CBSE Class 7 Maths Sample Paper PDF

CBSE BOARD Mathematics Sample Paper Class VII General Instructions 1. The paper is divided into 4 sections; A, B, C, and D 2. Section A consists of 10 ques...

CBSE BOARD Mathematics Sample Paper Class VII General Instructions 1. The paper is divided into 4 sections; A, B, C, and D 2. Section A consists of 10 questions of 1 mark each. 3. Section B consists of 10 questions of 2 marks each. 4. Section C consists of 10 questions of 3 marks each. 5. Section D consists of 8 questions of 5 marks each. SECTION A (1 Mark Questions) 1. Find the value of 5  6   2   3   5. Ans. The given expression 5  6   2   3   5 can be written as  5   6  2  3  5  5  (12)  60 2. If we subtract 5 from 2 we will move _________ on the number line. Ans. If we subtract 5 from 2 we get 2  (5)  2  5  3 , which is more than 2 so it will lie at its right. Hence, if we subtract 5 from 2 we will move towards right on the number line. 3. The absolute value of  8 is ____________. 1 Ans.  8 can be written as 1 8 that implies 1 8  8. The absolute value of  8 is 8. 4. Write the fraction for the unshaded region in the given figure Ans. The total number of boxes in the given figure are 6, out of which 2 boxes are not shaded. So, the fraction for unshaded region is: 2 1  6 3. 5. A helicopter is flying in the sky exactly over the submarine. The helicopter is at a height of 6000m above the sea level and the submarine is 1000m below the sea level. Find the vertical distance between the helicopter and the submarine. Ans. The vertical distance between the helicopter and submarine is 7000m. 1 6. Sushant reads 1/5th of a book in an hour and he reads the book for 2 hrs. What part of 3 the book is left to be read by him. 1 Ans. Part of the book read in an hour = 5 1 1 1 17 7 Part of book read in 2 hrs =  2      3 5  3  5  3  15 7 15  7 8 Part of book left to be read by him = 1    15 15 15 7. Find the average of 4.2, 3.8, 7.6 and 3.4. Ans. The average of the given numbers 4.2, 3.8, 7.6 and 3.4 is 2  4.2  3.8  7.6  3.4  4 19  4  4.75 8. Find the value of ‘l’ in the given equation: 3l  5  47 Ans. 3l  5  47 3l  5  5  47  5 3l  42 3l 42  3 3 l  14 9. Find the complement of the angle given below Ans. Complementary angles are the angles whose sum is 900. The given ABC  25 Therefore, the complement angle is of = 90  25  65 10. Find the exterior angle in the given figure: 3 Ans. The exterior angle is the sum of the opposite angles. ACD  ABC  BAC ∴ ACD  60  40 ACD  100 SECTION B (2 Marks Questions) 11. Find the median and range of the given data: - 24,36,40,30,35,20,38 Ans. For finding the median, at first we will arrange the data in increasing order as below: 20, 24, 30, 35, 36, 38, 40  n 1  th Median =   item  2  Where n = No. of items = 7  7 1  th  item  4 item  35 th Therefore, Median =   2  Range  L  S Where L = 40 and S = 20 Therefore, Range  40  20  20 12. Find the circumference and area of a circle whose diameter is 28 cm. 4 Ans. Diameter =28 cm 1 1 Radius = diameter = x 28 = 14 cm 2 2 22 Circumference of a circle  2 r  2  14  88cm 7 22 Area of circle   r ²   14 14  616 cm² 7 13. A Circular frame of diameter 16 cm is to be polished @ Rs 5/cm². Find the cost of polishing the frame? 