1st PUC Basic Maths Previous Year Question Paper March 2016 (North)

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2016 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. 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.
3. 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 imaginary part of the complex number \(\frac{-3+4 i}{5}\)
Answer:
Imaginary part of \(\frac{-3}{5}\) + i\(\frac{4}{5}\) is \(\frac{4}{5}\).

Question 2.
If set A has 16 subsets, find the number of elements in A.
Answer:
A has 4 elements.

Question 3.
If a function f(x) is defined by f(x) = 3x – 1, find f(\(\frac{1}{3}\)) .
Answer:

Question 4.
Find the 8th term of the sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\)………..
Answer:
8th term = T₈ = \(\frac{1}{16}\)

Question 5.
Find the value of log01 100
Answer:
Let log0_0.1 100 = x
100 = (0.1)x
102 = 10 – x ⇒ x = -2
log_0.1 100 = -2

Question 6.
Simplify \(\left(\frac{64}{81}\right)^{-1 / 4}\)
Answer:
\(\left(\frac{64}{81}\right)^{-1 / 4}=\left(\frac{81}{64}\right)^{1 / 4}=\frac{\left(3^{4}\right)^{1 / 4}}{\left(2^{6}\right)^{1 / 4}}=\frac{3}{2^{3 / 2}}=\frac{3}{2 \sqrt{2}}\)

Question 7.
Solve for x : x + a(x + b) = ax + b.
Answer:

x = b – ab
x = b(1 – a)

Question 8.
What is the present value of an income of Rs. 10,000 a year to be received forever at 10% p.a.?
Answer:
P_∞ = \(\frac{a}{r}=\frac{10,000}{0.1}\) = 1,00000

Question 9.
Convert 315° into radians.
Answer:

Question 10.
Find the slope of a line perpendicular to the line 3x – 4y + 5 = 0.
Answer:
Slope of the given line = \(\frac{3}{4}\)
Slope of the perpendicular line = –\(\frac{4}{3}\)

Question 11.
Find the coordinates of the centroid of the triangle formed by the points (1, 1), (-2, 1), and (-5, -5).
Answer:
Centroid = G(x,y) = \(\left[\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right]=\left[\frac{1-2-5}{3}, \frac{1+1-5}{3}\right]=\left[\frac{-6}{3}, \frac{-3}{3}\right]\) = [-2, -1]

Question 12.
Mr. Prem bought a cycle at Rs. 5000 and sold the same at a loss of 20% to Mr. Bheem. Find the cost price of Bheem.
Answer:
Cost price = 5000 – 20% of 5000 = 5000 – 1000 = 4000 Rs.

PART-B

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

Question 13.
Find the sum of positive divisors of 198.
Answer:
198 = 2¹ × 3² × 11¹
T(198) = (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12.

Question 14.
Find greatest number which can divide 60,72 and 84 without leaving any remainder.
Answer:
60 = 2² × 3¹ × 5¹
72 = 2³ × 3²
84 = 2² × 3¹ × 7¹
∴ greatest number is nothing but HCF = 22 × 31 = 12.

Question 15.
If A = {x / x ∈ N, x < 4}, B = {x – 25 = 0, x ∈ N} find B x A.
Answer:
A = (1, 2, 3} B = {5}
B × A = {(5, 1) (5, 2) (5, 3)

Question 16.
Prove that \(\frac{1}{1+\log _{b} a}+\frac{1}{1+\log _{a}}\) = 2
Answer:
L.H.S = \(\frac{1}{\log _{b}^{b}+\log _{b}^{a}}+\frac{1}{\log _{a}^{a}+\log _{a}^{b}}=\frac{1}{\log _{b}^{b a}}+\frac{1}{\log _{a}^{a b}}\) = logab b + laga ab = logab ab = 1 = R.H.S

