Kerala Syllabus Class 7 Maths Model Question Paper Set 1

During exam preparation, Std 7 Maths Question Paper Kerala Syllabus Set 1 guide students properly.

Kerala Syllabus 7th Standard Maths Question Paper Set 1

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) If 2³ × 2⁴ = 2ⁿ, then the value of n is:
(i) 7
(ii) 12
(iii) 81
(iv) 64
Answer:
(i) 7

Using x^m × xⁿ = x^m+n
2³ × 2⁴ = 2³⁺⁴
= 27

(b) Calculate the following quotients;
(i) \(\frac{1440}{120}\)
Answer:
1440 = 12² × 10
= (2² × 3)² × (2¹ × 5¹)
= 2⁵ × 3² × 5¹
120 = 12 × 10 (2² × 3) × (2 × 5)
120 = 2³ × 3¹ × 5¹
\(\frac{1440}{120}=\frac{2^5 \times 3^2 \times 5^1}{2^3 \times 3^1 \times 5^1}\) = 2^5-3 × 3^2-1 × 5^1-1
= 2² × 3¹ × 5⁰
= 4 × 3 × 1
= 12

(ii) \(\frac{729}{27}\)
Answer:
\(\frac{729}{27}\)
729 = 3⁶
27 = 3³
\(\frac{729}{27}=\frac{3^6}{3^3}\)
= 3^6-3 = 3³ = 27

(c) Write each number below as a product of powers of different primes:
i) 18² × 30²
Answer:
18² × 30² = (2¹ × 3²) × (2¹ × 3¹ × 5¹)²
= 2²⁺² × 3⁴⁺² × 5²
= 2⁴ × 3⁶ × 5²

ii) 20³ × 27¹
Answer:
20³ × 27¹ = (2² × 5¹)³ × (3³) = 2⁶ × 5³ × 3³

(d) The area of a square is 2⁶ square units. Find its side length.
Answer:
Area of square = (side)²
So, (side)² = 26
⇒ side = 2^6/2 = 2³ = 8
Therefore side = 8 units

OR

B. (a) Which of the following is correct?
(i) (x²)³ = x⁶
(ii) (x³)² = x⁵
(iii) x³ × y³ = (x + y)³
(iv) (xy)² = x² + y²
Answer:
(i) (x²)³ = x⁶ is correct

(b) Simplify: \(\frac{2^7 \times 2^3}{2^5}\)
Answer:
Numerator = 2⁷⁺³ = 2¹⁰
Now, \(\frac{2^{10}}{2^5}\) = 2^10-5
= 2⁵
= 32

(c) A bacteria population doubles every hour. If the initial count is 2⁵, what will be the count after 3 hours?
Answer:
Population after 3 hours = Initial × 2³.
= 2⁵ × 2³
= 2⁵⁺³
= 2⁸
= 256 bacteria

Question 2.
(a) i. The medians of a triangle intersect at a point called centroid.
ii. The centroid divides each median in the ratio 2:1.
iii. The perpendicular bisectors of a triangle always meet at the centroid. Which of the above statements are true?
A. i and ii correct
B. ii and iii correct
C. i and iii correct
D. All are correct
Answer:
A. i and ii correct

(b) Draw a triangle of sides 5, 6 and 7 centimetres. Draw three different right triangles of the same area.
Answer:

This is the required triangle with sides 5,6 and 7 cm.
Here are three different triangles, each with the same area:

(c) Draw squares of areas 5² square centimetres and 21 square centimetres.
Answer:
For the area of square 5² square centimetres
We can split 5² as
5² = 36 + 16 = 6² + 4²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 6 and 4 cm, then the area of the square on the hypotenuse is 52 sq.cm.

For the area of square 21 square centimetres We can split 21 as difference of two squares:
21 = 25 – 4 = 5² – 2²
So, if we draw a right triangle with a hypotenuse of 5 centimetres and another side of 2 centimetres then the area of the square on the third side would be 21 square centimetres

Question 3.
(a) A statement and reason is given below:
Statement (A): If the length and breadth of a rectangle are multiplied by the same number, the ratio of length to breadth remains the same.
Reason (R): Multiplying or dividing both terms of a ratio by the same non-zero number does not change the ratio.
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 true, but R is not the correct explanation of A.
Answer:
(c) Both A and R are true, and R is the correct explanation of A.

(b) If the sides of a rectangle are in the ratio 3:2 and each side is doubled, the new ratio of the sides is:
(a) 6:4
(b) 3:2
(c) 2:3
(d) 12:8
Answer:
(b) 3:2
New ratio = (3 × 2) : (2 × 2)
= 6 : 4 = 3 : 2

(c) In a class, the ratio of girls to boys is 3:5. If the total students are 160, find the number of girls and boys.
Answer:
Total ratio = 8 parts
That is, \(\frac{160}{3}\) = 20 per part
Girls = 3 × 20 = 60
Boys = 5 × 20 = 100

(d) A drink is made by mixing juice and water in the ratio 3:2. If 15 liters of juice is used, how much water should be added?
Answer:
Juice: Water = 3:2
For 15 L of juice,
Water = (\(\frac{2}{3}\)) × 15 = 10 L

Question 4.
(a) The fraction 7/20 expressed as a percent is:
(i) 20%
(ii) 25%
(iii) 35%
(iv) 40%
Answer:
(iii) 35%
(7/20) × 100 = 35%

(b) A shopkeeper gave a discount of 20% on a shirt. If the price of the shirt was ₹ 750, how much discount did the customer get?
Answer:
Discount = 20 % = \(\frac{20}{100}=\frac{1}{5}\)
Discount got by consumer = \(\frac{1}{5}\) × 750 = ₹ 150.

