Practicing with Class 8 Maths Question Paper Pdf Kerala Syllabus and Second Term Question Paper 2022-23 will help students prepare effectively for their upcoming exams.
Class 8 Maths Second Term Question Paper 2022-23 Kerala Syllabus
Time : 1½ Hours
Score : 40
Write any 4 questions from 1 to 5. Each question carries 2 scores. (4 × 2 = 8)
Question 1.
Fill in the blanks
(a) 51² = 50² + 50 + [1, 50, 51]
Answer:
51
(b) (x + 1)² = x² + x + [2x, x + 1, 1]
Answer:
x + 12
Question 2.
In the figure ABCD is an isosceles trapezium, ∠A = 80°
(a) ∠B = ___
(b) ∠C = ___
Answer:
(a) ∠B = 80°
(b) ∠C = 180° – 80° = 100°
Question 3.
Ramu deposited 1000 rupees in a bank, which gives simple interest. After one year, Ramu got 100 rupees as interest. What is the rate of interest?
Answer:
1000 × 1 × \(\frac{r}{100}\) = 100
r = \(\frac{100}{10}\) = 10%
Question 4.
(a) x² – y² = (x + y)(___)
Answer:
x – y
(b) Using the above principle, find 52² – 48².
Answer:
52² – 48² = (52 + 48)(52 – 48)
= 100 × 4
= 400
Question 5.
Which of the following statement is always true regarding a parallelogram?
(i) All angles right.
(ii) Diagonals bisect each other.
(iii) Diagonals are perpendicular bisectors of each other.
Answer:
(ii) Diagonals bisect each other.
Write any 4 questions from 6 to 11. Each question carries 3 scores. (4 × 3 = 12)
Question 6.
Four numbers forming a square in a calendar are given below:
| 4 | 5 |
| 11 | 12 |
(a) Find the difference of its diagonal products.
Answer:
(5 × 11) – (4 × 12) = 7
(b) Is it the same for all squares of four numbers in a calendar? Explain using algebra.
Answer:
Yes
| x | x + 1 |
| x + 7 | x + 8 |
Difference of diagonal product
= ((x + 1)(x + 7)) – (x ( x + 8))
= x² + 8x + 7 – (x² + 8x)
= 7
Question 7.
Draw a square of diagonal 7 centimetres.
Answer:
For drawing a line of length 7 cm
For drawing a 45° angles on its ends.
For drawing square.
or
For drawing a line of length 7 cm
For drawing the perpendicular bisector of the line.
For the marking two points 3.5 cm above and below on the perpendicular bisector from the midpoint of the line and completing the square.
Question 8.
Sumesh deposited 10000 rupees in a bank which gives 5% interest compounded annually. How much would he get back after two years?
Answer:
Interest for the first year
= 10000 × \(\frac{5}{100}\) = Rs.500
Amount gets after one year
= 10000 + 500 = Rs.10500
Interest for the second year
10500 × \(\frac{5}{100}\) = Rs.525
Amount gets after two years
= 10500 + 525
= Rs.11025
OR
A = 10000 × (1 + \(\frac{5}{100}\))2
= 10000 x \(\frac{105}{100} \times \frac{105}{100}\)
= Rs. 11025
Question 9.
(a) (x + 1)(y + 1) = xy + x + y + ___
Answer:
1
(b) Find 51 × 21 by using the above principle.
Answer:
51 × 21 = (50 + 1) (20 + 1)
= 50 × 20 + 50 + 20 + 1
= 1071
Question 10.
(a) ABCD is a rhombus. The diagonals AC and BD intersect at O.
What is the measure of ∠AOB?
Answer:
∠AOB = 90°
(b) Draw the rhombus ABCD in which AC = 7 centimetres, BD = 6 centimetres.
Answer:
For drawing a line of length 7 cm and drawing its perpendicular bisector.
For the marking two points B and D 3 cm above and below on the perpendicular bi-sector from the midpoint of the line and completing the rhombus.
Question 11.
The simple interest at 8% got for a certain amount after 2 years is 200 rupees. If the interest compounded annually, what would be the interest for the same amount at the same rate after 2 years?
