Category: 1st PUC

Students can Download Basic Maths Chapter 10 Averages Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 10 Averages

Question 1.
A student asked to find the arithmetic mean of the numbers 4, 19, 17, 12, 34,16, 7, – 23,18, 21, 25, and x. Find the mean table 18. What should be the number in place of x.
Answer:

18 = \(\frac{196+x}{2}\) = 216 = 196 + x
∴ x = 216 – 196 = 20

Question 2.
The average of 30 results is 20 and the average of other 20 results is 30. Find the average of results taken to gather.
Answer:
Given n₁ = 30, x̄₁ = 20, n₂ = 20 , x̄₂ = 30

Question 3.
If the average of daily wages of workers of two factories is Rs. 53 and average wages to factory ‘A’ with 250 employees is Rs. 50. Find the average wage of factoring ‘B’ with 200 employees.
Answer:
Given X̄ = 53, X̄_A = 50, n_A = 250, X̄_B = 7, n_B = 200

i.e., 200 X̄_B = 23850 – 12500 = 11,350 ∴ X̄_B = \(\frac{11,350}{200}\) = 56/75
∴ Average wage of factory B = Rs. 56.75

Question 4.
Ten years ago the average age of the family of 4 members was 24 years. Two children have been born. The average age of the family is same to day. Find the present age of two children assuming that the children’s age differ by 2 years.
Answer:
10 year ago x = 24, n = 4 ∴ x = 24 × 4 = 96 year
If 10 years ago, total age of 4 member = 96 year
total age of member now = (96 + 10 × 4) year = 136 year
Total age of 6 member = 74 × 6 = 144 year
i.e., Sum of age of 2 children = 144 – 136 = 8 year
so x + x + 2 = 8 = x = 3 year
∴ The ages of children 3 and 4 year.

Students can Download Basic Maths Chapter 9 Annuities Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 9 Annuities

Question 1.
A company needs Rs. 10,00,000 at the end of 5 years from now. It would like to set aside an equal amount each year out of its profits. If the present marker rate of interest is 16%, how much should be annual amount to be invested?
Answer:
It is under future value of annuity immediate
Given future value of annuity (F) = Rs. 10,00,000
Number of years = 5, r = 16%
∴ We have to find the value of annuity

⇒ 10,00,000 × 0.16 = A[(1.16)⁵ – 1]
1,60,000 = A[2.0990 – 1]
1.099 A = 1,60,000
∴ A = \(\frac{160000}{1.099}\) = 1,45,586.90
Calculate (1.16)⁵
(5 log to 6 = A^L (5 × 0.00644)
= A^L)0.322) = 2.990
The company set aside Rs. 1,45,586.90 each year to get Rs. 10,00,000 at the end of 5 years.

Question 2.
You have taken a loan of Rs. 10,000 from a finance company. Under the agreement you have to repay the loan in 5 equal instalments commencing from the end of year. If the finance company charge you interest of 18% p.a. What is the value of instalment? Also show that at the end of 5 years your loan account will be closed.
Answer:
It is under present value of annuity immediate Rs. 10,000 be the instalment which include principal and interest. Since the instalments are = equal.
(a) To find the instalment ‘A’, we have

(b) Loan repayment statement

Question 3.
Bank of India offers 12% p.a. interest on 12 months R.D a/c’s. If v deposit Rs. 1000 p.a. for 12 months how much can I get at the end of 12 months, assuming that each deposit is to be made of the begining of the month?
Answer:
It is under future value of annuity due deposits are monthly.

on simplifying we get = Rs. 12,809.30 total amount I get after 12 months.

Question 4.
A person working on salesman bought motor cycle under a hire purchase scheme. Where in he has to pay Rs. 1500 per month for 40 months. Each payment at payable at the begining of the month. If the dealer has charge interest at 18% p.m. What is cash price of the motor cycle.
Answer:
It is under present value annuity due cash price of the motor cycle is also known as the present value of the instlament payable.
r = \(\frac { 0.18 }{ 12 }\) = 0.15 n = 40 months A = Rs.1500

= Rs. 45,535.00 is the cash price of the motor cycle

Students can Download Basic Maths Chapter 8 Simple Interest, Compound Interest and Annuities Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 8 Simple Interest, and Compound Interest

Question 1.
Find simple Interest on Rs 7,300 from 15 may 2009 to 8^th October 2009 at 10% p.a.
Answer:
Given P = Rs.7,300, R = 10%
Time = May + June + July + Aug + Sept. + Oct.
= 16 + 30 + 31 + 31 + 30 + 8 = l46days.

