Chapterwise Revision Exercise

Find the amount of bill for the following intra-state transaction of goods/services. The rate of GST being 12 % :

Answer

If for the value of any transaction, discount = x %.

It's discounted value = (100 - x)% of its original value.

CGST = 6% of ₹ 1911 = $\dfrac{6}{100} \times 1938$ = ₹ 114.66

SGST = CGST = ₹ 114.66

Amount of bill = ₹ 1911 + ₹ 114.66 + ₹ 114.66 = ₹ 2140.32

Hence, amount of bill = ₹ 2140.32

Find the amount of bill for the following inter-state transaction of goods/services. The rate of GST being 5 % :

Answer

If for the value of any transaction, discount = x %.

It's discounted value = (100 - x)% of its original value.

IGST = 5% of ₹ 27780 = $\dfrac{5}{100} \times 27780 = \dfrac{138900}{100}$ = ₹ 1389

Amount of bill = ₹ 27780 + ₹ 1389 = ₹ 29169

Hence, amount of bill = ₹ 29169.

Find the amount of bill for the following transaction of goods/services from Patna (Bihar) to Ajmer (Rajasthan) :

Answer

This is a interstate transaction.

∴ GST = IGST.

Hence, the amount of bill = ₹ 29984.80

Find the amount for the following transaction of goods/services within Gujarat :

Answer

This is an intra-state transaction.

GST = CGST + SGST

We know, SGST = CGST

Hence, amount of bill = ₹ 86688.

A dealer in Kanpur (U.P.) supplies goods worth ₹ 5000 to a dealer in Meerut (U.P.). The dealer in Meerut supplies the same goods/services to a dealer in Delhi at a profit of ₹ 2000. Find the cost of goods/services in Delhi as per GST system. The rate of GST is 18%.

Answer

When product is sold from Kanpur to Meerut (It is an intra-state transaction)

For dealer in Kanpur :

S.P. = ₹ 5000

CGST = 9% of ₹ 5000 = $\dfrac{9}{100} \times 5000 = ₹ 450$.

SGST = CGST = ₹ 450.

When product is sold from Meerut to Delhi (It is an inter-state transaction)

For dealer in Meerut :

Input-tax credit (ITC) = ₹ 450 + ₹ 450 = ₹ 900.

C.P. (excluding GST) = ₹ 5000

Profit = ₹ 2000

S.P. = ₹ 5000 + ₹ 2000 = ₹ 7000

IGST = 18% of ₹ 7000 = $\dfrac{18}{100} \times 7000$ = ₹ 1260

C.P. in Delhi = S.P. in Meerut + IGST = ₹ 7000 + ₹ 1260 = ₹ 8260.

Hence, cost of goods/services in Delhi = ₹ 8260.

Ashok deposits ₹ 3200 per month in a cumulative deposit account for 3 years at the rate of 9% per annum. Find the maturity value of this account.

Answer

Given,

P = ₹ 3200

r = 9%

Maturity value of recurring deposit = Total sum deposited + Interest on it

= P × n + P × $\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

where n is time in months.

Time (n) = 3 years = 3 × 12 = 36 months.

Substituting values we get :

$= 3200 \times 36 + 3200 \times \dfrac{36 \times (36 + 1)}{2 \times 12} \times \dfrac{9}{100} \\[1em] = 115200 + 3200 \times \dfrac{36 \times 37}{24} \times \dfrac{9}{100} \\[1em] = 115200 + \dfrac{115200 \times 37 \times 9}{2400} \\[1em] = 115200 + 48 \times 37 \times 9 \\[1em] = 115200 + 15984 \\[1em] = 131184.$

Hence, maturity value = ₹ 131184.

Mrs. Karna has a recurring deposit account in Punjab National Bank for 3 years at 8% p.a. If she gets ₹ 9990 as interest at the time of maturity, find :

(i) the monthly instalment

(ii) the maturity value of the account.

Answer

(i) Given,

Time (n) = 3 years = 3 × 12 = 36 months.

Rate = 8%

Let monthly instalment be ₹ P.

By formula,

Interest = P × $\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get

$\Rightarrow 9990 = P \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} \\[1em] \Rightarrow 9990 = P \times \dfrac{3 \times 37}{2} \times \dfrac{8}{100} \\[1em] \Rightarrow 9990 = P \times \dfrac{3 \times 37}{25} \\[1em] \Rightarrow P = \dfrac{9990 \times 25}{3 \times 37} \\[1em] \Rightarrow P = \dfrac{249750}{111} \\[1em] \Rightarrow P = ₹ 2250.$

Hence, monthly installment = ₹ 2250.

(ii) Maturity value of recurring deposit = Total sum deposited + Interest on it

= P × n + 9990

= 2250 × 36 + 9990

= 81000 + 9990

= ₹ 90990.

Hence, the maturity value of this account = ₹ 90990.

A man has a 5 year recurring deposit account in a bank and deposits ₹ 240 per month. If he recieves ₹ 17,694 at the time of maturity, find the rate of interest.

Answer

Let rate of interest be r%.

Given,

P = ₹ 240

M.V. = ₹ 17694

Time (n) = 5 years = 5 × 12 = 60 months.

Maturity value = P × n + P × $\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

⇒ 17694 = 240 × 60 + 240 × $\dfrac{60 \times 61}{2 \times 12} \times \dfrac{r}{100}$

⇒ 17694 = 14400 + 6 × $\dfrac{305r}{5}$

⇒ 17694 - 14400 = 6 × 61r

⇒ 3294 = 366r

⇒ r = $\dfrac{3294}{366}$ = 9%.

Hence, rate of interest = 9%.

Sheela has a recurring deposit account in a bank of ₹ 2000 per month at the rate of 10% per annum. If she gets ₹ 83100 at the time of maturity, find the total time (in years) for which the account was held.

Answer

Let time = n months.

Given,

P = ₹ 2000

r = 10%

By formula,

M.V. = P × n + P × $\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 83100 = 2000 \times n + 2000 \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] \Rightarrow 83100 = 2000n + \dfrac{100n(n + 1)}{12} \\[1em] \Rightarrow 83100 = \dfrac{24000n + 100n^2 + 100n}{12} \\[1em] \Rightarrow 83100 \times 12 = 100(240n + n^2 + n) \\[1em] \Rightarrow 241n + n^2 = \dfrac{83100 \times 12}{100} \\[1em] \Rightarrow 241n + n^2 = 9972 \\[1em] \Rightarrow n^2 + 241n - 9972 = 0 \\[1em] \Rightarrow n^2 + 277n - 36n - 9972 = 0\\[1em] \Rightarrow n(n + 277) - 36(n + 277) = 0 \\[1em] \Rightarrow (n - 36)(n + 277) = 0 \\[1em] \Rightarrow n - 36 = 0 \text{ or } n + 277 = 0 \\[1em] \Rightarrow n = 36 \text{ or } n = -277.$

Since, time cannot be negative.

