Practice set 26 Q1
ANSWER:1
Q2 find the value
Practice set 27
Q1 simplify
+info
Chapter 6
Find square root of a number (prime factors)
Practice set 30
i) The prime factorization of 625 is,
625 = 5 × 5 × 5 × 5 To find the square root, we will take one number from each pair and multiply. 625 −− 5×5 = 25 ∴√625−−− = 25
(ii) The prime factorization of 1225 is,
1225 = 5 × 5 × 7 × 7 To find the square root, we will take one number from each pair and multiply. √1225−−−−5×7=35 ∴√1225−−−−= 35
(iii) The prime factorisation of 289 is, 289 = 17 × 17 To find the square root, we will take one number from each pair and multiply. √289−−−17 =17 ∴√ 289−−− =17
(iv) The prime factorization of 4096 is, 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 To find the square root, we will take one number from each pair and multiply. 4096−−−−2×2×2×2×2×2 = 64 ∴√4096−−−−64
(v) The prime factorizationn of 1089 is, 1089 = 3 × 3 × 11 × 11 To find the square root, we will take one number from each pair and multipy. √1089−−−−3×11=33 ∴√1089−−−−33
Chapter -8 Algebraic Expressions
Like terms and unlike terms (i)2x, (i) 7xy, 5x, 9y2, -2/3 x, -2xyz
Types of algebraic expression
Practice set 32
Question 1:
Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
ANSWER:1It is known that, expressions with one term is called monomial, expressions with two terms are binomials, expressions with three terms are trinomials and expression with more than three terms are polynomials. (i) 7x = Monomial (ii) 5y−7z = Binomial (iii) 3x3−5x2−11 = Trinomial (iv) 1−8a−7a2−7a3 = Polynomial (v) 5m−3 = Binomial (vi) 4 = Monomial (vii) 3y2−7y+5 = Trinomial
Chapter -8 Algebraic Expressions
Practice set 33
Question 1:
Add.
(i) 9p + 16q ; 13p + 2q ANSWER: i) (9p + 16q) + (13p + 2q) = 9p + 16q + 13p + 2q =(9p + 13p) + (16q + 2q) =22p + 18q
(ii) (2a + 6b + 8c) + (16a + 13c + 18b) = 2a + 6b + 8c + 16a + 13c + 18b = (2a + 16a) + (6b + 18b) + (8c + 13c) = 18a + 24b + 21c
(iii) (13x2 − 12y2) + (6x2 − 8y2) = 13x2 − 12y2 + 6x2 − 8y2 = (13x2 + 6x2) + (−12y2 − 8y2) =19x2 + (−20y2) =19x2 − 20y2
(iv) (17a2b2 + 16c) + (28c − 28a2b2) = 17a2b2 + 16c + 28c − 28a2b2 = (17a2b2 − 28a2b2) + (16c + 28c) = −11a2b2 + 44c
(v) (3y2 − 10y + 16) + (2y − 7) = 3y2 − 10y + 16 + 2y − 7 = 3y2 + (−10y + 2y) + (16 − 7) =3y2 + (−8y) + 9 =3y2 − 8y + 9
(vi) (−3y2 + 10y − 16) + (7y2 + 8) = −3y2 + 10y − 16 + 7y2 + 8 = (−3y2 + 7y2) + 10y + (−16 + 8) = 4y2 + 10y + (−8) = 4y2 + 10y − 8
THANKS!
Chapter 5 -Operation on Rational Numbers
If m is an integer and n is any non zero integer, then the number m is called a rational number. n
Practice set 22
Question 1: Carry out the following additions of rational numbers. (i) 5 + 6
36 42 LCM of 36 and 42 = 2 × 2 × 3 × 3 × 7 = 252 =5 x 7 + 6 x 6 = 35 + 36 = 71 36 x 7 42 x 6 252 252 252
= 1 x 3 +2 + 2 x 5 +4 3 5 =5 + 14 3 5 LCM of 3 and 5 is 15 =5 x 5 + 14 x 3 3 x 5 5 x 3 = 25 + 42 15 15 = 25+42 = 67 15 15
3. 11 + 13 17 19 LCM of 17 x 19= 323 =11 x 19 + 13 x 17 17 x 19 19 x 17 209 + 221 =209+221 323 323 323 =430 323
11 x 2 +3 + 77 x 1 + 3 11 77 LCM of 11 and 77 is 77 =25x7 + 80 = 175+80 = 255 11x7 77 77 77 =
Question 2: Carry out the following subtractions involving rational numbers. (i) 7 − 3 11 7 LCM of 11 and 7 is 77 7 x 7 - 3 x 11 = 49-33 = 16 11 x 7 7 x11 77 77
(ii) 13 − 2 36 40 LCM of 36 and 40 is 360 13 x10 - 2 x 9 = 130-18 = 112÷8 = 14 36 x10 40 x9 360 360÷8 45
(iii) =1×3+2 − 3×6+5 = 5 − 23 3 6 3 6 LCM of 3 and 6 is 6. = 5×2 − 23×1 3×2 6×1 = 10 − 23 6 6 = 10−23 6 = − 13 6
Practice set 22
Question 3: Multiply the following rational numbers. (i) 3 × 2 = 6 11 5 55 (ii) 12 × 4 = 48 ÷3 = 16 5 15 75 ÷3 25 iii) (−8) × 3 = -8 x 3 = -24 = -24 9 4 9 x 4 36 3 (iv) 0 × 3 = 0 = 0 6 4 24
Question 4: Write the multiplicative inverse. (i) 2 Ans. Multiplicative inverse is 5 5 2 (ii) −3 Ans. Multiplicative inverse is -8 8 3 (iii) −17 Ans. Multiplicative inverse is -39 39 17 (iv) 7 Ans. Multiplicative inverse is 1 7 (v) = 3 x7 +1 = - 22 Ans. Multiplicative inverse is -3 3 3 22
Question 5:
Carry out the divisions of rational numbers. (i) 40 ÷ 10 = 40 × 4 = 40×4 = 160 = 160÷40 12 4 12 10 12×10 120 120÷40 (Since, HCF of 160 and 120 is 40)= 4 3 (ii) −10÷ −11 = (−10)×(−10= (−10)×(−10)= 100 11 10 11 11 11×11 121 (iii) −7 ÷ −3 8 6 =(−7)×(−6)=(−7)×(−6)= 42 8 3 8×3 24 =42÷6 (Since, HCF of 42 and 24 is 6) = 7 24÷6 4
(iv) 2 ÷(−4) = 2 × (−1) = 2×(−1) = −2 = −2÷2 3 3 4 3×4 12 12÷2 (Since, HCF of 2 and 12 is 2) = −1 6 (v) = 2×5+1÷ 5×6+3 = 11÷ 33 = 11 × 6 = 11×6 = 66 = 66÷33 5 6 5 6 5 33 5×33 165 165÷33 (Since, HCF of 66 and 165 is 33) = 2 5
THANKS!
THANKS!
Chapter 3 - HCF And LCM
Question 1: Which number is neither a prime number nor a composite number? ANSWER:
1 is neither a prime number nor a composite number
Page No 15:
Question 2: Which of the following are pairs of co-primes?
(i) 8, 14
(ii) 4, 5
(iii) 17, 19
(iv) 27, 15 ANSWER: Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime We can write 17 as 17 × 1 and 19 as 17 × 1.
Hence, 17 and 19 is a pair of co-prime numbers.
Question 3: List the prime numbers from 25 to 100 and say how many they are. ANSWER: There are a total of 16 prime numbers between 25 and 100 which are 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Question 4: Write all the twin prime numbers from 51 to 100. ANSWER: If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers between 51 and 100 are 59 and 61, 71 and 73
Question 5: Write 5 pairs of twin prime numbers from 1 to 50. ANSWER: If the difference between two co-prime numbers is 2 then, the numbers are said to be twin prime numbers.
Hence, the twin prime numbers from 1 to 50 are (2,3), (5,7), (11,13), (17,19) and (29,31).
Question 6: Which are the even prime numbers? ANSWER: There is only even prime number which is 2.
Page No 17:
Question 1:
Factorise the following numbers into primes.
(i)
32 = 2 × 2 × 2 × 2 × 2
ii)
57 = 3 × 19(iii)
23 = 23 × 1 (iv)
150 = 2 × 3 × 5 × 5 (v)
216 = 2 × 2 × 2 × 3 × 3 × 3 (vi)
208 =2 × 2 × 2 × 2 × 13 (vii)
765 = 3 × 3 × 5 × 17 (viii)
342 = 2 × 3 × 3 × 19 (ix)
377 = 13 × 29 (x)
559 = 13 × 43
Find the HCF.
Question 2: Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
Question 1: Choose the right option. (i) The HCF of 120 and 150 is ................... . (1) 30 (2) 45 (3) 20 (4) 120
(ii) The HCF of this pair of numbers is not 1. (1) 13, 17 (2) 29, 20 (3) 40, 20 (4) 14, 15
Question 2:
Find the HCF and LCM
Question 3:
Find the LCM.
(i) 36, 42
Question 4: Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time
Question 9: Which two consecutive even numbers have an LCM of 180? ANSWER: Let us suppose the two consecutive even numbers be 2x and 2x + 2.
Now, product of two numbers = HCF × LCM
⇒ (2x)(2x + 2) = 2 × 180 (HCM of two even number is 2)
⇒ (x)(2x + 2) = 180
⇒ 2x2 + 2x = 180
⇒ 2x2 + 2x − 180 = 0
⇒ x2 + x − 90 = 0
⇒ (x − 9)(x + 10) = 0
⇒ x − 9 = 0 or x + 10 = 0
⇒ x = 9 or x = −10 (Neglecting) Hence, the two consecutive even numbers are18 and 20.
