English Medium Maths Modal Practice Paper Class 8th Session 2022-23

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English Medium Maths Modal Practice Paper Class 8th Session 2022-23

March 09, 2023 02:03AM 6806 Copy Share

Optional question
(Question No. 1 to 10)

Question 1. The numbers in which the decimal point never ends are called.............
(a) terminating decimal numbers
(b) Terminate decimal numbers
(c) both of the above
(d) not both
Answer– Terminating decimal numbers.

Question 2. -7 + (-5) = ................
(a) 12
(b) -12
(c) 2
(d) -2
Answer– -12

Q 3. Selling price – Cost price = ...........
(a) profit
(b) loss
(c) Discount
(d) Rate
Answer– Benefits.

Question 4. Additive inverse of a/b = ..............
(A) b/a
(b) a – b
(c) a + b
(d) -a/b
Answer– -a/b

Question 5. The number of times an observation occurs in the data is called the .......... of that observation.
(a) Distribution
(b) limit
(c) Frequency
(d) breadth
Answer– Frequency.

Q. 6. The highest degree of the variable used in a linear equation in one variable is ...........
(a) zero
(b) a
(c) two
(d) three
Answer– zero

Question 7. √400
(a) 20
(b) 40
(c) 80
(d) none of these
Answer– 20

Q.8. Cubes of negative natural numbers are.............
(a) only positive
(b) only negative
(c) both positive and negative
(d) not both
Answer– only negative.

Q.9. Surface area of .......... = 2(lb + bh + hl)
(a) Cube
(b) cuboid
(c) Cylinder
(d) Cone
Answer– cuboid.

Question 10. A quadrilateral whose all sides are equal is called ………….
(a) parallelogram
(b) Trapezium
(c) rectangle and square
(d) rhombus
Answer– Rhombus.

Short Answer Questions
(Question numbers 11 to 20)

Q.11. What is the value of 3 raised to the power of 10?
Solution– 10 raised to the power 3 = 10×10×10
= 1,000
Therefore, the value of 10 raised to the power of 3 is 1,000.

Question 12. (-4/5)×(-6/11) = ...........
Solution– (-4/5)×(-6/11)
= [(-4)×(-6)]/(5×11)
= 24/55
So it is clear that (-4/5)×(-6/11) = 24/55

Question 13. If the side of a cube is 5 m, find the volume of the cube.
Solution– Side of the cube = l = 5 m
We know that
Formula of volume of cube = l × l × l
Volume of that cube = 5 × 5 × 5 cubic metres
Hence, the volume of that cube will be 125 cubic metres.

Question 14. Is an equilateral triangle a regular polygon? Why?
Ans- Yes, an equilateral triangle is a regular polygon. Because in an equilateral triangle, the sides and angles are of equal measure.

Question 15. Solve the following equation–
17 + 6p = 9
Solution– 17 + 6p = 9
Subtracting 17 from both sides,
17-17+6p = 9-17
6p = -8
On dividing both sides by 6,
6p/6 = -8/6
p = -4/3
Hence, the required solution of the equation is p = -4/3.

Q.16. Find the cube root of 729.
Solution– 729 = 3×3×3×3×3×3
Cube root of 729 = 3×3 = 9
So the cube root of 729 is 9.

Q. 17. Factorise the following expression, q² - 10q + 21
Solution– q² - 10q + 21
= q² - (7+3)q + 21
= q² - 7q - 3q + 21
= q(q-7) - 3(q-7)
= (q-3)(q-7)
Therefore, factors of q² - 10q + 21 = (q-3)(q-7)

Question 18. Write the formula to find the simple interest.
Ans– Simple interest = [Principal × Rate × Time]/100
S.I. = [P × R × T]/100

Q. 19. Multiply the binomials, (y-8) and (3y-4)
Solution– (y-8)×(3y-4)
= y(3y-4) – 8(3y-4)
= (y×3y) – (y×4) – (8×3y) + (8×4)
= 3y^2 - 4y - 24y + 32
On adding like terms,
= 3y^2 - 28y + 32
So it is clear that (y-8)×(3y-4) = 3y^2 - 28y + 32

Q. 20. Find the square of 35.
Solution– Square of 35 = Square of (30+5)
= (30+5)(30+5)
= 30 (30+5) + 5 (30+5)
= square of 30 + 30×5 + 5×30 + square of 5
= 900 + 150 + 150 + 25
= 1225
So square of 35 = 1225

Long Answer Type Questions
(Question numbers 21 to 24)

Question 21. The ratio of two adjacent angles of a parallelogram is 3 : 2. Find the measure of all the angles of the parallelogram.
Solution– Let A and B be two adjacent angles in the parallelogram.
These angles are in the ratio 3 : 2.
Therefore angle A = 3x
angle B = 2x
We know that
Adjacent angles of a parallelogram are supplementary.
Therefore, angle A + angle B = 180°
3x + 2x = 180°
5x = 180°
Dividing both sides by 5,
5x/5 = 180°/5
x = 36°
Thus angle A = 3x = 3 ×36 =108°
Angle B = 2x = 2 × 36 = 72°
We know that
Opposite angles of a parallelogram are equal.
Therefore, angle A = angle C = 108°
Angle B = Angle D = 72°
Thus angle A = 108°, angle B = 72°, angle C = 108°, angle D = 72°

Q.22. A table marked at ₹ 15,000 is available for ₹ 14,400. Find the discount and discount percent.
Solution– Given–
Marked price = ₹ 15,000
Selling price = ₹ 14,400
We know that
Discount = Marked Price – Selling Price
Discount = ₹ 15000 - ₹ 14,400
Discount = ₹ 600
We know that
Discount % = (Discount × 100)/Marked Price
Discount % = (600 × 100)/15,000
Discount % = 60,000/15,000
Discount % = 4%
Hence the discount will be ₹ 600 and the discount percentage will be 4%.

Q.23. What is the probability of getting a prime number when a die is thrown?
Solution– A die can be thrown in 6 ways.
The outcomes of a die = {1, 2, 3, 4, 5, 6}
Total number of results = 6
Prime outcomes of all the outcomes of a die = {1, 3, 5}
Number of these prime outcomes (favourable events) = 3
Hence, required probability = (number of favorable events)/(total number of outcomes)
Required probability = 3/6 = 1/2
Hence, the probability of getting a prime number in a throw of a die is 1/2.

Q.24. The ratio of two numbers is 5 : 3. If their difference is 18, then find the numbers.
Solution– Let the numbers be 5x and 3x.
According to the question,
The difference between the two numbers is 18.
Therefore 5x-3x = 18
2x = 18
Dividing both sides by 2,
2x/2 = 18/2
x = 9
First number = 5x = 5 ×9 = 45
Second number = 3x = 3 × 9 = 27
Hence, the required numbers will be 45 and 27.

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वैकल्पिक प्रश्न कक्षा 8 वार्षिक परीक्षा सत्र 2022-23
1. 40 वैकल्पिक प्रश्न विषय विज्ञान कक्षा 8 वीं वार्षिक परीक्षा
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R F Temre
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