Question 1. What is the least number of planes that can enclose a solid ? What is the name of the solid ? Solution: The least number of planes that can enclose a solid is called a Tetrahedron.
Question 2. Can a polyhedron have for its faces : (i) three triangles ? (ii) four triangles ? (iii) a square and four triangles ? Solution: (i) No, polyhedron has three faces. (ii) Yes, tetrahedron has four triangles as its faces. (iii) Yes, a square pyramid has a square as its base and four triangles as its faces.
Question 3. Is it possible to have a polyhedron with any given number of faces ? Solution: Yes, it is possible if the number of faces is 4 or more.
Question 4. Is a square prism same as a cube ? Solution: Yes, a square prism is a cube.
Question 5. Can a polyhedron have 10 faces, 20 edges and 15 vertices ? Solution: No, it is not possible as By Euler’s formula F + V = E + 2 ⇒ 10 + 15 = 20 + 2 ⇒ 25 = 22 Which is not possible
Question 6. Verify Euler’s formula for each of the following polyhedrons : Solution: (i) In this polyhedron, Number of faces (F) = 7 Number of edges (E) = 15 Number of vertices (V) = 10 According to Euler’s formula, F + V = E + 2 ⇒ 7 + 10 = 15 + 2 ⇒ 17 = 17 Which is true. (ii) In this polyhedron, Number of faces (F) = 9 Number of edges (E) = 16 Number of vertices (V) = 9 According to Euler’s formula, F + V = E + 2 ⇒ 9 + 9 = 16 + 2 ⇒ 18 = 18 Which is true. (iii) In this polyhedron, Number of faces (F) = 9 Number of edges (E) =18 Number of vertices (V) = 11 According to Euler’s formula, F + V = E + 2 ⇒ 9 + 11 = 18 + 2 ⇒ 20 = 20 Which is true. (iv) In this polyhedron, Number of faces (F) = 5 Number of edges (E) = 8 Number of vertices (V) = 5 According to Euler’s formula, F + V = E + 2 ⇒ 5 + 5 = 8 + 2 ⇒ 10 = 10 Which is true. (v) In the given polyhedron, Number of faces (F) = 9 Number of edges (E) = 16 Number of vertices (V) = 9 According to Euler’s formula, F + V = E + 2 ⇒ 9 + 9 = 16 + 2 ⇒ 18 = 18 Which is true.
Question 7. Using Euler’s formula, find the unknown: Solution: We know that Euler’s formula is F + V = E + 2 (i) F + 6 = 12 + 2 ⇒ F + 6 = 14 ⇒ F = 14 – 6 = 8 Faces = 8 (ii) F + V = E + 2 ⇒ 5 + V = 9 + 2 ⇒ 5 + V = 11 ⇒ V = 11 – 5 = 6 Vertices = 6 (iii) F + V = E + 2 ⇒ 20 + 12 = E + 2 ⇒ 32 = E + 2 ⇒ E = 32 – 2 = 30 Edges = 30
Exercise 19.2
Question 1. Which among the following are nets for a cube ? Solution: Nets for a cube are (ii), (iv) and (vi)
Question 2. Name the polyhedron that can be made by folding each net: Solution: (i) This net is for a square (ii) This net is for triangular prism. (iii) This net is for triangular prism. (iv) This net is for hexagonal prism. (v) This net is for hexagon pyramid. (vi) This net is for cuboid.
Question 3. Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ? Solution: Figure (i) shows the net of cube or dice.
Question 4. Draw nets for each of the following polyhedrons: Solution: (i) Net for cube is given below : (ii) Net of a triangular prism is as under : (iii) Net of hexagonal prism is as under : (iv) The net for pentagonal pyramid is as under:
Question 5. Match the following figures: Solution: (a) (iv) (b) (i) (c) (ii) (d) (iii)
Question 1. Define the following terms: (i) Quadrilateral (ii) Convex Quadrilateral. Solution: (i) Quadrilateral: A closed figure made of four line segments is called a quadrilateral such that: (a) no three points of them are collinear (b) the line segments do not intersect except at their ends points. (ii) Convex quadrilateral: A quadrilateral is called a convex quadrilateral of the line containing any side of the quadrilateral has the remaining vertices on the same side of it. In the figure, quadrilateral ABCD is a convex quadrilateral.
Question 2. In a quadrilateral, define each of the following: (i) Sides (ii) Vertices (iii) Angles (iv) Diagonals (v) Adjacent angles (vi) Adjacent sides (vii) Opposite sides (viii) Opposite angles (ix) Interior (x) Exterior Solution: (i) Sides: In a quadrilateral ABCD, form line segments AB, BC, CD and DA are called sides of the quadrilateral. (ii) Vertices : The ends points are called the vertices of the quadrilateral. Here in the figure, A, B, C and D are its vertices. (iii) Angles: A quadrilateral has four angles which are at their vertices. In the figure, ∠A, ∠B, ∠C and ∠D are its angles. (iv) Diagonals: The line segment joining the opposite vertices is called diagonal. A quadrilateral has two diagonals. (v) Adjacent Angles : The angles having a common arm (side) are called adjacent angles. (vi) Adjacent sides : If two sides of a quadrilateral have a common end-point, these are called adjacent sides. (vii) Opposite sides: If two sides do not have a common end-point of a quadrilateral, they are called opposite sides. (viii) Opposite angles : The angles which are not adjacent are called opposite angles. (ix) Interior: The region which is surrounded by the sides of the quadrilateral is called its interior. (x) Exterior : The part of the plane made up by all points as the not enclosed by the quadrilateral, is called its exterior.
Question 3. Complete each of the following, so as to make a true statement: (i) A quadrilateral has ………… sides. (ii) A quadrilateral has ………… angles. (iii) A quadrilateral has ……….. vertices, no three of which are ………… (iv) A quadrilateral has …………. diagonals. (v) The number of pairs of adjacent angles of a quadrilateral is …………. (vi) The number of pairs of opposite angles of a quadrilateral is …………… (vii) The sum of the angles of a quadrilateral is ………… (viii) A diagonal of a quadrilateral is a line segment that joins two ………. vertices of the quadrilateral. (ix) The sum of the angles of a quadrilateral is …………. right angles. (x) The measure of each angle of a convex quadrilateral is …………. 180°. (xi) In a quadrilateral the point of intersection of the diagonals lies in ………….. of the quadrilateral. (xii) A point is in the interior of a convex quadrilateral, if it is in the ……….. of its two opposite angles. (xiii) A quadrilateral is convex if for each side, the remaining …………. lie on the same side of the line containing the side. Solution: (i) A quadrilateral has four sides. (a) A quadrilateral has four angles. (iii) A quadrilateral has four vertices, no three of which are collinear . (iv) A quadrilateral has two diagonals. (v) The number of pairs of adjacent angles of a quadrilateral is four . (vi) The number of pairs of opposite angles ot a quadrilateral is two. (vii) The sum of the angles of a quadrilateral is 360°. (viii) A diagonal of a quadrilateral is a line segment that join two opposite vertices of the quadrilateral. (ix) The sum of the angles of a quadrilateral is 4 right angles. (x) The measure of each angle of a convex quadrilateral is less than 180°. (xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral. (xii) A point is in the interior of a convex quadrilateral, if it is in the interior of its two opposite angles. (xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.
Question 4. In the figure, ABCD is a quadrilateral. (i) Name a pair of adjacent sides. (ii) Name a pair of opposite sides. (iii) How many pairs of adjacent sides are there? (iv) How many pairs of Opposite sides are there ? (v) Name a pair of adjacent angles. (vi) Name a pair of opposite angles. (vii) How many pairs of adjacent angles are there ? (viii) How many pairs of opposite angles are there ? Solution: In the figure, ABCD is a quadrilateral (i) Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB. (ii) Pairs of opposite sides are AB and CD; BC and AD. (iii) There are four pairs of adjacent sides. (iv) There are two pairs of opposite sides. (v) Pairs of adjacent angles are ∠A, ∠B; ∠B, ∠C; ∠C, ∠D; ∠D, ∠A. (vi) Pairs of opposite angles are ∠A and ∠C; ∠B and ∠D. (vii) There are four pairs of adjacent angles. (viii) There are two pairs of opposite angles.
Question 5. The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x. Solution: Sum of four angles of quadrilateral is 360° 110° + 12° + 55° + x° = 360° ⇒ 237° + x° = 360° ⇒ x° = 360° – 237° = 123° x = 123°
Question 6. The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle. Solution: The sum of four angles of a quadrilateral = 360° Three angles are 110°, 50° and 40° Let fourth angle = x Then 110° + 50° + 40° + x° = 360° ⇒ 200° + x° = 360° ⇒ x = 360° – 200° = 160° x = 160°
Question 7. A quadrilateral has three acute angles each measures 80°. What is the measure of fourth angle ? Solution: Sum of four angles of a quadrilateral = 360° Sum of three angles having each angle equal to 80° = 80° x 3 = 240° Let fourth angle = x Then 240° + x = 360° ⇒ x° = 360° – 240° ⇒ x° = 120° Fourth angle = 120°
Question 8. A quadrilateral has all its four angles of the same measure. What is the measure of each ? Solution: Let each equal angle of a quadrilateral = x 4x° = 360° ⇒ x° = 3604 = 90° Each angle will be = 90°
Question 9. Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles ? Solution: Measures of two angles each = 65° Sum of these two angles = 2 x 65°= 130° But sum of four angles of a quadrilateral = 360° Sum of the remaining two angles = 360° – 130° = 230° But these are equal to each other Measure of each angle = 2302 = 115°
Question 10. Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles ? Solution: Sum of four angles of a quadrilateral = 360° One angle = 150° Sum of remaining three angles = 360° – 150° = 210° But these three angles are equal Measure of each angle = 2103 = 70°
Question 11. The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles. Solution: Sum of four angles of a quadrilateral = 360° and ratio in angles = 3 : 5 : 7 : 9 Let first angles = 2x Then second angle = 5x third angle = 7x and fourth angle = 9x 3x + 5x + 7x + 9x = 360° ⇒ 24x = 369° ⇒ x = 36024 = 15° First angle = 3x = 3 x 15° = 45° second angle = 5x = 5 x 15° = 75° third angle = 7x = 7 x 15° = 105° and fourth angle = 9x = 9 x 15° = 135°
Question 12. If the sum of the two angles of a quadrilateral is 180°, what is the sum of the remaining two angles ? Solution: Sum of four angles of a quadrilateral = 360° and sum of two angle out of these = 180° Sum of other two angles will be = 360° – 180° = 180°
Question 13. In the figure, find the measure of ∠MPN. Solution: In the figure, OMPN is a quadrilateral in which ∠O = 45°, ∠M = ∠N = 90° (PM ⊥ OA and PN ⊥ OB) Let ∠MPN = x° ∠O + ∠M + ∠N + ∠MPN = 360° (Sum of angles of a quadrilateral) ⇒ 45° + 90° + 90° + x° = 360° ⇒ 225° + x° = 360° ⇒ x° = 360° – 225° ⇒x = 135° ∠MPN = 135°
Question 14. The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles ? Solution: The sides of a quadrilateral ABCD are produced in order, forming exterior angles ∠1, ∠2, ∠3 and ∠4. Now ∠DAB + ∠1 = 180° (Linear pair) ……(i) Similarly, ∠ABC + ∠2 = 180° ∠BCD + ∠3 = 180° and ∠CDA + ∠4 = 180° Adding, we get ∠DAB + ∠1 + ∠ABC + ∠2 + ∠BCD + ∠3 + ∠CDA + ∠4 = 180° + 180° + 180° + 180° = 720° ⇒ ∠DAB + ∠ABC + ∠CDA + ∠ADC + ∠1 + ∠2 + ∠3 + ∠4 = 720° But ∠DAB + ∠ABC + ∠CDA + ∠ADB = 360° (Sum of angles of a quadrilateral) 360° + ∠1 + ∠2 + ∠3 + ∠4 = 720° ⇒ ∠l + ∠2 + ∠3 + ∠4 = 720° – 360° = 360° Sum of exterior angles = 360°
Question 15. In the figure, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB. Solution: In quadrilateral ABCD, ∠D = 50°, ∠C = 100° PA and PB are the bisectors of ∠A and ∠B. In quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral) ⇒ ∠A + ∠B + 100° + 50° = 360° ⇒ ∠A + ∠B + 150° = 360°’ ⇒ ∠A + ∠B = 360° – 150° = 210° and 12 ∠A + 12 ∠B = 2102 = 105° (PA and PB are bisector of ∠A and ∠B respectively) ∠PAB + ∠PBA = 105° ⇒ ∠PAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle) ⇒ 105° + ∠APB = 180° ⇒ ∠APB = 180° – 105° = 75° ∠APB = 75°
Question 16. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral. Solution: Sum of angles A, B, C and D of a quadrilateral = 360° i.e. ∠A + ∠B + ∠C + ∠D = 360° But ∠A = ∠B = ∠C = ∠D = 1 : 2 : 4 : 5 Let ∠A = x, Then ∠B = 2x ∠C = 4x ∠D = 5x x + 2x + 4x + 5x = 360° ⇒ 12x = 360° ⇒ x = 36012 = 30° ∠A = x = 30° ∠B = 2x = 2 x 30° = 60° ∠C = 4x = 4 x 30° = 120° ∠D = 5A = 5 x 30° = 150°
Question 17. In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 12 (∠A + ∠B). Solution: In quadrilateral ABCD,
Question 18. Find the number of sides of a regular polygon when each of its angles has a measures of (i) 160° (ii) 135° (iii) 175° (iv) 162° (v) 150°. Solution: In a n-sided regular polygon, each angle
Question 19. Find the number of degrees in each exterior angle of a regular pentagon. Solution: In a pentagon or a polygon, sum of exterior angles formed by producing the sides in order, is four right angles or 360° Each exterior angle = 3605 = 72°
Question 20. The measure of angles of a hexagon are x°, (x – 5)° (x – 5)°, (2x – 5)°, (2x – 5)°, (2x + 20)°. Find the value of x. Solution: We know that the sum of interior angels of a hexagon = 720° (180° x 4) ⇒ x + x – 5 + x – 5 + 2x – 5 + 2x – 5 + 2x + 20 = 720° ⇒ 9x – 20 + 20 = 720 ⇒ 9x = 720 ⇒ x = 7209 = 80° x = 80°
Question 21. In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order. Solution: In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f ∠a + ∠b + ∠c + ∠d + ∠e + ∠f = 4 right angles (By definition) By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed In ∆ABC, ∠1 + ∠2 + ∠3 = 180° = 2 right angle (Sum of angles of a triangle) …… (i) Similarly, In ∆ACD, ∠4 +∠5 + ∠6 = 180° = 2 right angles In ∆ADE, ∠1 + ∠8 + ∠9 = 2 right angles …(iii) In ∆AEF, ∠10 + ∠11 + ∠12 = 2 right angles …(iv) Joining (i), (ii), (iii) and (iv) ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 8 right angles ⇒ ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 + ∠11 + ∠12 + ∠1 + ∠4 + ∠7 + ∠10 = 8 right angles ⇒ ∠B + ∠C + ∠D + ∠E +∠F + ∠A = 8 right angles ⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 2 (∠a + ∠b + ∠c + ∠d + ∠e + ∠f) Sum of all interior angles = 2(the sum of exterior angles) Hence proved.
