During exam preparation, Std 7 Maths Question Paper Kerala Syllabus Set 2 guide students properly.
Kerala Syllabus 7th Standard Maths Question Paper Set 2
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 the sides of a rectangle are in the ratio 3:2 and each side is doubled, the new ratio of the sides is:
(i) 6:4
(ii) 3:2
(iii) 2:3
(iv) 12:8
Answer:
(ii) 3:2
(b) If the width and height of a rectangle are 12 cm and 18 cm, find the ratio of width to height.
Answer:
12:18 = 2:3
(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 parts = 3 + 5 = 8 parts
That is per part \(\frac{160}{8}\) = 20 per part
Girls = 3 × 20 = 60
Boys = 5 × 20 = 100
(d) Consider the following statements:
(i) If both terms of a ratio are multiplied by the same number, the ratio remains unchanged.
(ii) If both terms of a ratio are divided by the same non-zero number, the ratio remains unchanged.
(iii) If only one term of a ratio is multiplied, the ratio does not change.
Options:
(a) (i) and (ii) are correct, (iii) is wrong
(b) (i) and (iii) are correct, (ii) is wrong
(c) (ii) and (iii) are correct, (i) is wrong
(d) All three are correct
Answer:
(a) (i) and (ii) are correct, (iii) is wrong
B. (a) Write the following statements using the language of algebra.
i. When 15 is added to a number, then it is equal to three times that number.
ii. If 25 is added to three times a number, the result is 70.
iii. One third of a number when added to 1 gives 15
Answer:
(i) n + 15 = 3n
(ii) 3n + 25 = 70
(iii) 1 + \(\frac{n}{3}\) = 15
(b) Find 24 + 16 + 34
Answer:
24 + 16 + 34 = 24 + (16 + 34)
= 24 + 50 = 74
(c) Find 3 × 13 + 3 × 7
Answer:
3 × 13 + 3 × 7 = 3(13 + 7)
= 3 × 20
= 60
Question 2.
(a) Which of the following is true in any triangle?
(a) The sum of two sides is smaller than the third side.
(b) The sum of two sides is equal to the third side.
(c) The sum of two sides is greater than the third side.
(d) The sum of two sides is unrelated to the third side.
Answer:
(c) The sum of two sides is greater than the third side.
(b) Find the third angle of a triangle if two angles are 65° and 75°.
Answer:
180° – (65°+ 75°) = 40°
(c) Can a triangle have sides 5cm, 7cm, and 15 cm ?
Answer:
5 + 7 = 12, which is not greater than 15.
So, No triangle possible
(d) Construct a triangle with sides 6 cm, 6 cm and 6 cm.
Answer:
That triangle is an equilateral triangle.
Question 3.
(a) A statement and reason is given below:
Statement (A): \(\frac{1}{2}\)
Reason (R): A fraction can be reduced to its simplest form by dividing numerator and denominator by their HCF. •
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 Rare 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) What is the area of a square of side 1\(\frac{1}{2}\) metre?
Answer:
Area = 1\(\frac{1}{2}\) × 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}=\frac{9}{4}\)
= 2\(\frac{1}{4}\) m
(c) If 4 strings of length \(\frac{1}{3}\) metre were laid end to end, what would be the total length?
Answer:
Total length = \(\frac{1}{3}\) × 4 = \(\frac{4}{3}\) = 1\(\frac{1}{3}\) metres
(d) Half the children in a class are girls. A third of them are in the Math Club. What fraction of the total children are they?
Answer:
Number of girls in Math club
= \(\frac{1}{3}\) of \(\frac{1}{2}\) of total children
= \(\frac{1}{3} \times \frac{1}{2}\) of total children
= \(\frac{1}{6}\) of total children
Question 4.
(a) Convert 0.07 to fraction in simplest form.
(a) \(\frac{7}{100}\)
(b) \(\frac{7}{10}\)
(c) \(\frac{7}{1000}\)
(d) \(\frac{70}{100}\)
Answer:
(a) \(\frac{7}{100}\)
(b) A rope is 27.6 m long. It is cut into 12 equal parts. What is the length of each part?
Answer:
Length of each part = \(\frac{27.6}{12}\) = 2.3 m
(c) Last year the ratio of the female teachers and male teachers of Ramapuram UP School was 6:1.
(i) If the number of male teachers is 6, what is the number of female teachers? Instead of the female teachers who got transfer from the school, male teachers joined there. Now the ratio of the female teachers and male teachers is 11:10.
(ii) How many female teachers got transfer this year?
(iii) This year some female teachers and male teachers will be 1:1. If so, how many female teachers will retire this year?