(c) A person who earns 45,000 rupees a month spends 10,500 rupees on rent. What percent of their earnings does this spending represent?
Answer:
Percentage = \(\left(\frac{10,500}{45,0000}\right)\) × 100
\(\frac{10,500}{45,000}=\frac{105}{450}=\frac{7}{30}\)

Percentage = \(\left(\frac{7}{30}\right)\) × 100
= 23\(\frac{1}{3}\)%
= 23.33 %

Question 5.
(a) In the figure, shaded part represents the product of two fractions, See the statements Babu wrote related to this.

(i) \(\frac{2}{5}\) parts of \(\frac{2}{3}\)
(if) \(\frac{1}{3}\) parts of \(\frac{2}{5}\)
(iii) \(\frac{4}{5}\) parts of \(\frac{2}{3}\)
(iv) \(\frac{4}{15}\) parts of the large rectangle
Which of the above statements are true?
(a) Only iv
(c) (i) and (iv)
(b) (i) and (ii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iv)

(b) A man travels 2\(\frac{1}{4}\) kilimetres in one hour. What is the distance he travelled in 2\(\frac{1}{3}\) hours?
Answer:
Speed = 2\(\frac{1}{4}\) km = \(\frac{9}{4}\)
Time = 2\(\frac{1}{3}\) = \(\frac{7}{3}\)
Distance = speed × time
= \(\frac{9}{4} \times \frac{7}{3}=\frac{63}{12}\)
= 5\(\frac{1}{4}\) km

(c) A 4 metre long rope is divided into 5 pices of equal length. What is the length of 3/5 part of its orie piece in metres? In centimetres?
Answer:
Length of one piece = \(\frac{4}{5}\) m
\(\frac{3}{5}\) of one piece = \(\frac{3}{5} \times \frac{4}{5}=\frac{12}{25}\) = 0.48m
Length = 0.48 m = 48 cm

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

Question 6.
A. (a) The division operations in column A are connected with the fractions in column B.

Which of the following is true?
(a) 1 → a, 2 → b, 3 → c
(b) 1 → c, 2 → a, 3 → d
(c) 1 → c, 2 → d, 3 → a
(d) 1 → c, 2 → a, 3 → b
Answer:
(a) 1 → a, 2 → b, 3 → c

(b) We want to fill 4\(\frac{1}{4}\) litres milk in identical bottles. A bottle can contain \(\frac{3}{4}\) litre milk.
How many bottles will be there with full of milk?
How many more litres of milk is needed to fill the bottle which is not completely filled?
Answer:
Total milk = 4\(\frac{1}{4}\)L = \(\frac{17}{4}\) L
Capacity of one bottle = \(\frac{13}{4}\) L
Number of bottles = \(\frac{17}{4} \div \frac{3}{4}=\frac{17}{4} \times \frac{4}{3}=\frac{17}{3}\) = 5\(\frac{2}{3}\)
So 5 full bottles can be filled, and there will be some milk left for the 6th.
Milk in 5 bottles = 5 × \(\frac{3}{4}=\frac{15}{4}\)
Left over milk = \(\frac{17}{4}-\frac{15}{4}=\frac{2}{4}=\frac{1}{2}\) liters
So the 6th bottle gets \(\frac{1}{2}\) liters
Extra milk needed to fill the 6th bottle
= \(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\) litre

(c) How many times of 1\(\frac{1}{2}\) is 5 ?
Answer:
1\(\frac{1}{2}\) = \(\frac{3}{2}\)
Number of times = 5 × \(\frac{2}{3}=\frac{10}{3}\) = 3\(\frac{1}{3}\)
5 is 3\(\frac{1}{3}\) times of 1\(\frac{1}{2}\)

OR

B. (a) A big vessel contains 25 litres of water and a small vessel contains 15 litres of water.
(i) Water in the smaller vessel is 3/5 part of the larger:
(ii) Water in the smaller vessel is 2/3 part of the larger.
(iii) Water in the larger vessel is 5/3 times the smaller.
Which of the statements given above are true?
(a) i, ii correct
(b) Only iii is correct
(c) i and iii correct
(d) ii and iii correct
Answer:
(c) i and iii correct

(b) How many pieces of 1\(\frac{1}{2}\) metres long rope can be cut off from 8\(\frac{1}{2}\) metres long rope? What is the length of the remaining rope?
Answer:
Total rope = 8\(\frac{1}{2}\) = \(\frac{17}{2}\)m
Length ofone piece = 1\(\frac{1}{2}\) = \(\frac{3}{2}\)m

Number of pieces = \(\frac{17}{2} \div \frac{3}{2}=\frac{17}{2} \times \frac{2}{3}=\frac{17}{3}=5 \frac{2}{3}\)
so 5 full pieces can be cut.

Length for 5pieces = 5 × \(\frac{3}{2}=\frac{15}{2}\) = 7\(\frac{1}{2}\)m

Remaining length = \(\frac{17}{2}-\frac{15}{2}=\frac{2}{2}\) = 1 m

(c) Among two numbers, larger is 1\(\frac{2}{3}\) times the smaller. Then, what part of the larger-number is the smaller?
Answer:
Larger number = 1\(\frac{1}{2}\) times the smaller
1\(\frac{1}{2}\) = \(\frac{5}{3}\)
Larger number = \(\frac{5}{3}\) × smaller
Therefore smaller = Larger \(\frac{5}{3}\)
= Larger × \(\frac{3}{5}\)
The smaller number is \(\frac{3}{5}\) part of the larger.