∴ n = 36 months = 3 years.

Hence, time = 3 years.

A man deposits ₹ 900 per month in a recurring account for 2 years. If he gets ₹ 1800 as interest at the time of maturity, find the rate of interest.

Answer

Let rate of interest be r%.

Given,

P = ₹ 900

Time (n) = 2 years = 24 months.

By formula,

Interest = P × $\dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 1800 = 900 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1800 = 225r \\[1em] \Rightarrow r = \dfrac{1800}{225} \\[1em] \Rightarrow r = 8$

Hence, the rate of interest = 8%.

What is the market value of $4\dfrac{1}{2}$ % (₹ 100) share, when an investment of ₹ 1800 produces an income of ₹ 72 ?

Answer

Given,

Nominal value (N.V.) = ₹ 100

Income (dividend) on one share = $4\dfrac{1}{2}$ % of N.V. = $\dfrac{4\dfrac{1}{2}}{100} \times 100 = 4\dfrac{1}{2}$.

Given,

Total income = ₹ 72

⇒ No. of shares × Income on one share = 72

⇒ No. of shares × 4.5 = 72

⇒ No. of shares = $\dfrac{72}{4.5} = 16$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 1800 = M.V. of each share × 16

⇒ M.V. of each share = $\dfrac{1800}{16}$ = ₹ 112.50

Hence, M.V. of each share = ₹ 112.50

By investing ₹ 10,000 in the shares of a company, a man gets an income of ₹ 800; the dividend being 10%. If the face-value of each share is ₹ 100, find :

(i) the market value of each share.

(ii) the rate percent which the person earns on his investment.

Answer

(i) Given,

Nominal value (N.V.) = ₹ 100

Income (dividend) on one share = 10 % of N.V. = $\dfrac{10}{100} \times 100 = 10$.

Given,

Total income = ₹ 800

⇒ No. of shares × Income on one share = 800

⇒ No. of shares × 10 = 800

⇒ No. of shares = $\dfrac{800}{10} = 80$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 10000 = M.V. of each share × 80

⇒ M.V. of each share = $\dfrac{10000}{80}$ = ₹ 125

Hence, M.V. of each share = ₹ 125

(ii) Given,

Man earns an income of ₹ 800 on investing ₹ 10000.

Rate percent = $\dfrac{\text{Income}}{\text{Investment}} \times 100$

$= \dfrac{800}{10000} \times 100 \\[1em] = \dfrac{80000}{10000} \\[1em] = 8$

Hence, person earns 8% on his investment.

A man holds 800 shares of ₹ 100 each of a company paying 7.5% dividend semi-annually.

(i) Calculate his annual dividend.

(ii) If he had bought these shares at 40% premium, what percentage return does he get on his investment ?

Answer

(i) Given,

A man holds 800 shares.

Nominal Value (N.V.) of each share = ₹ 100

Dividend paid semi-annually = 7.5%

Dividend paid annually = 7.5% × 2 = 15%

(i) Dividend on each share = 15% of N.V.

= $\dfrac{15}{100} \times 100$ = ₹ 15.

Total dividend = Dividend on each share × No. of shares

= ₹ 15 × 800 = ₹ 12000.

Hence, annual dividend of person = ₹ 12000.

(ii) Given,

The person had bought the shares at 40% premium.

Market Value (MV) = N.V. + Premium = 100 + $\dfrac{40}{100} \times 100$ = 100 + 40 = ₹ 140.

Total investment = M.V. × No. of shares

= 140 × 800 = ₹ 112000.

Rate of return = $\dfrac{\text{Income}}{\text{Investment}} \times 100 = \dfrac{12000}{112000} \times 100 = \dfrac{1200}{112}$ = 10.714 %.

Hence, there is a 10.714% percentage return on investment.

A man invests ₹ 10560 in a company, paying 9% dividend, at the time when its ₹ 100 shares can be bought at a premium of ₹ 32. Find :

(i) the number of shares bought by him;

(ii) his annual income from these shares and

(iii) the rate of return on his investment.

Answer

Given,

N.V. of share = ₹ 100

M.V. = N.V. + Premium = ₹ 100 + ₹ 32 = ₹ 132.

(i) No. of shares bought = $\dfrac{\text{Investment}}{\text{M.V.}} = \dfrac{10560}{132}$ = 80.

Hence, no. of shares = 80.

(ii) Income on 1 share = 9% of N.V.

= $\dfrac{9}{100} \times 100$ = ₹ 9.

Annual income = Income on 1 share × No. of shares

= ₹ 9 × 80 = ₹ 720.

Hence, annual income = ₹ 720.

(iii) Rate of return = $\dfrac{\text{Income}}{\text{Investment}} \times 100$

$= \dfrac{720}{10560} \times 100 \\[1em] = \dfrac{72000}{10560} \\[1em] = 6\dfrac{9}{11}$

Hence, rate of return = $6\dfrac{9}{11}$ %.

Find the market value of 12% ₹ 25 shares of a company which pays a dividend of ₹ 1875 on an investment of ₹ 20000.

Answer

Given,

Nominal value (N.V.) = ₹ 25

Income (dividend) on one share = 12 % of N.V. = $\dfrac{12}{100} \times 25 = 3$.

Given,

Total income = ₹ 1875

⇒ No. of shares × Income on one share = 1875

⇒ No. of shares × 3 = 1875

⇒ No. of shares = $\dfrac{1875}{3} = 625$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 20000 = M.V. of each share × 625

⇒ M.V. of each share = $\dfrac{20000}{625}$ = ₹ 32

Hence, M.V. of each share = ₹ 32.

A man invests ₹ 20,020 in buying shares of N.V. ₹ 26 at 10% premium. The dividend on the shares is 15% per annum. Calculate :

(i) the number of shares he buys.

(ii) the dividend he receives annually.

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

Answer

(i) M.V. = ₹ 26 + $\dfrac{10}{100} \times 26$ = ₹ 26 + ₹ 2.6 = ₹ 28.6

Investment = ₹ 20,020

No. of shares bought = $\dfrac{20020}{28.6}$ = 700.

Hence, no. of shares bought = 700.

(ii) Annual dividend = No. of shares × Rate of div. × N.V. of 1 share

= 700 × $\dfrac{15}{100} \times 26$

= ₹ 2,730.

Hence, dividend = ₹ 2,730.

(iii) Rate of interest = $\dfrac{2730}{20020} \times 100 = \dfrac{273000}{20020} = 13\dfrac{7}{11}$%.

Hence, the rate of interest = $13\dfrac{7}{11}$%.

Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the :

(i) original number of shares.

