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Samacheer Kalvi Class 7 Maths Practice Question Papers

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Brain Grain · braingrain.in
Maths — Practice Paper · Set 1
Class: 7Samacheer KalviMax Marks: 59
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 10 × 1 = 10

Choose the correct answer. (Answer all questions.)

1.Draw a line segment of given length and construct a perpendicular bisector to each line segment using scale and compass (e) 58 cmA. 8 cmB. 7 cmC. 5.6 cmD. 10.4 cm[1]
2.Which of the following rule is not sufficient to verify the congruency of two triangles. (i) SSS rule (ii) SAS rule (iii) SSA rule (iv) ASA rule[1]
3.Which of the following expressions is equal to -30. (i) -20 – (-5 × 2) (ii) (6 × 10) – (6× 5) (iii) (2 × 5)+ (4 × 5) (iv) (-6) × (+5)[1]
4.A straight angle measuresA. 45°B. 90°C. 180°D. 100°[1]
5.Round the following decimal numbers upto 3 place of decimalA. 24.4003B. 1251.2345C. 61.00203[1]
6.Which of the following can be the sides of a triangle? (i) 5.9.14 (ii) 7,7,15 (iii) 1,2,4 (iv) 3, 6, 8[1]
7.Which of the following statement is ALWAYS TRUE when parallel lines are cut by a transversal (i) corresponding angles supplementary (ii) alternate interior angles supplementary (iii) alternate exterior angles supplementary (iv) interior angles on the same side of the transversal are supplementary[1]
8.The corner of the A4 paper hasA. An acute angleB. A right angleC. StraightD. An obtuse angle[1]
9.Which of the following does not represent an integer? (i) 0 ÷ (-7) (ii) 20 ÷ (-4) (iii) (-9) ÷ 3 (iv) 12 ÷ 5[1]
10.A pool of fish translates from point F to point D.A. Describe the translation of the pool of fish.B. Can the fishing boat make the same translation? Explain.C. Describe a translation the fishing boat could make to get to point D.[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

11.The numerical co-efficient of the term -xy is ______ Hint: -x,y = (- 1 )xy.[1]
12.x+ 6 [ ] y + 6 __________[1]
13.(-62) ÷ (-62) = ____[1]
14.(-8) + 10 + (-2) = ____ (i) 2 (ii) 8 (iii) 0 (iv) 20[1]
15.-44 + ____ = -88[1]
Part III — True or False 5 × 1 = 5

Write True or False. (Answer all questions.)

16.The additive inverse of (-32) is -32[1]
17.An inequation, -3 < x < -1, where x is an integer, cannot be represented in the number line.[1]
18.(-33) + 8 = 8 + (-33)[1]
19.The quotient of two integers having opposite sign is a negative integer.[1]
20.15 – (-18) is the same as 15 + 18[1]
Part IV — Short Answer Questions 12 × 2 = 24

Answer briefly. (Answer all questions.)

21.A dozen bananas costs ₹ 20. What is the price of 48 bananas ?[2]
22.The area of parallelogram whose base 10 m and height 7 m is (i) 70 sq.m (ii) 35 sq.m (iii) 7 sq.m (iv) 10 sq.m[2]
23.Two parallel lines are cut by transversal. If one angle of a pair of corresponding angles can be represented by 42° less than three times the other. Find the corresponding angles.[2]
24.In a readymade shop there will be a board showing upto 50% off. Most of the people will realize that everything is half of its original price, Is that true?[2]
25.Read the following examples and group them in two categories.[2]
26.Reflect the shape with given line of reflection.[2]
27.Money spent on vegetables for five days is ₹ 120, ₹ 80, ₹ 75, ₹ 95 and ₹ 86.[2]
28.Which property is illustrated by the equation: (5 × 2) + (5 × 5) = 5 × (2 + 5) (i) commutative (ii) closure (iii) distributive (iv) associative[2]
29.The difference between the consecutive terms of the fifth slanting row containing four elements of a Pascal’s Triangle is (i) 3,6,10,… (ii) 4,10,20,… (iii) 1,4,10,… (iv) 1,3,6,…[2]
30.Simplify the following. (i) 19.2 ÷ 2.4 (ii) 4.95 ÷ 0.5 (iii) 19.11 ÷ 1.3 (iv) 0.399 ÷ 2.1 (v) 5.4 ÷ 0.6 (vi) 2.197 ÷ 1.3[2]
31.1.0 + 0.83 = ? (i) 0.17 (ii) 0.71 (iii) 1.83 (iv) 1.38[2]
32.Cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075. Find the area of the land.[2]
Part V — Long Answer Questions 3 × 5 = 15

Answer in detail. (Answer all questions.)

