7th Maths Past Paper PDF

Part One 1 Geometrical Constructions Let’s recall. In previous classes, we have learnt about the line, line segment, angle, angle bisector, etc. We measure an angle in degrees. If ∠ABC measures 40°, we write it as m∠ABC = 40°. A...

Part One 1 Geometrical Constructions Let’s recall. In previous classes, we have learnt about the line, line segment, angle, angle bisector, etc. We measure an angle in degrees. If ∠ABC measures 40°, we write it as m∠ABC = 40°. A Angle Bisector You see the figure of ∠ABC alongside. M An angle bisector divides an angle into two B equal parts. Is ray BM the bisector of ∠ABC? C Perpendicular Bisector of a Line Segment Draw a line segment PS of length 4 cm and C draw its perpendicular bisector. Name it line CD. M · How will you verify that CD is the S P perpendicular bisector? m∠CMS = ° · Is l(PM) = l(SM)? D Let’s learn. The Property of the Angle Bisectors of a Triangle Activity 1. Draw any DPQR. 2. Use a compass to draw the bisectors of all three of its angles. (Extend the bisectors, if P necessary, so that they intersect one another.) A C 3. These three bisectors pass through the same point. That is, they are concurrent. I Name the point of concurrence ‘I’. Note that the point of concurrence of the angle bisectors of a triangle is in the interior of the triangle. Q B R 4. Draw perpendiculars IA, IB and IC respectively from I on to the sides of the triangle PQ, QR and PR. Measure the lengths of these perpendiculars. Note that IA = IB = IC. 1  The Property of Perpendicular Bisectors of the Sides of a Triangle Activity 1. Use a ruler to draw an acute-angled triangle and an X obtuse-angled triangle. Draw the perpendicular bisectors of each C side of the two triangles. Y Z 2. In each triangle, note that the perpendicular bisectors of the sides are concurrent. X C 3. Name their point of concurrence ‘C’. Measure the distance between C and the vertices of the triangle. Note that CX = CY = CZ Y Z 4. Observe the location of the point of concurrence of the perpendicular bisectors. Something more (1) The angle bisectors of a triangle are concurrent. Their point of concurrence is called the incentre, I and is shown by the letter ‘I’. (2) The perpendicular bisectors of the sides of a triangle C are concurrent. Their point of concurrence is called the circumcentre and is shown by the letter ‘C’. Practice Set 1 1. Draw line segments of the lengths given each triangle. Where do the points of below and draw their perpendicular concurrence lie? bisectors. 4. Draw a right-angled triangle. Draw (1) 5.3 cm (2) 6.7 cm (3) 3.8 cm the perpendicular bisectors of its sides. Where does the point of 2. Draw angles of the measures given concurrence lie ? below and draw their bisectors. * 5. Maithili, Shaila and Ajay live in three (1) 105° (2) 55° (3) 90° different places in the city. A toy shop is equidistant from the three houses. 3. Draw an obtuse-angled triangle and a Which geometrical construction should right-angled triangle. Find the points of be used to represent this? Explain concurrence of the angle bisectors of your answer. 2  Let’s learn. Construction of a Triangle Activity Let us see if we can draw the triangles when the measures of some sides and Before constructing any building, the angles, are given. structure of the building is first drawn on Draw DABC such that paper. You might have even seen a small l(AB) = 4 cm, and l(BC) = 3 cm. model of some such building. It becomes easier to construct the building from a · Can this triangle be drawn? drawing. In a similar way, a rough sketch · A number of triangles can be drawn to of a geometrical construction makes it fulfil these conditions. Try it out. easier to draw the required figure. It helps · Which further condition must be placed to plan the sequence of the steps in if we are to draw a unique triangle the construction. using the above information? (I) To construct a triangle given the lengths of its three sides Example Draw DXYZ such that l(XY) = 6 cm, l(YZ) = 4 cm, l(XZ) = 5 cm · Let us draw a rough figure quickly and show the given information in it as accurately as possible. For example, side XY is the longest, so, in the rough figure, too, XY should be the longest side. Steps Z Rough figure 1. According to the rough figure, segment m 4 XY of length 6 cm is drawn as the base. 5c cm X Y 2. As l(XZ) is 5 cm, draw an arc on one 6 cm side of seg XY with the compass opened to 5 cm and with its point at X. Z 3. Next, with the point at Y and the compass m 4c opened to 4 cm, draw an arc to cut the 5c m first arc at Z. Draw segs XY and YZ. X Y 6 cm A similar construction can be drawn on 6 cm X Y the other side of the base as shown below. 5 cm m 6 cm 4c X Y 5c m cm 4 Rough figure Z Z 3  Practice Set 2 1. Draw triangles with the measures given 2. Draw an isosceles triangle with below. base 5 cm and the other sides 3.5 cm (a) In ∆ABC, l(AB) = 5.5 cm, ‌each. l(BC) = 4.2 cm, l(AC) = 3.5 cm 3. Draw an equilateral triangle with (b) In ∆STU, l(ST) = 7 cm, side 6.5 cm. l(TU) = 4 cm, l(SU) = 5 cm 4. Choose the lengths of the sides yourself and draw one equilateral, (c) In ∆PQR, l(PQ) = 6 cm, one isosceles and one scalene l(QR) = 3.8 cm, l(PR) = 4.5 cm triangle. (II) To construct a triangle given two sides and the angle included by them Example Draw ∆PQR such that l(PQ) = 5.5 cm, m∠P = 50°, l(PR) = 5 cm. (A rough figure has been drawn showing R Rough figure the given information. ∠P is an acute angle m and that is shown in the rough figure, too. ) 5c 50° P Q 5.5 cm Steps G 1. According to the rough figure, seg PQ forms the base of length 5.5 cm. R 2. Ray PG is drawn so that m∠GPQ = 50° cm 5 3. Open the compass to 5 cm. Placing the compass point on P, draw an arc to cut ray PG at R. Join points Q and R. ∆PQR is the 50° required triangle. P Q 5.5 cm The ray PG may be drawn on the other side of the seg PQ. Its rough figure will be as 5.5 cm Q P shown below. 50° Rough figure 5c P 5.5 cm m 50° Q 5c m R R G 4  Practice Set 3 ¤ Draw triangles with the measures given below. 1. In ∆MAT, l(MA) = 5.2 cm, 3. In ∆FUN, l(FU) = 5 cm, m∠A = 80°, l(AT) = 6 cm l(UN) = 4.6 cm, m∠U = 110° 2. In ∆NTS, m∠T = 40°, 4. In ∆PRS, l(RS) = 5.5 cm, l(NT) = l(TS) = 5 cm l(RP) = 4.2 cm, m∠R = 90° (III) To construct a triangle given two angles and the included side Example Construct ∆XYZ such that l(YX) = 6 cm, m∠ZXY = 30°, m∠XYZ = 100° ∠XYZ is an obtuse angle and Z Rough figure that is shown in the rough figure. Steps 100° 30° 1. According to the rough figure, Y X draw seg YX as base of length 6 cm 6 cm. 2. Draw ray YR such that R m∠XYR = 100° Z D 3. On the same side of seg YX as point R, draw ray XD so that m∠YXD = 30°. Name the point of intersection of rays YR and XD, Z. 100° ∆XYZ is the required triangle. 