(ii) sale proceeds.

(iii) new number of shares.

(iv) change in the two dividends.

Answer

(i) M.V. of first type of share = ₹ 100 + ₹ 20 = ₹ 120.

Investment = ₹ 9,600

No. of shares = $\dfrac{\text{Investment}}{\text{M.V.}} = \dfrac{9600}{120}$ = 80.

Hence, no. of shares = 80.

(ii) S.P. of 1 share = ₹ 160,

S.P. of 80 shares = 80 × ₹ 160 = ₹ 12,800.

Hence, sale proceeds = ₹ 12,800.

(iii) M.V. of second type of share = ₹ 40.

Investment = ₹ 12,800

No. of shares = $\dfrac{\text{Investment}}{\text{M.V.}} = \dfrac{12800}{40}$ = 320.

Hence, no. of shares = 320.

(iv) Annual dividend = No. of shares × Rate of div. × N.V. of 1 share

In first case :

Annual dividend = 80 × $\dfrac{8}{100} \times 100$ = ₹ 640.

In second case :

Annual dividend = 320 × $\dfrac{10}{100} \times 50$ = ₹ 1600.

Difference = ₹ 1600 - ₹ 640 = ₹ 960

Hence, there is an increase of ₹ 960 in dividend amount.

The given diagram represents two sets A and B on real number lines.

(i) Write down A and B in set-builder form.

(ii) Represent A ∪ B, A ∩ B, A' ∩ B, A - B and B - A on separate number lines.

Answer

(i) From graph,

A = {x ∈ R : -3 < x ≤ 2}, B = {x ∈ R : 0 ≤ x < 5}

(ii) From graph,

A ∪ B = {x ∈ R : -3 < x < 5}

A ∩ B = {x ∈ R : 0 ≤ x ≤ 2}

A' ∩ B = {x ∈ R : -2 < x < 5}

A - B = {x ∈ R : -3 < x < 0}

B - A = {x ∈ R : 2 < x < 5}

Find the values of x, which satisfy the inequation :

$-2 \le \dfrac{1}{2} - \dfrac{2x}{3} \lt 1\dfrac{5}{6}$; x ∈ N

Graph the solution set on the real number line.

Answer

Solving L.H.S. of the above inequation :

$\Rightarrow -2 \le \dfrac{1}{2} - \dfrac{2x}{3} \\[1em] \Rightarrow \dfrac{2x}{3} \le \dfrac{1}{2} + 2 \\[1em] \Rightarrow \dfrac{2x}{3} \le \dfrac{5}{2} \\[1em] \Rightarrow x \le \dfrac{5}{2} \times \dfrac{3}{2} \\[1em] \Rightarrow x \le \dfrac{15}{4} \\[1em] \Rightarrow x \le 3\dfrac{3}{4}\text{ ...........(1)}$

Solving R.H.S. of the inequation :

$\Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \lt 1\dfrac{5}{6} \\[1em] \Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \lt \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{1}{2} - \dfrac{11}{6} \lt \dfrac{2x}{3} \\[1em] \Rightarrow \dfrac{2x}{3} \gt \dfrac{3 - 11}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \gt -\dfrac{8}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \gt -\dfrac{4}{3} \\[1em] \Rightarrow x \gt -\dfrac{4}{3} \times \dfrac{3}{2} \\[1em] \Rightarrow x \gt -2 \text{ .........(2)}$

From (1) and (2),

-2 < x ≤ $3\dfrac{3}{4}$

Since, x ∈ N

∴ x = {1, 2, 3}.

Hence, x = {1, 2, 3}.

State for each of the following statements whether it is true or false :

(a) If (x - a)(x - b) < 0, then x < a and x < b.

(b) If a < 0 and b < 0, then (a + b)² > 0.

(c) If a and b are any two integers such that a > b, then a² > b².

(d) If p = q + 2, then p > q.

(e) If a and b are two negative integers such that a < b, then $\dfrac{1}{a} \lt \dfrac{1}{b}$

Answer

(a) False
Reason — If x < a and x < b, then (x - a)(x - b) will be greater than 0 as multiplication of two negative number generates a positive number.

(b) True
Reason — If a < 0 and b < 0, then (a + b) will be a negative number and square of any negative number is greater than 0, so (a + b)² is greater than 0.

(c) False
Reason — If a > b, then a² > b² is true only if a and b are positive numbers.

(d) True
Reason — If a positive number is added to any number (suppose x), then resultant number is greater than x.

(e) False
Reason — If, a < b and both are two negative integers then on recripocating the numbers the sign will be reversed, so $\dfrac{1}{a} \gt \dfrac{1}{b}$.

Given 20 - 5x < 5(x + 8), find the smallest value of x when :

(i) x ∈ I

(ii) x ∈ W

(iii) x ∈ N

Answer

Solving,

⇒ 20 - 5x < 5(x + 8)

⇒ 20 - 5x < 5x + 40

⇒ 5x + 5x > 20 - 40

⇒ 10x > -20

⇒ x > $-\dfrac{20}{10}$ = -2.

(i) As, x > -2 and x ∈ I.

∴ x = {-1, 0, 1 .......}

The smallest no. = -1.

Hence, smallest value of x = -1, when x is an integer.

(ii) As, x > -2 and x ∈ W.

∴ x = {0, 1, 2 .......}

The smallest no. = 0.

Hence, smallest value of x = 0, when x is a whole number.

(iii) As, x > -2 and x ∈ N.

∴ x = {1, 2 .......}

The smallest no. = 1.

Hence, smallest value of x = 1, when x is a natural number.

If x ∈ Z, solve : 2 + 4x < 2x - 5 ≤ 3x. Also, represent its solution on the real number line.

Answer

Given,

⇒ 2 + 4x < 2x - 5 ≤ 3x

Solving L.H.S. of the above inequation :

⇒ 2 + 4x < 2x - 5

⇒ 4x - 2x < - 5 - 2

⇒ 2x < -7

⇒ x < $-\dfrac{7}{2}$

⇒ x < -3.5 ..........(1)

Solving R.H.S. of the above inequation :

⇒ 2x - 5 ≤ 3x

⇒ 3x - 2x ≥ -5

⇒ x ≥ -5 .............(2)

From (1) and (2), we get :

-5 ≤ x < -3.5; Since, x ∈ Z

Hence, x = {-5, -4}.

If P = {x : 7x - 4 > 5x + 2, x ∈ R} and Q = {x : x - 19 ≥ 1 - 3x, x ∈ R}; find the range of set P ∩ Q and represent it on a number line.

Answer

Solving 7x - 4 > 5x + 2,

⇒ 7x - 5x > 2 + 4

⇒ 2x > 6

⇒ x > 3.