Question 22. The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon. Solution: Let number of sides of a regular polygon = n Each interior angle = 2n–4n right angles Sum of all interior angles = 2n–4n x n right angles = (2n – 4) right angles But sum of exterior angles = 4 right angles According to the condition, (2n – 4) = 3 x 4 (in right angles) ⇒ 2n – 4 = 12 ⇒ 2n = 12 + 4 = 16 ⇒ n = 8 Number of sides of the polygon = 8
Question 23. Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5. Solution: Ratio in exterior angle and interior angles of a regular polygon = 1 : 5 But sum of interior and exterior angles = 180° (Linear pair) By cross multiplication: 6n – 12 = 5n ⇒ 6n – 5n = 12 ⇒ n = 12 Number of sides of polygon is 12
Question 24. PQRSTU is a regular hexagon. Determine each angle of ∆PQT. Solution: In regular hexagon, PQRSTU, diagonals PT and QT are joined. In ∆PUT, PU = UT ∠UPT = ∠UTP But ∠UPT + ∠UTP = 180° – ∠U = 180° – 120° = 60° ∠UPT = ∠UTP = 30° ∠TPQ = 120° – 30° = 90° (QT is diagonal which bisect ∠Q and ∠T) ∠PQT = 1202 = 60° Now in ∆PQT, ∠TPQ + ∠PQT + ∠PTQ = 180° (Sum of angles of a triangle) ⇒ 90° + 60° + ∠PTQ = 180° ⇒ 150° + ∠PTQ = 180° ⇒ ∠PTQ = 180° – 150° = 30° Hence in ∆PQT, ∠P = 90°, ∠Q = 60° and ∠T = 30°
Question 1. Draw rough diagrams to illustrate the following: (i) Open curve (ii) Closed curve Solution:
Question 2. Classify the following curves as open or closed. Solution: Open curves : (i), (iv) and (v) are open curves. (ii) , (iii), and (vi) are closed curves.
Question 3. Draw a polygon and shade its interior. Also draw its diagonals, if any. Solution: In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.
Question 4. Illustrate, if possible, each one of the following with a rough diagram: (i) A closed curve that is not a polygon. (ii) An open curve made up entirely of line segments. (iii) A polygon with two sides. Solution: (i) Close curve but not a polygon. (ii) An open curve made up entirely of line segments. (iii) A polygon with two sides. It is not possible. At least three sides are necessary
Question 5. Following are some figures : Classify each of these figures on the basis of the following: (i) Simple curve (ii) Simple closed curve (iii) Polygon (iv) Convex polygon (v) Concave polygon (vi) Not a curve Solution: (i) It is a simple closed curve and a concave polygon. (ii) It is a simple closed curve and convex polygon. (iii) It is neither a curve nor polygon. (iv) it is neither a curve not a polygon. (v) It is a simple closed curve but not a polygon. (vi) It is a simple closed curve but not a polygon. (vii) It is a simple closed curve but not a polygon. (viii) It is a simple closed curve but not a polygon.
Question 6. How many diagonals does each of the following have ? (i) A convex quadrilateral (ii) A regular hexagon (iii) A triangle. Solution: (i) A convex quadrilateral Here n = 4
Question 7. What is a regular polygon ? State the name of a regular polygon of: (i) 3 sides (ii) 4 sides (iii) 6 sides. Solution: A regular polygon is a polygon which has all its sides equal and so all angles are equal, (i) 3 sides : It is an equilateral triangle. (ii) 4 sides : It is a square. (iii) 6 sides : It is a hexagon.
Question 1. Write each of the following as percent: Solution— (i) 725 (ii) 16625 (iii) 58 (iv) 0.8 (v) 0.005 (vi) 3 : 25 (vii) 11 : 80 (viii) 111 : 125 (ix) 13 : 75 (x) 15 : 16 (xi) 0.18 (xii) 7125 Solution:
Question 2. Convert the following percentages to fractions and ratios : (i) 25% (ii) 2.5% (iii) 0.25% (iv) 0.3% (v) 125% Solution:
Question 3. Express the following as decimal fractions : (i) 27% (ii) 6.3% (iii) 32% (iv) 0.25% (v) 7.5% (vi) 18 % Solution:
Exercise 12.2
Question 1. Find : (i) 22% of 120 (ii) 25% of Rs. 1000 (iii) 25% of 10 kg (iv) 16.5% of 5000 metres (v) 135% of 80 cm (vi) 2.5% of 10000 ml Solution:
Question 2. Find the number ‘a’, if (i) 8.4% of a is 42 (ii) 0.5% of a is 3 (iii) 12 % of a is 50 (iv) 100% of a is 100 Solution:
Question 3. x is 5% of y, y is 24% of z. If x = 480, find the values of y and z. Solution: x = 5% of y, y = 24% of z. x = 480
Question 4. A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income. Solution: Let his monthly income = Rs. x 15% of x = Rs. 150
Question 5. Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination. Solution: Let total marks of the examination = x 86.875% of x = 695 => 86.875 x 1100 x x = 695
Question 6. Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ? Solution: Let the school opened for = x days = 90% of x = 216
Question 7. A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees. Solution: Number of total trees = 2000 Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400 Number of orange trees = 1400
Question 8. Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take. Solution: Balance diet contains Protein = 12% Fats = 25% Carbohydrates = 63% Number of total calories = 2600 Number of calories of proteins = 12% of 2600 = 12100 x 2600 = 312 Number of calories of fats = 25% of 2600 = 25100 x 2600 = 650 Number of calories of carbohydrates = 63% of 2600 = 63100 x 2600 = 1638
Question 9. A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in : (i) Sixes (ii) Fours (iii) Twos (iv) Singles Solution: Total score of a cricketer = 62 runs (z) Number of sixes = 3 Run from 3 sixes = 3 x 6 = 18 Percentage = 1862 x 100 = 29.03% (ii) Number of fours = 8 Total run from 8 fours = 4 x 8 = 32 Percentage = 3262 x 100 = 51.61% (iii) Number of twos = 2 Total score from 2 twos = 2 x 2 = 4 Percentage = 462 x 100 = 40062 = 6.45% (iv) Number of single run = 8 Percentage = 862 x 100 = 80062 = 12.9%
Question 10. A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in : (i) 6’s (ii) 4’s (iii) 2’s (iv) singles What % of his shots were scoring ones ? Solution: Total runs scored by a cricketer =120 (i) Number of runs from sixes (6’s) = 20% of 120
Question 11. Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ? Solution: Total earning from investment = Rs. 187 Percent earning = 22% Let his investment = x Then 22% of x = Rs. 187
Question 12. Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ? Solution: Deposit in the bank = Rs. 1440 Percentage = 12% of his total income Let his total income = Rs. x
Question 13. Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ? Solution: (i) In gunpowder, Nitre = 75% Sulphur = 10% Let total amount of gunpowder = x kg Nitre = 9 kg
Question 14. An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ? Solution: In an alloy, Number of parts of tin = 15 and number of parts of copper = 105 Total parts = 15 + 105 = 120 Percentage of copper in the alloy = 105120 x 100 = 87.5%
Question 15. An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy. Solution: In an alloy, Copper = 32% Nickel = 40% Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28% Mass of zinc in 1 kg = 28% of 1 kg = 28100 x 100 gm = 280 gm.
Question 16. A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ? Solution: Distance travelled before first stop = 122 km Let total journey = x km 10% of x = 122
Question 17. A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ? Solution: Total numbers of teachers = 30 Number of male teachers = 12 Number of female teacher = 30 – 12 = 18 Percentage of female teachers = 18×10030 = 60%
Question 18. Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ? Solution: Let Anil’s income = Rs. 100 Then Aman’s income = Rs, 100 – 20 = Rs. 80 Now, difference of both’s incomes = 100 – 80 = Rs. 20 Anil income is Rs. 20 more than that of Aman’s Percentage = 20×10080 = 25%
Question 19. The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ? Solution: Present value of machine = Rs. 100000 Rate of depreciation per year = 5% Period = 2 years Value of machine after 2 years
Question 20. The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ? Solution: Present population of the town = 60000 Increase annually = 10% Period = 2 years Population after 2 years will be
Question 21. The population of a town increases 10% annually. If the present population is 22000, find its population a year ago. Solution: Let the population of the town a year ago was = x Increase in population = 10%
Question 22. Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ? Solution: Let the salary of Ankit before increment = x Increment given = 10% of the salary Salary after increment will be
Question 23. In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ? Solution: Let price of petrol before budged = Rs. 100 Increase = 10% Price after budget = Rs. 100 + 10 = Rs. 110 Let the consumption of petrol before budget = 100 l Price pf 100 l = Rs. 110 Now of new price is Rs. 110, consumption = 100 l are of new price will be 100, then
Question 24. Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ? Solution: Mohan’s income = Rs. 15500 We see that the savings is same There is no change in savings.