Question 2.
Determine the principal which will amount to Rs. 15,000 in 4 years at 8% p.a.
Answer:

Question 3.
In how many years will a sum of money double itself at 18.75% p.a simple interest.
Answer:
Let the principal be Rs. P, then amount = Rs. 2P
∴ SI = Amount – PrincipIe = Rs. 2P – P = P; R = 18.75% p.a

Question 4.
Kumar borrowed some many at the rate of 6% p.a for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond live years. If he pays a total interest of Rs. 11,400 at the end of nine years, how much money did he borrow.
Answer:
Let the sum borrowed be x. then

Hence sum borrowed = Rs. 12,000

Compound Interest

Question 1.
Find the CI amount and Cl on Rs. 12,000 for 3 years at 10% p.a. compounded annually.
Answer:
P = 12,000, n = 3,R = 10%p.a
We have amount A

∴ C.I = A.P = 15972 – 12000 = Rs. 3972

Question 2.
At what percent p.a. will a sum of Rs. 5,000 become Rs. 8,000 if the loan given for 4 years attract compound interest?
Answer:
Given A = Rs. 8000, P = Rs. 5000, n = 4, R = ?

Question 3.
Compute the CI. on Rs. 12,000 for 2 years at 20% p.a. when compounded half yearly.
Answer:
Given P = 12,000, R = 20% p.a. and n = 2

= 12000\(\left(\frac{11}{10}\right)^{4}\) = 12000 (1.4641) = 17569.20
A = Rs. 17,569.20
∴ Compound Interest = A – P = 17,569.20 – 12,000 = Rs. 5,569.20

Question 4.
A sum of money doubles itself at compound an Interest in 15 years. In how many years will it become eight times.
Answer:
Let the sum of the money be Rs. P is interested at the rate of R% p.a. It is given that the moñey double it self in 15 years

∴ the money will become 8 times in 45 years.

Question 5.
The population of a town in increasing at the rate of 5% p.a. What will be the population of the two as thus basic after two years if the present population is 20,0000.
Answer:
Given initial population = P = 20,000, R = 5%, n = 2
We have poplulation after 2 years = \(\mathrm{p}\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)

Question 6.
In a factory the production of cars rose to 36,300 from 30,000 in 2 years. Find the rate of growth p.a.
Answer:
Let the rate of growth be R% p.a. present production A = 36500, previa production P = 30,000, n = 2

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Karnataka 1st PUC Basic Maths Question Bank Chapter 7 Linear Inequalities

Question 1.
Solve graphically 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Answer:
3x + 4y = 60

x + 3y + 30

(0, 0) satisfies 3x + 4y ≤ 60 and x + 3 ≤ 30

Question 2.
Solve 2 (2x + 3) – 10 £ 6(x – 2)
Answer:
Given 2 (2x + 3) – 10 £ 6(x – 2)
⇒ 4x + 6 – 10 ≤ 6x – 12 ⇒ 4x – 4 ≤ 6x – 12
⇒ 4x – 6x ≤ – 12 + 4 ⇒ -2x ≤ -8
⇒ x ≥ \(\frac{8}{2}\) ⇒ x ∈ [4, ∞] is the solution

Question 3.
Solve the equation \(\left|\frac{2}{x-4}\right|>1\),  x ≠ 4.
Answer:
we have \(\left|\frac{2}{x-4}\right|>1\),  x ≠ 4
⇒ \(\frac{2}{|x-4|}>1\) ⇒ 2 > |x – 4|
⇒ 4 – 2 < x < 4 + 2
⇒ 2 < x < 6. ∴ x e (2, 6) But x 4

Question 4.
Find all pairs of consecutive even positive integers both of which are larger than 8, such that their sum is less than 25.
Answer:
Let x be the smaller of the two consecutive even positive integers, then the other even integer is x + 2.
Given x > 8 and x + (x + 2) < 25.
⇒ x > 8, and 2x + 2 < 25.
⇒ x > 8, 2x < 23 ⇒ x > 8, x < \(\frac{23}{2}\)
⇒ 8 < x < ⇒ \(\frac{23}{2}\) x = 10,
∴ the required parity even integers is (10, 12)

Question 5.
In the first four papers each of 100 marks, Ravi got 95, 72, 73, 83 marks. If he wants an average of greater than or equal to 75 marks and less than 80 marks, find the range of marks he should score in the fifth paper.
Answer:
Let score be x in the fifth paper, then

Hence Ravi must score between 52 and 77 marks.

