Prasanna

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Additional Questions

Question 1.
Color the part according to the given fraction.

Solution:
(i) Here \(\frac{3}{4}\) shows out of 4 parts 3 parts are shaded

(ii) Here \(\frac{2}{4}\) shows out of 4 parts 2 parts are shaded

Question 2.
Identify the error if any

Solution:
In the given figure, shaded portion is not equal to unshaded portion. So the fraction is not equal to \(\frac{1}{2}\).

Question 3.
What fraction of an hour is 20 minutes?
Solution:
We know that total minutes in an hour = 60 min
∴ Required fraction = \(\frac{20 \min }{60 \min }=\frac{20}{60}=\frac{1}{3}\)

Question 4.
Write a fraction equivalent to \(\frac{3}{5}\) with numerator 15.
Solution:
Given, numerator of an equivalent fraction = 15
Equivalent fraction of \(\frac{3}{5}=\frac{3 \times 5}{5 \times 5}=\frac{15}{25}\)

Question 5.
Which is the larger fraction \(\frac{6}{10}\) or \(\frac{7}{10}\)?
Solution:
Here the denominators, of both fractions are same.
Also 7 > 6 So \(\frac{7}{10}>\frac{6}{10}\)

Question 6.
Sona got one-fifth of the total marks and Mala got one-third of the total marks. Who got more?
Solution:
We know that if the numerators are same in two fractions, the fraction with smaller denominator is greater.
∴ \(\frac{1}{3}>\frac{1}{5}\)
∴ Mala got more marks.

Question 7.
A piece of rope \(\frac{7}{8}\) metre long is cut into two pieces. One piece was \(\frac{1}{4}\) m long. How long is the other?
Solution:

Question 8.
Meena travelled 3\(\frac{1}{2}\) km by bus, then she walked 1\(\frac{1}{8}\) km to reach a town. How much she travelled to reach-the town?
Solution:

She travelled 4\(\frac{5}{8}\) km

Question 9.
What should be subtracted from the sum of 2\(\frac{1}{4}\) and 3\(\frac{1}{7}\) to get 2\(\frac{3}{28}\) ?
Solution:

Question 10.
Compare 4\(\frac{2}{3}\) and 5\(\frac{3}{7}\)?
Solution:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Additional Questions

Question 1.
Find the figure which is different from others.

Hint:
D is different from other figures.
All other figures has 5 lines and D has 6 lines in it.
Solution:
(d)

Question 2.
If GARIMA is coded as 725432 and TINA as 6482, how will MARTINA be coded?
(a) 3256482
(b) 3265842
(c) 3645862
(d) 3658426
Hint: GARIMA and TINA both have the lettersT and A and they are coded as 4 and 2 repectively.
∴ In MARTINA, I coded as 4 and A coded as 2 in (A) 3256482
Solution:
(a) 3256482

Question 3.
If ‘+’ is ‘×’, is ‘+% ‘×’ is ‘÷’ and ‘÷’ is then answer the following 21 ÷ 8 8 + 2 – 12 × 3 = ?
(a) 14
(b) 9
(c) 13
(d) 11

Solution:
(a) 14

Question 4.
Which number will replace the question mark?

(a) 7
(b) 14
(c) 48
(d) 49
Hint: [∴ Left side numbers are square numbers] .
Solution:
(d) 49.

Question 5.
Find the correct figure which replaces the ‘?

Hint: Upside down
Solution:

Question 6.
Which comes next ?

Hint : Here the figures rotate clockwise and anticlockwise alternatively making an angle 90° and one arrow deleted in every successive block.
Solution:

Question 7.
Find the number which replaces the question mark in the following series.

(a) 25
(b) 49
(c) 97
(d) 193

Solution:
(d) 193

Question 8.
Identify the number which replaces the question mark?

(a) 82
(b) 124
(c) 100
(d) 64
Hint: (2 + 8)² = 10² = 100
Solution:
(c) 100

Question 9.

Hint: In each step each one of the existing elements moves to the clockwise adjacent
comer. Also in one step, the element that reaches the upper-right comer gets replaced by a new element and in the next step, the element that reaches the lower left comer gets replaced by a new element.
Solution:
(d) 5

Question 10.
1, 6, 15, ?, 45, 66, 91. Find the missing term.
(a) 25
(b) 28
(c) 33
(d) 38
Hint: 1 + 5 = 6, 6 + 9 = 15, so missing term = 15 + 13 = 28
Solution:
(b) 28