33.A group of 21 students paid ₹ 840 as the entry fee for a magic show. How many students entered the magic show if the total amount paid was ₹ 1680?[5]
34.Find the sum of the following expressions (i) 7p + 6q, 5p – q, q + 16p[5]
35.Solve the following inequalities. (i) 4n + 7 > 3n + 10, n is an integer (ii) 6(x + 6) > 5 (x – 3), x is a whole number. (iii) -13 < 5x + 2 < 32, x is an integer.[5]
🔑 Show Answer Key — Set 1
  1. 1. (a) 8 cm Construction :Step 1: Drawn a line. Marked two points A and B on it so that AB = 8 cm Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D. step 4: Joined C and D, CD intersect AB. Marked the point of intersection as ‘O’. CD is the required perpendicular bisector of AB.(b) 7 cm Construction :step 1: Drawn a line and marked points A and B on it so that AB = 7 cm. step 2: Using compass with A as centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB. step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D step 4: Joined C and D, CD is the required perpendicular bisector of AB.(c) 5.6 cm. Construction :Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6cm Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as ‘O’CD is the required perpendicular bisector of AB. Now ∠AOC = 90° AO = BO = 2.8 cm(d) 10.4 cm Construction :Step 1: Drawn a line and marked two points A and B on it so that AB = 10.4 cm. Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB. Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D. Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as O. CD is the required perpendicular bisector. Now ∠AOC = 90° ; AO = BO = 5.2 cm(e) 58 mmConstruction :Step 1: Drawn a line. Marked two points A and B on it so that AB = 5.8 cm = 58 mm. Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB. Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D. Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector. ∠AOC = 90° AO = BO = 2.9 cm About Us Privacy Policy Disclaimer Contact Us
  2. 2. (iii) SSA rule
  3. 3. (iv) (-6) × (+5) Hint: (i) -20 + (10) = -10 (ii) 60 – 30 = 30 (iii) 10 + 20 = 30 (iv) (-6) × (+5) = – 30
  4. 4. (c) 180° No, they are not adjacent pairs.
  5. 5. (a) 24.4003 Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.40 0 3. In 24.40 0 3 the digit next to the thousandths value is 3 which is less than 5. ∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400. (b) 1251.2345 Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.23 4 5, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235. (c) 61.00203 Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00 2 03 gives 61.00 2 03. In 61.00 2 03, the digit next to the thousandths place value is 0, which is less than 5. Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002. About Us Privacy Policy Disclaimer Contact Us
  6. 6. (iv) 3, 6, 8 (i) Here 5 + 9 = 14 = the measure of the third side. In a triangle the sum of the measures of any two sides must be greater than the third side. ∴ 5, 9, 14 cannot be the sides of a triangle. (ii) 7.7.15 Here sum of two sides 7 + 7 = 14 < the measures of the thrid side. So 1,1, 15 cannot be the sides of a triangles. (iii) 1,2,4 Here sum of two sides 1 + 2 = 3 < the measure of the third side. ∴ 1, 2, 4 cannot be the sides of a triangle. (iv) 3, 6, 8 Sum of two sides 3 + 6 = 9 > the third side. ∴ 3, 6, 8 can be the sides of a triangle.
  7. 7. (iv) Interior angles on the same side of the transversal are supplementary.
  8. 8. (b) a right angle
  9. 9. (iv) 12 ÷ 5
  10. 10. (a) Translation of pool of fish is 7 →, 2↓ (b) No, the fishing boat will be landed on the island if translated. (c) To get point D, the translation will be 5 →, 3↓
