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## Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2019 (North)

Time: 3.15 Hours

Max. Marks: 100

Instructions:

- The questions paper consists of five parts A, B, C, D, and E.
- Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
- Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.

Write the canonical representation of 360.

Answer:

360 = 2^{3} × 3^{2} × 5^{1}.

Question 2.

If A = {1, 2, 3, 4} and B = {2, 4, 6, 8} find A ∪ B

Answer:

A ∪ B = {1, 2, 3, 4, 6, 8}.

Question 3.

Simplify: \(\left(\frac{8}{27}\right)^{3 \times 1 / 3}\)

Answer:

\(\left(\frac{8}{27}\right)^{3 \times 1 / 3}=\left(\frac{2^{3}}{3^{3}}\right)^{1 / 3}=\frac{2}{3}\)

Question 4.

Define perpetuity.

Answer:

If annuity payments are made for an infinite period it is called a perpetuity.

Question 5.

Find the value of x, if log_{7} x = 2.

Answer:

x = 7^{2} ⇒ x = 49

Question 6.

Find the 12th term of A.P. 1, 4, 7, 10

Answer:

T_{n} = a + 11d = 1 +(11 × 3) = 1 +33 =34.

∴ 12th term is 34.

Question 7.

Solve \(\frac{x+2}{x-1}=\frac{5}{2}\)

Answer:

2(x + 2) = 5 (x – 1)

2x + 4 = 5x – 5

4 + 5 = 5x – 2x

3x = 9 ⇒ x = 3

Question 8.

Write the formula to find the present value of an annuity due.

Answer:

Present value of annuity due = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right](1+r)}{r(1+r)^{n}}\)

Question 9.

Convert 0.20 into a percentage.

Answer:

0.20 × 100 = 20%.

Question 10.

Convert \(\frac{7 \pi}{12}\) into degrees.

Answer:

\(\frac{7 \pi}{12}=\frac{7 \times 180}{12}\) = 105°

Question 11.

Find the value of sinA sec A.

Answer:

sinAsecA = \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) = tan A

Question 12.

If the slope of the line is \(\frac{2}{3}\), find the slope of its perpendicular line.

Answer:

Slope of perpendicular line = –\(\frac{3}{2}\)

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.

Find the L.C.M. of \(\frac{1}{3}, \frac{5}{6}, \frac{5}{7}\)

Answer:

LCM of Numerators 1, 5, 2 = 10

HCF of denominators 3, 6, 9 = 3

∴ LCM of fractions = \(\frac{\text { LCM of Numerators }}{\text { HCF of denominators }}=\frac{10}{3}\)

Question 14.

Find the product of two numbers is 216 and their L.C.M. is 36. Find the H.C.F.

Answer:

Question 15.

If A = {a,b,c, d}, B = (d, e, f, g). Find A – B and B – A.

Answer:

A – B = {a, b, c} B – A = {e, f, g, h}

Question 16.

If the first term of an A.P. is 3 and the common difference is -2. Find the 11th term.

Answer:

Given a = 3, d = -2, T_{n} = ?

T_{n} =a +10d = 3 + 10(-2) = 3 – 20 = —17

∴ 11th term = – 17.

Question 17.

Evaluate: sin^{2}\(\frac{\pi}{6}\) + cos^{2}\(\frac{\pi}{3}\)-tan2\(\frac{\pi}{4}\) + cot^{2}\(\frac{\pi}{4}\)

Answer:

LHS = sin^{2} 30 . cos^{2} 45 – cos^{2} 90 – sin^{2} 60.

= \(\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{1}{\sqrt{2}}\right)^{2}\) – (0)2\(\left(\frac{\sqrt{3}}{2}\right)^{2}\) = \(\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8}\)

Question 18.

Solve by method of elimination :

x + 2y = 4

3x + y = 7

Answer:

x + 2y = 4 ………..(i)

3x + y = 7 …………..(ii)

Multiply Eqn. 1 by 3 we get

Question 19.

Solve: 3(2 – x) ≥ 2(1 – x), x GR .

Answer:

3(2-x) ≥ 2(1 – x), x ∈ R

6-3x ≥ 2-2x ⇒ 6-2 = 3x-2x.

4 ≥ x ⇒ ∴ x ≤ 4

Question 20.

Simplify : \(\frac{a^{2 m+n} a^{3 m+n}}{a^{4 m+2 n}}\)

Answer:

\(\frac{a^{2 m+n} a^{3 m+n}}{a^{4 m+2 n}}\) = a^{2m+n+3m+n-4m-2n} = a^{n}

Question 21.