P = {x : x > 3, x ∈ R}

Solving x - 19 ≥ 1 - 3x

⇒ x + 3x ≥ 1 + 19

⇒ 4x ≥ 20

⇒ x ≥ 5.

Q = {x : x ≥ 5, x ∈ R}

P ∩ Q = Numbers common between P and Q

= {x : x ≥ 5, x ∈ R}

P ∩ Q on the number line is :

Find the values of x, which satisfy the inequation $-2\dfrac{5}{6} \lt \dfrac{1}{2} - \dfrac{2x}{3} \le 2$, x ∈ W. Graph the solution set on the number line.

Answer

Given,

$-2\dfrac{5}{6} \lt \dfrac{1}{2} - \dfrac{2x}{3} \le 2$

Solving L.H.S. of the inequation,

$\Rightarrow -2\dfrac{5}{6} \lt \dfrac{1}{2} - \dfrac{2x}{3} \\[1em] \Rightarrow -\dfrac{17}{6} \lt \dfrac{1}{2} - \dfrac{2x}{3} \\[1em] \Rightarrow \dfrac{2x}{3} \lt \dfrac{1}{2} + \dfrac{17}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \lt \dfrac{3 + 17}{6} \\[1em] \Rightarrow \dfrac{2x}{3} \lt \dfrac{20}{6} \\[1em] \Rightarrow x \lt \dfrac{20}{6} \times \dfrac{3}{2} \\[1em] \Rightarrow x \lt 5 .......(i)$

Solving R.H.S. of the inequation,

$\Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \le 2 \\[1em] \Rightarrow \dfrac{2x}{3} \ge \dfrac{1}{2} - 2 \\[1em] \Rightarrow \dfrac{2x}{3} \ge -\dfrac{3}{2} \\[1em] \Rightarrow x \ge -\dfrac{3}{2} \times \dfrac{3}{2} \\[1em] \Rightarrow x \ge -\dfrac{9}{4} \\[1em] \Rightarrow x \ge -2.25 ........(ii)$

From (i) and (ii) we get,

-2.25 ≤ x < 5

Since, x ∈ W,

∴ Solution set = {0, 1, 2, 3, 4}.

Solution on the number line is :

Solve :

$\dfrac{8}{x + 3} - 2 = \dfrac{3}{2 - x}$

Answer

Solving :

$\Rightarrow \dfrac{8}{x + 3} - 2 = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{8 - 2(x + 3)}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{8 - 2x - 6}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{2 - 2x}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow (2 - 2x)(2 - x) = 3(x + 3) \\[1em] \Rightarrow 4 - 2x - 4x + 2x^2 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x + 4 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x - 3x + 4 - 9 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow x = -\dfrac{1}{2} \text{ or } x = 5.$

Hence, x = $-\dfrac{1}{2}$ or 5.

Solve :

$\dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15}$

Answer

Solving :

$\Rightarrow \dfrac{x(6 - x) + 9}{3(6 - x)} = \dfrac{2(6 + x)}{15} \\[1em] \Rightarrow \dfrac{6x - x^2 + 9}{18 - 3x} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 + 216 - 135 - 90x = 0 \\[1em] \Rightarrow 9x^2 + 81 - 90x = 0 \\[1em] \Rightarrow 9(x^2 + 9 - 10x) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - x - 9x + 9 = 0 \\[1em] \Rightarrow x(x - 1) - 9(x - 1) = 0 \\[1em] \Rightarrow (x - 9)(x - 1) = 0 \\[1em] \Rightarrow x - 9 = 0 \text{ or } x - 1 = 0 \\[1em] \Rightarrow x = 9 \text{ or } x = 1.$

Hence, x = 1 or 9.

Find the value of k for which the roots of the following equation are real and equal.

k²x² - 2(2k - 1)x + 4 = 0

Answer

Given,

Equation : k²x² - 2(2k - 1)x + 4 = 0

The roots of the equation are real and equal, when D = 0.

⇒ b² - 4ac = 0

⇒ [-2(2k - 1)]² - 4 × k² × 4 = 0

⇒ [-4k + 2]² - 16k² = 0

⇒ 16k² + 4 - 16k - 16k² = 0

⇒ 16k = 4

⇒ k = $\dfrac{4}{16} = \dfrac{1}{4}$.

Hence, k = $\dfrac{1}{4}$.

Solve :

$\dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax}$, when x ≠ 0 and a ≠ 0.

Answer

Solving :

$\Rightarrow \dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a(a + b)}{ax} = \dfrac{ab + b^2}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a(a + b)}{ax} \times ax = ab + b^2 \\[1em] \Rightarrow x^2 - a^2 - ab = ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + ab + ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + 2ab + b^2 \\[1em] \Rightarrow x^2 = (a + b)^2 \\[1em] \Rightarrow x = \sqrt{(a + b)^2} \\[1em] \Rightarrow x = (a + b), -(a + b).$

Hence, x = (a + b), -(a + b).

If -5 is a root of the quadratic equation 2x² + px - 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Answer

As, -5 is the root of the quadratic equation 2x² + px - 15 = 0, so it will satisfy the equation.

∴ 2(-5)² + (-5)p - 15 = 0

⇒ 2 × 25 - 5p - 15 = 0

⇒ 50 - 5p - 15 = 0

⇒ 5p = 35

⇒ p = $\dfrac{35}{5}$ = 7.

Substituting value of p in p(x² + x) + k = 0, we get :

⇒ 7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

Since, above equation has real and equal roots.

∴ D = 0

⇒ b² - 4ac = 0

⇒ (7)² - 4 × 7 × k = 0

⇒ 49 - 28k = 0

⇒ 28k = 49

⇒ k = $\dfrac{49}{28} = \dfrac{7}{4} = 1\dfrac{3}{4}$.

Hence, k = $1\dfrac{3}{4}$.

x articles are bought at ₹ (x - 8) each and (x - 2) some other articles are bought at ₹ (x - 3) each. If the total cost of all these articles is ₹ 76, how many articles of first kind were bought ?

Answer

Since,

x articles are bought at ₹ (x - 8) each and (x - 2) some other articles are bought at ₹ (x - 3) each.

Total cost = x(x - 8) + (x - 2)(x - 3).

Given,

Total cost = ₹ 76.

∴ x(x - 8) + (x - 2)(x - 3) = 76

⇒ x² - 8x + x² - 3x - 2x + 6 = 76

⇒ 2x² - 13x + 6 - 76 = 0

⇒ 2x² - 13x - 70 = 0

⇒ 2x² - 20x + 7x - 70 = 0

⇒ 2x(x - 10) + 7(x - 10) = 0

⇒ (2x + 7)(x - 10) = 0

⇒ 2x + 7 = 0 or x - 10 = 0

⇒ 2x = -7 or x = 10

⇒ x = -$\dfrac{7}{2}$ or x = 10.

