7th Maths - Term 1 Exam 2024 - Original Question Paper with Solutions | Theni District
This post contains the fully solved question paper for the First Term Summative Examination 2024 for Standard VII Mathematics, as conducted in Theni District. Students can use this for reference and practice.
FIRST TERM SUMMATIVE EXAMINATION - 2024
Standard: VII
Subject: Mathematics
Time: 2.00 hrs
Marks: 60
I. Choose the correct answer (5 x 1 = 5)
1. (-10) + (+7) = ____
2. The area of a parallelogram whose base is 10m and height is 7m is ____
Solution: Area = base × height = 10 m × 7 m = 70 sq.m
3. The numerical co-efficient of -7mn is ____
4. The sum of all angles at a point is ____
5. If the cost of 3 books is Rs.90, then find the cost of 12 books.
Solution: Cost of 1 book = Rs.90 / 3 = Rs.30. Cost of 12 books = 12 × Rs.30 = Rs.360.
II. Fill in the blanks (5 x 1 = 5)
6. -44 + ____ = -88
7. The angle between the diagonals of a rhombus is ____.
8. The additive inverse of -37xyz is ____.
9. If the cost of 8 apples is Rs.56 then the cost of 12 apples is ____.
Solution: Cost of 1 apple = Rs.56 / 8 = Rs.7. Cost of 12 apples = 12 × Rs.7 = Rs.84.
10. A line which intersects two or more lines in different points is known as ____.
III. Match the following (5 x 1 = 5)
| Question | Correct Match |
|---|---|
| 11. 0 x 75 | 0 |
| 12. Area of the Trapezium | \(\frac{1}{2} \times h (a+b)\) |
| 13. Area of the parallelogram | bh |
| 14. x + 5 = 8 | 3 |
| 15. Right angle | 90° |
IV. True or False (5 x 1 = 5)
16. (-64) + (-64) = 0
17. If x is a natural number, then x+1 is its predecessor.
18. Area of rhombus is bh sq.units.
19. A tetromino is a shape obtained by 4 squares together.
20. An angle whose measure is exactly 180° is called a straight angle.
V. Answer any Ten questions (10 x 2 = 20)
21. Add the following (-3) and (-5) using number line.
1. Draw a number line.
2. Start at 0 and move 3 units to the left to represent -3.
3. From -3, move another 5 units to the left to represent adding -5.
4. You will land on -8. Thus, (-3) + (-5) = -8.
22. Find the product of (-2) x (+50) x (-25) x 4
(-2) × 50 × (-25) × 4
= (-100) × (-100)
= 10000
23. The product of two integers is -135. If one number is -15, Find the other integer.
Let the other number be x.
-15 × x = -135
x = -135 / -15
x = 9
24. One of the sides and the corresponding height of the parallelogram are 12m and 8m respectively. Find the area of the parallelogram.
Area of parallelogram = base × height
= 12 m × 8 m
= 96 sq.m
25. The area of a rhombus is 160sq.cm and the length of one of its diagonals is 8cm. Find the length of the other diagonal.
Area of rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
160 = \(\frac{1}{2} \times 8 \times d_2\)
160 = 4 × \(d_2\)
\(d_2\) = 160 / 4
\(d_2\) = 40 cm
26. Find the area of the trapezium whose height is 14cm and the parallel sides are 18cm and 9cm of length.
Area of trapezium = \(\frac{1}{2} \times h \times (a+b)\)
= \(\frac{1}{2} \times 14 \times (18+9)\)
= 7 × 27
= 189 sq.cm
27. If x=2 and y=3 then find the value of the following expressions i) x+y ii) x+1-y
i) x + y = 2 + 3 = 5
ii) x + 1 - y = 2 + 1 - 3 = 0
28. Subtract (m+n) from (3m-7n)
(3m - 7n) - (m + n)
= 3m - 7n - m - n
= 2m - 8n
29. A dozen bananas cost Rs.20, what is the price of 48 bananas?
1 dozen = 12 bananas.
Cost of 12 bananas = Rs.20.
Cost of 1 banana = Rs. 20 / 12.
Cost of 48 bananas = (20 / 12) × 48 = 20 × 4 = Rs.80.
