Prasanna

Students can download 12th Business Maths Chapter 1 Applications of Matrices and Determinants Ex 1.2 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Solve the following equations by using Cramer’s rule
(i) 2x + 3y = 7; 3x + 5y = 9
(ii) 5x + 3y = 17; 3x + 7y = 31
(iii) 2x + y – z = 3, x + y + z = 1, x – 2y – 3z = 4
(iv) x + y + z = 6, 2x + 3y – z = 5, 6x – 2y – 3z = -7
(v) x + 4y + 3z = 2, 2x – 6y + 6z = -3, 5x – 2y + 3z = -5
Solution:
(i) 2x + 3y = 7; 3x + 5y = 9

Question 2.
A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is ₹ 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is ₹ 56. What is the cost per unit of labour and capital? (Use determinant method).
Solution:
Let the cost per unit of labour be ₹ x and cost per unit of capital be ₹ y.

Question 3.
A total of ₹ 8,600 was invested in two accounts. One account earned 4\(\frac{3}{4}\)% annual interest and the other earned 6\(\frac{1}{2}\)% annual interest. If the total interest for one year was ₹431.25, how much was invested in each account? (Use determinant method).
Solution:
Let ₹ x and ₹ y be the amounts invested in the two accounts.

Question 4.
At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Benita spent ₹ 780 and ₹ 560 during the month of May.

Find the hourly charges for the two games (rides). (Use determinant method).
Solution:
Let hourly charges for horse riding be ₹ x and hourly charges for Quad bike riding be ₹ y.
According to the problem, for Keren, we have 3x + 4y as total amount and for Benita, we have 2x + 3y as the total amount.
That is 3x + 4y = 780
2x + 3y = 560

Hence hourly charges for horse riding and bike riding are ₹ 100 and ₹ 120 respectively.

Question 5.
In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Find the weights assigned to the three varieties by using Cramer’s Rule.
Solution:
Let the weights assigned to the three varieties be x, y and z respectively.
According to the problem,
For variety A, x + 2y + 3z = 11
For variety B, 2x + 4y + 5z = 21
For variety C, 3x + 5y + 6z = 27

Hence the weights assigned to the three varieties are 2, 3 and 1 units respectively.

Question 6.
A total of ₹ 8,500 was invested in three interest-earning accounts. The interest rates were 2%, 3% and 6% if the total simple interest for one year was ₹ 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule).

Solution:
Let the amounts invested in the three accounts be Rs. x, Rs. y and Rs. z
Interest for the three accounts are \(\frac{2}{100}\)x, \(\frac{3}{100}\)y and \(\frac{6}{100}\)z
According to the problem, x + y + z = 8500 ……. (1)
\(\frac{2}{100} x+\frac{3}{100} y+\frac{6}{100} z=380\)
(or) multiplying by 100,
2x + 3y + 6z = 38000 ……… (2)
z = x + y or x + y – z = 0 ………. (3)

Students can download 12th Business Maths Chapter 1 Applications of Matrices and Determinants Ex 1.1 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.
Find the rank of each of the following matrices.
Solution:
(i) Let A = \(\left(\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right)\)
Order of A is 2 × 2.
ρ(A) ≤ 2
Consider the second order minor
\(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\) = 40 – 42 = -2 ≠ 0
There is a minor of order 2, which is not zero.
ρ(A) = 2
(ii) Let A = \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Order of A is 2 × 2
ρ(A) ≤ 2
Consider the second order minor
\(\left|\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right|\) = -6 + 3 = -3 ≠ 0
ρ(A) = 2
(iii) Let A = \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Since A is of order 2 × 2, ρ(A) ≤ 2
Now \(\left|\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right|\) = 8 – 8 = 0
Since second order minor vanishes ρ(A) ≠ 2
But first order minors, |1|, |4|, |2|, |8| are non zero.
ρ(A) = 1
(iv) Let A = \(\left(\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right)\)
Order of A is 3 × 3
ρ(A) ≤ 3
Consider the third order minor
\(\left|\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right|\)
= 2(1 + 5) + 1 (3 + 5) + 1(3 – 1)
= 2(6) + 8 + 2
= 22 ≠ 0
There is a minor of order 3, which is non zero.
ρ(A) = 3
(v) Let A = \(\left(\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right)\)
Since order of A is 3 × 3, ρ(A) ≤ 3
Now,
\(\left|\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right|\)
= -1(12 – 16) -2(-16 + 8) – 2(16 – 6)
= 4 + 16 – 20
= 0
Since the third order minor vanishes, ρ(A) ≠ 3
Consider \(\left|\begin{array}{cc}
-1 & 2 \\
4 & -3
\end{array}\right|\) = 3 – 8 = -5 ≠ 0
There is a minor order 2, which is not zero
ρ(A) = 2
(vi) Let A = \(\left(\begin{array}{cccc}
1 & 2 & -1 & 3 \\
2 & 4 & 1 & -2 \\
3 & 6 & 3 & -7
\end{array}\right)\)
Let us transform the matrix A to an echelon form by using elementary transformations.

Question 2.
If A = \(\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right)\) and B = \(\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)\), then find the rank of AB and the rank of BA.
Solution:
Given A = \(\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right)\) and B = \(\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)\)

Question 3.
Solve the following system of equations by rank method.
x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0
Solution:
The given equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0
The matrix equation corresponding to the given system is

Question 4.
Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method.
Solution:
The given equations are,
5x + 3y + 7 = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given system is

Question 5.
Show that the following system of equations have unique solution:
x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 by rank method.
Solution:
The given system of equations can be written in matrix equation,

The last matrix is in echelon form. It has 3 non – zero rows,
ρ(A) = ρ([A, B]) = 3 = number of unknowns.
The given system is consistent and has a unique solution.
To find the solution, we write the echelon form into matrix form.
\(\left(\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
3 \\
1 \\
0
\end{array}\right)\)
x + y + z = 3 …… (1)
y + 2z = 1 …… (2)
2z = 0 …… (3)
(3) ⇒ z = 0
(2) ⇒ y = 1
(1) ⇒ x = 2
So the unique solution is x = 2, y = 1, z = 0

Question 6.
For what values of the parameter X, will the following equations fail to have unique solution: 3x – y + λz = 1, 2x + y + z = 2, x + 2y – λz = -1 by rank method.
Solution:
The given system can be written in matrix equation form as given below:


λ = \(\frac{-7}{2}\)
So when λ = \(\frac{-7}{2}\), the equations fail to have unique solution.
Note: The system cannot have an infinite number of solutions, since ρ([A, B]) = 3 = a number of unknowns.