Question 25. Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ? Solution: Let Shalu’s income = Rs. 100 Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160 Now difference in their incomes = Rs. 160 – 100 = Rs. 60 Shalu’s income is less than Shikha’s income by Rs. 60 Percentage less = 60×100160 = 752 % = 37.5%
Question 26. Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ? Solution: Let the third person gets = Rs. x Then second person will get
Question 27. After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ? Solution: Let the original price of the vase = Rs. x Hike in price = 20% Original price of the vase = Rs. 1666.66
Question 1. Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ? Solution: Rakesh can do it in 20 days = 1 his 1 day’s work = 120 and his 4 days work = 120 x 4 = 15 th work
Question 2. Rohan can paint 13 of a painting in 6 days. How many days will he take to complete the painting ? Solution: Rohan can paint 13 of painting in = 6 days he will complete the painting in = 6×31 = 18 days
Question 3. Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ? Solution: Anil’s 1 day’s work = 15
Question 4. Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ? Solution:
Question 5. Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together? Solution: Sita can do a work in 1 hour = 19
Question 6. A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ? Solution:
Question 7. A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ? Solution:
Question 8. A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ? Solution:
Question 9. A, B and C can reap a field in 1534 days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ? Solution:
Question 10. A and B can polish the floors of a building in 10 days A alone can do 14 th of it in 12 days. In how many days can B alone polish the floor ? Solution:
Question 11. A and B can finish a work in 20 days. A alone can do 15 th of the work in 12 days. In how many days can B alone do it ? Solution:
Question 12. A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ? Solution:
Question 13. A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ? Solution:
Question 14. Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ? Solution:
Question 15. A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work. Solution:
Question 16. 6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ? Solution: 6 men can complete the work in = 7 days 1 man will complete the same work in = 7 x 6 days (Less men, more days) 21 men will finish the work in = 7×621 days (More men, less days) = 2 days
Question 17. 8 men can do a piece of work in 9 days. In how many days will 6 men do it ? Solution: 8 men can do a work in = 9 days 1 men will do the work in = 9 x 8 days (Less men, more days) 6 men will do the work in = 9×86 days (More men, less days) = 726 = 12 days
Question 18. Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets? Solution: Reema can weave 35 baskets in = 25 days
Question 19. Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ? Solution: Neha types pages in 14 hours = 75 pages
Question 20. If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ? Solution: 12 boys in 7 days earn an amount of = Rs. 840
Question 21. If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ? Solution: 25 men can earn in 10 days = Rs. 1000
Question 22. Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ? Solution: Ashu can copy a book in 18 days working in a day = 8 hours He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day) He will copy the book in 12 days working in a day = 8×1812 hours (More days, less hours a day) = 14412 = 12 hours a day
Question 23. If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour. Solution: 135 garlands in 3 hours are prepared by = 9 girls 1 garland in 3 hours will be prepared by
Question 24. A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ? Solution: First tap’s 1 hour work to fill the cistern = 18
Question 25. Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ? Solution:
Question 26. A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak? Solution:
Question 27. A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ? Solution:
Question 28. A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ? Solution:
Question 1. Explain the concept of direct variation. Solution: If two quantifies a and b vary with each other in such a way that the ratio ab remains constant and is positive, then we say that a and b vary directly with each other or a and b are in direct variation.
Question 2. Which of the following quantities vary directly with each other ? (i) Number of articles (x) and their price (y). (ii) Weight of articles (x) and their cost (y). (iii) Distance x and time y, speed remaining the same. (iv) Wages (y) and number of hours (x) of work. (v) Speed (x) and time (y) (distance covered remaining the same). (vi) Area of a land (x) and its cost (y). Solution: (i) It is direct variation because more articles more price and less articles, less price. (ii) It is direct variation because, more weight more price, less weight, less price. (iii) It is not direct variation. The distance and time vqry indirectly or inversely. (iv) It is direct variation as more hours, more wages, less hours, less wages. (v) It is not direct variation, as more speed, less time, less speed, more time. (vi) It is direct variation, as more area more cost, less area, less cost. Hence (i), (ii), (iv) and (vi) are in direct variation.
Question 3. In which of the following tables x and y vary directly ? Solution: All are different. It is not in direct variation. Hence (i) and (ii) are in direct variation.
Question 4. Fill in the blanks in each of the following so as to make the statement true : (i) Two quantities are said to vary ……….. with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same. (ii) x and y are said to vary directly with each other if for some positive number k = k. (iii) If u = 3v, then u and v vary ……….. with each other. Solution: (i) Two quantities are said to vary directly with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same. (ii) x and y are said to vary directly with each other if for some positive number k, xy = k. (iii) If u = 3v, then u and v vary directly with each other.
Question 5. Complete the following tables given that x varies directly as y. Solution:
Question 6. Find the constant of variation from the table given below : Solution:
Set up a table and solve the following problems. Use unitary method to verify the answer. Question 7. Rohit bought 12 registers for Rs. 156, find the cost of 7 such registers. Solution: Price of 12 registers = Rs. 156 Let cost of 7 registers = Rs. x. Therefore
Question 8. Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes. Solution: For walking 100 m, time is taken = 125 minutes Let in 315 minutes, distance covered = m Therefore,
Question 9. If the cost of 93 m of a certain kind of plastic sheet is Rs. 1395, then what would it cost to buy 105 m of such plastic sheet. Solution: Cost of 93 m of plastic sheet = Rs. 1395 Let cost of 105 m of such sheet = Rs. x Therefore,
Question 10. Suneeta types 1080 words in one hour. What is GWAM (gross words a minute rate) ? Solution: 1080 words were typed in = 1 hour = 60 minutes Let x words will be typed in 1 minute Therefore,
Question 11. A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes. Solution: Speed of car = 50 km/hr = 50 km in 60 minutes Let it travel x km in 12 minutes. Therefore
Question 12. 68 boxes of a certain commodity require a shelf length of 13.6 m. How many boxes of the same commodity would occupy a shelf of 20.4 m ? Solution: For 68 boxes of certain commodity is required a shelf length of 13.6 m Let x boxes are require for 20.4 m shelf Then
Question 13. In a library 136 copies of a certain book require a shelf length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres ? Solution: For 136 copies of books require a shelf of length = 3.4 m For 5.1 m shelf, let books be required = x Therefore :
Question 14. The second class railway fare for 240 km of journey is Rs. 15.00. What would be the fare for a journey of 139.2 km ? Solution: Fare of second class for 240 km = Rs. 15.00 Let fare for 139.2 km journey = Rs. x Therefore :
Question 15. If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards. Solution: Thickness of a pile of 12 cardboards = 35 mm. Let the thickness of a pile of 294 cardboards = x mm Therefore :
Question 16. The cost of 97 metre of cloth is Rs. 242.50. What length of this can be purchased for Rs. 302.50 ? Solution: Cost of 97 m of cloth = Rs. 242.50 Let x m can be purchase for Rs. 302.50 Therefore :
Question 17. men can dig 634 metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type in one day ? Solution: 11 men can dig a trench = 634 m long Let x men will dig a trench 27 m long. Therefore,
Question 18. A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, for how many days did he work ? Solution: Payment for 6 day’s work = Rs. 210 Let payment for x day’s work = Rs. 875 Therefore :
Question 19. A worker is paid Rs. 200 for 8 days work. If he works for 20 days, how much will he get ? Solution: Labour for 8 days work = Rs. 200 Let x be the labour for 20 days work, then
Question 20. The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm ? Solution: 150 gm of weight produces an extension = 2.9 cm Let x gm of weight will produce an extension of 17.4 cm Therefore :
Question 21. The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm. Solution: A weight of 250 gm produces an extension of 3.5 cm. Let a weight of 700 gm will produce an extension of x cm. Therefore :
Question 22. In 10 days, the earth picks up 2.6 x 108 pounds of dust from the atmosphere. How much dust will it pick up in 45 days. Solution: In 10 days dust is picked up = 2.6 x 108 pounds Let x pounds of dust is picked up in = 45 days Therefore,
Question 23. In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4.8 x 10s kg of dust ? Solution: Dust of 1.2 x 108 kg is picked up in = 15 days Let the dust of 4.8 x 108 will be picked up in x days Therefore,
Exercise 10.2
Question 1. In which of the following tables x and y vary inversely : Solution: We see that it in 15 x 4 and 3 x 25 are not equal to 36 others are 72 In it x and y do not vary.
Question 2. It x and y vary inversely, fill in the following blanks : Solution:
Question 3. Which of the following quantities vary inversely as each other ? (i) The number of x men hired to construct a wall and the time y taken to finish the job. (ii) The length x of a journey by bus and price y of the ticket. (iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it. Solution: (i) Here x and’y var inversely More men less time and more time less men. (ii) More journey more price, less journey less price x and y do not vary inversely. (iii) More journey more petrol, less journey, less petrol x and y do not vary inversely. In (i) x and y, vary inversely.
Question 4. It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table : Solution:
Question 5. If 36 men can do a piece of work in 25 days, in how many days will 15 men do it ? Solution: Here less men, more days. Let in x days, 15 men can finish the work Therefore.
Question 6. A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men. Solution: Let in x months, the work will be completed by 125 men
Question 7. A work-force of 420 men with contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months? Solution: Let total x men can finish the work in 7 months. Therefore, Total men = 540 Number of men already employed = 420 Extra men required = 540 – 420 = 120
Question 8. 1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days ? Solution: Let x men can finish the stock, then Total men required = 1680 Already men working = 1200 More men required = 1680 – 1200 = 480
Question 9. In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel. How long will these provisions last ? Solution: Number of girls in the beginning = 50 More girls joined = 30 Total number of girls = 50 + 30 = 80 Let the provisions last for x days.
Question 10. A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance ? Solution: Let x km/hr be the speed. Then Speed required = 60 km/hr. Already speed = 48 km/hr Speed to be increase = 60 – 48 = 12 km/hr
Question 11. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort ? Solution: Period = 28 days After 4 day, the remaining period = 28 – 4 = 24 days In the beginning number of soldiers in the fort = 1200 Period for which the food lasted = 32 days Let for x soldier, the food was sufficient, then
Question 12. Three spraying machines working together can finish painting a house in 60 minutes. How long will it take 5 machines of the same capacity to do the same job ? Solution: Let in x minutes, 5 machines can do the work Now
Question 13. A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How new many numbers are there in this group now ? Solution: Let x members can finish the wheat in 18 day. Therefore : 5 member can consume the wheat Number of members already = 3 5 – 3 = 2 more member joined them.
Question 14. 55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days ? Solution: Let number of cows required = x Therefore :
Question 15. 18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required ? Solution: Let x men are required, Therefore,
Question 16. A person has money to buy 25 cycles worth Rs. 500 each. How many cycles he will be able to buy if each cycle is costing Rs. 125 more ? Solution: Price of one cycle = Rs. 500 Number of cycle purchased = 25 New price of the cycle = Rs. 500 + Rs. 125 = Rs. 625 Let number of cycle will be purchase = x
Question 17. Raghu has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine ? Solution: Price of each machine = Rs. 200 Price after given discount of Rs. 50 = Rs. 200 – 50 = Rs. 150 Let machine can be purchase = x Number of machines can be purchased = 100
Question 18. If x and y vary inversely as each other and (i) x = 3 when y = 8, find y when x = 4 (ii) x = 5 when y = 15, find x when y = 12 (iii) x = 30, find y when constant of variation = 900. (iv) y = 35, find x when constant of variation = 7. Solution: x and y vary inversely x x y is constant of variation (i) x = 3, y = 8 Constant = xy = 3 x 8 = 24
Question 1. Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by (i) 11 (ii) 15 Solution: Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9 (i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15 (ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11
Question 2. Without performing actual computations, find the quotient when 94 – 49 is divided by (i) 9 (ii) 5 Solution: Two given numbers are 94 and 49. Whose digits are reversed. (i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5 (ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9
Question 3. If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case. Solution: The given number is 985 The other two numbers by arranging its digits in cyclic order, will be 859, 598 of the form abc¯¯¯¯¯¯¯,bca¯¯¯¯¯¯¯,cba¯¯¯¯¯¯¯ Therefore, If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22 If this sum is divided by 22, then the quotient = 111 and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66
Question 4. Find the quotient when difference of 985 and 958 is divided by 9. Solution: The numbers of three digits are 985 and 958 in which tens and ones digits are reversed, then abc¯¯¯¯¯¯¯−acb¯¯¯¯¯¯¯ = 9 (b – c) 985 – 958 = 9 (8 – 5) = 9 x 3 i. e., it is divisible by 9, then quotient = b-c =8-5=3
Exercise 5.2
Question 1. Given that the number 35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3, where a is a digit, what are the possible volues of a ? Solution: The number 35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3 ∵The sum of its digits will also be divisible by 3 ∴ 3 + 5 + a + b + 4 is divisible by 3 ⇒ 18 + a is divisible by 3 ⇒ a is divisible by 3 (∵ 18 is divisible by 3) ∴ Values of a can be, 0, 3, 6, 9
Question 2. If x is a digit such that the number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3,’ find possible values of x. Solution: ∵ The number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3 ∴ The sum of its digits will also be divisible by 3 ⇒ l + 8+ x + 7 + 1 is divisible by 3 ⇒ 17 + x is divisible by 3 The sum greater than 17, can be 18, 21, 24, 27………… ∴ x can be 1, 4, 7 which are divisible by 3.