Solve and represent the following in equalities graphically 

Question 1.
x + y ≥ 4 : 2x – y > 0
Answer:

Question 2.
x + y ≤ 9, y > x, x ≥ 0
Answer:

Question 3.
2x – y > 1, x – 2y < – 1
Answer:

Question 4.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Answer:

Question 5.
2x + y ≥ 4, x + y ≤ 3, 2x – y ≤ 6.
Answer:

Question 6.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Answer:

Question 7.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, y ≥ 0
Answer:

Question 8.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Answer:

Question 9.
Solve the in equalities and represent the solutions graphically on number line.

Answer:
5(2x – 7) -3 (2x + 3) ≤ 0; 2x + 19 ≤ 6x + 47

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Karnataka 1st PUC Basic Maths Question Bank Chapter 6 Theory of Equations

Question 1.
Find the nature of roots of the equation 3x² – 4x + 9 = 0
Answer:
Nature of roots depends on b² – 4ac
⇒ a = 3, b = – 4, c = 9.
∴ b² – 4ac = (-4)² – 4 . (3) . 9 = 16 – 108 = – 92 < 0
∴ The roots are imaginary.

Question 2.
If a and B are the roots of 4x² + 3x – 5 = 0. Find the value of \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
a = 4, b = 3, c = – 5
Sum of roots = α + β = \(\frac{-b}{a}=\frac{-3}{4}\)
Product of roots = αβ = \(\frac{c}{a}=\frac{-5}{4}\)

Question 3.
Find three consecutive even natural numbers, such that the sum of their squares is 200.
Answer:
Let the required even numbers be x, x + 2, x + 4
By data x² + (x + 2)² + (x + 4)² = 200
⇒ x² + x² + 4x + 4 + x² + 8x + 16 – 200 = 0
3x² + 12x – 180 = 0 ⇒ x² + 4x – 60 = 0
⇒ (x + 10)(x – 6) =0
we require +ve the integers ∴ x = 6
⇒ 6,8, 10 be the numbers.

Question 4.
A piece of cloth costs Rs. 200. If the piece were 5mts longer and each meter of cloth costed Rs. 2 less, the cost of piece would have remained unchanged. How long is the piece and what is its original rate per meter.
Answer:
Let x be the length of the cloth, its cost = 200
∴ Rate of cloth per meter = \(\frac{200}{x}\)
Increased length of piece of cloth = (x + 5) mts
Now rate of cloth per meter = \(\frac{200}{21+5}\)
By date, old rate cloth – new rate of cloth = Rs 2.



⇒ x² + 5x – 500 = 0
⇒ x² + 25x – 20x – 500 = 0
⇒ x(x + 25) – 20(x + 25) = 0
⇒ x = 20 or -25 (neglected) [∴ x = 20mts]
length of cloth = 20 mts, rate per metre = Rs. 10.

Question 5.
Solve : x³ – 6x² + 11x – 6 = 0. since that ratio of two roots is 2 : 3
Answer:
By data α : β = 2 : 3
⇒ \(\frac{\alpha}{\beta}=\frac{2}{3}\) = α = 2β
from the given equation,

⇒ 5α³ – 12α² + 8 = 0
solving α = 2, ∴ α = 1
∴ The roots are 1,2,3

Question 6.
Solve by synthetic division that it has atleast one integral root between -3 and 3 for the equation x⁴ – 9x² + 4x + 12 = 0
Answer:
Let f(x) = x⁴ – 9X² + 4x + 12 = 0
Bu inspection x = ±1, ±2, ±3 are the roots or not
f(-3) = 0, f(-1) = 0, f(2) = 0
∴ -3, -1, 2 are the roots of the given equation
and
q
∴ x = – 3 is a root and x = – 2 is not a root
Try for x = – 1, x = -1 is a root
Now the quotient is x² – 4x + 4 = 0
⇒ (x – 2)² = 0 = x = 2, 2
Hence the form root are -3, -1,2, 2

Students can Download Basic Maths Chapter 5 Progressions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 5 Progressions

Arithmetic Progression (A.P)

