1 of 32

Compound Interest

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.]

2 of 32 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.

3 of 32 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.9. On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is Rs.180. Find the sum lent out, if the rate of interest in both 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) = 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.]

Q.10. A certain sum amounts to Rs.5292 in 2 years and to Rs.5556.60 in 3 years at compound interest. 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 (21×21)/(20×20) or, 5292 = x (441/400) or, x = (5292×400)/441 = 4800 Hence, rate = 5% and the sum = Rs.4800. [Ans.]

Q.11. A sum of money is lent out at compound interest for two years at 20% p.a. C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate percent per annum, C.I. being reckoned half-yearly. It would have fetched Rs482 more by way of interest. Calculate the sum of money lent out.

Solution :

Let P = x, r = 20%, Time = 2 years, n = 2. Using A = P(1 + r/100)n, A = x(1 + 20/100)2 = x(6/5)(6/5) = (36/25)x Secondly, P = x, r = 10% per conversion period, n = 2t = 4. Therefore, A = x(1 + 10/100)4 = x(11/10)(11/10)(11/10)(11/10) = (14641/10000)x As per question, (14641/10000)x – (36/25)x = 482 Or, {(14641 – 14400)/10000}x = 482 Or, 241x/10000 = 482

4 of 32 Or, x = 20000 Hence sum of money = Rs20,000. [Ans.]

Q.12. The cost of car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present 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 = ? Using V = V0(1 – r /100)n, 315600 = V0 (1 – 20/100)2 or, 315600 = V0 (80/100)2 = V0 (4×4) (5×5) =V0(16/25) or, V0 = 315600×25/16 = 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.]

Q.13. A person invests Rs10,000 for two years at a certain rate of interest compounded annually. At the end of one year this sum amounts to Rs12,000. Calculate :

i. The rate of interest per annum. ii. The amount at the end of second year.

Solution :

i. Principal = Rs10,000, Amount = Rs12,000, Time = 1 year, Interest = Rs12,000 – Rs10,000 = Rs2,000 Rate of interest = (100×I)/(P×T) = (100×2,000)/(10,000×1) = 20%. [Ans.]

ii. For second year, P = Rs12,000; R = 20%; T = 1 year. Interest = (PRT)/100 = (12,000×20×1)/100 = 2,400. Therefore, Amount at the end of second year = Rs12,000 + Rs2,400 = Rs14,400. [Ans.]

Q.14. If the interest is compounded half yearly, calculate the amount when the Principal is Rs7,400, the rate of interest is 5% per annum and the duration is one year.

Solution :

We have, P = Rs7,400, r = 5% half yearly = 5/2% per conversion period, t = 1 year , n = 2t = 2. Using formula A = P(1 + r/100)n, we get A = 7400 (1 + 5/200)2 = 7400×205/200×205/200 = Rs7774.63. [Ans.]

Q.15. A man invests Rs46,875 at 4% per annum compound interest for 3 years. Calculate :

i. The interest for the first year. ii. The amount standing to his credit at the end of the second year. iii. The interest for the third year.

Solution :

5 of 32 Do yourself Ans. =

i. Rs1,875. ii. Rs50,700. iii. Rs2,028.

Linear Inequation in One Unknown

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 :

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 :

6 of 32 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.7. Solve : 2 ≤ 2x – 3 ≤ 5, x ε R and mark it on a number line.

Solution :

We have, 2 ≤ 2x – 3 ≤ 5, x ε R. Or, 2 + 3 ≤ 2x ≤ 5 + 3 Or, 5 ≤ 2x ≤ 8 Or, 5/2 ≤ x ≤ 8/2 Or, 5/2 ≤ x ≤ 4 Or, x ε {2.5 ≤ x ≤ 4} [Ans.] Fig.

Q.8. Find the value of x which satisfies the inequation : – 2 ≤ 1/2 – 2x/3 ≤ 11/6, x ε N.