Question 3. If is a digit of the number 66784x¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ such that it is divisible by 9, find the possible values of x. Solution: ∵ The number 66784 x is divisible by 9 ∴ The sum of its digits will also be divisible by 9 ⇒ 6+6+7+8+4+x is divisible by 9 ⇒ 31 + x is divisible by 9 Sum greater than 31, are 36, 45, 54……… which are divisible by 9 ∴ Values of x can be 5 on 9 ∴ x = 5
Question 4. Given that the number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9, where y is a digit, what are the possible values of y ? Solution: ∵ The number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9 ∴The sum of its digits will also be divisible by 9 ⇒ 6 + 7+ y+ 1+ 9 is divisible by 9 ⇒ 23 + y is divisible by 9 ∴ The numbers greater than 23 are 27, 36, 45,…….. Which are divisible by 9 ∴y = A
Question 5. If 3×2¯¯¯¯¯¯¯¯ is a multiple of 11, where .v is a digit, what is the value of * ? Solution: ∵ The number 3×2¯¯¯¯¯¯¯¯ is multiple of 11 ∴ It is divisible by 11 ∴ Difference of the sum of its alternate digits is zero or multiple of 11 ∴ Difference of (2 + 3) and * is zero or multiple of 11 ⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0 Then x = 5
Question 6. If 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x. Solution: ∵ The number 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 4 ∴ The number formed by tens digit and units digit will also be divisible by 4 ∴ x2¯¯¯¯¯ is divisible by 4 ∴ Possible number can be 12, 32, 52, 72, 92 ∴ Value of x will be 1,3, 5, 7, 9
Question 7. If x denotes the digit at hundreds place of the number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ such that the number is divisible by 11. Find all possible values of x. Solution: ∵ The number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 11 ∴ The difference of the sums its alternate digits will be 0 or divisible by 11 ∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11 ⇒ 15+x-8 = 0, or multiple of 11, 7 + x = 0 ⇒ x = -7, which is not possible ∴ 7 + x = 11, 7 + x = 22 etc. ⇒ x=11-7 = 4, x = 22 – 7 ⇒ x = 15 which is not a digit ∴ x = 4
Question 8. Find the remainder when 981547 is divided by 5. Do this without doing actual division. Solution: A number is divisible by 5 if its units digit is 0 or 5 But in number 981547, units digit is 7 ∴ Dividing the number by 5, Then remainder will be 7 – 5 = 2
Question 9. Find the remainder when 51439786 is divided by 3. Do this without performing actual division. Solution: In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3. ∴ The sum of digits must by divisible by 3 ∴ Dividing 43 by 3, the remainder will be = 1 Hence remainder = 1
Question 10. Find the remainder, without performing actual division when 798 is divided by 11. Solution: Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6 ∴ Remainder = 6
Question 11. Without performing actual division, find the remainder when 928174653 is divided by 11. Solution: Let n = 928174653 = A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5) = A multiple of 11 + 33 – 12 = A multiple of 11 + 21 = A multiple of 11 + 11 + 10 = A multiple of 11 + 10 ∴ Remainder =10
Question 12. Given an example of a number which is divisible by : (i) 2 but not by 4. (ii) 3 but not by 6. (iii) 4 but not by 8. (iv) both 4 and 8 but not 32. Solution: (i) 2 but not by 4 A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4. ∴ The number can be 222, 342 etc. (ii) 3 but not by 6 A number is divisible by 3 if the sum of its digits is divisible by 3 But a number is divisible by 6, if it is divided by 2 and 3 both ∴ The numbers can be 333, 201 etc. (iii) 4 but not by 8 A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8. ∴ The number can be 244, 1356 etc. (iv) Both 4 and 8 but not by 32 A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also. But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.
Question 13. Which of the following statements are true ? (i) If a number is divisible by 3, it must be divisible by 9. (ii) If a number is divisible by 9, it must be divisible by 3. (iii) If a number is divisible by 4, it must be divisible by 8. (iv) If a number is divisible by 8, it must be divisible by 4. (v) A number is divisible by 18, if it is divisible by both 3 and 6. (vi) If a number is divisible by both 9 and 10, it must be divisible by 90. (vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately. (viii) If a number divides three numbers exactly, it must divide their sum exactly. (ix) If two numbers are co-priirie, at least one of them must be a prime number. (x) The sum of two consecutive odd numbers is always divisible by 4. Solution: (i) False, it is not necessarily that it must divide by 9. (ii) Trae. (iii) False, it is not necessarily that it must divide by 8. (iv) True. (v) False, it must be divisible by 9 and 2 both. (vi) True. (vii) False, it is not necessarily. (viii)True. (ix) False. It is not necessarily. (x) True.
Exercise 5.3
Solve each of the following cryptarithms. Question 1. Solution: Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9 ∴ A = 6 or less. ∴ 7 + B = A = 6 or less ∴ 7 + 9 or 8 = 16 or 15 ∴ But it is two digit number B = 8 Then A = 5
Question 2. Solution: Values of A and B can be between 0 and 9 In tens digit, A + 3 = 9 ∴ A = 9 – 3 = 6 or less than 6 In ones unit B + 7 = A = 6or less ∴ 7 + 9 or 8 = 16 or 15 But it is two digit number ∴ B = 8 and ∴ A = 5
Question 3. Solution: Value of A and B can be between 0 and 9 In units place. 1+B = 0 ⇒1+B = 10 ∴ B = 10 – 1 = 9 and in tens place 1 + A + 1 = B ⇒ A + 2 = 9 ⇒ A = 9 – 2 = 7
Question 4. Solution: Values of A and.B can be between 0 and 9 In units place, B+1 = 8 ⇒ B = 8-1=7 In tens place A + B= 1 or A + B = 11 ⇒ A + 7 = 11 ⇒ A =11-7 = 4
Question 5. Solution: Values of A and B can be between 0 and 9 In tens place, 2 + A = 0 or 2 + A=10 A = 10-2 = 8 In units place, A + B = 9 ⇒ 8 + B = 9 ⇒ B = 9- 8 = 1
Question 6. Solution: Values of A and B can be between 0 and 9 In hundreds place,
Question 7. Show that cryptarithm 4 x AB¯¯¯¯¯¯¯¯=CAB¯¯¯¯¯¯¯¯¯¯¯ does not have any solution. Solution: It means that 4 x B is a numebr whose units digit is B Clearly, there is no such digit Hence the given cryptarithm has no solution.
Question 1. Find the cubes of the following numbers: (i) 7 (ii) 12 (iii) 16 (iv) 21 (v) 40 (vi) 55 (vii) 100 (viii) 302 (ix) 301 Solution: (i) (7)3 = 7 x 7 x 7 = 343 (ii) (12)3 = 12 x-12 x 12 = 1728 (iii) (16)3 = 16 x 16 x 16 = 4096 (iv) (21)3 = 21 x 21 x 21 = 441 x 21 =9261 (v) (40)3 = 40 x 40 x 40 = 64000 (vi) (55)3 = 55 x 55 x 55 = 3025 x 55 = 166375 (vii) (100)3 = 100 x 100 x 100 =1000000 (viii)(302)3 = 302 x 302 x 302 = 91204 x 302 = 27543608 (ix) (301)3 = 301 x 301 x 301 = 90601 x 301 =27270901
Question 2. Write the cubes of all natural numbers between 1 and 10 and verify the following statements : (i) Cubes of all odd natural numbers are odd. (ii) Cubes of all even natural numbers are even. Solution: Cubes of first 10 natural numbers : (1)3 = 1 x 1 x 1 = 1 (2)3 = 2 x 2 x 2 = 8 (3)3 = 3 x 3 x 3 = 27 (4)3= 4 x 4 x 4 = 64 (5)3 = 5 x 5 x 5 = 125 (6)3 = 6 x 6 x 6 = 216 (7)3 = 7 x 7 x 7 = 343 (8)3 = 8 x 8 x 8 = 512 (9)3 = 9 x 9 x 9= 729 (10)3 = 10 x 10 x 10= 1000 We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.
Question 3. Observe the following pattern : Write the next three rows and calculate the value of 13 + 23 + 33 +…. + 93 + 103 by the above pattern. Solution: We see the pattern
Question 4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings : The cube of a natural number which is a multiple of 3 is a multiple of 27′ Solution: 5 natural numbers which are multiples of 3 3,6,9,12,15. (3)3 = 3 x 3 x 3 = 27 Which is multiple of 27 (6)3 = 6 x 6 x 6 = 216 ÷ 27 = 8 Which is multiple of 27 (9)3 = 9 x 9 x 9 = 729 + 27 = 27 Which is multiple of 27 (12)3= 12 x 12 x 12 = 1728 ÷ 27 = 64 Which is multiple of 27 (15)3 = 15 x 15 x 15 = 3375 ÷ 27 = 125 Which is multiple of 27 Hence, cube of multiple of 3 is a multiple of 27
Question 5. Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following : ‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’. Solution: 3n + 1 Let n = 1, 2, 3, 4, 5, then If n = 1, then 3n +1= 3 x 1+1= 3+1= 4 If n = 2, then 3n +1=3 x 2+1=6+1=7 If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10 If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13 If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16 Now (4)3 = 4 x 4 x 4 = 64 Which is 643=21, Remainder = 1 (7)3 = 7 x 7 x 7 = 343 Which is 3433 =114, Remainder = 1 (10)3 = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1 (13)3 = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1 (16)3 = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1 Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1
Question 6. Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following : ‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’. Solution: Natural numbers of the form 3n + 2, when n is a natural number i.e. 1, 2, 3, 4, 5,…………. If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5 If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8 If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11 If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14 and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17 Now (5)3 = 5 x 5 x 5 = 125 125 + 3 = 41, Remainder = 2 (8)2 = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2 (11)3 = 11 x 11 x 11 = 1331 1331 + 3 = 443, Remainder = 2 (14)3 = 14 x 14 x 14 = 2744 2744 + 3 = 914, Remainder = 2 (17)3 = 17 x 17 x 17 = 4913 4913 = 3 = 1637, Remainder = 2 We see the cube of the natural number of the form 3n + 2 is also a natural number of the form 3n + 2.
Question 7. Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following : ‘The cube of a multiple of 7 is a multiple of 73′. Solution: 5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35 (7)3 = (7)3 which is multiple of 73 (14)3 = (2 x 7)3 = 23 x 73, which is multiple of 73 (21)3 = (3 x 7)3 = 33 x 73, which is multiple of 73 (28)3 = (4 x 7)3 = 43 x 73, which is multiple of 73 (35)3 = (5 x 7)3 = 53 x 73 which is multiple of 73 Hence proved.
Question 8. Which of the following are perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533 Solution: (i) 64 = 2 x 2 x 2 x 2 x 2 x 2 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 64 is a perfect cube (ii) 216 = 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of equal factors, we see that no factor is left 216 is a perfect cube. (iii) 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, we see that two factors 3 x 3 are left ∴ 243 is not a perfect cube. (iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 1000 is a perfect cube. (v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of the equal factors, we see that no factor is left ∴ 1728 is a perfect cube, (vi) 3087 = 3 x 3 x 7 x 7 x 7 Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left ∴ 3087 is not a perfect cube. (vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left ∴ 4609 is not a perfect cube. (viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left ∴ 106480 is not a perfect cube. (ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 166375 is a perfect cube. (x) 456533 = 7 x 7 x 7 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, we see that no factor is left ∴ 456533 is a perfect cube.
Question 9. Which of the following are cubes of even natural numbers ? 216, 512, 729,1000, 3375, 13824 Solution: We know that the cube of an even natural number is also an even natural number ∴ 216, 512, 1000, 13824 are even natural numbers. ∴ These can be the cubes of even natural number.
Question 10. Which of the following are cubes of odd natural numbers ? 125, 343, 1728, 4096, 32768, 6859 Solution: We know that the cube of an odd natural number is also an odd natural number, ∴ 125, 343, 6859 are the odd natural numbers ∴ These can be the cubes of odd natural numbers.
