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Karnataka 1st PUC Basic Maths Model Question Paper 2 with Answers

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 4 – 5i
Answer:
-5

Question 2.
If A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7} find A – B.
Answer:
A – B = { }

Question 3.
If f: R → R is defined by f(x) = 3x + 5 then find f (-1)
Answer:
f(-1) = 3(- 1) + 5 = -3 + 5 = 2

Question 4.
Simplify (5°)2 + 5
Answer:
(1)2 + 5′ = 1 +5 = 6

Question 5.
Find the value of log₁₀.01
Answer:
log₁₀₀.01 = x ⇒ 10x = 0.01
10x = \(\frac{1}{100}\) ⇒ 10x = 10-2
x = -2

Question 6.
Find the 8th term of the progression -2, -4, -6……….
Answer:
a = -2, d = -4 + 2 = -2, n = 8, T₈ = ?
Tn = a + (n – l)d, T₈ = -2 + (8 – 1) (-2) = -2 + 7(-2) = -2 – 14 = -16.

Question 7.
Solve for x : 2(7 + x) – 10 = 16 – 2 (x – 24).
Answer:
14 + 2x – 10 = 16 – 2x + 48
2x + 2x = 16 + 48 – 14 + 10
4x = 60 ⇒ x = 15

Question 8.
Convert the ratio 3: 5 into percentages.
Answer:
3: 5= \(\frac{3}{5}\) × 100% = 3 × 20% = 60%.

Question 9.
Define perpetuity.
Answer:
If annuity payments are made for an infinite period then it is called a perpetuity.

Question 10.
Convert 450° into radians.
Answer:
450° = 450 × \(\frac{\pi}{180}=\frac{45 \pi}{18}=\frac{5 \pi}{2}\)

Question 11.
The average age of 10 boys in a class is 13 years. What is the sum of their ages?
Answer:
Sum of their ages =10 × 13 = 130

Question 12.
Find the slope of the line 2x + 5y – 11 = 0
Answer:
Slope of the line 2x + 5y – 11 = 0 is \(\frac{-a}{b}=\frac{-2}{5}\)

PART-B

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

Question 13.
Find the number which when divided by 36,40 and 48 leaves the same remainder 5,
Answer:
Let us find out the LCM of 36, 40, 48
36 = 2² × 3²
40 = 2³ × 5
48 = 2⁴ × 3
LCM = 2⁴ × 3² × 5¹
= 16 × 45
= 720
Since 5 has to be the remainder, we have to add 5 to 720 i.e. 720 + 5 = 725

Question 14.
If A = {1,23, 4) and B = (34, 5, 6) and U = {1,2, 3,4, 5,6, 7,8)
Verify (A ∪ B)’ =A’ ∩ B’
Answer:
A ∪ B = {1,2,3,4,5,6}
(A∪B)’ = (7,8)
A’ = {5, 6,7,8) B’ = {1,2, 7,8)
A’∩ B’ = {7, 8)
∴ (A∪B)’ = A’ ∩ B’

Question 15.
Find the number of positive divisors and the sum of all positive divisors of 1200.
Answer:
1200 = 2⁴ × 5² × 3¹ = P₁^α1 × P₂^α2 × P₃^α3
P₁ = 2; P₂ = 5; P₃ = 3
α₁ = 4; α₂ = 4; α₃ = 1
T(n) = (1 + α₁)(1 + α₂)(1 + α₃) = (1 + 4)(1 + 2)(1 + 1) = (5) (3) (2) = 30

S_n = \(\left(\frac{\mathrm{P}_{1}^{\alpha_{1}+1}-1}{\mathrm{P}_{1}-1}\right)\left(\frac{\mathrm{P}_{2}^{\alpha_{2}+1}-1}{\mathrm{P}_{2}-1}\right)\left(-\frac{\mathrm{P}_{3}^{\alpha_{3}+1}-1}{\mathrm{P}_{3}-1}\right)\)
= \(\left(\frac{2^{5}-1^{t}}{2-1}\right)\left(\frac{5^{3}-1}{5-1}\right)\left(\frac{3^{2}-1}{3-1}\right)=\left(\frac{31}{1}\right)\left(\frac{124}{4}\right)\left(\frac{8}{2}\right)\) = 31 × 124 = 3844

Question 16.
Simplify \(\frac{2^{n+1}+2^{n-1}}{2^{n}+2^{n+2}}\)
Answer:

Question 17.
If a, b, c are in G.P and a^x = b^y = c^z show that x, y and z are in H.P
Answer:
a,b,c are in GP
∴ b² = ac
a^x = b^y = c^z = k
a^x = k ⇒ a = k^1/x
b^y = k ⇒ b = k^1/y
c^z = k ⇒ c = k^1/z

b2 = ac
(k^1/y) = k^1/x+1/z
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\)
\(\frac{2}{y}=\frac{x+z}{x z}\)
y = \(\frac{2 x z}{x+z}\)
∴ x, y, z are in HP.

Question 18.
The sum of two numbers is 107 and their difference is 17. Find the numbers.
Answer:

a = 62
a + b =107
b = 107 – a = 107 – 62 = 45
∴ b = 45

Question 19.
Determine the principal which will amounts to ₹ 15000 in 8 years at 41% per annum simple interest
Answer:
H = ₹ 15,000, R = 11, T = 8
A = P + I = P + \(\frac{P+R}{100}\)
1500 = p(1 + \(\frac{\mathrm{TR}}{100}\)) = p(\(\frac{100+8 \times 11}{100}\)) = p\(\left(\frac{188}{100}\right)\) ⇒ p = \(\left(\frac{188}{100}\right)\)
∴ P = ₹ 7978.72

Question 20.
Solve 3x – 2 < 2x + 1 when x is an integer and x is a real number. Also represent on a number line.
Answer:
3x – 2 < 2x + 1
3x – 2x < 2 + 1
x < 3
(i) when x is an integer x = {………- 2, -1, 0, 1, 2}
(ii) when x is a real number x ∈ (-∞, 3)

Question 21.
The average score of 20 boys is 60% and the average score of 30 girls is 70%. Find the combined average.
Answer:
X₁ = 60%, N1 = 20
X₂ = 70%, N2 = 30
X̄ = \(\frac{X_{1} N_{1}+X_{2} N_{2}}{N_{1}+N_{2}}=\frac{(60)(20)+(70)(30)}{20+30}=\frac{3300}{50}\) = 66%

Question 22.
If the cost price of 10 articles is equal to the selling price of 9 articles, find the gain percent
Answer:
Let the cost of each article be x.
C.P of 10 articles = 10x
S.P of each article = \(\frac{10 x}{9}\)
S.P of articles = \(\frac{10 x}{9}\) × 10 = \(\frac{100 x}{9}\)

Profit = SP – CP = \(\frac{100 x}{9}\) – 10x = \(\frac{100 x-90 x}{9}=\frac{10 x}{9}\)
Proof% = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100 = \(\frac{\frac{10 x}{9}}{10 x}\) × 100 = \(\frac{10 x}{90 x}\) × 100 = 11\(\frac{1}{9}\)%

Question 23.
Find the value of sin² \(\frac{\pi}{6}\)cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\) + cot² \(\frac{\pi}{3}\)
Answer:
sin² \(\frac{\pi}{6}\)cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\) + cot² \(\frac{\pi}{3}\) = \(\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{1}{2}\right)^{2}-(1)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}\)

\(\frac{1}{4} \cdot \frac{1}{4}-1+\frac{1}{3}\)
\(\frac{1}{16}-1+\frac{1}{3}=\frac{3-48+16}{48}=\frac{-29}{48}\)

Question 24.
Prove that (SinA + Cos A)² + (Sin A – CosA)² = 2
Answer:
LHS (sin A + cos A)² + (SinA-cos A)²

Question 25.
Find the equation of the straight line passing through (2, 3) and (3, 4)
Answer:
(x₁, y₁) = (2, 3) (x2, y2) =(3,4)
\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow \frac{y-3}{x-2}=\frac{4-3}{3-2}\)
⇒ \(\frac{y-3}{x-2}=\frac{1}{1}\) ⇒ y – 3 = x – 2
x – y + 1 = 0

PART-C

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

Question 26.
In a group of 600 people, 150 students were found to be taking tea, 225 like Coffee, 100 like both tea and coffee. Find out how many were taking neither tea nor coffee?
Answer:
n(T) = 150
n(C) =225
n(C ∩ T) = 100
n(∪)=600
n(C ∪ T) = n(C) + n(T) – n(C ∩T) = 225 + 150 – 100 = 375 – 100 = 275
∴ Number of students taking neither coffee nor tea = 600 – 275 = 325

Question 27.
If R^-1 = {(2, 4), (1, 2) (3, 1), (3, 2) Find R. Also find its domain and range.
Sol. R = {(4, 2), (2, 1), (1, 3), (2,3)}
DomR= {1,2,4}
RanR = {1, 2; 3}

Question 28.
Prove \(\sqrt{5}\) is an irrational number.
OR
An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec,. They beeped together at 10 am. Find the time when they will next take a beep together at the earliest.
Answer:
If possible let \(\sqrt{5}\) is a rational number.
\(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\), p, q ∈ z, q ≠ 0
we shall also assume that p and q do not have any common factor
∴ 5 = \(\frac{\mathrm{p}}{\mathrm{q}}\)
⇒ p² = 5q²
∴ p² is a multiple of 5
∴ 5 divides p²
∴ 5 divides p
∴ p = 5k where k ∈ z, k ≠ 0
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.
OR

LCM = 22 × 31 × 51 × 31 = 1860 sec = 31 min.
They beeped together at 10am.
∴ After 31 minutes i.e., at 10.31 AM they will next make a beep together at the earliest.