30° Y X 4. See how an identical triangle can 6 cm be drawn on the other side of the base. Use your brain power. Example In ∆ABC, m∠A = 60°, m∠B = 40°, l(AC) = 6. Can you draw ∆ABC? What further information is required before it can be drawn? Which property can be used to get it? Draw the rough figure to find out. Recall the property of the sum of the angles of a triangle. Using that property, can we find the measures of two angles and an included side AC ? 5  Practice Set 4 ¤ Construct triangles of the measures given below. 1. In ∆SAT, l(AT) = 6.4 cm, 3. In ∆EFG, l(EG) = 6 cm, m∠A = 45°, m∠T = 105° m∠F = 65°, m∠G = 45° 2. In ∆MNP, l(NP) = 5.2 cm, 4. In ∆XYZ, l(XY) = 7.3 cm, m∠N = 70°, m∠P = 40° m∠X = 34°, m∠Y = 95° (IV) To construct a right-angled triangle given the hypotenuse and one side We know that a triangle with a right angle is called a right-angled triangle. In such a triangle, the side opposite the right angle is called the hypotenuse. Example Draw ∆LMN such that m∠LMN = 90°, hypotenuse = 5 cm, l(MN) = 3 cm. Rough figure Let us draw the rough figure using the given information. L As m∠LMN = 90°, we draw a right-angled 5c triangle approximately and mark the right angle. m Thus we show the given information in the rough figure. M N 3 cm Steps 1. As shown in the rough figure, draw the base seg MN of length 3 cm. T L 2. At point M of seg MN, draw ray MT to make an angle of 90° to seg MN. 5c 3. Opening the compass to 5 cm and with the m point at N, draw an arc to cut seg MT at L. ∆LMN is the required triangle. M N 4. Note that a similar figure can be drawn on the 3 cm other side of the base. Practice Set 5 ¤ Construct triangles of the measures given below. 1. In ∆MAN, m∠MAN = 90°, 3. In ∆ABC, l(AC) = 7.5 cm, l(AN) = 8 cm, l(MN) = 10 cm. m∠ABC = 90° , l(BC) = 5.5 cm. 2. In the right-angled ∆STU, hypotenuse 4. In ∆PQR, l(PQ) = 4.5 cm, SU = 5 cm and l(ST) = 4 cm. l(PR) = 11.7 cm, m∠PQR = 90°. 5. Students should take examples of their own and practise construction of triangles. 6 Activity Try to draw triangles with the following data. 1. ∆ABC in which m∠A = 85°, m∠B = 115° l(AB) = 5 cm 2. ∆PQR in which l(QR) = 2 cm, l(PQ) = 4 cm, l(PR) = 2 cm Could you draw these triangles? If not, look for the reasons why you could not do so. An activity for learning something more Example Draw ∆ABC such that l(BC) = 8 cm, l(CA) = 6 cm, m∠ABC = 40°. Draw a ray to make an angle of 40° with the base BC, l(BC) = 8 cm. We have to obtain point ‘A’ on the ray. With ‘C’ as the centre, draw an arc of radius 6 cm to do so. What do we observe? The arc intersects the ray in two different points. Thus, we get two triangles of two different shapes having the given measures. Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn? Let’s learn. Congruence of Segments Activity I Take a rectangular paper. Place two opposite sides one upon the other. B They coincide exactly. Activity II Using the ruler measure the lengths of seg AB and seg PQ. A l(AB) =............. l(PQ) =............. P Q Are they of the same length? You cannot pick up and place one segment over the other. Trace the seg AB along with the names of the points on a sheet of transparent paper. Place this new segment on seg PQ. Verify that if point A is placed on point P, then B falls on Q. It means that seg AB is congruent with seg PQ. We can infer from this that if two line segments have the same lengths, they coincide exactly with each other. That is, they are congruent. If seg AB and seg PQ are congruent, it is written as seg AB ≅ seg PQ. Now I know ! If given line segments are equal in length, they are congruent. { If seg AB ≅ seg PQ it means that seg PQ ≅ seg AB. { Note also that if seg AB ≅ seg PQ and seg PQ ≅ seg MN, then seg AB ≅ seg MN. In other words, if one line segment is congruent to another and that segment is congruent to a third, then the first segment is also congruent to the third. 7 Activity I Take any box. Measure the lengths of each of its edges. Which of them are congruent? Activity II From the shape shown below, write the names of the pairs of congruent line segments. A D (1) seg AB ≅ seg DC B C (2) seg AE ≅ seg BH E F (3) seg EF ≅ seg....... H G (4) seg DF ≅ seg....... Practice Set 6 1. Write the names of pairs of congruent line segments. (Use a divider to find them.) N B (i)........................... G (ii)........................... M R (iii)........................... (iv)........................... D C E 2. On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks. Q P Z Y X W A B C (i) seg AB ≅ seg....... (ii) seg AP ≅ seg....... (iii) seg AC ≅ seg....... (iv) seg....... ≅ seg BY (v) seg....... ≅ seg YQ (vi) seg BW ≅ seg....... Let’s learn. Congruence of Angles Observe the given angles and write the names of those having equal measures. P A N 100° T 90° B 35° C I T Q 8  S R S 90° R 100° 35° I P M T S Activity L X 40° 40° M N Y Z Draw two angles ∠LMN and ∠XYZ of 40° each as shown in the figure. Trace the arms of ∠LMN and the names of the points on a transparent paper. Now lift the transparent paper and place the angle you obtain on ∠XYZ. Observe that if point M is placed on Y and ray MN on ray YZ, then ray ML falls on ray YX. We can infer that angles of equal measure are congruent. The congruence of angles does not depend on the length of their arms. It depends upon the measures of those angles. That ∠LMN is congruent with ∠XYZ is written as ∠LMN ≅ ∠XYZ. Now I know ! Two angles with equal measures are congruent to each other. { If ∠LMN ≅ ∠XYZ then ∠XYZ ≅ ∠LMN. { If ∠LMN ≅ ∠ABC and ∠ABC ≅ ∠XYZ then ∠LMN ≅ ∠XYZ. Let’s discuss. (1) What time does this clock show? (2) What is the measure of the angle between its two hands? (3) At which other times is the angle between the hands congruent with this angle? 9  Practice Set 7 ¤ Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures. S R L A B P 45° 30° M 45° D 90° N T 30° Q R O C Let’s learn. Congruence of Circles (a) 1 cm (b) 2 cm 1 cm 1.3 cm (c) (d) Activity I Observe the circles in the figures above. Draw similar circles of radii 1 cm, 2 cm, 1 cm and 1.3 cm on a paper and cut out these circular discs. Place them one upon the other to find out which ones coincide exactly. Observations : 1. The circles in (a) and (c) coincide. 