given below : Date | Particulars | Withdrawal | Deposit | Balance | 01-01-07 | B/F | | | Rs.14000 | 01-02-07 | By cash | | Rs.11500 | Rs.25500 | 12-02-07 | To cheque | Rs.5000 | | Rs.20500 | 05-04-07 | By cash | | Rs.3750 | Rs.24250 | 15-04-07 | To cheque | Rs.4250 | | Rs.20000 | 09-05-07 | By cash | | Rs.1500 | Rs.21500 | 04-06-07 | By cash | | Rs.1500 | Rs.23000 | xii. Calculate the interest for six months (January to June) at 4% per annum on the minimum balance on or after the tenth day of each month. xiii. Solution : xiv. As per passbook entries, we have the following information : Month | Balance of the month | Minimum balance between 10th to the last day of the month | Qualifying 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 | xv. 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.] xvi. Q.4. Mr. Siv Kumar has a Saving Bank account in Punjab National Bank. His pass book has the following entries : Date | Particulars | Withdrawal (in Rs) | Deposit (in Rs) | Balance (in Rs) | April 1, 2007 | B/F | | | 3220.00 | April 15 | By transfer | | 2010.00 | 5230.00 | May 8 | To cheque | 298.00 | | 4932.00 | July 15 | By clearing | | 4628.00 | 9560.00 | July 29 | By cash | | 5440.00 | 15,000.00 | Sept. 10 | To self | 698.00 | | 8020.00 | Jan. 10, 2008 | By cash | | 8000.00 | 16020.00 | xvii. Calculate the interest due to him at the end of financial year (March 31st 2008) at the rate of 6% per annum. xviii. Solution : xix. Do yourself. [Ans. = Rs565.78] xx. Q.5. Bharti has a recurring deposit account in a bank for 5 years at 9% per annum simple interest. If she gets Rs.51607.50 at the time of maturity, find the monthly instalment. xxi. Solution : xxii. 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.] xxiii. xxiv. Q.6. Mira Kumar has an account with a bank. The following entries are from her pass-book : Date | Particulars | Withdrawals Rs. p. | Deposits Rs. 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 | − | xxv. Compute the above page of her pass book and calculate the interest for the six months. February to July 2008, at 4.5% per annum. xxvi. Solution : xxvii. Mira Kumar’s Pass-Book. Month | Balance of the month | Minimum 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. 6740.00 | June | 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 | xxviii. 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. [Ans.]