Question 17.
Insert 3 geometric means between \(\frac{1}{4}\) and \(\frac{1}{64}\)
Answer:
Insert 3 Gms G₁, G₂, G₃. between \(\frac{1}{4}\) & \(\frac{1}{64}\)
\(\frac{1}{4}\), G₁, G₂, G₃, \(\frac{1}{64}\) are in G.P
a = \(\frac{1}{4}\), ar⁴ = \(\frac{1}{64}\) ⇒ \(\frac{1}{4}\)r⁴ = \(\frac{1}{64}\) ⇒ r⁴ = \(\frac{4}{64}\) ⇒ r⁴ = \(\frac{1}{16}\) = \(\left(\frac{1}{2}\right)^{4}\) ⇒ r = \(\frac{1}{2}\)

Question 18.
If sum of 2 numbers is 107 and their difference is 17, find the numbers.
Answer:
Let the two number be x and y
Given x + y = 107 …. (1)
x – y = 17 …. (2) solving we get
2x = 124 ⇒ x = 62, y = 107 – 62,y = 45
∴ the 2 numbers are x = 62 and y = 45.

Question 19.
Solve : 3x – 2 ≤ 2x + 1, x ∈ R and represent on the number line.
Answer:
3x – 2 ≤ 2x + 1, x ∈ R .

3x – 2x ≤ 2 +1
x ≤ 3

Question 20.
In how many years will a sum of money double itself at 10% simple interest p.a.?
Answer:
Let principle = 100 Rs. ∴ Amount = 200 Rs.
P = 100, F = 200, r = 10% = 0.1 T = ?
SI = F – P= 100 Rs.
w.k.t S.I = \(\frac{\text { PTR }}{100}\)

∴ In 10 years sum of money double itself.

Question 21.
The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th Student.
Answer:
Average age of 10 students = 6yrs
∴ Total age of 10 students = 10 × 6 = 60yrs
Total age of 9 students = 52 yrs
∴ Age of the 10th students = 60 – 52 = 8yrs.

Question 22.
The difference between the cost price and the selling price is Rs. 225. If the profit percentage is 15% find the selling price.
Answer:
Let the cost price be x
As there is profit SP > CP
SPx + 225
SP = \(\frac{100+\text { Profit } \%}{100}\) × CP
x + 225 = \(\left(\frac{100+15}{100}\right)\)x
100x + 22500 = 100x + 15x
15x = 22500
x = = 1500
SP = x + 225 = 1500+225 = 1725

Question 23.
Prove that cotθ + tanθ = secθ cosecθ.
Answer:
LHS = \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cdot \cos \theta}\) = cosecθ sec θ = RHS

Question 24.
IfA, B, C are angles of a triangle, then prove that cos\(\left(\frac{\mathbf{A}+\mathbf{B}}{\mathbf{2}}\right)\) = sinC/2.
Answer:
W.K.T A + B + C = 180°
\(\frac{A+B+C}{2}\) = 90 ⇒ \(\frac{A+B}{2}\) = 90 – \(\frac{\mathrm{C}}{2}\)
Taking sin on both sides
We get sin\(\left(\frac{A+B}{2}\right)\) = sin \(\left(90-\frac{\mathrm{C}}{2}\right)\)
sin\(\left(\frac{A+B}{2}\right)\) = cos\(\frac{\mathrm{C}}{2}\)

Question 25.
Derive two point form of equation of straight line in the form y – y₁ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x₁)
Answer:
Let A(x₁, y₁), B(x₂, y₂) be the two given points, let p(x, y) be any point on the line. Here A, B, and P tie on the same line
∴ The slope of AP = Slope of AB

i.e., \(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
(y – y₁) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x₁)
This is called two point formula.

