7th Standard Maths - Term 1 Exam 2024 (Thoothukudi) | Solutions
PART - A (10 x 1 = 10)
I. Choose the best answer:
1. $11 \times (-1)$ = ___
a) -1 b) 0 c) +1 d) -11
2. Which of the following does not represent an Integer?
a) $0 \div (-7)$ b) $20 \div (-4)$ c) $(-9) \div 3$ d) $12 \div 5$
3. In a parallelogram, if all the sides are equal then it is called
a) Rectangle b) Trapezium c) Rhombus d) Triangle
4. The area of a parallelogram whose base 12m and height 8m is
a) 69 sq.m b) 99 sq.m c) 66 sq.m d) 96 sq.m
5. An algebraic expression which is equivalent to the verbal statement, "Five times the sum of a and b" is
a) 5(a+b) b) 5 + a + b c) 5a+b d) a + 5b
6. When we subtract 'a' from '-a', we get
a) 0 b) 2a c) -2a d) -a
7. If Mani buys 5kg of potatoes for ₹75 then he can buy ___ kg of potatoes for ₹105.
a) 6 b) 7 c) 8 d) 5
8. 12 cows can graze a field for 10 days. 20 cows can graze the same field for ___ days.
a) 15 b) 18 c) 6 d) 8
9. Vertically opposite angles are ___
a) not equal in measure b) complementary c) supplementary d) equal in measure
10. A line which intersects two or more lines in different points is known as
a) parallel lines b) transversal c) non-parallel lines d) intersecting line
II. Fill in the blanks: (5 x 1 = 5)
11. $71 + \underline{\hspace{1cm}} = 0$
12. When the non-parallel sides of a trapezium are equal then it is known as
13. The additive inverse of -15xy is
14. The sum of all angles at a point is
15. A tetromino is a shape obtained by
III. Match the following: (5 x 1 = 5)
| Question | Correct Match |
|---|---|
| 16. $0 \div 3$ | 0 |
| 17. $(-3) + 3$ | 0 |
| 18. The numerical coefficient of 3xy | 3 (The OCR has '6' and '-1' as possible answers, but the correct answer is 3. Assuming a typo in the options provided in the image.) |
| 19. Area of a Rhombus | $\frac{1}{2} \times d_1 \times d_2$ |
| 20. Area of a Trapezium | $\frac{1}{2} h(a+b)$ |
PART - B (Answer any 10 questions) (10 x 2 = 20)
21. Find the value using number line 7 - (-10).
$7 - (-10) = 7 + 10 = 17$.
On a number line, starting at 7 and subtracting a negative number (-10) means moving 10 units to the right. So, we move from 7 to 17.
22. Find 4 pairs of integers that add up to 2.
Here are four possible pairs:
- (1, 1) $\implies 1 + 1 = 2$
- (0, 2) $\implies 0 + 2 = 2$
- (-1, 3) $\implies -1 + 3 = 2$
- (4, -2) $\implies 4 + (-2) = 2$
23. Prove that $(-7) \times (+8)$ is an integer and mention the property.
$(-7) \times (+8) = -56$.
Since -56 is an integer, the product of the two integers (-7 and 8) is also an integer.
Property: This demonstrates the Closure Property of Multiplication for Integers, which states that the product of any two integers is always an integer.
24. (-400) divided into 10 equal parts gives?
Dividing -400 into 10 equal parts is the same as calculating $(-400) \div 10$.
$(-400) \div 10 = -40$.
Each part is -40.
25. Find the height 'h' of the parallelogram whose area and base are 368 sq.cm and 23cm respectively.
Given: Area = 368 sq.cm, Base (b) = 23 cm.
Formula for the area of a parallelogram: Area = base × height.
$368 = 23 \times h$
$h = \frac{368}{23}$
$h = 16$ cm.
The height is 16 cm.
26. Find the area of Rhombus. (Given diagonals are 16cm and 8cm).
Given: Diagonal 1 ($d_1$) = 16 cm, Diagonal 2 ($d_2$) = 8 cm.
Formula for the area of a rhombus: Area = $\frac{1}{2} \times d_1 \times d_2$.