Since, no. of articles cannot be negative.

∴ x = 10.

Hence, there are 10 articles of first kind.

In a two digit number, the unit's digit exceeds its ten's digit by 2. The product of the given number and the sum of its digits is equal to 144. Find the number.

Answer

Let ten's digit be x.

Unit's digit = x + 2

Number = 10x + x + 2 = 11x + 2

Sum of digits = x + x + 2 = 2x + 2

According to question,

Number × Sum of digits = 144

⇒ (11x + 2)(2x + 2) = 144

⇒ 22x² + 22x + 4x + 4 = 144

⇒ 22x² + 26x + 4 = 144

⇒ 22x² + 26x = 140

⇒ 2(11x² + 13x) = 140

⇒ 11x² + 13x = 70

⇒ 11x² + 13x - 70 = 0

⇒ 11x² + 35x - 22x - 70 = 0

⇒ x(11x + 35) - 2(11x + 35) = 0

⇒ (x - 2)(11x + 35) = 0

⇒ x - 2 = 0 or 11x + 35 = 0

⇒ x = 2 or x = -$\dfrac{35}{11}$

Since, digit cannot be negative.

∴ x = 2.

Number = 11x + 2 = 11(2) + 2

= 22 + 2 = 24.

Hence, number = 24.

The time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey. If he returned at a speed of 10 km/hour more than the speed of going, what was the speed per hour in each direction ?

Answer

Let speed while going be x km/hour and speed while returning will be (x + 10) km/hour.

Time taken while going = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{150}{x}$ hours.

Time taken while returning = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{150}{x + 10}$ hours.

Given,

Time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey.

$\therefore \dfrac{150}{x} - \dfrac{150}{x + 10} = 2.5 \\[1em] \Rightarrow \dfrac{150(x + 10) - 150x}{x(x + 10)} = \dfrac{25}{10} \\[1em] \Rightarrow \dfrac{150x + 1500 - 150x}{x(x + 10)} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{1500}{x^2 + 10x} = \dfrac{5}{2} \\[1em] \Rightarrow 3000 = 5(x^2 + 10x) \\[1em] \Rightarrow x^2 + 10x = \dfrac{3000}{5} \\[1em] \Rightarrow x^2 + 10x = 600 \\[1em] \Rightarrow x^2 + 10x - 600 = 0 \\[1em] \Rightarrow x^2 + 30x - 20x - 600 = 0 \\[1em] \Rightarrow x(x + 30) - 20(x + 30) = 0 \\[1em] \Rightarrow (x - 20)(x + 30) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 30 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -30.$

Since, speed cannot be negative.

∴ x = 20 km/hour and (x + 10) = 30 km/hour.

Hence, speed while going = 20 km/hour and speed while returning = 30 km/hour.

A takes 9 days more than B to do a certain piece of work. Together they can do the work in 6 days. How many days will A alone take to do the work ?

Answer

Let A alone takes x days and B alone takes y days.

Given,

A takes 9 days more than B to do a certain piece of work

∴ x = y + 9 .........(1)

As, A takes x days to do a work

So, one day work of A = $\dfrac{1}{x}$

As, B takes y days to do a work

So, one day work of B = $\dfrac{1}{y}$

As, together they can do work in 6 days.

∴ In one day both of them do $\dfrac{1}{6}$ th part of work.

$\therefore \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{1}{y + 9} + \dfrac{1}{y} = \dfrac{1}{6} \space [\text{From (1)}] \\[1em] \Rightarrow \dfrac{y + y + 9}{y(y + 9)} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{2y + 9}{y^2 + 9y} = \dfrac{1}{6} \\[1em] \Rightarrow 6(2y + 9) = y^2 + 9y \\[1em] \Rightarrow 12y + 54 = y^2 + 9y \\[1em] \Rightarrow y^2 + 9y - 12y - 54 = 0 \\[1em] \Rightarrow y^2 - 3y - 54 = 0\\[1em] \Rightarrow y^2 - 9y + 6y - 54 = 0 \\[1em] \Rightarrow y(y - 9) + 6(y - 9) = 0 \\[1em] \Rightarrow (y + 6)(y - 9) = 0 \\[1em] \Rightarrow y + 6 = 0 \text{ or } y - 9 = 0 \\[1em] \Rightarrow y = -6 \text{ or } y = 9.$

Since, days cannot be negative so, y ≠ -6.

x = y + 9 = 18.

Hence, A alone will take 18 days to complete the work.

A man bought a certain number of chairs for ₹ 10000. He kept one for his own use and sold the rest at the rate ₹ 50 more than he gave for one chair. Besides getting his own chair for nothing, he made a profit of ₹ 450. How many chairs did he buy ?

Answer

Given,

Man bought a certain number of chairs for ₹ 10000.

Let no. of chairs bought be x.

∴ Cost of one chair = ₹ $\dfrac{10000}{x}$

S.P. of each chair = ₹ $\dfrac{10000}{x} + 50$

Since, he kept one chair for himself so he sold (x - 1) chairs.

Given,

⇒ Profit = ₹ 450

⇒ S.P. - C.P. = ₹ 450

$\Rightarrow (x - 1) \times \Big(\dfrac{10000}{x} + 50\Big) - 10000 = 450 \\[1em] \Rightarrow 10000 + 50x - \dfrac{10000}{x} - 50 - 10000 = 450 \\[1em] \Rightarrow 50x - \dfrac{10000}{x} = 450 + 50 \\[1em] \Rightarrow \dfrac{50x^2 - 10000}{x} = 500 \\[1em] \Rightarrow 50x^2 - 10000 = 500x \\[1em] \Rightarrow 50(x^2 - 200) = 500x \\[1em] \Rightarrow x^2 - 200 = 10x \\[1em] \Rightarrow x^2 - 10x - 200 = 0 \\[1em] \Rightarrow x^2 - 20x + 10x - 200 = 0 \\[1em] \Rightarrow x(x - 20) + 10(x - 20) = 0 \\[1em] \Rightarrow (x + 10)(x - 20) = 0 \\[1em] \Rightarrow x + 10 = 0 \text{ or } x - 20 = 0 \\[1em] \Rightarrow x = -10 \text{ or } x = 20.$

Since no. of chairs cannot be negative.

∴ x = 20.

Hence, no. of chairs bought = 20.

In the given figure; the area of unshaded portion is 75% of the area of the shaded portion. Find the value of x.

Answer

From figure,

Length of larger portion (L) = (80 + x) meters

Breadth of larger portion (B) = (50 + x) meters

Area of larger portion = L × B = (80 + x)(50 + x)

= 4000 + 80x + 50x + x²

= (x² + 130x + 4000) m².