2. Circles in fig (b) and (c) and in fig (a) and (d) do not coincide. Are there other pairs like these ? Circles which coincide exactly are said to be congruent circles. Activity II Get bangles of different sizes but equal thickness and find the congruent ones among them. Activity III Find congruent circles in your surroundings. Activity IV Take some round bowls and plates. Place their edges one upon the other to find pairs of congruent edges. Now I know ! Circles of equal radii are congruent circles. ICT Tools or Links Use the construction tools in the Geogebra software to draw triangles and circles. qqq 10  2 Multiplication and Division of Integers Let’s recall. · In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below. (1) 5 + 7 = (2) 10 + ( - 5) = (3) - 4 + 3 = (4) ( - 7) + ( - 2) = (5) (+ 8) - (+ 3) = (6) (+ 8) - ( - 3) = · Write a number in each bracket to obtain the answer ‘3’ in each operation. ( - 6) + ( ) 4 - ( ) 7+( ) 3 ( - 5) - ( ) -8 + ( ) 9 - ( ) Let’s learn. Multiplication of Integers Mayuri’s bicycle got punctured on the way back from school and she did not have enough money to get it repaired. Sushant, Snehal and Kalpana lent her five rupees each. Thus she borrowed 15 rupees altogether and got the bicycle repaired. We show borrowed money, or a debt, using the ‘ - ’ (minus) sign. That is, Mayuri had a debt of 15 rupees or Mayuri had - 15 rupees. We see here that ( - 5) + ( - 5) + ( - 5) = - 15 Hence note that ( - 5) × 3 = 3 × ( - 5) = - 15 Of course, Mayuri paid back her debt the next day. We have learnt the multiplication and division of whole numbers. We have even made tables to carry out the multiplication. Now let us learn to multiply integers i.e. multiplication of numbers in the set that includes negative numbers, positive numbers and zero. ( - 3) + ( - 3) + ( - 3) + ( - 3) This addition is the addition of ( - 3) taken 4 times. It equals - 12. It can be written as ( - 3) × 4 = - 12. Similarly, ( - 5) × 6 = - 30, ( - 7) × 2 = - 14, 8 × ( - 7) = - 56 11  Now, let us make the table of ( - 4). ( - 4) × 0 = 0 Observe the pattern here. As the multiplier of ( - 4) ( - 4) × 1 = - 4 increases by 1, the product is reduced by 4. ( - 4) × 2 = -8 ( - 4) × 3 = - 12 Keeping the same pattern, if we extend the table upwards, decreasing the multiplier, this is what we will get. ( - 4) × ( - 2) = 8 As the multiplier of ( - 4) decreases by one unit, ( - 4) × ( - 1) = 4 the product increases by 4. ( - 4) × 0 = 0 The table for ( - 5) is given below. Complete the tables of ( - 6) and ( - 7). ( - 5) × ( - 3) = 15 ( - 6) × ( - 3) = ( - 7) × ( - 3) = ( - 5) × ( - 2) = 10 ( - 6) × ( - 2) = ( - 7) × ( - 2) = ( - 5) × ( - 1) = 5 ( - 6) × ( - 1) = ( - 7) × ( - 1) = ( - 5) × 0 = 0 ( - 6) × 0 = ( - 7) × 0 = ( - 5) × 1 = - 5 ( - 6) × 1 = ( - 7) × 1 = ( - 5) × 2 = - 10 ( - 6) × 2 = ( - 7) × 2 = ( - 5) × 3 = - 15 ( - 6) × 3 = ( - 7) × 3 = ( - 5) × 4 = - 20 ( - 6) × 4 = ( - 7) × 4 = Now I know ! · The product of two positive (+ve) (+ve number) × (+ve number) = (+ve number) integers is a positive (+ve) integer. · The product of one positive (+ve) and (+ve number) × ( - ve number) = ( - ve number) one negative ( - ve) integer is a negative integer. ( - ve number) × (+ve number) = ( - ve number) · The product of two negative ( - ve) ( - ve number) × ( - ve number) = (+ve number) integers is a positive (+ve) integer. Practice Set 8 ¤ Multiply. (i) ( - 5) × ( - 7) (ii) ( - 9) × (6) (iii) (9) × ( - 4) (iv) (8) × ( - 7) (v) ( - 124) × ( - 1) (vi) ( - 12) × ( - 7) (vii) ( - 63) × ( - 7) (viii) ( - 7) × (15) 12  Let’s learn. Division of Integers We have learnt how to divide one positive integer by another. We also know that the quotient of such a division may be an integer or a fraction. 6 5 2 Example 6 ÷ 2 = = 3, 5 ÷ 3 = = 1 + 2 3 3 On the number line, we can show negative integers on the left of the zero. We can show parts of integers also in the same way. -5 -3 3 5 2 2 2 2 -4 -3 -2 -1 0 1 2 3 4 5 3 3 5 Here, the numbers - , - , , are shown on the number line. 2 2 2 2  1 1   3 −3  −5 5 Note that  − 2 , 2 ,  2 , 2  ,  2 , 2 are mutually opposite numbers.       1 -1 3 ( −3) 5 5 That is, + 2 = 0, + = 0, - + =0 2 2 2 2 2 Pairs of opposite numbers are also called pairs of additive inverse numbers. We have seen that (- 1) × (- 1) = 1. If the two sides of this equation are divided by (-1) we get 1 1 the equation (- 1) =. Therefore, the quotient of the division ( − 1) ( − 1) is (-1). 1 6 Hence, we see that 6 × (- 1) = 6 × − 1 = − 1. ( ) ( ) To divide any positive integer by a negative integer 7 7 ×1 1 1 7 1 ( 7 ) × ( −1) -7 = =7× × = × (-1) × = = -2 ( − 1) × 2 ( − 1) 2 1 2 2 2 To divide any negative integer by a negative integer - 13 ( − 1) ×13 ( − 1) ×13 × 1 ( −1) × 13 = 1× 13 = 13 = −1 × 2 = = ( −1) × -2 ( ) ( −1) 2 1 2 2 2 - 25 25 - 18 18 Similarly, verify that = , -2 = = 9 etc. -4 4 2 This explains the division of negative integers. When one integer is divided by another non-zero integer, it is customary to write the denominator of the quotient as a positive integer. 7 -7 -11 11 Hence we write = , = 3 -2 2 -3 13  Now I know ! The rules of division of integers are like the rules of multiplication 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 (ii) 98 ÷ (- 28) (iii) (- 51) ÷ 68 (iv) 38 ÷ (- 57) (v) (- 85) ÷ 20 (vi) (-150) ÷ (- 25) (vii) 100 ÷ 60 (viii) 9 ÷ (- 54) (ix) 78 ÷ 65 (x) (- 5) ÷ (- 315) 24 2*. Write three divisions of integers such that the fractional form of each will be 5. -5 3*. Write three divisions of integers such that the fractional form of each will be. 7 4. The fish in the pond below, carry some numbers. Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers. For example, −24 −8 1. (-13) × (-15) = 195 2. (-24) ÷ 9 = = 9 3 - 13 - 24 +9 - 27 + 41 + 12 + 13 - 15 -8 - 37 - 18 - 28 qqq 14  3 HCF and LCM Let’s recall. · Which is the smallest prime number? · List the prime numbers from 1 to 50. How many are they? · Circle the prime numbers in the list below. 17, 15, 4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97 Co-prime numbers : Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime or mutually prime numbers. For example, 10 and 21 are co-primes, because the divisors of 10 are 1, 2, 5, 10 while the divisiors of 21 are 1, 3, 7, 21 and the only factor common to both 10 and 21 is 1. Some other co-prime numbers are (3, 8) ; (4, 9); (21, 22) ; (22, 23) ; (23, 24). Verify that any two consecutive natural numbers are co-primes. Let’s learn. Twin Prime Numbers If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers. For example : (3, 5) ; (5, 7) ; (11, 13) ; (29, 31) etc. Practice Set 10 1. Which number is neither a prime number nor a composite number? 