7th Maths Question

 👉Malayalam

Chapter 1: Parallel Lines

Main Concepts
This chapter introduces the fundamental properties of parallel lines and the angles formed when lines intersect them.

• Definition of Parallel Lines: Lines that do not meet and maintain the same distance between them. They can be drawn using a scale and a set square or by drawing two lines at the same slant to a given line.
• Angles Formed by Intersecting Lines: When one line crosses another, four angles are formed.
  • The two small angles are of the same measure.
  • The two large angles are of the same measure.
  • The sum of a small angle and a large angle is 180°.
• Angles Formed by a Transversal Line Intersecting Two Parallel Lines: When a line intersects two parallel lines, eight angles are formed.
  • Corresponding Angles: Pairs of angles in the same position relative to the transversal and the parallel lines (e.g., top-right, bottom-right). Angles in each such pair measure the same.
  • Alternate Angles: Pairs of angles in opposite positions (e.g., top-right and bottom-left). Angles in each such pair measure the same.
  • Co-interior Angles: Pairs of interior angles on the same side of the transversal. Their sum is 180°.
  • Co-exterior Angles: Pairs of exterior angles on the same side of the transversal. Their sum is 180°.
• Angles in a Parallelogram: A parallelogram is a quadrilateral where opposite sides are parallel. The properties of parallel lines apply to its angles.
• Triangle Sum Property: An important application of parallel lines is demonstrating that the sum of all angles of a triangle is 180°. If one angle of a triangle is subtracted from 180°, the result is the sum of the other two angles.

Types of Problems & Approach
The problems in this chapter primarily involve calculating unknown angles using the properties of parallel lines and the triangle sum property. You will often be given one angle and asked to find others.

• Calculating Angles: Apply the relationships between small/large angles, corresponding, alternate, co-interior, and co-exterior angles.
• Parallelogram Angles: Use the fact that opposite sides are parallel to identify angle relationships.
• Triangle Angles: Use the 180° sum property.

Example Problem & Solution Method

(1) "In each of the pictures below, can you calculate the other seven angles which the parallel blue lines make with the green line?" (Assuming one angle is given as 50°)

• Approach: Identify the type of angle given and use the relationships to find the others.
  • If a small angle is 50°:
    • The large angle adjacent to it on the same intersection point will be 180° - 50° = 130°.
    • The small angle vertically opposite to it will also be 50°.
    • The large angle vertically opposite to the 130° angle will also be 130°.
  • Now, consider the second intersection point (with the other parallel line):
    • The corresponding small angle (same position) will also be 50°.
    • The alternate small angle (opposite position) will also be 50°.
    • Using these, you can find the remaining large angles at this intersection point (180° - 50° = 130°).
    • Alternatively, the co-interior angle to the given 50° would be 180° - 50° = 130°.

(2) "One angle of a right triangle is 40°. What is the measure of the angle other than the right angle?"

• Approach: A right triangle has one angle that is 90°. The sum of all angles in a triangle is 180°.
• Method:
  1. You know two angles: 90° (right angle) and 40° (given).
  2. Sum of these two angles = 90° + 40° = 130°.
  3. The third angle = 180° (total sum) - 130° (sum of two angles) = 50°.
  • Alternative Method: The sum of the two non-right angles in a right triangle is 180° - 90° = 90°. So, the other angle = 90° - 40° = 50°.

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Chapter 2: Fractions

Main Concepts
This chapter builds on basic fraction concepts to cover multiplication involving fractions and applying these to various real-world scenarios.

• Multiplication (Number × Fraction): Multiplying a whole number by a fraction means adding the fraction that many times. For example, 4 × (1/2) = (1/2) + (1/2) + (1/2) + (1/2) = 2. Generally, N × (a/b) = (N × a) / b.
• Share and Fraction (Division): Division can be expressed as a fraction. If 4 litres are divided among 3 persons, each gets 4/3 litres.
• Part and Multiplication (Fraction of a Number): Finding a fraction of a number is also a multiplication operation. For example, (1/2) of 6 is 3, which can be written as (1/2) × 6 = 3. Generally, (a/b) of N = (a/b) × N = (a × N) / b.
• Part of a Part (Multiplication of Fractions): To find a fraction of another fraction, you multiply the numerators and the denominators. For example, (1/3) of (1/2) is (1/3) × (1/2) = 1/6. Generally, (a/b) × (c/d) = (a × c) / (b × d).
• Mixed Fractions in Multiplication: Convert mixed fractions to improper fractions before multiplying.

Types of Problems & Approach
Problems require performing arithmetic operations with fractions, often in word problem contexts involving quantities like weight, length, volume, or area.

• Direct Calculation: Multiply whole numbers by fractions, fractions by fractions.
• Word Problems: Translate the problem into a fractional multiplication or division, then calculate. Pay attention to "times," "of," "part," and "share."
• Area Problems: Apply the concept that the area of a rectangle is length × breadth, even when dimensions are fractions.