Question 11. What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ? (i) 675 (ii) 1323 (iii) 2560 (iv) 7803 (v) 107311 (vi) 35721 Solution: (i) 675 = 3 x 3 x 3 x 5 X 5 Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet So, by multiplying by 5, the triplet will be completed. ∴ Least number to be multiplied = 5 (ii) 1323 = 3 x 3 x 3 x 7 x 7 Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left So, multiplying by 7, we get a triplet ∴ The least number to be multiplied = 7 (iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 Grouping the factors in triplet of equal factors, 5 is left. ∴ To complete a triplet 5 x 5 is to multiplied ∴ Least number to be multiplied = 5 x 5 = 25 (iv) 7803 = 3 x 3 x 3 x 17 x 17 Grouping the factors in triplet of equal factors, we find the 17 x 17 are left So, to complete the triplet, we have to multiply by 17 ∴ Least number to be multiplied = 17 (v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11 Grouping the factors in triplet of equal factors, factor 3 is left So, to complete the triplet 3 x 3 is to be multiplied ∴ Least number to be multiplied = 3 x 3 = 9 (vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7 Grouping the factors in triplet of equal factors, we find that 7 x 7 is left So, in order to complete the triplets, we have to multiplied by 7 ∴ Least number to be multiplied = 713&ifi=8&uci=a!8&btvi=7&fsb=1&xpc=6d2Qjo9kVP&p=https%3A//www.learninsta.com&dtd=6298
Question 12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube ? (i) 675 (ii) 8640 (iii) 1600 (iv) 8788 (v) 7803 (vi) 107811 (vii) 35721 (viii) 243000 Solution: (i) 675 = 3 x 3 x 3 x 5 x 5 Grouping the factors in triplet of equal factors, 5 x 5 is left 5 x 5 is to be divided so that the quotient will be a perfect cube. ∴ The least number to be divided = 5 x 5 = 25 (ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets of equal factors, 5 is left ∴ In order to get a perfect cube, 5 is to divided ∴ Least number to be divided = 5 (iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 Grouping the factors in triplets of equal factors, we find that 5 x 5 is left ∴ In order to get a perfect cube 5 x 5 = 25 is to be divided. ∴ Least number to be divide = 25 (iv) 8788 = 2 x 2 x 13 x 13 x 13 Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left ∴ In order to get a perfect cube, 2 x 2 is to be divided ∴ Least number to be divided = 4 (v) 7803 = 3 x 3 x 3 x 17 x 17 Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left. So, in order to get a perfect cube, 17 x 17 is be divided ∴ Least number to be divided = 17 x 17 = 289 (vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11 Grouping the factors in triplets of equal factors, 3 is left ∴ In order to get a perfect cube, 3 is to be divided ∴ Least number to be divided = 3 (vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7 Grouping the factors in triplets of equal factors, we see that 7 x 7 is left So, in order to get a perfect cube, 7 x 7 = 49 is to be divided ∴ Least number to be divided = 49 (viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5 Grouping the factors in triplets of equal factors, 3 x 3 is left ∴ By dividing 3 x 3, we get a perfect cube ∴ Least number to be divided = 3 x 3=9
Question 13. Prove that if a number is trebled then its cube is 27 times the cube of the given number. Solution: Let x be the number, then trebled number of x = 3x Cubing, we get: (3x)3 = (3)3 x3 = 27x3 27x3 is 27 times the cube of x i.e., of x3
Question 14. What happenes to the cube of a number if the number is multiplied by (i) 3 ? (ii) 4 ? (iii) 5 ? Solution: number (x)3 = x3 (i) If x is multiplied by 3, then the cube of ∴ (3x)3 = (3)3 x x3 = 27x3 ∴ The cube of the resulting number is 27 times of cube of the given number (ii) If x is multiplied by 4, then the cube of (4x)3 = (4)3 x x3 = 64x3 ∴ The cube of the resulting number is 64 times of the cube of the given number (ii) If x is multiplied by 5, then the cube of (5x)3 = (5)3 x x3 = 125x3 ∴ The cube of the resulting number is 125 times of the cube of the given number
Question 15. Find the volume of a cube, one face of which has an area of 64 m2. Solution: Area of one face of a cube = 64 m2 ∴ Side (edge) of cube = √64 = √64 = 8 m ∴ Volume of the cube = (side)3 = (8 m)3 = 512 m3
Question 16. Find the volume of a cube whose surface area is 384 m2. Solution: Surface area of a cube = 384 m2 Let side = a Then 6a2 = 384 ⇒ a2 = 3846= 64 = (8)2 ∴ a = 8 m Now volume = a3 = (8)3 m3 = 512 m3
Question 17. Evaluate the following : Solution: Question 18. Write the units digit of the cube of each of the following numbers : 31,109,388,833,4276,5922,77774,44447, 125125125. Solution: We know that if unit digit of a number n is = 1, then units digit of its cube = 1 = 2, then units digit of its cube = 8 = 3, then units digit of its cube = 7 = 4, then units digit of its cube = 4 = 5, then units digit of its cube = 5 = 6, then units digit of its cube = 6 = 7, then the units digit of its cube = 3 = 8, then units digit of its cube = 2 = 9, then units digit of its cube = 9 = 0, then units digit of its cube = 0 Now units digit of the cube of 31 = 1 Units digit of the cube of 109 = 9 Units digits of the cube of 388 = 2 Units digits of the cube of 833 = 7 Units digits of the cube of 4276 = 6 Units digit of the cube of 5922 = 8 Units digit of the cube of 77774 = 4 Units digit of tl. cube of 44447 = 3 Units digit of the cube of 125125125 = 5
Question 19. Find the cubes of the following numbers by column method : (i) 35 (ii) 56 (iii) 72 Solution:
Question 20. Which of the following numbers are not perfect cubes ? (i) 64 (ii) 216 (iii) 243 (iv) 1728 Solution: (i) 64 = 2 x 2 x 2 X 2 x 2 x 2 Grouping the factors in triplets, of equal factors, we see that no factor is left ∴ 64 is a perfect cube. (ii) 216 = 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that no factor is left ∴ 216 is a perfect cube. (iii) 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left ∴ 243 is not a perfect cube. (iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 Grouping the factors m triplets, of equal factors, we see that no factor is left. ∴ 1728 is a perfect cube.
Question 21. For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be (a) multiplied so that the product is a perfect cube. (b) divided so that the quotient is a perfect cube. Solution: In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3 Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left. (a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet. (b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided.
Question 22. By taking three different values of n verify the truth of the following statements : (i) If n is even, then n3 is also even. (ii) If n is odd, then n3 is also odd. (iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3. (iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type. Solution: (i) n is even number. Let n = 2, 4, 6 then (a) n3 = (2)3 = 2 x 2 x 2 = 8, which is an even number. (b) (n)3= (4)3 = 4 x 4 x 4 = 64, which is an even number. (c) (n)3 = (6)3 = 6 x 6 x 6 = 216, which is an even number.
(ii) n is odd number. Letx = 3, 5, 7 (a) (n)3 = (3)3 = 3 x 3 x 3 = 27, which is an odd number. (b) (n)3 = (5)3 = 5 x 5 x 5 = 125, which is an odd number. (c) (n)3 = (7)3 = 7 x 7 x 7 = 343, which is an odd number.
(iii) If n leaves remainder 1 when divided by 3, then n3 is also leaves 1 as remainder, Let n = 4, 7, 10 If n = 4, then «3 = (4)3 = 4 x 4 x 4 = 64 = 64 + 3 = 21, remainder = 1 If n = 7, then n3 = (7)3 = 7 x 7 x 7 = 343 343 + 3 = 114, remainder = 1 If n – 10, then (n)3 = (10)3 = 10 x 10 x 10 = 1000 1000 + 3 = 333, remainder = 1
(iv) If the natural number is of the form 3p + 2, then n3 is also of the same type Let p =’1, 2, 3, then (a) If p = 1, then n = 3p + 2 = 3 x 1+2=3+2=5 ∴ n3 = (5)3 = 5 x 5 x 5 = 125 125 = 3 x 41 + 2 = 3p +2
(b) If p = 2, then n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8 ∴ n3 = (8)3 = 8 x 8 x 8 = 512 ∴ 512 = 3 x 170 + 2 = 3p + 2
(c) If p = 3, then n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11 ∴ n3 = (11)3 = 11 x 11 x 11 = 1331 and 1331 =3 x 443 + 2 = 3p + 2 Hence proved.
Exercise 4.2
Question 1. Find the cubes of: (i) -11 (ii) -12 (iii) – 21 Solution: (i) (-11)3=(-11)3=(11 x 11 x 11) =-1331 (ii) (-12)3=(-12)3=(12 x 12 x 12) = -1728 (iii) (-21)3=(-21)3=(21 x 21 x 21) = -9261
Question 2. Which of the following numbers are cubes of negative integers. (i) -64 (ii) -1056 (iii) -2197 (iv) -2744 (v) -42875 Solution: ∴ All factors of 64 can be grouped in triplets of the equal factors completely. ∴ -64 is a perfect cube of negative integer. All the factors of 1056 can be grouped in triplets of equal factors grouped completely ∴ 1058 is not a perfect cube of negative integer. All the factors of -2197 can be grouped in triplets of equal factors completely ∴ 2197 is a perfect cube of negative integer, All the factors of -2744 can be grouped in triplets of equal factors completely ∴ 2744 is a perfect cube of negative integer All the factors of -42875 can be grouped in triplets of equal factors completely ∴ 42875 is a perfect cube of negative integer.
Question 3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer : (i) -5832 (ii) -2744000 Solution: Grouping the factors in triplets of equal factors, we see that no factor is left ∴ -5832 is a perfect cube Now taking one factor from each triplet we find that -5832 is a cube of – (2 x 3 x 3) = -18 ∴ Cube root of-5832 = -18 Grouping the factors in tuplets of equal factors, we see that no factor is left. Therefore it is a perfect cube. Now taking one factor from each triplet, we find that. -2744000 is a cube of – (2 x 2 x 5 x 7) ie. -140 ∴ Cube root of -2744000 = -140
Question 4. Find the cube of : Solution:
Question 5. Which of the following numbers are cubes of rational numbers : Solution:
Exercise 4.3
Question 1. Find the cube roots of the following numbers by successive subtraction of numbers : 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, (i) 64 (ii) 512 (iii) 1728 Solution: (i) 64 64 – 1 = 63 63 – 7 = 56 56 – 19 = 37 37 – 37 = 0 ∴ 64 = (4)3 ∴ Cube root of 64 = 4
Question 2. Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes : (i) 130 (ii) 345 (iii) 792 (iv) 1331 Solution: (i) 130 130 – 1 = 129 129 -7 = 122 122 -19 = 103 103 -37 = 66 66 – 61 = 5 We see that 5 is left ∴ 130 is not a perfect cube.
(ii) 345 345 – 1 = 344 344 – 7 = 337 337 – 19 = 318 318 – 37 = 281 81 – 61 =220 220 – 91 = 129 129 – 127 = 2 We see that 2 is left ∴ 345 is not a perfect cube.
(iii) 792 792 – 1 = 791 791 – 7 = 784 784 – 19 = 765 765 – 37 = 728 728 – 61 = 667 667 – 91 = 576 576 – 127 = 449 449 – 169 = 280 ∴ We see 280 is left as 280 <217 ∴ 792 is not a perfect cube.
Question 3. Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots ? Solution: We have examined in Question 2, the numbers 130, 345 and 792 are not perfect cubes. Therefore (i) 130 130 – 1 = 129 129 -7= 122 122 -19 = 103 103 – 37 = 66 66 – 61 = 5 Here 5 is left ∴ 5 < 91 5 is to be subtracted to get a perfect cube. Cube root of 130 – 5 = 125 is 5
(ii) 345 345 – 1 = 344 344 -7 = 337 337 – 19 = 318 318 – 37 = 281 281 – 61 =220 220 – 91 = 129 129 – 127 = 2 Here 2 is left ∵ 2 < 169 ∴ Cube root of 345 – 2 = 343 is 7 ∴ 2 is to be subtracted to get a perfect cube.