Answer:
i) Ratio of female teachers to male teachers = 6:1
Number of male teachers = 6
Number of female teachers = 6 × 6 =36
ii) Total number of teachers = 36 + 6 = 42
Ratio = 11:10
Number of female teachers
= 42 × \(\frac{11}{21}\) = 22
So, number of female teachers transferred = 36 – 22= 14
iii) Number of male teachers = 42 × \(\frac{10}{21}\) = 20
So, inorder to become ratio 1:1, number of male teachers = number of female teachers
The number of female teachers retire this year = 22 – 20 = 2
Question 5.
(a) If two triangles have the same base and their third vertices lie on a line parallel to the base, then:
(i) Both triangles have equal area
(ii) The bigger triangle has more area
(iii) The smaller triangle has more area
(iv) Cannot be determined
Answer:
(i) Both triangles have equal area
(b) A flag is triangular in shape with base 24 cm and height 18 cm. Find its area. If cloth costs ₹ 2 per cm2, find the total cost.
Answer:
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 24 × 18 = 216 cm2
Cost = 216 × 2 = ₹ 432
(c) Find the areas of the triangles shown below:
Answer:
(i) Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 8 × 6
= 24 sq.cm
(ii) Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 10 × 12
= 60 sq cm
The sixth question has 2 questions ( 6 A and 6 B). You only need to answer one of them.
Question 6.
A. (a) Take a set of five consecutive natural numbers and add all five.
i) Check if the sum has any relation with any one of the five numbers added.
Answer:
Let’s take an example of consecutive natural numbers and add them together:
34 + 35 + 36 + 37 + 38= 180
For the consecutive natural numbers 34, 35, 36, 37,38.
34 + 35 + 36 + 37 + 38 = 180 and 5 × 36 = 180
That is, the sum of five consecutive numbers is always five times the middle number.
ii) Explain why this relation holds for any five consecutive natural numbers.
Answer:
Consider the five consecutive numbers 34, 35, 36,37 and 38. These can be expressed in relation to the middle term as 36 – 2, 36 – 1, 36, and 36+ 1,36 + 2.
Thus, the sum of these five numbers is:
(36 – 2) + (36 – 1) + 36 + (36 + 1) + (36 + 2) = 180
On simplifying this expression, we get
(36 – 2) + (36- 1) + 36 + (36 + 1) + (36 + 2)
= 5 × 36 = 180
This is because, in the addition of the five numbers the 2, 1 and -1,-2 cancel each other out, resulting in a sum that is five times the middle number.
iii) Write this relation, first in ordinary language and then using algebra.
Answer:
In ordinary language, we can say it as:
The sum of any five consecutive natural numbers is five times the middle number. Algebraically, if the five consecutive numbers are x – 2, x – 1, x, and x + 1, x + 2 the sum is: (x – 2) + (x – 1) + x + (x + 1) + (x + 2) = 5x, for any natural numbers x.
(b) a) Three number pyramids are given here. Find the relation between the first and second pyramid and complete the third pyramid accordingly.
Answer:
b) Which will be the starting number to get 100 as the topmost number?
Answer:
Beginning with 10, the top most number be 100
c) Find the relation between the topmost number and the first number of the bottom row.
Write the algebraic form of it.
Answer:
The first number of the bottom can be taken as x.
Then the rest of the numbers became 2x, 3x, 4x respectively.
For the second row x + 2x = 3x
3x + 4x = 7x
Top most number is 3x + 7x = 10x
From this we can say that the top most number will be 10 times of the first number of the bottom row.
OR
B. A kudumbasree unit took a loan of rupees 480000 from a Co-operative bank for a period of one year to start
a computer institute. The bank collects 6\(\frac{1}{4}\)% interest for one year. They spend 60% of the amount to buy computers and the rest to buy furniture.
(i) What is the fraction corresponding to \(\frac{1}{4}\) % ?
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{16}\)
Answer:
(d) \(\frac{1}{16}\)
(ii) How much amount is spent on buying computers?
Answer:
Amount spent for buying Computers
= 4,80,000 × \(\frac{60}{100}\)
= ₹ 2,88,000
(iii) What is the difference between the amounts used for buying computers and furniture?
Answer:
Difference between the amount percents
= 60-40 = 20%
Difference between the amounts
= 480000 × \(\frac{20}{100}\)
= ₹ 96,000
(iv) What is the amount to be paid in the bank after one year?
Answer:
480000 × \(\frac{1}{16}\) = 30000
Amount to be repaid to the bank = 480000 + 30000
= ₹ 5,10,000