Question 7.
The price of three commodities, X,Y and Z are and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of X and 3 units of Y. Mr.Amar purchases a unit of Y and sells 3 units of X and 2 units of Z. Mr. Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn ₹ 5,000/-, ₹2,000/- and ₹ 5,500/- respectively. Find the prices per unit of three commodities by rank method.
Solution:

The price of three commodities X, Y, Z are given as x, y, z.
We form the following system of equations from the given conditions.
Anand → 2x + 3y – 6z = 5000
Amar → 3x – y + 2z = 2000
Amit → -x + 3y + z = 5500
The matrix equation is given by



The prices per unit of the three commodities are Rs.1000, Rs.2000 and Rs.500

Question 8.
An amount of ₹ 5,000/- is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is ₹ 358/-. If the income from the first two investments is ₹ 70/- more than the income from the third, then find the amount of investment in each bond by the rank method.
Solution:
Let the amount of investment in the three different bonds be Rs. x, Rs. y and Rs. z respectively.
We get the following equations according to the given conditions,
x + y + z = 5000


The above equivalent matrix is in echelon form with 3 non-zero rows.
So ρ(A) = ρ([A, B]) = 3 = number of unknowns.
the system has a unique solution.
The matrix equation is given by
\(\left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right)\)
x + y + z = 5000 …(1)
y + 2z = 5800 …(2)
-16z = -28800 …(3)
(3) ⇒ z = 1800
(2) ⇒ y = 5800 – 2(1800) = 2200
(1) ⇒ x = 5000 – 2200 – 1800 = 1000
The amount invested in the three bonds are ₹ 1000, ₹ 2200 and ₹ 1800.

Students can Download Tamil Nadu 11th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The number of relations on a set containing 3 elements is………..
(a) 9
(b) 81
(c) 512
(d) 1024
Solution:
(c) 512

Question 2.
If n[(A × B) ∩ (A × C)] = 12 and n(B ∩ C) = 2 then n(A) is……….
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(d) 6

Question 3.
If |x – 3| ≤ 5 then x belongs to………
(a) [-2, 8]
(b) (-2, 8)
(c) [-2, ∞]
(d) (-∞, 8)
Solution:
(a) [-2, 8]

Question 4.
The number of solutions of x² + |x – 1| = 1 is
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(c) 2

Question 5.
If a, 8, b are in A.P. a, 4, b are in G.P. and a, x, b are in H.P then x = ………..
(a) 2
(b) 1
(c) 4
(d) 16
Solution:
(a) 2

Question 6.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are………..
(a) 45
(b) 40
(c) 10!
(d) 210
Solution:
(a) 45

Question 7.
The value of e^2logx………..
(a) 2x
(b) x²
(c) √2
(d) \(\frac{x}{2}\)
Solution:
(b) x²

Question 8.
The nth term of the sequence 1, 2, 4, 7, 11 …. is ………..
(a) n³ + 3n² + 2n
(b) n³ – 3n² + 3n
(c) \(\frac{n(n+1)(n+2)}{3}\)
(d) \(\frac{n²-n+2}{2}\)
Solution:
(d) \(\frac{n²-n+2}{2}\)

Question 9.
The last term in the expansion (2 + √3)⁸ is
(a) 81
(b) 27
(c) 9
(d) 3
Solution:
(a) 81

Question 10.
A line perpendicular to the line 5x – y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq.units, then its equation is………….
(a) x + 5y ± 5√2 = 0
(b) x – 5y ± 5√2 = 0
(c) 5x + y ± 5√2 = 0
(d) 5x – y ± 5√2 = 0
Solution:
(a) x + 5y ± 5√2 = 0

Question 11.
A factor of the determinant  is ……….
(a) x + 3
(b) 2x – 1
(c) x – 2
(d) x – 3
Solution:
(a) x + 3

Question 12.
If λ\(\vec {i}\) + 2λ\(\vec {j}\) + 2λ\(\vec {k}\) is a unit vector then the value of λ is…………
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Solution:
(a) \(\frac{1}{3}\)

Question 13.
One of the diagonals of parallelogram ABCD with \(\vec {a}\) and \(\vec {b}\) are adjacent sides is The other diagonal BD is…………
(a) \(\vec {a}\) + \(\vec {b}\).
(b) \(\vec {a}\) – \(\vec {b}\)
(c) \(\vec {b}\) – \(\vec {a}\)
(d) \(\frac{\vec a+\vec b}{2}\)
Solution:
(b) \(\vec {a}\) – \(\vec {b}\)

Question 14.
If (1, 2, 4) and (2, -3λ, -3) are the initial and terminal points of the vector \(\vec {i}\) + 5\(\vec {j}\) – 7\(\vec {k}\) then the value of λ ………….
(a) \(\frac{7}{3}\)
(b) –\(\frac{7}{3}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{-5}{b}\)
Solution:
(b) –\(\frac{7}{3}\)

Question 15.
If y = mx + c and f(0) = f'(0) = 1 then f(2) =…………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 16.
The derivative of (x + \(\frac{1}{x}\))² w.r.to. x is…………
(a) 2x – \(\frac{2}{x³}\)
(b) 2x + \(\frac{2}{x³}\)
(c) 2(x + \(\frac{1}{x}\))
(d) 0
Solution:
(a) 2x – \(\frac{2}{x³}\)

Question 17.
If f(x) =  is differentiable at x = 1, then………
(a) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(b) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(c) a = –\(\frac{1}{2}\), b = –\(\frac{3}{2}\)
(d) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Solution:
(c) a = –\(\frac{1}{2}\), b = –\(\frac{3}{2}\)

Question 18.
∫sin 7x cos 5x dx =…………
(a) \(\frac{1}{2}\) [\(\frac{cos 12x}{2}\)+\(\frac{cos 2x}{2}\)] + c
(b) –\(\frac{1}{2}\) [\(\frac{cos 12x}{2}\)+\(\frac{cos 2x}{2}\)] + c
(c) –\(\frac{1}{2}\) [\(\frac{cos 6x}{6}\) + cos x] + c
(d) –\(\frac{1}{2}\) [\(\frac{sin 12x}{2}\)+\(\frac{sin 2x}{2}\)] + c
Solution:
(b) –\(\frac{1}{2}\) [\(\frac{cos 12x}{2}\)+\(\frac{cos 2x}{2}\)] + c

Question 19.
∫ \(\frac{1}{e^x}\) dx = ………..
(a) log e^x + c
(b) x + c
(c) \(\frac{1}{e^x}\) + c
(d) \(\frac{-1}{e^x}\)
Solution:
(d) \(\frac{-1}{e^x}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is…………
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Solution:
(a) \(\frac{19}{33}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Prove that \(\frac{tanθ+secθ-1}{tanθ-secθ+1}\) \(\frac{1+sinθ}{cosθ}\)
Solution:

Question 22.
Prove that the relation ‘friendship’ is not an equivalence relation on the set-of all people in chennai.
Solution:
S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflextive.
aRb ⇒ bRa so it is symmetric
aRb, bRc does not ⇒ aRc
so it is not transitive
⇒ it is not an equivalence relation.