  11. 11. -1
  12. 12. x+ 6 [ > ] y + 6
  13. 13. 1
  14. 14. (iii) 0
  15. 15. -44
  16. 16. False
  17. 17. True
  18. 18. True, by commutative property on intergers
  19. 19. True
  20. 20. True
  21. 21. Let the required price be ₹ x. As the number of bananas increases price also increases ∴ Number of bananas and cost are in direct proportion.
  22. 22. (i) 70 sq. m Hint: = base × height = 10m × 7m = 70 sq.m
  23. 23. We know that the corresponding angles are equal. Let one of the corresponding angles be x. Then the other will be 3x – 42°.
  24. 24. No. Only some of them are half of its original price. Exercise 2.2 Try These (Text book Page No. 33)
  25. 25. About Us Privacy Policy Disclaimer Contact Us
  26. 26. About Us Privacy Policy Disclaimer Contact Us
  27. 27. Arithmetic Mean = 91.2 Think (Text book Page No. 99) Check the properties of arithmetic mean for the example given below:
  28. 28. (iii) distributive
  29. 29. (ii) 4,10,20,…
  30. 30. (i) 19.2 ÷ 2.4(ii) 4.95 ÷ 0.5(iii) 19.11 ÷ 1.3(iv) 0.399 ÷ 2.1(v) 5.4 ÷ 0.6(vi) 2.197 ÷ 1.3
  31. 31. (iii) 1.83
  32. 32. Cost of levelling the entire land = ₹ 10,075 Cost of levelling 1 sq. ft = ₹ 15.50∴ Area of the land = 650 sq.ft.
  33. 33. Let the required number of students be x.As the number of students increases the entry fees also increases. ∴ They are in direct proportion .∴ The number of students entered magic show = 42
  34. 34. (7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p = (7p + 5p + 16p) + (6q – q + q) = (7 + 5 + 16) p + (6 – 1 + 1) q = (12 + 16) p + 6q = 28p + 6q (ii) a + 5b + 7c, 2a + 106 + 9c (a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c = a + 2a + 5b + 10b + 7c + 9c = (1 + 2)a + (5 + 10)b + (7 + 9)c = 3a + 15b + 16c (iii) mn + t, 2mn – 2t, – 3t + 3mn (mn + t) + (2mn – 2t) + (-3t + 3mn) = mn + t + 2mn – 2t + (-3t) + 3mn = (mn + 2mn + 3mn) + (t – 2t – 3t) = (1 + 2 + 3) mn + (1 – 2 – 3) t = 6mn + (1 – 5)t = 6mn + (- 4) t = 6mn – 4t (iv) u + v, u – v, 2u + 5v, 2u – 5v (u + v) + (u – v) + (2u + 5v) + (2u – 5v) = u + v + u – v + 2u + 5v + 2u – 5v = u + u + 2u + 2u + v – v + 5v – 5v = (1 + 1 + 2 + 2) u +(1 – 1 + 5 – 5)v = 6u + 0v = 6u (v) 5xyz – 3xy, 3zxy – 5yx 5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy = (5 + 3) xyz + [(-3) + (-5)] xy = 8xyz + (-8) xy = 8xyz – 8xy
  35. 35. (i) 4n + 7 > 3n + 10, n is an integer. 4n + 7 – 3n > 3n + 10 – 3n n(4 – 3) + 7 > 3n + 10 – 3n n (4 – 3) + 7 > n (3 – 3) + 10 n + 7 > 10 Subtracting 7 on both sides n + 7 – 7 > 10 – 7 n > 3 Since the solution is an integer and is greater than or equal to 3, the solution will be 3, 4, 5, 6, 7, ….. n = 3, 4, 5, 6,7, …. (ii) 6 (x + 6) > 5 (x – 3), x is a whole number. 6x + 36 > 5x – 15 Subtracting 5x on both sides 6x + 36 – 5x > 5x – 15 – 5x x (6 – 5) + 36 > x(5 – 5) – 15 x + 36 > -15 Subtracting 36 on both sides x + 36 – 36 > -15 -36 x > -51 The solution is a whole number and which is greater than or equal to -51 ∴ The solution is 0, 1, 2, 3, 4,… x = 0,1,2, 3,4,…(iii) -13 < 5x + 2 < 32, x is an integer. Subtracting throughout by 2 -13 – 2 < 5x + 2 – 2 < 32 – 2 -15 < 5x < 30 Dividing throughout by 5 \(\frac { -15 }{ 5 } \) < \(\frac { 5x }{ 5 } \) < \(\frac { 30 }{ 5 } \) – 3 < x < 6 ∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution as -3, -2, -1,0, 1,2, 3, 4, 5, 6. i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6. About Us Privacy Policy Disclaimer Contact Us
Brain Grain · braingrain.in
Maths — Practice Paper · Set 2
Class: 7Samacheer KalviMax Marks: 59
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 10 × 1 = 10