2. Which of the following are pairs of co-primes? (i) 8, 14 (ii) 4, 5 (iii) 17, 19 (iv) 27, 15 3. List the prime numbers from 25 to 100 and say how many they are. 4. Write all the twin prime numbers from 51 to 100. 5. Write 5 pairs of twin prime numbers from 1 to 50. 6. Which are the even prime numbers? Let’s learn. Factorising a Number into its Prime Factors A simple but important rule given by Euclid is often used to find the GCD or HCF and LCM of numbers. The rule says that any composite number can be written as the product of prime numbers. 15  Let us learn how to find the prime factors of a number. Example Write the number 24 in the form of the product of its prime factors. Method for finding prime factors Vertical arrangement Horizontal arrangement 2 24 24 = 2 × 12 2 12 =2×2×6...12 is factorised. 2 6 =2×2×2×3...6 is factorised. 3 3 The prime factors are 2 and 3. 1 24 = 2 × 2 × 2 × 3 Remember : To write a given number as a product of its prime factors is to factorise it into primes. Example Write each of the given numbers as a product of its prime factors. 63 45 20 7 × 9 2 × 10 5 × 9 7 × 3 × 3 2 × 2 × 5 5 × 3 × 3 63 = 7 × 3 × 3 45 = 5 × 3 × 3 20 = 2 × 2 × 5 Example Factorise into primes : 117. Example Factorise into primes : 250. 3 117 2 250 3 39 117 = 13 × 9 5 125 250 = 2 × 125 13 13 = 13 × 3 × 3 5 25 = 2 × 5 × 25 1 5 5 =2×5×5×5 1 117 = 3 × 3 × 13 250 = 2 × 5 × 5 × 5 16 Example Find the prime factors of 40. Vertical arrangement Horizontal arrangement 2 40 2 20 40 = 10 × 4 40 = 8 × 5 2 10 =5×2×2×2 =2×2×2×5 5 5 1 40 = 2 × 2 × 2 × 5 Practice Set 11 ¤ Factorise the following numbers into primes. (i) 32 (ii) 57 (iii) 23 (iv) 150 (v) 216 (vi) 208 (vii) 765 (viii) 342 (ix) 377 (x) 559 Let’s recall. Greatest Common Divisor (GCD) or Highest Common Factor (HCF) We are familiar with the HCF and LCM of positive integers. Let us learn something more about them. The HCF or the GCD of given numbers is their greatest common divisor or factor. In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (i) 28, 42 (ii) 51, 27 (iii) 25, 15, 35 Let’s learn. Prime Factors Method It is easy to find the HCF of numbers by first factorising all the numbers. Example Find the HCF of 24 and 32 by the prime factors method. 2 24 24 = 4 × 6 2 32 32 = 8 × 4 2 12 = 2 × 2 × 2 × 3 2 16 = 2 × 2 × 2 × 2 × 2 2 6 2 8 3 3 2 4 1 2 2 1 The common factor 2 occurs thrice in each number. Therefore, the HCF = 2×2×2 = 8. 17 Example Find the HCF of 195, 312, 546. 195 = 5 × 39 312 = 4 × 78 546 = 2 × 273 = 5 × 3 × 13 = 2 × 2 × 2 × 39 = 2 × 3 × 91 = 2 × 2 × 2 × 3 × 13 = 2 × 3 × 7 × 13 The common factors 3 and 13 each occur once in all the numbers. ∴ HCF = 3 × 13 = 39 Example Find the HCF of 10, 15, 12. 10 = 2 × 5 15 = 3 × 5 12 = 2 × 2 × 3 No number except 1 is a common divisor. Hence, HCF = 1 Example Find the HCF of 60, 12, 36. 60 = 4 × 15 12 = 2 × 6 36 = 3 × 12 =2×2×3×5 = 2 × 2 × 3 = 3 × 3 × 4 = 2 × 2 × 3 × 3 ∴ HCF = 2 × 2 × 3 = 12 Let us work out this example in the vertical arrangement. We write all the numbers in one line and find their factors. 2 60 12 36 ∴ HCF = 2 × 2 × 3 = 12 2 30 6 18 3 15 3 9 Note that 12 is a divisor of 36 and 60. 5 1 3 Now I know ! · If one of the given numbers is a divisor of all the others, then it is the HCF of the given numbers. · If no prime number is a common divisor of all the given numbers, then 1 is their HCF because it is the only common divisor. * Something more 2 is the HCF of any two consecutive even numbers and 1 is the HCF of any two consecutive odd numbers. Verify the rule, by taking many different examples. 18 The Division Method for Finding the HCF Example Find the HCF of 144 and 252. 1. Divide the bigger number by the smaller one. 144)   252( 1 2. Divide the previous divisor by the remainder in this -144 division.     108)144( 1 3. Divide the divisor of step 2 by the remainder - obtained in the division in step 2.         108 4. Continue like this till the remainder becomes zero.              36)108(3 The divisor in the division in which the remainder -   108 is zero is the HCF of the given numbers. 000 ∴ The HCF of 144 and 252 = 36 209 Example Reduce 247 to its simplest form. 209)   247 ( 1 -209 To reduce the number to its simplest form, we will find the common factors of 209 and 247.      38)209 (5 Let us find their HCF by the division method. -        190 Here, 19 is the HCF. That is, the numerator and denominator are both divisible by 19.             19)38(2 - 209 209 ÷19 11   38 ∴ = = 00 247 247 ÷19 13 Practice Set 12 1. Find the HCF. (i) 25, 40 (ii) 56, 32 (iii) 40, 60, 75 (iv) 16, 27 (v) 18, 32, 48 (vi) 105, 154 (vii) 42, 45, 48 (viii) 57, 75, 102 (ix) 56, 57 (x) 777, 315, 588 2. Find the HCF by the division method and reduce to the simplest form. 275 76 161 (i) 525 (ii) (iii) 133 69 Let’s recall. Least Common Multiple (LCM) The Least Common Multiple of the given numbers is the smallest number that is divisible by each of the given numbers. · Write the tables of the given numbers and find their LCM. (i) 6, 7 (ii) 8, 12 (iii) 5, 6, 15 19  Let’s learn. Example Find the LCM of 60 and 48. Let us find the prime factors of each number. 60 = 2 × 2 × 3 × 5 48 = 2 × 2 × 2 × 2 × 3 Let us consider each prime number in these multiplications. 2 occurs a maximum of 4 times. (in the factors of 48) 3 occurs only once (in the factors of 60) 5 occurs only once (in the factors of 60) ∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 = 10 × 24 = 240 Example Find the LCM of 18, 30, 50. 18 = 2 × 9 30 = 2 × 15 50 = 2 × 25 = 2 × 3 × 3 = 2 × 3 × 5 =2×5×5 2, 3, 5 are the prime numbers that occur in the multiplications above. In the products above, the number 2 occurs a maximum of times, 3 occurs a maximum of times and 5 a maximum of times. ∴ LCM = 2 × 3 × 3 × 5 × 5 = 450 ∴ The LCM of 18, 30, 50 is 450. Example Find the LCM of 16, 28, 40. Vertical arrangement · Use the tests of divisibility to find the prime number that divides all the numbers and then divide the given numbers. 2 16 28 40 Repeat this process for the quotients as many times 2 8 14 20 as possible. 