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points = 15
To draw a Triangle we need 3 points
∴Selecting 3 from 15 points can be done in ¹⁵C³ ways.
∴ No. of Triangle formed = ¹⁵C³
= \(\frac{15×14×13}{3×2×1}\) = 455

Question 24.
Expand (2x + 3)⁵
Solution:
By taking a = 2x, b = 3 and n = 5 in the binomial expansion of (a + b)ⁿ
we get (2x + 3)⁵ = (2x)⁵ + 5(2x)⁴3 + 10(2x)³3² + 10(2x)²3³ + 5(2x)3⁴ + 3⁵
= 32x⁵ + 240x⁴ + 720x³ + 1980x² + 810x + 243.

Question 25.
If λ = -2, determine the value of
Solution:
Given λ = -2
2λ = -4; λ² = (-2)² = 4; 3λ² + 1 = 3(4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13

expanding along R,
0(0) + 4 (0 + 13) +1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew symmetric matrix is zero.

Question 26.
Compute \(\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1}\)
Solution:
Here \(\lim _{x \rightarrow 1}\)(x – 1) = 0. In such cases, rationalise the numerator.

Question 27.
Differentiate the following \(\frac{x²}{a²}\)+ \(\frac{y²}{b²}\) = 1
Solution:
Given \(\frac{x²}{a²}\)+ \(\frac{y²}{b²}\) = 1
Differentiating w.r.to x

Question 28.
Evaluate \(\frac{1}{(x+1)²-25}\)
Solution:

Question 29.
Given that P(A) = 0.52, P(B) = 0.43, and P(A ∩ B) = 0.24, find p(A ∩ \(\bar { B }\))
Solution:
P(A ∩ \(\bar { B }\)) = P(A) – P(A ∩ B)
= 0.52 – 0.24 = 0.28
P(A ∩ \(\bar { B }\)) = 0.28

Question 30.
Show that 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines
Solution:
4x² + 4xy + y² – 6x – 3y – 4 = 0
a = 4, b = 1, h = 4/2 = 2
h² – ab = 2² – (4) (1) = 4 – 4 = 0
⇒ The given equation represents a pair of parallel lines.

PART-III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
If A and B are two sets so that n(B – A) = 2n(A – B) = 4n(A ∩ B) and if n(A ∪ B) = 14 then find n(PA)
Solution:
To find n(P(A)), we need n(A).
Let n( A ∩ B) = t. Then n( A – B) = 2k and n(B – A) = 4k.
Now n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B) = 7k.
It is given that n(A ∪ B) = 14. Thus 7k = 14 and hence k = 2.
So n(A – B) = 4 and n(B – A) = 8. As n(A) = n(A – B) + n(A ∩ B), we get n(A) = 6 and hence n(P(A)) = 2⁶ = 64.

Question 32.
Resolve \(\frac{1}{x²-a²}\) into partial fraction.
Solution:
Factorizing the denominator
Dr = x² – a² = (x – a) (x + a)

Equating the numerator we get
x – a = 0
⇒ x = a
x + a = 0
⇒ x = -a
1 = A (x + a) + B (x – a)
This equation is true for any value of x

Question 33.
Count the number of positive integers greater than 7000 and less than 8000 which are divisible by 5 provided that no digits are repeated.
Solution:
It should be a 4-digit number greater than 7000 and less than 8000. Then the thousand’s place will be the digit 7. Further, as the number must be divisible by 5 the unit place should be either 0 or 5.

As repetition is not permitted, the 100th place can be filled in 8 ways using remaining numbers and 10th place can be filled in 7 ways. Hence, the required number of numbers is 1 × 8 × 7 × 2 = 112.

Question 34.
Find the \(\sqrt[3]{126}\) approximately to two decimal places.
Solution:

Question 35.
Find the equation of the line through the intersection of the lines
3x + 2y + 5 = 0 and 3x – 4y + 6 = 0 and the point (1, 1)
Solution:
The family of equations of straight lines through the point of intersection of the lines is of the form (a₁x + b₁y + c₁) + (a₂x + b₂y + c₂) = 0
That is, (3x + 2y + 5) + λ (3x – 4y + 6) = 0
Since the required equation passes through the point (1, 1), the point satisfies the above equation Therefore {3 + 2(1) + 5} + λ(3 (1) – 4(1) + 6} = 0 ⇒ λ = -2
Substituting λ = -2 in the above equation we get the required equation as 3x – 10y + 7 = 0 (verify the above problem by using two points form)

Question 36.
Show that
Solution:

Question 37.
Complete the following table using calculator and use the result to estimate \(\lim _{x \rightarrow 2}\) \(\frac{x-2}{x²-x-2}\)

Solution:

Limit is 0.333…. = 0.\(\bar{3}\)

Question 38.
Differentiate \(\frac{e^{3x}}{1+e^x}\) with respect to x
Solution:

Question 39.
Evaluate: e^x (tan x + log sec x)
Solution:
Let I = ∫e^x (tan x + log sec x) dx
Take f(x) = log sec x
f(x) = \(\frac{1}{sec x}\) × sec x tan x = tan x
This is of the form ∫e^x[f(x) + f'(x)] dx = e^x f(x) + c
∴ ∫e^x(log sec x + tan x) dx = e^x log |sec x| + c

Question 40.
The position vectors of the vertices of a triangle are \(\vec{i}\) + 2\(\vec{j}\) + 3\(\vec{k}\), 3\(\vec{i}\) – 4\(\vec{j}\) + 5\(\vec{k}\) and -2\(\vec{i}\) + 3\(\vec{j}\) – 7\(\vec{k}\) Find the perimeter of a triangle.
Solution:
Let A, B, C be the vertices of triangle ABC,

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
If f : R → R is defined by f(x) = 3x – 5, prove that f is a bijection and find its inverse.
Solution:
p(x) = 3x – 5
Let g(y) = 3x – 5 ⇒ 3x = y + 5
x = \(\frac{y+5}{3}\)
Let g(y) = \(\frac{y+5}{3}\)
Now g o f(y) = g[(f(x))] = g(3x-5)

Thus g o f = I_x and f o g = I_y
f and g are bi-jections and inverse to each other. Hence f is a bi-jection and f^-1(y) = \(\frac{y+5}{3}\)
Replacing y by x we get f^-1(x) = \(\frac{x+5}{3}\)

[OR]

(b) Prove that tan^-1(\(\frac{m}{n}\)) – tan^-1(\(\frac{m-n}{m+n}\)) = \(\frac{π}{4}\)
Solution:

Question 42 (a).
Find the values of k so that the equation x² = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x² – x(2) (1 + 3k) – 7 (3 + 2k) = 0
The roots are real and equal
⇒ Δ = 0 (i.e.,) b² – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b² – 4ac = 0 ⇒ [-2 (1 + 3k)]² – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4(1+ 3k)² – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)² – 7(3 + 2k) = 0
1 + 9k² + 6k – 21 – 14k = 0
9k² – 8k – 20 = 0
(k – 2) (9k + 10) = 0
⇒ k – 2 > 0 or 9k + 10 = 0
⇒ k = 2 or k = \(\frac{-10}{9}\)
To solve the quadratic inequalities ax² + bx + c < 0 (or) ax² + bx + c > 0