Question 4. Find the cube root of each of the following natural numbers : (i) 343 (ii) 2744 (iii) 4913 (iv) 1728 (v) 35937 (vi) 17576 (vii) 134217728 (viii) 48228544 (ix) 74088000 (x) 157464 (xi) 1157625 (xii) 33698267 Solution:
Question 5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product. Solution: Grouping the factors in triplets of equal factors, we see that 2, 3 x 3 and 5 x 5 are left ∴ In order to complete the triplets, we have to multiply it by 2, 3 and 5. ∴ The smallest number to be multiplied = 2×2 x 3 x 5 = 60 Now product = 3600 x 60 = 216000 and cube root of 216000
Question 6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product. Solution: Grouping the factors in triplets of equal factors, we see that 41 x 41 is left ∴ In order to complete the triplet, we have to multiply it by 41 ∴ Smallest number to be multiplied = 41 ∴ Product = 210125 x 41 = 8615125 ∴ Cube root of 8615125
Question 7. What is the smallest number by which 8192 must be divided so that quotient is a perfect cube ? Also, find the cube root of the quotient so obtained. Solution: Grouping the factors in triplets of equal factors, we see that 2 is left ∴ Dividing by 2, we get the quotient a perfect cube ∴ Perfect cube = 8192 + 2 = 4096
Question 8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers. Solution: Ratio in numbers =1:2:3 Let first number = x Then second number = 2x and third number = 3x ∴ Sum of cubes of there numbers = (x)3 + (2x)3+(3x)3
Question 9. The volume of a cube is 9261000 m3. Find the side of the cube. Solution:
Exercise 4.4
Question 1. Find the cube roots of each of the following integers : (i) -125 (ii) -5832 (iii) -2744000 (iv) -753571 Solution:
Question 2. Show that : Solution:
Question 3. Find the cube root of each of the following numbers : (i) 8 x 125 (ii) -1728 x 216 (iii) -27 x 2744 (iv) -729 x -15625 Solution:
Question 4. Evaluate : Solution:
Question 5. Find the cube root of each of the following rational numbers. Solution:
Question 6. Find the cube root of each of the following rational numbers : (i) 0.001728 (ii) 0.003375 (iii) 0.001 (iv) 1.331 Solution:
Question 7. Evaluate each of the following : Solution:
Question 8. Show that : Solution:
Question 9. Fill in the Blanks : Solution:
Question 10. The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box. Solution:
Question 11. Three numbers are to one another 2:3: 4. The sum of their cubes is 0.334125. Find the numbers. Solution:
Question 12. Find side of a cube whose volume is Solution:
Question 13. Evaluate : Solution:
Question 14. Find the cube root of the numbers : 2460375,20346417,210644875,57066625 using the fact that (i) 2460375 = 3375 x 729 (i) 20346417 = 9261 x 2197 (iii) 210644875 = 42875 x 4913 (iv) 57066625 = 166375 x 343 Solution:
Question 15. Find the units digit of the cube root of the following numbers ? (i) 226981 (ii) 13824 (iii) 571787 (iv) 175616 Solution: (i) 226981 In it unit digit is 1 ∴The units digit of its cube root will be = 1 (∵ 1 x 1 x 1 = 1) ∴Tens digit of the cube root will be = 6 (ii) 13824 ∵ The units digit of 13824 = 4 (∵ 4 X 4 X 4 = 64) ∴Units digit of the cube root of it = 4 (iii) 571787 ∵ The units digit of 571787 is 7 ∴The units digit of its cube root = 3 (∵ 3 x 3 x 3 = 27) (iv) 175616 ∵ The units digit of 175616 is 6 ∴The units digit of its cube root = 6 (∵ 6 x 6 x 6=216)
Question 16. Find the tens digit of the cube root of each of the numbers in Question No. 15. Solution: (i) In 226981 ∵ Units digit is 1 ∴Units digit of its cube root = 1 We have 226 (Leaving three digits number 981) 63 = 216 and 73 = 343 ∴63 ∠226 ∠ T ∴The ten’s digit of cube root will be 6 (ii) In 13824 Leaving three digits number 824, we have 13 ∵ (2)3 = 8, (3)3 = 27 ∴23 ∠13 ∠3′ ∴Tens digit of cube root will be 2 (iii) In 571787 Leaving three digits number 787, we have 571 83 = 512, 93 = 729 ∴ 83 ∠571 ∠93 Tens digit of the cube root will be = 8 (iv) In 175616 Leaving three digit number 616, we have 175 ∵ 53 = 125, 63 = 216 ∴53 ∠175 ∠63 ∴Tens digit of the cube root will be = 5
Exercise 4.5
Making use of the cube root table, find the cubes root of the following (correct to three decimal places) Question 1. 7 Solution: 7–√3 =1.913 (From the table)
Question 2. 70 Solution: 70−−√3 =4.121 (From the table)
Question 1. Which of the following numbers are perfect squares ? (i)484 (ii) 625 (iii) 576 (iv) 941 (v) 961 (vi) 2500 Solution: Grouping the factors in pairs, we have left no factor unpaired ∴ 484 is a perfect square of 22 ∴ Grouping the factors in pairs, we have left no factor unpaired ∴ 625 is a perfect square of 25. Grouping the factors in pairs, we see that no factor is left unpaired ∴ 576 is a perfect square of 24 (iv) 941 has no prime factors ∴ 941 is not a perfect square. (v) 961 =31 x 31 Grouping the factors in pairs, we see that no factor is left unpaired ∴ 961 is a perfect square of 31 Grouping the factors in pairs, we see that no factor is left impaired ∴ 2500 is a perfect square of 50 .
Question 2. Show that each of the following* numbers is a perfect square. Also find the number whose square is the given number in each case : (i) 1156 (ii) 2025 (iii) 14641 (iv) 4761 Solution: Grouping the factors in pairs, we see that no factor is left unpaired ∴ 1156 is a perfect square of 2 x 17 = 34 Grouping the factors in pairs, we see that no factor is left unpaired 2025 is a perfect square of 3 x 3 x 5 =45 Grouping the factors in pairs, we see that no factor is left unpaired ∴ 14641 is a perfect square of 11×11 = 121 Grouping the factors in pairs, we see that no factor is left unpaired ∴ 4761 is a perfect square of 3 x 23 = 69
Question 3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square. (i) 23805 (ii) 12150 (iii) 7688 Solution: Grouping the factors in pairs of equal factors, we see that 5 is left unpaird ∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square. Requid smallest number = 5 (ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5 Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired ∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square. ∴ Required smallest number = 6 Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired ∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square ∴ Required smallest number = 2
Question 4. Find the smallest number by which the given number must be divided so that the resulting number is a perfect square. (i) 14283 (ii) 1800 (iii) 2904 Solution: Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired Deviding by 3, the quotient will the perfect square. Grouping the factors in pair of equal factors, we see that 2 is left unpaired. ∴ Dividing by 2, the quotient will be the perfect square. Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired ∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.
Question 5. Which of the following numbers are perfect squares ? 11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121 Solution: 11 is not a perfect square as 11 = 1 x 11 12 is not a perfect square as 12 = 2×2 x 3 16 is a perfect square as 16 = 2×2 x 2×2 32 is not a perfect square as 32 = 2×2 x 2×2 x 2 36 is a perfect square as 36 = 2×2 x 3×3 50 is not a perfect square as 50 = 2 x 5×5 64 is a perfect square as 64 = 2×2 x 2×2 x 2×2 79 is not a perfect square as 79 = 1 x 79 81 is a perfect square as 81 = 3×3 x 3×3 111 is not a perfect square as 111 = 3 x 37 121 is a perfect square as 121 = 11 x 11 Hence 16, 36, 64, 81 and 121 are perfect squares.
Question 6. Using prime factorization method, find which of the following numbers are perfect squares ? ∴ 189,225,2048,343,441,2916,11025,3549 Solution: Grouping the factors in pairs, we see that are 3 and 7 are left unpaired ∴ 189 is not a perfect square Grouping the factors in pairs, we see no factor left unpaired ∴ 225 is a perfect square Grouping the factors in pairs, we see no factor left unpaired ∴ 2048 is a perfect square Grouping the factors in pairs, we see that one 7 is left unpaired ∴ 343 is not a perfect square. Grouping the factors in pairs, we see that no factor is left unpaired ∴ 441 is a perfect square. Grouping the factors in pairs, we see that no factor is left unpaired ∴ 2916 is a perfect square. Grouping the factors in pairs, we see that no factor is left unpaired ∴ 11025 is a perfect square. Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired ∴ 3549 is a perfect square.
Question 7. By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number. (i) 8820 (ii) 3675 (iii) 605 (iv) 2880 (v) 4056 (vi) 3468 Solution: Grouping the factors in pairs, we see that 5 is left unpaired ∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be = 2 x 3 x 5 x 7 = 210 Grouping the factors in pairs, we see that 3 is left unpaired ∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be = 3 x 5 x 7 = 105 Grouping the factors in pairs, we see that 5 is left unpaired ∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be = 5 x 11 =55 Grouping the factors in pairs, we see that 5 is left unpaired ∴ Multiplying 2880 by 5, we get the perfect square. Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120
Grouping the factors in pairs, we see that 2 and 3 are left unpaired ∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square. and square root of the product will be = 2 x 2 x 3 x 13 = 156
Grouping the factors in pairs, we see that 3 is left unpaired ∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102
Grouping the factors in pairs, we see that 2 and 3 are left unpaired ∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be = 2 x 2 x 2 x 3 x 3 x 3 = 216
Question 8. By what numbers should each of the following be .divided to get a perfect square in each case ? Also find the number whose square is the new number. (i) 16562 (ii) 3698 (iii) 5103 (iv) 3174 (v) 1575 Solution: Grouping the factors in pairs, we see that 2 is left unpaired ∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91 Grouping the factors in pairs, we see that 2 is left unpaired, ∴ Dividing 3698 by 2, the quotient is a perfect square and square of quotient will be = 43 Grouping the factors in pairs, we see that 7 is left unpaired ∴ Dividing 5103 by 7, we get the quotient a perfect square. and square root of the quotient will be 3 x 3 x 3 = 27 Grouping the factors iq pairs, we see that 2 and 3 are left unpaired ∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23 Grouping the factors in pairs, we find that 7 is left unpaired i ∴ Dividing 1575 by 7, the quotient is a perfect square and square root of the quotient will be = 3 x 5 = 15
Question 9. Find the greatest number of two digits which is a perfect square. Solution: The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100 ∴ 81 is the greatest two digit number which is a perfect square.
Question 10. Find the least number of three digits which is perfect square. Solution: The smallest three digit number =100 We know that 92 = 81, 102 = 100, ll2 = 121 We see that 100 is the least three digit number which is a perfect square.
Question 11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square. Solution: By factorization: Grouping the factors in pairs, we see that 11 is left unpaired ∴ The least number is 11 by which multiplying 4851, we get a perfect square.
Question 12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square. Solution: By factorization, Grouping the factors in pairs, we see that 13 is left unpaired ∴ Dividing 28812 by 3, the quotient will be a perfect square.
Question 13. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number. Solution: By factorization, Grouping the factors in pairs, we see that one 2 is left unpaired. ∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24
Exercise 3.2
Question 1. The following numbers are not perfect squares. Give reason : (i) 1547 (ii) 45743 (iii) 8948 (iv) 333333
Solution: We know that if the units digit is 2, 3, 7 or 8 of a number, then the number is not a perfect square.
(i) ∴ 1547 has 7 as units digit. ∴ It is not a perfect square.
(ii) 45743 has 3 as units digit ∴ It is not a perfect square.
(iii) ∴ 8948 has 8 as units digit ∴ It is not a perfect square.
(iv) ∴ 333333 has 3 as units digits ∴ It is not a perfect square.
Question 2. Show that the following numbers are not perfect squares : (i) 9327 (ii) 4058 (iii) 22453 (iv) 743522
Solution: (i) 9327 ∴ The units digit of 9327 is 7 ∴ This number can’t be a perfect square.
(ii) 4058 ∴ The units digit of 4058 is 8 ∴ This number can’t be a perfect square.
(iii) 22453 ∴ The units digit of 22453 is 3 .∴ This number can’t be a perfect square.
(iv) 743522 ∴ The units digit of 743522 is 2 ∴ This number can’t be a perfect square.
Question 3. The square of which of the following numbers would be an odd number ? (i) 731 (ii) 3456 (iii) 5559 (iv) 42008 Solution: We know that the square of an odd number is odd and of even number is even. Therefore (i) Square of 731 would be odd as it is an odd number. (ii) Square of 3456 should be even as it is an even number. (iii) Square of 5559 would be odd as it is an odd number. (iv) The square of 42008 would be an even number as it is an even number. Therefore suqares of (i) 731 and (ii) 5559 will be odd numbers.
Question 4. What will be the units digit of the squares of the following numbers ? (i) 52 (ii) 977 (iii) 4583 (iv) 78367 (v) 52698 (vi) 99880 (vii) 12796 (viii) 55555 (ix) 53924
Solution:
(i) Square of 52 will be 2704 or (2)2 = 4 ∴ Its units digit is 4.