THANKS!
Ch2 Multiplication and Division of integers
page 11 1. 5 + 7 = 12 2. 10 + ( - 5) = 5 3. - 4 + 3 = -1 4. (-7) + (- 2 ) = -9 5. ( + 8 ) - (+ 3 ) = 5 6. ( + 8 ) - (- 3 ) = 11
+ x + = + - x - = + - x + = - + x - = -
Practice set 8 1. (-5) x (-7) = 35 2.(-9) x (6) = - 54 3. 9 x (-4) = - 36 4.(8) x (-7) = - 56 5.(-124) x ( - 1 ) =124
6. (-12) x (-7) = 84 7. (-63) x (-7) = 441 8. ( - 7 ) x ( 15 ) = -105
practice set 9 (i) (− 96 ) ÷ 16 = −6 ii) 98 ÷ (− 28 ) = − 7/2 (iii) (− 51 ) ÷ 68 = − 3/4 (iv) 38 ÷ (−57 ) = − 2/3 (v) (− 85 ) ÷ 20 = − 17/4
(vi) (− 150 ) ÷ (− 25 ) = 6 vii) 100 ÷ 60 = 5/3 (viii) 9 ÷ (− 54) = − 1/6 ix) 78 ÷ 65 = 6/5 x) (−5)÷(−315)=1/63
Write three divisions of integers such that the fractional form of each will be 245 . ANSWER: The three divisions of integers are −24/−5, 48/10 and −48/−10
Write three divisions of integers such that the fractional form of each will be −57 . ANSWER:
ch 4 Angles and pairs of Angles
Practice set 17 continued....
Q4 The difference between the measures of the two angles of a complementary pair is 40° . Find the measure of the two angles. solution- Let X and Y be the two complementary angles. X + Y = 90° X - Y = 40° (Given) X + Y + X - Y = 90 + 40° (adding both) 2 X = 130 X= 130 ÷ 2 = 65 Y = 65 - 40 = 25 Ans Two angles are 65° and 25°
Q5 □ PTNM is a rectangle .Write the names of the pairs of supplementary angles? Ans: (i) ∟P and ∟ T (ii) ∟P and ∟N (iii) ∟P and ∟M (iv) ∟ T and ∟N, (v) ∟T and ∟M, (vi) ∟N and ∟M
P T M N
Q6 If m∟A =70 °,What is the measure of the supplement of the complement of ∟A ? Solution : m∟A = 70° complement of ∟A = 90° - 70° = 20° Supplement of the complement of ∟A = 180 - 20 = 160° Ans -Supplement of the complement of ∟A = 160°
Q7∟A and ∟B are supplementary angles and m∟B =(x+20) ,then what would be m∟A ? Solution : ∟A + ∟B = 180 ° m∟B =(x +20)° ∟A = 180 - (x +20) =180- x - 20 =(160-x)° Ans m∟A=(160-x)°
ch 4 Angles and pairs of Angles
Supplementary angles
If the sum of the measure of two angles is 180° they are known as supplementary angles
Practice set 17
Q1 Write the measures of the supplements of the angles given below.(i)15° Let the measure of the supplementary angle be x ° . The sum of the measure of two supplementary angles is 180° 15 + x = 180
∴ x = 180 - 15 = 165° Hence, the measure of the supplement of 15° is 165°.
(ii) 85° Let the measure of the supplementary angle be x° . The sum of the measure of two supplementary angles is 180° 85 + x = 180
∴ x = 180 - 85 = 95° Hence, the measure of the supplement of 85° is 95°.
(iii)120° Let the measure of the supplementary angle be x°. The sum of the measure of two supplementary angles is 180° 120 + x = 180 ∴ x = 180-120=60° Hence, the measure of the supplement of 120 is 60°.
(vi) 0° Let the measure of the supplementary angle be x° . The sum of the measure of two supplementary angles is 180° 0 + x = 180
∴ x = 180 - 0 = 180° Hence, the measure of the supplement of an angle of measure 0° is 180°. (vii)-a° Let the measure of the supplementary angle be x. The sum of the measure of two supplementary angles is 180° a + x = 180 ∴ x = (180 − a )° Hence, the measure of the supplement of an angle of measure a° is (180 − a )°
Question 2:
The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.
m∠B = 60° m∠N = 30° m∠Y = 90° m∠J = 150°
m∠D = 75° m∠E = 0° m∠F = 15° m∠G = 120°
solution-If the sum of the measures of two angles is 90° they are known as complementary angles.
Hence,the pairs of complementary angles are ∠B and ∠N,( m∠B = 60° m∠N = 30° )∠D and ∠F, ( m∠D = 75° m∠F = 15° ) ∠Y and ∠E. (m∠Y = 90° m∠E = 0°)If the sum of the measures of two angles is 180° they are known as supplementary angles. Hence, the pairs of supplementary angles are ∠B and ∠G, ( m∠B = 60° m∠G = 120°) ∠N and ∠J (m∠N = 30° m∠J = 150°)
Q3 .In ∆ XYZ ,m∟Y=90°,what kind of a pair do ∟X and ∟Z make ? Solution : ∟X + ∟Y + ∟Z =180°-- (sum of the angles of a ∆ is 180°)
∟X + ∟Z= 180°-90° =90° ∟X and ∟Z are complementary
(iv)37° Let the measure of the supplementary angle be x 37 + x = 180
∴ x = 180-37=143° Hence, the measure of the supplement is 143°.
(v) 108°- (v)
Let the measure of the supplementary angle be x . 108 + x = 180
∴ x= 72°
Hence, the measure of the supplement of an angle of measure 108° is 72°
Chapter 3 HCF and LCM
LCM-Least Common Multiple
Practice set 13 continued.... Q2
GCD ( HCF ) x LCM = Product of given numbers
Q2 Find the HCF and LCM of the numbers given below.verify that their product is equal to the product of the given numbers. (i) 32 ,37 here, 1 is the only common factor for 32 and 37. HCF = 1, LCM = 32 x 37 = 1184 HCF x LCM = 1 x 1184 = 1184 product of the given numbers = 32 x 37 =1184 thus verified that product of HCF and LCM is equal to the product of the given numbers .
(ii) 46 , 51 Here, 1 is the only common factor for 46 and 51. HCF =1, LCM = 46 x 51= 2346 HCF x LCM =1 x 2346= 2346 product of the numbers =46 x 51 =2346 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
(iii) 15 , 60 3 15 , 60 5 5 , 20 1 , 4 HCF= 3 x 5 = 15 LCM=3 x 5 x 4 = 60 HCF x LCM =15 x 60 = 900 Product of the numbers = 15 x 60 = 900 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
(iv) 18 , 63 3 18 , 63 3 6 , 21 2, 7 HCF= 3 x 3 =9 LCM = 3 x 3 x 2 x 7 =126 HCF x LCM = 9 x 126 =1134 product of the numbers= 18 x 63 =1134 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
Least Common Multiple (LCM)
"The Least common Multiple of the given number is the smallest number that is divisible by each of the given numbers . Eg. Find the LCM of 16,28,40(vertical arrangement)
2 16 , 28 , 40 2 8 , 14 , 20 2 4 , 7, 10 2 , 7 , 5
Ans. LCM of 16,28,40 is 2 x 2 x 2 x 2 x 7 x 5 = 560
LCM - Practice set 13
Q1 Find the LCM (i) 12 , 15
(ii) 6 , 8, 10 2 6 , 8, 10 3 3, 4, 5 1, 4, 5 LCM = 2 x 3 x 4 x 5 = 120
2 12 , 15 2 6, 15 3 3, 15 1, 5 LCM = 2 x 2 x 3 x 5 = 60
(iv) 2 6 , 8 , 10 3 3 , 4 , 5 1 , 4 , 5 LCM = 2 x 3 x 4 x 5 = 120
(iii) 18 , 32 2 18 , 32 9 , 16 LCM = 2 x 9 x 16 = 288
(v) 45 , 86 here ,the only common divisor for 45 and 86 is 1 LCM = 45 x 86 = 3870
(vii) 105 , 195 5 105 ,195 3 21, 39 7, 13 LCM= 5 x 3 x 7 x 13 = 1365
(vi) 15 , 30 , 90 5 15 , 30 , 90 3 3, 6 , 18 2 1, 2, 6 1 , 1 , 3 LCM= 5 x 3 x 2 x 3 = 90
While dividing a positive integer by a negative integer or dividing a negative integer by a positive integer the quotient will be negative.
Eg (-6) ÷ 2= (-3)28 ÷ (-7)= (-4)
while dividing a negative integer by a negative integer the quotient will be positive.