Solution :

7 of 32 We have, – 2 ≤ 1/2 – 2x/3 ≤ 11/6, x ε N. Or, – 2 – 1/2 ≤ 1/2 – 2x/3 – 1/2 ≤ 11/6 – 1/2 Or, – 5/2 ≤ – 2x/3 ≤ 8/6 Or, – 15 ≤ – 4x ≤ 8 Or, – 8 ≤ 4x ≤ 15 Or, – 2 ≤ x ≤ 15/4 But, x ε N, hence possible values of x are 1, 2, 3. [Ans.]

Q.9. Solve the inequation : – 3 ≤ 3 – 2x < 9, x ε R. Represent your solution on a number line.

Solution :

Do yourself. [Ans. = {x : – 3 < x ≤ 3, x ε R}]

Q.10. Solve the inequation : 12 + 1 5/6 x ≤ 5 + 3x, x ε R. Represent the solution on a number line.

Solution :

Do yourself. [Solution set = {x : x ≥ 6 x ε R}]

Factor Theorem

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.]

8 of 32 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.9. Find the value of q if the polynomial f(x) = x3 + 2x2 – 13x + q is divisible by g(x) = x – 2. Hence find all the factors.

Solution :

As, f(x) = x3 + 2x2 – 13x + q is divisible by g(x) = x – 2, Therefore, f(2) = 0

9 of 32 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) = (x – 2)(x2 + 5x – x – 5) = (x – 2){x(x + 5) – 1(x + 5)} = (x – 2)(x – 1)(x + 5). [Ans.]

Q.10. (x – 2) is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the value of a and b.

Solution :

As (x – 2) is a factor of f(x) = x3 + ax2 + bx + 6, Therefore, f(2) = 0 Or, (2)3 + a(2)2 + b(2) + 6 = 0 Or, 8 + 4a + 2b + 6 = 0 Or, 4a + 2b = – 14 Or, 2a + b = – 7 -------------------- (i) On dividing f(x) by (x – 3) the remainder is 3 Therefore, f(3) = 3 Or, (3)3 + a(3)2 + b(3) + 6 = 3 Or, 27 + 9a + 3b + 6 = 3 Or, 9a + 3b = – 30 Or, 3a + b = – 10 -------------------- (ii) Subtracting (ii) from (i) we get a = – 3 Substituting value of a = – 3 in (i) we get 2×(– 3) + b = – 7 Or, – 6 + b = – 7 Or, b = – 1. Therefore, a = – 3, b = – 1. [Ans.]

Q.11. Use the factor theorem to factorize completely. x3 + x2 – 4x – 4.

Solution :

We have, x3 + x2 – 4x – 4. Let x + 1 is a factor, then x + 1 = 0 => x = – 1. Putting x = – 1, we get (– 1)3 +(– 1)2 – 4 (– 1) – 4 = – 1 + 1 + 4 – 4 = 0. Therefore, x + 1 is a factor. To factorize completely x3 + x2 – 4x – 4 = x2(x + 1) – 4(x + 1) = (x + 1)(x2 – 4) = (x + 1)(x + 2)(x – 2). [Ans.]

Q.12. Using Factor theorem, show that (x – 3) is a factor of x3– 7x2 + 15x – 9. Hence, factorise the given expression completely.

Solution :

Do yourself. [Ans. = (x – 1)(x – 3)2]

Sales Tax and Value added Tax

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.

10 of 32 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.]

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.?

Solution :

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.]

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.

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 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.]

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.]

11 of 32 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.]

Q.8. The catalogue price of a computer set is Rs45,000. The shopkeeper gives a discount of 7% on the listed price. He gives a further off-season discount of 4% on the balance. However, sales tax at 8% is charged on the remaining amount. Find :

i. The amount of sales tax a customer has to pay. ii. The final price he has to pay for the computer set.

Solution :

Do yourself [Ans. = Rs3214.08 ; Rs43390.08]

Q.8. Kiran purchases an article for Rs5,400 which includes 10% rebate on the marked price and 20% sales tax on the remaining price. Find the marked price of the article.