PART-C

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

Question 26.
Prove that \(\sqrt{5}\) is irrational.
Answer:
If possible let \(\sqrt{5}\) is a rational number.
∴ .\(\sqrt{5}\) = p, q∈z, q ≠ 0
we shall also assume that p and q do not have any common factor
∴ 5 = \(\frac{\mathrm{p}^{2}}{\mathrm{q}^{2}}\)
∴ p² is a multiple of 5
∴ 5 divides p²
∴ 5 divides p
∴ p = 5k where k∈z, k≠O
p² = 25k²
5q² = 25k² ∵ p² = 5q²
q² = 5k²
∴ q² is a multiple of 5
∴ 5 divides q²
∴ 5 divides q
∴ 5 divides both p and q
This is a contradiction
∴ Our assumption is wrong
∴ \(\sqrt{5}\) is irrational.

Question 27.
Let f {(1, 1), (2, 3), (0, -1)) be a function from Z to Z defined by f(x) = ax + b a, b ∈ Z determine a and b.
Answer:
Given f(x) = ax + b
When x = 1, f(x) = 1 ⇒ a + b = 1
When x = 0, f(x) = -1 ⇒ 0 + b = -1 ⇒ b = -1
if b = 1 then a – 1 = 1 ⇒ a = 2
∴ a = 2, b = -1.

Question 28.
If A = {0, 1,2,3), B = {2, 3, 4), C = {4,5,6) then prove that A∩(B∪C)=(A∩B)∪(A∩C)
Answer:
A∩(B∩C) = {0, 1, 2, 3} n {2, 3, 4, 5, 6} {2, 3} ………………(1)
(A ∩ B) ∪ (A ∩ C) = {2, 3} ∪ {Q} = {2, 3) …………………..(2)
From 1 and 2 we have
A∩(B∪C) = (A∩B)∪(A∩C)

Question 29.
If x² + y² = 12xy. Show that 2 log (x – y) = log 10 + log (xy)
Answer:
Given x² + y² = 12xy
x² + y² – 2xy = 10xy
(x – y)² = 10xy
Taking logm both sides
log(x – y)² = log(10xy)
2 log (x – y) = log 10 + log xy

Question 30.
Find three numbers in A.P, whose sum is 24 and product is 440.
Answer:
Let the 3 number in AP is {a – d) a, a + d
Given Sum = 24

3a = 24 ⇒ a = 8
Also given product = 440
(a – d) (a) (a + d) = 440
a² – d² =55 ⇒ d² =64 – 65 = 9 ⇒d = ±3
∴ the 3 number are 5, 8,11 of a = 8, d = 3 the 3 numbers are 11,8, 5 if a = 8, d = -3

Question 31.
If α and β are the roots of the equation 3x² – 6x + 4 = 0 find the value of
(i) α³ + β³
(ii) α/β² + β/α²
Answer:
Given 3x² – 6x + 4 = 0
Here α + β= \(\frac{-b}{a}=\frac{6}{3}\) = 2
αβ = \(\frac{c}{a}=\frac{4}{3}\)

Question 32.
Solve the system of inequalities 2x + y ≥ 8, and x + y ≥ 10 graphically.
Answer:
Consider 2x + y = 8
Put x = 0, y = 8 (0, 8)
Put y = 0, x = 4 (4, 0)

x + y = 10
put x = 0, y = 10 (0, 10)
put y =0, x = 10 (10, 0)

Plot the above point on the graph.

Question 33.
Mr X bought a Walkman for Rs. 1800. If it depreciates at the rate of 15% per year, how much is it worth after 3 years?
Answer:
C = Rs 1800, r = 15% = 0.15, n = 3yrs, B = ?
B = C(1 – r)n
B = 1800 (1 – 0.15)3 = (1800) × (0.85)3 = 1105.425
worth after 3 years = 1105.425

Question 34.
A batsman finds that, by scoring a century in the 11th innings of his test matches, he has bettered his average of preview test innings by 5 runs. What is the average after the 11 innings?
Answer:
Let the average of 10 inninngs = x
∴ Total runs after 10 inninngs = lOx
Average after 11 inninngs x +5
∴ Total runs after 11 inninngs = 11 (x + 5)
11 (x + 5) – 10x = 100
11x + 55 – 10x = 100
x + 55 = 100
x = 100 – 55 = 45
Average after the 11th inninngs = x + 5 = 45 + 5 = 50 runs