Eg (-18) ÷ (-2)= (9)
When one integer is divided by another non zero integer, it is customary to write the denominator of the quotient as a positive integerEg 7 = - 7 - 2 2
Rules for Division of integers
- we cannot divide any number by zero
- the quotient of two positive integers is a positive number
- the quotient of two negative integers is a positive number
- the quotient of a positive integer and a negative integer is always a negative number
(+) ÷ (+) = (+) (+) ÷ (-) = ( - ) (-) ÷ (+) = ( - ) (-) ÷ (-) = (+)
Practice set 9
1.Solve (i) (-96) 16 = (-6) (ii) 98 (-28)= -7 98 14 = 7 =-7 2 -28) 14 -2 2
(-51) 17 = -363 17 4
(iii) (-51) 68 = -3 4
(iv) 38 (-57) = 38 19 = -2 (-57) 19 3
(v) (-85) 20 = (-85) 5 = -17 20 5 4
(vi) (-150) (-25) = (-150) 25 = -6 = 6 (-25) 25 -1
(vii) 100 ÷ 60 = 10 =10 ÷ 2 =5
6 6 2 3
2. Write three divisions of integers such that the fractional form of each will be 24 48 , (-72) etc 5 10 (-15) 3. Write three divisions of integers such that the fractional form of each will be -5 = -10 15 , - 20 7 14 -21 28
ch 16-Quadrilaterals
Take four points A, B, C, D, on a paper, such that any three
of them will be non-collinear. These points are to be joined to make
a closed figure, but in such a way that when any two points are
joined the other two must lie on the same side of that line.
The figure obtained by following the given rule is called a
quadrilateral. Observe the figures below and say which of them are quadrilaterals.
Like a triangle, quadrilateral ABCD is a closed figure. The four line segments that form
a quadrilateral are called its sides. Seg AB, seg BC, seg CD and seg AD are the four sides
of this quadrilateral. Points A, B, C and D are the vertices of the quadrilateral.
Reading and Writing of a Quadrilateral
ü A quadrilateral can be named by starting at any vertex and going serially either
clockwise or anti-clockwise around the figure.
When writing the name of a quadrilateral a sign like this ‘ ’ is put in place of the word
‘quadrilateral’.
Write the names of this quadrilateral starting at any vertex and going anti-clockwise
around the figure.
Adjacent Sides of a Quadrilateral The sides AB and AD of ABCD have a common vertex A. Sides AB and AD are adjacent sides. Name the pairs of adjacent sides in the figure alongside. (1) ...seg AB.... and ....seg AD.... (2) ........ segBC and ........ Seg AB (3) ........ Seg AD and ........Seg DC (4) ........ Seg DC and ........ Seg BC Every quadrilateral has four pairs of adjacent sides.
Adjacent sides of the quadrilateral have a common vertex.
Opposite Sides of a Quadrilateral In ABCD the sides AB and DC have no common vertex.
Side AB and side DC are opposite sides of the quadrilateral.
Name the pairs of opposite sides of this quadrilateral. Pairs of opposite sides : (1) ...Seg AB..... and ...Seg DC..... (2) ...seg AD..... and .......seg BC...
Opposite sides of the quadrilateral do not have a common vertex
Adjacent Angles of a Quadrilateral
The quadrilateral DEFG. The two angles ∠DEF and ∠GFE
have a common arm EF. These angles are neighbouring or adjacent angles.
Name the adjacent angles of the quadrilateral DEFG. (1) .∠DEF...... and .∠GFE........ (2) ...∠EDG..... and ..∠DGF....... (3) ..∠GDE...... and ..∠DEF....... (4) ..∠GFE...... and ..∠DGF.......
The angles of a quadrilateral which have one common arm are called
adjacent angles of the quadrilatera
Opposite Angles of a Quadrilateral In DEFG, the angles ∠DEF and ∠DGF do not have any common arm. ∠DEF and
∠DGF lie opposite to each other. Hence they are the opposite angles of a quadrilateral.
Name the other opposite angles in the figure. 1. Angle opposite to ∠EFG is .∠EDG 2. Angle opposite to ∠FGD is ∠DEF The angles of a quadrilateral which do not have a common arm are called
opposite angles of a quadrilateral.
Diagonals of a Quadrilateral In ABCD, the line segments that join the vertices of the
opposite angles ∠A and ∠C, as also of ∠B and ∠D, have been
drawn. The segments AC and BD are the diagonals of the
quadrilateral ABCD. The diagonal AC joins the vertices of the
opposite angles ∠A and ∠C.
The line segments which join the vertices of the opposite angles of a
quadrilateral are the diagonals of the quadrilateral.
In the figure above, name the angles whose vertices are joined by the diagonal BD.
BAD and BCD
Practice Set 38
1. Draw XYZW and name the following.
(1) The pairs of opposite angles. (2) The pairs of opposite sides.
(3) The pairs of adjacent sides. (4) The pairs of adjacent angles.
(5) The diagonals of the quadrilateral.
(6) The name of the quadrilateral in different ways.
(1) ∠X and ∠Z, ∠Y and ∠W (2) seg XY and seg ZW, seg XW and seg YZ (3) seg XY and seg YZ, seg YZ and seg WZ; seg WZ and seg XW, seg XW and seg XY (4) ∠X and ∠Y, ∠Y and ∠Z, ∠Z and ∠W, ∠X and ∠W
(5) Diagonal XZ and Diagonal YW (6) □YZWX, □ZWXY, □XYZW, □XWZY, □WZYX, □WXYZ etc.
1) In the given figure, there are five sides.
Hence, the given figure is a Pentagon. (2) In the given figure, there are six sides.
Hence, the given figure is a Hexagon. (3) In the given figure, there are seven sides.
Hence, the given figure is a Heptagon. (4) In the given figure, there are eight sides.
Hence, the given figure is a Octagon.
Question 2:
In the table below, write the number of sides the polygon has.
Question 3:
Look for examples of polygons in your surroundings. Draw them.
Question 4:
We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Question 4:
We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Question 5:
Draw any polygon and divide it into triangular parts as shown here. Thus work out the sum of the measures of the angles of the polygon.
In the figure alongside, some points
and some line segments joining them
have been drawn.
Which of these figures is a triangle?
Which figure is not a triangle? Why
not?
∆ABC has three sides. Line segment AB is one side. Write the names of the other two
sides. ∆ABC has three angles. ∠ABC is one of them. Write the names of the other angles.
Points A, B and C are called the vertices of the triangle.
A triangle is a closed figure made by joining three non-collinear points by line segments.
The vertices, sides and angles of a triangle are called the parts of the triangle. Types of Triangles - Based on Sides
Measure the sides of the following triangles in centimetres, using a divider and ruler.
Enter the lengths in the table below. What do you observe?
‘Length of line segment AB’ is written as l(AB).
4cm
3cm
3cm
3cm
4cm
3cm
5cm
3cm
4cm
In the table above, the lengths of all sides of ∆ABC are equal. Therefore, this triangle
is an equilateral triangle. ‘Lateral’ refers to the sides of a figure.
A triangle with all three sides equal is called an equilateral triangle. In ∆PQR, the length of the two sides PQ and PR are equal. ∆PQR is called an
isosceles triangle.
A triangle with two equal sides is called an isosceles triangle. The lengths of the sides of ∆XYZ are all different. Such a triangle is called a scalene
triangle.
A triangle with no two sides equal is called a scalene triangle.
Types of Triangles - Based on Angles
Measure all the angles of the triangles given below. Enter them in the following table.
scalene
equilateral triangle
isosceles triangle.
25
4 5
25
30
120
90
30
40
m/_F
45
In the figures above, ∆DEF is an acute angled triangle.
A triangle with all three acute angles is called an acute angled triangle.
∆PQR is a right angled triangle.
A triangle with one right angle is a right angled triangle.
∆LMN is an obtuse angled triangle.
A triangle with one obtuse angle is called an obtuse angled triangle.
Practice Set 36
1. Observe the figures below and write the type of the triangle based on its angles.
∆XYZ is .....obtuse. triangle.
∆PQR is ..right.... triangle.
∆LMN is ...acute... triangle.
2. Observe the figures below and write the type of the triangle based on its sides.
∆ABC is equilateral triangle.
∆UVW is .isosceles triangle.
∆DEF is ..scalene triangle
Chapter 15
Triangles and their Properties
Recap Parts of triangle Types of triangle based on sides and angles
The sum of the measures of the three angles of a triangle is 180°.
The sum of the lengths of any two sides of a triangle
is always greater than the length of the third side.
3+4
4+5
5+3
3. As shown in the figure, Avinash is standing
near his house. He can choose from two
roads to go to school. Which way is
shorter? Explain why.
In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side.
Therefore, in △ABC, AC < AB + BC.
Hence, the shortest way to reach school is via AC.
4. The lengths of the sides of some triangles are given. Say what types of triangles they
are. (1) 3 cm, 4 cm, 5 cm
(2) 3.4 cm, 3.4 cm, 5 cm
(2)
Two sides are equal.
Therefore, it is an Isosceles triangle
1)
All sides are different.
Therefore, it is a Scalene triangle.
.
.
(3) 4.3 cm, 4.3 cm, 4.3 cm
(4) 3.7 cm, 3.4 cm, 4 cm
(3)
All sides are equal.
Therefore, it is an Equilateral triangle
.
(4)
All sides are different.
Therefore, it is a Scalene triangle
5. The lengths of three segments are given for constructing a triangle. Say whether a
triangle with these sides can be drawn. Give the reason for your answer. (1) 17 cm, 7 cm, 8 cm
1) 17 cm, 7 cm, 8 cm Here, 17 > 7 + 8 17 > 15In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(2) 7 cm, 24 cm, 25 cm
(3) 9 cm, 6 cm, 16 cm
Here, 7 + 24 > 25 31 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side.
Therefore, the triangle is possible.
Here, 16 > 9 + 6 16 > 15 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side.
Therefore, the triangle is not possible.
