Kerala Syllabus Class 7 Maths Model Question Paper Set 4

During exam preparation, Std 7 Maths Question Paper Kerala Syllabus Set 4 guide students properly.

Kerala Syllabus 7th Standard Maths Question Paper Set 4

Time : 2 Hrs 15 Min

Instructions:

• 15 minutes is given as cool – off time. Use this time to read and understand the questions and plan answers accordingly.
• Answer all the 6 questions
• Question number 1 and 6 have internal choices (1 A and 1B;6A and 6 B ). Answer to any one of the sub questions A or B under it.

The first question has 2 questions (1A and 1B). You only need to answer one of them.

Question 1.
A. (a) Two parallel lines are cut by a transversal. Given that one of the angles measures 60°, calculate all the other angles formed by the intersection. Provide a diagram for reference.
Answer:
Given that one of the angles formed by a transversal cutting two parallel lines is 60°:

• Corresponding angles: 60°
• Alternate interior angles: 60°
• Co-interior angles: 120°
• Vertical opposite angles: 60°
• The other angles are 120° each.

(b) The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle

Answer:
Consider the figure below,

Since, ABCD is rectangle ∠P, ∠Q, ∠R, ∠S equals to 90°
Now, it is given that one part of ∠Q is 50°, so the remaining part of ∠A is 40°
Similarly, one part of ∠R is 30°, so the remaining part of ∠R is 60°.
It implies that two angles of the triangle are 40° and 60°.
Therefore, the third angle of the triangle is
= 180° – (40° + 60°)
= 180° – (100°) = 80°
So, the angles of triangles are 40°, 60°, 80°

(c) A transversal cuts two parallel lines. One angle measures 65°.
(i) Its vertically opposite angle is also 65°.
(ii) Its adjacent angle is 115°.
(iii) Its corresponding angle is 115°.
Which of the above statements are true?
(A) i and ii correct
(B) Only iii correct
(C) i and iii correct
(D) i and ii correct
Answer:
(A) i and ii correct

B. (a) Two vertical lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Consider the figure below

Given that AB and CD are parallel line
Draw a line PQ parallel to both AB and CD
∠A and ∠AQP are alternate angles
Therefore, ∠A = ∠AQP = 30°
Similarly,
∠C and ∠CQP are alternate angles
Therefore, ∠C = ∠CQP = 40°
But, ∠Q = ∠AQP + ∠CQP = 30° + 40° = 70°

(b) A triangle is drawn inside a parallelogram. Calculate the angles of the triangle.

Answer:
Consider the figure below

∠D and angle ∠B are opposite angles of paral-lelogram.
Therefore, ∠D = ∠B= 105°
But, one part of angle ∠D is 50°, so remaining part of ∠D = 105° – 50° = 55°
Now, considering the angles angle ∠A and ∠D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180°- ∠D = 180° – 105° = 75°
But, one part of ∠A is 20°, so remaining part of ∠A = 75° – 20°= 55°
It means, we get two angles of the triangle which are 55° and 55°
So, third angle of the triangle = 180° – (55° + 55°)
= 180° – (110°)
= 70°
Hence, all three angles of the triangles are 55°, 55°, 70°.

(c) Look at the statements below:
i. If a transversal is perpendicular to one parallel line, it is also perpendicular to the other.
ii. If a transversal is perpendicular to one line, all the angles formed are
iii. If a transversal is perpendicular to one parallel line, then alternate interior angles are not equal.
Which of the above statements are true?
A. i and ii correct
B. ii and iii correct
C. Only iii correct
D. All are correct
Answer:
A. i and ii correct

Question 2.
(a) A statement and reason is given below:
Statement (A): \(\frac{7}{12}>\frac{2}{3}\)
Reason (R): To compare fractions, we convert them to like denominators.
Examine the statement and reason, find which is correct.
(a) A is true, but R is false
(b) A is false, but R is true
(c) Both A and R are true, and R is the correct explanation of A
(d) Both A and R are ture, but R is not the correct explanation of A
Answer:
(b) A is false, but R is true

(b) If a man can complete a work in 12 days and his friend can complete the same work in 6 days, then:
(i) The faster worker is how many times faster?
(ii) The slower worker is what part of the faster worker?
Answer:
(i) Times = \(\frac{12}{2}\) = 2 times
Part = \(\frac{6}{12}=\frac{1}{2}\)