[OR]

(b) If the roots of the equation (q – r) x² + (r – p)x + (p – q) = 0 are equal then show that p, q and r are in A.P.
Solution:
The roots are equal ⇒ Δ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r; b = r – p; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr +4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 43 (a).
Find the sum of all 4 digit-numbers that can be formed using the digits 1, 2, 3, 4 and 5 repetition not allowed?
Solution:
The given digits are 1, 2, 3, 4, 5
The no. of 4 digit numbers

= 5 × 4 × 3 × 2 = 120
(i.e) ⁵P₄ = 120
Now we have 120 numbers
So each digit occurs \(\frac{120}{5}\) = 24 times
Sum of the digits =1+2 + 3 + 4 + 5 = 15
Sum of number’s in each place = 24 × 15 = 360
Sum of numbers = 360 × 1 = 360
360 × 10 = 3600
360 × 100 = 36000
360 × 1000 = 360000
Total = 399960

[OR]

(b) Three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are such that |\(\vec{a}\)| = 2,|\(\vec{b}\)| = 3, |\(\vec{c}\)| = 4 and \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0. Find 4\(\vec{a}\).\(\vec{b}\) + 3\(\vec{b}\).\(\vec{c}\) + 3\(\vec{c}\).\(\vec{a}\).
Solution:
Given \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0
⇒ \(\vec{a}\) + \(\vec{a}\) = –\(\vec{a}\)
so (\(\vec{a}\) + \(\vec{a}\))² = \(\vec{c}\)²
(i.e.) a² + b²+ 2\(\vec{a}\).\(\vec{b}\) = \(\vec{c}\)²
⇒ 4 + 9 + 2\(\vec{a}\).\(\vec{b}\) = 16
⇒ 2\(\vec{a}\).\(\vec{b}\) = 16 – 4 – 9 = 3
\(\vec{a}\) \(\vec{b}\) = 3/2
Again \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0
⇒ \(\vec{a}\) + \(\vec{c}\) = –\(\vec{b}\)
(\(\vec{a}\) + \(\vec{c}\))² = \(\vec{b}\)²
\(\vec{a}\)² + \(\vec{c}\)² + 2\(\vec{a}\).\(\vec{c}\) = \(\vec{b}\)²
4 + 16 + 2\(\vec{a}\) – \(\vec{a}\) = 9
2\(\vec{a}\) – \(\vec{c}\) = 9 – 4 – 16 = -11
\(\vec{a}\).\(\vec{c}\) = \(\frac{-11}{2}\) (i.e.,) \(\vec{c}\) – \(\vec{a}\) = \(\frac{-11}{2}\) (∵\(\vec{a}\).\(\vec{c}\) = \(\vec{c}\).\(\vec{a}\))
Also \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0
\(\vec{b}\) + \(\vec{c}\) = –\(\vec{a}\)
(\(\vec{b}\) + \(\vec{c}\))² = \(\vec{a}\)²
9 + 16 + 2\(\vec{b}\) – \(\vec{c}\) = 4
2\(\vec{b}\) – \(\vec{c}\) = 4 – 9 – 16 = -21
\(\vec{b}\) – \(\vec{c}\) = \(\frac{21}{2}\)
Here \(\vec{a}\).\(\vec{b}\) = 3/2 \(\vec{b}\).\(\vec{c}\) = \(\frac{-21}{2}\) and \(\vec{c}\).\(\vec{a}\) = \(\frac{-11}{2}\)

Question 44 (a).
If a, b, c are respectively the p^th, q^th and r^th terms of a G.P. show that (q – r) log a+ (r – p) log b + (p – q) log c = 0.
Solution:
Let the G.P. be l, lk, lk²,…
We are given t_p = a, t_q = b, t_r = c
⇒ a = lk^p-1; b = l k^q-1; c = l k^r-1
a = lk^p-1 ⇒ log a = log l + log k^p-1 = log l + (p – 1) log k
b = lk^q-1 ⇒ log b = log l + log k^q-1 = log l + (q – 1) log k
c = lk^r-1 ⇒ log r = log l + log k^r-1 = log l + ( r – 1) log k
LHS = (q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log l + (p – 1) log k ] + (r – p) [log l + (q – 1) log k]
(p – q) [log l + (r – 1) log k]
= log l[p – q + q – r + r – p] + log k[(q – r)(p – 1) + (r – p) (q – 1) + (p – q)(r – 1)]
= log l (0) + log k[p (q – r) + q (r – p) + r(p – q) – (q – r + r – p + p – q)]
= 0 = RHS.

[OR]

(b) If A = \(\left[\begin{array}{ll} \frac{1}{2} & \alpha \\ 0 & \frac{1}{2} \end{array}\right]\), Prove that \(\sum_{k=1}^{n}\) det(A_k) = \(\frac{1}{3}\) (1 – \(\frac{1}{4^n}\))
Solution:

Which is a G.P with a = \(\frac{1}{4}\) and r = \(\frac{1}{4}\)

Question 45 (a).
Find the equation of the straight line passing through intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and perpendicular to the straight line 3x – 5y + 11 = 0.
Solution:
Equation of line through the intersection of straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 is 5x – 6y – 1 + k (3x + 2y + 5) = 0 x(5 + 3k) + y(-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x – 5y + 11 = 0
That is, the product of their slopes is -1
–\(\frac{5+3k}{-6+2k}\) (-\(\frac{3}{-5}\)) = -1
⇒ \(\frac{15+9k}{-30+10k}\) = 1
⇒ 15 + 9k = -30 + 10k
45 = k
Required equation is 5x – 6y – 1 + 45 (3x + 2y + 5) = 0
140x + 84y + 224 = 0
20x + 12y + 32 = 0
5x + 3y + 8 = 0

[OR]

(b) Integrate the following \(\frac{√x}{1+√x}\) dx
Solution:

Question 46 (a).
If u = tan^-1(\(\frac{\sqrt{1+x^{2}}-1}{x}\)) and v = tan^-1x, find \(\frac{dx}{dy}\)
Solution:

[OR]

(b)
If y = Ae^6x + Be^-x prove that \(\frac{d²y}{dx²}\) – 5\(\frac{dx}{dy}\) – 6y = 0
Solution:
y = Ae^6x + Be^-x …. (1)
y₁ = \(\frac{dx}{dy}\) = Ae^6x(6) + Be^-x (-1)
= 6Ae^6x – Be^-x…. (2)
y₂ = \(\frac{dx}{dy}\) = 6Ae^6x (6) – Be^-x (-1)
= 36Ae^6x + Be^-x ……(3)
eliminating A and B from (1), (2) and (3) we get

y (6 + 36) – y₁ (1 – 36) + y₂ (-1 – 6) = 0
42y + 35y₁ – 7y₂ = 0
(÷ by -7) y₂ – 5y₁ – 6y = 0
(i.e.,) \(\frac{d²y}{dx²}\) – 5\(\frac{dx}{dy}\) – 6y = 0

Question 47 (a).
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+16}-4}\)
Solution:

[OR]
(b) Urn-I contains 8 red and 4 blue balls and urn-II contains 5 red and 10 blue balls. One urn is chosen at random and two balls are drawn from it. Find the probability that both balls are red.
Solution:
Let A₁ be the event of selecting um-I and A₂ be the event of selecting um-II.
Let B be the event of selecting 2 red balls.