(2) 7 cm, 24 cm, 25 cm
(3) 9 cm, 6 cm, 16 cm
Here, 7 + 24 > 25 31 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
Here, 16 > 9 + 6 16 > 15 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(5) 15 cm, 20 cm, 25 cm
(4) 8.4 cm, 16.4 cm, 4.9 cm
Here, 16.4 > 8.4 + 4.9 16.4 > 13.3 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(5) 15 cm, 20 cm, 25 cm
Here, 15 + 20 > 25 35 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
(6) 12 cm, 12 cm, 16 cm
Here, 12 + 12 < 16 24 < 28 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
THANKS!
Chapter 8
90,220,435,450,72,336,108,960
72:-7+2=9 and 9 is divisible by 3 so 72 is also divisible by 3
72:-the digits in units and tens place is 72 which is divisible by 4
72:-7+2=9 and 9 is divisible by 9 so 72 is divisible by 9.
72
72
72
90 :-9+0=9and 9 is divisible by 3 so 90 is also divisible by 3
90:-the digits in units and tens place is 90 which is not divisible by 4 .
90:- 9+0=9 and 9 is divisible by 9 so 90 is divisible by 9.
90
90
435:-4+3+5=12 and 12 is divisible by 3 therefore 435 is divisible by 3.
435:-the number formed by the digit in the tens and units place is 35 which is not divisible by 4 ,therefore 435 is not divisible by 4 .
435:-4+3+5=12 and 12 is not divisible by 9 therefore 435 is not divisible by 9.
435
336:-3+3+6=12 and 12 is divisible by 3 so 336 is also divisible by 3
336:-the digits in units and tens place is 36 which is divisible by 4 therefore 336 is divisible by 4
336:- 3+3+6=12 and 12 is not divisible by 9 so 336 is not divisible by 9
336
336
9 .
220:- 2+2+0=4and 4 is not divisible by 3 therefore 220 is not divisible by 3
220:-the number formed by the digits in the tens and units place is 20 and 20 is divisible by 4 therefore 220 is divisible by 4
220:-2+2+0=4 and 4 is not divisible by 9 therefore 220 is not divisible by 9
220
450:-4+5+0=9 and 9is divisible by 3 therefore 450 is divisible by 3.
450:-the number formed by the digits in the tens and units place is 50 and 50 is not divisible by 4 therefore 450 is not divisible by 4
450:-4+5+0=9 and 9 is divisible by 9 therefore 450 is divisible by 9.
450
450
108:-1+0+8=9 is divisible by 3 therefore 108 is divisible by 3
108:-the number formed by the digits in the tens and units place is 08 and 8 is divisible by 4 therefore 220 is divisible by 4
108:-1+0+8=9 and 9 is divisible by 9 therefore 108 is divisible by 9.
108
108
108
960:-9+6+0=15 is divisible by 3 therefore 960 is divisible by 3
960:-the number formed by the digits in the tens and units place is 60 and 60 is divisible by 4 therefore 960 is divisible by 4
960:-9+6+0=15 and 15 is not divisible by 9 therefore 450 is not divisible by 9.
960
960
Test of Divisibility by 2,5,10
page 43
125-- the digit in the units place is 5 so 125 is divisible by 5 364- the digit in the units place is 4 which is even so 364 is divisible by 2. 475- the digit in the units place is 5 so the number 475 is divisible by 5. 750- the digit in the units place is 0so 750 is divisible by 2,5&10
800-the digit in the units place is 0 so 800 is divisible by 2,5,10. 628-the digit in the units place is 8 which is even so 628 is divisible by 2. 206-the digit in the units place is 6 which is even so 206 is divisible by 2. 508- the digit in the units place is 8 which is even so 508 is divisible by 2
7009-the digit in the units place is 9 , so 7009 is not divisible by 2,5&10. 5345- the digit in the units place is 5 so 5345 is divisible by 5. 8710-the digit in the units place is 0 which is even so 8710 is divisible by 2,5&10.
7009
page 43
750 , 800, 8710
364, 750 ,800 ,628, 206, 508 ,8710
125, 475, 750, 800, 5345, 8710
Divisibility by 2
Example a) 62 =31 2 62 - the digit in the units place is 2 so 62 is divisible by 2
b)264 264/2=132 The digit in the units place is 4 which is divisible by 2 so 264 is divisible by 4
Example. 230, 125, 475, 800
Divisibility by 5 a) 65 = 65/5=13 the digit in units place is 5 so 65 is divisible by 5 b) 760 =760/5=132 The digit in the unit place is 0 so the entire number is divisible by 5
Example. 50, 800, 9000
Divisibility test by 10 a)600= 600/10=60 The digit in the units place is 0 so the number 600 is divisible by 10. b)1560= 1560/10=156 the digit in the units place is 0 so the number 1560 is divisible by 10
108
72
450
435
90
960
960
336
Practice set 22
1)
72 The digits in tens and units place is 72 which is divisible by 4. 72 4= 18 7+2=9 which is divisible by 3 .so 72 is divisible by 3 and 9.
90 The digit in tens and units place is not divisible by 4 so 90 is not divisible by 4 The sum of the digits is 9+0=9 so 90 is divisible by 3 & 9
108
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THANKS!
Lorem ipsum dolor
Lets Recall
- Addition and subtraction of decimal fraction.
- Showing decimal fraction on a number line.
Converting a decimal fraction into a common fraction.
Multiplication of decimal fraction
1.
First multiply ignoring decimal point.
In the product starting from the right hand side we count as many places as the total decimal places in the multiplicand and multiplier and place the decimal point before them.
2.
Example multiply 4 . 3 x 5 4 3 x 5 2 1 . 5
Practice set 16
1. If 3 1 7 x 4 5 = 1 4 2 6 5 then 3 . 1 7 x 4 . 5 = ? =1 4 . 2 6 5
2. If 5 0 3 x 2 1 7 = 1 0 9 1 5 1 then 5 . 0 3 x 2 . 1 7 = ? =1 0 . 9 1 5 1
Multiply 1) 2 . 7 x 1 . 4 2 7 x 1 4 1 0 8 2 7 0 3 . 7 8
2) 6. 1 7 x 3.9 6 1 7 x 3 9 5 5 5 3 1 8 5 1 0 2 4 . 0 6 3
5 . 0 4 x 0 . 7 5 0 4 x 0 . 7 3 . 5 2 8
0 . 5 7 x2 0 5 7 x 2 1 . 1 4
Virendra bought 18 bags of rice ,each bag weighing 5.250kg . How much rice did he buy altogether? if the rice cost 42 rupees per kg, how much did he pay for it
one bag of rice weighs 5.250kg 18 bags will weigh ?
1 kg rice cost Rs 42 94.500kgs will cost 9 4 . 5 x 4 2 1 8 9 0 3 7 8 0 0 3 9 6 9. 0
5 . 2 5 0 x 1 8 4 2 0 0 0 5 2 5 0 0 9 4 . 5 0 0
18 bags of rice will weigh 94.500kg
Virendra paid Rs 3969
Vedika has 23.50 metres of cloth .she used it to make 5 curtains of equal size .If each curtain required 4 metres 25 cm to make, how much cloth is left over?
4.25 m for 1 curtain x 5 curtains 21.25 m for 5 curtains 23.50 m vedika had -21.25 m used for 5 curtains 2.25 m left. cloth left with vedika 2.25 m
THANKS!
Chapter 5
Decimal Fractions
Lets Recall
- Addition and subtraction of decimal fraction.
- Showing decimal fraction on a number line.
*Converting a decimal fraction into a common fraction.
*Multiplication of decimal fraction
Practice set 17
Q1 Carry out the following division
4 . 8 2 4 8 2 1 0 1 48 x 1 10 2 = 24 =2.4 10
2 )
1 )
20.6 2 2 0 6 2 1 0 1 206 x 1 10 2 =103 =10.3 10
4 )
3 )
17.5 5 175 5 10 1 175 x 1 10 5 =35 = 3.5 10
32.5 25 325 25 10 1 325 x 1 10 25 13 = 1.3 10
A road is 4 km 800 m long. If trees are planted on both of its sides at intervals of 9.6 m,how many trees were planted?
Q 2
A road is 4 km 800m long 1 km = 1000 m 4 km = 4000 m + 800 m 4800 m 4800 9.6 = 4800 96 1 10 =4800 x 10 = 100 x 5 = 500 trees are planted on 1 side 96 Ans : 1000 trees are planted on both sides of the road
Q3 Pradnya exercises regularly by walking alond a circular path on a field .if she walks a distance of 3.825 km in 9 round of the path , how much doesshe walk in one round
In 9 rounds distance walked is 3.825km, in 1 round distance walked is ? in 1 round distance walked = 3.825 9 3.825 x 1 = 3825 x 1 9 1000 9 = 3825 x 1 9 1000 = 425 = 0.425 km 1000
Q4 A Pharmaceutical manufacturer bought 0.25 quintal of hirada , amedicinal plant , for 9500 rupees . what is the cost per quintal of hirada?1 quintal = 100 kg
0.25 quintal hirada for Rs 95001 quintal hirada for Rs 9500 0.25 9500 25 = 9500 x 100 1 100 25 = 9500 x 4 =Rs 38,000 The cost per quintal of hirada is Rs 38,000
THANKS!