Example Problem & Solution Method

(1) "What is the total length of four pieces of ribbons, each of length three fourths of a metre?"

• Approach: This is a direct multiplication of a whole number (4 pieces) by a fraction (3/4 metre per piece).
• Method:
  • Total length = 4 × (3/4) metres.
  • 4 × (3/4) = (4 × 3) / 4 = 12 / 4 = 3 metres.

(2) "We have to cut off 3/5 of a 7 metre long string. How long is this piece?"

• Approach: This is finding a "part of" a whole, which is a fractional multiplication.
• Method:
  • Length of piece = (3/5) of 7 metres = (3/5) × 7.
  • (3/5) × 7 = (3 × 7) / 5 = 21 / 5.
  • Convert the improper fraction to a mixed number: 21 / 5 = 4 1/5 metres.
  • (This can also be expressed as 4 meters and 20 centimeters, as 1/5 of a meter is 20 cm).

(3) "What is the area of a rectangle of length 2 1/5 centimetres and breadth 3 1/3 centimetres?"

• Approach: Convert mixed fractions to improper fractions, then multiply them.
• Method:
  1. Convert 2 1/5 to an improper fraction: (5 × 2 + 1) / 5 = 11/5.
  2. Convert 3 1/3 to an improper fraction: (3 × 3 + 1) / 3 = 10/3.
  3. Area = (11/5) × (10/3).
  4. Multiply numerators and denominators: (11 × 10) / (5 × 3) = 110 / 15.
  5. Simplify the fraction by dividing by 5: 22 / 3.
  6. Convert to a mixed number: 22 / 3 = 7 1/3 square centimetres.

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Chapter 3: Triangles

Main Concepts
This chapter focuses on the geometric construction of triangles and the fundamental properties related to their sides and angles.

• Drawing Triangles with Given Side Lengths: Use a compass and ruler. Draw one side, then use the given lengths as radii to draw arcs from each end of the first side. The intersection point is the third vertex.
  • Equilateral Triangle: A triangle where all three sides are equal.
• Relationship Between Angles and Opposite Sides: In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. The angles and their opposite sides have sizes in the same order.
• Triangle Inequality Theorem: For three lengths to form a triangle, the length of the greatest side must be less than the sum of the lengths of the other two sides. Equivalently, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
• Drawing Triangles with Specified Angles: Knowing two angles means the third is also known (sum is 180°). To draw a unique triangle, you need a side length and the two angles on that side.
• Drawing Triangles with Two Sides and the Angle Between Them: If two side lengths and the included angle are specified, the triangle is uniquely determined.
• Drawing Triangles with Two Sides and an Angle NOT Between Them: This can sometimes lead to two possible triangles, one unique triangle (e.g., right-angled case), or no triangle at all.

Types of Problems & Approach
Problems involve determining if a triangle can be formed from given side lengths, drawing triangles with specific properties, and applying the relationship between sides and angles.

• Feasibility of Triangle Construction: Use the triangle inequality theorem.
• Angle-Side Relationship: Identify the largest/smallest angle and its opposite side.
• Construction Challenges: Visualize or attempt to draw triangles based on various given conditions.

Example Problem & Solution Method

(1) "The sides of a triangle are natural numbers. If the lengths of two sides are 5 centimetres and 8 centimetres, what are the possible numbers which can be the length of the third side?"

• Approach: Apply the triangle inequality theorem. Let the unknown third side be 'x'.
• Method:
  1. Sum of any two sides > third side:
     • 5 + 8 > x => 13 > x (or x < 13)
     • 5 + x > 8 => x > 8 - 5 => x > 3
     • 8 + x > 5 (This is always true since x must be a positive length).
  2. Combining the inequalities: x must be greater than 3 and less than 13.
  3. Since 'x' must be a natural number, the possible lengths for the third side are 4, 5, 6, 7, 8, 9, 10, 11, and 12 centimetres.

(2) "Which of the following sets of three lengths can be used to draw a triangle? (i) 4, 6, 10 cm (ii) 3, 4, 5 cm (iii) 10, 5, 4 cm"

• Approach: Apply the triangle inequality theorem: "the length of the greatest side is less than the sum of the length of the other two sides".
• Method:
  • (i) 4 cm, 6 cm, 10 cm:
    • Greatest side is 10 cm. Sum of other two is 4 + 6 = 10 cm.
    • Is 10 < 10? No. Cannot form a triangle.
  • (ii) 3 cm, 4 cm, 5 cm:
    • Greatest side is 5 cm. Sum of other two is 3 + 4 = 7 cm.
    • Is 5 < 7? Yes. Can form a triangle.
  • (iii) 10 cm, 5 cm, 4 cm:
    • Greatest side is 10 cm. Sum of other two is 5 + 4 = 9 cm.
    • Is 10 < 9? No. Cannot form a triangle.