Solution :

Let marked price of the article be x, rebate = 10%, S.T. = 20%. Selling Price = x – 10% of x = 9x/10. S.T. 20% of 9x/10 = 9x/50, Price including S.T. = 9x/10 + 9x/50 = 54x/50. As per question, 54x/50 = 5400 => x = 5000, Hence, marked price = Rs5,000. [Ans.]

12 of 32 Q.9. [VAT] A manufacturer produces a good which cost him Rs.500. He sells it to a wholesaler at a price of Rs.500 and wholesaler sells it to retailer at a price of Rs.600. The retailer sells it to the customer at 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. by manufacturer

Tax paid to Govt. by wholesaler

Tax paid to Govt. by retailer

Tax paid to Govt. by customer

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

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

Trigonometrical Identities

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

13 of 32 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.]

14 of 32 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.]

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.

15 of 32

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 ∆ 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.]

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

Solution :

The following figure is drawn according to the question :

16 of 32 C to O and D to B are joined. L ADB = 90º [angle in a semi-circle is 90º, AB is diameter] L ACO = 90º [same reason, OA is diameter] In ∆s ACO and ADB, L ACO = L ADB = 90º 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.]

Q.6. In the figure given below, PT touches a circle with centre O at R. Diameter SQ when produced meets PT at P. If L SPR = xº and L QRP = yº, show that xº + 2yº = 90º.

Solution :

PRT is tangent at R and RQ is a chord from the contact point, Therefore, L PRQ = L QSR = yº [Angle in the alternate segment] and L QRS = 90º [angle in semi-circle is 90º as, QS is a diameter.] In ∆ PRS, L SPR +L PRS + LRSP = 180º [sum of angles of a ∆ = 180º] Or, xº + yº + 90º + yº = 180º

Or, xº + 2yº = 180º – 90º = 90º. [Ans.]

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

Solution :

By arc property of the circle, Lx = 1/2(LAOC) = 1/2(160º) = 80º The vertices of the quadrilateral lie on the circle, so it is a cyclic quadrilateral. Therefore, Lx + Ly = 180º Or, 80º + Ly = 180º Or, Ly = 100º Hence, 3Ly – 2Lx = 3×100º – 2×80º = 140º. [Proved.]

Q.8. A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that L ATC = 36º and that the L ACT = 48º. 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 = 180º– (48º + 36º) = 96º, L CAB = 48º + 36º = 84º, Now OC is perpendicular to TC, [C being point of contact] L OAC = L OCA = 90º – 48º = 42º [OA = OC, being radii]

L OAB = L CAB – L OAC = 84º – 42º = 42º, Again ∆ AOB is isosceles, L OBA = L OAB = 42º Therefore, L AOB = 180º – (42º + 42º) = 180º – 84º = 96º. [Ans.]

17 of 32 Q.9. In the given figure below, find TP if AT = 16 cm and AB = 12 cm.

Solution :

PA and PB are joined as shown in the figure below :

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 = AT×BT = AT×(AT - AB) = 16(16 – 12) = 64 Or, PT = √64 = 8 cm. [Ans.]

Q.10. In the figure, PM is a tangent to the circle and PA = AM. Prove that :

i. ∆PMB is isosceles. ii. PA×PB = MB2.

Fig.

Solution :

i. In ∆ PAM, PA = AM [Given] Therefore, LAPM = LAMP --------------------- (i) Also, LABM = LAMP ---------------------- (ii) [Alternate segment property of tangent] Therefore, from (i) and (ii) we get, LAPM = LABM Therefore, PM = MB i.e. ∆ PMB is an isosceles triangle. [Proved.]

ii. By rectangle property of tangent and chord, PA×PB = PM2 As, PM = MB proved in part (i) Therefore, PA×PB = MB2. [Proved.]

Q.11. In the given figure, O is the centre of the circle and LPBA = 45º. Calculate the value of LPQB.

Solution :

Given – Fig.

LPBA = 45º, AOB is diameter of the circle. Therefore, LAPB = 90º [Angle in a semi-circle is right angle] Hence, in ∆ APB, LPAB = 180º – (90º + 45º) = 45º. As, LPAB = LPQB [Angle in the same segment are equal] Therefore, LPQB = 45º. [Ans.]