Prove that tanA + cotA = secA cosecA

Answer:

LHS = tanA + cotA = \(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\)

⇒ \(\frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A}=\frac{1}{\cos A \sin A}\) = secAcosecA = RHS

Question 22.

The average age of 7 members of a family is 18 years. If the head of the family is excluded, the average age of the rest of the members would fall to 13 years. What is the age of the head of the family?

Answer:

Total age of 7 members = 18 × 7 = 126 years

Total age of 6 members = 13 × 6 = 78

∴ the age of head of the family = 126 – 78 = 48 years

Question 23.

Insert 3 geometric means between \(\frac{1}{4}\) and \(\frac{1}{64}\).

Answer:

Insert 3 Gms G_{1}, G_{2}, G_{3} between \(\frac{1}{4}\) & \(\frac{1}{64}\)

∴ \(\), G_{1}, G_{2}, G_{3}, are in GP

a = \(\frac{1}{4}\), ar^{4} = \(\frac{1}{64} \Rightarrow \frac{1}{4}\)r^{4} =\(\frac{1}{64}\) ⇒ r^{4} = \(\frac{4}{64}\) ⇒ r^{4} = \(\frac{1}{16}=\left(\frac{1}{2}\right)^{4}\) ⇒ r = \(\frac{1}{2}\)

Question 24.

Prove that (1 +tan^{2} θ)(1 – sin^{2} θ) = 1

Answer:

LHS = (1 + tan^{2}θ)(1 – sin^{2}θ) sec2θ.cos2θ = \(\frac{1}{\cos ^{2} \theta}\).cos^{2} θ = 1 = RHS

Question 25.

Find the value of k, if the distance between (2k, 5) and ( -k, -4). If the distance between the points (3, -2) and (-1, a) is 5 units find the values of a.

Answer:

AB = \(\sqrt{(-1-3)^{2}+(a+2)^{2}}\) = 5

Squaring and simplifying we get

a = 1 and a = – 5

Part – C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.

Prove that \(\sqrt{2}\) is an irrational number.

Answer:

We shall prove it by the method ofcontradiction.

If possible Let \(\sqrt{2}\) be a rational number

Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.

Further let p and q are coprime i.e. H.C.F. of p and q = 1.

\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p

⇒ 2q^{2} = p^{2}

⇒ 2 divides p^{2} ⇒ 2 divides p

⇒ p is even

Let p = 2k where k is an integer p^{2} = 4k^{2}

2q^{2} = 4k^{2}

q^{2} = 2k^{2} ⇒ q^{2} is even

⇒ q is even.

Now p is even and q is even which implies p and q have a common factor 2. which is a

contradiction of the fact that p and q are co-prime.

∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.

If U = {1, 2, 3, 4, 6, 8, 9} A = (2, 3, 6, 8} and B = {1, 3, 6, 9} verify (A∪B) – A’ ∪ B’.

Answer:

Let U = {1, 2, 3, 4, 6, 8, 9}

A = {2, 3, 6, 8}

B = {1, 3, 6, 9}

∴ A ∪ B = {1, 2, 3, 6 ,8, 9}

A’ = U – A = {1, 2, 3, 4, 6, 8, 9} – {2, 3, 6, 8}

= {1, 4, 9}

B’ = U – B = {1, 2, 3, 4, 6, 8, 9} – {1, 3, 6, 9}

= {2, 4, 8}

∴ A’ ∪ B’ = {1, 4, 9} ∪ {2, 4, 8}

= {1, 2, 4, 8, 9}

∴ (A∪B) – A’ ∪ B’ = {1, 2, 3, 6 ,8, 9} – {1, 2, 4, 8, 9}

= {3, 6, 4, 9}

Question 28.

Solve log_{2} x + log_{4} x = 3.

Answer:

\(\frac{\log x}{\log 2}+\frac{\log x}{\log 4}\) = 3

\(\frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}\) = 3

\(\frac{2 \log x+\log x}{2 \log 2}\) = 3

log x =2 log2 = log 2^{2}= log 4

log x = log 4

∴ x = 4

Question 29.

If a^{x} = b^{y} = c^{x} and b^{2} = ac show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Answer:

Let a^{x} = b^{y} = c^{z} = k (say)

∴ a^{x} = k ⇒ a = k^{1/x}, b^{y} = k ⇒ b = k^{1/y}, c^{z} = k ⇒ c = k^{1/z}

Now, b^{2} = ac

∴ (k^{1/y})^{2} = k^{1/x}. k^{1/z}

∴ k^{2/y} = k^{1/x+1/z}

Bases are the same ∴ Equating powers on both sides, we get.