Question 35.
By how many percent should the use of coffee be increased if the price of coffee is decreased by 10% so that the expenditure remains unchanged.
Answer:
Let the price of coffee =100
Let the quantity used be 100
∴ Total expenditure = 100 x 100 = 10,000
∴ New price = 100 – 10 = 90
Let the new quantity used = x
the total expenditure = 90x
For unchanged expenditure 90x = 10000

∴ the increase in consanption of coffee =111.11 – 100 = 11.11

Question 36.
Find x, if xsin\(\frac{\pi}{4}\) xtan \(\frac{\pi}{3}\) = \(\frac{\sin \frac{\pi}{6} \times \cot \frac{\pi}{6}}{3 \cos \frac{\pi}{6} \times \cos s e \frac{\pi}{4}}\)
Answer:

Question 37.
Find the equation of the locus of the point which moves such that its distance from the line 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
Answer:
A = (1,-1)
Let p(x, y) be any point on the locus
Given: PA = distance of P from the line 3x- 4y + 1 = 0
\(\sqrt{(x-1)^{2}+(y+1)^{2}}=\left|\frac{3 \dot{x}-4 y+1}{\sqrt{3^{2}+(-4)^{2}}}\right|\)
\(\sqrt{(x-1)^{2}+(y+1)^{2}}=\frac{3 x-4 y+1}{\sqrt{25}}\) S.B.S
(x – 1)² + (y – 1)² = \(\frac{(3 x-4 y+1)^{2}}{25}\)
25[(x – 1)² + (y + 1)²] = (3x – 4y + 1)²
25[x² + y² – 2x + 2y + 2] = 9x² + 16y² +1 – 34xy – 8y + 6x
⇒ 25x² + 25y² – 50x + 50y + 50 = 9x² + 16y² + 1 – 24xy – 8y + 6x
16x² + 9y² + 24xy – 42y – 44x + 49 – 0 is the required equation.

Question 38.
Find the future value of an annuity of Rs. 5000 at 12% p.a. for 6 years.
Answer:
A = Rs. 5,000 r = 0.12 n = 6
F = \(\frac{\mathrm{A}\left[(1+\mathrm{r})^{\mathrm{n}}-1\right]}{\mathrm{r}}=\frac{5000\left[(1+0.12)^{6}-1\right]}{0.12}=\frac{4869.11}{0.12}\) = 40,575.9

PART-D
IV. Answer any six questions : (6 × 5 = 30)

Question 39.
In a class of 50 students, 15 do not participate in any game, 25 play cricket, and 20 play football. Find the number of students who play.
(i) both cricket and football.
(ii) Only football.
Represent the results through the Venn diagram.
Answer:
Given n(u) = 50

n(C ∪ F) = 50 – 15 = 35, r(C) = 25, n(F) = 20
n(C ∩ F) = n(C) + r(F) – r(C ∪ F) = 25 + 20 – 35 = 10
From venn diagram we get only
Number of students who play only foot ball = 10.

Question 40.
Evaluate \(\frac{123.4 \times 54.8}{0.562 \times 7346}\) using logarithmic tables.
Answer:
Let x = \(\frac{123.4 \times 54.8}{0.562 \times 7346}\)
Then log x = log\(\left[\frac{123.4 \times 54.8}{0.562 \times 7540}\right]\)
= log123.4 + log54.8- log0.562 – log7346
x = Anti log (value)

Question 41.
The first term of G.P exceeds the second term by \(\frac{1}{2}\) and the sum to infinity is 2. Find the GP.
Answer:
a, ar, ar² ……………. GP
a – ar = 1/2
S_∞ = 2 …….(2)

a² = 1 ⇒ a = 1
\(\frac{\mathrm{a}}{1-\mathrm{r}}\) = 2 \(\frac{1}{1-r}\) = 2
1 – r = 1/2
1 – 1/2 = r
r = 1/2