stay home stay safe
Q3 convert the decimal fractions into common fraction
1. 27.5 = 275 10 2. 0.007 = 7 1000 4. 39.15 = 3915 100
6. 70.400 = 70400 1000
23.7.20
St.Annes Primary school
Std 3A,B,C
Number Work
3 Digit Numbers Part 3
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St.Annes High School,pune
Appeared 200
100 percent Results Distinction -156 Ist class-140 2nd class-4 Highest -Vidhisha solanki 96.80
Introduction here
This is a paragraph- of text waiting to be awesome subtitle
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Using symbols to show smaller and bigger ... < >
Activity 1
Put the correct sign < , > 427-----267 150----501 813----791 300---624
ACTIVITY-2
The expanded form of numbers example :
H T U
537=500+30+7
- 467=400+ 60+ 7
- 211=200+ 10+ 1
- 306=300+ 00+ 6
- 150=100 + 50+ 0
6c 10 dec
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Transcript
Practice set 26 Q1
ANSWER:1
Q2 find the value
Practice set 27
Q1 simplify
+info
Chapter 6
Find square root of a number (prime factors)
Practice set 30
i) The prime factorization of 625 is, 625 = 5 × 5 × 5 × 5 To find the square root, we will take one number from each pair and multiply. 625 −− 5×5 = 25 ∴√625−−− = 25
(ii) The prime factorization of 1225 is, 1225 = 5 × 5 × 7 × 7 To find the square root, we will take one number from each pair and multiply. √1225−−−−5×7=35 ∴√1225−−−−= 35
(iii) The prime factorisation of 289 is, 289 = 17 × 17 To find the square root, we will take one number from each pair and multiply. √289−−−17 =17 ∴√ 289−−− =17
(iv) The prime factorization of 4096 is, 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 To find the square root, we will take one number from each pair and multiply. 4096−−−−2×2×2×2×2×2 = 64 ∴√4096−−−−64
(v) The prime factorizationn of 1089 is, 1089 = 3 × 3 × 11 × 11 To find the square root, we will take one number from each pair and multipy. √1089−−−−3×11=33 ∴√1089−−−−33
Chapter -8 Algebraic Expressions
Like terms and unlike terms (i)2x, (i) 7xy, 5x, 9y2, -2/3 x, -2xyz
Types of algebraic expression
Practice set 32
Question 1: Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
ANSWER:1It is known that, expressions with one term is called monomial, expressions with two terms are binomials, expressions with three terms are trinomials and expression with more than three terms are polynomials. (i) 7x = Monomial (ii) 5y−7z = Binomial (iii) 3x3−5x2−11 = Trinomial (iv) 1−8a−7a2−7a3 = Polynomial (v) 5m−3 = Binomial (vi) 4 = Monomial (vii) 3y2−7y+5 = Trinomial
Chapter -8 Algebraic Expressions
Practice set 33
Question 1: Add. (i) 9p + 16q ; 13p + 2q ANSWER: i) (9p + 16q) + (13p + 2q) = 9p + 16q + 13p + 2q =(9p + 13p) + (16q + 2q) =22p + 18q
(ii) (2a + 6b + 8c) + (16a + 13c + 18b) = 2a + 6b + 8c + 16a + 13c + 18b = (2a + 16a) + (6b + 18b) + (8c + 13c) = 18a + 24b + 21c
(iii) (13x2 − 12y2) + (6x2 − 8y2) = 13x2 − 12y2 + 6x2 − 8y2 = (13x2 + 6x2) + (−12y2 − 8y2) =19x2 + (−20y2) =19x2 − 20y2
(iv) (17a2b2 + 16c) + (28c − 28a2b2) = 17a2b2 + 16c + 28c − 28a2b2 = (17a2b2 − 28a2b2) + (16c + 28c) = −11a2b2 + 44c
(v) (3y2 − 10y + 16) + (2y − 7) = 3y2 − 10y + 16 + 2y − 7 = 3y2 + (−10y + 2y) + (16 − 7) =3y2 + (−8y) + 9 =3y2 − 8y + 9
(vi) (−3y2 + 10y − 16) + (7y2 + 8) = −3y2 + 10y − 16 + 7y2 + 8 = (−3y2 + 7y2) + 10y + (−16 + 8) = 4y2 + 10y + (−8) = 4y2 + 10y − 8
THANKS!
Chapter 5 -Operation on Rational Numbers
If m is an integer and n is any non zero integer, then the number m is called a rational number. n
Practice set 22
Question 1: Carry out the following additions of rational numbers. (i) 5 + 6 36 42 LCM of 36 and 42 = 2 × 2 × 3 × 3 × 7 = 252 =5 x 7 + 6 x 6 = 35 + 36 = 71 36 x 7 42 x 6 252 252 252
= 1 x 3 +2 + 2 x 5 +4 3 5 =5 + 14 3 5 LCM of 3 and 5 is 15 =5 x 5 + 14 x 3 3 x 5 5 x 3 = 25 + 42 15 15 = 25+42 = 67 15 15
3. 11 + 13 17 19 LCM of 17 x 19= 323 =11 x 19 + 13 x 17 17 x 19 19 x 17 209 + 221 =209+221 323 323 323 =430 323
11 x 2 +3 + 77 x 1 + 3 11 77 LCM of 11 and 77 is 77 =25x7 + 80 = 175+80 = 255 11x7 77 77 77 =
Question 2: Carry out the following subtractions involving rational numbers. (i) 7 − 3 11 7 LCM of 11 and 7 is 77 7 x 7 - 3 x 11 = 49-33 = 16 11 x 7 7 x11 77 77
(ii) 13 − 2 36 40 LCM of 36 and 40 is 360 13 x10 - 2 x 9 = 130-18 = 112÷8 = 14 36 x10 40 x9 360 360÷8 45
(iii) =1×3+2 − 3×6+5 = 5 − 23 3 6 3 6 LCM of 3 and 6 is 6. = 5×2 − 23×1 3×2 6×1 = 10 − 23 6 6 = 10−23 6 = − 13 6
Practice set 22
Question 3: Multiply the following rational numbers. (i) 3 × 2 = 6 11 5 55 (ii) 12 × 4 = 48 ÷3 = 16 5 15 75 ÷3 25 iii) (−8) × 3 = -8 x 3 = -24 = -24 9 4 9 x 4 36 3 (iv) 0 × 3 = 0 = 0 6 4 24
Question 4: Write the multiplicative inverse. (i) 2 Ans. Multiplicative inverse is 5 5 2 (ii) −3 Ans. Multiplicative inverse is -8 8 3 (iii) −17 Ans. Multiplicative inverse is -39 39 17 (iv) 7 Ans. Multiplicative inverse is 1 7 (v) = 3 x7 +1 = - 22 Ans. Multiplicative inverse is -3 3 3 22
Question 5: Carry out the divisions of rational numbers. (i) 40 ÷ 10 = 40 × 4 = 40×4 = 160 = 160÷40 12 4 12 10 12×10 120 120÷40 (Since, HCF of 160 and 120 is 40)= 4 3 (ii) −10÷ −11 = (−10)×(−10= (−10)×(−10)= 100 11 10 11 11 11×11 121 (iii) −7 ÷ −3 8 6 =(−7)×(−6)=(−7)×(−6)= 42 8 3 8×3 24 =42÷6 (Since, HCF of 42 and 24 is 6) = 7 24÷6 4
(iv) 2 ÷(−4) = 2 × (−1) = 2×(−1) = −2 = −2÷2 3 3 4 3×4 12 12÷2 (Since, HCF of 2 and 12 is 2) = −1 6 (v) = 2×5+1÷ 5×6+3 = 11÷ 33 = 11 × 6 = 11×6 = 66 = 66÷33 5 6 5 6 5 33 5×33 165 165÷33 (Since, HCF of 66 and 165 is 33) = 2 5
THANKS!
THANKS!
Chapter 3 - HCF And LCM
Question 1: Which number is neither a prime number nor a composite number? ANSWER: 1 is neither a prime number nor a composite number
Page No 15:
Question 2: Which of the following are pairs of co-primes? (i) 8, 14 (ii) 4, 5 (iii) 17, 19 (iv) 27, 15 ANSWER: Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime We can write 17 as 17 × 1 and 19 as 17 × 1. Hence, 17 and 19 is a pair of co-prime numbers.
Question 3: List the prime numbers from 25 to 100 and say how many they are. ANSWER: There are a total of 16 prime numbers between 25 and 100 which are 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Question 4: Write all the twin prime numbers from 51 to 100. ANSWER: If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers. Hence, the twin prime numbers between 51 and 100 are 59 and 61, 71 and 73
Question 5: Write 5 pairs of twin prime numbers from 1 to 50. ANSWER: If the difference between two co-prime numbers is 2 then, the numbers are said to be twin prime numbers. Hence, the twin prime numbers from 1 to 50 are (2,3), (5,7), (11,13), (17,19) and (29,31).
Question 6: Which are the even prime numbers? ANSWER: There is only even prime number which is 2.
Page No 17:
Question 1: Factorise the following numbers into primes.
(i) 32 = 2 × 2 × 2 × 2 × 2
ii) 57 = 3 × 19(iii) 23 = 23 × 1 (iv) 150 = 2 × 3 × 5 × 5 (v) 216 = 2 × 2 × 2 × 3 × 3 × 3 (vi) 208 =2 × 2 × 2 × 2 × 13 (vii) 765 = 3 × 3 × 5 × 17 (viii) 342 = 2 × 3 × 3 × 19 (ix) 377 = 13 × 29 (x) 559 = 13 × 43
Find the HCF.