We have to find the total probability of event B. That is, P(B).
Clearly A₁ and A₂A₁ are mutually exclusive and exhaustive events.
We have

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.1

Question 1.
A School purchases some furniture and gets the following bill.

(i) What is the name of the store?
(ii) What is the serial number of the bill?
(iii) What is the cost of a blackboard?
(iv) How many sets of benches and desks does the school buy?
(v) Verify whether the total bill amount is correct.
Solution:
(i) Mullai Furniture Mart, Thanjavur
(ii) Serial No. 728
(iii) Cost of a blackboard is ₹ 3000
(iv) 50 sets
(v) Yes, the total bill amount is correct.

Question 2.
Prepare a bill for the following books of biographies purchased from Maruthu Book Store, Chidambaram on 12.04.2018 bearing the bill number 507.
10 copies of Subramanya Bharathiar @ ₹ 55 each, 15 copies of Thiruvalluvar @ ₹ 75 every 12 copies of Veeramamunivar @ ₹ 60 each and 12 copies of Thiru. Vi.Ka @ ₹ 70 each.
Solution:

Question 3.
Fillup the appropriate boxes in the following table.

Solution:
(i) Here Cost price < Selling price
Profit = S.P – C.P = 120 – 100 = ₹ 20
Here S.P > C.P
Profit = S.P – C.P = 120 – 110 = ₹ 10
(iii) Profit = ₹ 20
Profit = S.P. – C.P
⇒ 20 = S.P. – 120
⇒ 20 + 120 = S.P
⇒ S.P = ₹ 140
(iv) C.P = ₹ 100
S.P = ₹ 90
Here S.P < C.P
Loss = 100 – 90 = ₹ 10
(v) Profit = S.P – C.P
⇒ 25 = S.P – 120
⇒ 25 + 120 = S.P
⇒ S.P = ₹ 145

Question 4.
Fill up the appropriate boxes in the following table.

Solution:
(i) S.P. = M.P – Discount
S.P = ₹ 130
Here S.P > C.P
Profit = S.P – C.P = 130 – 110 = ₹ 20
(ii) S.P = M.P – Discount = 130 – 10 = ₹ 120
Here S.P > C.P
Profit = S.P – C.P = ₹ 120 – ₹ 110 = ₹ 10
(iii) S.P = M.P – Discount = 130 – 30 = ₹ 100
Here S.P < C.P
Loss = 110 – 100 = ₹ 10
(iv) C.P = ₹ 110
M.P = ₹ 120
Loss = ₹ 10
Loss = C.P – S.P
⇒ 10 = 110 – SP
⇒ SP = 110 – 10 = ₹ 100
Discount = M.P – S.P = 120 – 100 = ₹ 20
(v) M.P = ₹ 120
Discount = ₹ 10
Discount = M.P – S.P
⇒ 10 = 120 – S.P
⇒ S.P = 120 – 10 = ₹ 110
Profit = ₹ 20
Profit = S.P – C.P
⇒ 20 = 110 – C.P
⇒ C.P = 110 – 20 = 90

Question 5.
Rani bought a set of bangles for Rs 310. Her neighbour liked it most. So, Rani sold it to her for Rs 325. Find the profit or loss to Rani.
Solution:
CP = Rs 310
SP = Rs 325
Profit = SP – CP = Rs 325 – Rs 310 = Rs 15

Question 6.
Sugan bought a Jeans pant for ₹ 750. It did not fit him. He sold it to his friend for ₹ 710. Find the profit or loss to sugan.
Solution:
C.P of the Jeans pant = ₹ 750
S.P of the Jeans pant = ₹ 710
Here S.P < C.P
Loss = C.P – S.P = 750 – 710 = ₹ 40
Loss = ₹ 40

Question 7.
Somu bought a second hand bike for Rs 28,000 and spent Rs 2,000 on its repair. He sold it for Rs 30,000. Find his profit or loss.
Solution:
CP = Rs 28,000 + Rs 2,000
CP = Rs 30,000
SP = Rs 30,000
CP = SP
No profit / Loss

Question 8.
Muthu has a car worth ₹ 8,50,000 and he wants to sell it at a profit of ₹ 25,000. What should be the selling price of the car? Solution:
Cost price of the car = ₹ 8,50,000
Expected profit = ₹ 25,000
We know that profit = S.P – C.P
25000 = S.P – 8,50,000
⇒ 25,000 + 8,50,000 = S.P
⇒ S.P = 8,75,000
Selling Price of the car should be ₹ 8,75,000

Question 9.
Valarmathi sold her pearl set for Rs 30,000 at profit of Rs 5,000. Find the cost price of the pearl set.
Solution:
SP = Rs 30,000
Profit = Rs 5,000
CP = SP – Profit
= Rs 30,000 – Rs 5,000
= Rs 25,000

Question 10.
If Guna marks his product to be sold for ₹ 325 and gives a discount of ₹ 30, then find the S.P.
Solution:
Marked Price of the Product = ₹ 325
Discount = ₹ 30
Selling Price = M.P – Discount = 325 – 30 = ₹ 295
S.P. = ₹ 295

Question 11.
A man buys a chair for 1,500. He wants to sell it at a profit of Rs 250 after making a discount of Rs 100. What is the M.P. of the chair?
Solution:
CP = Rs 1,500
Profit = Rs 250
SP = CP + Profit
= Rs 1,500 + Rs 250
= Rs 1,750
Discount = Rs 100
SP = MP – Discount
MP = SP + Discount
= Rs 1,750 + Rs 100
= Rs 1,850

Question 12.
Amutha marked her home product of pickle as ₹ 300 per pack. But she sold it for only ₹ 275 per pack. What was the discount offered by her per pack?
Solution:
M.P of the pickle = ₹ 300
S.P = ₹ 275
S.P = M.P – Discount
⇒ 275 = 300 – Discount
⇒ Discount = 300 – 275 = ₹ 25
Discount per pack = ₹ 25

Question 13.
Valavan bought 24 eggs for Rs 96. Four of them were broken and also he had a loss of Rs 36 on selling them. What is the selling price of one egg?
Solution:
Cost of 24 eggs = Rs 96
Since 4 of the eggs were broken, the number of remaining eggs = 24 – 4 = 20
Since the loss is Rs 36
Selling price of 20 eggs
SP = CP – Loss
= Rs 96 – Rs 36
= Rs 60
∴ Cost of 1 egg = Rs 60 / 20 = Rs 3