\(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Question 30.

The 3rd and 5th element of a GP are 3 and 27 respectively. Find the 8th element.

Answer:

Given T_{3} = 3 ⇒ ar^{2} = 3 ………………(1)

T_{5} = 27 ⇒ or = 27 …………………(2)

Divide \(\frac{(2)}{(1)}\) gives r^{2} = 9 ⇒ r = 3

ar^{2} = 3 ⇒ a = \(\frac{3}{r^{2}}=\frac{3}{9}=\frac{1}{3}\)

∴ T_{8} = ar^{7} = \(\frac{1}{3}\).37 = 36

Question 31.

If a and b are the roots of the equation 3×2 – 6x + 4 = 0, then find the value of \(\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+\left[2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3(\alpha+\beta)\right]\)

Answer:

α + β = \(\frac{-b}{a}=\frac{6}{3}\) = 2

αβ = \(\frac{c}{a}=\frac{4}{3}\)

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) + 2\(\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) + 3(α + β)

= \(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\) + 2\(\left(\frac{\alpha+\beta}{\alpha \beta}\right)\) + 3(α + β) = \(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\) + 2\(\left(\frac{\alpha+\beta}{\alpha \beta}\right)\) + 3(α + β)

= \(\frac{(2)^{2}-2(4 / 3)}{4 / 3}\) + 2\(\left(\frac{2}{4 / 3}\right)\) + 3(2) = \(\frac{4-8 / 3}{4 / 3}+\frac{4}{4 / 3}\) + 6 = \(\frac{\frac{12-8}{3}}{\frac{4}{3}}\) + 3 + 6 = 1 + 3 + 6 = 10

Question 32.

Solve the system of inequalities 2x – y < 1 and x – 2y < -1 graphically.

Answer:

Consider (i) 2x – y = 1

x = 0 ⇒ y = -1 ∴ A(0, 1)

(ii)x – 2y = -1

x = 0 ⇒ y = \(\frac{1}{2}\) ∴ C(0, \(\frac{1}{2}\))

y = 0 ⇒ x = -1 ∴ D(-1, 0)

Question 33.

Find the compound interest on ₹ 22,000 for 21/2 years at 6% p.a.

Answer:

Given P = 22,000, i = \(\frac{6}{100}\) = 0.06

A = P(1 + r)^{n} = 22000(1 + 6.06)^{2} (1 +0.06 × \(\frac{1}{2}\)) = ₹ 25460.78

C.I = A – P = 3460.78

Question 34.

A bookseller bought 228 notebooks at an average price of ₹ 8.50 of which 80 books he bought at ₹ 7.50 each and 84 at ₹ 10.50 each. Find the price of the remaining books per unit.

Answer:

Total price of 228 books = 228 × 8.50 = 1938

Total price of 164 books = 1482

∴ Total price of 64 books = 1938 – 1482 = 456

∴ Average price of 64 books = \(\frac{456}{64}\) = 7.12

Question 35.

A watch is sold for ₹ 150 at a profit of 25%. At what price should it be sold in order to have a 50% profit?

Answer:

Let C.P be 100, then S.P is 125

If SP is 150, then CP =?

\(\frac{100 \times 150}{125}\) = 120 ⇒ S.P. to get a profit of 50% is \(\frac{120 \times 150}{100}\) = ₹ 180

Question 36.

Find the value of 3tan^{2} 30 +4cos^{2} 30 – \(\frac{1}{2}\)cot^{2} 45 – \(\frac{2}{3}\)sin^{2} 60 + \(\frac{1}{8}\)sec^{4} 60

Answer:

\(3\left(\frac{1}{\sqrt{3}}\right)^{2}+4\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{1}{2}(1)^{2}-\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{8}\left(2^{4}\right)\)

= 1 + 3 – \(\frac{1}{2}\) – \(\frac{1}{2}\) + 2 = 6 – 1 = 5

Question 37.

Find the points of trisection of the line joining (3, 4) and (5, -2).

Answer:

Let A = (3, 4) B = (5, -2)

Let P and Q be the points of trisection of AB

then P divides AB internally in the ratio 1 : 2 & Q is the midpoint of PB.

P = \(\left(\frac{1 \times 5+2 \times 3}{1+2}, \frac{1 \times-2+2 \times 4}{1+2}\right)\) = (\(\frac{11}{3}\), 2)

Q = \(\left(\frac{\frac{11}{3}+5}{2}, \frac{2+(-2)}{2}\right)\) = (\(\frac{13}{3}\), 0)

∴ the points of trisection = (\(\frac{11}{3}\), 2) (\(\frac{13}{3}\), 0)

Question 38.