Question 2: Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
Question 1: Choose the right option. (i) The HCF of 120 and 150 is ................... . (1) 30 (2) 45 (3) 20 (4) 120
(ii) The HCF of this pair of numbers is not 1. (1) 13, 17 (2) 29, 20 (3) 40, 20 (4) 14, 15
Question 2: Find the HCF and LCM
Question 3: Find the LCM. (i) 36, 42
Question 4: Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time
Question 9: Which two consecutive even numbers have an LCM of 180? ANSWER: Let us suppose the two consecutive even numbers be 2x and 2x + 2. Now, product of two numbers = HCF × LCM ⇒ (2x)(2x + 2) = 2 × 180 (HCM of two even number is 2) ⇒ (x)(2x + 2) = 180 ⇒ 2x2 + 2x = 180 ⇒ 2x2 + 2x − 180 = 0 ⇒ x2 + x − 90 = 0 ⇒ (x − 9)(x + 10) = 0 ⇒ x − 9 = 0 or x + 10 = 0 ⇒ x = 9 or x = −10 (Neglecting) Hence, the two consecutive even numbers are18 and 20.
THANKS!
Ch2 Multiplication and Division of integers
page 11 1. 5 + 7 = 12 2. 10 + ( - 5) = 5 3. - 4 + 3 = -1 4. (-7) + (- 2 ) = -9 5. ( + 8 ) - (+ 3 ) = 5 6. ( + 8 ) - (- 3 ) = 11
+ x + = + - x - = + - x + = - + x - = -
Practice set 8 1. (-5) x (-7) = 35 2.(-9) x (6) = - 54 3. 9 x (-4) = - 36 4.(8) x (-7) = - 56 5.(-124) x ( - 1 ) =124
6. (-12) x (-7) = 84 7. (-63) x (-7) = 441 8. ( - 7 ) x ( 15 ) = -105
practice set 9 (i) (− 96 ) ÷ 16 = −6 ii) 98 ÷ (− 28 ) = − 7/2 (iii) (− 51 ) ÷ 68 = − 3/4 (iv) 38 ÷ (−57 ) = − 2/3 (v) (− 85 ) ÷ 20 = − 17/4
(vi) (− 150 ) ÷ (− 25 ) = 6 vii) 100 ÷ 60 = 5/3 (viii) 9 ÷ (− 54) = − 1/6 ix) 78 ÷ 65 = 6/5 x) (−5)÷(−315)=1/63
Write three divisions of integers such that the fractional form of each will be 245 . ANSWER: The three divisions of integers are −24/−5, 48/10 and −48/−10
Write three divisions of integers such that the fractional form of each will be −57 . ANSWER:
ch 4 Angles and pairs of Angles
Practice set 17 continued....
Q4 The difference between the measures of the two angles of a complementary pair is 40° . Find the measure of the two angles. solution- Let X and Y be the two complementary angles. X + Y = 90° X - Y = 40° (Given) X + Y + X - Y = 90 + 40° (adding both) 2 X = 130 X= 130 ÷ 2 = 65 Y = 65 - 40 = 25 Ans Two angles are 65° and 25°
Q5 □ PTNM is a rectangle .Write the names of the pairs of supplementary angles? Ans: (i) ∟P and ∟ T (ii) ∟P and ∟N (iii) ∟P and ∟M (iv) ∟ T and ∟N, (v) ∟T and ∟M, (vi) ∟N and ∟M
P T M N
Q6 If m∟A =70 °,What is the measure of the supplement of the complement of ∟A ? Solution : m∟A = 70° complement of ∟A = 90° - 70° = 20° Supplement of the complement of ∟A = 180 - 20 = 160° Ans -Supplement of the complement of ∟A = 160°
Q7∟A and ∟B are supplementary angles and m∟B =(x+20) ,then what would be m∟A ? Solution : ∟A + ∟B = 180 ° m∟B =(x +20)° ∟A = 180 - (x +20) =180- x - 20 =(160-x)° Ans m∟A=(160-x)°
ch 4 Angles and pairs of Angles
Supplementary angles
If the sum of the measure of two angles is 180° they are known as supplementary angles
Practice set 17
Q1 Write the measures of the supplements of the angles given below.(i)15° Let the measure of the supplementary angle be x ° . The sum of the measure of two supplementary angles is 180° 15 + x = 180 ∴ x = 180 - 15 = 165° Hence, the measure of the supplement of 15° is 165°.
(ii) 85° Let the measure of the supplementary angle be x° . The sum of the measure of two supplementary angles is 180° 85 + x = 180 ∴ x = 180 - 85 = 95° Hence, the measure of the supplement of 85° is 95°.
(iii)120° Let the measure of the supplementary angle be x°. The sum of the measure of two supplementary angles is 180° 120 + x = 180 ∴ x = 180-120=60° Hence, the measure of the supplement of 120 is 60°.
(vi) 0° Let the measure of the supplementary angle be x° . The sum of the measure of two supplementary angles is 180° 0 + x = 180 ∴ x = 180 - 0 = 180° Hence, the measure of the supplement of an angle of measure 0° is 180°. (vii)-a° Let the measure of the supplementary angle be x. The sum of the measure of two supplementary angles is 180° a + x = 180 ∴ x = (180 − a )° Hence, the measure of the supplement of an angle of measure a° is (180 − a )°
Question 2: The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles. m∠B = 60° m∠N = 30° m∠Y = 90° m∠J = 150° m∠D = 75° m∠E = 0° m∠F = 15° m∠G = 120°
solution-If the sum of the measures of two angles is 90° they are known as complementary angles. Hence,the pairs of complementary angles are ∠B and ∠N,( m∠B = 60° m∠N = 30° )∠D and ∠F, ( m∠D = 75° m∠F = 15° ) ∠Y and ∠E. (m∠Y = 90° m∠E = 0°)If the sum of the measures of two angles is 180° they are known as supplementary angles. Hence, the pairs of supplementary angles are ∠B and ∠G, ( m∠B = 60° m∠G = 120°) ∠N and ∠J (m∠N = 30° m∠J = 150°)
Q3 .In ∆ XYZ ,m∟Y=90°,what kind of a pair do ∟X and ∟Z make ? Solution : ∟X + ∟Y + ∟Z =180°-- (sum of the angles of a ∆ is 180°)
∟X + ∟Z= 180°-90° =90° ∟X and ∟Z are complementary
(iv)37° Let the measure of the supplementary angle be x 37 + x = 180 ∴ x = 180-37=143° Hence, the measure of the supplement is 143°.
(v) 108°- (v) Let the measure of the supplementary angle be x . 108 + x = 180 ∴ x= 72° Hence, the measure of the supplement of an angle of measure 108° is 72°
Chapter 3 HCF and LCM
LCM-Least Common Multiple
Practice set 13 continued.... Q2
GCD ( HCF ) x LCM = Product of given numbers
Q2 Find the HCF and LCM of the numbers given below.verify that their product is equal to the product of the given numbers. (i) 32 ,37 here, 1 is the only common factor for 32 and 37. HCF = 1, LCM = 32 x 37 = 1184 HCF x LCM = 1 x 1184 = 1184 product of the given numbers = 32 x 37 =1184 thus verified that product of HCF and LCM is equal to the product of the given numbers .
(ii) 46 , 51 Here, 1 is the only common factor for 46 and 51. HCF =1, LCM = 46 x 51= 2346 HCF x LCM =1 x 2346= 2346 product of the numbers =46 x 51 =2346 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
(iii) 15 , 60 3 15 , 60 5 5 , 20 1 , 4 HCF= 3 x 5 = 15 LCM=3 x 5 x 4 = 60 HCF x LCM =15 x 60 = 900 Product of the numbers = 15 x 60 = 900 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
(iv) 18 , 63 3 18 , 63 3 6 , 21 2, 7 HCF= 3 x 3 =9 LCM = 3 x 3 x 2 x 7 =126 HCF x LCM = 9 x 126 =1134 product of the numbers= 18 x 63 =1134 Thus ,verified that product of HCF and LCM is equal to the product of the given numbers.
Least Common Multiple (LCM)
"The Least common Multiple of the given number is the smallest number that is divisible by each of the given numbers . Eg. Find the LCM of 16,28,40(vertical arrangement)
2 16 , 28 , 40 2 8 , 14 , 20 2 4 , 7, 10 2 , 7 , 5
Ans. LCM of 16,28,40 is 2 x 2 x 2 x 2 x 7 x 5 = 560
LCM - Practice set 13
Q1 Find the LCM (i) 12 , 15
(ii) 6 , 8, 10 2 6 , 8, 10 3 3, 4, 5 1, 4, 5 LCM = 2 x 3 x 4 x 5 = 120
2 12 , 15 2 6, 15 3 3, 15 1, 5 LCM = 2 x 2 x 3 x 5 = 60
(iv) 2 6 , 8 , 10 3 3 , 4 , 5 1 , 4 , 5 LCM = 2 x 3 x 4 x 5 = 120
(iii) 18 , 32 2 18 , 32 9 , 16 LCM = 2 x 9 x 16 = 288
(v) 45 , 86 here ,the only common divisor for 45 and 86 is 1 LCM = 45 x 86 = 3870
(vii) 105 , 195 5 105 ,195 3 21, 39 7, 13 LCM= 5 x 3 x 7 x 13 = 1365
(vi) 15 , 30 , 90 5 15 , 30 , 90 3 3, 6 , 18 2 1, 2, 6 1 , 1 , 3 LCM= 5 x 3 x 2 x 3 = 90
While dividing a positive integer by a negative integer or dividing a negative integer by a positive integer the quotient will be negative.
Eg (-6) ÷ 2= (-3)28 ÷ (-7)= (-4)
while dividing a negative integer by a negative integer the quotient will be positive.