16 Ans. Radius of frame   8 cm 2 22 Area of frame   r ²   8  8  201 cm² 7 Cost of polishing = 201 5  Rs1005 14. In a circular park of radius 20m there is a circular garden of radius 5m. Find the area of park excluding the garden. Ans. Let R be radius of park 20m And r be radius of garden 5m Area of park secluding garden 5    R²  r ²  22 =  20  ²   5  ²  7  22 = x  400  25   7  22 = x375  1178m² 7 15. The measurement of a rectangular box is 7 cm by 10 cm. Find its a) Area b) Perimeter Ans. a) Area of Rectangle = L× B = 7× 10 70 = cm ² b) Perimeter of Rectangle = 2 (L+ B) = 2 ( 7 +10 ) = 2(17) = 34 cm 16. The area of a rectangular park, which is 40 m long is same as the area of square park which is 60m wide. Find the perimeter of the rectangular park. Ans. Area of Square park = Side × Side 6  60  60  3600 m² Area of Rectangular park =Area of square park Length  width  3600m² 40 m  w  3600m² 3600 w   90 40 Perimeter of rectangular park = 2(L+B) = 2 (40 +90) = 2 (130) = 260 m 17. A window of length 3m and width 2m is fitted in a wall whose length is 6m and width is 5m. Find the cost of while washing the wall @Rs 25/m². Ans. Area of wall with window  6 m  5m  30m ² Area of window  3m  2 m  6 m² Area of wall to be white washed = 30m² − 6m² = 24 m² Cost of white washing the wall =24 ×25 = Rs 600. 18. Simplify by using law exponents 7 a 5  a 6  a3  a 2    a 2  a3  a3  Ans. Using the rule a m  a n  a m  n a 5  a 6  a 3  a 2  a 5 63 2  a16 a 2  a 3  a 3  a 2  3 3  a 8 Using the rule a m  a n  a m n a16  a8  a168  a 8 19. Express 108 × 270 as a product of prime factors is exponential form. Ans. Prime Factors of 180 = 2  2  3 3 3  2²  3³ Prime Factors of 270  2  3 3 3 5  2  3³  5 Therefore, 108  270  2²  3³  2  3³  5  23365 3  2 5 24 20. Simplify:- × 4  9 4 3 Ans. 3  x  2 2 4 5 310 24  x  4  9  2  3  4 3 4 3 2 2 310 24 3106 34 = x  8 4  4 28 36 2 2 81 = 16 SECTION C 8  (3 Marks Questions) 21. For the given values of a & b, verify that a   b   a  b a =16 b=15 Ans. a   b   16 –  15   16  15  31 a  b  16  15  31  a –   b  a  b 22. Find the product using suitable properties 15   18   15  13   5  3  2  3  5  4 Ans. The given repression 15   18  15  13  5  3  2  3  5  4 Can be written as  15   18  15 13  15  2   15  4  Using distributive property  15  18  13  2  4   15  22  15   15  7    105 23. In an entrance examination containing 50 questions, 4 marks are awarded for every correct answer and (−1) mark for every wrong answer and 0 mark for unattempted 9 questions. Ajit attempted only 40 questions of which 25 were correct and 15 were wrong. Find his score. Ans. Marks for correct answers =4×25=100 Marks for wrong answers =1×15 =15 Marks for unattempted questions = 0× 10 =0 Total marks scored by Ajit =100 −15 + 0 =85 24. Solve 3 7 2 3 + + + 10 15 5 20 Ans. To make the denominators same, we will take the L.C.M of 10, 15,5,20 L.C.M of 10, 15, 5 and 20 =120 3 7 2 3 + + + 10 15 5 20 can be written as 3 x12 7 x6 2 x 24 3 x6 + + + 120 120 120 120 36 56 48 18 158 79 19 + + + = = =1 120 120 120 120 120 120 60 3 25. Waheeda has a bottle of 2 liters of cold drinks; she drank th of it and gave 4 remaining to Jharna. How much cold drinks did Waheeda drink and what part did she give to jharan ? Ans. Total Quantity of cold drink Waheeda had 2liters. 10  3 3 3 She drank th of it, so she drank=2× = =15 liters. 5 4 2 3 43 1 She Gove to Jharna = 1  = = th part of cold drinks. 4 4 4 26. Find the value of missing angle “x” in the figure given below: Ans. ABE  EBD  DBC  180 30  x  25  180 55  x  180 x  180  55  125 27. Find the value of:  25  210    28  24  Ans. 1 Using the rule a m  a n  a m n and a  m  am 2 5  210   2510  25  1 25 28  24   284  24 11 2 5  210    28  24   1 2 5  24 1 24 2  5 4 25 2 am 1 Using the rule n  a m n and a  m  m a a 24 1 5  245  21  2 2 28. Find the value of: 3 p 2  2 p  1 for p  3 Ans. The given expression is 3 p 2  2 p  1 Substituting p=3, we get 3 p2  2 p 1  3(3) 2  2(3)  1  3(9)  2(3)  1  27  6  1  32 29. Find the radius and area of the circular sheet whose circumference is 157cm. (𝛑=3.14) Ans. The circumference of a circle is = 2 r 2 r  157 2  3.14  