Question 29.
Prove that \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = 4
Answer:
LHS = \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = log₄ 2 + log₄ 8 + log₄ 16 = log₄(2 × 8 × 16) = log₄ 256 = log₄ 44 = 4 log₄ 4 = 4(1) = 4

Question 30.
The sum of three numbers in an A.P is 15 and their product is 105. Find the numbers.
Answer:
Let the numbers be a – d, a, a + d
sum = 15
a-d + a + a + d = 15
3a = 15 ⇒ a = 15
Product = 105
(a-d) a(a + d) = 105
(5 – d). 5. (5 + d) = 105
(5-d)(5+d) = \(\frac{105}{5}\) = 21
5² – d² = 21
25 – 21 = d²
d² = 4
d = 2
∴ The number are a – d, a, a + d,
5-2, 5, 5 + 2
3, 5, 7

Question 31.
Find the difference between the compound interest and the simple interest on ₹ 5000 invested for 4 years at 8% p.a.
Answer:
P = ₹ 5,000, T = 4yrs, R = 8%
SI = \(\frac{\mathrm{PTR}}{100}=\frac{5000 \times 4 \times 8}{100}\) = 50 × 4 × 8 = 1600
A = P(1 + R)n = 5000(1 + 0.08)8 = 5000 (1.08)4 = 5000 × 1.3604 = 6802.44
∴ CI = 6802.44 – 5000
= 1802.44
Difference between Cl and SI
= CI – SI = 1802.44 – 1600 = ₹ 202.44

Question 32.
Solve the linear inequalities graphically:
5x + 4y ≥ 40, x ≥ 2, y ≥ 3, x, y ≥ 0
Answer:
5x + 4y = 40
y = 0, 5x = 40 ⇒ x = 8, A = (8, 0)
x = 0, 4y = 40 ⇒ y = 10, B= (0, 10)

The solution set of each equation is indicated by the arrow mark in the figure. The intersection of these solution sets is in the feasible region shaded as shown in the figure.

Question 33.
Sachin finds by getting out for a duck (0 runs) in the lith innings of his test matches his average of the previous innings is decreased by 5 runs. What is his average after the 11th innings?
Answer:
Let the average after 10 inninngs = x
∴ Total runs after 10 inninngs = 10x
Average after 11 inninngs = x – 5
∴ Total runs after 11 inninngs = 11 (x – 5)
11(x- 5) – 10x = 0
11x- 55 – 10x = 0
x- 55 = 0
x = 55
∴ Average after 11 inninngs = x – 5
= 55 – 5 = 50

Question 34.
Find the ratio in which the line joining the points (2,5) and (1,9) is divided by the x-axis. Also, find the point of division.
Answer:
P(x, 0) A = (2, 5,) = (x1, y1)
B = (1, 9) = (x2, y2)
P = \(\left[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right]=\left[\frac{m(1)+n(2)}{m+n}, \frac{m(9)+n(5)}{m+n}\right]\)
(x, 0) = \(\left[\frac{m+2 n}{m+n}, \frac{9 m+5 n}{m+n}\right]\)
\(\frac{9 m+5 n}{m+n}\) = 0
\(\frac{m}{n}=\frac{-5}{9}\) ⇒ 9m + 5n = 0 ⇒ 9m = – 5n
m :n = 5:9 externally

Taking m = 5, n = 9 to find x use section formula for external division namely,
x = \(\frac{m x_{2}-n x_{1}}{m-n}=\frac{5(1)-9(2)}{5-9}=\frac{5-18}{-4}=\frac{-13}{-4}=\frac{13}{4}\)
P = (\(\frac{13}{4}\), 0) → Point of division.

Question 35.
If an article is sold at ₹ 24 there is a profit of 4 and if it is sold at a loss of 10% find the selling price of the article.
Answer:
SP of the article = ₹ 24
Profit = ₹ 4
CP = SP – Profit = 24 – 4 = ₹ 20
Loss = 10%
We have SP = \(\frac{100-\operatorname{loss} \%}{100}\) × CP = \(\frac{100-10}{100}\) × 20 = \(\frac{90}{100}\) × 20
∴ SP = ₹ 18

Question 36.
If Cot A = \(\frac{5}{12}\) and A is acute, show that 2 cosec A – 4sec A = – \(\frac{247}{50}\)
Answer:

Cosec A = \(\frac{13}{2}\)
Sec A = \(\frac{13}{5}\)
2 cosec A – 4 sec A = 2\(\left(\frac{13}{12}\right)\) – 4\(\left(\frac{13}{5}\right)\) = – \(\frac{494}{60}=-\frac{247}{30}\)

Question 37.
Show that the points A(2, 2) B(6, 3) and C(4, 11) form a right angled triangle.
Answer:
AB = \(\sqrt{(6-2)^{2}+(3-2)^{2}}=\sqrt{16+1}=\sqrt{17}\)
BC = \(\sqrt{(4-6)^{2}+(11-3)^{2}}=\sqrt{4+64}=\sqrt{68}=\sqrt{4 \times 17}\) = 2\(\sqrt{17}\)
CA= \(\sqrt{(4-2)^{2}+(11-2)^{2}}=\sqrt{4+81}=\sqrt{85}\)
AB² =17, BC² = 68, CA² = 85
∴ AB² + BC² = CA²
∴ ABC form a right angled triangle.

Question 38.
If a train travels 15 km / hr faster it would take 1 hour less to travel 180 km, find the original speed of the train.
Answer:
Let s be the actual speed
t be the actual time d = 180 km
s = \(\frac{d}{t}=\frac{180}{t}\) ⇒ t = \(\frac{180}{s}\) ⇒ st = 180
If speed is 15km/hr more then speed becomes s + 15 and time taken will be 1 hour less it becomes t – 1.

∴ Actual speed of the train = 45 km/hr
OR
If α and β are the roots of the equation 2x² + 5x + 5 = O then find the value of \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\)
Answer:
α + β = \(\frac{-5}{2}\) and αβ = \(\frac{5}{2}\)
∴ \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{\left(\frac{-5}{2}\right)^{2}-2\left(\frac{5}{2}\right)}{\left(\frac{5}{2}\right)^{2}}=\frac{\frac{25}{4}-5}{\frac{25}{4}}\)

PART – D

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

Question 39.
Let f = {(1, 1), (2, 3), (0, -1)} be a function from Z to Z defined by f(x) = ax + b some integers a and b. (i) Determine a & b. (ii) If f(x) = 2x + 1 g(x) = x² + 2x + 1 Find (a) fog(2)(b) gof (3)
Answer:
f(x) = ax + b
when x – 1 f(x) = 1
∴ a + b = 1
x = 0 f(x) = -1
a(0) + b = -1 ⇒ b = -1
a + b = 1
a – 1 = 1 ⇒ a = 1 + 1 = 2
∴ a = 2
∴ a = 2, b = -1
(a) fog (2) = f [g(2)] = f [22 + 2(2) + 1]
= f [4 + 4 + 1] = f [9] = 2(9) + 1 = 18 + 1 = 19

(b) gof (3) = g [f(3)] = g [2(3) + 1] – g(7)
= 72 + 2(7) + 1 = 49 + 14 + 1 = 64

Question 40.
Find the sum of all integers between 60 and 400, which are divisible by 13.
Answer:
S_n = 65 + 78 + 91 +……….+ 390
a = 5 d= 13 n = ? T_n = 390
T_n = a + (n – l)d
390 = 65 + (n- 1)13
390 = 65+ 13n-13
13n= 338
∴ n = 26
S_n = \(\frac{\mathrm{n}}{2}\)[a + l] = \(\frac{26}{2}\)(65 + 390)= 13 (455) = 5,915
∴ S = 5,9.15

Question 41.
The age of the father is 5 times that of the son’s 3 years if the age of the father was 8 times that of his son. Find their present ages.
Answer:
Let the present age of the son = x
Present age of the father = 5x
Three years-agey sons age = x – 3 and fathers age = -5x – 3
Given 5x – 3 = 8(x – 3)
5x – 3 = 8x – 24
8x – 5x = 24 – 3
3x = 21
x = 7
Sons age = x = 7 years
Fathers age 5x = 5(7) = 35 years

Question 42.
A machine depreciates at 10% of its value at the beginning of a year. The cost and scrap value realized at the time of sale are 23240 and 9000 respectively. For how many years the machine was put to use?
Answer:
r= 10% = 0.10 C =23,240 B=9,000 n=?
B = C(1 – r)ⁿ
9000 = 23240(1 – 0. 10)
9000 = 23240 (0.90)ⁿ
(0.9)ⁿ = \(\frac{9000}{23240}\) =0.3872
log (0.9) = log 0.3 872
n log 0.9 = log 0.3872
n = \(\frac{\log 0.3872}{\log 0.9}=\frac{1.5879}{1.9542}=\frac{-1+0.5879}{-1+09542}=\frac{-0.4121}{-0.0458}\)
= 8.99 ≈ 9 years

Question 43.
Calculate the future value of the annuity immediate of ₹ 1000 p.a for 12 years at 16% p.a compounded quarterly?
Answer:
A = ₹ 1,000 p.a, n = 12, R = 16%= 0.6, r = ?, q = 4
r = \(\left(1+\frac{\mathrm{R}}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.16}{4}\right)^{4}\) – 1 = 0.1698 = 16.98%
F = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right]}{r}=\frac{1000\left[(1+0.1698)^{12}-1\right]}{0.1698}\)
F = 32,783.17

Question 44.
Find the equation of the straight line passing through (-2,6) and the sum of the intercepts on the co-ordinate axes is 5.
Answer:
a + b = 5 ⇒ b = 5 – a, Let the equation of the straight line be
\(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{x}{a}+\frac{y}{5-a}\) = 1

It passes through (-2. 6) put x = -2, y = 6
\(\frac{-2}{a}+\frac{6}{5-a}\) = 1 ⇒ \(\frac{-10+2 a+6 a}{a(5-a)}\) = 1
⇒ – 10 + 8a = a(5 – a) ⇒ – 10 + 8a = 5a – a²
⇒ a²(a + 5) – 2(a + 5) = 0 ⇒ (a + 5)(a + 2) = 0
⇒ a = 2 ∴ b = 5 – a = 5 – 2 = 3
Case (i)
\(\frac{x}{2}+\frac{y}{3}\) = 1
3x + 2y = 6
3x + 2y – 6 = 0
a = -5
∴ b = 5 – a = 5 – (-5) = 10

Case (ii)
\(\frac{x}{-5}+\frac{y}{10}\) = 1
\(\frac{-2 x+y}{10}\) = 1
-2x + y – 10 = 0

\(\frac{-x}{5}+\frac{y}{10}\) = 1
-2x + y = 10
2x – y + 10 = 0

Question 45.
(a) Form a quadratic equation whose roots are 2 + \(\sqrt{3}\) & 2 – \(\sqrt{3}\)
(b) Evaluate using log. tables \(\frac{(0.5634)^{2} \times(0.0635)}{(2.563)^{2}}\)
Answer:
(a) α = 2 + \(\sqrt{3}\)
β = 2 – \(\sqrt{3}\)
α + β = (2 + \(\sqrt{3}\)) + (2 – \(\sqrt{3}\)) =4
αβ = (2 + \(\sqrt{3}\))(2 – \(\sqrt{3}\)) = 4 – 3 = 1
∴ Quadratic equation’s
x² – (α + β)x + αβ = 0 ⇒ x² – 4x+l=0