Eg (-18) ÷ (-2)= (9)
When one integer is divided by another non zero integer, it is customary to write the denominator of the quotient as a positive integerEg 7 = - 7 - 2 2
Rules for Division of integers
(+) ÷ (+) = (+) (+) ÷ (-) = ( - ) (-) ÷ (+) = ( - ) (-) ÷ (-) = (+)
Practice set 9
1.Solve (i) (-96) 16 = (-6) (ii) 98 (-28)= -7 98 14 = 7 =-7 2 -28) 14 -2 2
(-51) 17 = -363 17 4
(iii) (-51) 68 = -3 4
(iv) 38 (-57) = 38 19 = -2 (-57) 19 3
(v) (-85) 20 = (-85) 5 = -17 20 5 4
(vi) (-150) (-25) = (-150) 25 = -6 = 6 (-25) 25 -1
(vii) 100 ÷ 60 = 10 =10 ÷ 2 =5
6 6 2 3
2. Write three divisions of integers such that the fractional form of each will be 24 48 , (-72) etc 5 10 (-15) 3. Write three divisions of integers such that the fractional form of each will be -5 = -10 15 , - 20 7 14 -21 28
ch 16-Quadrilaterals
Take four points A, B, C, D, on a paper, such that any three of them will be non-collinear. These points are to be joined to make a closed figure, but in such a way that when any two points are joined the other two must lie on the same side of that line. The figure obtained by following the given rule is called a quadrilateral. Observe the figures below and say which of them are quadrilaterals.
Like a triangle, quadrilateral ABCD is a closed figure. The four line segments that form a quadrilateral are called its sides. Seg AB, seg BC, seg CD and seg AD are the four sides of this quadrilateral. Points A, B, C and D are the vertices of the quadrilateral. Reading and Writing of a Quadrilateral ü A quadrilateral can be named by starting at any vertex and going serially either clockwise or anti-clockwise around the figure. When writing the name of a quadrilateral a sign like this ‘ ’ is put in place of the word ‘quadrilateral’.
Write the names of this quadrilateral starting at any vertex and going anti-clockwise around the figure.
Adjacent Sides of a Quadrilateral The sides AB and AD of ABCD have a common vertex A. Sides AB and AD are adjacent sides. Name the pairs of adjacent sides in the figure alongside. (1) ...seg AB.... and ....seg AD.... (2) ........ segBC and ........ Seg AB (3) ........ Seg AD and ........Seg DC (4) ........ Seg DC and ........ Seg BC Every quadrilateral has four pairs of adjacent sides. Adjacent sides of the quadrilateral have a common vertex.
Opposite Sides of a Quadrilateral In ABCD the sides AB and DC have no common vertex. Side AB and side DC are opposite sides of the quadrilateral. Name the pairs of opposite sides of this quadrilateral. Pairs of opposite sides : (1) ...Seg AB..... and ...Seg DC..... (2) ...seg AD..... and .......seg BC...
Opposite sides of the quadrilateral do not have a common vertex
Adjacent Angles of a Quadrilateral
The quadrilateral DEFG. The two angles ∠DEF and ∠GFE have a common arm EF. These angles are neighbouring or adjacent angles. Name the adjacent angles of the quadrilateral DEFG. (1) .∠DEF...... and .∠GFE........ (2) ...∠EDG..... and ..∠DGF....... (3) ..∠GDE...... and ..∠DEF....... (4) ..∠GFE...... and ..∠DGF.......
The angles of a quadrilateral which have one common arm are called adjacent angles of the quadrilatera
Opposite Angles of a Quadrilateral In DEFG, the angles ∠DEF and ∠DGF do not have any common arm. ∠DEF and ∠DGF lie opposite to each other. Hence they are the opposite angles of a quadrilateral. Name the other opposite angles in the figure. 1. Angle opposite to ∠EFG is .∠EDG 2. Angle opposite to ∠FGD is ∠DEF The angles of a quadrilateral which do not have a common arm are called opposite angles of a quadrilateral.
Diagonals of a Quadrilateral In ABCD, the line segments that join the vertices of the opposite angles ∠A and ∠C, as also of ∠B and ∠D, have been drawn. The segments AC and BD are the diagonals of the quadrilateral ABCD. The diagonal AC joins the vertices of the opposite angles ∠A and ∠C. The line segments which join the vertices of the opposite angles of a quadrilateral are the diagonals of the quadrilateral.
In the figure above, name the angles whose vertices are joined by the diagonal BD.
BAD and BCD
Practice Set 38
1. Draw XYZW and name the following. (1) The pairs of opposite angles. (2) The pairs of opposite sides. (3) The pairs of adjacent sides. (4) The pairs of adjacent angles. (5) The diagonals of the quadrilateral. (6) The name of the quadrilateral in different ways.
(1) ∠X and ∠Z, ∠Y and ∠W (2) seg XY and seg ZW, seg XW and seg YZ (3) seg XY and seg YZ, seg YZ and seg WZ; seg WZ and seg XW, seg XW and seg XY (4) ∠X and ∠Y, ∠Y and ∠Z, ∠Z and ∠W, ∠X and ∠W (5) Diagonal XZ and Diagonal YW (6) □YZWX, □ZWXY, □XYZW, □XWZY, □WZYX, □WXYZ etc.
1) In the given figure, there are five sides. Hence, the given figure is a Pentagon. (2) In the given figure, there are six sides. Hence, the given figure is a Hexagon. (3) In the given figure, there are seven sides. Hence, the given figure is a Heptagon. (4) In the given figure, there are eight sides. Hence, the given figure is a Octagon.
Question 2: In the table below, write the number of sides the polygon has.
Question 3: Look for examples of polygons in your surroundings. Draw them.
Question 4: We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Question 4: We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Question 5: Draw any polygon and divide it into triangular parts as shown here. Thus work out the sum of the measures of the angles of the polygon.
In the figure alongside, some points and some line segments joining them have been drawn. Which of these figures is a triangle? Which figure is not a triangle? Why not? ∆ABC has three sides. Line segment AB is one side. Write the names of the other two sides. ∆ABC has three angles. ∠ABC is one of them. Write the names of the other angles. Points A, B and C are called the vertices of the triangle.
A triangle is a closed figure made by joining three non-collinear points by line segments. The vertices, sides and angles of a triangle are called the parts of the triangle. Types of Triangles - Based on Sides Measure the sides of the following triangles in centimetres, using a divider and ruler. Enter the lengths in the table below. What do you observe? ‘Length of line segment AB’ is written as l(AB).
4cm
3cm
3cm
3cm
4cm
3cm
5cm
3cm
4cm
In the table above, the lengths of all sides of ∆ABC are equal. Therefore, this triangle is an equilateral triangle. ‘Lateral’ refers to the sides of a figure. A triangle with all three sides equal is called an equilateral triangle. In ∆PQR, the length of the two sides PQ and PR are equal. ∆PQR is called an isosceles triangle. A triangle with two equal sides is called an isosceles triangle. The lengths of the sides of ∆XYZ are all different. Such a triangle is called a scalene triangle. A triangle with no two sides equal is called a scalene triangle. Types of Triangles - Based on Angles Measure all the angles of the triangles given below. Enter them in the following table.
scalene
equilateral triangle
isosceles triangle.
25
4 5
25
30
120
90
30
40
m/_F
45
In the figures above, ∆DEF is an acute angled triangle. A triangle with all three acute angles is called an acute angled triangle. ∆PQR is a right angled triangle. A triangle with one right angle is a right angled triangle. ∆LMN is an obtuse angled triangle. A triangle with one obtuse angle is called an obtuse angled triangle.
Practice Set 36
1. Observe the figures below and write the type of the triangle based on its angles.
∆XYZ is .....obtuse. triangle.
∆PQR is ..right.... triangle.
∆LMN is ...acute... triangle.
2. Observe the figures below and write the type of the triangle based on its sides.
∆ABC is equilateral triangle.
∆UVW is .isosceles triangle.
∆DEF is ..scalene triangle
Chapter 15
Triangles and their Properties
Recap Parts of triangle Types of triangle based on sides and angles
The sum of the measures of the three angles of a triangle is 180°.
The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
3+4
4+5
5+3
3. As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.
In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, in △ABC, AC < AB + BC. Hence, the shortest way to reach school is via AC.
4. The lengths of the sides of some triangles are given. Say what types of triangles they are. (1) 3 cm, 4 cm, 5 cm
(2) 3.4 cm, 3.4 cm, 5 cm
(2) Two sides are equal. Therefore, it is an Isosceles triangle
1) All sides are different. Therefore, it is a Scalene triangle. . .
(3) 4.3 cm, 4.3 cm, 4.3 cm
(4) 3.7 cm, 3.4 cm, 4 cm
(3) All sides are equal. Therefore, it is an Equilateral triangle
. (4) All sides are different. Therefore, it is a Scalene triangle
5. The lengths of three segments are given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer. (1) 17 cm, 7 cm, 8 cm
1) 17 cm, 7 cm, 8 cm Here, 17 > 7 + 8 17 > 15In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(2) 7 cm, 24 cm, 25 cm
(3) 9 cm, 6 cm, 16 cm
Here, 7 + 24 > 25 31 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, the triangle is possible.
Here, 16 > 9 + 6 16 > 15 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, the triangle is not possible.
(2) 7 cm, 24 cm, 25 cm
(3) 9 cm, 6 cm, 16 cm
Here, 7 + 24 > 25 31 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
Here, 16 > 9 + 6 16 > 15 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(5) 15 cm, 20 cm, 25 cm
(4) 8.4 cm, 16.4 cm, 4.9 cm
Here, 16.4 > 8.4 + 4.9 16.4 > 13.3 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we cannot draw the triangle .
