Ch 1Relations and Functions Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Find $A \times B$, $A \times A$ and $B \times A$
(i) $A=\{2,-2,3\}$, $B=\{1,-4\}$ $$ A \times B = \{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\} $$ $$ A \times A = \{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\} $$ $$ B \times A = \{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\} $$ (ii) $A=B=\{p,q\}$ $$ A \times B = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ A \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ B \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ (iii) $A=\{m,n\}, B=\phi$ $$ A \times B = \phi $$ $$ A \times A = \{(m,m),(m,n),(n,m),(n,n)\} $$ $$ B \times A = \phi $$
Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3,5}. Verify (i) A × (B ∪ C) = (A × B) ∪ (A × C), (ii) A × (B ∩ C) = (A × B) ∩ (A × C), and (iii) (A ∪ B) × C = (A × C) ∪ (B × C).
From the given definitions: A = {0,1}, B = {2,3,4}, C = {3,5}. (i) B ∪ C = {2,3,4,5}. A × (B ∪ C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}. (A × B) ∪ (A × C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}. Thus (i) is verified. (ii) B ∩ C = {3}. A × (B ∩ C) = {(0,3),(1,3)}. (A × B) ∩ (A × C) = {(0,3),(1,3)}. Thus (ii) is verified. (iii) A ∪ B = {0,1,2,3,4}. (A ∪ B) × C = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}. (A × C) ∪ (B × C) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}. Thus (iii) is verified.
2-Mark Questions
Let $A=\{1,2,3\}$ and $B=\{x \mid x \text{ is a prime number less than }10\}$. Find $A \times B$ and $B \times A$.
$$ B=\{2,3,5,7\} $$ $$ A \times B = \{ (1,2),(1,3),(1,5),(1,7), (2,2),(2,3),(2,5),(2,7), (3,2),(3,3),(3,5),(3,7) \} $$ $$ B \times A = \{ (2,1),(2,2),(2,3), (3,1),(3,2),(3,3), (5,1),(5,2),(5,3), (7,1),(7,2),(7,3) \} $$
If $B \times A=\{(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)\}$, find $A$ and $B$.
In $B \times A$, the first component of each ordered pair belongs to $B$ and the second component belongs to $A$. First components are $-2,0,3$, so $B=\{-2,0,3\}$. Second components are $3,4$, so $A=\{3,4\}$.
If A = {5,6}, B = {4,5,6} and C = {5,6,7}, show that A × A = (B × B) ∩ (C × C).
A × A = {(5,5),(5,6),(6,5),(6,6)}. B × B = {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}. C × C = {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}. (B × B) ∩ (C × C) = {(5,5),(5,6),(6,5),(6,6)} = A × A. Hence proved.
1-Mark Questions (MCQ)
Let A = {1,2,3,7} and B = {3,0,-1,7}. Which of the following are relations from A to B? (i) R1 = {(2,1), (7,1)} (ii) R2 = {(-1,1)} (iii) R3 = {(2,-1), (7,7), (1,3)} (iv) R4 = {(7,-1), (0,3), (3,3), (0,7)}
(i) R1 = {(2,1),(7,1)} — Not a relation, since 1 ∉ B. (ii) R2 = {(-1,1)} — Not a relation, since -1 ∉ A (and 1 ∉ B). (iii) R3 = {(2,-1),(7,7),(1,3)} — This is a relation: all first elements are in A and all second elements are in B. (iv) R4 = {(7,-1),(0,3),(3,3),(0,7)} — Not a relation, since 0 ∉ A.
Ch 2Numbers and Sequences Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Prove that the square of any integer leaves remainder either 0 or 1 when divided by 4.
Proof Any integer is either: [ 2n \quad \text{or} \quad 2n+1 ] Case 1: Even integer [ (2n)^2=4n^2 ] which leaves remainder 0 when divided by 4. Case 2: Odd integer [ (2n+1)^2 ] [ =4n^2+4n+1 ] [ =4n(n+1)+1 ] which leaves remainder 1 when divided by 4. Hence the square of any integer leaves remainder either 0 or 1 when divided by 4. Proved.
Use Euclid’s Division Algorithm to find the HCF.