(ii) Square of 977 will be 954529 or (7)2 = 49 . ∴ Its units digit is 9
(iii) Square of 4583 will be 21003889 or (3)2 = 9 ∴ Its units digit is 9
(iv) IS 78367, square of 7 = 72 = 49 ∴ Its units digit is 9
(v) In 52698, square of 8 = (8)2 = 64 ∴ Its units digit is 4
(vi) In 99880, square of 0 = 02 = 0 ∴ Its units digit is 0
(vii) In 12796, square of 6 = 62 = 36 ∴ Its units digit is 6
(viii) In In 55555, square of 5 = 52 = 25 ∴ Its units digit is 5
(ix) In 53924, square pf 4 = 42 = 16 ∴ Its units digit is 6
Question 5. Observe the following pattern 1 + 3 = 22 1 + 3 + 5 = 32 1+34-5 + 7 = 42 and write the value of 1 + 3 + 5 + 7 + 9 +…………upto n terms. Solution: The given pattern is 1 + 3 = 22 1 + 3 + 5 = 32 1+3 + 5 + 7 = 42 1+3 + 5 + 7 + 9 +……………… upto n terms (number of terms)2 = n2
Question 6. Observe the following pattern : 22 – 12 = 2 + 1 32 – 22 = 3 + 2 42– 32 = 4 + 3 52 – 42 = 5 + 4 Find the value of (i) 1002 – 992 (ii) 1112 – 1092 (iii) 992 – 962 Solution: From the given pattern, 22 – 12 = 2 + 1 32 – 22 = 3 + 2 42 – 32 = 4 + 3 52 – 42 = 5 + 4 Therefore (i) 1002-99° = 100 + 99
Question 7. Which of the following triplets are Pythagorean ? (i) (8, 15, 17) (ii) (18, 80, 82) (iii) (14, 48, 51) (iv) (10, 24, 26) (vi) (16, 63, 65) (vii) (12, 35, 38) Solution: A pythagorean triplet is possible if (greatest number)2 = (sum of the two smaller numbers)
(i) 8, 15, 17 Here, greatest number =17 ∴ (17)2 = 289 and (8)2 + (15)2 = 64 + 225 = 289 ∴ 82 + 152 = 172 ∴ 8, 15, 17 is a pythagorean triplet
(ii) 18, 80, 82 Greatest number = 82 ∴ (82)2 = 6724 and 182 + 802 = 324 + 6400 = 6724 ∴ 182 + 802 = 822 ∴ 18, 80, 82 is a pythagorean
(iii) 14, 48, 51 Greatest number = 51 ∴ (51)2 = 2601 and 142 + 482 = 196 + 2304 = 25 00 ∴ 512≠ 142 + 482 ∴ 14, 48, 51 is not a pythagorean triplet
(iv) 10, 24, 26 Greatest number is 26 ∴ 262 = 676 and 102 + 242 = 100 + 576 = 676 ∴ 262 = 102 + 242 ∴ 10, 24, 26 is a pythagorean triplet
(vi) 16, 63, 65 Greatest number = 65 ∴ 652 = 4225 and 162 + 632 = 256 + 3969 = 4225 ∴ 652 = 162 + 632 ∴ 16, 63, 65 is a pythagorean triplet
(vii) 12, 35, 38 Greatest number = 38 ∴ 382 = 1444 and 122 + 352 = 144 + 1225 = 1369 ∴ 382 ≠122 + 352 ∴ 12, 35, 38 is not a pythagorean triplet.
Question 8. Observe the following pattern Solution: From the given pattern
Question 9. Observe the following pattern and find the values of each of the following : (i) 1 + 2 + 3 + 4 + 5 +….. + 50 (ii) 31 + 32 +… + 50 Solution: From the given pattern,
Question 10. Observe the following pattern and find the values of each of the following : (i) 12 + 22 + 32 + 42 +…………… + 102 (ii) 52 + 62 + 72 + 82 + 92 + 102 + 122 Solution: From the given pattern, Question 11. Which of the following numbers are squares of even numbers ? 121,225,256,324,1296,6561,5476,4489, 373758 Solution: We know that squares of even numbers is also are even number. Therefore numbers 256, 324,1296, 5476 and 373758 have their units digit an even number. ∴ These are the squares of even numbers.
Question 12. By just examining the units digits, can you tell which of the following cannot be whole squares ?
1026
1028
1024
1022
1023
1027
Solution: We know that a perfect square cam at ends with the digit 2, 3, 7, or 8 ∴ By examining the given number, we can say that 1028, 1022, 1023, 1027 can not be perfect squares.
Question 13. Write five numbers for which you cannot decide whether they are squares. Solution: A number which ends with 1,4, 5, 6, 9 or 0 can’t be a perfect square 2036, 4225, 4881, 5764, 3349, 6400
Question 14. Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit. Solution: A number which does not end with 2, 3, 7 or 8 can be a perfect square ∴ The five numbers can be 2024, 3036, 4069, 3021, 4900
Question 15. Write true (T) or false (F) for the following statements. (i) The number of digits in a square number is even. (ii) The square of a prime number is prime. (iii) The sum of two square numbers is a square number. (iv) The difference of two square numbers is a square number. (vi) The product of two square numbers is a square number. (vii) No square number is negative. (viii) There is not square number between 50 and 60. (ix) There are fourteen square number upto 200.
Solution: (i) False : In a square number, there is no condition of even or odd digits. (ii) False : A square of a prime is not a prime. (iii) False : It is not necessarily. (iv) False : It is not necessarily. (vi) True. (vii) True : A square is always positive. (viii) True : As 72 = 49, and 82 = 64. (ix) True : As squares upto 200 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 which are fourteen in numbers..
Exercise 3.3
Question 1. Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication : (i) 25 (ii) 37 (iii) 54 (iv) 71 (v) 96 Solution: (i) (25)2
Question 2. Find the squares of the following numbers using diagonal method : (i) 98 (ii) 273 (iii) 348 (iv) 295 (v) 171 Solution:
Question 3. Find the squares of the following numbers : (i) 127 (ii) 503 (iii) 451 (iv) 862 (v) 265 Solution: (i) (127)2 = (120 + 7)2 {(a + b)2 = a2 + lab + b2} = (120)2 + 2 x 120 x 7 + (7)2 = 14400+ 1680 + 49 = 16129
Question 4. Find the squares of the following numbers (i) 425 (ii) 575 (iii) 405 (iv) 205 (v) 95 (vi) 745 (vii) 512 (viii) 995 Solution: (i) (425)2 Here n = 42 ∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806 ∴ (425)2 = 180625
(ii) (575)2 Here n = 57 ∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306 ∴ (575)2 = 330625
(iii) (405)2 Here n = 40 ∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640 ∴ (405)2 = 164025
(iv) (205)2 Here n = 20 ∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420 ∴ (205)2 = 42025
(v) (95)2 Here n = 9 ∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90 ∴ (95)2 = 9025
(vi) (745)2 Here n = 74 ∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550 ∴ (745)2 = 555025
(vii) (512)2 Here a = 1, b = 2 ∴ (5ab)2 = (250 + ab) x 1000 + (ab)2 ∴ (512)2 = (250 + 12) x 1000 + (12)2 = 262 x 1000 + 144 = 262000 + 144 = 262144
(viii) (995)2 Here n = 99 ∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900 ∴ (995)2 = 990025
Question 5. Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1 (i) 405 (ii) 510 (iii) 1001 (iv) 209 (v) 605 Solution: a + b)2 = a2 + lab + b2
Question 1. Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 Solution: (i) In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9 (ii) In 799356−−−−−−√ ∴ the units digit is 6 ∴ The units digit of the square root can be 4 or 6 (iii) In 7998001−−−−−−−√ ∴ the units digit is 1 ∴ The units digit of the square root can be 1 or 9 (iv) In 657666025 ∴ The unit digit is 5 ∴ The units digit of the square root can be 5
Question 2. Find the square root of each of the following by prime factorization. (i) 441 (ii) 196 (iii) 529 (iv) 1764 (v) 1156 (vi) 4096 (vii) 7056 (viii) 8281 (ix) 11664 (x) 47089 (xi) 24336 (xii) 190969 (xiii) 586756 (xiv) 27225 (xv) 3013696 Solution:
Question 3. Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained. Solution: Factorising 180, 180 = 2 x 2 x 3 x 3 x 5 Grouping the factors in pairs we see that factor 5 is left unpaired. ∴ Multiply 180 by 5, we get the product 180 x 5 = 900 Which is a perfect square and square root of 900 = 2 x 3 x 5 = 30
Question 4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorising 147, 147 = 3 x 7×7 Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired ∴ Multiplying 147 by 3, we get the product 147 x 3 = 441 Which is a perfect square and its square root = 3×7 = 21
Question 5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number. Solution: Factorising 3645 3645 = 3 x 3 3 x 3 x 3 x 3 x 5 Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired ∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27
Question 6. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorsing 1152, 1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired. ∴ Dividing by 2, the quotient 576 is a perfect square . ∴ Square root of 576, it is 24
Question 7. The product of two numbers is 1296. If one number is 16 times the others find the numbers. Solution: Product of two numbers = 1296 Let one number = x Second number = 16x ∴ First number = 9 and second number = 16 x 9 = 144
Question 8. A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents. Solution: Total donation collected = Rs. 202500 Let number of residents = x Then donation given by each resident = Rs. x ∴ Total collection = Rs. x x x
Question 9. A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute? Solution: Total amount collected = Rs. 92.16 = 9216 paise Let the number of members = x Then amount collected by each member = x paise ∴ Number of members = 96 and each member collected = 96 paise
Question 10. A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ? Solution: Total fee collected = Rs. 2304 Let number of students = x Then fee paid by each student = Rs. x ∴ x x x = 2304 => x2 = 2304 ∴ x = 2304−−−−√
Question 11. The area of a square field is 5184 m2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field. Solution: The area of a square field = 5184 m2 Let side of the square = x ∴ side of square= 72 m ∴ Perimeter, of square field = 72 x 4 m = 288 m Perimeter of rectangle = 288 m Let breadth of rectangular field (b) = x Then length (l) = 2x ∴ Perimeter = 2 (l + b) = 2 (2x + x) = 2 x 3x = 6x = 2 (2x + x) = 2 x 3x = 6x ∴ Length of rectangular field = 2x = 2 x 48 = 96 m and breadth = 48 m and area = l x b = 96 x 48 m2 = 4608 m2
Question 12. Find the least square number, exactly divisible by each one of the numbers : (i) 6, 9,15 and 20 (ii) 8,12,15 and 20 Solution: LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180 =2 x 2 x 3 x 3 x 5 We see that after grouping the factors in pairs, 5 is left unpaired ∴ Least perfect square = 180 x 5 = 900 We see that after grouping the factors, factors 2, 3, 5 are left unpaired ∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600
Question 13. Find the square roots of 121 and 169 by the method of repeated subtraction. Solution:
Question 14. Write the prime factorization of the following numbers and hence find their square roots. ^ (i) 7744 (ii) 9604 (iii) 5929 (iv) 7056 Solution: Factorization, we get: (i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11 Grouping the factors in pairs of equal factors, Question 15. The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. Solution: Total amount of donation = 2401 Let number of students in VIII = x ∴ Amount donoted by each student = Rs. x
Question 16. A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement. Solution: Number of students = 6000 Students left out = 71 ∴ Students arranged in a field = 6000 – 71=5929
Exercise 3.5
Question 1. Find the square root of each of the long division method. (I) 12544 (ii) 97344 (iii) 286225 (iv) 390625 (v) 363609 (vi) 974169 (vii) 120409 (viii) 1471369 (ix) 291600 (x) 9653449 (xi) 1745041 (xii) 4008004 (xiii) 20657025 (xiv) 152547201 (jcv) 20421361 (xvi) 62504836 (xvii) 82264900 (xviii) 3226694416 (xix)6407522209 (xx) 3915380329 Solution: <
Question 2. Find the least number which must be subtracted from the following numbers to make them a perfect square : (i) 2361 (ii) 194491 (iii) 26535 (iv) 16160 (v) 4401624 Solution: (i) 2361 Finding the square root of 2361 We get 48 as quotient and remainder = 57 ∴ To make it a perfect square, we have to subtract 57 from 2361 ∴ Least number to be subtracted = 57 (ii) 194491 Finding the square root of 194491 We get 441 as quotient and remainder = 10 ∴ To make it a perfect square, we have to subtract 10 from 194491 ∴ Least number to be subtracted = 10 (iii) 26535 Finding the square root of 26535 We get 162 as quotient and 291 as remainder ∴ To make it a perfect square, we have to subtract 291 from 26535 ∴ Least number to be subtracted = 291 (iv)16160 Finding the square root of 16160 We get 127 as quotient and 31 as remainder ∴ To make it a perfect square, we have to subtract 31 from 16160 ∴ Least number to be subtracted = 31 (v) 4401624 Find the square root of 4401624 We get 2098 as quotient and 20 as remainder ∴ To make it a perfect square, we have to subtract 20 from 4401624 ∴ Least number to be subtracted = 20