r  157 6.28r  157 r  25cm The area of a circle is =  r 2  r 2  3.14  25  25  1962.5 12 30. ABC is a right angle triangle as given below: Find BC. Ans. Using Pythagoras theorem, we get AC 2  AB 2  BC 2 (25) 2  (7) 2  BC 2 625  49  BC 2 625  49  BC 2 576  BC 2 BC  576  24 SECTION D (5 Marks Questions) 31. Find the area of a quadrilateral ABCD given below AC=11m, BM=1.5m, DN=1.5m 13 Ans. It is given that: AC=11m, BM=1.5m, DN=1.5m Area of Quadrilateral = Area of ∆ ABC + Area of ∆ ADC Area of triangle 1 = × Base × height 2 Area of ∆ ABC 22  111.5  8.25m² 7 Area of ∆ ADC 22  111.5  8.25m² 7 Area of ABCD  8.25  8.25  16.50m² 32. A Square park of 700m length has two cross roads of 20m width; find the area of the park excluding the cross roads. Ans. Area of park ABCD 14  Side  Side  700  700  490000m² Area of IJKL  L B  700  20  14000m² Area of EFGH  L B  700  20  14000m² Area of MNOP  side  side  20  20  400m² Area of Cross Road  Area of IJKL  Area of EFGH  Area of MNOP  14000  14000  400  27600m² Area of Park excluding the Cross roads  Area of ABCD – Area of cross roads  490000  27600  462, 400m² 15 33. Find the area of the shaded region. Ans. Area of ABCD will be =L×B Where, L=length=15m B=width =10m Area of ABCD  15  10  150 m ² For EFGH The length will be =15+3+3=21m And width will be = 10+3+3=16m Therefore, Area of EFGH  L B  21 16  336 m² So Area of the Shaded region will be 16  Area of EFGH  Area of ABCD  336 m²  150 m²  186 m² 34. Find the mean of the given data:- Marks 0-10 10-20 20-30 30-40 40-50 Total No. of 2 4 18 12 4 40 Students Ans. To find the mean of the given data, we will reduce it in a tabular form Marks No. of Students (f) Mid Value (X) fX 0-10 2 5 10 10-20 4 15 60 20-30 18 25 450 30-40 12 35 420 40-50 4 45 180 40 1120  fX Mean = f 1120  40  28 35. ∆ ABCD is an isosceles triangle where AB=AC=8 cm and BC =10 cm.The height of AD from A to BC is 7 cm. Find the area of the given triangle and also find the length of BE drawn from B to AC. Ans. The above figure will be. 17 Area of triangle ABC 1 = XBase × Height 2 1 = × BC × AD 2 1 = × 10× 7 2 =35 cm² Now to find the height of BE we will take AC as the base. 1 Area of triangle = × Base × Height 2 1 35   8  h 2 35  4  h 35 h  4 h  8.75 cm 36. In the last two overs of a cricket match 30 runs were required to win the match. Rahul and Jadeja were batting, Rahul scored double than Jadeja in the last two overs but they lost the match by 3 runs. Find the score of the two batsman in the last two overs. 18 Ans. Let the score of Jadeja in last two overs be x So, the score of Rahul will be 2x They lost match by 3 runs and for winning the runs required were 30 ∴ The score of the team = 30  3  27runs Hence, Rahul and Jadega scored 27 runs in the last 2 overs, it can be written as: 2 x  x  27 3 x  27 27 x 3 x9 Therefore, the runs scored in last two overs by Jadega = 9 Rahul = 18 3  2  3  7  5 37. Find the value of:     4  3  2  12  6 Ans. 3  2  3  11  5 The given expression is     4  3  2  12  6 The L.C.M of 4, 3, 2, 12, and 6 = 12 3  2  3  11  5 So,      can be written as: 4  3  2  12  6 19 3  2  3  11  5     4  3  2  12  6  3  3   2  4   3  6   11 1   5  2         12   12   12   12   12  9  8   18   11   10        12  12   12   12   12  9 8 18 11 10      12 12 12 12 12 9  8  18  11  10  12 22  18  12 4  12 1  3 38. In the given figure, find the value of the missing angles, given that the base angles are equal. Ans. The sum of all the angles of a triangle = 1800 It is given that: Q  R  x P  50 P  Q  R  180 50  x  x  180 Subtracting 50 from both sides 20 50  x  x  50  180  50 2 x  130 Dividing both sides by 2, we get 2 x 130  2 2 x  65 Therefore, Q  R  65 21