Length of smaller portion (l) = 80 meters

Breadth of smaller portion (b) = 50 meters

Area of smaller portion = l × b = 80 × 50 = 4000 m²

Area of unshaded portion = Area of larger portion - Area of smaller portion

= x² + 130x + 4000 - 4000

= (x² + 130x) m².

Given,

Area of unshaded portion is 75% of the area of the shaded portion.

$\therefore 75$

Since, length cannot be negative.

∴ x = 20 meters.

Hence, x = 20 meters.

Solve for x : $\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = b$.

Answer

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{(\sqrt{a + x} + \sqrt{a - x}) - (\sqrt{a + x} - \sqrt{a - x})} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{a + x}}{2\sqrt{a - x}} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{b + 1}{b - 1} \\[1em]$

Squaring both sides we get :

$\Rightarrow \dfrac{a + x}{a - x} = \dfrac{(b + 1)^2}{(b - 1)^2}$

Again applying componendo and dividendo, we get :

$\Rightarrow \dfrac{a + x + a - x}{a + x - (a - x)} = \dfrac{(b + 1)^2 + (b - 1)^2}{(b + 1)^2 - (b - 1)^2} \\[1em] \Rightarrow \dfrac{a + a + x - x}{a - a + x - (-x)} = \dfrac{b^2 + 1 + 2b + b^2 + 1 - 2b}{b^2 + 1 + 2b - (b^2 + 1 - 2b)} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{2(b^2 + 1)}{4b} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{b^2 + 1}{2b} \\[1em] \Rightarrow x = \dfrac{2ab}{b^2 + 1}.$

Hence, x = $\dfrac{2ab}{b^2 + 1}$.

If a : b = 2 : 3, b : c = 4 : 5 and c : d = 6 : 7, find a : b : c : d.

Answer

Given,

a : b = 2 : 3

$\Rightarrow \dfrac{a}{b} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{2 \times 8}{3 \times 8} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{16}{24} ...........(1)$

b : c = 4 : 5

$\Rightarrow \dfrac{b}{c} = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{b}{c} = \dfrac{4 \times 6}{5 \times 6} \\[1em] \Rightarrow \dfrac{b}{c} = \dfrac{24}{30} ...........(2)$

c : d = 6 : 7

$\Rightarrow \dfrac{c}{d} = \dfrac{6}{7} \\[1em] \Rightarrow \dfrac{c}{d} = \dfrac{6 \times 5}{7 \times 5} \\[1em] \Rightarrow \dfrac{c}{d} = \dfrac{30}{35} ...........(3)$

From (1), (2) and (3), we get :

a : b : c : d = 16 : 24 : 30 : 35

Hence, a : b : c : d = 16 : 24 : 30 : 35.

If $\dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$ = x, prove that x² - 4ax + 1 = 0.

Answer

Given,

$\dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$ = x.

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a+ 1} - \sqrt{2a - 1})} = \dfrac{x + 1}{x - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \dfrac{x + 1}{x - 1} \\[1em] \Rightarrow \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}} = \dfrac{x + 1}{x - 1}$

Squaring both sides we get,

$\Rightarrow \dfrac{2a + 1}{2a - 1} = \dfrac{(x + 1)^2}{(x - 1)^2}$

Applying componendo and dividendo we get :

$\Rightarrow \dfrac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)} = \dfrac{(x + 1)^2 + (x - 1)^2}{(x + 1)^2 - (x - 1)^2} \\[1em] \Rightarrow \dfrac{4a}{2} = \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow 2a = \dfrac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow 8ax = 2(x^2 + 1) \\[1em] \Rightarrow 4ax = x^2 + 1 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.$

Hence, proved that x² - 4ax + 1 = 0.

Find the compounded ratio of :

(i) (a - b) : (a + b) and (b² + ab) : (a² - ab)

(ii) (x + y) : (x - y), (x² + y²) : (x + y)² and (x² - y²)² : (x⁴ - y⁴)

(iii) (x² - 25) : (x² + 3x - 10); (x² - 4) : (x² + 3x + 2) and (x + 1) : (x² + 2x).

Answer

(i) Calculating the compounded ratio :

$\Rightarrow \dfrac{a - b}{a + b} \times \dfrac{b^2 + ab}{a^2 - ab} \\[1em] \Rightarrow \dfrac{a - b}{a + b} \times \dfrac{b(a + b)}{a(a - b)} \\[1em] \Rightarrow \dfrac{b}{a} \\[1em] \Rightarrow b : a.$

Hence, compounded ratio of (a - b) : (a + b) and (b² + ab) : (a² - ab) = b : a.

(ii) Calculating the compounded ratio :

$\Rightarrow \dfrac{x + y}{x - y} \times \dfrac{x^2 + y^2}{(x + y)^2} \times \dfrac{(x^2 - y^2)^2}{(x^4 - y^4)} \\[1em] \Rightarrow \dfrac{1}{x - y} \times \dfrac{x^2 + y^2}{(x + y)} \times \dfrac{(x^2 - y^2)^2}{(x^2 - y^2)(x^2 + y^2)} \\[1em] \Rightarrow \dfrac{1}{x - y} \times \dfrac{\cancel{x^2 + y^2}}{(x + y)} \times \dfrac{(x^2 - y^2)^{\cancel{2}}}{\cancel{(x^2 - y^2)}\cancel{(x^2 + y^2)}} \\[1em] \Rightarrow \dfrac{(x^2 - y^2)}{(x - y)(x + y)} \\[1em] \Rightarrow \dfrac{(x^2 - y^2)}{(x^2 - y^2)} \\[1em] \Rightarrow 1 : 1.$

Hence, the compounded ratio = 1 : 1.

(iii) Calculating the compounded ratio :

$\Rightarrow \dfrac{x^2 - 25}{x^2 + 3x - 10} \times \dfrac{x^2 - 4}{x^2 + 3x + 2} \times \dfrac{x + 1}{x^2 + 2x} \\[1em] \Rightarrow \dfrac{x^2 - 25}{x^2 + 5x - 2x - 10} \times \dfrac{(x - 2)(x + 2)}{x^2 + x + 2x + 2} \times \dfrac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \dfrac{(x - 5)(x + 5)}{x(x + 5) - 2(x + 5)} \times \dfrac{(x - 2)(x + 2)}{x(x + 1) + 2(x + 1)} \times \dfrac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \dfrac{(x - 5)(x + 5)}{(x - 2)(x + 5)} \times \dfrac{(x - 2)(x + 2)}{(x + 1)(x + 2)} \times \dfrac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \dfrac{(x - 5)\cancel{(x + 5)}}{\cancel{(x - 2)}\cancel{(x + 5)}} \times \dfrac{\cancel{(x - 2)}\cancel{(x + 2)}}{\cancel{(x + 1)}\cancel{(x + 2)}} \times \dfrac{\cancel{x + 1}}{x(x + 2)} \\[1em] \Rightarrow \dfrac{(x - 5)}{x(x + 2)} \\[1em] \Rightarrow (x - 5) : x(x + 2).$

Hence, the compounded ratio = (x - 5) : x(x + 2).