2 4 7 10 · Now find the number that divides at least two of the numbers obtained and divide those numbers by the number you find. 2 7 5 Do this as many times as possible. If a number cannot be divided, leave it as it is. · Stop when the only common divisor you get is 1. · Find the product of the numbers in the column on the left. Multiply this product by the numbers in the last row. LCM = 2 × 2 × 2 × 2 × 5 × 7 = 560 Example Find the LCM and HCF of 18 and 30. Compare the 2 18 30 product of the LCM and HCF with the product of the given numbers. HCF = 2 × 3 = 6 3 9 15 LCM = 2 × 3 × 3 × 5 = 90 3 5 HCF × LCM = 6 × 90 = 540 Product of the two given numbers = 18 × 30 = 540 Product of the two given numbers = HCF × LCM 20  We see that the product of two numbers is equal to the product of their GCD and LCM. Verify this statement for the following pairs of numbers : (15, 48), (14, 63), (75, 120) Example Find the LCM and HCF of 15, 45 and 105. 3 15 45 105 15 = 3 × 5 45 = 3 × 3 × 5 5 5 15 35 105 = 3 × 5 × 7 1 3 7 GCD = 3 × 5 = 15 LCM = 3 × 3 × 5 × 7 = 315 Example The product of two 2-digit numbers is 1280 and the GCD = 4. What is their LCM? GCD × LCM = Product of given numbers 4 × LCM = 1280 1280 ∴ LCM = = 320 4 Practice Set 13 1. Find the LCM. (i) 12, 15 (ii) 6, 8, 10 (iii) 18, 32 (iv) 10, 15, 20 (v) 45, 86 (vi) 15, 30, 90 (vii) 105, 195 (viii) 12, 15, 45 (ix) 63, 81 (x) 18, 36, 27 2. 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 (ii) 46, 51 (iii) 15, 60 (iv) 18, 63 (v) 78, 104 The Use of LCM and HCF Example A shop sells a 450 g bottle of jam for 96 rupees and a bigger bottle of 600 g for 124 rupees. Which bottle is it more profitable to buy? Solution : We have learnt the unitary method. Using that we can find the cost of 1 gm jam in each bottle and compare. However, the calculation is easier if we use a bigger common factor rather than a smaller one. Let us use 150, the HCF of 450 and 600 to compare. 450 = 150 × 3, 600 = 150 × 4 21  96 ∴ In the small bottle, 150 g of jam costs = 32 rupees. 3 124 In the large bottle, 150 g of jam costs = 31 rupees. 4 ∴ Thus, it is more profitable to buy the 600 g bottle of jam. 17 11 Example Add + Method 1 In order to add, let us make the 28 35 denominators of the fractions equal. 17 11 17 × 35 + 11× 28 595 + 308 903 903 129 Solution : + = = = 28 × 35 = = 28 35 28 × 35 28 × 35 980 140 Method 2 Let us find the LCM of 28 and 35 in order to add the fractions. LCM = 7 × 4 × 5 = 140 17 11 17 × 5 11× 4 85 + 44 129 Solution : + = + = = 140 28 35 28 × 5 35 × 4 140 Taking the LCM rather than multiplying the denominators made our calculations so much easier! Example On dividing a certain number by 8, 10, 12, 14 the remainder is always 3. Which is the smallest 2 8 10 12 14 such number? 2 4 5 6 7 Solution : To find this multiple, let us find the LCM 2 5 3 7 of the given divisors. LCM = 2 × 2 × 2 × 5 × 3 × 7 = 840 To the LCM we add the remainder obtained every time. Hence, that number = LCM + remainder = 840 + 3 = 843 Example Find the LCM of the numbers 16, 20, 80. Solution : 16 = 2 × 2 × 2 × 2 4 16 20 80 20 = 2 × 2 × 5 80 = 2 × 2 × 2 × 2 × 5 4 4 5 20 LCM = 4 × 4 × 5 = 80 5 1 5 5 Did you notice that here 80 is one of the given numbers and that the other numbers 16 and 20, 1 1 1 are its divisors. Remember : If the greatest of the given numbers is divisible by the other numbers, then that greatest number is the LCM of the given numbers. In order to verify the above rule, examine these groups of numbers (18,90) (35,140,70). 22 Example Shreyas, Shalaka and Snehal start running from the same point on a circular track at the same time and complete one lap of the track in 16 minutes, 24 minutes and 18 minutes respectively. What is shortest period of time in which they will all reach the starting point together ? Solution : The number of minutes they will take to reach together will be a multiple of 16, 24 and 18. To find out the smallest such number, we will find the LCM. 16 = 2 × 2 × 2 × 2 24 = 2 × 2 × 2 × 3 18 = 2 × 3 × 3 LCM = 2 × 2 × 2 × 2 × 3 × 3 = 144 They will come together in 144 minutes or 2 hours 24 minutes. Practice Set 14 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 2. Find the HCF and LCM. (i) 14, 28 (ii) 32, 16 (iii) 17, 102, 170 (iv) 23, 69 (v) 21, 49, 84 3. Find the LCM. (i) 36, 42 (ii) 15, 25, 30 (iii) 18, 42, 48 (iv) 4, 12, 20 (v) 24, 40, 80, 120 4. Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time. 348 221 437 5. Reduce the fractions , , to the lowest terms. 319 247 551 6. The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other? 7. The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM? 8. A trader has three bundles of string 392 m, 308 m and 490 m long. What is the greatest length of string that the bundles can be cut up into without any left over string? 9*. Which two consecutive even numbers have an LCM of 180? qqq 23 4 Angles and Pairs of Angles Let’s recall. A · Write the name of the angle shown alongside..................... · Write the name of its vertex................. · Write the names of its arms..................... B · Write the names of the points marked C on its arms...................... Let’s learn. The Interior and Exterior of an Angle In the plane of the figure alongside, the group of points like point N, point M, point T which are not on the arms of the angle, form the interior of the ∠PQR. The group of points in the plane of the angle like M point G, point D, point E, which are neither on P N T the arms of the angle nor in its interior, form the G exterior of the angle. O D Q R S E Adjacent Angles Look at the angles in the figure alongside. The ray MQ is a common arm of the angles ∠BMQ and ∠QMD while M is their common vertex. The interiors of these angles do not have a single common point. B They may be said to be neighbouring angles. Such angles are called adjacent angles. Adjacent angles have one common arm and the Q other arms lie on opposite sides of the common arm. M They have a common vertex. Adjacent angles have D separate interiors. In the given figure, MB is the common arm of the angles ∠BMD and ∠BMQ. However, they are not adjacent angles because they do not have separate interiors. 24  Now I know ! Two angles which have a common vertex, a common arm and separate interiors are said to be adjacent angles. Practice Set 15 1. Observe the figure and complete the table for ∠AWB. N A Points in the interior R X T Points in the exterior C U W G Q B Y Points on the arms of the angles V 2. Name the pairs of adjacent angles in the figures below. B P R N Q T A C S 3. Are the following pairs adjacent angles? R S If not, state the reason. N Q (i) ∠PMQ and ∠RMQ (ii) ∠RMQ and ∠SMR T (iii) ∠RMS and ∠RMT (iv) ∠SMT and ∠RMS M P Let’s learn. Complementary Angles P S · Draw ∠PQR, a right angle. · Take any point S in its interior. · Draw ray QS. · Add the measures of the angles ∠PQS and ∠SQR. What will be the Q R sum of their measures? If the sum of the measures of two angles is 90° they are known as complementary angles. Here, ∠PQS and ∠SQR are mutually complementary angles. 