During exam preparation, Std 7 Maths Question Paper Kerala Syllabus Set 5 guide students properly.
Kerala Syllabus 7th Standard Maths Question Paper Set 5
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) Simplify \(\frac{3}{4} \div \frac{5}{6}\)
Answer:
\(\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{18}{20}=\frac{9}{20}\)
(b) A recipe requires \(\frac{3}{4}\) cup of sugar to make ¡2 cookies. How much sugar is needed to make 36 cookies?
Answer:
\(\frac{3}{4}\) cup of sugar gives 12 cookies
12 × 3 gives 36
So, for 36 cookies we need \(\frac{3}{4}\) × 3 = \(\frac{9}{4}\) cups of sugar.
(c) If \(\frac{2}{5}\) of the cake is eaten, what fraction of the whole cake remains?
Answer:
The whole cake is \(\frac{5}{5}\) = 1
Remaining cake = \(\frac{5}{5}-\frac{2}{5}=\frac{3}{5}\)
Thus, \(\frac{3}{5}\) part of the whole cake remains.
OR
B. (a) \(\frac{1.234}{0.01234}\) = ?
Answer:
(b) \(\frac{2.3 \times 3.2}{0.4}\) = ?
Answer:
(c) 6.5 kilograms of chilli powder was packed in 025 kilogram packets. How many packets are there?
Answer:
Total chilli power = 6.5 = \(\frac{65}{10}\) kg
Amount of chilli power in a small packet
= 0.25 = \(\frac{25}{100}\) kg
Number of small packets
= \(=\frac{\text { Total chilli powder }}{\text { Amount of chilli powder in a small packet }}\)
Here, \(\frac{25}{100}\) is the dividing fraction. Its reciprocal is \(\frac{100}{25}\)
Therefore, the number of packets = \(\frac{65}{10} \times \frac{100}{25}\)
= \(\frac{650}{25}\)
= 26
Question 2.
(a) A statement and reason is given below
Statement (A): ¡ If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line as well.
Reason (R): All right angles are equal in measure.
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 B are true, but R is not the correct explanation of A.
Answer:
(d) Both A and B are true, but R is not the correct explanation of A.
(b) The top and bottom lines in the figure are parallel. Find the unknown angle shown in the figure.
Answer:
Draw a horizontal line parallel to the other two parallel lines
From the figure, we get
∠EAB = ∠ABD = 30°
∠DBC = ∠BCF = 60°
∴ ∠ABC = ∠ABD + ∠DBC
= 30°+ 60° = 90°
(c) Draw the parallelogram below with the given measures.
Answer:
Draw a horizontal line segment AB that is 6 cm long. This will be one side of the parallelogram
At point A, use a protractor to measure and mark an angle of 50° from line AB.
From point A, draw a line segment AC that is 4 cm long, making sure it forms the 50° angle with AB.
From point B, draw a line segment BD that is 4 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.
From point C, draw a line segment CD that is 6 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.
Question 3.
(a) Angles of a linear pair are in the ratio 4:5. What is the measure of each angle?
Answer:
Ratio of angles in the linear pair = 4:5
Sum of angles =180
In 180, \(\frac{4}{9}\) is one angle and \(\frac{5}{9}\) is the other angle.
ISo, one angle = 180 × \(\frac{4}{9}\) = 80°
Other angle = 180 × \(\frac{5}{9}\) = 100°
(b) What does it mean to say that the width to length ratio of a rectangle is 1:1? What sort of rectangle is it?
Answer:
The width to length ratio of a rectangle is 1:1,
Which means, length = width
When the sides of a rectangle are equal, it will be a square.
Question 4.
(a) Which of the following is correct?
(A) 40 % = \(\frac{2}{3}\)
(B) 62.5% = \(\frac{5}{8}\)
(C) 12.5% = \(\frac{1}{6}\)
(D) 80 % = \(\frac{3}{5}\)
Answer:
(B) 62.5% = \(\frac{5}{8}\)
(b) A jacket is marked at ₹ 1600 and sold at a discount of 12.5%. Find the selling price.