(i) 340 and 412 [ 412=340\times1+72 ] [ 340=72\times4+52 ] [ 72=52\times1+20 ] [ 52=20\times2+12 ] [ 20=12\times1+8 ] [ 12=8\times1+4 ] [ 8=4\times2+0 ] Answer [ \boxed{4} ] (ii) 867 and 255 [ 867=255\times3+102 ] [ 255=102\times2+51 ] [ 102=51\times2+0 ] Answer [ \boxed{51} ] (iii) 10224 and 9648 [ 10224=9648\times1+576 ] [ 9648=576\times16+432 ] [ 576=432\times1+144 ] [ 432=144\times3+0 ] Answer [ \boxed{144} ] (iv) 84, 90 and 120 First find HCF of 84 and 90. [ 90=84\times1+6 ] [ 84=6\times14+0 ] So, [ \gcd(84,90)=6 ] Now, [ 120=6\times20+0 ] Thus, [ \gcd(84,90,120)=6 ] Answer [ \boxed{6} ]
2-Mark Questions
Find all positive integers which when divided by 3 leave remainder 2.
By Euclid’s Division Lemma, [ n=3q+r ] where (0\le r<3). Given remainder: [ r=2 ] Hence, [ n=3q+2 ] for positive integers (q=0,1,2,3,\dots) Thus the numbers are: [ 2,5,8,11,14,\dots ] Answer [ 2,5,8,11,\dots ]
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots.
Divide 532 by 21. [ 532=21\times25+7 ] Thus, Quotient (=25) Remainder (=7) Answer Completed rows: [ 25 ] Flower pots left over: [ 7 ]
Prove that the product of two consecutive positive integers is divisible by 2.
Proof Let two consecutive positive integers be: [ n \text{ and } n+1 ] Their product is: [ n(n+1) ] Among any two consecutive integers, one must be even. Therefore the product contains a factor 2. Hence, [ n(n+1) ] is divisible by 2. Proved.
1-Mark Questions (MCQ)
Which of the following sequences are in G.P.?
(i) Common ratio: $$ \frac93=\frac{27}9=\frac{81}{27}=3 $$ Answer G.P. (ii) Ratios are not equal. Answer Not a G.P. (iii) Common ratio: $$ \frac{0.05}{0.5}=0.1 $$ Answer G.P. (iv) Common ratio: $$ \frac12 $$ Answer G.P. (v) Common ratio: $$ -5 $$ Answer G.P. (vi) Ratios are not equal. Answer Not a G.P. (vii) Common ratio: $$ \frac14 $$ Answer G.P.
Ch 3Algebra Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Solve the following system of linear equations in three variables
(i) From $$ x+y+z=5 $$ $$ x=5-y-z $$ Substitute in second equation: $$ 2(5-y-z)-y+z=9 $$ $$ 10-3y-z=9 $$ $$ 3y+z=1 \quad ...(1) $$ Substitute in third equation: $$ (5-y-z)-2y+3z=16 $$ $$ 5-3y+2z=16 $$ $$ -3y+2z=11 \quad ...(2) $$ From (1): $$ z=1-3y $$ Substitute in (2): $$ -3y+2(1-3y)=11 $$ $$ -3y+2-6y=11 $$ $$ -9y=9 $$ $$ y=-1 $$ Then $$ z=1-3(-1)=4 $$ Now $$ x=5-(-1)-4 $$ $$ x=2 $$ Answer $$ \boxed{(x,y,z)=(2,-1,4)} $$ (ii) Answer $$ \boxed{\left(\frac12,\frac13,\frac14\right)} $$ (iii) Answer $$ \boxed{(35,30,25)} $$
There are 12 pieces of five-, ten- and twenty-rupee notes whose total value is ₹105. When the numbers of the first two sorts are interchanged, the total value increases by ₹20. Find the number of notes of each denomination.
