2026MATH108721
Rajya Shiksha Kendra M.P. Bhopal
Annual Examination Session – 2025-26
Subject – Mathematics
Multiple Choice Questions (Q.1 to 5)
Instructions – Read the question and choose the correct option from the given alternatives.
Q.1 The solution of the equation 3x – 5 = 10 will be:
(1)
Solution: 3x – 5 = 10 ⇒ 3x = 10 + 5 ⇒ 3x = 15 ⇒ x = 5
Q.2 By adding 10 to four times of a number, 30 is obtained. That number is:
(1)
Solution: Let the number be x. 4x + 10 = 30 ⇒ 4x = 20 ⇒ x = 5
Q.3 The diagonals of a rhombus are mutually:
(1)
Q.4 Which of the following is a perfect square number?
(1)
Solution: 81 is a perfect square because 9 × 9 = 81.
Q.5 The interest charged on the principal along with the accumulated interest is called –
(1)
Fill in the blanks (Q.6 to 10)
Instructions – Fill in the blanks. Each question carries 1 mark.
Q.6 The distributive law of multiplication over addition is a(b + c) = ab + ac
(1)
Q.7 The opposite angles of a parallelogram are equal.
(1)
Q.8 The graph showing data in intervals is called a histogram.
(1)
Q.9 The factors of x2 – y2 will be (x – y) and (x + y).
(1)
Q.10 The formula for the circumference of a circle is 2πr.
(1)
Very Short Answer Type Questions (Q.11 to 16)
Instructions – Solve the following questions. Each question carries 2 marks.
Q.11 Simplify: 23 × 24 (2)
Solution: Using the law of exponents am × an = am+n
23 × 24 = 23+4 = 27 (or 128)
23 × 24 = 23+4 = 27 (or 128)
Q.12 Give an example of a monomial expression. (2)
Solution: An expression containing only one term is called a monomial. Example: 5x or -3y2
Q.13 Write a Pythagorean triplet whose one member is 6. (2)
Solution: A Pythagorean triplet is of the form 2m, m2-1, m2+1.
Let 2m = 6 ⇒ m = 3.
m2 – 1 = 32 – 1 = 9 – 1 = 8
m2 + 1 = 32 + 1 = 9 + 1 = 10
Therefore, the Pythagorean triplet is 6, 8, 10.
Let 2m = 6 ⇒ m = 3.
m2 – 1 = 32 – 1 = 9 – 1 = 8
m2 + 1 = 32 + 1 = 9 + 1 = 10
Therefore, the Pythagorean triplet is 6, 8, 10.
Q.14 What will be the probability of getting an even number when a die is thrown? (2)
Solution:
Total possible outcomes = 6 (1, 2, 3, 4, 5, 6)
Favorable outcomes (Even numbers: 2, 4, 6) = 3
Probability = Favorable outcomes / Total outcomes = 3 / 6 = 1/2
Total possible outcomes = 6 (1, 2, 3, 4, 5, 6)
Favorable outcomes (Even numbers: 2, 4, 6) = 3
Probability = Favorable outcomes / Total outcomes = 3 / 6 = 1/2
Q.15 Factorize: x2 + 7x + 12 (2)
Solution:
= x2 + 4x + 3x + 12
= x(x + 4) + 3(x + 4)
= (x + 3)(x + 4)
= x2 + 4x + 3x + 12
= x(x + 4) + 3(x + 4)
= (x + 3)(x + 4)
Q.16 Find the area of a rectangle whose length is 10m and breadth is 5m. (2)
Solution:
Area of a rectangle = Length × Breadth
Area = 10m × 5m = 50 m2
Area of a rectangle = Length × Breadth
Area = 10m × 5m = 50 m2
Short Answer Type Questions (Q.17 to 22)
Instructions – Solve the following questions. Each question carries 3 marks.
Q.17 Find the square root of 729 by prime factorization method. (3)
Solution:
Prime factors of 729:
729 = 3 × 243
243 = 3 × 81
81 = 3 × 27
27 = 3 × 9
9 = 3 × 3
So, 729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3 × 3)2
√729 = 3 × 3 × 3 = 27
Prime factors of 729:
729 = 3 × 243
243 = 3 × 81
81 = 3 × 27
27 = 3 × 9
9 = 3 × 3
So, 729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3 × 3)2
√729 = 3 × 3 × 3 = 27
Q.18 Find the number of sides of a regular polygon if each of its exterior angles is 45°. (3)
Solution:
The sum of exterior angles of a regular polygon is always 360°.
Number of sides = 360° / Measure of each exterior angle
Number of sides = 360° / 45° = 8
The sum of exterior angles of a regular polygon is always 360°.
Number of sides = 360° / Measure of each exterior angle
Number of sides = 360° / 45° = 8
Q.19 Find the simple interest on ₹2000 at the rate of 10% per annum for 2 years. (3)
Solution:
Principal (P) = ₹2000, Rate (R) = 10%, Time (T) = 2 years
Simple Interest (SI) = (P × R × T) / 100
SI = (2000 × 10 × 2) / 100 = ₹400
Principal (P) = ₹2000, Rate (R) = 10%, Time (T) = 2 years
Simple Interest (SI) = (P × R × T) / 100
SI = (2000 × 10 × 2) / 100 = ₹400
Q.20 Find the value of (102)2 using the identity (a+b)2. (3)
Solution:
Using the identity: (a + b)2 = a2 + 2ab + b2
(102)2 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2
= 10000 + 400 + 4
= 10404
Using the identity: (a + b)2 = a2 + 2ab + b2
(102)2 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2
= 10000 + 400 + 4
= 10404
Q.21 If a die is thrown once, what will be the probability of getting the outcome 2? (3)
Solution:
Total possible outcomes = 6 (1, 2, 3, 4, 5, 6)
Favorable outcome (getting a 2) = 1
Probability = Favorable outcome / Total outcomes = 1/6
Total possible outcomes = 6 (1, 2, 3, 4, 5, 6)
Favorable outcome (getting a 2) = 1
Probability = Favorable outcome / Total outcomes = 1/6
Q.22 If Chameli had ₹600 left which is 25% of her total money, how much money did she have in the beginning? (3)
Solution:
Let the total money Chameli had in the beginning be x.
According to the question, 25% of x = ₹600
(25 / 100) × x = 600
(1 / 4) × x = 600
x = 600 × 4
x = ₹2400
Let the total money Chameli had in the beginning be x.
According to the question, 25% of x = ₹600
(25 / 100) × x = 600
(1 / 4) × x = 600
x = 600 × 4
x = ₹2400
Long Answer Type Questions (Q.23 to 26)
Instructions – Write the answers to the questions in 75 or 100 words. Each question carries 5 marks.
Q.23 Simplify the expression: (a + b)(2a – 3b + c) – (2a – 3b)c (5)
Solution:
= [a(2a – 3b + c) + b(2a – 3b + c)] – [c(2a) – c(3b)]
= [2a2 – 3ab + ac + 2ab – 3b2 + bc] – [2ac – 3bc]
= 2a2 – ab + ac – 3b2 + bc – 2ac + 3bc
Grouping like terms:
= 2a2 – 3b2 – ab + (ac – 2ac) + (bc + 3bc)
= 2a2 – 3b2 – ab – ac + 4bc
= [a(2a – 3b + c) + b(2a – 3b + c)] – [c(2a) – c(3b)]
= [2a2 – 3ab + ac + 2ab – 3b2 + bc] – [2ac – 3bc]
= 2a2 – ab + ac – 3b2 + bc – 2ac + 3bc
Grouping like terms:
= 2a2 – 3b2 – ab + (ac – 2ac) + (bc + 3bc)
= 2a2 – 3b2 – ab – ac + 4bc
Q.24 The radius of a cylindrical milk tank is 1.5m and its length is 7m. Find the quantity of milk in liters that can be stored in it. (1m3 = 1000 liters) (5)
Solution:
Radius (r) = 1.5 m
Length/Height (h) = 7 m
Volume of the cylindrical tank = πr2h
= (22/7) × (1.5)2 × 7
= 22 × 2.25
= 49.5 m3
Since 1 m3 = 1000 liters,
Quantity of milk = 49.5 × 1000 = 49500 liters
Radius (r) = 1.5 m
Length/Height (h) = 7 m
Volume of the cylindrical tank = πr2h
= (22/7) × (1.5)2 × 7
= 22 × 2.25
= 49.5 m3
Since 1 m3 = 1000 liters,
Quantity of milk = 49.5 × 1000 = 49500 liters
Q.25 A rectangular piece of paper of width 21 cm is rolled along its width to form a cylinder of radius 30 cm. Find the volume of the cylinder. (5)
Solution:
When a rectangular sheet is rolled along its width, the width of the paper becomes the height of the cylinder.
Height of cylinder (h) = 21 cm
Radius of cylinder (r) = 30 cm
Volume of a cylinder = πr2h
= (22/7) × (30)2 × 21
= (22/7) × 900 × 21
= 22 × 900 × 3
= 66 × 900
= 59400 cm3
When a rectangular sheet is rolled along its width, the width of the paper becomes the height of the cylinder.
Height of cylinder (h) = 21 cm
Radius of cylinder (r) = 30 cm
Volume of a cylinder = πr2h
= (22/7) × (30)2 × 21
= (22/7) × 900 × 21
= 22 × 900 × 3
= 66 × 900
= 59400 cm3
Q.26 Find the cube root of the number 3375 by prime factorization method. (5)
Solution:
Let’s find the prime factors of 3375:
3375 = 3 × 1125
1125 = 3 × 375
375 = 3 × 125
125 = 5 × 25
25 = 5 × 5
So, 3375 = 3 × 3 × 3 × 5 × 5 × 5
Making groups of 3 (triplets) for cube root:
³√3375 = ³√(33 × 53)
= 3 × 5
= 15
Let’s find the prime factors of 3375:
3375 = 3 × 1125
1125 = 3 × 375
375 = 3 × 125
125 = 5 × 25
25 = 5 × 5
So, 3375 = 3 × 3 × 3 × 5 × 5 × 5
Making groups of 3 (triplets) for cube root:
³√3375 = ³√(33 × 53)
= 3 × 5
= 15
Class - 8
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