(b) x = \(\frac{(0.5634)^{2} \times(0.0635)}{(2.563)^{2}}\)
log x = 2log 0.5634 + log 0.0635 – 2log 2.563
= 2 (1.7508) + (2.8027)- 2(0.4087) = 2(-1 + 0.7508) + (-2+ 0.8027) – 2(0.4087)
= -2 + 1.5016-2 + 0.8027-0.8174
= 2.5131 =-2-0.5131
= -2 – 1 + 1-0.5131
= -3 + 0.4869
log x = 3.4869
x = AL[34869] =0.00306

Question 46.
Find the reflection of the point P(2, 1) in the line x + y = 5

Slope of x + y – 5 = 0 is
PQ – x + y – 5 =0
∴ Slope of PQ = +1 =m

∴ Equation of PQ y – y₁ =m(x – x₁)
y – 1 = 1(x – 2)
y – 1 = x – 2
x – y – 1 = 0
x + y – 5 =0
Add ⇒ 2x – 6 = 0
⇒ x = 3

x – y – 1 = 0
x + y – 5 =0
3 + y – 5 = 0
A = (3, 2) A is the mid point of PQ

∴ A = \(\left(\frac{2+a}{2}, \frac{1+b}{2}\right)\)
(3, 2) = \(\left(\frac{2+a}{2}, \frac{1+b}{2}\right)\)
⇒ \(\frac{2+a}{2}\) = 3
⇒ 2 + a = 6 ⇒ a = 6 – 2
a = 4

⇒ \(\frac{1+b}{2}\) = 3
⇒ 1 + b = 4 ⇒ b = 4 – 1
b = 3
∴ Q = (4, 3)

Question 47.
If x = r cos A cos B, y = r cos A sin B and z = r sin A Then P.T. x2 + y2 + z2 = r2
Answer:
LHS = x² + y² + z² = r² cos² A cos² B + r² cos² A sin² B + r² sin² A = r² cos² A (cos² B + sin² B) + r² sin² A= r² cos² A(1) + r² sin² A = r² (cos² A+ sin² A) = r² (1) = r² = RHS

Question 48.
Find the locus of a point equidistant from (2, 0) and (-2, 0)
Answer:
A = (2, 0), B = (-2, 0)
Let P(x, y) be any point on the locus
∴ PA = PB
PA² = PB²

– 4x – 4x + 4 – 4 = 0
– 8x = 0
x = 0

PART-E

IV. Answer any one question. (1 × 10 = 10)

Question 49.
(a) Find the equation of the straight line passing through the intersection of the line 2x + 3y = 5, 7x – y = 6 is perpendicular to the line 3x + 4y + 1 = 0.
(b) If tan A + sin A = m and tanA – sin A = n show that m² – n² = ±4\(\sqrt{\mathrm{mn}}\) .
(c) Find the number of digits in 320?
Answer:
(a) 2x + 3y = 5 × 1
7x – y = 6 × 3

23x = 23 ⇒ x = 1
7x – y = 6
7(1) – y = 6
7 – 6 = y
y = 6
∴ (x1, y1) = (1, 1)
Slope of the given line = \(\frac{-a}{b}=\frac{-3}{4}\)
Required line is perpendicular to given line
∴ slope of the required line = \(\frac{+4}{3}\) = m
∴ Equation of the required line
y – y₁ = m(x – x₁)
y – 1 = \(\frac{4}{3}\)(x – 1)
3y – 3 = 4x – 4
4x – 3y – 1 = 0

(b) m² – n² – (m – n) (m + n)
= (tanA+ sin A- tan A + sin A) (tanA + sin A + tanA- sinA)
= 2 sin A. 2 tan A = 4 sin A tan A = 4 sin A. tan A
(m² – n²)² = 16 sin² A. tan² A …………..(1)
mn = (tan A + sin A)(tan A – sin A) = tan² A – sin² A
= \(\frac{\sin ^{2} A}{\cos ^{2} A}\) – sin² A = sin²A(\(\frac{1}{\cos ^{2} A}\) – 1) = sin²A\(\frac{1-\cos ^{2} A}{\cos ^{2} A}=\frac{\sin ^{2} A}{\cos ^{2} A}\)sin2A
mn = tan² A. sin² A ……….(2)
substitue the value of mn from (2) in (1) we get
(m² – n²)² = 16mn

(c) Let x = 3²⁰
log x = log 3²⁰ = 20log3 = 20[0.4771]
log x = 9.5424
As the characteristic of log x is 9, the number of digits = 9 + 1 = 10.

Question 50.
(a) Find the sum to n terms of the series: 5 + 55 + 555 +…. n term.
(b) A confectioner makes and sells biscuits. Does he sell one pack of biscuits at ₹ 80? His cost of manufacturing is ₹ 40 per packet as variable cost and 3000 as a fixed cost. Find the
(i) Revenue function
(ii) Cost function
(iii) Profit function
(iv) If he limits his production to 100 packets can he make a profit?
(v) What will be the number of boxes he must sell to make a profit so that he does not incur a loss?
(e) If the product of two numbers is 216 and their LCM is 36. Find their HCF.
Answer:
(a) Let S_n = 5 + 55 + 555 + …….n terms
S_n = 5(1 + 11 + 111 +…….. )
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = 1 + 11 + 111 + ………
Multiply both the sides by 9,
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = 9 + 99 + 999 + …………
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = (10 – 1) + (10² – 1) + (10³ – 1) +
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = (10 + 10² + 10³ + ……….. ) – (1 + 1 + 1 + ………..)
\(\frac{9 S_{n}}{5}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}=\frac{10\left(10^{\mathrm{n}}-1\right)}{9}\) – n
S_n = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]=\frac{5}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-n\right]\), n ∈ N

(b) (a) Revenue function = R (x) = Selling price X quantity
= 80 x (x = No. of pack of biscuit)
(b) Cost function = C (x) = ax + b
= 40x + 3000
(c) Profit function = P (x)
= R(x) – C(x)
= 8Ox – (40x + 3000)
= 80x – 40x – 3000
= 40x – 3000

(d) If x = 100 packets
∴ P(x) = 40x – 3000
P(100) =40 (100)- 3000
= ₹ 1,000

(e) No. of packets required to ensure no Joss (at BEP)
C (x) =R(x)
40x + 3000 = 80 x
3000 = 80x – 40x
3000 = 40x
x = \(\frac{3000}{40}\)
x =75 packet

(c) ab = 216 LCM = 36 HCF = ?
HCF × LCM = ab = 216
HCF = \(\frac{216}{\mathrm{LCM}}\)
HCF = \(\frac{216}{36}\) = 6

Students can Download 1st PUC Basic Maths Model Question Paper 1 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 1 with Answers

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.
Define an imaginary number.
Answer:
A number of the form a + ib where a, b ∈ R, the set of real numbers and i = \(\sqrt{-1}\) is called a complex number or an imaginary number.
Example : 2 + 3i

Question 2.
If A = {1, 2} and B = {a, b} then find B × A.
Answer:
B × A={(a, 1),(a, 2),(b, 1 ),(b, 2)}

Question 3.
If f: R → R is defined by f(x) = 2x + 3 then find f(\(\frac{1}{2}\))
Answer:
f(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) + 3 = 1 + 3 = 4

Question 4.
Simplify \(\left(\frac{81}{256}\right)^{\frac{1}{4}}\)
Answer:
\(\left(\frac{8}{256}\right)^{1 / 4}=\left(\frac{3^{4}}{4^{4}}\right)^{1 / 4}=\left\{\left(\frac{3}{4}\right)^{4}\right\}^{1 / 4}=\frac{3}{4}\)

Question 5.
Find the value of log_0.1 10
Answer:
Let log_0.1 10 = x ⇒(0.1)^x ⇒ \(\left(\frac{1}{10}\right)^{x}\) = 10
⇒ 10^-x = 10¹ ⇒ -x = 1 ⇒ x = -1
∴ log_0.1 = 10 = -1

Question 6.
Find the nature of the roots without solving the equation x² – x + 1 = 0
Answer:
a = 1, b = -1, c = 1
Δ = b² – 4a = (-1)² – 4(1)(1) = 1 – 4 = -3 < 0
Δ < 0
∴ The roots are complex.

Question 7.
What is the present value of an income of 1000 a year to be received forever? Assume the discount rate to be 12%.
12%
Answer:
A = ₹ 1,000, r = 12% = \(\frac{12}{100}\) = 0.12,
P_∞ = A/r = \(\frac{1000}{0.12}\) = ₹ 8333.33

Question 8.
Convert \(\frac{1}{4}\) into percentage
Answer:
\(\frac{1}{4}=\frac{1}{4}\) × 100% = 25%

Question 9.
Define a radian
Answer:
A radian is an angle subtended at the center of a circle by an arc of the circle such that the length of the arc is equal to the radius of the die circle.

Question 10.
Convert 315° into radians
Answer:
315° = 315 × \(\frac{\pi}{180}=\frac{7 \pi^{c}}{4}\)

Question 11.
If the slope of the line AB is \(\) and AB ⊥ CD then find the slope of CD.
Answer:
m₁ = \(\frac{3}{4}\), m₂ = ?
AB ⊥ CD
∴ m₁ × m₂ = -1 ⇒ \(\frac{3}{4}\) × m₂ = 1 ⇒ m₂ = – \(\frac{4}{3}\) = slope of CD

Question 12.
Find the centroid of the triangle formed by the points (2, 4), (5, 3), and (8, 3)
Answer:
G = \(\left(\frac{2+5+8}{3}, \frac{4+3+3}{3}\right)=\left(\frac{15}{3}, \frac{10}{3}\right)=\left(5, \frac{10}{3}\right)\)

PART-B

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

Question 13.
Find the number of positive divisors of 960
Answer:
960
960 = 26 × 31 × 51
p₁ = 2, α₁ = 6, p₂ = 3, α₂ = 1, p₃ = 5, α₃ = 1
T(n) = (1 + α₁)(1 + α₂)(1 + α₃)
T(960) = (1 + 6)(1 + 1)(1 + 1) = (7)(2)(2) = 28

Question 14.
If A =(1, 3,5, 7,9) and B = (2,4,6,8, 10, 12) define a function f: A → B by f(x) = x + 1 ∀ x ∈ A, verfiy that the function is one – one and onto
Answer:
f: A → B is one-one but not onto as 12 has no pre-image.