Find the point of intersection of the lines 3x + 2y – 5 = 0 and 4x -y – 3 = 0

Answer:

3x + 2y – 5 = 0 ……….. (1)

4x – y – 3 = 0 ………………(2)

Multiply equation (2) by 2 and we get

Add 11x – 11 – 0 ⇒ x = 1

Put x = 1 in 3x + 2y – 5 = 0 ⇒ y = 1

∴ the point of intersection is (1, 1)

Part – D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.

In a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazine A and B, 10 read magazine A and C, 5 read magazine B and C while 3 read all the 3 magazines?

Find (i) How many read none of the 3 magazines, (ii) How many read only magazine C?

Answer:

Question 40.

Evaluate \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\) using log tables.

Answer:

Let x = \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)

log x = log\(\left[\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\right]\)

= log 0.5634 + log 0.0635 – log 2.563 – log 12.5

= 1.7508 + 2.8028 – 0.4087 – 1.0969

= 2.5536 – 1.5056 = 2+0.5536 – 1.5056 = -2 – 0.952 = -2.952

= -3 + 3 – 2.952 = -3 + 0.048 = 1048

Question 41.

The sum of four numbers which are in A.P. is 28 and 10 times the least number is 4 times the greatest number. Find the numbers.

Answer:

Let the four numbers are (a – 3d), (a – d)(a + d)(a + 3d).

Given sum = -28

Also given 10(a – 3d) = 4(a + 3d) ⇒ 10a – 30d = 4a + 12

6a = 42d

∴ 42d = 6 × 7 = 42 ⇒ d = 1

∴ the four numbers are 4, 6, 8, 10

Question 42.

Find an integral root between -3 and 3 by inspection and they using synthetic division solve the equation x^{3} – 3x^{2} – 5x + 6 = 0

Answer:

Letf(x) = x^{3} – 2x^{2} – 5x + 6

f(1)= 1 – 2 – 5 + 6 = 7 – 7 = 0

∴ x = 1 is a root of the given equation by using Synthetic division.

∴ The resulting equation is x2 – x – 6 = 0 is the quotient and remainder = 0

– x – 6 = 0

(x – 3)(x + 2) = 0

x = 3 or -2

Thus x = 1, -2, 3 are the roots of the given equation.

Question 43.

Find the difference between simple interest and compound interest and compound interest on ₹ 6400 at 9% p.a. for 5 years.

Answer:

Simple Interest

P = 18,000 T = 4 R = 4

SI = \(\frac{\mathrm{PTR}}{100}\)

= \(\frac{18,000 \times 4 \times 8}{100}\)

= ₹5,760

Compound Interest

P = 18,000 n = 4 I = 008

A = P(1 + i)^{n}

= 18,000 (1 + 0.08)^{4}

= 24,488.8

CI = A – P

= 24,488.8 – 18,000

CI = 6488.8

Difference between CI and SI is

CI – SI = 6488.8, – 5760

= 728.8.

Question 44.

Preritha wants to buy a house after 5 years when it is expected to cost ₹ 50 lakhs. How much should she save annually if her savings earn a compound interest of 12%?

Answer:

Given: F = 50,00,000; n =5, i = 0.12

F = \(\frac{\mathrm{A}\left[(1+\mathrm{i})^{\mathrm{n}}-1\right]}{\mathrm{i}}\)

50,00,000 = A(6.3 528)

A = \(\frac{50,00,000}{6.3528}\)

A = ₹ 787054.5

Question 45.

A person gets ₹ 1216 more when selling a product for a profit of 15% instead of a loss of 4%. What would be the percentage profit or loss if it is sold for ₹7552?

Answer:

Let the C.P be ₹ 100

SP making profit of 15% = 115

SP while making a loss of 4% = 96

Difference = 19

If the CP is 100, the difference is 19.

If the CP is ?, difference is 1216.

∴ C.P = \(\frac{100 \times 1216}{19}\) = 6400

If SP = 7552 ⇒ Profit = 7552 – 6400 = 1152

Profit% = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 ⇒ Proflt% = \(\frac{1152}{6400}\) × 100 = 18%

Question 46.

If cosθ = \(\frac{4}{5}\) and θ is acute, show that \(\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{24}{7}\)

Answer:

LHS = \(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\)

If cosθ = \(\frac{4}{5}\) then tanθ = \(\frac{3}{4}\)

L.H.S = \(\frac{2 \cdot \frac{3}{4}}{1-\frac{9}{16}}=\frac{3}{2} \times \frac{16}{16-9}=\frac{3}{2} \times \frac{16}{7}=\frac{24}{7}\) = RHS

Question 47.