25 Example Observe the angles in the figure and enter the proper number in the box. A m∠ABC = ° P m∠PQR = ° 40° C 50° m∠ABC + m∠PQR = ° Q B R The sum of the measures of ∠ABC and ∠PQR is 90°. Therefore, they are complementary angles. Example Find the measure of the Example Angles of measures (a + 15)° and complement of an angle of (2a)° are complementary. What is measure 70°? the measure of each angle? Solution : Let the measure of the Solution : a + 15 + 2a = 90 complementary angle be x. 3a + 15 = 90 70 + x = 90 3a = 75 ∴ 70 + x - 70 = 90 - 70 a = 25 x = 20° ∴ a + 15 = 25 + 15 = 40° The measure of the complement and 2a = 2 × 25 = 50° of an angle of measure 70° is 20°. Practice Set 16 1. The measures of some angles are given below. Write the measures of their complementary angles. (i) 40° (ii) 63° (iii) 45° (iv) 55° (v) 20° (vi) 90° (vii) x° 2. (y - 20)° and (y + 30)° are the measures of complementary angles. Find the measure of each angle. Let’s recall. T is a point on line AB. · What kind of angle is ∠ATB ? A T B · What is its measure? Let’s learn. Supplementary Angles · A line AC is shown in the figure alongside. A ray BD stands on it. How many angles are D formed here? m∠ABD = °, m∠DBC = ° m∠ABD + m∠DBC = ° A B C If the sum of the measures of two angles is 180° they are known as supplementary angles. Here ∠ABD and ∠DBC are supplementary angles. 26 Example Observe the angles in the figure below and enter the proper number in the box. M m∠PQR = ° m∠MNT = ° R m∠PQR + m∠MNT = ° 110° 70° ∠PQR and ∠MNT are supplementary angles. N P Q T Example Find the measure of the Example (a + 30)° and (2a)° are the supplement of an angle of 135°. measures of two supplementary Solution : Let the supplementary angle angles. What is the measure of measure p°. each angle? The sum of the measures of two Solution : a + 30 + 2a = 180 supplementary angles is 180°. ∴ 3a = 180 - 30 135 + p = 180 ∴ 3a = 150 ∴ 135 + p - 135 = 180 - 135 ∴ a = 50 ∴ p = 45 ∴a + 30 = 50 + 30 = 80° ∴ The measure of the supplement ∴ 2a = 2 × 50 = 100° of an angle of 135° is 45°. ∴ The measures of the angles are 80° and 100°. Practice Set 17 1. Write the measures of the supplements of the angles given below. (i) 15° (ii) 85° (iii) 120° (iv) 37° (v) 108° (vi) 0° (vii) a° 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° 3. In ∆XYZ, m∠Y = 90°. What kind of a pair do ∠X and ∠Z make? 4. The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles. P T 5. PTNM is a rectangle. Write the names of the pairs of supplementary angles. M N 6*. If m∠A = 70°, what is the measure of the supplement of the complement of ∠A? 7. If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A? 27  Let’s discuss. Discuss the following statements. If a statement is right, give an example. If it is wrong, state why. · Two acute angles can make a pair of · Two acute angles can form a pair of complementary angles. supplementary angles. · Two right angles can make a pair of · Two right angles can form a pair of complementary angles. supplementary angles. · One acute angle and one obtuse angle · One acute angle and one obtuse angle can make a pair of complementary can form a pair of supplementary angles. angles. Let’s learn. Opposite Rays B (1) Name the rays in the figure alongside. A C (2) Name the origin of the rays. Figure (i) (3) Name the angle in figure (i). (1) Name the angle in figure (ii) alongside. A B C (2) Name the rays whose origin is point B. Figure (ii) In figure (i), ray BC and ray BA meet to form an obtuse angle while in figure (ii) ray BC and ray BA meet to form a straight angle and we get a straight line. Here, ray BC and ray BA are opposite rays. Now I know ! Two rays which have a common origin and form a straight line are said to be opposite rays. Practice Set 18 1. Name the pairs of opposite rays in the figure L alongside. T 2. Are the ray PM and PT opposite rays? P Give reasons for your answer. N M 28  Let’s learn. Angles in a Linear Pair · Write the names of the angles in the R figure alongside. · What type of a pair of angles is it? · Which arms of the angles are not the common arms? P Q S · m∠PQR = ° · m∠RQS = ° · m∠PQR + m∠RQS = 180° The angles ∠PQR and ∠RQS in the figure above are adjacent angles and are also supplementary angles. The arms that are not common to both angles form a pair of opposite rays i.e. these arms form a straight line. We say that these angles form a linear pair. The sum of the measures of the angles in a linear pair is 180°. Now I know ! Angles which have a common arm and whose other arms form a straight line are said to be angles in a linear pair. Angles in a linear pair are supplementary angles. Activity : Use straws or sticks to make all the kinds of angles that you have learnt about. Practice Set 19 Draw the pairs of angles as described below. If that is not possible, say why. (i) Complementary angles that are not adjacent. (ii) Angles in a linear pair which are (iii) Complementary angles that do not form a not supplementary. linear pair. (iv) Adjacent angles which are not in a linear pair. (v) Angles which are neither complementary (vi) Angles in a linear pair which are nor adjacent. complementary. Let’s learn. Vertically Opposite Angles In the figure alongside, line PT and P S line RS intersect each other at point Q. Thus, four angles are formed. ∠PQR is formed by the rays QP and QR. The Q rays opposite to ray QP and QR are QT and QS respectively. These opposite rays form the angle ∠SQT. Hence, ∠SQT is R T called the opposite angle of ∠PQR. 29  Now I know ! The angle formed by the opposite rays of the arms of an angle is said to be its opposite angle. Let’s learn. The Property of Vertically Opposite Angles · Name the angle opposite to ∠PQS in the figure. As shown in the figure, m∠PQS = a, m∠SQT = b, m∠TQR = c, m∠PQR = d. ∠PQS and ∠SQT are the angles in a linear pair. ∴ a + b = 180° Also m∠SQT and m∠TQR are two angles in a linear pair. R P ∴ b + c = 180° d ∴a+b=b+c c Q a ∴a=c (Subtracting b from both sides.) b ∴ ∠PQS and ∠TQR are congruent angles. T S Also, m∠PQR = m∠SQT i.e. ∠PQR and ∠SQT are congruent angles. Now I know ! The vertically opposite angles formed when two lines intersect, are of equal measure. Practice