Question 15.
Simplify \(\left(\frac{x^{2}}{x^{b}}\right)^{n+b}\left(\frac{x^{b}}{x^{c}}\right)^{b+c}\left(\frac{x^{c}}{x^{2}}\right)^{c+x}\)
Answer:
(x^a-b)^a+b(x^b-c)^b+c(x^c-a)^c+a = x^a2-b2.x^b2-c2.x^c2-a2
= \(x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\) = x° = 1

Question 16.
Prove that log₂ [log₂ [log₂ 16]] = 1
Answer:
Log₂ [log₂[log₂16]] = log₂ [log₂[log₂ 24]] = log₂[log₂[4log₂ 2]] = log₂ [log₂ [4(1)]]
= log₂ [log2 4] = log₂ [log₂ 22] = log₂ [2 log₂ 2] = log₂ [2(1)] = log₂ 2 = 1 [∵ log_a o = 1]

Question 17.
Is -300 a term of the A.P. 10, 7, 4….. ?
Answer:
a =10, d = 7 – 10 = -3
T_n = a + (n- 1) ⇒ -300= 10 + (n – 1) (-3)
⇒ -300 = 10 – 3n + 3 ⇒ 3n = 10 + 3 + 300
∴ 3n = 313
∴ n = \(\frac{313}{3}\) ∉ N
∴ n being number of terms cannot be a fraction.
∴ -300 is not a term of the A.P

Question 18.
If k + 9, k – 6, 4 are in GP then find the value of K.
Answer:
(k – 6)² = (k + 9)4
k² – 12k + 36 = 4k + 36 ⇒ k² – 12k – 4k + 36 – 36 = 0
k² – 16k =0 ⇒ k(k – 16) = 0
k =0, k= 16

Question 19.
If a and p are the roots of the equation x² – x + 2 = 0 then show that α²β + β²α = 2
Answer:
α + β = \(\frac{-b}{a}=\frac{-(-1)}{1}\) = 1 and αβ = \(\frac{c}{a}=\frac{2}{1}\) = 2
α²β + αβ² = αβ(α + β) = 1(2) = 2

Question 20.
Sove 5x – 3 < 3x + 1 when x is an integer and x is a real number
Answer:
5x – 3 < 3x + 1
2x <4

5x – 3x < 3 + 1
⇒ x < 2

1. x = {…………..-2, -1,0, 1} x ∈ Z
2. x ∈ (-h, 2) x ∈ R

Question 21.
Sowmya invested ₹ 1500 for 8 years and Anisha invested ₹ 7500 for 3 years at the same rate of interest. If altogether they received ₹ 1725 as interest find the rate of simple interest charged.
Answer:
Sowmya,
P₁ = 1500, T₁ = 8, R₁ = R
I₁ = \(\frac{1500 \times 8 \times \mathrm{R}}{100}\) = 120R

Anisha,
P₂ = 7500 T₂ = 3 R, = R
I₂ = \(\frac{7500 \times 3 \times R}{100}\) = 225R
I₁ + I₂ = 1725
120 R + 225 R = 1725
345 R = 1725 ⇒ R = \(\frac{1725}{345}\) = 5%

Question 22.
If the cost price of a machine is ₹ 150 and selling price is ₹ 100 find the loss percentage?
Answer:
CP = ₹ 150 . SP = ₹ 100
CP > SP
∴ loss
loss = CP – SP = 150- 100 = ₹ 50
loss% = \(\frac{\text { loss }}{\mathrm{CP}}\) × 100 = \(\frac{50}{150}\) × 100 = 33.3%

Question 23.
The average score of 20 boys is 60% and the average score of 30 girls is 70%. Find the combined average of boys and girls?
Answer:
X₁ = 60%, N₁ = 20; X₂ = 70%, N₂ = 30
X̄₁₂ = \(\frac{\mathrm{X}_{1} \mathrm{~N}_{1}+\mathrm{X}_{2} \mathrm{~N}_{2}}{\mathrm{~N}_{1}+\mathrm{N}_{2}}=\frac{(60)(20)+(70)(30)}{20+30}=\frac{3300}{50}\) = 66%

Question 24.
Derive the equation of the line in one point form. i.e. y – y₁ = m (x – x₁) geometrically, where m is the slope and p(x₁ y₁) is the given point
Answer:

Slope of AP = m = tan θ = \(\frac{\mathrm{PL}}{\mathrm{AL}}=\frac{\mathrm{PN}-\mathrm{NL}}{\mathrm{MN}}=\frac{\mathrm{PN}-\mathrm{AM}}{\mathrm{ON}-\mathrm{OM}}=\frac{y-y_{1}}{x-x_{1}}\)

∴ y – y₁ = m(x – x₁)

Question 25.
Find the value of x if the distance between (x, 3) and (4, 5) is “5 units”.
Answer:
A = (x, 3), B = (4, 5)
AB =5
\(\sqrt{(x-4)^{2}+(3-5)^{2}}\) =5
(x- 4)² + 4= 25
x² – 8x + 16 + 4 – 25 =0
x² – 8x – 5 = 0
x = \(\frac{-(-8) \pm \sqrt{64-4(1)(-5)}}{2(1)}=\frac{8 \pm \sqrt{64+20}}{2}=\frac{8 \pm \sqrt{84}}{2}=\frac{8 \pm 2 \sqrt{21}}{2}=\frac{2(4 \pm \sqrt{21})}{2}\)
x = 4 ± \(\sqrt{21}\)

PART-C

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

Question 26.
Prove the \(\sqrt{2}\) is an irrational number. Sol. We shall prove it by contradiction.
Answer:
We shall prove it by contradiction
If possible, Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\) 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.
\(\sqrt{2}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\) ⇒ \(\sqrt{2}\)q = p ,
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p ⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² q² 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.
Define an equivalence relation with an example. Also, give an example of a relationship that is only symmetric.
Answer:
A relation R on a non-empty set A is called an equivalence relation if it is reflexive, symmetric, and transitive.
Example: On a set of straight lines in a plane the relation “parallel to” is an equivalence relation. Let A= {1, 2, 3} R = {(1, 1), (3, 3), (2, 1), (1, 2)}
R is symmetric but not reflexive and not transitive.

Question 28.
If a^x = b^y = c^z and b² = ac. Show that \(\frac{1}{x}+\frac{1}{x}=\frac{2}{y}\)
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’*
Now, b² = ac
∴ (k ^1/y)² = ,k^1/x. k^1/z
∴ k^2/y = k^{1/x + 1/z}.
Bases are same
∴ Equating powers on both the sides, we get.
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\)

Question 29.
Find the number of zeroes in (0.2)100 after the decimal point and the first significant figure.
Answer:
log (0.2)¹⁰⁰ = 100 log 0.2= 100 (T. 0.3010)= 100(-1 + 0.3010) = – 100 + 30.10
= -100 + 30 + 0.10 = -70 + 0.10 = 70.10
characteristic = 70
∴ number of zeroes = 70 – 1 = 69.

Question 30.
Find the sum of the following series : 7 + 77 + 777 + 7777 +……… n terms.
Answer:
Let S_n = 7 + 77 + 777…………..n terms
= 7 [1 + 11+ 111 +………..to n terms]
= \(\frac{7}{9}\) [9 + 99 + 999 + …………..to n terms]
= \(\frac{7}{9}\) [(10- 1) + (100 – 1) + (1000 – 1)…………..to n terms]
= \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]\) [∵ 10 + 10² + 10³ ……… to n terms is in G.P
where a = 10, r = 10]

S_n = \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-\mathrm{n}\right]\)

Question 31.
A father is 28 years older than a son. After 5 years the father’s age will be 7 years more than twice that of the son. Find their present ages.
Answer:
Let father’s age be x years and son’s age be y years.
Given: x – y = 28……………….(1)
After five years father’s age is x + 5 and son’s age is y + 5.
Also given: x + 5 = 2(y + 5) + 7
x + 5 = 2y + 10 + 7
Solving (1) and (2), we get: x = 44 and y = 16
Thus father’s age is 44 years and the son’s age is 16.

Question 32.
Solve the linear inequalities graphically: x +3y ≥ 3, 2x + y ≥ 2, x ≥ 10, y ≥ 0
Answer:

Question 33.
In what time a sum of ₹ 500 will earn ₹ 975 at the rate pf 6% p.a if the compound interest is payable half yearly?
Answer:
R = 6% = \(\frac{6}{100}\) = 0.06
q = 4
r = \(\left(1+\frac{R}{q}\right)^{q}\) – 1 = 0.0607
A = P(1 + r)ⁿ
log A= log P + n log (1 + r)
n = \(\frac{\log A-\log P}{\log (1+r)}=\frac{\log 975-\log 500}{\log (1+0.0607)}=\frac{2.9890-2.6990}{0.0256}\) = 11.33 years

Question 34.
Insert 4 arithmetic means between 5 and 10.
Answer:
Let A₁, A₂, A₃, A₄ be the 4 A.M’s between 5 and 10
∴ 5, A₁, A₂, A₃, A₄, 10 are in A.P,
Last element is 6th element a + 5d = 10
5 + 5d =10
5d = 10-5
5d = 5
∴ A. M’s are 6,7, 8, 9

Question 35.
By how many percent should the use of tea be increased if the price of tea is decreased by 10% so that the expenditure remains unchanged.
Answer:
Let Price be 100 Let Quantity be 100
Total expenditure = 100 × 100 = 10,000
Then New Price = 100 – 10 = 90
Let New Qty = y
Total expenditure = 90 y
90 y = 10,000
Increase = 111.11 – 100 = 11.11
∴ % increase = \(\frac{\text { increase }}{\text { Original value }}\) × 100 = \(\frac{11.11}{100}\) × 100 = 11.11%

Question 36.
If tan A = \(\frac{12}{13}\) and 180 < A < 270. Find the value of \(\frac{3 \sin A-2 \cos A}{9 \cos A+4 \sin A}\)
Answer:
OP² = OM² + MP² = 169 + 144 = 313
OP = \(\sqrt{313}\)
SinA = \(\frac{-12}{\sqrt{313}}\)
CosA = \(\frac{-13}{\sqrt{313}}\)

Question 37.
Show that the straight lines 2x – 3y = 7,3x – 4y =13, 8x – 11y = 33 are concurrent. Also find the point of concurrency. .
Answer:
2x – 3y = 7 × 3
3x – 4y = 7 × 2

y = 5
2x-3y = 7 ⇒ 2x – 3(5) = 7 ⇒ 2x – 15 = 7 ⇒ 2x = 22 ⇒ x = 11
Put x =11 y = 5 in 8x- 11y = 33
8 (11) – 11 (5) = 88 – 55 =33 .
∴ The three lines are concurrent. Point of concurrency = (11, 5)

Question 38.
Find the distance between two parallel lines 3x + 4y + 5 = 0 and 6x + 8y + 20 = 0
Answer:
3x + 4y + 5 = 0 × 2
6x + 8y + 10 = 0 ……(1)
6x + 8y + 20 = 0 ………(2)

PART – D

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

Question 39.
In a group of 150 people, 70 like Cricket, 30 like hockey and cricket both. How many like Cricket only and not hockey? How many like hockey? Show the result using the Venn diagram.
Answer:
n(C∪A)=150 n(C) = 70 n(H) = ? n(C∩H) = 30
n(C∪H) = n(C) + n(H) – n(C∩H)
150 = 70 + n(H) – 30
150 = 40 + n(H) [∵ n(H) = 110]
n(C – H) = n(C) – n(C∩H) = 70 – 30 = 40 .