Question 42.
Two years ago, a father’s age was three times the square of his son’s age. After 3 years, his age will be four times his son’s age. Find the present age of father and son.
Answer:
Let the age of the father be x yrs and age of the son be y yrs.
Given: x-2 = 3(y-2)² ⇒ x-2 = 3(y² + 4 – 4y)
Also given x + 3 = 4(y + 3)
x + 3 = 4y + 12

3y² – 16y + 5 = 0
3y² – 15y – y + 5 = 0
3y(y – 5) – 1(y – 5) = 0
(3y – 1)(y – 5) = 0
y = 51

Sons age is 5 yrs and fathers age is
x = 4y + 9 = 20 + 9
x = 29yrs

Question 43.
A certain amount invested at 4% p.a. compounded semi-annually amounts to Rs. 78,030 at the end of one year. Find the sum.
Answer:
F = 78030, i = 0.04, n= 1
F = P\(\left(1+\frac{i}{2}\right)^{n \times 2}\)
78030 = P\(\left(1+\frac{0.04}{2}\right)^{1 \times 2}\)
P = \(\frac{78030}{(1.02)^{2}}\) = 75,000
sum = 75,000.

Question 44.
If a person wants to have Rs. 80,000 after 5 years, how much should he deposit every year if the bank offers 12% p.a. interest compounded quarterly.
Answer:
F = 80,000, n = 5yrs, r = 12% = 0.12, i = \(\left(1+\frac{0.12}{4}\right)^{4}\) – 1 = 0.125
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)
a = \(\frac{\mathrm{F} \times i}{(1+i)^{n}-1}=\frac{80,000 \times 10.125}{(1.125)^{5}-1}=\frac{80,000}{6.416}\) = ₹12,468.8

Question 45.
In a dance competition, 70% of the participants were girls, 35% of the boys, and 65% of the girls got qualified for the next round. If 49 girls were eliminated find the number of boys who were eliminated.
Answer:
Let the tota! number of participants = x
Total number of girls = \(\frac{70}{100}\)x = 0.7x ⇒ Total number of boys = \(\frac{30}{100}\) × x = 0.3x
If 65% of girls got qualified then 35% of girls got eliminated
∴ The total girls eliminated = \(\frac{35}{100}\) × 0.7x = 0.245x
Hence 0.245. x = 49
x = \(\frac{49}{0.245}\)
Hence total number of boys eleminated = \(\frac{35}{1000}\) × 0.3 × 200 = 2

Question 46.
Prove that \(\frac{\operatorname{cosec}\left(180^{\circ}+0\right) \sin \left(360^{\circ}-0\right) \tan \left(36^{\circ}+\theta\right)}{\sin \left(90^{\circ}+\theta\right) \cos \left(180^{\circ}-\theta\right) \tan (-\theta)}\) = sec2 θ
Answer:

\(\frac{1}{\cos ^{2} \theta}\) = sec2θ = RHS

Question 47.
Show that the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle. Also, find its area.
Answer:
Let A= (4, 4) B = (3, 5) and C = (-1, -1)
AB = \(\sqrt{(4-3)^{2}+(4-5)^{2}}=\sqrt{1^{2}+(-1)^{2}}\) ⇒ AB² = 2
BC = \(\sqrt{(4-3)^{2}+(4-5)^{2}}=\sqrt{1^{2}+(-1)^{2}}\) ⇒ AB² = 2
CA = \(\sqrt{(3+1)^{2}+(5+1)^{2}}=\sqrt{16+36}=\sqrt{50}\) ⇒ CA² = 50