Question 14.
Mangai bought a cell phone for ₹ 12,585. It fell down. She spent ₹ 500 on it repair. She sold it for ₹ 7,500. Find her profit or loss.
Solution:
Cost of the cell phone = ₹ 12,585
Spent on repairs = ₹ 500
Cost price = Cost of cell phone + repair charge = 12,585 + 500 = ₹ 13,085
S.P. = 7,500
Here C.P > S.P
∴ It is loss
Loss = C.P – S.P = 13,085 – 7500 = ₹ 5,585
Loss = ₹ 5,585

Objective Type Questions

Question 15.
Discount is subtracted from ……… to get S.P.
(a) M.P
(b) C.P
(c) Loss
(d) Profit
Solution:
(a) M.P

Question 16.
‘Overhead expenses’ is always included in _____.
(a) S.P
(b) C.P.
(c) Profit
(d) Loss
Solution:
(b) C.P.

Question 17.
There is no profit or loss when
(a) C.P = S.P
(b) C.P > S.P
(c) C.P < S.P
(d) M.P = Discount
Solution:
(a) C.P = S.P

Question 18.
Discount = M.P _____
(a) Profit
(b) S.P
(c) Loss
(d) C.P
Solution:
(b) S.P

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.2

Question 1.
Fill in the blanks.

(iii) Representation of data by using pictures is known as _______
Solution:
(i) 150 balls
(ii)
(iii) Pictograph

Question 2.
Draw a pictograph for the give data.

(Choose your own suitable scales)
Solution:
Pictograph of Number of computers sold.

Question 3.
The following table shows the number of tourists who visited the places in the month of May. Draw a pictograph.

(Choose your own suitable scale)
Solution:
The pictograph for the number of tourists who visited various places:

Question 4.
The following pictograph shows the number of students playing different games in a school

Answer the following questions.
(i) Which is the most popular game among the students?
(ii) Find the number of students playing kabaddi?
(iii) Which two games are played by an equal number of students?
(iv) What is the difference between the number of students playing Kho-Kho and Hockey?
(v) Which is the least popular game among the students?
Solution:
(i) Kabaddi is the most popular game among students.
(ii) There are 11 × 10 = 110 students playing kabaddi.
(iii) Kho-Kho and Hockey are played by an equal number of students.
(iv) Difference is 90 – 90 = 0.
(v) Basketball is the least popular game among students.

Objective Type Questions

Question 5.
The representation of‘one picture to many objects’ in a pictograph is called
(a) Tally mark
(b) Pictoword
(c) Scaling
(d) Frequency
Solution:
(c) Scaling

Question 6.
The representation of ‘one picture too many objects’ in a Pictograph is called _____
(a) Tally mark
(b) Pictoword
(c) Scaling
(d) Frequency
Solution:
(c) Scaling

Question 7.
A Pictograph is also known as ______
(a) Pictoword
(b) Pictogram
(c) Pictophrase
(d) Pictograph
Solution:
(b) Pictogram

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.1

Question 1.
Fill in the blanks.
(i) The collected information is called _____
(ii) An example of Primary data is ______
(iii) An example of a Secondary data is ______
(iv) The tally marks for number 8 in standard form is _____
Solution:
(i) Data
(ii) List of Absentees in a class
(iii) Cricket scores gathered from a website
(iv)

Question 2.
Viji threw a die 30 times and noted down the result each time as follows. Prepare a table on the numbers shown using Tally Marks.

Solution:
We prepare a table using tally marks from the given information

Question 3.
The following list tells colours liked by 25 students. Prepare a table using Tally Marks.

Solution:
We prepare the table using Tally marks as

Question 4.
The following are the marks obtained by 30 students in a class test out of 20 in Mathematics subject.

Prepare a table using Tally Marks.
Solution:
We prepare a table using Tally Marks as

Question 5.
The table shows the number of calls recorded by a Fire Service Station in one year.

Complete the table and answer the following questions.
(i) Which type of call was recorded the most?
(ii) Which type of call was recorded the least?
(iii) How many calls were recorded in all?
(iv) How many calls were recorded as False Alarm?
Solution:
The completed table is given below.

(i) The call for “Other Fires” was recorded the most
(ii) The call for “Rescues” was recorded the least
(iii) The total of 35 calls was recorded
(iv) There are 7 calls were recorded as False alarm.

Objective Type Questions

Question 6.
The tally marks for the number 7 in standard form is ________
(a) 7
(b)
(c) ✓✓✓✓✓✓✓
(d)
Solution:
(b)

Question 7.
The tally marks represent the number count
(a) 5
(b) 8
(c) 9
(d) 10
Solution:
(c) 9

Question 8.
The plural form of ‘datum’ is
(a) datum
(b) datums
(c) data
(d) datas
Solution:
(c) data

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.6

Miscellaneous Practice Problems

Question 1.
Try to open my locked suitcase which has the biggest 5 digits odd number as the password comprising the digits 7, 5, 4, 3 and 8. Find the password.
Solution:
Using a place value chart
The number should be the biggest odd

The password is 87543

Question 2.
As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population.

Solution:
All the four values have 8 digits
Comparing the leftmost digits we have 91276115, 72626809, 72147030, 68548437
Descending order: 91276115 > 72626809 > 72147030 > 68548437
Ascending order: 68548437 < 72147030 < 72626809 < 91276115
Ascending order: Rajasthan < Tamil Nadu < Madhy Pradesh < West Bengal
Descending order: West Bengal > Madhya Pradesh > TamilNadu > Rajasthan

Question 3.
Study the following table and answer the questions.

(i) How many tigers were there in 2011?
(ii) How many tigers were less in 2008 than in 1990?
(iii) Did the number of tigers increase or decrease between 2011 and 2014? If yes, by how much?
Solution:
(i) There are 1706 tigers in 2011

(ii) No. of tigers in 2008 = 1400
No. of tigers in 1990 = 3500
There were 2100 lesser tigers

(iii) No. of tigers in 2014 = 2226
No. of tigers in 2011 = 1706
Difference = 520
2226 > 1706
The number of tigers increased from 2011 to 2014.
Yes, the number of tigers increased, 520 more tigers are there in 2014.

Question 4.
Mullaikodi has 25 bags of apples. In each bag there are 9 apples. She shares them equally amongst her 6 friends. How many apples do each get? Are there any apples left over?
Solution:
No of apple bags = 25
Apples in each bag = 9
Total no of apples = 25 × 9 = 225
Apples shared among her 6 friends = 225 ÷ 6
So, among her 6 friends, each of them get 37 apples and 3 apples are left over.