(c) A ribbon is cut into 12 equal parts. What fraction of the ribbon is 3 parts?
Answer:
Part (or fraction) = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)

Question 3.
(a) 12 metre long rope cut into 4 equal pieces.
What is the length of each piece?
What if it is cut into 5 equal pieces?
Answer:
(i) Length of each piece = \(\frac{12}{4}\) = 3 metres
(ii) If it is cut into 5 equal pieces,
Length of each piece = \(\frac{12}{5}\) = 2\(\frac{2}{5}\) metres

(b) The length and breadth of some rectangles are given below. Calculate their areas.
(i) 4\(\frac{1}{2}\)cm, 3\(\frac{1}{4}\)cm
Answer:
Area = 4\(\frac{1}{2}\)cm, 3\(\frac{1}{4}\)cm
= \(\frac{9}{2} \times \frac{13}{4}=\frac{9 \times 13}{2 \times 4}=\frac{117}{8}\)
= 14\(\frac{5}{8}\) cm

(ii) 6\(\frac{3}{4}\)cm, 5\(\frac{1}{3}\)cm
Answer:
6\(\frac{3}{4}\)cm, 5\(\frac{1}{3}\)cm
Area = 6\(\frac{3}{4}\) × 5\(\frac{1}{3}\) = \(\frac{27}{4} \times \frac{16}{3}\)
= \(\frac{27}{3} \times \frac{16}{4}\)
= 9 × 4
= 36 cm²

Question 4.
(a) If the breadth of a rectangle is 2.5 metres and the area is 5.6 square metres, what is its length?
Answer:
Given, breadth = 2.5m, area = 5.6 m2
Length = Area ÷ Breadth
= 5.6 ÷ 2.5
= 2.24 m

(b) What is the decimal form of \(\frac{3.2}{16}\)
(a) 0.2
(b) 0.02
(c) 2
(d) 4
Answer:
0.2

(c) If the length of a box is 1.2 metres, breadth is 0.8 m and heigtht is 0.2 m, what is the volume of that box?
Answer:
Length = 1.2 m, Breadth = 0.8 m,
Height = 0.2 m
Volume = 1.2 × 0.8 × 0.2
= 0.96 × 0.2
= 0.192 m³

Question 5.
(a) The number of factors of 3⁴ × 5^-2 is:
(a) 8
(b) 12
(c) 15
(d) 20
Answer:
(c) 15
(4 + 1)(2 + 1) = 5 × 3 = 15

(b) Find the common factors of 48 and 60.
Answer:
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹
Common primes = 2, 3
Lowest powers = 2¹ x 3¹ = 4 × 3 =12
Factors of 12 = 1,2, 3, 4, 6, 12
Common factors = {1, 2, 3, 4, 6, 12}

(c) Two friends have 72 chocolates and 90 chocolates. They want to distribute them into equal groups without any chocolate left.
(i) What is the maximum number of chocolates in each group?
(ii) How many such groups will be formed?
Answer:
72 = 2³ × 3²
90= 2¹ × 3² × 5¹
Common = 2¹ × 3² = 18
(i) Maximum chocolates in each group = 18
(ii) Number of groups = (72 + 90) /18
= 162/18 = 9
∴ 18 chocolates per group and 9 groups.

The sixth question has 2 questions ( 6A and 6B). You only need to answer one of them.

Question 6.
A. (a) A person spends 5000 rupees a month for food. It is 20% of his monthly earnings. What is his monthly earnings?
Answer:
20% = \(\frac{20}{100}=\frac{1}{5}\) of the monthly earning spendfor food.
So the total monthly is \(\frac{5}{1}\) times the money spent for food
So, monthly earning = \(\frac{5}{1}\) × 5000 = 2500 rupees

(b) The bar graph shows the number of girls in the three divisions of class 7 in a school.

Draw a pie chart of this.

Answer:

OR

B. (a) Convert 3/5 into a percent.
(a) 30%
(b) 40%
(c) 50%
(d) 60%
Answer:
(d) 60%

(b) A student scored 36 marks out of 50 in a test.
(i) What fraction of marks did he get?
(ii) What percent of marks did he score?
Answer:
(i) Fraction = \(\frac{36}{50}=\frac{18}{25}\)
(ii) Percent = \(\frac{36}{50}\) × 100 = 72%