From the above AB2 + CA2 = BC2

Question 48.
Find the equation of line passing through (3, 4) and having intercepts on the axes whose sum is 14.
Answer:
Req. Eqn is \(\frac{x}{a}+\frac{y}{b}\) = 1
Give a + 6=14 ⇒ b = 14 – a
eqn becomes \(\frac{x}{a}+\frac{y}{14-a}\)
(14 – a)x + ay = a(14 – a)
By data this must pass the (3, 4)
(14 – a) 3 + 4a = a(14 – a)
42 – 3a + 4a = 14a – a²
a² + a – 14a + 42 = 0
a² -13a + 42 = 0
(a – 7)(a – 6) = 0 ⇒ a = 7
If a = 7 then b = 7 If a = 6 then b = 8.
∴ the required equations are
\(\frac{x}{7}+\frac{y}{7}\) = 1 & \(\frac{x}{6}+\frac{y}{8}\) = 1

PART-E

V. Answer any one question : (1 × 10 = 10)

Question 49.
(a) Find the equation of a line passing through the point of intersection of the lines x – 2y + 4 = 0 and 4x – 3y + 1 = 0 inclined at angle of 135° with positive x-axis. Also write the x intercept of the line. (4M)
Answer:
Point of inter section of x – 2y + 4 = 0 and 4x – 3y + 1 = 0
Multiply equation one by 4 we get

y = \(\frac{15}{-5}\) = 3
x – 2y + 4 = 0
x = 2y – 4
x = 6 – 4 = 2
∴ point of intersection is (2,3)
Slope = m = tan (135°) = – 1
The required equation of the line with slope = -1 and the point (2,3) is given by
y – y₁ = m(x – x₁).
y- 3 = -1(x – 2)
∴ x + y – 5 = 0

(b) If sin θ = -8/17, π< θ < \(\frac{3 \pi}{2}\) find the value of \(\frac{\tan \theta-\cos \theta}{\sec \theta+{cosec} \theta}\)
Answer:
If sinθ = \(\frac{-8}{17}=\frac{O p p}{1+y p}\) Here adj = \(\sqrt{17^{2}-8^{2}}=\sqrt{225}\) = 15
θ lies in Quadrant

(c) Two numbers are in the ratio 7:5 and their difference is 17. Find the numbers. 2
Answer:
Let the two numbers be 7x & 5x
Given 7x – 5x = 17
2x = 17
x = 8.5
∴ the two number are 7 × 8.5 & 5 × 8.5 = 59.5 and 92.5.

Question 50.
(a) A confectioner makes and sells biscuits. 11e sells one pack of biscuits at 80. His cost of manufacturing is 40 per pack as a variable cost and 3000 as a fixed cost.
Find (i) his profit function.
(ii) break-even point. also
(iii) if he limits his Production o 100 packets, can be.make profit? 4
Answer:
T.R (x) = Price x Quantity = 80x
T.C C(x) = ax + b = 40x + 3000
Profit function = P(x) = R(x) – C(x) = 80x – 40x – 3000 = 40x – 3000
If x = 100 then P(100) = 40 × 100 – 3000 = ₹1 ,000

(b) Find the sum of n terms the series 5 + 35 + 555 + ………….
Answer:
Let S = 5 + 55 + 555 +…………… to n terms
\(\frac{\mathrm{S}}{5}\) = 1+ 11 + 111+………..to n terms
\(\frac{9 \mathrm{~S}}{5}\) = 9 + 99 + 999 +…………… to n terms
=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – 1 × n
S = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) If 3^x = 5^y = 15^z, Show that z(x + y) = xy.
Answer:
Let 3^x = 5^y = 15^z = k(say)
3^x = k ⇒ 3 = k^1/x, 5^y = k ⇒ 5 = k^1/y, 15^z = k⇒ 15 = k^1/z
Now 3 x 5 = 15
k^1/x. k^1/y = k^1/z ⇒ K^1/x+1/y = K^1/z
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\)
⇒ \(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{xy}}=\frac{1}{\mathrm{z}}\)
⇒ z (y + x) = xy ⇒ z (x + y) = xy