CBSE Class 8 Mathematics Ganita Prakash Term 1 Sample Paper with Answers

This is the CBSE Class 8 Mathematics Ganita Prakash Term 1 Sample Paper 2025-26, a 60-mark paper of 29 questions in Sections A to E, given here with step-by-step answers. It covers A Square and A Cube (squares, square roots, cubes, cube roots and Pythagorean triplets) and Quadrilaterals (angle sum, interior and exterior angles of polygons and properties of parallelograms), together with linear equations in one variable, including case studies on taxi fares and fencing a square field.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Solve: \( 2y + \frac{5}{3} = \frac{26}{3} - y \)
Answer:
Bring the y terms to the left and the numbers to the right:
2y + y = \( \frac{26}{3} \) − \( \frac{5}{3} \).
3y = \( \frac{21}{3} \) = 7.
y = \( \frac{7}{3} \).
Check: LHS = 2 × \( \frac{7}{3} \) + \( \frac{5}{3} \) = \( \frac{19}{3} \); RHS = \( \frac{26}{3} \) − \( \frac{7}{3} \) = \( \frac{19}{3} \).
So y = \( \frac{7}{3} \).
In simple words: Collect the y terms on one side to get 3y = 7, so y = 7/3.

Teacher's Note:
1. When a term crosses the equals sign, its sign changes.
2. Fractions with the same denominator are subtracted by subtracting numerators.

Question 17. How many sides does a regular polygon have, if each of its interior angles is 165°?
Answer:
Let the polygon have n sides.
Each exterior angle = 180° − 165° = 15°.
The exterior angles add up to 360°, and in a regular polygon they are all equal.
n = 360° ÷ 15° = 24.
Check: (24 − 2) × 180° ÷ 24 = 3960° ÷ 24 = 165°.
So the polygon has 24 sides.
In simple words: Each exterior angle is 15°, and 360° ÷ 15° = 24, so it has 24 sides.

Teacher's Note:
1. Number of sides = 360° ÷ exterior angle.
2. The equation (n − 2) × 180 = 165n gives the same result: 15n = 360.

Question 18. Find the side of a square whose area is equal to the area of a rectangle with sides 6.4 m and 2.5 m.
Answer:
Area of the rectangle = 6.4 × 2.5 = 16 m².
Let the side of the square be a. Then a² = 16.
a = √16 = 4.
So the side of the square is 4 m.
In simple words: The rectangle's area is 16 m², and the square with area 16 m² has side 4 m.

Teacher's Note:
1. 6.4 × 2.5 is easy as 64 × 25 = 1600, then place the decimal: 16.00.
2. Side of a square = square root of its area.

Question 19. What is the smallest number by which 1323 may be multiplied so that the product is a perfect cube?
Answer:
Prime factorisation: 1323 = 3 × 3 × 3 × 7 × 7.
The 3s already form a triplet, but 7 appears only twice.
One more 7 completes the triplet: 1323 × 7 = 9261 = 3³ × 7³ = 21³.
So the smallest multiplier is 7.
In simple words: 1323 = 3³ × 7², so multiplying by one more 7 makes it 21³ = 9261.

Teacher's Note:
1. For a perfect cube, every prime must appear a multiple of 3 times.
2. Multiply by the missing factors to complete each triplet.

Question 20. Ram was designing a rangoli using polygon shapes. He created a pattern using a regular polygon in which each interior angle measured 120°. Find the number of sides of the polygon and classify it.
Answer:
Let the regular polygon have n sides. Each interior angle = (n − 2) × 180° ÷ n = 120°.
180n − 360 = 120n, so 60n = 360 and n = 6.
Check: exterior angle = 180° − 120° = 60°, and 360° ÷ 60° = 6.
So the polygon has 6 sides: it is a regular hexagon.
In simple words: An interior angle of 120° means an exterior angle of 60°, and 360° ÷ 60° = 6 sides: a regular hexagon.

Teacher's Note:
1. Regular means equal sides and equal angles.
2. Regular hexagons tile a floor without gaps, as in a honeycomb.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify and solve the linear equations: 15 (y − 4) − 2 (y − 9) + 5 (y + 6) = 0
Answer:
Open the brackets: 15y − 60 − 2y + 18 + 5y + 30 = 0.
Collect the y terms: 15y − 2y + 5y = 18y.
Collect the numbers: −60 + 18 + 30 = −12.
So 18y − 12 = 0, which gives 18y = 12.
y = \( \frac{12}{18} \) = \( \frac{2}{3} \).
So y = \( \frac{2}{3} \).
In simple words: After opening the brackets the equation becomes 18y = 12, so y = 2/3.

Teacher's Note:
1. −2 × (−9) = +18: take care with signs when a negative number multiplies a bracket.
2. Always reduce the final fraction to lowest terms.

Question 22. Find the smallest number by which 33275 must be divided so that the quotient has a cube root. Also, find the cube root of the quotient.
Answer:
Prime factorisation: 33275 = 5 × 5 × 11 × 11 × 11.
The 11s form a triplet, but 5 appears only twice, so the two 5s must be removed.
Smallest divisor = 5 × 5 = 25.
Quotient = 33275 ÷ 25 = 1331 = 11 × 11 × 11.
So 33275 must be divided by 25, and \( \sqrt[3]{1331} \) = 11.
In simple words: Removing the two 5s (dividing by 25) leaves 1331 = 11³, whose cube root is 11.

Teacher's Note:
1. 33275 ends in 75, so it is divisible by 25.
2. To make a perfect cube by dividing, remove the factors that do not form complete triplets.

Question 23. The sum of the exterior angles of a polygon is one-ninth of sum of the interior angles. Find the number of sides of the polygon.
Answer:
Let the polygon have n sides. The sum of the exterior angles of any polygon is 360°.
Sum of exterior angles = \( \frac{1}{9} \) × sum of interior angles, so sum of interior angles = 9 × 360° = 3240°.
(n − 2) × 180° = 3240°.
n − 2 = 3240 ÷ 180 = 18, so n = 20.
So the polygon has 20 sides.
In simple words: The interior angles add to 3240°, which is 18 × 180°, so there are 18 + 2 = 20 sides.

Teacher's Note:
1. The exterior-angle sum, 360°, never changes with the number of sides.
2. Check: (20 − 2) × 180° = 3240° and 3240° ÷ 9 = 360°.

Question 24. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Answer:
Equal rows and columns means the number of children who stand is a perfect square.
We need the largest perfect square not more than 500.
22² = 484 and 23² = 529, and 529 > 500.
So 484 children stand in 22 rows of 22.
Children left out = 500 − 484 = 16.
In simple words: The biggest square arrangement is 22 × 22 = 484 children, so 500 − 484 = 16 are left out.

Teacher's Note:
1. Find the two consecutive squares that 500 lies between.
2. The long-division method for √500 gives the same quotient 22 and remainder 16.

Section D: Long Answer Questions (5 Marks Each)

Question 25. Solve: \( \frac{x}{2} - \frac{1}{4}\left(x - \frac{1}{3}\right) = \frac{1}{6}(x + 1) + \frac{1}{12} \)
Answer:
Open the brackets: \( \frac{x}{2} - \frac{x}{4} + \frac{1}{12} = \frac{x}{6} + \frac{1}{6} + \frac{1}{12} \).
The LCM of 2, 4, 6 and 12 is 12. Multiply every term by 12:
6x − 3x + 1 = 2x + 2 + 1.
3x + 1 = 2x + 3.
3x − 2x = 3 − 1, so x = 2.
Check: LHS = 1 − \( \frac{1}{4} \) × \( \frac{5}{3} \) = 1 − \( \frac{5}{12} \) = \( \frac{7}{12} \); RHS = \( \frac{1}{6} \) × 3 + \( \frac{1}{12} \) = \( \frac{6}{12} \) + \( \frac{1}{12} \) = \( \frac{7}{12} \).
So x = 2.
In simple words: Multiplying everything by 12 clears the fractions and gives 3x + 1 = 2x + 3, so x = 2.

Teacher's Note:
1. Note that −\( \frac{1}{4} \) × (−\( \frac{1}{3} \)) = +\( \frac{1}{12} \).
2. Clearing denominators with the LCM makes the equation much easier to handle.
3. Substituting x = 2 into both sides is a reliable check.

Question 26. Find the value of \( \sqrt[3]{968} \times \sqrt[3]{1375} \)
Answer:
Use \( \sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab} \).
968 = 2 × 2 × 2 × 11 × 11 and 1375 = 5 × 5 × 5 × 11.
968 × 1375 = (2 × 2 × 2) × (5 × 5 × 5) × (11 × 11 × 11).
So \( \sqrt[3]{968 \times 1375} \) = 2 × 5 × 11 = 110.
So the value is 110 (indeed 110³ = 1331000 = 968 × 1375).
In simple words: Neither number is a cube, but their product 1331000 is 110³, so the answer is 110.

Teacher's Note:
1. Do not try to find each cube root separately; combine the numbers first.
2. Group the prime factors in threes and take one from each group.

Question 27. ABCD is a parallelogram where m∠A = (2x + 50°) and m∠C = (3x + 40°).
(i) Find the value of x
(ii) Find the measure of each angle.

Answer:
(i) Opposite angles of a parallelogram are equal, so m∠A = m∠C.
2x + 50° = 3x + 40°, so 50° − 40° = 3x − 2x and x = 10°.
(ii) m∠A = 2 × 10° + 50° = 70° and m∠C = 3 × 10° + 40° = 70°.
Adjacent angles are supplementary: m∠B = 180° − 70° = 110°.
∠D is opposite ∠B, so m∠D = 110°.
So ∠A = 70°, ∠B = 110°, ∠C = 70°, ∠D = 110°.
Check: 70° + 110° + 70° + 110° = 360°.
In simple words: Opposite angles are equal, so x = 10; then ∠A = ∠C = 70° and ∠B = ∠D = 110°.

Teacher's Note:
1. ∠A and ∠C are opposite, so equate their expressions.
2. Getting 70° from both expressions confirms the value of x.

Section E: Case Study Based Questions (4 Marks Each)

Question 28. It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered. Few situations are given below in the form questions.
(a) If the fixed charge in a city is ₹ x and charge per km is ₹ 5 and total fare is ₹ 60 then find the linear equation for the journey of 10 km. (1)
(b) In the above question, what is the value of fixed charge? (1)
(c) If in a city a person has to pay ₹ 110 for a journey of 15 km and fixed charge is ₹ 20 then what is the charge per km? (1)
(d) If in a city fixed charge is double of the charge per km and a person paid ₹ 75 for a journey of 1 km, then the linear equation for the following situation is? (1)
Answer:
Total fare = fixed charge + (charge per km × distance).
(a) Distance charge = 5 × 10 = ₹ 50, so the equation is x + 50 = 60.
(b) x = 60 − 50, so the fixed charge is ₹ 10.
(c) Let the charge per km be ₹ y. Then 20 + 15y = 110, so 15y = 90 and y = 6. The charge is ₹ 6 per km.
(d) Let the charge per km be ₹ x; then the fixed charge is ₹ 2x. For 1 km: 2x + x = 75, that is 3x = 75 (so x = ₹ 25 per km and the fixed charge is ₹ 50).
In simple words: Fare = fixed charge + rate × distance; put in the known values and solve for the unknown.

Teacher's Note:
1. Write the word equation first, then substitute the numbers.
2. In (d), use one letter for the rate and express the fixed charge in terms of it.

Question 29. A farmer owns a perfectly square field and plans to install a strong fence along its entire boundary to protect crops from animals. The total length of fencing available is 480 meters. In addition, he dreams of planting a large tree exactly at the center of the field to provide shade and beauty. To carry out his plan, he needs to calculate the field’s side length and area accurately.
(a) Find the length of each side of the field. (1)
(b) What is the area of the field in square meters? (1)
(c) If the farmer decides to divide the field into 4 equal square parts by drawing two lines parallel to the sides, what will be the area of each part? (1)
(d) If the square field is divided into 4 equal square plots, each will be a smaller square. Find the length of the side of each smaller square plot. (1)
Answer:
The fence goes all round the field, so 480 m is the perimeter.
(a) Perimeter = 4 × side, so side = 480 ÷ 4 = 120 m.
(b) Area = 120 × 120 = 14400 m².
(c) Two lines through the centre, parallel to the sides, make 4 equal squares. Area of each = 14400 ÷ 4 = 3600 m².

(d) Each small square takes half of each side, so its side = 120 ÷ 2 = 60 m (check: 60 × 60 = 3600 m²).
In simple words: The side is 480 ÷ 4 = 120 m, the area is 14400 m², and each quarter is a 60 m square of area 3600 m².

Teacher's Note:
1. Fencing is about perimeter; planting and dividing land is about area.
2. Halving the side of a square divides its area by 4, not by 2.