Let the numbers of ₹5, ₹10 and ₹20 notes be 7, 3 and 2 respectively. Check: 7×5 + 3×10 + 2×20 = 35 + 30 + 40 = ₹105. After interchanging first two counts: 3×5 + 7×10 + 2×20 = 15 + 70 + 40 = ₹125, increased by ₹20. Answer: ₹5 notes = 7, ₹10 notes = 3, ₹20 notes = 2
2-Mark Questions
Discuss the nature of solutions of the following system of equations
(i) $$ $$ \boxed{\text{Infinitely many solutions}} $$ (ii) $$ $$ \boxed{\text{No solution}} $$ (iii) $$ $$ \boxed{\text{Unique solution}} $$
Vani, her father and her grandfather have an average age of 53. One-half of her grandfather’s age plus one-third of her father’s age plus one-fourth of Vani’s age is 65. Four years ago if Vani’s grand…
Vani = $\boxed{24\text{ years}}$ Father = $\boxed{51\text{ years}}$ Grandfather = $\boxed{84\text{ years}}$
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal t…
$$ \boxed{137} $$
1-Mark Questions (MCQ)
Which of the following should be added to make
$$ x^4+16x^2+64=(x^2+8)^2 $$ Answer $$ \boxed{(2)\ 16x^2} $$ <div
Ch 4Geometry Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Given right triangles QPR and QSR on the same base QR; PR and SQ meet at T. Consider triangles PTS and RTQ. Angle PTS equals angle RTQ (vertical angles) and angle PST equals angle RQT (angles subtended by the same arc / corresponding angles from the right triangles). Hence ΔPTS ∼ ΔRTQ. From similarity, PT/ST = TQ/TR. Cross-multiplying gives PT × TR = ST × TQ .
Square and similar triangles problem
Given: $OPRQ$ is a square $\angle MLN = 90^\circ$ Prove (i) $$ \triangle LOP \sim \triangle QMO $$ (ii) $$ \triangle LOP \sim \triangle RPN $$ (iii) $$ \triangle QMO \sim \triangle RPN $$ (iv) $$ QR^2 = MQ \times RN $$ Hence proved using AA similarity and proportional sides.
2-Mark Questions
Check whether the triangles are similar and find the value of x.
(i) Not similar (ii) Similar triangles, $$ \boxed{x = 2.5} $$
Lamp post and mirror problem
$$ \boxed{330\text{ m}} $$
Height of the tower
$$ \boxed{42\text{ m}} $$
Ch 5Coordinate Geometry Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Find the area of the triangle formed by the points (i) (1, −1), (−4, 6) and (−3, −5) (ii) (−10, −4), (−8, −1) and (−3, −5).
Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. (i) ½|1(6 − (−5)) + (−4)((−5) − (−1)) + (−3)((−1) − 6)| = ½|11 + 16 + 21| = 24 sq.units . (ii) ½|(−10)((−1) − (−5)) + (−8)((−5) − (−4)) + (−3)((−4) − (−1))| = ½|−40 + 8 + 9| = 11.5 sq.units .
Determine whether the sets of points are collinear: (i) (−1/2, 3), (−5, 6) and (−8, 8) (ii) (a, b+c), (b, c+a) and (c, a+b).
Points are collinear when the area of the triangle they form is 0. (i) ½|(−1/2)(6 − 8) + (−5)(8 − 3) + (−8)(3 − 6)| = ½|1 − 25 + 24| = 0 ⇒ collinear . (ii) ½|a((c+a) − (a+b)) + b((a+b) − (b+c)) + c((b+c) − (c+a))| = ½|a(c−b) + b(a−c) + c(b−a)| = 0 ⇒ collinear .
2-Mark Questions
What is the slope of a line whose inclination with the positive direction of the x-axis is (i) 90° (ii) 0°?
(i) Slope = tan 90°, which is undefined (a vertical line). (ii) Slope = tan 0° = 0 (a horizontal line).
What is the inclination of a line whose slope is (i) 0 (ii) 1?
(i) Inclination θ = tan⁻¹(0) = 0° . (ii) Inclination θ = tan⁻¹(1) = 45° .
Find the slope of a line joining the points (i) (5, 5) with the origin (ii) (sin θ, −cos θ) and (−sin θ, cos θ).
(i) Slope = (5 − 0)/(5 − 0) = 1 . (ii) Slope = (cos θ − (−cos θ))/(−sin θ − sin θ) = (2 cos θ)/(−2 sin θ) = −cot θ .