Show that the points (1,1)(4,1) and (1,4) are the vertices of a square. Also And the area.

Answer:

AB = \(\sqrt{(4-1)^{2}+(1-1)^{2}}=\sqrt{9+0}\) = 3

BC = \(\sqrt{(4-4)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3

CD = \(\sqrt{(1-4)^{2}+(4-4)^{2}}=\sqrt{9+0}\) = 3

DA = \(\sqrt{(1-1)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3

AB = BC = CD + AD

AC = \(\sqrt{(4-1)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)

BD = \(\sqrt{(1-4)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)

∴ AC = BD

∴ ABCD Δ is a Square.

Question 48.

Derive the equation of a straight line in the form x cosα + y sinα = p

Answer:

Let the line cut the x-axis at A and y axis at B.

Let ON be the perpendicular to AB and ON = P & ∠AON = a

∴ ∠NOB = 90 – α

From Δ ONA

cos α = \(\frac{\mathrm{ON}}{\mathrm{OA}}\) ⇒ OA = \(\frac{p}{\cos \alpha}\) = a(say)

cos(90 – α) = \(\frac{p}{\mathrm{OB}}\) ⇒ sin α = \(\frac{p}{\mathrm{OB}}\)

⇒ OB = \(\frac{p}{\sin \alpha}\) = b (say)

Using equation of the line

\(\frac{x}{a}+\frac{y}{b}\) = 1

⇒ \(\frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}\) = 1 ⇒ \(\frac{x \cos \alpha}{P}+\frac{y \sin \alpha}{P}\) = 1

∴ xcosα + ysinα = p

This is called normal form of the equation.

Part – E

V. Answer any ONE question (1 × 10 = 10)

Question 49.

(a) If f(x) = x and g(x) = x1 + 1, find (i) fog(1) (ii) fog(2) (iii) gof (1), (iv) gof(2)

Answer:

(i) fog(1) = f[g(1)] = f[1^{3 }+ 1] = f(2) = 2

(ii) fog (2)= f[g(2)] = f[2^{3} + 1] = f(9) = 9

(iii) gof(1) = g[f(1)] = g [1] = 1 + 1 = 2

(iv) gof(-1) = g[f(-1)] = g[-1] = (-1)^{3} + 1 = -1 + 1 = 0

(b) Find the sum to n terms of a G.P 4 + 44 + 444 +………….

Answer:

S = 4 + 44 + 444 + ………………. to n terms

S_{n} = 4(1 + 11 + 111 + ……………to n terms)

\(\frac{9 \mathrm{~S}_{n}}{4}\) = 9 + 99 + 999 + ……………… to n terms

\(\frac{9 \mathrm{~S}_{n}}{4}\) = (10 – 1) + (10^{2} – 1) + (10^{3} – 1) +………………

= (10 + 10^{2} + 10^{3}+ ……………. ) – (1 + 1 + 1 + …………..)

\(\frac{9 \mathrm{~S}_{n}}{4}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n

Sn = \(\frac{4}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) The average age of 12 boys is 8 years. Another boy of 21 years joins this group. Find the average age of the new group.

Answer:

The average age of 12 boys 8 years

∴ Total age of boys = 8 × 12 = 96 years

when a boy of 21 yrs joins the group,

the total age of 13 boys 96 + 21 = 117years

∴ Average age of the new group = \(\frac{117}{13}\) = 9 years.

Question 50.

(a) A manufacturer produces and sells balloons at ₹8 per unit. His fixed cost is 6500 and the variable cost/balloon is ₹3.50.

Calculate (i) Revenue function (ii) Cost function (iii) Profit function (iv) Break Even point.

Answer:

i.e. (i) R(x) = 8x

(ii) C(x) = 3.5x + 6500

(iii) P(x) = 4.5x – 6500

i.e.,P(x) = 6500 = 0

(iv) At BEP P(A) = 0

4.5x – 6500 = 0

4.5x = 6500

x = 144.4 units

(b) For what values of k are the 3 lines x – 2y + 1 =0, 2x + 5y + 3 = 0 and 5x – 9y + k 0 are concurrent.

Answer:

(c) Find the H.C.F. of 16, 24, and 48.

Answer:

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

48 = 2 × 2 × 2 × 2 × 3

H.C.F of (16, 24, 48) = 2 × 2 × 2 = 10