Set 20 1. Lines AC and BD intersect at point P. m∠APD = 47° A B Find the measures of ∠APB, ∠BPC, ∠CPD. 47° P 2. Lines PQ and RS intersect at point M. m∠PMR = x° P D S C What are the measures of ∠PMS, ∠SMQ and ∠QMR? x° M Let’s learn. Interior Angles of a Polygon R Q A Interior Angles of a Triangle ∠A, ∠B, ∠C are the interior angles of ∆ABC. m∠ABC + m∠BAC + m∠ACB = ° B C 30 Observe the table given below and draw your conclusions. Number Number Name of the Polygon of Sum of interior angles of sides polygon triangles 3 Triangle 1 180° × 1 = 4 Quadrilateral 2 180° × 2 = 5 Pentagon 3 180° × 3 = 6 Hexagon 4 180° × = 7 Heptagon 5 8 Octagon 6.................... n A figure with n sides (n-2) 180° × (n-2) Note that the number of triangles formed in a polygon as shown above is two less than the number of sides the polygon has. Now I know ! The sum of the measures of the interior angles of a polygon is = 180° × (n-2) 31  Let’s learn. The Exterior Angle of a Triangle P If the side BC of ∆ABC is extended as A shown in the figure, an angle ∠ACD is formed which lies outside the triangle. B C D Q ∠ACD is an exterior angle of ∆ABC. ∠ACD and ∠ACB are angles in a linear pair. ∠PAB and ∠QBC are also exterior angles of ∆ABC. Now I know ! On extending one side of a triangle, the angle obtained which forms a linear pair with the adjacent interior angle of the triangle is called an exterior angle of that triangle. Example In the figure alongside, all exterior P b a angles of a triangle are shown. a, b, c, d, e, f are the exterior angles c e of ∆PQR. In the same way, every triangle has six exterior angles. Q R d f Let’s learn. The Property of an Exterior Angle of a Triangle In the figure alongside, ∠PRS is an exterior P angle of ∆PQR. The interior angle adjacent to it is ∠PRQ. The other two interior angles, ∠P and ∠Q are further away from ∠PRS. They are called the remote interior angles of ∠PRS. Q R S m∠P + m∠Q + m∠PRQ = °...........(sum of the three angles of a triangle) m∠PRS + m∠PRQ = °...........(angles in a linear pair) \ m∠P + m∠Q + m∠PRQ = m∠PRS + m∠PRQ \ m∠P + m∠Q = m∠PRS....... (subtracting m∠PRQ from both sides) 32  Now I know ! The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles. Practice Set 21 1. ∠ACD is an exterior angle of DABC. A The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B. 140° B 2. Using the measures of the angles C D B given in the figure alongside, find A the measures of the remaining three 8y C angles. F 4y O 6y D E 3*. In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x - 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B. ICT Tools or Links · With the help of Geogebra, draw two rays with a common origin. Using the ‘Move’ option, turn one ray and observe the position where the two rays become opposite rays. · Form angles in a linear pair. By ‘moving’ the common arm, form many different pairs of angles all of which are linear pairs. · Using Polygon Tools from Geogebra, draw many polygons. Verify the property of the interior angles of a polygon. qqq 33  5 Operations on Rational Numbers Let’s learn. Rational Numbers In previous standards, we have learnt that the counting numbers 1, 2, 3, 4,..... are called natural numbers. We know that natural numbers, zero, and the opposite numbers of natural 7 2 1 numbers together form the group of integers. We are also familiar with fractions like , ,. 11 5 7 Is there then, a group that includes both integers and fractions? Let us see. 12 7 -3 0 4= ,7= ,-3= , 0 = Thus, we also know that all integers can be written 3 1 1 2 m m in the form n. If m is any integer and n is any non-zero integer, then the number n is called a rational number. This group of rational numbers includes all types of numbers mentioned before. 5 3 -17 Complete the table given below. Rational Numbers , , 28 , 7 4 2.17 etc. Integers - 3, - 8, - 1 etc. 3 5 5 -3 -17 - Whole Numbers 0 5 11 Natural Numbers Natural Number × 1, 2, 3,..... Integers Rational Number Operations on Rational Numbers Rational numbers are written like fractions using a numerator and a denominator. That is why, operations on rational numbers are carried out as on fractions. 5 9 55 + 63 118 1 3 4 - 21 -17 (1) + = = (2) - = = 7 11 77 77 7 4 28 28 1 8 15 50 9 4 9× 4 36 (3) 2 +3 = + (4) 13 × 7 = = 91 7 14 7 14 13 × 7 30 50 3 (-4) 3 × ( − 4) -12 = + (5) × = = 14 14 5 5 5× 5 25 80 40 9 26 3× 2 6 = = (6) 13 × = = 14 7 3 1 1 34  Let’s recall. To divide one number by another is to multiply the first by the multiplicative inverse of the other. 5 6 2 11 We have seen that and 5 , 11 and 2 are pairs of multiplicative inverses. 6  −5   −4   −7   −2   −5   −4  Similarly,   ×  5  = 1 ;   ×  7  = 1 Thus   and  5  as also  4     2   4     −7   −2  -5 -4 -7   and  7  are pairs of multiplicative inverses. Similarly, and or and  2  4 5 2 -2 -5 -4 are pairs of multiplicative inverses. That is, and are each other’s multiplicative 7 4 5 -7 -2 inverses and so are and. 2 7 Take care! -11 9 -11 9 Example The product of and is -1. Therefore, , is not a pair of 9 11 9 11 multiplicative inverses. Let’s discuss. Let us look at the characteristics of various groups of numbers. Discuss them in class to help you complete the following table. Consider the groups of natural numbers, integers and rational numbers. In front of each group, write the inference you make after carrying out the operations of addition, subtraction, multiplication and division, using a ( ) or a (×). Remember that you cannot divide by zero. · If natural numbers are added, their sum is always a natural number. So, we put a tick ( ) under ‘addition’ in front of the group of natural numbers. · However, if we subtract one natural number from another, the answer is not always a natural number. There are numerous examples like 7-10 = -3. So, under subtraction, we put a (×). If there is a cross in the table, explain why. To explain the reason for a (×), giving only one of the numerous examples is sufficient. Group of numbers Addition Subtraction Multiplication Division Natural numbers × × 3 (7 - 10 = - 3) (3 ÷ 5 = ) 5 Integers Rational numbers 35  Now I know ! · The group of natural numbers is closed for the addition and multiplication but not for the subtraction and division. In other words, the difference of any two natural numbers or the quotient obtained on dividing one natural number by another will not always be natural number. · The group of integers is closed for addition, subtraction and multiplication but not for division. · The group of rational numbers is closed for all operations - addition, subtraction, multiplication and division. However, we cannot divide by zero. Practice Set 22 1. Carry out the following additions of rational numbers. 