30. Given that AB is a straight line. Calculate the value of x° in the following case.
Angles on a straight line sum to 180°.
72° + x° = 180°
x° = 180° - 72°
x° = 108°
So, x = 108.
31. Mention two real - life situations where we use parallel lines.
1. Railway tracks.
2. Opposite edges of a ruler, door, or window.
32. If 30 men can reap a field in 15 days, then in how many days can 20 men reap the same field.
This is a case of inverse proportion.
Total work = Men × Days = 30 × 15 = 450 man-days.
Days required for 20 men = Total work / Number of men
= 450 / 20 = 22.5 days.
33. Draw a tetromino which passes symmetry.
A tetromino is a shape made of four squares. A tetromino with symmetry is the 'I' shape (a 1x4 rectangle), which has two lines of symmetry.
[][][][]
VI. Answer any five (5 x 3 = 15)
34. Find all possible pairs of integers that give a product of -50.
The pairs are: (1, -50), (-1, 50), (2, -25), (-2, 25), (5, -10), (-5, 10).
35. If P=-15 and Q=5 find (P-Q) + (P+Q)
(P - Q) + (P + Q) = (-15 - 5) + (-15 + 5)
= (-20) + (-10)
= -30
36. The area of a trapezium is 1080sq.cm. If the length of its parallel sides are 55.6cm and 34.4 cm find the distance between them.
Area = \(\frac{1}{2} \times h \times (a+b)\)
1080 = \(\frac{1}{2} \times h \times (55.6 + 34.4)\)
1080 = \(\frac{1}{2} \times h \times (90)\)
1080 = 45 × h
h = 1080 / 45
h = 24 cm. The distance between them is 24 cm.
37. What should be added to 3x+6y to get 5x+8y?
Required expression = (5x + 8y) - (3x + 6y)
= 5x + 8y - 3x - 6y
= 2x + 2y
38. If the cost of 7kg of onions is Rs.84. Find the following.
(i) Weight of the onions bought for Rs.180.
(ii) The cost of 3kg of onions.
Cost of 1 kg of onions = Rs.84 / 7 = Rs.12.
(i) Weight bought for Rs.180 = Rs.180 / Rs.12 per kg = 15 kg.
(ii) Cost of 3 kg of onions = 3 × Rs.12 = Rs.36.
39. Find the value of x, y and z.
Angles on the straight line AD add up to 180°.
x + (3x + 40°) = 180°
4x + 40° = 180°
4x = 140°
x = 35°
Vertically opposite angles are equal.
y + 30° = 3x + 40°
y + 30° = 3(35) + 40° = 105° + 40° = 145°
y = 145° - 30°
y = 115°
Also, vertically opposite angles x and z+10° are equal.
x = z + 10°
35° = z + 10°
z = 35° - 10°
z = 25°
40. Find the shortest route to Vivekandar Memorial Hall from the Mandapam using the given map.
There are two possible routes:
Route 1: Mandapam → Pallivasal Island → Vivekandar Memorial Hall
Distance = 6 km + 2 km = 8 km.
Route 2: Mandapam → Krusadai Island → Vivekandar Memorial Hall
Distance = 7 km + 1.5 km = 8.5 km.
Comparing the distances, the shortest route is Route 1 with a distance of 8 km.
VII. Answer any one of the following (1 x 5 = 5)
41. a) Draw a line segment of AB=8cm and construct a perpendicular bisector.
- Draw a line segment AB of length 8 cm.
- With A as the center and a radius more than half of AB, draw arcs above and below the line segment AB.
- With B as the center and the same radius, draw two more arcs that intersect the previous arcs at points P and Q.
- Join P and Q. The line PQ is the perpendicular bisector of AB. It intersects AB at its midpoint M.
(OR)
b) Construct bisector of the ∠ABC with the measure 80°.
- Draw a ray BC.
- Using a protractor, with B as the vertex, draw an angle ∠ABC = 80°.
- With B as the center, draw an arc of any suitable radius that cuts the arms BA and BC at points E and D respectively.
- With D as the center and a radius more than half of DE, draw an arc in the interior of the angle.
- With E as the center and the same radius, draw another arc to intersect the previous arc at F.
- Join B and F and extend it. The ray BF is the required angle bisector of ∠ABC.