Choose the correct answer. (Answer all questions.)

1.Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).A. ______ sq. unitsB. ______ sq. unitsC. ______ sq. unitsD. ______ sq. units[1]
2.Find the missing values in the following table for the circles with radius (r), diameter (d) and Circumference (C).[1]
3.Construct the following angles using protractor and draw a bisector to each of the angle using ruler and compass. (e) 110°.A. 60°B. 100°C. 90°D. 48°[1]
4.If a perpendicular line is bisecting the given line, you would have twoA. right anglesB. obtuse anglesC. acute anglesD. reflex angles[1]
5.An angle with measure 128° is called ___ angle.A. a straightB. an obtuseC. an acuteD. Right[1]
6.Which of the following methods are used to check the congruence of plane figures? (i) translation method (ii) superposition method (iii) substitution method (iv) transposition method[1]
7.An angle that measure 0° is calledA. right angleB. obtuse angleC. acute angleD. Zero angle.[1]
8.Find the area o the following parallelograms by measuring their base and height, using formula. (e) _____ sq. unitsA. _____ sq. unitsB. _____ sq. unitsC. _____ sq. unitsD. _____ sq. units[1]
9.Draw a line segment of given length and construct a perpendicular bisector to each line segment using scale and compass (e) 58 cmA. 8 cmB. 7 cmC. 5.6 cmD. 10.4 cm[1]
10.Which of the following rule is not sufficient to verify the congruency of two triangles. (i) SSS rule (ii) SAS rule (iii) SSA rule (iv) ASA rule[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

11.x – y [ ] 0 __________[1]
12.20 + (-9) + 9 = ____ (i) 20 (ii) 29 (iii) 11 (iv) 38[1]
13.____ × (-9) = -45[1]
14.When we subtract ‘a’ from ‘-a’, we get ______ (i) a (ii) 2a (iii) -2a (iv) -a[1]
15.12 cows can graze a field for 10 days. 20 cows can graze the same field for ____ days (i) 15 (ii) 18 (iii) 6 (iv) 8[1]
Part III — True or False 5 × 1 = 5

Write True or False. (Answer all questions.)

16.(-30) ÷ (-6) = -6[1]
17.-7 + 2 = 2 + (-7)[1]
18.(-100) × 0 × 20 = 0[1]
19.(-11) + (-8) = (-8) + (-11)[1]
20.The sum of two integers can never be zero[1]
Part IV — Short Answer Questions 12 × 2 = 24

Answer briefly. (Answer all questions.)

21.Circumference of a circle is always (i) three times of its diameter (ii) more than three times of its diameter (iii) less than three times of its diameter (iv) three times of its radius[2]
22.In 8 hours duration, with uniform decrease in temperature, the temperature dropped 24°C. How many degrees did the temperature drop each hour?[2]
23.During summer, the level of the water in a pond decreases by 2 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?[2]
24.Find the order of rotational symmetry for an equilateral triangle.[2]
25.The solution of 3x + 5 = x + 9 is t (i) 2 (ii) 3 (iii) 5 (iv)4[2]
26.Is it the only way to decompose the numbers representing length and breadth? Discuss.[2]
27.The decimal representation of 30 kg and 43 g is ____ kg. (i) 30.43 (ii) 30.430 (iii) 30.043 (iv) 30.0043[2]
28.Find the diameter of your bicycle wheel?[2]
29.Find two pairs of integers whose product is +15.[2]
30.Name the transformation that will map footprint A onto the indicated footprint. (i) Footprint B (ii) Footprint (iii) Footprint D (iv) Footprint E[2]
31.How mean is unchanged in the series A, B and C ?[2]
32.Will the figure be symmetric about both the diagonals?[2]
Part V — Long Answer Questions 3 × 5 = 15

Answer in detail. (Answer all questions.)