Q.12. In the given figure, if LACE = 43º and LCAF = 62º , find the value of a, b and c.

Fig.

Solution :

ABDE is a cyclic quadrilateral. Therefore, LABD + LAED = 180º and LEAB + LBDE = 180º

18 of 32 Now in ∆ ACE, LA + LC + LE = 180º Or, 62º + 43º + LE = 180º Therefore, LE = 180º – 105º = 75º Also, LABD + LAED = 180º Therefore, a + 75º = 180º Therefore, a = 105º LEDF = LBAC [Exterior angle of cyclic quadrilateral] Therefore, c = 62º In ∆ABF, LABF + LBAF + LBFA = 180º Or, 105º + 62º + b = 180º Or, b = 180º – 167º = 13º Hence, a = 105º, b = 13º and c = 62º. [Ans.]

Q.13. In the given figure below, LBAD = 65º, LABD = 70º and LBDC = 45º. Find :

i. LBCD. ii. LADB. Hence show that AC is a diameter.

. Fig.

Solution :

Given : – LBAD = 65º, LABD = 70º and LBDC = 45º.

i. ABCD is a cyclic quadrilateral. Therefore, LDAB + LBCD = 180º

Or, 65º + LBCD = 180º Or, LBCD = 180º – 65º = 115º [Ans.]

ii. In ∆ADB, LDAB + LABD + LADB = 180º Or, 65º + 70º + LADB = 180º Or, LADB = 180º – 135º = 45º. [Ans.] In ∆DBC, LDBC + LBCD LBDC = 180º Or, LDBC + 115º + 45º = 180º Or, LDBC = 180º – 160º = 20º Therefore, LABC = LABD + LDBC = 70º + 20º = 90º. Hence, AC is a diameter. [Proved.]

Q.14. In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If LCAB = 34º, find :

i. LCBA, ii. LCQA.

Fig.

Solution :

AB is a diameter, hence LACB = 90º [Angle in semicircle is right angle] Therefore, in ∆ ACB, LA + LC + LB = 180º Or, 34º + 90º + LB = 180º Or, LB = 180º 0 – 124º = 56º

19 of 32 Therefore, LCBA = 56º Now CQ is a tangent at C, Therefore, LQCB = LCAB [Angle in the alternate segment are equal] = 34º. [Ans.] And LCBQ = 180º – LCBA = 180º – 56º = 124º Therefore, LCQA = 180º – (LQCB + LCBQ) = 180º – (34º + 124º) = 180º – 158º = 22º. [Ans.]

Q.15. PQR is a right angled triangle with PQ = 3 cm and QR = 4 cm. A circle which touches all the sides of the triangle is inscribed in the triangle. Calculate the radius of the circle.

Solution :

Fig.

Let O be the centre of the circle touching all the sides of the triangle at A, B and C. Let AO = OB = AQ = QB = r, Therefore, PA = 3 – r , BR = 4 – r. As, PA = PC and RB = RC [Length of tangants from external point is equal] Therefore, PR = PC + RC = 3 – r + 4 – r = 7 – 2r ------------------------ (i)

Also, PR = √(PQ2 + QR2) = √(32 + 42) = √(9 + 16) = √25 = 5 cm. Therefore, 7 – 2r = 5 => 2r = 2 => r = 1 cm. [Ans.]

Q.16. In the figure, AB is a diameter and AC is a chord of a circle such that LBAC = 30º. The tangent at C intersects AB produced at D. Prove that BC = BD.

Fig.

Solution :

AC is joined. LACB = 90º [Angle in the semi-circle is right angle] LABC = 60º [Angle sum property of triangle]

LCBD = 120º [Adj. to LCBA = 30º] LOCD = 90º [CD is tangent at C] LCOB = 60º [LCOB = 2LCAB] LOCB = 60º [Angle sum property of a triangle] LLBCD = LOCD – LOCB = 90º – 60º = 30º Therefore, LBCD = LBDC = 30º Therefore, BD = BC [Proved.]

Q.17. P and Q are centres of circles of radius 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cms, which touches the above circles externally. Given that LPRQ = 90º, write an equation in x and solve it.