Question 40.
Evaluate using log, tables \(\frac{42.15 \times(0.2713)^{2}}{0.8932}\)
Answer:
Let x = \(\frac{42.15(0.2713)^{2}}{0.8932}\)
log x = log 42.15+ 21og 0.2713-log 0.8932
= 1.6247+ 2(1.433)-(). 9509)= 1.6247+ 2(-l + 0.433)-(1.9509)
= 1.6247-2 +0.866-(-1 + 0.9509)= 1.6247-2+ 0.866-(-1 + 0.9509)
= 1.6247-2 + 0:866+ 1 -0,9509
log x = 0.5398 =AL [0.5398] = 3.4733
∴ x = 3.4733

Question 41.
Find the sum of the series : 4 + 44 + 444 + ….. n terms,
Answer:
Let S_n = 4 + 44 + 444 + n terms
S_n = 4 (1 + 11 + 111 + ……….)
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{4}\) = 1 + 11 + 111 + ……
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{4}\) = 9 + 99 + 999
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{4}\) = (10 – 1) + (102 – 1) + (103 – 1) + ………
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{4}\) = (10 + 102 + 103 + ………) – (1 + 1 + 1 + …………….)
\(\frac{9 S_{n}}{4}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n
\(\frac{9 S_{n}}{4}=\frac{10\left(10^{n}-1\right)}{9}\) – n
S_n = \(\frac{4}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-n\right]\), n∈N

Question 42.
The cost of 2 kgs of sugar and 5 bags of dhal is ₹ 90. The cost of 5 kgs of sugar and 2 kgs of dhal is ₹ 120. Find the cost of sugar and dhal per kg.
Answer:
Let the cost of 1kg of sugar = x and cost of 1 kg of dhal = y
2x + 5y = 90 × 2
5x + 2y – 120 × 5

2x + 5 = 90
2(20) + 5y = 90
5y = 90 – 40 = 50
y = 10

∴ cost of 1 kg of sugar = ₹ 20
cost of 1 kg of dhal = ₹ 10

Question 43.
Govindappa bought 51 bags in the wholesale market at an average price of ₹ 318 each. The price of 33 leather bags was ₹ 426 each. The price of the remaining cotton bags forms an increasing Arithmetic progression. If the price of the costliest cotton bag was ₹ 150/ – find the price of the cheapest cotton bag.
Answer:
Average price of 11 bags = Rs.450
Total price of 11 bages = 11 × 450 = 4950
Average price of 7 leather bags = Rs. 575
Total price of 7 leather bags = 575 × 7 = 4025
The price of the remaining 4 cotton bags from an increasing A.P, Whose first term be a and common difference d,
1 = 300, n = 4, Sn = 925.
Sum of remaining 4 bags = 4950 – 4025 = 925
Sn =925
\(\frac{\mathrm{n}}{2}\) (a + 1) = 925
\(\frac{4}{2}\) (a + 300) = 925
2 (a + 300) = 925
2a + 600 = 925
2a = 325
a = \(\frac{325}{2}\) =162.50
Price of the cheapest cotton bag = Rs. 162.50

Question 44.
The daily cost of production ‘C is 1₹ and ‘x’ unit of an assembly is C(x) = 3.5x + 1200. If each unit is sold for 6 then find the minimum number of units that should be produced and sold to ensure no loss. If the selling price is increased by half a rupee a unit then what would be the Break-Even Point.
Answer:
C(x) = 3.5x + 1200
R(x) = C(x) for a break even point
R(x) = 6x
6x = 3.5x + 1200
6x – 3.5x = 1200
2.5x = 1200
x = \(\frac{1200}{2.5}\) = 480 units

If the S.P is increased by ₹ 0.5 a unit then
R(x) = 6.5x
R(x) = C(x)
6.5x = 3.5x + 1200
6.5x – 3.5x = 1200
3x = 1200
x = 400 units

Question 45.
Prove that sec²\(\frac{5 \pi}{4}\) cosec²\(\frac{5 \pi}{4}\) – sin²\(\frac{3 \pi}{4}\)cos²\(\frac{4 \pi}{3}\) = 31/8
Answer:
sec\(\frac{5 \pi}{4}\) = sec 225 = sec(180+45°) = Sec45° = – \(\sqrt{2}\)
cosec\(\frac{5 \pi}{4}\) = coscc 225° = cosec (180 + 45) = cosec 45° = – \(\sqrt{2}\)
sin \(\frac{3 \pi}{4}\) = sin 135°= sin(180 – 45°)sin45° = \(\frac{1}{\sqrt{2}}\)
cos \(\frac{4 \pi}{3}\) = cos 240 = cos (180 + 60) = – cos 60 = –\(\frac{1}{2}\)
∴ LHS = sec²\(\frac{5 \pi}{4}\) cosec²\(\frac{5 \pi}{4}\) – sin²\(\frac{3 \pi}{4}\) cos²\(\frac{4 \pi}{3}\)

= \((\sqrt{2})^{2} \cdot(\sqrt{2})^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2} \cdot\left(\frac{-1}{2}\right)^{2}\)
= 2.2 – \(\frac{1}{2}\).\(\frac{1}{4}\) = 4 – \(\frac{1}{8}\) = \(\frac{32-1}{8}=\frac{31}{8}\) = R.H.S

Question 46.
Derive the section formula for the internal division if A (x₁, y₁) and B (x₂, y₂) are the two given points and P(x, y) be the point that divides the join of AB in the ratio m:n Sol. Let X’ OX and YOY’ be the axes.
Answer:
Let P (x₁, y₁), Q (x₂, y₂) be the given points?
Let R (x, y) be the point that divides PQ internally in the ratio m:n.

The Δles PRT and RQS are similar
∴ \(\frac{\mathrm{PT}}{\mathrm{RS}}=\frac{\mathrm{RT}}{\mathrm{QS}}=\frac{\mathrm{PR}}{\mathrm{RQ}}\)
\(\frac{\mathrm{PT}}{\mathrm{RS}}=\frac{\mathrm{PR}}{\mathrm{RQ}}\)
⇒ \(\frac{\mathrm{PT}}{\mathrm{RS}}=\frac{\mathrm{m}}{\mathrm{n}}\) (from (1))
⇒ \(\frac{\mathrm{ML}}{\mathrm{LN}}=\frac{\mathrm{m}}{\mathrm{n}}\)
⇒ \(\frac{\mathrm{OL}-\mathrm{OM}}{\mathrm{ON}-\mathrm{OL}}=\frac{\mathrm{m}}{\mathrm{n}}\)
\(\frac{x-x_{1}}{x_{2}-x}=\frac{m}{n}\)

n (x – x₁) = m (x₂ – x)
⇒ nx – nx₁ = mx₂ – mx
⇒ mx – nx = mx₂ – nx₁
⇒ (m + n) x = mx₂ – nx₁
⇒ x = \(\frac{m x_{2}+n x_{1}}{m+n}\)
\(\frac{\mathrm{RT}}{\mathrm{QS}}=\frac{\mathrm{PR}}{\mathrm{RQ}}\)
⇒ \(\frac{\mathrm{RT}}{\mathrm{QS}}=\frac{\mathrm{m}}{\mathrm{n}}\)
⇒ \(\frac{\mathrm{RL}-\mathrm{TL}}{\mathrm{QN}-\mathrm{SN}}=\frac{\mathrm{m}}{\mathrm{n}}\)
⇒ \(\frac{\mathrm{RL}-\mathrm{PM}}{\mathrm{QN}-\mathrm{RL}}=\frac{\mathrm{m}}{\mathrm{n}}\)
⇒ \(\frac{y-y_{1}}{y_{2}-y}=\frac{m}{n}\)
⇒ ny- ny₁ = my₂ – my
⇒ my – ny = my₂ – ny₁
⇒ (m – n)y = my₂ – ny₁
⇒ y = \(\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\)
The coordinates of R are
(x, y) = \(\left(\frac{m y_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
This is called the section formula for the internal point of division.