Answer:
Discount = 12.5 % of 1600
= 0.125 × 1600 = ₹ 200
Selling Price = 1600 – 200 = ₹ 1400
(c) A shopkeeper marks a shirt at 2,000 rupees. He allows a 20% discount on the marked price. If the cost price of the shirt is 1,400 rupees, find his profit percentage.
Answer:
Marked Price (MP) = 2,000 Rs
Discount = 20 % of 2,000 = 400 Rs
Selling Price (SP) = 2,000 – 400 =1,600 Rs
Cost Price (CP) = 1,400 Rs
Profit = SP – CP = 1,600 – 1,400 = 200 Rs
Profit % = (Profit ÷ CP) x 100
= (200 – 1,400) x 100 = 14.28%
Question 5.
(a)The algebraic expression for the number obtained by adding 10 repeatedly to 9 is:
(A) 10n
(B) 9n
(C) 9 + 10n
(d) 10 + 9n
Answer:
(d) 10 + 9n
(b) The numbers 25, 36,47… are got by starting with 25 and adding 11 again and again
i) What is the remainder if any such number is divided by 11?
Answer:
The remainder will be 3. Thus, the number can be written as:
25 = (2 × 11) + 3
36 = (3 × 11) + 3
47 = (4 × 11) + 3
ii) Write the general algebraic form of all these numbers.
Answer:
Algebraically we can write in the form of 11n + 3 where n is one of the numbers 0, 1, 2, 3….
iii) Is 100 among these numbers? What about 1000?
Answer:
For the number 100 we can write it as:
100 = (9 × 11) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10
The Sixth question has 2 questions (6A and 6B). You only need to answer one of them.
Question 6.
A (a) Consider a square of side 5 cm .
(i) Its area is 52 = 25 cm2.
(ii) If we double the side to 10 cm , the area becomes double, i.e., 50 cm2.
(iii) If we take the diagonal of the first square as side, the new square will have double the area. Which of the above statements are true?
(a) i, ii correct
(b) ii and iii correct
(c) i and iii correct
(d) Only i correct
Answer:
(c) i and iii correct
(b) The area of a square is 196 sq. cm. Then find the length of the sides?
Answer:
It is given that area of the square = 196 Sq.cm
We know that area of the square = side × side = 14 × 14 = 196 Sq. cm
Length of the sides = = \(\sqrt{196}\) = 14 cm
(c) Find the sides of the triangle in the figure
Answer:
In the given figure the one perpendicular side is 4 cm,
This side is parallel to the side of the rectangle and hence it is 4 cm.
And the other side is 8 – 5 = 3 cm
Therefore third side of the triangle
= \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5 cm
Or
(B) (a) Laws and concepts of exponents are given below:
| Column 1 | Column 2 |
| 1. xm × xn | (a) xmn |
| 2. \( \frac{x^m}{x^n} \) | (b) (xy)n |
| 3. xn × yn | (c) xm + xn |
| 4. (xm)n | (d) xm – xn |
Which of the following is true:
(A) 1 → a, 2 → c, 3 → b, 4 → d
(B) 1 → c, 2 → d, 3 → b, 4 → a
(C) 1 → b, 2 → c, 3 → d, 4 → a
(D) 1 → d, 2 → a, 3 → c, 4 → b
Answer:
(B) 1 → c, 2 → d, 3 → b, 4 → a
(b) Calculate the powers below as fractions:
(a) \(\left(\frac{3}{2}\right)^3\)
(b) \(\left(\frac{3}{5}\right)^2\)
(c) \(\left(2 \frac{3}{2}\right)^2\)
Answer:
(c) Write each product below as the product of powers of different primes: .
i) 75 × 45
Answer:
75 × 45
75 = 3 × 52
45 = 32 × 5
5 × 45 = (3 × 52) × (32 × 5)
= 33 × 53
ii) 96 × 144
Answer:
96 × 144
96 = 25 × 31
144 = 122 = (22 × 3)2 = 24 × 32
96 × 144 = (25 × 31) × (24 × 32)
= 29 × 23
iii) 72 × 175
Answer:
72 × 175
72 = 8 × 9 = 23 × 32
175 = 25 × 7 = 52 × 71
72 × 175 = (23 × 32) × (52 × 71)