20 of 32 Solution :

Fig.

∆ POQ is a right angled ∆. Therefore, by Pythagoras Theorem, (9 + x)2 + (2 + x)2 = (17)2 Or, 81 + 18x + x2 + 4 + 4x + x2 = 289 Or, 2x2 + 22x – 204 = 0 Or, x2 + 11x – 102 = 0 Or, x2 + 17x – 6x – 102 = 0

Or, x(x + 17) – 6(x + 17) = 0 Or, (x + 17)(x – 6) = 0 Or, x = 6, – 17 But x = – 17 is not possible. Therefore, value of x = 6 cm. [Ans.]

Q.18. In the given figure, AB is the diameter of a circle with centre O. LBCD is 120º. Find : (i) LDBA and (ii) LBAD.

Solution :

Given – Ab is a diameter of the circle with centre O and LBCD = 120º. To find – LDBA and LBAD. Construction – BD is joined. Proof – LADB = 90º [Angle in a semi-circle] ABCD is a cyclic quadrilateral Therefore, LA + LC = 180º Or, LA = LBAD = 180º – LC = 180º – 120º = 60º. [Ans.] Now in ∆ ABD, LDAB + LADB + LDBA = 180º Or, 60º + 90º + LDBA = 180º

Or, 150º + LDBA = 180º Or, LDBA = 180º – 150º = 30º. [Ans.]

Q.19. In the figure, chord AB and CD when extended meet at X. Given AB = 4 cm, BX = 6 cm, XD = 5 cm, calculate the length of CD.

Fig.

Solution :

We know that XB.XA = XD.XC [from the figure] Or, XB.(XB + BA) = XD.(XD + DC) Or, 6(6 + 4) = 5(5 + DC) Or, 60 = 5(5 + DC)

Or, 5 + DC = 60/5 = 12 Or, DC = 12 – 5 = 7 cm. [Ans.]

Q.20. In the figure, AB is a common tangent to two circles intersecting at C and D. Write down the measure of (LACB + LADB). Justify your answer.

21 of 32 Solution :.

CD is joined. LCAB = LADC and LCBA = LBDC [Angles in the alternate segments] Therefore, LACB = 180º – (LCAB + LCBA) = 180º – (LADC + LBDC) = 180º 0 – LADB Therefore, LACB + LADB = 180º. [Ans.]

Shares and Dividends

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

22 of 32 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.]

23 of 32 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.]

Q.9. Mr. Ram Gopal invested Rs8,000 in 7% Rs100 shares at Rs80. After a year he sold these shares at Rs75 each and invested the proceeds (including his dividend) in 18%, Rs25 shares at Rs41. Find :

i. His dividend for the first year. ii. His annual income in the second year. iii. The percentage increase in his return on his original investment.

Solution :

Investment = Rs8,000; Market value of one share = Rs80 Therefore, No. of shares = Rs8,000/Rs80 = 100. Dividend per share = 7% of Rs100 = Rs7. Therefore, dividend for 100 shares = 100×Rs7 = Rs700.

i. Or, dividend for the first year = Rs700. [Ans.]

24 of 32 ii. S.P. of 100 shares @ Rs75 each = 100×Rs75 = Rs7,500.

Investment = Rs7,500 + Rs700 = Rs8,200. Market value of one share = Rs41. Therefore, No. of shares = 8,200/41 = 200. Dividend per share = 18% of Rs25 = Rs4.50. Therefore, dividend for 200 shares = 200×Rs4.50 = Rs900. His annual income for the second year = Rs900. [Ans.]

iii. Increase in income = Rs900 – Rs700 = Rs200. Therefore, percentage increase in his return on his original investment = (Rs200/Rs8,000)×100% = 2.5%. [Ans.]

Q.9. Mr. Tiwari invested Rs29,040 in 15% Rs100 shares quoted at a premium of 20%. Calculate :

i. The number of shares bought by Mr. Tiwari. ii. Mr. Tiwari’s income from the investment. iii. The percentage return on his investment.