33.Observe the figure. There are two angles namely ∠PQR = 150° and ∠QPS = 30° Is all this pair of supplementary angles a linear pair? Discuss[5]
34.Give an algebraic equation for the following statement: “The difference between the area and perimeter of a rectangle is 20”.[5]
35.Identify the degree of the expression, 2a 3 be + 3a 3 b + 3a 3 c – 2a 2 b 2 c 2[5]
🔑 Show Answer Key — Set 2
  1. 1. Converting the given parallelograms into rectangles we get.(a) 10 sq. units (b) 18 sq. units (c) 16 sq. units (d) 5 sq. units
  2. 2. (i) Given radius r = 15cm ∴ diameter d = 2 × 15 = 30 cm Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 30 = \(\frac { 660 }{ 7 } \) = 94.28 cm (ii) Given circumference C = 1760 cm 2πr = 1760 2 × \(\frac { 22 }{ 7 } \) × r = 1760 r = \(\frac{1760 \times 7}{2 \times 22}\) = \(\frac{160 \times 7}{2 \times 2}\) = 40 × 7 = 280 cm diameter = 2 × r = 2 × 280 = 560 cm (iii) diameter d = 24m radius r = \(\frac { d }{ 2 } \) = \(\frac { 24 }{ 2 } \) = 12 m Circumference C = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 12 = \(\frac { 528 }{ 7 } \) = 75.4 m Tabulating the results
  3. 3. (a) 60° Construction:Step 1: Drawn the given angle ∠ABC with the measure 60° using protractor. Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC. Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of same measure with centre at F to cut the previous arc. Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given ∠ABC(b) 100°Construction : Step 1: Drawn the given angle ∠ABC with the measure 100° c using protractor. Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC. Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of the same measure with centre at F to cut the previous arc. Step 4: Marked the point of intersection at G. Drawn a ray BX through G. BG is the required bisector of angle ∠ABC ∠ABG = ∠GBC = 50°(c) 90° Construction :Step 1: Drawn the given angle ∠ABC with the measure 90° using protractor. Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC. Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc. Step 4: Mark the point of interaction as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC ∠ABG = ∠GBC = 45° (d) 48°Construction : Step 1: Drawn the given angle ∠ABC with the measure 48° using protractor. Step 2: With B as center and convenient radius, drawn an arc to cut BA and to cut BA and BC. Marked the points of intersection as E on BA and F on BC. Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of the same measure with center at F to cut the previous arc. Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC Now ∠ABC = ∠GBC = 24°(e) 110° Construction:Step 1: Drawn the given angle ∠ABC with the measure 110° using protractor. Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as E on BA and F BC. Step 3: With the same radius and E as center, drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc. Step 4: Mark the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC ∠ABG = ∠GBC = 55° About Us Privacy Policy Disclaimer Contact Us
  4. 4. (a) right angle
  5. 5. (b) an obtuse
  6. 6. (ii) superposition method
  7. 7. (d) Zero angle Try this (Text Book Page No. 86)
  8. 8. (a) Area of the rectangle = (base × height) sq. units base = 5 units height = 5 units ∴ Area = (5 × 5 ) = sq. units = 25 sq. units (b) Area of the rectangle = (base × height) sq. units base = 4 units height = 1 units ∴ Area = (4 × 1 ) = sq. units = 4 sq. units (c) Area of the rectangle = (base × height) sq. units base = 2 units height = 3 units ∴ Area = (2 × 3 ) = sq. units = 6 sq. units (d) Area of the rectangle = (base × height) sq. units base = 4 units height = 4 units ∴ Area = (4 × 4 ) = sq. units = 16 sq. units (e) Area of the parallelogram = (base × height) sq. units base = 7 units height = 5 units = 7 × 5 = 35 sq. units
  9. 9. (a) 8 cm Construction :Step 1: Drawn a line. Marked two points A and B on it so that AB = 8 cm Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D. step 4: Joined C and D, CD intersect AB. Marked the point of intersection as ‘O’. CD is the required perpendicular bisector of AB.(b) 7 cm Construction :step 1: Drawn a line and marked points A and B on it so that AB = 7 cm. step 2: Using compass with A as centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB. step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D step 4: Joined C and D, CD is the required perpendicular bisector of AB.(c) 5.6 cm. Construction :Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6cm Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as ‘O’CD is the required perpendicular bisector of AB. Now ∠AOC = 90° AO = BO = 2.8 cm(d) 10.4 cm Construction :Step 1: Drawn a line and marked two points A and B on it so that AB = 10.4 cm. Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB. Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D. Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as O. CD is the required perpendicular bisector. Now ∠AOC = 90° ; AO = BO = 5.2 cm(e) 58 mmConstruction :Step 1: Drawn a line. Marked two points A and B on it so that AB = 5.8 cm = 58 mm. Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB. Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D. Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector. ∠AOC = 90° AO = BO = 2.9 cm About Us Privacy Policy Disclaimer Contact Us
  10. 10. (iii) SSA rule
  11. 11. x – y [ > ] 0
  12. 12. (i) 20 About Us Privacy Policy Disclaimer Contact Us
  13. 13. 5
  14. 14. (iii) -2a Hint: – a – a = – 2a
  15. 15. (iii) 6 Hint:
  16. 16. False
  17. 17. True, by commutative property on intergers
  18. 18. True
  19. 19. True, because addition is commutative for intergers
  20. 20. False
  21. 21. (ii) more than three times of its diameter About Us Privacy Policy Disclaimer Contact Us
  22. 22. In 8 hours the drop in temperature = 24 In 1 hour the drop in temperature = \(\frac{24}{8}\) = 3° The temperature dropped 3°C every hour.
  23. 23. Level of water decreases a week = 2 inches. Level of water decreases in 6 weeks = 6 × 2 = 12 inches
  24. 24. For an equilateral triangle order of rotational symmetry is 3.Think (Text book Page No. 74)
  25. 25. (i) 2 Hint: 3x + 5 = x + 9 ⇒ 3x – x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2
  26. 26. No, for example 15 can be decompose into 1 × 15, 3 × 5, 5 × 3, 15 × 1Try These (Text book Page No. 52)
  27. 27. (iii) 30.043 Hint: 30 kg and 43 g = 30 kg + \(\frac { 43 }{ 1000 } \) kg = 30 + 0.043 = 30.043
  28. 28. Diameter of my bicycle wheel is 700 mm
  29. 29. (i) (+3) × (+5) (ii) (-3) × (-5)
  30. 30. (i) It is translation (ii) Reflection about horizontal line. (iii) Reflection about vertical line. (iv) Rotation about the heel.
  31. 31. The difference between the given numbers are equal.
  32. 32. Yes, it is symmetric about both the diagonals.Try These (Text book Page No. 74)
  33. 33. Given ∠PQR =150° ∠QPS = 30° They are supplementary angles, But they are not adjacent angles as they don’t have common vertex or common arm. ∴ They are not a linear pair. Try this (Text book Page No. 90)
  34. 34. Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b) Area – Perimeter = 20 ∴ lb – 2(l + b)
  35. 35. The terms of the given expression are 2a 3 bc, 3a 3 b + 3a 3 c – 2a 2 b 2 c 2 Degree of each of the terms: 5,4,4,6. Terms with the highest degree: – 2a 2 b 2 c 2 Therefore degree of the expression is 6.
Brain Grain · braingrain.in
Maths — Practice Paper · Set 3
Class: 7Samacheer KalviMax Marks: 59
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 10 × 1 = 10