Question 5.
A Poultry has produced 15472 eggs and fits 30 eggs in a tray. How many trays do they need?
Solution:
Total eggs = 15472
No. of eggs in 1 tray = 30
No. of trays needed = 15472 ÷ 30 = 516
No. of trays needed = 516 [515 + 1 for remaining 22 eggs]

Challenging Problems (Text book Page No.36 & 37)

Question 6.
Read the table and answer the following questions.

Solution:
(i) Write the Canopus star’s diameter in words in the Indian and the International System.
(ii) Write the sum of the place values of 5 in Sirius star’s diameter in Indian System.
(iii) Eight hundred sixty-four million seven hundred thirty. Write in Indian System
(iv) Write the diameter in words of Arcturus star in the International System.
(v) Write the difference of the diameters of Canopus and Arcturus star in the Indian and the International Systems.
Solution:
(i) Canopus star’s diameter is 25941900 miles
Indian System: Two crores Fifty-Nine Lakh Forty-one thousand Nine Hundred
International System: Twenty-Five Million Nine Hundred Forty-One Thousand Nine Hundred.

(ii) Sirus star’s diameter = 1556500 miles
Sum of place values of 5 is 5 × 100000 + 5 × 10000 + 5 × 100
= 500000 + 50000 + 500
= 5,50,500

(iii) Given value is 864,000,730
In Indian System 86,40,00,730
Eighty-six crore forty lakhs seven hundred and thirty.

(iv) The diameter of the Arcturus Star is 19,888,800 miles
Nineteen Million Eight Hundred and Eighty-Eight Thousand Eight Hundred.

(v) The diameter of Canopus = 25941900
Diameter of Arcturus = 19888800
Difference = 6053100
In Indian System 60,53,100
Sixty lakh fifty-three thousand one hundred.
In International System 6,053,100
Six Million fifty-three thousand one hundred.

Question 7.
Anbu asks Arivu Selvi to guess a five digit odd number. He gives the following hints.
i. The digit in the 1000s place is less than 5
ii. The digit in the 100s place is greater than 6
iii. The digit in the 10s place is 8
What is Arivu Selvi answer? Does she give more than one answer?
Solution:
63785, 53781

Question 8.
A Music concert is taking place in a stadium, A total of 7,689 chairs are to be put in rows of 90.
(i) How many rows will there be?
(ii) Will there be any chairs left over?
Solution:
(i) There will be 85 rows
(ii) Yes, there are 39 chairs left over.

Question 9.
Round off the seven-digit number 29,75,842 to the nearest lakhs and ten lakhs. Are they the same?
Solution:

To the nearest lakhs = 30,00,000
To the nearest ten lakhs = 30,00,000
Yes, they are the same.

Question 10.
Find the 5 or 6 or 7 digit numbers from a newspaper or a magazine to get a rounded number to the nearest ten thousand.
Solution:
(i) 14276 \(\simeq\) 10000
(ii) 1,86945 \(\simeq\) 1,90000

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.5

Question 1.
Fill in the blanks.
(i) The difference between the smallest natural number and the smallest whole number is _____
(ii) 17 × ___ = 34 × 17
(iii) When _____ is added to a number, it remains the same.
(iv) Division by ____ is not defined.
(v) Multiplication by ____ leaves a number unchanged.
Solution:
(i) 1
Hint: 1 – 0 = 1
(ii) 34
(iii) 0
(iv) 0
(v) 1

Question 2.

Say True or False.

1. 0 is the identity for multiplication of whole numbers.
2. Sum of two whole numbers is always less than their product.
3. Both addition and multiplication are associative for whole numbers.
4. Both addition and multiplication are commutative for whole numbers.
5. Multiplication is distributive over addition for whole numbers.

Solution:

1. False
2. False
3. True
4. True
5. True

Question 3.
Name the property being illustrated in each of the cases given below:

1. 75 + 34 = 34 + 75
2. (12 × 4) × 8 = 12 × (4 × 8)
3. 50 + 0 = 50
4. 50 × 1 = 50
5. 50 × 42 = 50 × 40 + 50 × 2

Solution:

1. Addition is commutative
2. Multiplication is associative
3. 0 is the additive identity
4. 1 is the multiplicative identity.
5. Distributivity of multiplication over addition

Question 4.
Use the properties of whole numbers and simplify.
(i) 50 × 102
(ii) 500 × 689 – 500 × 89
(iii) 4 × 132 × 25
(iv) 196 + 34 + 104
Solution:
(i) 50 × 102
= 50 × (100 + 2)
= (50 × 100) + (50 × 2)
= 5000 + 100 = 5100

(ii) 500 × 689 – 500 × 89
= 500 × (689 – 89)
= 344500 – 44500
= 300000
= 500 × (689 – 89)
= 500 × 600
= 3,00000

(iii) (4 × 132) × 25
= 4 × (132 × 25)
= (4 × 132) × 25
= 528 × 25
= 13200
= 4 × (132 × 25)
= 4 × 3300
= 13200

(iv) (196 + 34) + 104 = 196 + (34 + 104)
(196 + 34) + 104 = 230 + 104 = 334
196 + (34 + 104) = 196 + 138 = 334

Objective Type Questions

Question 5.
(53 + 49) × 0 is
(a) 102
(b) 0
(c) 1
(d) 53 + 49 + 0
Solution:
(b) 0
53 × 0 + 49 × 0 = 0 + 0 = 0

Question 6.
\(\frac{59}{1}\) is
(a) 1
(b) 0
(c) \(\frac{1}{59}\)
(d) 59
Solution:
(d) 59

Question 7.
The product of a non-zero whole number and its successor is always
(a) an even number
(b) an odd number
(c) zero
(d) none of these
Solution:
(a) an even number

Question 8.
The whole number that does not have a predecessor is
(a) 10
(b) 0
(c) 1
(d) none of these
Solution:
(b) 0
0 is the smallest whole number

Question 9.
Which of the following expressions is not zero?
(a) 0 × 0
(b) 0 + 0
(c) 2/0
(d) 0/2
Solution:
(c) 2/0

Question 10.
Which of the following is not true?
(a) (4237 + 5498) + 3439 = 4237 + (5498 + 3439)
(b) (4237 × 5498) × 3439 = 4237 × (5498 × 3439)
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439
(d) 4237 × (5498 + 3439) = (4237 × 5498) + (4237 × 3439)
Solution:
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.2

Question 1.
Find in the blanks.
(i) The algebraic statement of ‘f’ decreased by 5 is _______
(ii) The algebraic statement of ‘s’ divided by 5 is _______
(iii) The verbal statement of ‘2m – 10’ is _______
(iv) If A’s age is ‘n’ years now, 7 years ago A’s age was ______
(v) If ‘p – 5’ gives 12 then ‘p’ is ______
Solution:
(i) f – 5
(ii) \(\frac { s }{ 5 }\)
(iii) 10 less than 2 times m (or) Take away 10 from the product of 2 and m
(iv) n – 7
(v) 17
Hint: n – 5 = 12⇒ n = 12 + 5 = 17