Solution :

i. Market value of one share = Rs100 + 20% of Rs100 = Rs120. No. of shares = Rs29,040/Rs120 = 242. [Ans.]

ii. Dividend per share = 15% of Rs100 = Rs15. Therefore, dividend income = 242×Rs15 = Rs3.630. [Ans.]

iii. Percentage return on his investment = (Rs15/Rs120)×100% = 12.5%. [Ans.]

Q.10. A man invest Rs1,680 in buying shares of nominal value Rs24 and selling at 12% premium. The dividend on the shares is 15% per annum.

i. Calculate the number of shares he buys; ii. Calculate the dividend he receives annually.

Solution :

i. Nominal value per share = Rs24; Premium = 12%; Investment = Rs1,680. Marked value of one share = Rs24 + 12% of Rs24 = Rs24(1 + 12/100) = Rs24×1.12 = Rs26.88. number of shares bought = Rs1,680/Rs26.88 = 62.5, which is a fraction. Hence only 62 shares is bought. [Ans.]

ii. Dividend per share = 15% of Rs24 = Rs3.60. Therefore, dividend for 62.5 shares = 62.5×Rs3.60 = Rs225. [Ans.]

Banking

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 ?

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.

25 of 32 = Rs. (5400 × 1/12 × 8/100) = Rs. 36. The maturity value = Rs. 1200 + Rs. 36 = Rs. 1236. [Ans.]

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

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.

Solution :

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.2690

26 of 32

Rs.2687.83

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

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.]

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

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.

Solution :

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

27 of 32

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.]

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

Calculate the interest due to him at the end of financial year (March 31st 2008) at the rate of 6% per annum.

Solution :

Do yourself. [Ans. = Rs565.78]

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.

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.]

Ratio and Proportion

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

28 of 32 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 :

29 of 32 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.10. In a regiment, the ratio of number of officers to the number of soldiers was 3:31 before a battle. In the battle 6 officers and 22 soldiers were killed. The ratio between the number of officers and the number of soldiers now is 1:13. Find the number of officers and soldiers in the regiment before the battle.

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 Or, 8x = 56 Or, x = 7 Hence, no. of officers = 3 ×7 = 21, and no. of soldiers = 31 ×7 = 217. [Ans.]

Q.11. 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.]

Q.12. If x : y : : y :z , show that x :z = x2 : y2.

Solution :

30 of 32 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.]

Q.13. 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.]

Q.14. If (3a+ 4b)/(3c + 4d) = (3a – 4b)/(3c – 4d), prove that a/b = c/d.

Solution :

Do yourself.

Q.15. Given a/b = c/d, prove that (3a – 5b)/(3a + 5b) = (3c – 5d)/(3c + 5d).

Solution :

We have, a/b = c/d Multiplying both sides by 3/5 , we get 3a/5b = 3c/5d By Componenedo and Dividendo, we get (3a + 5b)/(3a – 5b) = (3c + 5d)/(3c – 5d) By invertendo, we get (3a – 5b)/(3a + 5b) = (3c – 5d)/(3c + 5d) [Proved.]

Q.16. 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.]

Q.17. The work done by (x – 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio of 3 : 10. Find the value of x.

Solution :

31 of 32 We have, (x – 3)(2x + 1) : (2x + 1)(x + 4) = 3 : 10 Or, 10(x – 3)(2x + 1) = 3(2x + 1)(x + 4) Or, 10x – 30 = 3x + 12 Or, 10x – 3x = 12 + 30 Or, 7x = 42 Or, x = 6. [Ans.]

Height and Distance

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 ∆ ABC, tan 45º = AB/BC => 1 = h/x => x = h ------------- (1) In right angled ∆ 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 ∆ADC, tan60º = h/CD Or, √3 = h/CD Or, CD = h/√3 In ∆ADB, 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 :

32 of 32 i. the height of the tower TF

ii. the angle ’y’ correct to nearest degree.

Solution :

i. In ∆ 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 ∆ 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 ∆ SPR, tan θ = h/a Or, h = a tan θ --------------------------------- (i) In ∆SQR, 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.]