(5) 15 cm, 20 cm, 25 cm Here, 15 + 20 > 25 35 > 25 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
(6) 12 cm, 12 cm, 16 cm
Here, 12 + 12 < 16 24 < 28 In any triangle, the sum of the lengths of any two sides of a triangle is always greater than the third side. Therefore, we can draw the triangle .
THANKS!
Chapter 8
90,220,435,450,72,336,108,960
72:-7+2=9 and 9 is divisible by 3 so 72 is also divisible by 3
72:-the digits in units and tens place is 72 which is divisible by 4
72:-7+2=9 and 9 is divisible by 9 so 72 is divisible by 9.
72
72
72
90 :-9+0=9and 9 is divisible by 3 so 90 is also divisible by 3
90:-the digits in units and tens place is 90 which is not divisible by 4 .
90:- 9+0=9 and 9 is divisible by 9 so 90 is divisible by 9.
90
90
435:-4+3+5=12 and 12 is divisible by 3 therefore 435 is divisible by 3.
435:-the number formed by the digit in the tens and units place is 35 which is not divisible by 4 ,therefore 435 is not divisible by 4 .
435:-4+3+5=12 and 12 is not divisible by 9 therefore 435 is not divisible by 9.
435
336:-3+3+6=12 and 12 is divisible by 3 so 336 is also divisible by 3
336:-the digits in units and tens place is 36 which is divisible by 4 therefore 336 is divisible by 4
336:- 3+3+6=12 and 12 is not divisible by 9 so 336 is not divisible by 9
336
336
9 .
220:- 2+2+0=4and 4 is not divisible by 3 therefore 220 is not divisible by 3
220:-the number formed by the digits in the tens and units place is 20 and 20 is divisible by 4 therefore 220 is divisible by 4
220:-2+2+0=4 and 4 is not divisible by 9 therefore 220 is not divisible by 9
220
450:-4+5+0=9 and 9is divisible by 3 therefore 450 is divisible by 3.
450:-the number formed by the digits in the tens and units place is 50 and 50 is not divisible by 4 therefore 450 is not divisible by 4
450:-4+5+0=9 and 9 is divisible by 9 therefore 450 is divisible by 9.
450
450
108:-1+0+8=9 is divisible by 3 therefore 108 is divisible by 3
108:-the number formed by the digits in the tens and units place is 08 and 8 is divisible by 4 therefore 220 is divisible by 4
108:-1+0+8=9 and 9 is divisible by 9 therefore 108 is divisible by 9.
108
108
108
960:-9+6+0=15 is divisible by 3 therefore 960 is divisible by 3
960:-the number formed by the digits in the tens and units place is 60 and 60 is divisible by 4 therefore 960 is divisible by 4
960:-9+6+0=15 and 15 is not divisible by 9 therefore 450 is not divisible by 9.
960
960
Test of Divisibility by 2,5,10
page 43
125-- the digit in the units place is 5 so 125 is divisible by 5 364- the digit in the units place is 4 which is even so 364 is divisible by 2. 475- the digit in the units place is 5 so the number 475 is divisible by 5. 750- the digit in the units place is 0so 750 is divisible by 2,5&10
800-the digit in the units place is 0 so 800 is divisible by 2,5,10. 628-the digit in the units place is 8 which is even so 628 is divisible by 2. 206-the digit in the units place is 6 which is even so 206 is divisible by 2. 508- the digit in the units place is 8 which is even so 508 is divisible by 2
7009-the digit in the units place is 9 , so 7009 is not divisible by 2,5&10. 5345- the digit in the units place is 5 so 5345 is divisible by 5. 8710-the digit in the units place is 0 which is even so 8710 is divisible by 2,5&10.
7009
page 43
750 , 800, 8710
364, 750 ,800 ,628, 206, 508 ,8710
125, 475, 750, 800, 5345, 8710
Divisibility by 2
Example a) 62 =31 2 62 - the digit in the units place is 2 so 62 is divisible by 2
b)264 264/2=132 The digit in the units place is 4 which is divisible by 2 so 264 is divisible by 4
Example. 230, 125, 475, 800
Divisibility by 5 a) 65 = 65/5=13 the digit in units place is 5 so 65 is divisible by 5 b) 760 =760/5=132 The digit in the unit place is 0 so the entire number is divisible by 5
Example. 50, 800, 9000
Divisibility test by 10 a)600= 600/10=60 The digit in the units place is 0 so the number 600 is divisible by 10. b)1560= 1560/10=156 the digit in the units place is 0 so the number 1560 is divisible by 10
108
72
450
435
90
960
960
336
Practice set 22
1)
72 The digits in tens and units place is 72 which is divisible by 4. 72 4= 18 7+2=9 which is divisible by 3 .so 72 is divisible by 3 and 9.
90 The digit in tens and units place is not divisible by 4 so 90 is not divisible by 4 The sum of the digits is 9+0=9 so 90 is divisible by 3 & 9
108
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THANKS!
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Lets Recall
Converting a decimal fraction into a common fraction.
Multiplication of decimal fraction
1.
First multiply ignoring decimal point.
In the product starting from the right hand side we count as many places as the total decimal places in the multiplicand and multiplier and place the decimal point before them.
2.
Example multiply 4 . 3 x 5 4 3 x 5 2 1 . 5
Practice set 16
1. If 3 1 7 x 4 5 = 1 4 2 6 5 then 3 . 1 7 x 4 . 5 = ? =1 4 . 2 6 5
2. If 5 0 3 x 2 1 7 = 1 0 9 1 5 1 then 5 . 0 3 x 2 . 1 7 = ? =1 0 . 9 1 5 1
Multiply 1) 2 . 7 x 1 . 4 2 7 x 1 4 1 0 8 2 7 0 3 . 7 8
2) 6. 1 7 x 3.9 6 1 7 x 3 9 5 5 5 3 1 8 5 1 0 2 4 . 0 6 3
5 . 0 4 x 0 . 7 5 0 4 x 0 . 7 3 . 5 2 8
0 . 5 7 x2 0 5 7 x 2 1 . 1 4
Virendra bought 18 bags of rice ,each bag weighing 5.250kg . How much rice did he buy altogether? if the rice cost 42 rupees per kg, how much did he pay for it
one bag of rice weighs 5.250kg 18 bags will weigh ?
1 kg rice cost Rs 42 94.500kgs will cost 9 4 . 5 x 4 2 1 8 9 0 3 7 8 0 0 3 9 6 9. 0
5 . 2 5 0 x 1 8 4 2 0 0 0 5 2 5 0 0 9 4 . 5 0 0
18 bags of rice will weigh 94.500kg
Virendra paid Rs 3969
Vedika has 23.50 metres of cloth .she used it to make 5 curtains of equal size .If each curtain required 4 metres 25 cm to make, how much cloth is left over?
4.25 m for 1 curtain x 5 curtains 21.25 m for 5 curtains 23.50 m vedika had -21.25 m used for 5 curtains 2.25 m left. cloth left with vedika 2.25 m
THANKS!
Chapter 5
Decimal Fractions
Lets Recall
*Converting a decimal fraction into a common fraction.
*Multiplication of decimal fraction
Practice set 17
Q1 Carry out the following division
4 . 8 2 4 8 2 1 0 1 48 x 1 10 2 = 24 =2.4 10
2 )
1 )
20.6 2 2 0 6 2 1 0 1 206 x 1 10 2 =103 =10.3 10
4 )
3 )
17.5 5 175 5 10 1 175 x 1 10 5 =35 = 3.5 10
32.5 25 325 25 10 1 325 x 1 10 25 13 = 1.3 10
A road is 4 km 800 m long. If trees are planted on both of its sides at intervals of 9.6 m,how many trees were planted?
Q 2
A road is 4 km 800m long 1 km = 1000 m 4 km = 4000 m + 800 m 4800 m 4800 9.6 = 4800 96 1 10 =4800 x 10 = 100 x 5 = 500 trees are planted on 1 side 96 Ans : 1000 trees are planted on both sides of the road
Q3 Pradnya exercises regularly by walking alond a circular path on a field .if she walks a distance of 3.825 km in 9 round of the path , how much doesshe walk in one round
In 9 rounds distance walked is 3.825km, in 1 round distance walked is ? in 1 round distance walked = 3.825 9 3.825 x 1 = 3825 x 1 9 1000 9 = 3825 x 1 9 1000 = 425 = 0.425 km 1000
Q4 A Pharmaceutical manufacturer bought 0.25 quintal of hirada , amedicinal plant , for 9500 rupees . what is the cost per quintal of hirada?1 quintal = 100 kg
0.25 quintal hirada for Rs 95001 quintal hirada for Rs 9500 0.25 9500 25 = 9500 x 100 1 100 25 = 9500 x 4 =Rs 38,000 The cost per quintal of hirada is Rs 38,000
THANKS!
stay home stay safe
Q3 convert the decimal fractions into common fraction
1. 27.5 = 275 10 2. 0.007 = 7 1000 4. 39.15 = 3915 100
6. 70.400 = 70400 1000
23.7.20
St.Annes Primary school
Std 3A,B,C
Number Work
3 Digit Numbers Part 3
LOREM IPSUM DOLOR SIT AMET
St.Annes High School,pune
Appeared 200
100 percent Results Distinction -156 Ist class-140 2nd class-4 Highest -Vidhisha solanki 96.80
Introduction here
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Using symbols to show smaller and bigger ... < >
Activity 1
Put the correct sign < , > 427-----267 150----501 813----791 300---624
ACTIVITY-2
The expanded form of numbers example :
H T U
537=500+30+7