The ratio of the prices of first two fans was 16 : 23. Two years later, when the price of the first fan had risen by 10% and that of the second by ₹ 477, the ratio of their prices became 11 : 20. Find the original prices of the two fans.

Answer

Let price of first fan be f₁ and second fan be f₂.

Given,

Ratio of the prices of first two fans was 16 : 23.

$\Rightarrow \dfrac{f_1}{f_2} = \dfrac{16}{23} \\[1em] \Rightarrow f_1 = \dfrac{16f_2}{23} ..........(1)$

First fan price after 2 years = $f_1 + \dfrac{10}{100}f_1 = f_1 + \dfrac{f_1}{10} = \dfrac{11f_1}{10}$.

Second fan price after two years = $f_2 + 477$.

Given,

Ratio of fan's prices becomes 11 : 20 after 2 years.

$\therefore \dfrac{\dfrac{11}{10}f_1}{f_2 + 477} = \dfrac{11}{20} \\[1em] \Rightarrow \dfrac{11f_1}{10(f_2 + 477)} = \dfrac{11}{20} \\[1em] \Rightarrow \dfrac{f_1}{f_2 + 477} = \dfrac{11}{20} \times \dfrac{10}{11} \\[1em] \Rightarrow \dfrac{f_1}{f_2 + 477} = \dfrac{1}{2} \\[1em] \Rightarrow f_1 = \dfrac{f_2 + 477}{2}$

Substituting value of f₁ from (1) in above equation we get :

$\Rightarrow \dfrac{16f_2}{23} = \dfrac{f_2 + 477}{2} \\[1em] \Rightarrow 16f_2 \times 2 = 23(f_2 + 477) \\[1em] \Rightarrow 32f_2 = 23f_2 + 10971 \\[1em] \Rightarrow 32f_2 - 23f_2 = 10971 \\[1em] \Rightarrow 9f_2 = 10971 \\[1em] \Rightarrow f_2 = \dfrac{10971}{9} \\[1em] \Rightarrow f_2 = 1219.$

Substituting value of f₂ in equation (1), we get :

$\Rightarrow f_1 = \dfrac{16f_2}{23} \\[1em] = \dfrac{16 \times 1219}{23} \\[1em] = 16 \times 53 \\[1em] = 848.$

Hence, original price of fans are ₹ 848 and ₹ 1219.

If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x² + y² and y² + z².

Answer

Given,

y is the mean proportional between x and z

$\therefore \dfrac{x}{y} = \dfrac{y}{z} \Rightarrow y^2 = xz.$

To prove,

xy + yz is the mean proportional between x² + y² and y² + z²

$\therefore \dfrac{x^2 + y^2}{xy + yz} = \dfrac{xy + yz}{y^2 + z^2} \\[1em] \Rightarrow (xy + yz)(xy + yz) = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow x^2y^2 + xy^2z + xy^2z + y^2z^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 ....[\text{i}]$

Substituting y² = xz in L.H.S. of (i)

⇒ x²(xz) + x(xz)z + x(xz)z + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(ii)

Substituting y² = xz in R.H.S. of (i)

⇒ x²(xz) + x²z² + (xz)² + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(iii)

Since, (ii) = (iii)

Hence, proved that xy + yz is the mean proportional between x² + y² and y² + z².

Using componendo and dividendo, find value of x :

$\dfrac{\sqrt{3x + 4} + \sqrt{3x - 5}}{\sqrt{3x + 4} - \sqrt{3x - 5}} = 9.$

Answer

Applying componendo and dividendo we get,

$\Rightarrow \dfrac{\sqrt{3x + 4} + \sqrt{3x - 5} + \sqrt{3x + 4} - \sqrt{3x - 5}}{\sqrt{3x + 4} + \sqrt{3x - 5} - (\sqrt{3x + 4} - \sqrt{3x - 5})} = \dfrac{9 + 1}{9 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{3x + 4}}{2\sqrt{3x - 5}} = \dfrac{10}{8} \\[1em] \Rightarrow \dfrac{\sqrt{3x + 4}}{\sqrt{3x - 5}} = \dfrac{10}{8} \\[1em] \Rightarrow \dfrac{3x + 4}{3x - 5} = \dfrac{100}{64} \\[1em]$

Again applying componendo and dividendo we get,

$\Rightarrow \dfrac{3x + 4 + 3x - 5}{3x + 4 - (3x - 5)} = \dfrac{100 + 64}{100 - 64} \\[1em] \Rightarrow \dfrac{6x - 1}{9} = \dfrac{164}{36} \\[1em] \Rightarrow 36(6x - 1) = 1476 \\[1em] \Rightarrow 216x - 36 = 1476 \\[1em] \Rightarrow 216x = 1512 \\[1em] \Rightarrow x = \dfrac{1512}{216} \\[1em] \Rightarrow x = 7.$

Hence, x = 7.

Given that x + 2 and x - 3 are factors of x³ + ax + b. Calculate the values of a and b. Also, find the remaining factor.

Answer

By factor theorem,

If (x - a) is a factor of polynomial f(x), then remainder f(a) = 0.

⇒ x + 2 = 0

⇒ x = -2.

Since, x + 2 is a factor of x³ + ax + b

∴ Remainder = 0.

⇒ (-2)³ + (-2)a + b = 0

⇒ -8 - 2a + b = 0

⇒ b = 2a + 8 ................(1)

x - 3 = 0

⇒ x = 3.

Since, x - 3 is a factor of x³ + ax + b

∴ Remainder = 0.

⇒ (3)³ + (3)a + b = 0

⇒ 27 + 3a + b = 0

⇒ b = -27 - 3a ................(2)

From (1) and (2), we get:

⇒ 2a + 8 = -27 - 3a

⇒ 2a + 3a = -27 - 8

⇒ 5a = -35

⇒ a = $\dfrac{-35}{5}$

⇒ a = -7.

Substituting value of a in equation (1), we get :

⇒ b = 2a + 8 = 2 × -7 + 8 = -14 + 8 = -6.

f(x) = x³ + ax + b = x³ - 7x - 6.

Substituting x = -1 in f(x), we get :

⇒ (-1)³ - 7(-1) - 6

⇒ -1 + 7 - 6

⇒ -1 + 1

⇒ 0.

∴ (x + 1) is a factor of x³ - 7x - 6.

Hence, a = -7, b = -6 and (x + 1) is the remaining factor.