Area = $\frac{1}{2} \times 16 \times 8$
Area = $8 \times 8 = 64$ sq.cm.
The area of the rhombus is 64 sq.cm.
27. The parallel sides of a trapezium are 23cm and 12cm. The distance between the parallel sides is 9cm. Find the area of the trapezium.
Given: Parallel sides a = 23 cm, b = 12 cm. Height (h) = 9 cm.
Formula for the area of a trapezium: Area = $\frac{1}{2} \times h \times (a+b)$.
Area = $\frac{1}{2} \times 9 \times (23 + 12)$
Area = $\frac{1}{2} \times 9 \times 35$
Area = $\frac{315}{2} = 157.5$ sq.cm.
The area of the trapezium is 157.5 sq.cm.
28. Subtract 3m - 7n from m+n.
We need to calculate $(m+n) - (3m - 7n)$.
= $m + n - 3m + 7n$
= $(m - 3m) + (n + 7n)$
= $-2m + 8n$.
The result is -2m + 8n.
29. Solve: 2x + 10 = 30
$2x + 10 = 30$
$2x = 30 - 10$
$2x = 20$
$x = \frac{20}{2}$
$x = 10$.
The solution is x = 10.
30. Find the value of 3m + 2n given that m = 2 and n = -1.
Substitute m = 2 and n = -1 into the expression $3m + 2n$.
= $3(2) + 2(-1)$
= $6 - 2$
= 4.
The value is 4.
31. A dozen bananas costs ₹20. What is the price of 48 bananas?
A dozen = 12 bananas.
Cost of 12 bananas = ₹20.
Cost of 1 banana = ₹$\frac{20}{12}$.
Cost of 48 bananas = $\frac{20}{12} \times 48$
= $20 \times 4$ = ₹80.
The price of 48 bananas is ₹80.
32. Two angles are in the ratio 3:2. If they are linear pair, Find them.
Let the angles be $3x$ and $2x$.
A linear pair of angles adds up to 180°.
$3x + 2x = 180^\circ$
$5x = 180^\circ$
$x = \frac{180}{5} = 36^\circ$.
The angles are:
First angle = $3x = 3 \times 36 = 108^\circ$
Second angle = $2x = 2 \times 36 = 72^\circ$
The angles are 108° and 72°.
33. i) The angle that corresponds to $\angle 8$ is ___. ii) The angle that is alternate interior to $\angle 3$ is ___.
Based on the standard diagram of a transversal intersecting two parallel lines:
i) The angle corresponding to $\angle 8$ is $\angle 4$.
ii) The angle that is alternate interior to $\angle 3$ is $\angle 6$.
34. i) Rotation of Tetromino in 180° is ___. ii) Rotation of Tetromino in 360° is ___.
i) Rotation of a Tetromino by 180° results in a shape that looks the same as the original, but is oriented upside down (Point Symmetry).
ii) Rotation of a Tetromino by 360° results in the shape being in its original position.
PART - C (Answer any 5 questions) (5 x 3 = 15)
35. Add: i) $(-48) + (-15)$ ii) $0 + (-95)$ iii) $20 + (-72)$
i) $(-48) + (-15) = -48 - 15 = -63$
ii) $0 + (-95) = -95$
iii) $20 + (-72) = 20 - 72 = -52$
36. Prove that $[(-2) \times 3] \times (-4) = (-2) \times [3 \times (-4)]$.
This question demonstrates the Associative Property of Multiplication.
Left Hand Side (LHS):
$[(-2) \times 3] \times (-4)$
= $[-6] \times (-4)$
= 24
Right Hand Side (RHS):
$(-2) \times [3 \times (-4)]$
= $(-2) \times [-12]$
= 24
Since LHS = RHS (24 = 24), the property is proved.
37. A ground is in the shape of parallelogram. The height of the parallelogram is 14m and corresponding base is 8m longer than its height. Find the cost of levelling the ground at the rate of ₹15 per sq.m.
Given: Height (h) = 14 m.
Base (b) is 8m longer than height, so $b = 14 + 8 = 22$ m.
Area of the parallelogram = base × height
Area = $22 \times 14 = 308$ sq.m.