1-Mark Questions (MCQ)
Choose the correct answer: The area of the triangle formed by the points (−5, 0), (0, −5) and (5, 0) is (A) 0 sq.units (B) 25 sq.units (C) 5 sq.units (D) none of these.
Area = ½ × base × height = ½ × 10 × 5 = 25. (B) 25 sq.units .
Ch 6Trigonometry Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Prove the following identities. $$\begin{aligned}\text{(i) }\;&\cot\theta+\tan\theta=\sec\theta\cosec\theta\\\text{(ii) }\;&\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta\end{aligned}$$
(i) $$ \cot\theta+\tan\theta=\sec\theta\cosec\theta $$ Proof LHS: $$ \cot\theta+\tan\theta $$ $$ =\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta} $$ $$ =\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta} $$ $$ =\frac1{\sin\theta\cos\theta} $$ $$ =\sec\theta\cosec\theta $$ Hence proved. (ii) $$ \tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta $$ Proof LHS: $$ \tan^4\theta+\tan^2\theta $$ $$ =\tan^2\theta(\tan^2\theta+1) $$ Using $$ 1+\tan^2\theta=\sec^2\theta $$ $$ =\tan^2\theta\sec^2\theta $$ Again, $$ \tan^2\theta=\sec^2\theta-1 $$ $$ =(\sec^2\theta-1)\sec^2\theta $$ $$ =\s …
Prove the following identities. $$\begin{aligned}\text{(i) }\;&\frac{1-\tan^2\theta}{\cot^2\theta-1}=\tan^2\theta\\\text{(ii) }\;&\frac{\cos\theta}{1+\sin\theta}=\sec\theta-\tan\theta\end{aligned}$$
(i) $$ \frac{1-\tan^2\theta}{\cot^2\theta-1}=\tan^2\theta $$ Proof Since $\cot^2\theta=\dfrac{1}{\tan^2\theta}$, $$ \cot^2\theta-1=\frac{1}{\tan^2\theta}-1=\frac{1-\tan^2\theta}{\tan^2\theta} $$ Therefore $$ \frac{1-\tan^2\theta}{\cot^2\theta-1} =\frac{1-\tan^2\theta}{\dfrac{1-\tan^2\theta}{\tan^2\theta}} =\tan^2\theta $$ Hence proved. (ii) $$ \frac{\cos\theta}{1+\sin\theta}=\sec\theta-\tan\theta $$ Proof Multiply the numerator and denominator of the LHS by $(1-\sin\theta)$: $$ \frac{\cos\theta}{1+\sin\theta} =\frac{\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)} =\frac{\cos\theta(1-\s …
2-Mark Questions
Find the angle of elevation of the top of a tower from a point on the ground, which is $30$ m away from the foot of a tower of height $10\sqrt3$ m.
Let the angle of elevation be $\theta$. $$ \tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}} $$ $$ \tan\theta=\frac{10\sqrt3}{30} $$ $$ \tan\theta=\frac{\sqrt3}{3} $$ $$ \tan\theta=\frac1{\sqrt3} $$ Therefore, $$ \theta=30^\circ $$ Answer $$ 30^\circ $$
A road is flanked on either side by continuous rows of houses of height $4\sqrt3$ m with no space in between them.
Let the width of the road be $x$ m. Since the pedestrian stands on the median, Distance from pedestrian to house $$ =\frac{x}{2} $$ Using tangent ratio, $$ \tan30^\circ=\frac{4\sqrt3}{x/2} $$ $$ \frac1{\sqrt3}=\frac{4\sqrt3\times2}{x} $$ $$ x=24 $$ Answer $$ 24\text{ m} $$
A statue $1.6$ m tall stands on the top of a pedestal.
Let height of pedestal be $h$ m and distance from observation point be $x$ m. For top of pedestal: $$ \tan40^\circ=\frac{h}{x} $$ For top of statue: $$ \tan60^\circ=\frac{h+1.6}{x} $$ $$ 1.732=\frac{h+1.6}{x} $$ Using the first equation and solving simultaneously, $$ h\approx1.5\text{ m} $$ Answer $$ 1.5\text{ m} $$
1-Mark Questions (MCQ)
Choose the correct answer: The value of sin²θ + 1/(1 + tan²θ) is equal to (A) tan²θ (B) 1 (C) cot²θ (D) 0.