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Chapter 4: Reciprocals

Main Concepts
This chapter introduces the concept of reciprocals as a way to "reverse" the relationship between numbers expressed as "times" and "part," and then extends this to define division of fractions.

• Times and Part: Quantities can be compared by saying one is "X times" the other, or one is "1/X part of" the other. For example, 8 is 4 times 2, and 2 is 1/4 of 8.
• Reciprocals: Fractions obtained by "turning the fraction upside down" (interchanging numerator and denominator) are called reciprocals. For example, 2/3 and 3/2 are reciprocals.
• Using Reciprocals to Reverse Relationships: If Quantity A is (X/Y) times Quantity B, then Quantity B is (Y/X) part of Quantity A.
• Fraction Division: Division by a number is equivalent to multiplication by its reciprocal. For example, (1/2) ÷ (3/4) = (1/2) × (4/3).

Types of Problems & Approach
Problems involve converting "times" relationships to "part" relationships and vice versa, and performing division with fractions by converting them into multiplication by the reciprocal.

• Comparison Problems: Express one quantity as a fraction/multiple of another.
• Inverse Problems: If A is a fraction of B, find B from A using reciprocals.
• Direct Division Problems: Apply the rule of multiplying by the reciprocal.

Example Problem & Solution Method

(1) "The price of 1 1/2 kilograms of tomato is 30 rupees. What is the price of one kilogram tomato?"

• Approach: Recognize that 30 rupees is "1 1/2 times" the price of one kilogram. To find the price of one kilogram, you need to find "1 / (1 1/2) part of" 30 rupees, which means multiplying by the reciprocal.
• Method:
  1. Convert the mixed fraction to an improper fraction: 1 1/2 = 3/2.
  2. The price of 3/2 kg is 30 rupees.
  3. To find the price of 1 kg, multiply 30 by the reciprocal of 3/2, which is 2/3.
  4. Price of 1 kg = 30 × (2/3) = (30 × 2) / 3 = 60 / 3 = 20 rupees.

(2) "The area of a rectangle is 1/2 square metre and the length of one side is 3/4 metre; What is the length of the other side?"

• Approach: Area = Length × Width. So, Width = Area / Length. This is a division problem.
• Method:
  1. Width = (1/2) ÷ (3/4).
  2. To divide by a fraction, multiply by its reciprocal. The reciprocal of 3/4 is 4/3.
  3. Width = (1/2) × (4/3).
  4. Multiply numerators and denominators: (1 × 4) / (2 × 3) = 4 / 6.
  5. Simplify the fraction: 4 / 6 = 2/3 metres.

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Chapter 5: Decimal Methods

Main Concepts
This chapter details operations with decimal numbers, emphasizing their relationship with fractions (specifically those with powers of 10 as denominators).

• Decimal Forms (Fractions to Decimals): Fractions with denominators of 10, 100, 1000, etc., can be easily written as decimals by shifting the decimal point. For example, 43/10 = 4.3.
• Decimal Place Value: Each shift of the decimal point to the right represents multiplication by 10, and to the left, division by 10.
• Fractional Forms (Decimals to Fractions): Any decimal can be written as a fraction with a denominator of 10, 100, or 1000, depending on the number of decimal places. For example, 327.45 = 32745/100.
• Multiplication of Decimals by Whole Numbers: Convert the decimal to a fraction, multiply by the whole number, then convert back to a decimal. For example, 3 × 4.25 = 3 × (425/100) = 1275/100 = 12.75.
• Multiplication of Decimals by Decimals: Convert both decimals to fractions, multiply the fractions, then convert the resulting fraction back to a decimal. The number of decimal places in the product is the sum of the decimal places in the numbers being multiplied. For example, 8.5 × 6.5 = (85/10) × (65/10) = 5525/100 = 55.25.
• Division of Decimals by Whole Numbers: Convert the decimal to a fraction, divide by the whole number, then convert back to a decimal. For example, 10.4 ÷ 2 = (104/10) ÷ 2 = (104/10) × (1/2) = 5.2.
• Division of Decimals by Decimals: Convert both decimals to fractions, then use the reciprocal method for fraction division. For example, 5.25 ÷ 0.75 = (525/100) ÷ (75/100) = (525/100) × (100/75) = 7.

Types of Problems & Approach
This chapter includes problems for converting between fractions and decimals, performing arithmetic operations with decimals, and solving word problems related to measurement.

• Conversions: Practice shifting decimal points and writing decimals as fractions.
• Arithmetic Operations: Apply the methods (conversion to fractions or rules of decimal places) for multiplication and division.
• Word Problems: Set up the calculation based on the problem statement and perform it using decimal methods.