Question 1.
Which term of A.P 4\(\left(5 \frac{1}{3}-4\right), 6 \frac{2}{3} \ldots \ldots\) is 104?
Answer:
In given A.P: a = 4, d = 5^1/3 – 4 = \(\frac { 16 }{ 3 }\) – 4 = \(\frac { 4 }{ 3 }\)
Let n^th term be 104 = T_n = a + (n – 1) d
⇒ 4 + (n – 1) \(\frac { 4 }{ 3 }\) = 104
⇒ 12 + 4n – 4 = 312 ⇒ 4n – 4 = 312
⇒ 4n = 312 – 8
⇒ 4n = 304
∴ n = 76

Question 2.
How many terms of series 24 + 20 + 16 + ……. must be taken so that their sum is 72.
Answer:
In the given A.P; a = 24, d = 4, S_n = 72
we have S_n = \(\frac { n }{ 2 }\){2a + (n – 1)d}
72 = \(\frac { n }{ 2 }\){2(24) + (n – 1)(-4)}
= 144 = n{48 – 4n + 4}
⇒ 144 = n{52 – 4n}
⇒ 144 = 52n – 4n²
⇒ 4n² – 52n + 144 = 0
⇒ n² – 13n + 36 = 0
⇒ (n – 9)(n – 4) = 0
∴ n = 9,4

Question 3.
The first and the last term of an A.P are -4 and 146 and the sum of the A.P is 7171. Find the number of terms in the A.P and the A.P.
Answer:
Given a = -4, l = 146.
We have l = a + (n – 1 )d = 146 = -4 + (n – 1) d
⇒ (n – 1)d = 150
Also we have Sn = \(\frac { n }{ 2 }\){2a + (n – 1)d}
7171 = \(\frac { n }{ 2 }\){2(-4) + 150}
14342 = n{-8 + 150}
⇒ 142n = 14342
∴ n = \(\frac{14342}{142}\) = 101
So number of terms is A.P = 101
consider (n – 1)d = 150.

Question 4.
Find the sum of all numbers between 100 and 1000 which are divisible by 11.
Answer:
The 1^st greatest No. greater than 100 is 110. the last which is less than 1000 is, 990.
110,121, ………990 be the number
T_n = 90 ⇒ 110 + (n – 1) 11 = 990
11n – 11 = 990 – 110
11n = 880 + 11 = 891
∴ n = \(\frac { 891 }{ 11 }\) = 81
Now S₈₁ = 8\(\frac { 1 }{ 2 }\)(110 + 990) = \(\frac { 81 }{ 2 }\)(550) = 4450

Question 5.
The first, second and the last terms of an A.P are a, b, c respectively prove that the sum is \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)
Answer:
1^st term is a, common difference = d = b – a.
Let ‘n’ be the number of term in the series.
∴ c = a + (n – 1)d
⇒ (n – 1)d = c – a
⇒ (n – 1 )(b – a) = c – a
⇒ n – 1 = \(\frac{c-a}{b-a}\)
⇒ n = \(\frac{c-a}{b-a}\) + 1
= \(\frac{n}{2}\){2a + c – a} = \(\frac{n}{2}\){a + c} = \(\left(\frac{b+c-2 a}{b-a}\right)\left(\frac{a+c}{2}\right)\)

Geometric Progression (G.P)

Question 1.
Find the 9^th term and sum of 6 term of the geometric sequence \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27} \ldots \ldots\)
Answer:
a = \(\frac{1}{3}\) r = \(\frac{1}{3}\)

Question 2.
The second and fifth term of GP are 3 and \(\frac{81}{8}\) respectively. Find the GP
Answer:
Given T₂ = ar = 3, T₅ = ar⁴ = \(\frac{81}{8}\)

∴ r = \(\sqrt[3]{\frac{33}{23}}=\frac{3}{2}\)
If ar = 3 ⇒ a. \(\frac{3}{2}\) = 3 ⇒ a = 2
G.p is 2, 3, \(\frac{9}{2}\), …….

Question 3.
The sum of three numbers which are in G.P is \(\frac{39}{10}\) and their product is 1. Find the numbers.
Answers:
Let the number be \(\frac{a}{2}\), a, ar. which are in G.P.
i.e., \(\frac{a}{2}\) + a + ar = \(\frac{39}{10}\)

⇒ a³ = 1
∴ a = 1
put a = 1 in (1),
\(\frac{1}{r}\) + 1 + r = \(\frac{39}{10}\)
⇒ \(\frac{1+r+r^{2}}{r}=\frac{39}{10}\)
⇒ 10 + 10r + 10r³ = 39
r ⇒ 10r² – 29r + 10 = 0
⇒ r = \(\frac{5}{2}\) and \(\frac{2}{5}\)