Question 3. Find the least number which must be added to the following numbers to make them a perfect square : (i) 5607 (ii) 4931 (iii) 4515600 (iv) 37460 (v) 506900 Solution: (i) 5607 Finding the square root of 5607, we see that 742 = 5607- 131 =5476 and 752 = 5625 ∴ 5476 < 5607 < 5625 ∴ 5625 – 5607 = 18 is to be added to get a perfect square ∴ Least number to be added = 18 (ii) 4931 Finding the square root of 4931, we see that 702= 4900 ∴ 712 = 5041 4900 <4931 <5041 ∴ 5041 – 4931 = 110 is to be added to get a perfect square. ∴ Least number to be added =110 (iii) 4515600 Finding the square root of 4515600, we see that 21242 = 4511376 and 2 1 252 = 45 1 56 25 ∴ 4511376 <4515600 <4515625 ∴ 4515625 – 4515600 = 25 is to be added to get a perfect square. ∴ Least number to be added = 25 (iv) 37460 Finding the square root of 37460 that 1932 = 37249, 1942 = =37636 ∴ 37249 < 37460 < 37636 ∴ 37636 – 37460 = 176 is to be added to get a perfect square. ∴ Least number to be added =176 (v) 506900 Finding the square root of 506900, we see that 7112 = 505521, 7122 = 506944 ∴ 505521 < 506900 < 506944 ∴ 506944 – 506900 = 44 is to be added to get a perfect square. ∴ Least number to be added = 44
Question 4. Find the greatest number of 5 digits which is a perfect square. Solution: Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder ∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits ∴ Greatest square 5-digits perfect square = 99856
Question 5. Find the least number of four digits which is a perfect square. Solution: Least number of 4-digits = 10000 Finding square root of 1000 We see that if we subtract 39 From 1000, we get three digit number ∴ We shall add 124 – 100 = 24 to 1000 to get a perfect square of 4-digit number ∴ 1000 + 24 = 1024 ∴ Least number of 4-digits which is a perfect square = 1024
Question 6. Find the least number of six-digits which is a perfect square. Solution: Least number of 6-digits = 100000 Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits. So we shall add 4389 – 3900 = 489 to 100000 to get a perfect square Past perfect square of six digits= 100000 + 489 =100489
Question 7. Find the greatest number of 4-digits which is a perfect square. Solution: Greatest number of 4-digits = 9999 Finding the square root, we see that 198 has been left as remainder ∴ 4-digit greatest perfect square = 9999 – 198 = 9801
Question 8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row. Solution: Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60 ∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100
Question 9. The area of a square field is 60025 m2. A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point. Solution: Area of a square field = 60025 m2
Question 10. The cost of levelling and turfing a square lawn at Rs. 250 per m2 is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ? Solution: Cost of levelling a square field = Rs. 13322.50 Rate of levelling = Rs. 2.50 per m2 and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m ∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460
Question 11. Find the greatest number of three digits which is a perfect square. Solution: 3-digits greatest number = 999 Finding the square root, we see that 38 has been left ∴ Perfect square = 999 – 38 = 961 ∴ Greatest 3-digit perfect square = 961
Question 12. Find the smallest number which must be added to 2300 so that it becomes a perfect square. Solution: Finding the square root of 2300 We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square ∴ Smallest number to be added = 4
Exercise 3.6
Question 1. Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 Solution: (i) In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9 (ii) In 799356−−−−−−√ ∴ the units digit is 6 ∴ The units digit of the square root can be 4 or 6 (iii) In 7998001−−−−−−−√ ∴ the units digit is 1 ∴ The units digit of the square root can be 1 or 9 (iv) In 657666025 ∴ The unit digit is 5 ∴ The units digit of the square root can be 5
Question 2. Find the square root of each of the following by prime factorization. (i) 441 (ii) 196 (iii) 529 (iv) 1764 (v) 1156 (vi) 4096 (vii) 7056 (viii) 8281 (ix) 11664 (x) 47089 (xi) 24336 (xii) 190969 (xiii) 586756 (xiv) 27225 (xv) 3013696 Solution:
Question 3. Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained. Solution: Factorising 180, 180 = 2 x 2 x 3 x 3 x 5 Grouping the factors in pairs we see that factor 5 is left unpaired. ∴ Multiply 180 by 5, we get the product 180 x 5 = 900 Which is a perfect square and square root of 900 = 2 x 3 x 5 = 30
Question 4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorising 147, 147 = 3 x 7×7 Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired ∴ Multiplying 147 by 3, we get the product 147 x 3 = 441 Which is a perfect square and its square root = 3×7 = 21
Question 5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number. Solution: Factorising 3645 3645 = 3 x 3 3 x 3 x 3 x 3 x 5 Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired ∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27
Question 6. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained. Solution: Factorsing 1152, 1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired. ∴ Dividing by 2, the quotient 576 is a perfect square . ∴ Square root of 576, it is 24
Question 7. The product of two numbers is 1296. If one number is 16 times the others find the numbers. Solution: Product of two numbers = 1296 Let one number = x Second number = 16x ∴ First number = 9 and second number = 16 x 9 = 144
Question 8. A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents. Solution: Total donation collected = Rs. 202500 Let number of residents = x Then donation given by each resident = Rs. x ∴ Total collection = Rs. x x x
Question 9. A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute? Solution: Total amount collected = Rs. 92.16 = 9216 paise Let the number of members = x Then amount collected by each member = x paise ∴ Number of members = 96 and each member collected = 96 paise
Question 10. A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ? Solution: Total fee collected = Rs. 2304 Let number of students = x Then fee paid by each student = Rs. x ∴ x x x = 2304 => x2 = 2304 ∴ x = 2304−−−−√
Question 11. The area of a square field is 5184 m2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field. Solution: The area of a square field = 5184 m2 Let side of the square = x ∴ side of square= 72 m ∴ Perimeter, of square field = 72 x 4 m = 288 m Perimeter of rectangle = 288 m Let breadth of rectangular field (b) = x Then length (l) = 2x ∴ Perimeter = 2 (l + b) = 2 (2x + x) = 2 x 3x = 6x = 2 (2x + x) = 2 x 3x = 6x ∴ Length of rectangular field = 2x = 2 x 48 = 96 m and breadth = 48 m and area = l x b = 96 x 48 m2 = 4608 m2
Question 12. Find the least square number, exactly divisible by each one of the numbers : (i) 6, 9,15 and 20 (ii) 8,12,15 and 20 Solution: LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180 =2 x 2 x 3 x 3 x 5 We see that after grouping the factors in pairs, 5 is left unpaired ∴ Least perfect square = 180 x 5 = 900 We see that after grouping the factors, factors 2, 3, 5 are left unpaired ∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600
Question 13. Find the square roots of 121 and 169 by the method of repeated subtraction. Solution:
Question 14. Write the prime factorization of the following numbers and hence find their square roots. ^ (i) 7744 (ii) 9604 (iii) 5929 (iv) 7056 Solution: Factorization, we get: (i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11 Grouping the factors in pairs of equal factors, Question 15. The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. Solution: Total amount of donation = 2401 Let number of students in VIII = x ∴ Amount donoted by each student = Rs. x
Question 16. A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement. Solution: Number of students = 6000 Students left out = 71 ∴ Students arranged in a field = 6000 – 71=5929
Exercise 3.7
Find the square root of the following numbers in decimal form :
Question 1. 84.8241 Solution:
Question 2. 0.7225 Solution:
Question 3. 0.813604 Solution:
Question 4. 0.00002025 Solution:
Question 5. 150.0625 Solution:
Question 6. 225.6004 Solution:
Question 7. 3600.720036 Solution:
Question 8. 236.144689 Solution:
Question 9. 0.00059049 Solution:
Question 10. 176.252176 Solution:
Question 11. 9998.0001 Solution:
Question 12. 0.00038809 Solution:
Question 13. What is that fraction which when multiplied by itself gives 227.798649 ? Solution:
Question 14. square playground is 256.6404 square metres. Find the length of one side of the playground. Solution: Area of square playground = 256.6404 sq. m
Question 15. What is the fraction which when multiplied by itself gives 0.00053361 ? Solution:
Question 16. Simplify : Solution:
Question 17. Evaluate 50625−−−−−√ and hence find the value of 506.25−−−−−√+5.0625−−−−−√. Solution:
Question 18. Find the value of 103.0225−−−−−−−√ and hence And the value of Solution:
Exercise 3.8
Question 1. Find the square root of each of the following correct to three places of decimal : (i) 5 (ii) 7 (iii) 17 (iv) 20 (v) 66 (vi) 427 (vii) 1.7 (vii’) 23.1 (ix) 2.5 (x) 237.615 (xi) 15.3215 (xii) 0.9 (xiii) 0.1 (xiv) 0.016 (xv) 0.00064 (xvi) 0.019 (xvii) 78 (xviii) 512 (xix) 2 12 (xx) 287 88 Solution:
Question 2. Find the square root of 12.0068 correct to four decimal places. Solution:
Question 3. Find the square root of 11 correct to five decimal places. Solution: https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&outp/www.learninsta.com&dtd=1834
Question 4. Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following : Solution:
Question 5. Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following : Solution:
Exercise 3.9
Using square root table, find the square roots of the following :
Question 1. 7 Solution:
Question 2. 15 Solution:
Question 3. 74 Solution:
Question 4. 82 Solution:
Question 5. 198 Solution:
Question 6. 540 Solution:
Question 7. 8700 Solution:
Question 8. 3509 Solution:
Question 9. 6929 Solution:
Question 10. 25725 Solution:
Question 11. 1312 Solution:
Question 12. 4192 Solution:
Question 13. 4955 Solution:
Question 14. 99144 Solution:
Question 15. 57169 Solution:
Question 16. 101169 Solution:
Question 17. 13.21 Solution:
Question 18. 21.97 Solution:
Question 19. 110 Solution:
Question 20. 1110 Solution:
Question 21. 11.11 Solution:
Question 22. The area of a square field is 325 m2. Find the approximate length of one side of the field. Solution:
Question 23. Find the length of a side of a square, whose area is equal to the area bf the rectangle with sides 240 m and 70 m. Solution:
Question 1. Express the following numbers in standard form : (i) 6020000000000000 (ii) 0.00000000000942 (iii) 0.00000000085 (iv) 846 X 107 (v) 3759 x 10-4 (vi) 0.00072984 (vii) 0.000437 x 104 (Viii) 4 + 100000 Solution:
Question 2. Write the following numbers in the usual form : (i) 4.83 x 107 (ii) 3.02 x 10-6 (iii) 4.5 x 104 (iv) 3 x 10-8 (v) 1.0001 x 109 (vi) 5.8 x 102 (vii) 3.61492 x 106 (viii) 3.25 x 10-7 Solution:
Exercise 2.2
Question 1. Write each of the following in exponential form : Solution:
Question 2. Evaluate : Solution:
Question 3. Express each of the following as a rational number in the form pq: Solution:
Question 4. Simplify : Solution:
Question 5. Express each of the following rational numbers with a negative exponent : Solution:
Question 6. Express each of the following rational numbers with a positive exponent : Solution:
Question 7. Simplify : Solution:
Question 8. By what number should 5-1 be multiplied so that the product may be equal to (-7)-1 ? Solution:
Question 9. By what number should (12)−1 be multiplied so that the product may be equal to (−47)−1 ? Solution:
Question 10. By what number should (-15)-1 be divided so that the quotient may be equal to (-5)-1 ? Solution:
Question 11. By what number should (53)−2 be multiplied so that the product may be (73)−1 ? Solution:
Question 12. Find x, if Solution:
Question 13. Solution:
Question 14. Find the value of x for which 52x + 5-3 = 55. Solution:
Exercise 2.3
Question 1. Express the following numbers in standard form : (i) 6020000000000000 (ii) 0.00000000000942 (iii) 0.00000000085 (iv) 846 X 107 (v) 3759 x 10-4 (vi) 0.00072984 (vii) 0.000437 x 104 (Viii) 4 + 100000 Solution:
Question 2. Write the following numbers in the usual form : (i) 4.83 x 107 (ii) 3.02 x 10-6 (iii) 4.5 x 104 (iv) 3 x 10-8 (v) 1.0001 x 109 (vi) 5.8 x 102 (vii) 3.61492 x 106 (viii) 3.25 x 10-7 Solution:
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