Question 2.
Say True or False.
(i) 10 more to three times ‘c’ is ‘3c + 13’.
(ii) If the cost of 10 rice bags is Rs ‘t’, then the cost of 1 rice bag is Rs \(\frac{t}{10}\)
(iii) The statements ‘x’ divided by 3 and 3 divided by ‘x’ are the same.
(iv) The product of ‘q’ and 20 is ‘20q’.
(v) 7 less to 7 times ‘y’ is ‘7 – 7y’.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
Express the following verbal statement to algebraic statement.
(i) ‘t’ is added to 100
(ii) 4 times ‘q’
(iii) 8 is reduced by ‘y’
(iv) 56 added to 2 times ‘x’
(v) 4 less to 9 times of ‘y’
Solution:
(i) t + 100
(ii) 4q
(iii) 8 – y
(iv) 2x + 56
(v) 9y – 4

Question 4.
Express the following algebraic statement to verbal statement.
(i) x ÷ 3
(ii) 5n – 12
(iii) 11 + 10x
(iv) 70s
Solution:
(i) x divided by 3
(ii) 12 less to 5 times n
(iii) 11 added to 10 times x
(iv) 70 times s

Question 5.
The teacher asked two students to write the algebraic statement for the verbal statement “8 more than a number” on the board. Vetri wrote ‘8 + x’ but Maran wrote ‘8x’. Who gave the correct answer?
Solution:
Let the number be x; 8 more than the number = 8 + x.
Vetri gave the correct answer as 8 + x.

Question 6.
Answer the following questions:
(i) If ‘n’ takes the value 3 then find the value of ‘n + 10’
(ii) If ‘g’ is equal to 300. What is the value of ‘g – 1’ and ‘g + 1’?
(iii) What is the value of s; if ‘2s – 6’ gives 30?
Solution:
(i) Given n = 3
then n + 10 = 3 + 10 = 13
n + 10 = 13

(ii) Given g = 300
g – 1 = 300 – 1 = 299
g + 1 = 300 + 1 = 301
g – 1 = 299; g + 1 = 301

(iii) Given 2s – 6 = 30
From the table:
2 × 18 – 6 = 30
s = 18

Question 7.
Complete the table and find the value of ‘k’ for which \(\frac { k }{ 3 }\) gives 5.
Solution:
Given: k ÷ 3 = 5 From the above table we find that
15 ÷ 3 = 5
k = 5

Objective Type Questions

Question 8.
The value of ‘y’ in y + 7 = 13 is
(a) y = 5
(b) y = 6
(c) y = 7
(d) y = 8
Solution:
(b) y = 6

Question 9.
6 less to ‘n’ gives 8 is represented as
(a) n – 6 = 8
(b) 6 – n = 8
(c) 8 – n = 6
(d) n – 8 = 6
Solution:
(a) n – 6 = 8

Question 10.
The value of ‘c’ for which \(\frac{3c}{4}\) gives 18 is
(a) c = 15
(b) c = 21
(c) c = 24
(d) c = 27
Solution:
(c) c = 24

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.4

Question 1.
Fill in the blanks.
(i) The nearest 100 of 843 is _____
(ii) The nearest 1000 of 756 is ______
(iii) The nearest 10,000 of 85654 is ______
Solution:
(i) 800.
The digit in tens place is 4 < 5
(ii) 1000.
The digit in hundred places is 7 ≥ 5
(iii) 90,000.
The digit in a thousand places is 5 ≥ 5

Question 2.
Say True or False.
(i) 8567 is rounded off as 8600 to the nearest 10.
(ii) 139 is rounded off as 100 to the nearest 100.
(iii) 1,70,51,972 is rounded off as 1,70,00,000 to the nearest lakh.
Solution:
(i) False
(ii) True
(iii) False

Question 3.
Round off the following to the given nearest place.
(i) 4065; hundred
(ii) 44,555; thousand
(iii) 86943; ten thousand
(iv) 50,81,739; lakh
(v) 33,75,98,482; Ten crore
Solution:
(i) We have to round off 4065 to hundreds
The place value is a hundred
The digit in hundreds place is 0
The digit to the right is 6 > 5
Adding 1 to 0 ⇒ 0 + 1 = 1
Changing the digits to the right of 41 to zeros
4065 \(\simeq\) 4100 ⇒ 4100

(ii) To round off 44555 to thousands
The digit in the thousands place is 4
digit to the right is 5 = 5.
Adding 1 to the thousand place value digit 4 ⇒ 4 + 1 = 5
Changing the digits to the right of 45 to zeros we get 45000
44555 \(\simeq\) 45000

(iii) To round off 86,943 to ten thousand
The digit in ten thousand places is 8
The digit to the right is 6 > 5
Adding 1 to 8 ⇒ 8 + 1 = 9.
Changing the digits to the right of 9 to zeroes we get 90,000
86943 \(\simeq\) 90,000

(iv) To round off 5081739 to lakhs
The digit in the lakhs place is 0
The digit to the right is 8 > 5
Adding 1 to 0 ⇒ 1 + 0 = 1
Changing the digits right off 51 to zeros
We get 51,00,000
5081739 \(\simeq\) 51,00,000

(v) To round off 337598482 to ten crore
The digit in the ten crores place is 3
The digit to the right is 3 < 0
The digit in ten crore place remains the same 3.
Changing the digit to the right of 3 zeros we get 30,00,00,000
33,75,98,482 \(\simeq\) 30,00,00,000

Question 4.
Estimate the sum of 157826 and 32469 rounded off to the nearest ten thousand.
Solution:

Question 5.
Estimate by rounding off each number to the nearest hundred.
(i) 8074 + 4178
(ii) 1768977 + 130589
Solution:

Question 6.
The population of a city was 43,43,645 in the year 2001 and 46,81,087 in the year 2011. Estimate the increase in population by rounding off to the nearest thousand.
Solution:
Population in 2001 = 43,43,645
Population in 2011 = 46,81,087
Increase in population = 46,81,087 – 43,43,645 = 3,37,442
When rounded off to the nearest thousand = 3,37,000

Objective Type Questions

Question 7.
The number which on rounding off to nearest thousands given 11000 is
(a) 10345
(b) 10855
(c) 11799
(d) 10056
Solution:
(b) 10855
In 10855, the digit in a hundred places is 8 ≥ 5. So 11000

Question 8.
The estimation to the nearest hundred of 76812 is
(a) 77000
(b) 76000
(c) 76800
(d) 76900
Solution:
(c) 76800

Question 9.
The number 9785764 is rounded off to nearest lakh as
(a) 9800000
(b) 9786000
(c) 9795600
(d) 9795000
Solution:
(a) 9800000
In ten thousand places, the digit is 8 5. So 9800000

Question 10.
The estimated difference of 167826 and 2765 rounded off to the nearest thousand is
(a) 180000
(b) 165000
(c) 140000
(d) 155000
Solution:
(b) 165000