5 6 2 4 11 13 3 3 (i) + (ii) 1 +2 (iii) + (iv) 2 +1 36 42 3 5 17 19 11 77 2. Carry out the following subtractions involving rational numbers. 7 3 13 2 2 5 1 1 (i) - (ii) - (iii) 1 - 3 (iv) 4 - 3 11 7 36 40 3 6 2 3 3. Multiply the following rational numbers. 3 2 12 4 ( −8) 3 0 3 (i) × (ii) × (iii) × (iv) × 11 5 5 15 9 4 6 4 4. Write the multiplicative inverse. 2 -3 -17 1 (i) (ii) (iii) (iv) 7 (v) - 7 5 8 39 3 5. Carry out the divisions of rational numbers. 40 10 -10 -11 -7 -3 2 (i) ÷ (ii) ÷ (iii) ÷ (iv) ÷ ( - 4) 12 4 11 10 8 6 3 1 3 -5 7 9 2 (v) 2 ÷5 (vi) ÷ (vii) ÷ ( −8 ) (viii) 5 ÷ 5 6 13 26 11 5 Let’s learn. Numbers in between Rational Numbers · Write all the natural numbers between 2 and 9. · Write all the integers between - 4 and 5. 1 3 · Which rational numbers are there between 2 and ? 4 36  1 4 Example Let us look for rational numbers between the two rational numbers and. 2 7 To do that, let us convert these numbers into fractions with equal denominators. 1 1× 7 7 4 4× 2 8 = = , = = 2 2× 7 14 7 7×2 14 7 8 7 and 8 are consecutive natural numbers. But, are and consecutive rational numbers? 14 14 The denominator of any number can be increased. Then the numerator also increases the same number of times. 7 70 8 80 = , = 140..... (Multiplying the numerator and denominator by 10) 14 140 14 70 71 79 80 7 8 Now, < 140..... < < 140 So, how many numbers do we find between and ? 140 140 14 14 7 700 8 800 Also, = , =..... (Multiplying the numerator and denominator by 100) 14 1400 14 1400 700 701 799 800 Hence, <...... < < 1400 1400 1400 1400 Thus, when rational numbers are converted into equivalent fractions with increasingly bigger denominators, more and more rational numbers which lie between them can be expressed. 1 3 For example, let us find numbers between the rational numbers and. 2 5 Let us first convert each of the numbers into their equivalent fractions. 1 5 3 6 For example, = , and = 2 10 5 10 11 20 0 1 1 5 6 - 10 10 10 5 6 On the number line, there are points showing the numbers and. Let us find the 10 10 mid-point of the line segment joining these two points and see what number it shows. 1  5 6  11  + = This is the mid-point of the line segment. 2  10 10  20 6 11 12 - 11 1 11 5 11 - 10 1 Because - = = and also - = = 10 20 20 20 20 10 20 20 11 5 6 Thus, is the number that lies exactly in the middle of and. It means that, 20 10 10 11 1 3 20 is a number that lies between 2 and 5. Similarly, we can find numbers that lie between 1 11 11 3 2 and 20 and between 20 and 5. 37  Now I know ! There are innumerable rational numbers between any two rational numbers. Practice Set 23 ¤ Write three rational numbers that lie between the two given numbers. 2 6 4 2 2 4 7 5 (i) , (ii) , (iii) - , (iv) , - 7 7 5 3 3 5 9 9 -3 +5 7 -5 (v) , (vi) , (vii) 5 , 11 (viii) 0, -3 4 4 8 3 7 7 4 Something more If m is an integer, then m + 1 is the next bigger integer. There is no integer between m and m + 1. The integers that lie between any two non-consecutive integers can be counted. However, there are infinitely many rational numbers between any two rational numbers. Let’s recall. We have learnt to multiply and divide decimal fractions. 35.1 1 351 1 351 = 35.1 × 10 = × 10 = = 3.51 10 10 100 35.1 35.1 1 351 1 351  = × = × =   = 0.351 100 1 100 10 100  1000  351 35.1 × 10 = × 10 = 351.0 10 351  351000  35.1 × 1000 = × 1000 =   = 35100.0 10  10  Thus we see that we can divide a decimal fraction by 100, by moving the decimal point two places to the left. To multiply by 1000, we move the point three places to the right. The following rules are useful while multiplying and dividing. No matter how many zeros we place at the end of the fractional part of a decimal fraction, and no matter how many zeros we place before the integral part of the number, it does not change the value of the given fraction. 135 100 13500 1.35 = 100 × 100 = 10000 = 1.3500 38  135 1000 135000 1.35 = 100 × 1000 = 100000 = 1.35000 etc. Also, see how we use : 1.35 = 001.35. 1.35 001.35 = = 0.0135 100 100 Let’s learn. Decimal Form of Rational Numbers 7 Example Write the rational number in decimal form. 1.75 4 (1) 7 = 7.0 = 7.000 (Any number of zeros can be added after the 4) 7.000 fractional part.) -   4 30 (2) 1 is the quotient and 3 the remainder after dividing 7 by 4. -  28 Now we place a decimal point after the integer 1. Writing the 0 from the dividend after the remainder 3, we divide 30 by 4. 20 As the quotient we get now is fractional, we write 7 after the - 20 decimal point. Again we bring down the next 0 from the 00 dividend and complete the division. 1 Example Write 2 in decimal form. 5 1 11 We shall find the decimal form of 2 = in three different ways. 5 5 1 Find the decimal form of. 5 (I) (II) (III)   2.2   0.2 5) 11.000 11 11× 2 5) 1.0 -   10 5 = 5× 2 -   0 1 22 5 = 0.2   010 =   10 -    10 10 -    10 = 2.2 00 00 1 11 \ 2 5 = 2.2 5 = 2.2 -5 Example Write the rational number in decimal form. 8 5 -5 The decimal form of 8 obtained by division is 0.625. ∴ = - 0.625 8 In all the above examples, we have obtained zero as the remainder. This type of decimal form of a rational number is called the terminating decimal form. 39 Example Let us see how the decimal form of some rational numbers is different. 5 2 (i) Write the number in decimal form. (ii) Write the number 11 in decimal form. 3 1.66 0.18 3) 5.00 11) 2.00 -   3 -   0 20 5    20 2 - 1 8 ∴ - = 1.666.......   11 ∴ 11 = 0.1818....... 3 20 5 · 90 2 -   18 ∴ = 1.6 - ∴ 11 = 0.18 3      88 2 20 1 1 7 5 (iii) Find the decimal form of 2. 2 = (iv) Work out the decimal form of. 3 3 3 6 2.33 0.833 3) 7.00 6) 50 -   6 -   48 1 5 10 2 = 2.33...   020 = 0.833... -   9 3 -    18 6 1 · 5 ·   10 ∴ 2 = 2.3 020 ∴ = 0.83 3    -    18 6 -   9    01 02 In all the above examples, the division does not come to an end. Here, a single digit or a group of digits occurs repeatedly on the right of the decimal point. This type of decimal form of a rational number is called the recurring decimal form. If in a decimal fraction, a single digit occurs repeatedly on the right of the decimal 1 · point, we put a point above it as shown here. 2 = 2.33... = 2.3, and if a group of 3 digits occurs repeatedly, we show it with a horizontal line above the digits. 2 5 · Thus, 11 = 0.1818........= 0.18 and = 0.83 6 Now I know ! So