Question 4.
Find three numbers in GP. Whose sum is 35 and sum of whose squares is 525.
Answer:
Let the three numbers be a, ar, ar² which are is GP.
Given a + ar + ar² = 35 ⇒ a( 1 + r + r²) = 35
and a² + a²r² + a²r⁴ = 525
⇒ a²( 1 + r² + r⁴) = 525
\(\frac{a^{2}\left(1+r^{2}+r^{4}\right)}{a\left(1+r+r^{2}\right)}=\frac{525}{35}=15\)
a(1 + r² + r⁴) = 15(1 + r + r²)

⇒ a(1 – r + r²) = 15 …….. (2)
Add (1) and (2)
2a + 2ar² = 50
all (2) – (1) ⇒ 2ar = -20

\(\frac{1}{r}\) + r = \(\frac{-5}{2}\)
⇒ 2(1 + r²) = -5r
⇒ 2r² + 5r + 2 = 0
⇒ r = \(\frac{-5 \pm \sqrt{25-4(2)(2)}}{4}\)
⇒ \(\frac{-5 \pm 3}{4}=\frac{-1}{2},-2\)
when r = -2 2ar = -20 ⇒ -4a = -20 [∴ a = 5]
when r = \(-\frac{1}{2}\) , 2ar = -20 ⇒ -a = -20 ⇒ [a = 20]
Hence the three number which are in G.P are 5, -10, 20 (or) 20, – 10, 5

Question 5.
Find the sum of n terms of 5 + 55 + 555 + ……..
Answer:
Let s = 5 + 55 + 555 + …….. n terms
⇒ s = 5[1 + 11 + 111 + …… to term
= \(\frac{5}{9}\)(9 + 99 + 999 + ….. to n term) \(\frac{5}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) + …… to terms]
= \(\frac{5}{9}\)[10 + 10² + 10³ + …… to n terms] + (1 + 1 + ……. to n terms)

Harmonic Progression (H.P)

Question 1.
If the 12^th element of H.P is \(\frac { 1 }{ 5 }\) and the 19th term is \(\frac { 3 }{ 32 }\). Find the 10^th term of H.P.
Answer:
12^th term = \(\frac { 1 }{ 15 }\)
19^th term = \(\frac{3}{2^{2}}\)
∴ 12^th term, AP is 5 and
19^th termof A.P is \(\frac { 22 }{ 3 }\)

Question 2.
If 1, x, 3 are in H.P . Find x.
Answer:
Given 1, x, 3 are in H.P

Question 3.
If the m^th term of H.P is ‘n’ and n^th term is m, prove that (m + n)^th is \(\frac{m n}{m+n}\)
Answer:
Given m^th term of H.P is n,
∴ m^th term of A.P = \(\frac{1}{n}\) = a + (m – 1) d
Also n^th term of HP is m.
∴ n^th term of A.P = \(\frac{1}{m}\) = a + (n – 1)d
a + (m – 1)d = \(\frac{1}{n}\) …… (1)
a + (n – 1)d = \(\frac{1}{m}\) …… (2)
Solving (1) and (2)

Arithmetic, Geometric and Hormonic Means.

Question 1.
Find six geometrical mean between 27 and \(\frac { 1 }{ 81 }\)
Answer:
Let the required GM is g₁, g₂, g₃, g₄, g₅, g₆
= 27, g₁, g₂, g₃, g₄, g₅, g₆, \(\frac{1}{81}\) are in GP
Now a = 27 and \(\frac{1}{81}\) = 8^th element
⇒ \(\frac{1}{81}\) = ar⁷ ⇒ \(\frac{1}{81}\) = 27.r⁷
⇒ \(\frac{1}{81 \times 27}\) = r⁷
⇒ r⁷ = \(\frac{1}{3^{7}}\)
⇒ r = \(\frac{1}{3}\)
∴ the six GM are 9, 3, 1,\(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\)

Question 2.
If a, b, c are in A.P and a, mb, c are in GP then show that a, m²b, c are in H.P
Answer:
a, b, c, are in AP .
⇒ b = \(\frac{a+c}{2}\)
a, mb, c are in GP,
⇒ mb = \(\sqrt{a c}\)
⇒ m²b² = ac
⇒ m²b. b. = ac
⇒ m²b \(\left(\frac{a+c}{2}\right)\) = ac \(\left(\because b=\frac{a+c}{2}\right)\)
⇒ m²b = \(\frac{2 a c}{a+c}\)
⇒ a, m²b, c are in H.P.