Choose the correct answer. (Answer all questions.)

1.Which of the following expressions is equal to -30. (i) -20 – (-5 × 2) (ii) (6 × 10) – (6× 5) (iii) (2 × 5)+ (4 × 5) (iv) (-6) × (+5)[1]
2.A straight angle measuresA. 45°B. 90°C. 180°D. 100°[1]
3.Round the following decimal numbers upto 3 place of decimalA. 24.4003B. 1251.2345C. 61.00203[1]
4.Which of the following can be the sides of a triangle? (i) 5.9.14 (ii) 7,7,15 (iii) 1,2,4 (iv) 3, 6, 8[1]
5.Which of the following statement is ALWAYS TRUE when parallel lines are cut by a transversal (i) corresponding angles supplementary (ii) alternate interior angles supplementary (iii) alternate exterior angles supplementary (iv) interior angles on the same side of the transversal are supplementary[1]
6.The corner of the A4 paper hasA. An acute angleB. A right angleC. StraightD. An obtuse angle[1]
7.Which of the following does not represent an integer? (i) 0 ÷ (-7) (ii) 20 ÷ (-4) (iii) (-9) ÷ 3 (iv) 12 ÷ 5[1]
8.A pool of fish translates from point F to point D.A. Describe the translation of the pool of fish.B. Can the fishing boat make the same translation? Explain.C. Describe a translation the fishing boat could make to get to point D.[1]
9.Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).A. ______ sq. unitsB. ______ sq. unitsC. ______ sq. unitsD. ______ sq. units[1]
10.Find the missing values in the following table for the circles with radius (r), diameter (d) and Circumference (C).[1]
Part II — Fill in the Blanks 5 × 1 = 5

Fill in the blanks. (Answer all questions.)

11.(-40) ÷ ___ =40[1]
12.___ – 75 = -45[1]
13.If the integers -15, 12, -17, 5, -1, -5, 6 are marked on the number line then the integer on the extreme left is _____ .[1]
14.If a = 5, the value of 2a + 5 is _______.[1]
15.2.08 × 10 = ______ (i) 20.8 (ii) 208.0 (iii) 0.208 (iv) 280.0[1]
Part III — True or False 5 × 1 = 5

Write True or False. (Answer all questions.)

16.The product of two negative integers is a positive integer.[1]
17.The sum of a positive integer and a negative integer is always a positive integer.[1]
18.(-90) + (-30) = 60[1]
19.When y = -1, the value of the expression 2y – 1 is 3. Hint: 2(-1) – 1 = -2 – 1 = – 3[1]
20.8 × (-4) = 32[1]
Part IV — Short Answer Questions 12 × 2 = 24

Answer briefly. (Answer all questions.)