Question 47.
Find the equation of the locus of the point which moves such that the ratio of its distances (2, -3) and (4, -2) is 2 : 3.
Answer:
A = (2,-3), B = (4,-2)
Let P(x, y) be any point on the locus
\(\frac{\mathrm{PA}}{\mathrm{PB}}=\frac{2}{3}\) ⇒ 3PA = 2PB

3\(\sqrt{(x-2)^{2}+(y+3)^{2}}\) = 2\(\sqrt{(x-4)^{2}+(y+2)^{2}}\)
9[(x – 2)² + (y + 3)²] = 4[(x – 4)² + (y + 2)²]
9[x² – 4x + 4 + y² + 6y + 9] = 4[x² – 8x + 16 + y² + 4y + 4]
9[x² – y² – 4x + 6y + 13] = 4[x² + y² – 8x + 4y + 20]
9x² + 9y² – 36x + 54y + 117 = 4x² + 4y² – 32x + 16y + 80
9x² – 4x² + 9y² – 4y² — 36x + 32x + 54y – 16y + 117 – 80 = 0
5x² + 5y² – 4x + 38y + 37 = 0

Question 48.
Find the equation of the line which passes through the point of intersection of the lines x – 2y + 4 = O and 4x – 3y + 1 = 0 and is inclined at an angle of 135° with the x-axis.
Answer:
x – 2y + 4 = 0 × 1
4x – 3y + 1 = 0 × 1

5y = 15
y = 3
x – 2y + 4 = 0
x – 2 = 0

x – 6 + 4=O
x = 2

∴ The point of intersection = (2, 3)
θ = 135°
m = tanθ = tan 135 = tan(180 – 45) = -tan45°= -1

∴ Equation of the required line ¡s
y – y₁ = m(x – x₁) ⇒ y – 3 = -1(x – 2)
y – 3 = -x + 2
x + y – 5 = 0

PART – E

IV. Answer any One Question. (1 × 10 = 10)

Question 49.
(a) Find the domain and Range of the function.
f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x-12}\) where x ∈ R
(b) What is the future value of 1000 deposited annually years for 12 gathering compound interest at 16%?
(c) Form the cubic equation whose roots are 3, 5 and 7
Answer:
(a) f(1) = \(\frac{4}{-19}=\frac{4}{19}\)
f(3) = \(\frac{9+6+1}{9-24+12}\)
= \(\frac{16}{-3}\)
f(5) = \(\frac{5^{2}+10+1}{5^{2}-8(5)+12}\)
= \(\frac{36}{-3}\) = – 12
Domain = { 1, 3, 5 ………….}
Range = {-4/19, -16/3, -12 …………..}

(b) A = 1000, n = 12, r = 0.16
F = \(\frac{\mathrm{A}\left[(1+r)^{n}-\right]}{r}=\frac{100\left[(1+0.16)^{12}-1\right]}{0.16}=\frac{1000\left[(1.16)^{12}-1\right]}{0.16}\) = 30.850.16

(c) (x – 3)(x – 5)(x – 7) =0 ⇒ (x – 3)(x² – 5x – 7x + 35) =0
⇒ (x – 3)(x² – 12x + 35) 0 ⇒ x³ – 12x² + 35x – 3x² + 36x – 105 =0
x³ – 15x² + 71x – 105 = 0

Question 50.
(a) Find the value of x
If \(\frac{x \cdot \sin ^{2} 300 \cdot \sec ^{2} 240}{\cos ^{2} 225 \cdot {cosec}^{2} 240}\) = cot2 315°. tan 300
(b) Find the equation of the line perpendicular to 3x – 2y + 1 = O and passing through the point (1, -2)
(c) Insert two geometric means between 1 and 1//27
Answer:
sin 300 = sin (360 – 60) = – sin 60 = \(\frac{\sqrt{3}}{2}\)
sec 240 = sec(180 + 60)= – sec60°= – 2
cos 225 = cos(180 + 45°) – cos45°= – \(\frac{1}{\sqrt{2}}\)
cosec 240 = cosec (180 + 60°) = – cosec 60 = – \(\frac{2}{\sqrt{3}}\)
cot 315° = cot (360 – 45°) = – cot 45° = – 1
tan 300 = tan (360 – 60°) = – tan 600 = – \(\sqrt{3}\)

∴ The given equation becomes:
\(\frac{x \cdot \frac{3}{4} \cdot 4}{\frac{1}{2}, \frac{4}{3}}\) = 1.3
⇒ x = \(\frac{6}{9}=\frac{2}{3}\)

(b) Slope of the given line 3x – 2y + 1 = 0 is \(\frac{3}{2}\) = m1
Since the two lines are perpendicular m₁ m₂ = – 1
∴ m₂ = \(\frac{-1}{m_{1}}=\frac{-1}{3 / 2}=\frac{-2}{3}\)

Thus the equation of the ime vith slope m = \(\) and passing through
(x₁ y₁) (1, -2) is given by
⇒ y -y₁ = m(x – x₁)
⇒ y – 2 = (x – 1)
⇒ 3y – 6 = – 2x + 2
⇒ 2x + 3y + 4 = 0

(c) 1, g, g₂, \(\frac{1}{27}\)
a = 1, T_n = \(\frac{1}{27}\), r =?, n = 4
T_n = ar^n-1 ⇒ \(\frac{1}{27}\) = 1.r³
⇒ \(\left(\frac{1}{3}\right)^{3}\) = r³
⇒ r = \(\frac{1}{3}\)
∴ 1, 1/3, 1/9, 1/21
∴ Geometric means are 1/3, 1/9

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तुलसीदास के दोहे Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिएः

Tulsidas Ke Dohe Class 11 Notes KSEEB Solutions प्रश्न 1.
तुलसीदास किस पर विश्वास करते हैं?
उत्तर:
तुलसीदास भगवान राम (रघुनाथ) पर विश्वास करते हैं।

Tulsidas Ke Dohe Notes KSEEB Solutions प्रश्न 2.
तुलसीदास किसको आराध्य देव मानते हैं?
उत्तर:
तुलसीदास श्रीराम जी को आराध्य देव मानते हैं।

Tulsi Ke Dohe Class 11 Notes KSEEB Solutions प्रश्न 3.
संत का स्वभाव कैसा होता है?
उत्तर:
संत का स्वभाव आम के पेड़ की तरह होता है।

Tulsidas Ke Dohe Class 11 KSEEB Solutions प्रश्न 4.
तुलसीदास काया और मन की उपमा किससे देते हैं?
उत्तर:
तुलसीदास काया को खेत और मन को किसान की उपमा देते हैं।

Tulsidas Ke Dohe 1st Puc KSEEB Solutions प्रश्न 5.
मधुर वचन से क्या मिटता है?
उत्तर:
मधुर वचन से अभिमान मिट जाता है।

Tulsi Ke Dohe Notes KSEEB Solutions प्रश्न 6.
पंडित और मूर्ख एक समान कब लगते हैं?
उत्तर:
काम, क्रोध, मद और लोभ को मन में धारण करने पर पंडित और मूर्ख एक समान लगते हैं।

Tulsidas Ke Dohe Question Answer KSEEB Solutions प्रश्न 7.
तुलसीदास कहाँ जाने के लिए मना करते हैं?
उत्तर:
तुलसीदास कहते हैं कि जहाँ आदर और स्नेह नहीं मिलता हो वहाँ नहीं जाना चाहिए।

Tulsidas Notes KSEEB Solutions प्रश्न 8.
बिना तेज के पुरुष की अवस्था कैसी होती है?
उत्तर:
बिना तेज के पुरुष की अवस्था अवज्ञा अथवा आग के बुझने पर राख की तरह उपेक्षित होती है।

अतिरिक्त प्रश्नः

Tulsidas Kisko Aaradhya KSEEB Solutions प्रश्न 9.
तुलसीदास के अनुसार जग से लोग किस संख्या की तरह नाता रखना चाहिए?
उत्तर:
तुलसीदास के अनुसार जग से लोग छत्तीस की तरह नाता रखना चाहिए।

प्रश्न 10.
सुअंब तरू दूसरों के हित के लिए क्या देता है?
उत्तर:
सुअंब तरू दूसरों के हित के लिए फल देता है।

प्रश्न 11.
तुलसीदास के अनुसार कौन से दो बीज हैं?
उत्तर:
तुलसीदास के अनुसार पाप और पुण्य दो बीज हैं।

प्रश्न 12.
दूध के उफान को किससे मिटाया जाता है?
उत्तर:
दूध के उफान को शीतल जल से मिटाया जाता है।

प्रश्न 13.
इस संसार में किस प्रकार के लोग हैं?
उत्तर:
इस संसार में भाँति-भाँति के लोग हैं।

प्रश्न 14.
तुलसीदास किसका त्याग न करने के लिए कहते हैं?
उत्तर:
तुलसीदास अपने कुल की रीति त्याग न करने के लिए कहते हैं।

प्रश्न 15.
ब्याह और प्रीति किसके साथ करनी चाहिए?
उत्तर:
ब्याह और प्रीति उसी से करनी चाहिए जो लायक हो।

प्रश्न 16.
किसकी अवज्ञा अवश्य होती है?
उत्तर:
बिना तेज के पुरुष की अवज्ञा अवश्य होती है।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
तुलसीदास की रामभक्ति का वर्णन कीजिए।
उत्तर:
तुलसीदास जी कहते हैं कि मुझे तो केवल मेरे इष्ट राम पर भरोसा है, उसी की आशा, उसी का बल है; जैसे स्वाति नक्षत्र में बारिश के पानी की पहली बूंद के लिए चातक पक्षी तरसता रहता है। अर्थात् मेरे राम यदि स्वाति-सलिल हैं, तो मैं चातक हूँ।

प्रश्न 2.
तुलसीदास के अनुसार संत के स्वभाव का वर्णन कीजिए।
उत्तर:
तुलसीदास जी के अनुसार संत का स्वभाव फलदार आम के वृक्ष की तरह होना चाहिए। आम का वृक्ष दूसरों की भलाई के लिए ही फलता और फूलता है। मनुष्य आम को पाने के लिए पेड़ को पत्थर मारता है, बदले में पेड़ मनुष्य को फल (आम) देता है। उसी प्रकार संतो को चाहिए कि वे लोक निंदा कि परवाह न कर समाज को सुधारने के कार्य में लगे रहें।

प्रश्न 3.
तुलसीदास ने मधुर वचन के महत्व का कैसे वर्णन किया है?
उत्तर:
मधुर वचन के बारे में तुलसीदास जी कहते हैं कि मीठा बोलने से हमारे मन में रहनेवाला अहंकार मिट जाता है। ठीक वैसे ही, जैसे कि उफनते हुए दूध पर थोड़ा-सा ही शीतल (ठंडा) जल छिड़कने से उफनता हुआ दूध नीचे बैठ जाता है।

प्रश्न 4.
तुलसीदास के अनुसार मनुष्य के जीवन में संतोष-धन का क्या महत्व हैं?
उत्तर:
दुनिया में कई प्रकार के धन हैं। जैसे – गोधन, गजधन, वाजिधन और रतनधन आदि। इन सबके रहते हुए भी अथवा इनको प्राप्त करने पर भी मनुष्य को संतुष्टि नहीं होती। तुलसीदास जी कहते हैं कि जब सन्तोष रूपी धन हमें मिल जाता है, तो ये सभी प्रकार के धन धूल के समान हो जाते हैं। अर्थात् ‘सन्तोष’ ही सबसे बड़ा धन है।