Answer:
If the amount is taken as x rupees,
x × 2 × \(\frac{8}{100}\) = 200
⇒ x = \(\frac{200 \times 100}{2 \times 8}\) = Rs. 1250
Interest for the first year
= 1250 × \(\frac{8}{100}\) = Rs.100
Amount gets after one year
= 1250 + 100 = Rs.1350
Interest for the second year
= 1350 × \(\frac{8}{100}\) = Rs.108
Compound interest
= 100+ 108 = Rs. 208
OR
A = 1250 × (1 + \(\frac{8}{100}\))2
= 1250 × \(\frac{108}{100} \times \frac{108}{100}\)
= Rs 1458
Compound interest = 1458 – 1250
= Rs.208
Another method
Simple interest for two years = Rs.200
Simple interest for one years = Rs.100
Compound interest for the first year = Rs. 100
Compound interest for the second year
= 100 = 100 + 100 × \(\frac{8}{100}\) = Rs. 108
Total compound interest =100 + 108
= Rs. 208
Write any 5 questions from 12 to 18. Each question carries 4 scores. (5 × 4 = 20)
Question 12.
Look at this pattern.
1 × 4 = (2 × 3) – 2
2 × 5 = (3 × 4) – 2
3 × 6 = (4 × 5) – 2
____________
____________
(a) Find the difference of its diagonal products.
Answer:
4 × 7 = (5 × 6) – 2
5 × 8 = (6 × 7) – 2
(b) If a, b, c, d are four consecutive natural numbers, then find the relation between a × d and b × c.
Answer:
a × d = (b × c) – 2
(c) Using the above principle, find 98 × 101
Answer:
98 × 101 = (99 × 100) – 2 = 9898
Question 13.
A laptop manufacturing company decreases the price of a particular model by 10% each year. The current price of this model is 40000 rupees. What would be the price after 2 years?
Answer:
4000 × (1 – \(\frac{10}{100}\))2
= 4000 × \(\frac{90}{100} \times \frac{90}{100}\)
= Rs. 32400
Question 14.
Draw the quadrilateral shown below with given measurement.
Answer:
Question 15.
Some natural numbers can be, written as the difference of two perfect squares. For example,
8 = 4 × 2 × 1 = 3² – 1²
12 = 4 × 3 × 1 = 4² – 2²
16 = 4 × 4 × 1 = 5² – 3²
(a) Write 20 as the difference of two perfect squares.
Answer:
20 = 4 × 5 × 1 = 6² – 4²
(b) Explain using algebra, the method of writing all multiples of 4, starting with 8 as the difference of two perfect squares.
Answer:
4 × x × 1 = (x + 1)² – (x – 1)² (x + 1)² – (x – 1)²
= (x + 1 + x – 1) (x + 1 – [x – 1])
= 2x × 2
= 4 × x × 1
Question 16.
Ajayan deposited 20000 rupees in a bank which gives interest compounded annually and Vijayan deposited the same amount in another bank which gives interest com-pounded half yearly. The annual rate of in-terest is 8% in both the banks. After 2 years they withdrew their deposits.
(a) How much amount did Ajayan get?
Answer:
Interest for the first year = 20000 ×
= Rs. 1600
Amount gets after one year = 20000 + 1600
= Rs.21600
Interest for the second year
= 21600 × \(\frac{8}{100}\) = Rs. 1728
Amount gets after second year = 21600 + 1728
= Rs. 23328
OR
A = 20000 × (1 + \(\frac{8}{100}\))2
= 20000 × \(\frac{108}{100} \times \frac{108}{100}\)
= Rs. 23328
(b) How much amount did Vijayan get?
Answer:
A = 20000 × (1 + \(\frac{8}{2 \times 100}\))2
= 20000 × \(\left(\frac{104}{100}\right)^4\)
= 20000 × \(\frac{104}{100} \times \frac{104}{100} \times \frac{104}{100} \times \frac{104}{100}\)
= Rs. 23397.17
(c) Who got more amount? By how much?
Answer:
Vijayan got Rs 69.7 more
Question 17.
Three equal rhombuses are joined together in the figure given below. Draw the figure in the given measurements.
Answer:
Question 18.
Read the mathematical concept given below and answer the following questions. Observe the number pyramids:-
Do you see any relation between the numbers?
“Pairs of adjacent numbers in the bottom row are added to get the two numbers in the middle row: and these two numbers are added to get the topmost number. Also the numbers in the bottom row are consecutive natural numbers”.
(a) Write the missing number in this number pyramid.
Answer:
20
(b) Identify the relation between the middle number in the bottom row and the topmost number.
Answer:
4 × 5
(c) Write the missing numbers in this number pyramid:
Answer:
