Differentiation (Part-38) | S N De

Class 8 Mathematics Part-II Examination 2025 – Questions and Detailed Solutions

Ramakrishna Mission Boys' Home
Junior High School & High School (H.S.)
Part-II Examination – 2025
Class VIII | Mathematics | English Medium
Time: 3 Hours | Full Marks: 100

About this Solution: This post presents Questions 1–21 from the Class VIII Mathematics Part-II Examination 2025 in clean, typed English, followed by step-by-step solutions. The mathematical expressions have been formatted for easy reading on desktop and mobile devices.

Question 1 — Choose the Correct Answer

Q1. The percentage of 85 g out of 17 kg is:

(a) 0.25% (b) 0.5% (c) 0.75% (d) 1.25%

Solution

First convert kilograms into grams:

17 kg = 17 × 1000 = 17000 g

Therefore, the required percentage is:

85 17000 × 100 = 0.5%

Answer: (b) 0.5%

Q2. The ratio of hydrogen and oxygen in water is 2 : 1. What is the percentage of hydrogen in water?

(a) 20% (b) 33⅓% (c) 66⅔% (d) 80%

Solution

Total parts in the ratio:

2 + 1 = 3

Hydrogen occupies 2 parts out of 3 parts.

Percentage of hydrogen = 2 3 × 100 = 66⅔%

Answer: (c) 66⅔%

Q3. Express the following fraction in its lowest form:

18a⁴b⁵c² 21a⁷b²

(a) 2a³b³c²3 (b) 6b³c²7a³ (c) 9a³c²7b³ (d) 7a³6b³c²

Solution

Reduce the numerical coefficient:

18 21 = 6 7

For the variables:

a⁴ ÷ a⁷ = 1/a³

b⁵ ÷ b² = b³

Therefore:

= 6b³c² 7a³

Answer: (b)

Q4. Find the G.C.D. of 3a²b²c, 12a²b⁴c² and 9a⁵b⁴.

(a) 3a²b² (b) 2a²b²c (c) 36a⁵b⁴c² (d) 36a²b²

Solution

Consider the numerical coefficients:

G.C.D. of 3, 12 and 9 = 3

For the variables, take the lowest powers common to all terms:

• Lowest power of a = a²
• Lowest power of b = b²
• c is absent from the third term, so c is not included.

G.C.D. = 3a²b²

Answer: (a) 3a²b²

Q5. In ΔABC, AB = AC. If ∠BAC = 80°, find ∠ABC.

(a) 40° (b) 80° (c) 20° (d) 50°

Solution

Since AB = AC, ΔABC is an isosceles triangle.

Therefore, the base angles are equal:

∠ABC = ∠ACB

Sum of the angles of a triangle is 180°.

∠ABC + ∠ACB + ∠BAC = 180°

2∠ABC + 80° = 180°

2∠ABC = 100°

∠ABC = 50°

Answer: (d) 50°

Q6. In ΔABC, if AC > AB, then:

(a) x = 2y (b) x = y (c) x = 3/5 y (d) None of these

Solution

The photographed question refers to the quantities x and y, but the page does not define what x and y represent. No accompanying diagram is visible with this question.

Therefore, an exact relation such as x = 2y, x = y, or x = 3y/5 cannot be established from the supplied question alone.

Answer based strictly on the supplied image: (d) None of these

Note: If the original examination contained a diagram or definitions of x and y that are missing from this photograph, the answer should be checked against that original material.

Section 2 – Short Answer Questions

Question 7

Q7. 12½% of what amount is Rs. 320?

Solution

Let the required amount be Rs. x.

12½% = 25 2 % = 1 8

Therefore:

x/8 = 320

x = 320 × 8 = 2560

Answer: Rs. 2,560

Question 8

Q8. Express 0.1̅6 in percentage.

Solution

The recurring decimal printed in the question represents 0.16666... .

0.16666... = 1 6

To convert it into percentage, multiply by 100:

1 6 × 100 = 16⅔%

Answer: 16⅔%

Question 9

Q9. Simplify:

a − b − c a + a + b + c a

Solution

Both fractions have the same denominator a.

= (a − b − c) + (a + b + c) a

Combine the terms in the numerator:

a − b − c + a + b + c = 2a

= 2a a = 2

Answer: 2

Question 10

Q10. Find the G.C.D. of (x³ − 3x²y) and (x² − 9y²).

Solution

Factorise the first expression:

x³ − 3x²y = x²(x − 3y)

Factorise the second expression as a difference of squares:

x² − 9y² = x² − (3y)²

= (x − 3y)(x + 3y)

The common factor is:

x − 3y

Answer: x − 3y

Question 11

Q11. In ΔABC, AB = AC. If ∠BAC = 70°, which side is greatest in length?

Solution

Since AB = AC, the triangle is isosceles and the two base angles are equal.

∠ABC = ∠ACB

Using the angle-sum property:

∠ABC + ∠ACB + 70° = 180°

2∠ABC = 110°

∠ABC = ∠ACB = 55°

The greatest angle is ∠A = 70°. The side opposite the greatest angle is the greatest side.

The side opposite ∠A is BC.

Answer: BC is the greatest side.

Question 12

Q12. In ΔABC, BC is produced to D. If ∠ACD = 112° and ∠ABC = 60°, find ∠BAC.

Solution

Since BC is produced to D, ∠ACD is an exterior angle of ΔABC.

By the exterior-angle theorem:

∠ACD = ∠BAC + ∠ABC

Substituting the given values:

112° = ∠BAC + 60°

∠BAC = 112° − 60° = 52°

Answer: ∠BAC = 52°

Section 3 – Application and Algebra Questions

Question 13

Q13. The length of each side of a square was increased by 10%. Find the percentage increase in its area.

Solution

Let the original side of the square be 100 units.

After a 10% increase:

New side = 100 + 10 = 110 units

Original area:

100 × 100 = 10,000 square units

New area:

110 × 110 = 12,100 square units

Increase in area:

12,100 − 10,000 = 2,100

Percentage increase:

2,100 10,000 × 100 = 21%

Answer: 21% increase in area

Question 14

Q14. When water freezes into ice, its volume increases by 10%. Find, in percentage, how much the volume decreases when the ice melts back into water.

Solution

Assume the original volume of water is 100 units.

After freezing, the volume increases by 10%:

Volume of ice = 100 + 10 = 110 units

When the ice melts, it returns to 100 units.

Decrease in volume:

110 − 100 = 10 units

The percentage decrease must be calculated with respect to the volume of ice, i.e. 110 units.

Percentage decrease = 10 110 × 100

= 9.09...% = 9 1 11 %

Answer: 9 1/11% decrease approximately 9.09%

Question 15

Q15. In a certain type of brass, the ratio of copper to zinc is 5 : 2. What will be the ratio of copper to zinc in 28 kg of brass if 4 kg of copper is added to it?

Solution

Total parts:

5 + 2 = 7

Therefore, in 28 kg of brass:

Copper = 5 7 × 28 = 20 kg

Zinc = 2 7 × 28 = 8 kg

4 kg of copper is added:

New copper = 20 + 4 = 24 kg

New zinc = 8 kg

Therefore:

Copper : Zinc = 24 : 8 = 3 : 1

Answer: 3 : 1

Question 16

Q16. Resolve into factors: 6x² − x − 15

Solution

We need two numbers whose product is:

6 × (−15) = −90

and whose difference is −1.

The required numbers are −10 and +9.

Split the middle term:

6x² − 10x + 9x − 15

Group the terms:

2x(3x − 5) + 3(3x − 5)

Take the common factor:

(3x − 5)(2x + 3)

Answer: (3x − 5)(2x + 3)

Question 17

Q17. Find the G.C.D. of 2ax(a − x)² and 4a²x(a − x)³.

Solution

Compare the numerical coefficients:

G.C.D. of 2 and 4 = 2

Lowest power of a:

a¹

Lowest power of x:

x¹

Lowest power of (a − x):

(a − x)²

G.C.D. = 2ax(a − x)²

Answer: 2ax(a − x)²

Question 18

Q18. Express in reduced form:

a + 1 a − 2 × a² − a − 2 a² + a

Solution

Factorise the numerator and denominator:

a² − a − 2 = (a − 2)(a + 1)

a² + a = a(a + 1)

Therefore:

a + 1 a − 2 × (a − 2)(a + 1) a(a + 1)

Cancel the common factors:

= a + 1 a

Answer: (a + 1)/a

Question 19

Q19. Simplify:

a a² + ab − b (a + b)²

Solution

Factorise the first denominator:

a² + ab = a(a + b)

Thus:

a a(a + b) − b (a + b)²

Simplify the first fraction:

= 1 a + b − b (a + b)²

Take the common denominator (a + b)²:

= a + b − b (a + b)²

= a (a + b)²

Answer: a/(a + b)²

Question 20

Q20. Resolve into factors by expressing as the difference of two squares: x² − 2x − 3.

Solution

Rewrite the expression by completing the square:

x² − 2x − 3

= x² − 2x + 1 − 4

= (x − 1)² − 2²

Using a² − b² = (a − b)(a + b):

= [(x − 1) − 2][(x − 1) + 2]

= (x − 3)(x + 1)

Answer: (x − 3)(x + 1)

Question 21

Q21. Find the L.C.M. of (x²y² − x²) and (xy² − 2xy + x).

Solution

First factorise the first expression:

x²y² − x² = x²(y² − 1)

Since y² − 1 is a difference of squares:

= x²(y − 1)(y + 1)

Now factorise the second expression:

xy² − 2xy + x = x(y² − 2y + 1)

= x(y − 1)²

Therefore, the factorised forms are:

x²(y − 1)(y + 1)

x(y − 1)²

For the L.C.M., take the highest power of every factor:

• Highest power of x = x²
• Highest power of (y − 1) = (y − 1)²
• Highest power of (y + 1) = (y + 1)

L.C.M. = x²(y − 1)²(y + 1)

Answer: x²(y − 1)²(y + 1)

Class 8 Mathematics – Detailed Solutions
Part-II Examination 2025 | Questions 1–21