Example Problem & Solution Method

(1) "Find the area in square metres of a rectangle of length 6.25 metres and width 4.2 metres."

• Approach: Area = Length × Width. Perform decimal multiplication.
• Method (Fraction Conversion):
  1. Convert to fractions: 6.25 = 625/100 and 4.2 = 42/10.
  2. Multiply: (625/100) × (42/10) = (625 × 42) / (100 × 10) = 26250 / 1000.
  3. Convert back to decimal: 26250 / 1000 = 26.25 square metres.
• Method (Decimal Place Rule):
  1. Multiply 625 by 42 (ignoring decimal points): 625 × 42 = 26250.
  2. 6.25 has 2 decimal places and 4.2 has 1 decimal place, for a total of 3.
  3. Place the decimal point in the product to have 3 places: 26.250 or 26.25.

(2) "A vessel contains 4.05 litres of coconut oil. It is to be used to fill 0.45 litre bottles. How many bottles are needed?"

• Approach: This is a division problem: Total volume / Volume per bottle.
• Method (Fraction Conversion):
  1. Number of bottles = 4.05 ÷ 0.45.
  2. Convert to fractions: (405/100) ÷ (45/100).
  3. Divide using reciprocals: (405/100) × (100/45).
  4. Cancel the 100s and divide: 405 / 45 = 9 bottles.
• Method (Shifting Decimals):
  1. Since both numbers have two decimal places, the division is equivalent to dividing the whole numbers: 405 ÷ 45 = 9 bottles.

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Chapter 6: Ratio

Main Concepts
This chapter introduces the concept of ratio as a comparison between two quantities, emphasizing simplification and application.

• Definition of Ratio: A comparison of two quantities, often expressed as "A to B" or A : B.
• Simplifying Ratios: Ratios are usually expressed using the smallest possible natural numbers. To simplify, divide both parts of the ratio by their greatest common divisor. For example, 15:25 simplifies to 3:5.
• Aspect Ratio: In rectangles, the ratio of height to width (or width to height) is called the aspect ratio. Maintaining this ratio is crucial for scaling images without distortion.
• Ratios for Any Measures: Ratios can be used to compare any two measures, such as capacities, counts, weights, etc.
• Ratio of Mixtures: Ratios are used in mixing ingredients to ensure consistent quality. If a ratio is A:B, and quantity A is scaled, quantity B must be scaled by the same factor.
• Division Problems with Ratios: When a total quantity is to be divided in a given ratio (e.g., A:B), the total parts are A+B. Each person gets their share as a fraction of the total.

Types of Problems & Approach
Problems involve finding, simplifying, and applying ratios in diverse scenarios, including geometric shapes, mixtures, and sharing.

• Direct Ratio Calculation: Given two quantities, find their ratio and simplify it.
• Scaling Quantities with Ratios: Given a ratio and one quantity, find the corresponding other quantity.
• Dividing a Total by a Ratio: Given a total amount and a ratio, distribute the amount accordingly.

Example Problem & Solution Method

(1) "Write down the ratio of the height to width of a rectangle with height 20 centimetres and width 1 metre."

• Approach: First, ensure both quantities are in the same unit. Then, express their comparison as a ratio and simplify.
• Method:
  1. Convert 1 metre to centimetres: 1 metre = 100 centimetres.
  2. The height is 20 cm, the width is 100 cm.
  3. Ratio of height to width = 20 : 100.
  4. Simplify the ratio by dividing both sides by 20.
  5. The simplified ratio is 1 : 5.

(2) "To plaster the walls of a house, cement and sand are mixed in the ratio 1 : 5. If 45 sacks of cement were bought, how many sacks of sand are needed?"

• Approach: The ratio means that for every 1 part of cement, there are 5 parts of sand. Determine the scaling factor for cement and apply it to sand.
• Method:
  1. The ratio of cement to sand is 1 : 5.
  2. 45 sacks of cement correspond to the '1' part of the ratio.
  3. The scaling factor is 45.
  4. To find the amount of sand, multiply the '5' part of the ratio by the same scaling factor: 5 × 45 = 225 sacks of sand.

(3) "Suhara and Sita started a business. Suhara invested 40,000 rupees and Sita 50,000 rupees. They made a profit of 9,000 rupees, divided in the ratio of their investments. How much did each get?"

• Approach: First, find and simplify the ratio of their investments. Then, divide the total profit according to this ratio.
• Method:
  1. Find the ratio of investments: Suhara : Sita = 40,000 : 50,000.
  2. Simplify by dividing both by 10,000: 4 : 5.
  3. The total parts in the ratio are 4 + 5 = 9 parts.
  4. Each part represents: 9,000 rupees ÷ 9 parts = 1,000 rupees/part.
  5. Suhara's share (4 parts) = 4 × 1,000 = 4,0