प्रश्न 5.
तुलसीदास कुल रीति के पालन करने के सम्बन्ध में क्या कहते हैं?
उत्तर:
तुलसीदास जी कहते हैं कि हमें अपने कुल के रीति-रिवाज अथवा कुल की परंपरा का कभी त्याग नहीं करना चाहिए। जो लायक हो उसी से ब्याह कीजिए, जो लायक हो उसी से बैर कीजिए और जो लायक हो, उसी से प्रीति या प्रेम कीजिए।

अतिरिक्त प्रश्नः

प्रश्न 6.
तुलसीदास ने छत्तीस अंक को लेकर अपना कौन सा मत प्रकट किया है?
उत्तर:
तुलसीदास जी कहते हैं कि मनुष्य को इस संसार अर्थात् सांसारिक मोह माया से 36 का रिश्ता रखना चाहिए। जिस प्रकार छत्तीस में 3 और 6 के अंक एक दूसरे के विपरीत होते हैं उसी तरह हमें भी सांसारिकता चीजों के विपरीत रहना चाहिए।

प्रश्न 7.
“जैसा कर्म वैसा फल” इस बात को तुलसीदास ने कैसे स्पष्ट किया है?
उत्तर:
तुलसी ‘जैसा कर्म वैसा फल’ को एक दृष्टांत के माध्यम से समझाते हुए कहते हैं- हमारा शरीर एक खेत है और मन किसान है तथा पाप व पुण्य दोनों बीज हैं। इस प्रकार खेत में जिस प्रकार का बीज बोकर खेती होती है, वैसी ही फसल भी प्राप्त होती है। अर्थात् जीवन में जैसा काम करेंगे हमें फल भी उसके अनुरूप मिलेगा।

प्रश्न 8.
तुलसीदास ने पंडित और मूर्ख की अलग पहचान कैसे बताई है?
उत्तर:
तुलसी कहते हैं कि जब तक मनुष्य काम (वासना), क्रोध, मद, लोभ से ग्रसित रहता है तब तक वह विद्वान होकर भी मूर्ख के समान है। इसलिए मनुष्य को काम, क्रोध, मद तथा लोभ का शीघ्र ही त्याग कर देना चाहिए।

प्रश्न 9.
तुलसीदास ने स्वाभिमान की रक्षा और स्नेह स्वागत के बारे में क्या कहा है?
उत्तर:
तुलसीदास कहते हैं कि तेजहीन मनुष्य का इस संसार में सदैव ही अपमान होता है। अतः हर एक मनुष्य को अपना तेज अर्थात स्वाभिमान को बचाएं रखना चाहिए। तेजवान व्यक्ति का कोई भी अपमान नहीं कर सकता।

प्रश्न 10.
तुलसीदास ने बुझी हुई आग का उदाहरण देकर क्या समझाया है?
उत्तर:
तुलसी बुझी हुई आग का उदाहरण देकर कहते हैं कि आग के बुझ जाने पर, राख को न केवल छूना आसान हो जाता है, बल्कि उसे लोग कुचल भी देते हैं। उसी तरह तेजहीन मनुष्य को भी लोग अपमानित करते हैं। उसे कुचल देते हैं।

III. ससंदर्भ भाव स्पष्ट कीजिए:

प्रश्न 1.
तुलसी संत सुअंब तरु, फूलि फलहिं पर-हेत।
इतते ये पाहन हनत, उतते वे फल देत॥
उत्तर:
प्रसंग : प्रस्तुत दोहा हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘तुलसीदास के दोहे’ नामक संग्रह से लिया गया है। इसके रचयिता तुलसीदास हैं।
संदर्भ : यहाँ तुलसीदास संत लोगों की विशेषता पर प्रकाश डाल रहे हैं।
स्पष्टीकरण : तुलसीदास कहते हैं कि संत पुरुष उस आम के पेड़ की तरह होते हैं जो हमेशा दूसरों को मीठे फल देता रहता है। बदले में वह कुछ भी नहीं लेता। वह तो दूसरों का उपकार करने के लिए ही फलता-फूलता है। लोग धरती से पत्थर भी फेंकते हैं तो भी वह उन्हें फल ही देता है। संत पुरुष भी ठीक वैसे ही होते है। उनका कितना भी अपमान किया जाए, वे विनम्रता नहीं त्यागते हैं। क्रोध नहीं करते हैं। उनके जीवन का उद्देश्य दूसरों का हित करना ही होता है। एक जगह तुलसी यह भी कहते हैं कि ‘परहित सरिस धर्म नहिं भाई’ अर्थात् दूसरों की भलाई से बड़ा और कोई धर्म नहीं हो सकता। संत पुरुषों के लिए परहित ही धर्म होता है।
विशेष : अवधी भाषा का प्रयोग। लोक-कल्याण की महत्ता पर प्रकाश डाला गया है।

प्रश्न 2.
तुलसी काया खेत है, मनसा भयौ किसान।
पाप पुण्य दोउ बीज हैं, बुवै सौ लुनै निदान॥
उत्तर:
प्रसंग : प्रस्तुत दोहा हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘तुलसीदास के दोहे’ नामक संग्रह से लिया गया है। इसके रचयिता तुलसीदास हैं।
संदर्भ : यहाँ तुलसीदास जी कहते हैं कि हम जैसे कर्म करते हैं, हमें वैसा ही फल मिलता है, इसलिए हमें हमेशा अच्छे कर्म करने चाहिए।
स्पष्टीकरण : कवि कहते हैं कि हमारा शरीर खेत है और हमारा मन किसान है। पाप और पुण्य ये दो प्रकार के बीज हैं। अब हम इनमें से जो बीज बोएँगे, हमें वैसा ही फल मिलेगा।
अतः अच्छा या बुरा फल पाना हमारे ही हाथ में है।

प्रश्न 3.
तुलसी एहि संसार में, भाँति भाँति के लोग।
सब सो हिल मिल बोलिए, नदी नाव संयोग।।
उत्तर:
प्रसंग : प्रस्तुत दोहा हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘तुलसीदास के दोहे’ नामक संग्रह से लिया गया है। इसके रचयिता तुलसीदास हैं।
संदर्भ : तुलसीदास कहते हैं कि इस जगत में भिन्न भिन्न विचारों के लोग रहते हैं। हमें उनसे मेल मिलाप के साथ रहना चाहिए।
स्पष्टीकरण : तुलसीदास कहते हैं कि इस संसार में भिन्न-भिन्न विचारों को, धर्मों को मानने वाले लोग रहते है। हमें सबके साथ सहिष्णुता के साथ रहना चाहिए। हमें उनके विचारों, मान्यताओं का सम्मान करना चाहिए। सबसे प्रेमपूर्वक बोलना चाहिए। भिन्न विचारों के साथ समन्वय जरूरी होता है।
विशेष : अवधी भाषा और ब्रज भाषा का प्रयोग। भिन्न विचारों के बीच समन्वय पर जोर दिया गया है।

प्रश्न 4.
काम, क्रोध, मद, लोभकी, जौ लौं मन में खान।
तौं लौ पंडित मूरखौ, तुलसी एक समान।।
उत्तर:
प्रसंग : प्रस्तुत दोहा हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘तुलसीदास के दोहे’ नामक संग्रह से लिया गया है। इसके रचयिता तुलसीदास हैं।
संदर्भ : तुलसीदास जी कहते है कि व्यक्ति चाहें जितना बड़ा विद्वान क्यों न हो, जब तक उसके पास अहंकार और लोभ जैसे संस्कार उसमें विद्यमान रहते हैं तब तक वह मूर्ख के समान ही होता है।
स्पष्टीकरण : तुलसीदास कहतें हैं- जब तक हमारे मनरूपी खजाने में काम-क्रोधा आदी अरिषड्वर्ग स्थिर रहते हैं तब तक मूर्ख और पंड़ित एक समान होते हैं। एक साधारण मनुष्य जो उपरोक्त मनोविकारों से बचकर रहता है, वही महान होता है। नहीं तो बड़ा पंडित भी मूर्ख बन जाता है।

तुलसीदास के दोहे कवि परिचयः

महाकवि तुलसीदासजी का जन्म सन् 1532 में हुआ था। आप रामभक्ति शाखा के सर्वश्रेष्ठ रचनाकार माने जाते हैं। आपके पिता का नाम आत्माराम दुबे और माता का नाम हुलसी था। आपके गुरु बाबा नरहरिदास थे। आपके आराध्य देव श्री रामचन्द्र थे और लोकमंगल भावना आपके काव्य का सिद्धांत रहा। आपने अन्यान्य काव्यशैलियों का प्रयोग करते हुए कई रचनाएँ हिन्दी साहित्य जगत को समर्पित की। आपकी मृत्यु सन् 1623 में हुई। ‘रामचरित मानस’, ‘विनय पत्रिका’, ‘कवितावली’, ‘गीतावली’, ‘जानकी मंगल’ और ‘दोहावली’ आपकी सर्वश्रेष्ठ रचनाएँ मानी जाती है। आपका अवधी और ब्रजभाषा पर समान अधिकार था।
प्रस्तुत दोहों में रामभक्ति, नीति, सदाचार, विनय, दान एवं जीवन के समस्त पक्षों पर प्रकाश डाला गया हैं।

दोहे का भावार्थः

1) एक भरोसो, एक बल, एक आस, विश्वास।
स्वाति-सलिल रघुनाथ-जस, चातक तुलसीदास ॥१॥

तुलसीदास जी कहते हैं कि मुझे अपने इष्ट श्रीराम पर पूरा भरोसा है। एक ही बल, एक ही आशा और एक उन्हीं पर अटूट विश्वास है; ठीक वैसे ही जैसे कि भगवान श्री रघुनाथजी यदि स्वाति नक्षत्र का जल है, तो तुलसीदास चातक पक्षी की तरह है।

The poet, Tulasidasji, says that he has complete faith in his beloved Lord Rama. He has only one source of strength, only one desire, and only one unbreakable faith in his master Lord Rama. He says that if Lord Rama is akin to rain (of the swati nakshatra), then he is like a ‘chatak’ bird (a kind of cuckoo which is supposed to drink only drops of rain).