Cost of levelling = Area × Rate per sq.m
Cost = 308 × ₹15 = ₹4620.
The total cost of levelling the ground is ₹4620.
38. Simplify: $(x+y-z) + (3x - 5y + 7z) - (14x + 7y - 6z)$.
= $x + y - z + 3x - 5y + 7z - 14x - 7y + 6z$
Combine the 'x' terms: $x + 3x - 14x = 4x - 14x = -10x$
Combine the 'y' terms: $y - 5y - 7y = -4y - 7y = -11y$
Combine the 'z' terms: $-z + 7z + 6z = 6z + 6z = 12z$
The simplified expression is $-10x - 11y + 12z$.
39. 9 added to thrice a whole number gives 45. Find the number.
Let the whole number be 'x'.
Thrice the whole number is $3x$.
9 added to it gives 45. So, the equation is: $3x + 9 = 45$.
$3x = 45 - 9$
$3x = 36$
$x = \frac{36}{3} = 12$.
The number is 12.
40. 60 workers can spin a bale of cotton in 7 days. In how many days will 42 workers spin it?
This is an inverse proportion problem. (Fewer workers will take more days).
Let the number of days for 42 workers be 'x'.
Workers₁ × Days₁ = Workers₂ × Days₂
$60 \times 7 = 42 \times x$
$420 = 42x$
$x = \frac{420}{42} = 10$ days.
42 workers will take 10 days to spin the bale of cotton.
41. The angles at a point are $x^\circ, 2x^\circ, 3x^\circ, 4x^\circ, 5x^\circ$. Find the value of the largest angle.
The sum of all angles at a point is 360°.
$x^\circ + 2x^\circ + 3x^\circ + 4x^\circ + 5x^\circ = 360^\circ$
$15x = 360$
$x = \frac{360}{15} = 24$.
The largest angle is $5x^\circ$.
Largest angle = $5 \times 24 = 120^\circ$.
The value of the largest angle is 120°.
42. Observe the picture and answer the following:
i) Find all the possible routes from A to D.
ii) Find the shortest distance between E and C.
i) Possible routes from A to D:
- Route 1: A -> F -> G -> D (Distance: $100 + 300 + 200 = 600$m)
- Route 2: A -> B -> C -> D (Distance: $400 + 120 + 200 = 720$m)
- Route 3: A -> F -> E -> B -> C -> D (Distance: $100 + 250 + 100 + 120 + 200 = 770$m)
ii) Shortest distance between E and C:
- Route 1: E -> B -> C (Distance: $100 + 120 = 220$m)
- Route 2: E -> F -> G -> D -> C (Distance: $250 + 300 + 200 + 200 = 950$m)
The shortest distance between E and C is via point B, which is 220m.
PART - D (Answer any 1 of the following) (1 x 5 = 5)
43. Construct bisector of the $\angle ABC$ with the measure 80°.
Steps of Construction:
- Draw a ray BC.
- Using a protractor, place its center at point B and the baseline along BC. Measure 80° and mark a point A.
- Draw the ray BA. Now $\angle ABC = 80^\circ$.
- With B as the center and any convenient radius, draw an arc that intersects the rays BA and BC at points, say P and Q, respectively.
- With P as the center and a radius more than half of PQ, draw an arc inside the angle.
- With Q as the center and the same radius, draw another arc that intersects the previous arc at a point, say D.
- Join B and D and extend it. The ray BD is the required angle bisector of $\angle ABC$.
- On measuring, you will find that $\angle ABD = \angle DBC = 40^\circ$.
44. Construct a perpendicular bisector of the line segment CD of length 10.4cm.
Steps of Construction:
- Draw a line segment CD of length 10.4 cm using a ruler.
- With C as the center and a radius more than half the length of CD (e.g., more than 5.2 cm), draw arcs on both sides of the line segment CD.
- With D as the center and the same radius, draw arcs that intersect the previous arcs at two points, say P and Q.
- Join the points P and Q with a straight line using a ruler.
- The line PQ is the perpendicular bisector of the line segment CD. It intersects CD at a point M.
- By construction, PQ is perpendicular to CD, and CM = MD = 5.2 cm.