Summation of Series

Question 1.
1² + 3² + 5² + ………  to ’n’ terms.
Answer:
1, 3, 5, ………. are in A.P
T_n = a + (n – 1)d = 1 + (n – 1)2 = 1 + 2x – 2 = 2x – 1
∴ n^th term = (2n – 1)²
T_n = (2x – 1)² = 4n² – 4x + 1
required sum = S_n = ΣT_n = Σ4n² – 4x + 1 = 4Σn² – 4Σn + Σ1.
\(=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n\)
= \(\frac{n}{3}\) [2(n + 1)(2n + 1) – 6(n + 1) + 3] = \(\frac{n}{3}\)[4n² – 1]

Question 2.
Find the sum to ‘n’ terms the series 1.2 + 4.5 + 7.8 + ….. to term.
Answer:
1st factors are 1, 4, 7,
∴ n^thterm = 1 + (n – 1) 3 = 3n – 2
2nd factor are 2, 5, 8,
∴ n^th term = 2 + (n – 1) 3 = 3n – 1
⇒ n^th term of the series is (3n – 2)(3n – 1)
∴ The sum = Σ(3n – 2)(3n – 1)
= Σ(9x² – 9x + 2) = 9Σn² -9Σn + 2Σ1
\(=9 \frac{n(n+1)(2 n+1)}{6}-9 \frac{n(n+1)}{2}+2 n\)
= \(\frac{1}{2}\)n(n + 1)[3(2n + 1) – 9] + 2n = n(3n² – 1).

Question 3.
Sum the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots \ldots .16 \text { terms }\) 16terms
Answer:

Students can Download Basic Maths Chapter 4 Logarithms Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 4 Logarithms

Logarithms

Question 1.
\(\log _{5} \frac{\sqrt[4]{25}}{625}\)
Answer:

Question 2.
Find x if log^x_0.4 = -3
Answer:

Question 3.
If x = 27, b = log⁴₃ then flow that x^b = 64
Answer:
x^b = 27log⁴₃
= 3^3log4₃ = 3^log(4)3₃
= 3^{log64₃ = 64}

Question 4.
If log₁₀\(\sqrt{1+x}+3\) log₁₀ \(\sqrt{1-x}\) = log₁₀ \(\sqrt{1-x^{2}}+2\) . Find x.
Answer:
log₁₀ (x + 1)^1/2 + log₁₀(1 – x)^3/2 = log₁₀(1 – x²)^1/2 + log¹⁰⁰₁₀
⇒ (x + 1)^1/2 (1 – x)^3/2 = (1 – x²)^1/2 – 100
⇒ \(\sqrt{1-x^{2}}\) (1 – x – 100) = 0
⇒ x = ±1. x = -99 But x ≠ ± 1, -99
∴ There is no value of x exists.

Question 5.
If x = log^a_2a, y = log^2a_3a, z = log^3a_4a. Show that xyz + 1 = 2yz.
Answer:

= 2log^2a_3a . log^3a_4a = 2yz = RHS

Question 6.
If a, b, c are GP. Prove that logⁿ_a logⁿ₆ logⁿ_c are in HP.
Answer:
Given, a, b, c are in GP = b² = ac
Taking log on both sides
2 log 6 = log a + log c
⇒ log^a_n log^b_n log^c_n are inn AP
⇒ log^a_n log^b_n log^c_n are in HP.

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Karnataka 1st PUC Basic Maths Question Bank Chapter 3 Theory of Indices

Question 1.
3(5^-2) + \(\left(\frac{5}{3}\right)^{-4}\) + (a + 5)⁰
Answer:

Question 2.
Show that \(\left(\frac{x^{b}}{x^{c}}\right)^{a}-\left(\frac{x^{c}}{x^{a}}\right)^{b} \cdot\left(\frac{x^{a}}{x^{b}}\right)^{c}=1\)
Answer:
LHS = (x^b-c)^a (x^c-a)³ (x^a-b)^c
= x^ab-bc .x^bc-ab .x^ac-bc
= x^{ab-bc+bc+-ab+ac-bc}
= x⁰
= 1
= RHS

Question 3.
If pqr = 1 Show that \(\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1\)
Answer:

Question 4.
If a^x = b^y = and b² = 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, b^y = k, C^z = k
⇒ a = k^1/x; b = k^1/y; c = k^1/z
By data b² = ac
(k^1/y)² = k^1/x .k^1/z
k^2/y = k^1/x+1/z
⇒ \(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\) (∵bases are same)