शब्दार्थ :

• भरोसो – भरोसा
• स्वाति-सलिल – स्वाति वर्षा का जल;
• जस – कीर्ति;
• चातक – चातक पक्षी।

2) जग ते रह छत्तीस है, राम चरन छ: तीन।
तुलसी देखु विचार हिय, है यह मतौ प्रबीन ॥२॥

तुलसीदास जी कहते हैं कि मनुष्य को इस संसार से अर्थात् सांसारिक प्रपंच से छत्तीस (३६) जैसा सम्बन्ध रखना चाहिए और परमात्मा के साथ तिरसठ (६३) का सम्बन्ध रखना चाहिए। इसका अभिप्राय यह है कि संसार में रहते हुए भी मनुष्य को संसार के प्रति उदासीनता दिखानी चाहिए और परमात्मा के प्रति हमेशा प्रेम प्रकट करना चाहिए।

Tulasidasji says that man must have a relationship of ’36’ with the world, i.e. with the worldly universe. With the almighty, however, man must establish a relationship of ’63’. What this means is that, although living in a materialistic world like ours, one must show indifference towards the world and towards worldly pleasures. Towards the almighty, however, one must always show love.

शब्दार्थ :

• छत्तीस – 36;
• छः तीन – 63;
• हिय – हृदय;
• मतौ – अभिप्राय, राय;
• प्रबीन – प्रवीण, दक्ष।

3) तुलसी संत सुअंब तरु, फूलि फलहिं पर-हेत।
इतते ये पाहन हनत, उतते वे फल देत ॥३॥

तुलसीदास जी कह रहे हैं कि संत पुरुष उस आम के वृक्ष की भाँति होते हैं। जो आम का वृक्ष दूसरों का उपकार करने के लिए ही फलता-फूलता है। लोग धरती से उसकी ओर पत्थर फेंकते हैं, तो वह आम का फल उन्हें प्रदान करता है। उसी प्रकार संतों का चाहे जितना ही अपमान क्यों न किया जाय, पर वे हम लोगों का उपकार ही करते हैं।

Tulasidasji says that saints and godly men are like a mango tree. They are like the mango tree which flowers and bears fruits only for the sake of others. From the ground, when people throw stones at the mango tree, the tree showers them with fruit. Similarly, no matter how much a saint or godly man is insulted on this earth, he will always bless and wish well for the person who insults.

शब्दार्थ :

• सुअंब तरु – फल से भरा हुआ आम का वृक्ष;
• पर-हेत – दूसरों के हित के लिए;
• इतते – जितने;
• हनत – मारने से;
• उतते – उतने ही;
• पाहन – पत्थर।

4) तुलसी काया खेत है, मनसा भयौ किसान।
पाप पुण्य दोउ बीज हैं, बुवै सौ लुनै निदान ॥४॥

तुलसीदास जी कहते हैं कि हमारा यह शरीर एक खेत है और मन किसान है तथा पाप व पुण्य दोनों बीज हैं। इस प्रकार खेत में जिस प्रकार का बीज बो कर खेती होती है, वैसी फसल प्राप्त होती है। उसी प्रकार शरीर रूपी खेत में मन रूपी किसान पाप-पुण्य दोनों के बीज डालता है। इनमें से जो भी बीज किसान बोता है, उसी की फसल भी काटता है। अर्थात् मनुष्य का मन ही पाप व पुण्य दोनों में से किसी एक की ओर आकर्षित होता है। हम जैसा बोयेंगे, वैसा ही फल पायेंगे अर्थात् अच्छे कर्मों का फल अच्छा और बुरे कर्मों का फल बुरा मिलेगा।

Tulasidasji says that our body is like a farm, whereas our mind is like a farmer and sin and virtue are like seeds. Thus, whatever type of seed is sown in the farm, the produce of the farm will be of that quality. Similarly, in our body (which is like a farm), our mind (the farmer) sows seeds of both sin and virtue. Whichever seeds the mind nourishes and nurtures will grow and yield produce. What the poet is implying is that the human mind is attracted to either sin or virtue. Whichever one our mind is attracted to, we have to reap the results of that quality. Good deeds always produce good results and bad deeds always produce bad results.

शब्दार्थ :

• काया – शरीर;
• भयौ – होता है;
• दोऊ – दोनों;
• निदान – अंत में

5) मधुर बचन तें जात मिटै, उत्तम जन अभिमान|
तनक सीत जलसों मिटै, जैसे दूध उफान ॥५॥

तुलसीदास जी कहते हैं कि मधुर वचनों से या मीठी वाणी से अच्छे लोगों का अभिमान भी मिट जाता है। जैसे कि उफनते हुए गरम दूध में थोड़ा-सा ही शीतल (ठंडा) जल डाल देंगे, तो वह शांत हो जाएगा अर्थात् उफनेगा नहीं। इस प्रकार मधुर वचन का लाभ है।

Tulasidasji says that through a soulful song or a sweet voice, the pride (ego) of even good persons is easily erased. It is similar to the act of adding cool water to boiling milk. When we add cool water to boiling milk, the milk which was bubbling and frothing, suddenly becomes calm. Such is the benefit of a touching song or a sweet voice.

शब्दार्थ :

• जात मिटै – मिट जाता है;
• अभिमान – अहंकार, गर्व;
• तनक – थोड़ा
• सीत जलसो – ठंडा पानी से;
• उफान – उबराना

6) गो-धन, गज-धन, वाजि-धन और रतन-धन खान।
जब आवत संतोष-धन, सब धन धूरि समान ॥६॥

तुलसीदास जी कहते हैं कि मनुष्य के पास भले ही गौ रूपी धन हो, गज (हाथी) रूपी धन हो, वाजि (घोड़ा) रूपी धन हो और रत्न रूपी धन का भंडार हो, वह कभी संतुष्ट नहीं हो सकता। जब उसके पास सन्तोष रूपी धन आ जाता है, तो बाकी सभी धन उसके लिए धूल या मिट्टी के बराबर है। अर्थात् सन्तोष ही सबसे बड़ी सम्पत्ति है।

Tulasidasji says that even if a man has a wealth of corps, or a wealth of elephants, or a wealth of horses or even a wealth of gems and diamonds, he will never be satisfied. When a man comes to possess wealth in the form of satisfaction, then all other forms of wealth are just like dust or sand, i.e., they appear to be worthless.This means that satisfaction and happiness are the.. greatest wealth in the world.

शब्दार्थ :

• वाजि – घोड़ा;
• धूरि – धूल।

7) काम, क्रोध, मद, लोभकी, जौ लौं मन में खान।
तौं लौ पंडित मूरखौ, तुलसी एक समान ॥७॥

तुलसीदास जी कहते हैं कि जब तक मनुष्य के मन में काम (वासना), क्रोध, घमंड, लोभ आदि विकार उत्पन्न होते रहते हैं, तब तक वह विद्वान होते हुए भी मूर्ख के समान है। इसलिए मनुष्य को काम, क्रोध, मद तथा लोभ का शीघ्र ही त्याग कर देना चाहिए।

Tulasidasji says that as long as a person’s mind is under the influence of lust (fancy), anger, arrogance, greed and other such bad qualities, till then even if he is a learned person, he is equal to a fool. Thus, the poet warns us that we must stay away from, or immediately abandon qualities like lust, anger, intoxication etc.

शब्दार्थ :

• जौ लौं – जब तक;
• तौं लौ – तब तक।

8) आवत ही हर्षे नहीं, नैनन नहीं सनेह।
तुलसी तहाँ न जाइए, कञ्चन बरसे मेह ॥८॥

तुलसीदास जी कहते हैं कि यदि किसी के घर जाने पर वह प्रसन्न नहीं होता और उसके नेत्रों में स्नेह नहीं झलकता, तो उसके घर में कभी नहीं जाना चाहिए, भले ही वहाँ सोने की वर्षा क्यों न होती हो। अर्थात् कोई व्यक्ति कितना ही बड़ा अमीर क्यों न हो, उसके घर जाने पर अतिथि सत्कार नहीं होता हो, तो वहाँ नहीं जाना चाहिए।

Tulasidasji says that upon visiting someone’s house, if they are not happy to see us, or if we cannot see love reflected in their eyes, then we must never enter that house even if gold is showered in that house. This means that no matter how rich a person is, if we are not welcome in their house, or if guests are not honoured in their house, then we must never go there.

शब्दार्थ :

• नैनन – आँखों में, नयनों में;
• आवत ही – आते समय;
• कञ्चन – सोना;
• मेह – मेघ, बादल

9) तुलसी कबहुँ न त्यागिए, अपने कुल की रीति।
लायक ही सो कीजिए, ब्याह, बैर अरु प्रीति ॥९॥

तुलसीदास जी कहते हैं कि अपने कुल की रीति को या परंपरा को कभी भी त्यागना नहीं चाहिए। कहा गया है कि जो योग्य हो अथवा लायक हो उसी से विवाह रचाया जाना चाहिए। बैर भी सोच-समझ कर जो उसके लायक हो उसी से करना चाहिए तथा लायक से ही प्रीति करनी चाहिए।

Tulasidasji says that we must never abandon the ways or traditions of our ancestors. It has been said that one must only marry a person who deserves or is worth marrying. Even as our enemies, we must only choose those who are worthy of us. Similarly, one must also love a person who is deserving or worthy of our love.

शब्दार्थ :

• कबहुँ – कभी;
• लायक – उपयुक्त, ठीक

10) बिना तेज के पुरुष की, अवशि अवज्ञा होय।
आगि बुझे ज्यों राखको, आपु छुवै सब कोय ॥१०॥

तुलसीदास जी कहते हैं कि तेजहीन मनुष्य का संसार में निश्चय ही अपमान होता है। अतः हर एक मनुष्य को अपना तेज, स्वाभिमान को बचाए रखना चाहिए। जिस प्रकार अग्नि के बुझ जाने पर, राख को न केवल स्पर्श करना सरल हो जाता है, अपितु उसे तो लोग रौंद भी देते हैं। वही जलती हुई आग का स्पर्श करना सहज नहीं होता है। उसी प्रकार तेजहीन मनुष्य का अपमान करना सरल है, परन्तु तेजयुक्त व्यक्ति का अपमान कोई नहीं कर सकता।

Tulasidasji says that in this world it is certain that a person without knowledge and education will be insulted. It becomes quite easy to touch the ashes when the fire is extinguished. In fact, the ashes are also trampled upon by people. The same fire, before it is extinguished, is not so easy to touch. In the same manner, an uneducated person is easy to insult, but no one can insult an enlightened person.

शब्दार्थ :

• तेज – तेजस्, स्वाभिमान, कीर्ति;
• अवशि – अवश्य;
• अवज्ञा – अनादर, अपमान।

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