21.The pre-image and the image after a translation coincide. What can you say about the translation[2]
22.The sum of all angles at a point is (i) 360° (ii) 180° (iii) 90° (iv) 0°[2]
23.2.01 ÷ 0.03 = ? (i) 6.7 (ii) 67.0 (iii) 0.67 (iv) 0.067[2]
24.Find the percentage of children whose scores fall in different categories given in table below.[2]
25.Observe the sequence of numbers obtained in the 3rd and 4th slanting rows of Pascal’s Triangle and find the difference between the consecutive numbers and complete the table given below.[2]
26.Which average will be most aprropriate for the companies producing the following goods? why? (i) Diaries and notebooks (ii) School bags (iii) Jeans and T-shirts.[2]
27.A tetromino is a shape obtained by …… squares together.[2]
28.In a trapezium if the sum of the parallel sides is 10 m and the area is 140 sq.m, then the height is (i) 7 cm (ii) 40 cm (iii) 14 cm (iv) 28 cm[2]
29.The addition of 3mn, -5mn, 8mn and – 4mn is (i) mn (ii) – mn (iii) 2mn (iv) 3mn[2]
30.Shade the figure completely, by using five Tetromino shapes only once.[2]
31.An exterior angle of a triangle is 70° and two interior opposite angles are equal. Then measure of each of these angle will be (i) 110° (ii) 120° (iii) 35° (iv) 60°[2]
32.Ages of 15 students in 8th standard is 13, 12, 13, 14, 12, 13, 13, 14, 12, 13, 13, 14, 13, 12, 14. Find the mean age of the students.[2]
Part V — Long Answer Questions 3 × 5 = 15

Answer in detail. (Answer all questions.)