Question 5.
If 65^1/x = 13^1/y = 5^1/z show that x=y+z.
Answer:
Let 65^1/x ⇒ k = 65 k; 13^1/y = k ⇒ 13 = k; 5^1/z = k ⇒ 5 = k
We know that 65 = 13 × 5 k^x = k^y. k^z ⇒ x = y + z

Students can Download Basic Maths Chapter 2 Sets, Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions

sets

Question 1.
If A = {1, 2, 3, B = {2, 3, 4}, C = {3, 4, 5, 6} and U = {1, 2, 3, 4, 5, 6, 7}. Find
(i) (A ∪ B)’
(ii) Verity (A ∩ B)’ = A’ ∪ B’
Answer:
(i) A ∪ B = {1,2, 3, 4} ∴ (A ∪ B)’ = ∪ – (A ∪ B) = {5, 6, 7}
(ii) (A ∩ B) = {2,3}
(A ∩ B)’ = ∪ – (A ∩ B) = {1,4, 5,6} …… (1)
⇒ A’ = ∪ – A = {4, 5, 6, 7}
B’ = ∪ – B = {1, 5, 6, 7}
∴ A’ ∪ B’ = (1,4, 5, 6, 7} ……. (2)
From (1) and (2) (A ∩ B)’ = A’ ∪ B’

Question 2.
n(u) = 700, n(A) = 200, n(B).= 300. n(A ∩ B) = 100. Findn(A’ ∩ B’)
Answer:
n(A’ ∩ B’) = n(A ∪ B)’
= n(U) – n(A ∪ B) = 700 – n(A ∪ B)
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 200 + 300 – 100 = 400
∴ n(A’ ∩ B’) = 700 – 400 = 300

Question 3.
In a town three daily news papers X, Y, Z are published. 52% of the people read paper X, 61 % read Y and 78% read Z. 40% read X and Y; 38% read Y and Z, 46% read X and Z; 18% do not read any of the three news papers. Find the percentage of persons who read all the three papers.
Answer:
Let the number of person is the town be 100.
Now it is given n(X) = 52, n(Y) = 61, n(Z) = 78, n(X ∩ Y) = 40, n(Y ∩ Z) = 38
n(X ∩ Z) = 46, n(X ∪ Y ∪ Z) = 100 – 18 = 82
We have
n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) – n(X ∩ Y) – n(Y ∩ Z) – n(X ∩ Z) + n(X ∩ Y ∩ Z).
⇒ 82 = 52 + 61 + 78 – 40 – 38 – 46 + n(X ∩ Y ∩ Z)
= 191 – 124 + n(X ∩ Y ∩ Z) = 67 + (X ∩ Y ∩ Z)
∴ n(X ∩ Y ∩ Z) = 82 – 67 = 15
Here 15% of the people read all the three papers.

Question 4.
How many integers between 1 and 500 are divisible by 2, 3, or 5.
Answer:
\(\frac { 500 }{ 2 }\) = 250 No. of integer divisible by 2 is 250.
∴ n(A) = 250
\(\frac { 500 }{ 3 }\) = 16666 ∴ No. of integer divisible by 3 is 166. i.e., n(B) = 166
\(\frac { 500 }{ 5 }\) = 100 i.e., No. of integer divisibility 5 is 100 is n(C) = 100
Now \(\frac{500}{2 \times 3}=\frac{500}{6}\) = 83.33 divisible 2 and 3 is 83.
n(A ∩ B) = 83.

We have
n(A ∩ B ∩ C) = n(A) + n(B) + n(c) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 250 + 166 + 100 – 83 – 33 – 50 + 17 = 533 – 166 = 367

Question 5.
If A, B and C are three subsets of the universal set U, draw the venn diagram of (A’ ∩ B’)∩C’
Answer:

(A’ ∩ B’)∩C’ – A’ ∩ B’ ∩ C’ = {x/x∈A’ and x∈B’ and x∈C’}
= {x/x∉A, x∉B, x∉C}
The shaded region is the compliment of the set.