Use the remainder theorem to factorise the expression 2x³ + 9x² + 7x - 6. Hence, solve the equation 2x³ + 9x² + 7x - 6.

Answer

For x = -2, the value of the given expression

= 2(-2)³ + 9(-2)² + 7(-2) - 6

= 2 × -8 + 9 × 4 - 14 - 6

= -16 + 36 - 14 - 6

= -36 + 36

= 0.

⇒ x + 2 is a factor of 2x³ + 9x² + 7x - 6.

Now dividing 2x³ + 9x² + 7x - 6 by (x + 2),

$\begin{array}{l} \phantom{x + 2)}{2x^2 + 5x - 3} \\ x + 2\overline{\smash{\big)}2x^3 + 9x^2 + 7x - 6} \\ \phantom{x + 2}\underline{\underset{-}{}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x + 2}2x^3+2}5x^2 + 7x - 6 \\ \phantom{{x + 2}x^3+2}\underline{\underset{-}{}5x^2 \underset{-}{+} 10x} \\ \phantom{{x + 2}x^3+2-5x^2}-3x - 6 \\ \phantom{{x + 2}x^3+2-5x^23}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2}x^3+2-5x^23-3}\times \end{array}$

we get quotient = 2x² + 5x - 3

∴ 2x³ + 9x² + 7x - 6 = (x + 2)(2x² + 5x - 3)

= (x + 2)(2x² + 6x - x - 3)

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

= (x + 2)(2x - 1)(x + 3).

∴ (x + 2), (2x - 1) and (x + 3) are the factors of 2x³ + 9x² + 7x - 6.

⇒ x + 2 = 0

⇒ x = -2

⇒ 2x - 1 = 0

⇒ x = $\dfrac{1}{2}$

⇒ x + 3 = 0

⇒ x = -3.

Hence, 2x³ + 9x² + 7x - 6 = (x + 2)(2x - 1)(x + 3) and x = -2, -3, $\dfrac{1}{2}$.

When 2x³ + 5x² - 2x + 8 is divided by (x - a) the remainder is 2a³ + 5a². Find the value of a.

Answer

By remainder theorem,

If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a).

∴ On dividing 2x³ + 5x² - 2x + 8 by (x - a)

Remainder = 2a³ + 5a² - 2a + 8

Given, remainder = 2a³ + 5a².

∴ 2a³ + 5a² - 2a + 8 = 2a³ + 5a²

⇒ -2a + 8 = 0

⇒ 2a = 8

⇒ a = 4.

Hence, a = 4.

What number should be added to x³ - 9x² - 2x + 3 so that the remainder may be 5 when divided by (x - 2) ?

Answer

Let a be added to x³ - 9x² - 2x + 3 so that the remainder may be 5 when divided by (x - 2).

By remainder theorem,

If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a).

∴ On dividing x³ - 9x² - 2x + 3 + a by (x - 2)

Remainder = (2)³ - 9(2)² - 2(2) + 3 + a

Since, remainder = 5.

∴ 8 - 9(4) - 4 + 3 + a = 5

⇒ 8 - 36 - 1 + a = 5

⇒ a - 29 = 5

⇒ a = 5 + 29 = 34.

Hence, 34 must be added to x³ - 9x² - 2x + 3 so that the remainder may be 5 when divided by (x - 2).

Let R₁ and R₂ are remainders when the polynomial x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 are divided by (x + 1) and (x - 2) respectively. If 2R₁ + R₂ = 6; find the value of a.

Answer

On dividing x³ + 2x² - 5ax - 7 by (x + 1), we get remainder R₁ :

∴ R₁ = (-1)³ + 2(-1)² - 5a(-1) - 7

⇒ R₁ = -1 + 2 + 5a - 7

⇒ R₁ = 5a - 6.

On dividing x³ + ax² - 12x + 6 by (x - 2), we get remainder R₂

∴ R₂ = (2)³ + a(2)² - 12(2) + 6

⇒ R₂ = 8 + 4a - 24 + 6

⇒ R₂ = 4a - 10.

Given,

⇒ 2R₁ + R₂ = 6

⇒ 2(5a - 6) + 4a - 10 = 6

⇒ 10a - 12 + 4a - 10 = 6

⇒ 14a - 22 = 6

⇒ 14a = 28

⇒ a = $\dfrac{28}{14}$ = 2.

Hence, a = 2.

Find matrix B, if matrix A = $\begin{bmatrix*}[r] 1 & 5 \\ 1 & 2 \end{bmatrix*}$, matrix C = $\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*}$ and AB = 3C.

Answer

Let order of matrix B = m × n

Given,

AB = 3C

$\begin{bmatrix*}[r] 1 & 5 \\ 1 & 2 \end{bmatrix*}_{2 \times 2}B_{m \times n} = 3\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*}_{2 \times 1}$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ m = 2

Also, the no. of columns of product (resulting) matrix is equal to no. of columns of second matrix.

∴ n = 1

Order of matrix B = m × n = 2 × 1.

Let matrix B = $\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$

Substituting matrix in AB = 3C we get,

$\Rightarrow \begin{bmatrix*}[r] 1 & 5 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = 3\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times a + 5 \times b \\ 1 \times a + 2 \times b \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 5b \\ a + 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 3 \end{bmatrix*}.$

∴ a + 5b = 6 ........(1)

a + 2b = 3 ........(2)

Subtracting equation (2) from (1), we get :

⇒ a + 5b - (a + 2b) = 6 - 3

⇒ a - a + 5b - 2b = 3

⇒ 3b = 3

⇒ b = 1.

Substituting value of b in (1), we get :

⇒ a + 5(1) = 6

⇒ a + 5 = 6

⇒ a = 1.

B = $\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 1 \end{bmatrix*}$

Hence, B = $\begin{bmatrix*}[r] 1 \\ 1 \end{bmatrix*}$.

Find the matrices A and B, if 2A + B = $\begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*}$ and A - 2B = $\begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*}$

Answer

Given,

⇒ 2A + B = $\begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*}$ .........(1)

⇒ A - 2B = $\begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*}$ .........(2)

Multiplying equation (1) by 2 and adding in equation (2), we get :

$\Rightarrow 2(2A + B) + A - 2B = 2\begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 4A + 2B + A - 2B = \begin{bmatrix*}[r] 6 & -8 \\ 4 & 14 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 5A = \begin{bmatrix*}[r] 6 + 4 & -8 + 3 \\ 4 + 1 & 14 + 1 \end{bmatrix*} \\[1em] \Rightarrow 5A = \begin{bmatrix*}[r] 10 & -5 \\ 5 & 15 \end{bmatrix*} \\[1em] \Rightarrow A = \dfrac{1}{5}\begin{bmatrix*}[r] 10 & -5 \\ 5 & 15 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 3 \end{bmatrix*}.$