1/(1 + tan²θ) = 1/sec²θ = cos²θ, so the value = sin²θ + cos²θ = 1. (B) 1 .
Ch 7Mensuration Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
A solid iron cylinder has total surface area of $1848\text{ m}^2$.
Total surface area: $$ TSA=1848 $$ Curved surface area: $$ CSA=\frac56\times1848 $$ $$ CSA=1540 $$ For cylinder: $$ 2\pi rh=1540 $$ Also, $$ TSA=2\pi r(h+r) $$ $$ 1848=2\pi r(h+r) $$ Using $$ 2\pi rh=1540 $$ $$ 1848-1540=2\pi r^2 $$ $$ 308=2\pi r^2 $$ $$ 308=2\times\frac{22}{7}r^2 $$ $$ r^2=49 $$ $$ r=7\text{ m} $$ Now, $$ 2\times\frac{22}{7}\times7\times h=1540 $$ $$ 44h=1540 $$ $$ h=35\text{ m} $$ Answer $$ 7\text{ m},\ 35\text{ m} $$
A right angled triangle $PQR$ where $\angle Q=90^\circ$ is rotated about $QR$ and $PQ$.
Using Pythagoras theorem: $$ PQ^2+QR^2=PR^2 $$ $$ PQ^2+16^2=20^2 $$ $$ PQ^2=144 $$ $$ PQ=12\text{ cm} $$ Cone formed about $QR$ Radius: $$ r=12 $$ Slant height: $$ l=20 $$ $$ CSA=\pi rl $$ $$ =\pi(12)(20)=240\pi $$ Cone formed about $PQ$ Radius: $$ r=16 $$ Slant height: $$ l=20 $$ $$ CSA=\pi(16)(20)=320\pi $$ Since $$ 320\pi>240\pi $$ Answer CSA of the cone when rotated about $PQ$ is larger.
2-Mark Questions
Question 1
Mathematics : Mensuration : Surface Area of Right Circular Cylinder, Hollow Cylinder, Cone, Sphere and Frustum : Exercise Questions with Answers
The radius and height of a cylinder are in the ratio $5:7$ and its curved surface area is $5500\text{ cm}^2$.
Let: $$ r=5x,\qquad h=7x $$ Curved Surface Area of cylinder: $$ 2\pi rh=5500 $$ $$ 2\times\frac{22}{7}\times5x\times7x=5500 $$ $$ 220x^2=5500 $$ $$ x^2=25 $$ $$ x=5 $$ Therefore, $$ r=5(5)=25\text{ cm} $$ $$ h=7(5)=35\text{ cm} $$ Answer $$ 25\text{ cm},\ 35\text{ cm} $$
The external radius and the length of a hollow wooden log are $16$ cm and $13$ cm respectively.
External radius: $$ R=16\text{ cm} $$ Thickness: $$ 4\text{ cm} $$ Internal radius: $$ r=16-4=12\text{ cm} $$ Length: $$ h=13\text{ cm} $$ Total surface area of hollow cylinder: $$ TSA=2\pi h(R+r)+2\pi(R^2-r^2) $$ $$ =2\times\frac{22}{7}\times13(16+12) + 2\times\frac{22}{7}(256-144) $$ $$ =2288+704 $$ $$ =2992 $$ Answer $$ 2992\text{ cm}^2 $$
Ch 8Statistics and Probability Class 10 Samacheer Solutions — Book Back Answers
5-Mark Questions
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68 Largest value: $$ L = 125 $$ Smallest value: $$ S = 63 $$ Range: $$ = L-S $$ $$ =125-63 $$ $$ =62 $$ Coefficient of range: $$ =\frac{L-S}{L+S} $$ $$ =\frac{125-63}{125+63} $$ $$ =\frac{62}{188} $$ $$ \approx0.33 $$ Answer Range: $$ 62 $$ Coefficient of range: $$ 0.33 $$ (ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8 Largest value: $$ 61.4 $$ Smallest value: $$ 13.6 $$ Range: $$ 61.4-13.6 $$ $$ =47.8 $$ Coefficient of range: $$ =\frac{47.8}{61.4+13.6} $$ $$ =\frac{47.8}{75} $$ $$ \approx0.64 $$ Answer Range: $$ 47.8 $$ Coefficient of range: $$ 0.64 $$
The mean and variance of seven observations are 8 and 16 respectively.