Relations and Functions

Question 1.
If A = {a, b}, B = {b, c}, C = {c, d). Find (i) (A × B) ∪ (A × C) (ii) A × (B ∩ C) .
Answer:
(i) A × B = {(a, b), (a, c),(b, b), (b, c)}
A × C = {(a,c), (a,d), (b,c) (b,d)}
(A × B) ∪ (A × C) = {(a, b) (a, c) (b, b) (b, c) (a, d) (b, d)}
(ii) B ∩ C = {c}, A × (B ∩ C) = {a, b} × {c} = {(a, c), (b, c}

Question 2.
If (x, x + y) = (6,2). Find x and y.
Answer:
2x = 6 x + y = 2
∴ x = 3 y = 2 – 3 = – 1.

Question 3.
If A = {x/x ∈ N and x < 3} and B = {x/x² – 16 = 0 and x < 0}. FindB x A,
Answer:
⇒ A = {1,2), since n² – 16 = 0, x = + 4, B = {-4}
∴ B × A = {(-4,1) (-4,2)}

Question 4.
If f(n) = 7x + 10. Find f^-1(x) and f(4), f^-1(5).
Answer:
By data f(x) = 7x + 10
Let f^-1(x) = y ⇒ f(y) = x ⇒ 7y + 10 = x
⇒ y = \(\frac{x-10}{7}\)
Then f^-1(x) = \(\frac{x-10}{7}\) Now f^-1(5) = \(\frac{5-10}{10}=\frac{5}{7}\)
Also, (x) = 7(4) + 10 = 28 + 10 = 38
f(4) . f^-1(5) = 38. \(\left(\frac{-5}{7}\right)=\frac{-190}{7}\)

Question 5.
If f: R → R, 5 defined by f(n) = 9x – 11. Show that f is bijective and find the inverse function.
Answer:
We have f(x) ⇒ 9x – 11
Let f(x₁) = f(x₂) = x₁ = x₂ ⇒ 9x₁ – 1 = 9x₂ – 11
⇒ 9x₁ = 9x₂ ⇒ x₁ = x₂ f is one-one
To show that f is on to, we have to show that for every y ∈ Q (codomain) there exists x ∈ Q (domain) such that f(x) = y.
Let f(x) = y ⇒ 9x – 11 = y
⇒ 9x = y + 11 ∴ x = \(\frac{b+11}{9}\)
Then for any y ∈ Q, we have x ∈ Q, such that f(x) = y. ie., x = \(\frac{b+11}{9}\). Hence f is on to.

Question 6.
If f(x) = x² + 2x + 1, g(x) = 2x – 3. Find (i) fog, (ii) gof.
Answer:
By data f(x) = x²(2x + 1), gx = 2x – 3
(i) fog fg(x) = f(2x – 3) where k = 2x – 3
= k² + 2k + 1 = (2x – 3)² + 2(2x – 3) + 1
= 4x² 9 – 12x + 4x – 6 + 1 = 4x² – 8x + 4

(ii) gof = gf(x) = g(n² + 2x + 1) = g(k)
= 2k – 3 = 2(n² + 2x + 1) – 3
= 2x² + 4x + 2 – 3 = 2x² + 4x – 1
∴ gof(2) = 2(2²) + 4(2) – 1 = 8 + 8 – 1 = 15

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Karnataka 1st PUC Basic Maths Question Bank Chapter 1 Number Theory

Question 1.
Find the GCD of 1819 and 3587
Answer:

The GCD of 1819 and 3587 is 17
⇒ (1819,3587) = 17.

Question 2.
Find GCD of 91 and 175 and write is the form 91x + 175y.
Answer:
84 = 175 – 91(1) ………. (1)
7 =91 – 84(1) ……….. (2)
∴ (91, 175) = 7

Question 3.
Find number of positive divisions and the sum of the any five division of 4896.
Answer:
4896 = 2⁵ × 3² × 17¹
T_n = (1 + L₁)(1 + L₂)(1 + L₃)
= (1 + 5)(1 + 2)(1 + 1) = 6.3 .2 = 36

Question 4.
Find the HCF of 180, 252, and 576
Answer:
HCF of 180 and 252

∴ HCF of 180 and 252 is 36

Now find HCF of 36 and 576

∴ HCF of 36 and 576 is 36
∴ ACP of 180,252 and 576 is 36

Question 5.
Find the greatest number which divides 989 and 1327 learn remainder 5 and 7 respectively.
Answer:
Required greatest number is HCF of (989 – 5) and(1327 – 7) i.e., to find HCF of 984 and 1320.

∴ HCF of 984 and 1320 is 24.

Question 6.
If the product of number is 216 and their LCM is 36. Find their HCF.
Answer:
ab = 216, LCM = 36, HCF = ?