33.Draw circles for the following measurements of radius (r)/ diameters(d). (i) r = 4 cm (ii) d = 12 cm (iii) r = 3.5 cm (iv) r = 6.5 cm. (v) d = 6 cm[5]
34.Factorise the following algebraic expressions by using the identity a 2 – b 2 = (a + b)(a – b). (i) z 2 – 16 (ii) 9 – 4y 2 (iii) 25a 2 – 49b 2 (iv) x 4 – y 4[5]
35.Express the following using decimal notation. (i) 8 m 30 cm in metres (ii) 24 km 200 m in kilometres[5]
🔑 Show Answer Key — Set 3
  1. 1. (iv) (-6) × (+5) Hint: (i) -20 + (10) = -10 (ii) 60 – 30 = 30 (iii) 10 + 20 = 30 (iv) (-6) × (+5) = – 30
  2. 2. (c) 180° No, they are not adjacent pairs.
  3. 3. (a) 24.4003 Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.40 0 3. In 24.40 0 3 the digit next to the thousandths value is 3 which is less than 5. ∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400. (b) 1251.2345 Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.23 4 5, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235. (c) 61.00203 Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00 2 03 gives 61.00 2 03. In 61.00 2 03, the digit next to the thousandths place value is 0, which is less than 5. Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002. About Us Privacy Policy Disclaimer Contact Us
  4. 4. (iv) 3, 6, 8 (i) Here 5 + 9 = 14 = the measure of the third side. In a triangle the sum of the measures of any two sides must be greater than the third side. ∴ 5, 9, 14 cannot be the sides of a triangle. (ii) 7.7.15 Here sum of two sides 7 + 7 = 14 < the measures of the thrid side. So 1,1, 15 cannot be the sides of a triangles. (iii) 1,2,4 Here sum of two sides 1 + 2 = 3 < the measure of the third side. ∴ 1, 2, 4 cannot be the sides of a triangle. (iv) 3, 6, 8 Sum of two sides 3 + 6 = 9 > the third side. ∴ 3, 6, 8 can be the sides of a triangle.
  5. 5. (iv) Interior angles on the same side of the transversal are supplementary.
  6. 6. (b) a right angle
  7. 7. (iv) 12 ÷ 5
  8. 8. (a) Translation of pool of fish is 7 →, 2↓ (b) No, the fishing boat will be landed on the island if translated. (c) To get point D, the translation will be 5 →, 3↓
  9. 9. Converting the given parallelograms into rectangles we get.(a) 10 sq. units (b) 18 sq. units (c) 16 sq. units (d) 5 sq. units
  10. 10. (i) Given radius r = 15cm ∴ diameter d = 2 × 15 = 30 cm Circumference C = π d units = \(\frac { 22 }{ 7 } \) × 30 = \(\frac { 660 }{ 7 } \) = 94.28 cm (ii) Given circumference C = 1760 cm 2πr = 1760 2 × \(\frac { 22 }{ 7 } \) × r = 1760 r = \(\frac{1760 \times 7}{2 \times 22}\) = \(\frac{160 \times 7}{2 \times 2}\) = 40 × 7 = 280 cm diameter = 2 × r = 2 × 280 = 560 cm (iii) diameter d = 24m radius r = \(\frac { d }{ 2 } \) = \(\frac { 24 }{ 2 } \) = 12 m Circumference C = 2 π r units = 2 × \(\frac { 22 }{ 7 } \) × 12 = \(\frac { 528 }{ 7 } \) = 75.4 m Tabulating the results
  11. 11. -1
  12. 12. 30
  13. 13. The least number will be on the extreme left. ∴ -17 will be on the extreme left.
  14. 14. 15
  15. 15. (i) 20.8 Hint: 208 × 10 = 2080 2.08 × 10 = 20.80 = 20.8
  16. 16. True
  17. 17. False
  18. 18. False
  19. 19. False
  20. 20. False
  21. 21. There is no right, left, up or down movement took place. Try These (Text book Page No. 80)
  22. 22. (i) 360°
  23. 23. (ii) 67.0 Hint: \(\frac { 2.01 }{ 0.03 } \) = \(\frac { 201 }{ 3 } \) = 67
  24. 24. Try These (Text book Page No. 29)
  25. 25. Try These (Text book Page No. 96)
  26. 26. for all the above data mode will be more appropriate.Exercise 5.3 Try These (Text book Page No. 106)
  27. 27. 4
  28. 28. (iv) 28 cm Hint: Area = \(\frac{1}{2}\) × h × (a + b) = 140 = \(\frac{1}{2}\) × h × 10 ⇒ h = 28
  29. 29. (iii) 2mn Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn
  30. 30. (refer textbook)
  31. 31. (iii) 35°
  32. 32. = \(\frac { 195 }{ 15 } \) = 13 Mean age of the students = 13
  33. 33. (i) r = 4 cmStep 1 : Market a point ‘O’ on the paper. Step 2 : Extended the compass distance equal to radius 4 cm. Step 3 : At center ‘O’, helded the compass firmly and placed the pointed end of the compass. Step 4 : Slowly rotated the compass around to get the circle.(ii) d = 12 cm given d= 12 cm ∴ radius r = \(\frac { d }{ 2 } \) = \(\frac { 12 }{ 2 } \) = 6 cmStep 1: Marked a point ‘O’ on the paper. Step 2: Extended the compass distance equal to radius 6 cm. Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass. Step 4: Slowly rotated the compass around to get the circle. (iii) r = 3.5 cmStep 1: Market a point ‘O’ on the paper. Step 2: Extended the compass distance equal to radius 3.5 cm. Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass. Step 4: Slowly rotated the compass around to get the circle.(iv) r = 6.5 cmStep 1: Market a point ‘O’ on the paper. Step 2: Extended the compass distance equal to radius 6.5 cm. Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass. Step 4: Slowly rotated the compass around to get the circle. (v) d = 6 cm ∴ radius r = \(\frac { d }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3 cmStep 1: Market a point ‘O’ on the paper. Step 2: Extended the compass distance equal to radius 3 cm. Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass. Step 4: Slowly rotated the compass around to get the circle.
  34. 34. (i) z 2 – 16 z 2 – 16 = z 2 – 4 2 We have a 2 – b 2 = (a + b) (a – b) let a = z and b = 4, z 2 – 4 2 = (z + 4) (z – 4) (ii) 9 – 4y 2 9 – 4y 2 = 3 2 – 2 2 y 2 = 3 2 – (2y) 2 let a = 3 and b = 2y, then a 2 – b 2 = (a + b) (a – b) ∴ 3 2 – (2y) 2 = (3 + 2y) (3 – 2y) 9 – 4y 2 = (3 + 2y) (3 – 2y) (iii) 25a 2 – 49b 2 25a2 – 49b2 = 52 – a2 – 72 = (5a)2 – (7b)2 let A = 5a and B = 7b A 2 B 2 (5a) 2 – (7b) 2 = (5a + 7b) (5a – 7b) (iv) x 4 – y 4 Let x 4 – y 4 = (x 2 ) 2 – (y 2 ) 2 We have a 2 – b 2 = (a + b) (a – b) (x 2 ) 2 – (y 2 ) 2 = (x 2 + y 2 ) (x 2 – y 2 ) x 4 – y 4 = (x 2 + y 2 ) (x 2 – y 2 ) Again we have x 2 – y 2 = (x + y) (x – y) ∴ x 4 – y 4 = (x 2 + y 2 ) (x + y) (x – y)
  35. 35. (i) 8 m 30 cm in metres 8 m + \(\frac { 30 }{ 100 } \) m = 8 m + 0.30 m = 8.30 m (ii) 24 km 200 m in kilometres 24 km + \(\frac { 200 }{ 1000 } \) km = 24 km + 0.200 km = 24.200 km

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