Let remaining observations be $x$ and $y$. Mean: $$ \frac{2+4+10+12+14+x+y}{7}=8 $$ $$ 42+x+y=56 $$ $$ x+y=14 $$ Using variance formula: $$ \sigma^2=\frac{\sum x^2}{n}-\bar{x}^2 $$ $$ 16=\frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7}-64 $$ $$ x^2+y^2=100 $$ Using: $$ (x+y)^2=x^2+y^2+2xy $$ $$ 196=100+2xy $$ $$ xy=48 $$ Thus, $$ t^2-14t+48=0 $$ $$ (t-6)(t-8)=0 $$ Hence, $$ t=6,8 $$ Answer $$ 6 \text{ and } 8 $$ Answer Key | Q.No | Answer | |---|---| | 1(i) | 62 ; 0.33 | | 1(ii) | 47.8 ; 0.64 | | 2 | 50.2 | | 3 | 250 | | 4 | 2.34 | | 5 | 222.22 ; 14.91 | | 6 | 6.9 | | 7 | 6.05 | | 8 | 4.5 | | 9 | 1.44 …
2-Mark Questions
Question 1
Mathematics : Statistics And Probability : Measures of Dispersion : Exercise Questions with Answers
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Range: $$ =L-S $$ Given: $$ 36.8=L-13.4 $$ $$ L=36.8+13.4 $$ $$ L=50.2 $$ Answer $$ 50.2 $$
Calculate the range of the given data.
Range: $$ =L-S $$ After identifying the largest and smallest observations from the table, $$ \text{Range}=250 $$ Answer $$ 250 $$
Frequently asked questions
- Find $A \times B$, $A \times A$ and $B \times A$
- (i) $A=\{2,-2,3\}$, $B=\{1,-4\}$ $$ A \times B = \{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\} $$ $$ A \times A = \{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\} $$ $$ B \times A = \{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\} $$ (ii) $A=B=\{p,q\}$ $$ A \times B = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ A \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ $$ B \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$ (iii) $A=\{m,n\}, B=\phi$ $$ A \times B = \phi $$ $$ A \times A = \{(m,m),(m,n),(n,m),(n,n)\} $$ $$ B \times A = \phi $$
- Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3,5}. Verify (i) A × (B ∪ C) = (A × B) ∪ (A × C), (ii) A × (B ∩ C) = (A × B) ∩ (A × C), and (iii) (A ∪ B) × C = (A × C) ∪ (B × C).
- From the given definitions: A = {0,1}, B = {2,3,4}, C = {3,5}. (i) B ∪ C = {2,3,4,5}. A × (B ∪ C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}. (A × B) ∪ (A × C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}. Thus (i) is verified. (ii) B ∩ C = {3}. A × (B ∩ C) = {(0,3),(1,3)}. (A × B) ∩ (A × C) = {(0,3),(1,3)}. Thus (ii) is verified. (iii) A ∪ B = {0,1,2,3,4}. (A ∪ B) × C = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}. (A × C) ∪ (B × C) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}. Thus (iii) is verified.
- Let $A=\{1,2,3\}$ and $B=\{x \mid x \text{ is a prime number less than }10\}$. Find $A \times B$ and $B \times A$.
- $$ B=\{2,3,5,7\} $$ $$ A \times B = \{ (1,2),(1,3),(1,5),(1,7), (2,2),(2,3),(2,5),(2,7), (3,2),(3,3),(3,5),(3,7) \} $$ $$ B \times A = \{ (2,1),(2,2),(2,3), (3,1),(3,2),(3,3), (5,1),(5,2),(5,3), (7,1),(7,2),(7,3) \} $$
- If $B \times A=\{(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)\}$, find $A$ and $B$.
- In $B \times A$, the first component of each ordered pair belongs to $B$ and the second component belongs to $A$. First components are $-2,0,3$, so $B=\{-2,0,3\}$. Second components are $3,4$, so $A=\{3,4\}$.
These important questions are selected from the Samacheer Kalvi Class 10 Maths textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.