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Ex 1.2

Miscellaneous Practice Problems

Question 1.
Sankari purchased 2\(\frac{1}{2}\) m cloth to stitch a long skirt and 1\(\frac{3}{4}\) m cloth to stitch blouse. If the cost is ₹ 120 per metre then find the cost of cloth purchased by her
Solution:
Cloth to stitch a long skirt = 2\(\frac{1}{2}\) m
Cloth to stitch a blouse = 1\(\frac{3}{4}\) m
Total length of the cloth = \(2 \frac{1}{2}+1 \frac{3}{4} m=2+\frac{1}{2}+1+\frac{3}{4}\) m

Cost of cloth purchased by Sankari = ₹ 510

Question 2.
From his office, a person wants to reach his house on foot which is at a distance of 5\(\frac{3}{4}\) km. If he had walked 2\(\frac{1}{2}\) km, how much distance still he has to walk to reach his house?
Solution:

Distance still he has to be walked = 3\(\frac{1}{4}\) km

Question 3.
Which is smaller? The difference between \(\frac{1}{2}\) and 3\(\frac{2}{3}\) or the sum of 1\(\frac{1}{2}\) and 2\(\frac{1}{4}\)
Solution:

Question 4.
Mangai bought 6\(\frac{3}{4}\) kg of apples. If Kalai bought 1\(\frac{1}{2}\) times as Mangai bought, then how many kilograms of apples did Kalai buy?
Solution:
Weight of apples Mangai bought = 6\(\frac{3}{4}\) kg

Weight of apples Kalai bought = 10\(\frac{1}{8}\) kg

Question 5.
The length of the staircase is 5 \(\frac{1}{2}\) m. If one step is set at \(\frac{1}{4}\) m, then how many steps will be there in the staircase?
Solution:
Length of the staircase = 5\(\frac{1}{2}\) m
Distance between each step = \(\frac{1}{4}\) m
∴ Number of steps in the staircase = \(5 \frac{1}{2} \div \frac{1}{4}=\frac{11}{2} \div \frac{1}{4}=\frac{11}{2} \times \frac{4}{1}\) = 22
There will be 22 steps in the staircase

Challenge Problems

Question 6.
By using the following clues, find who am I?
(i) Each of my numerator and denominator is a single digit number.
(ii) The sum of my numerator and denominator is a multiple of 3.
(iii) The product of my numerator and denominator is a multiple of 4
Solution:
The numerator may be any one of!, 2, 3,4, 5, 6, 7, 8, 9 and the denominator may be any one of 1, 2,3,4, 5,6, 7,8,9. Sum of numerator and denominator is a multiple of 3.
∴ Possible proper fractions are \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{2}{4}, \frac{2}{7}, \frac{3}{6}, \frac{3}{9}, \frac{4}{5}, \frac{4}{8}, \frac{5}{7}, \frac{6}{9}\)
Also given the product of numerator and denominator is a multiple of 4.
∴ Possible fractions are \(\frac{1}{8}, \frac{2}{4}, \frac{4}{5}, \frac{4}{8}\)

Question 7.
Add the difference between 1\(\frac{1}{3}\) and 3\(\frac{1}{6}\) and the difference between 4\(\frac{1}{6}\) and 2\(\frac{1}{3}\)
Solution:

Question 8.
What fraction is to be subtracted from 9\(\frac{3}{7}\) to get 3\(\frac{1}{5}\) ?
Solution:

Question 9.
The sum of two fractions is 5\(\frac{3}{9}\). If one of the fractions is 2\(\frac{3}{4}\), find the other fraction.
Solution:

The other number is 2\(\frac{7}{12}\)

Question 10.
By what number should 3\(\frac{1}{16}\) be multiplied to get 9\(\frac{3}{16}\) ?
Solution:

The number to be multiplied is 3.

Question 11.
Complete the fifth row in the Leibnitz triangle which is based on subtraction.

Solution:

Question 12.
A painted \(\frac{3}{8}\) of the wall of which one third is painted in yellow colour. What fraction is the yellow colour of the entire wall.

Solution:

\(\frac{1}{8}\) of the wall is painted yellow.

Question 13.
A rabbit has to cover 26\(\frac{1}{4}\) m to fetch its food. If it covers 1\(\frac{3}{4}\) m in one jump, thenhow many jumps will it take to fetch its food?

Solution:
Total distance to be covered by the rabbit = 26\(\frac{1}{4}\)m
Distance covered in one jump = 1\(\frac{3}{4}\) m

∴ The rabbit jumps 15 times to fetch its food.

Question 14.
Look at the picture and answer the following questions :
(i) What is the distance from School to library via bus stop?
(ii) What is the distance between school and library via Hospital?
(iii) Which is the shortest distance between (i) and (ii)?
(iv) The distance between School and Hospital is ____ times the distance between school and Bus stop.

Solution:


The distance between school and Hospital is 6 times the distance between school and bus stop.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Intext Questions

Try These (Textbook Page No. 1)

Question 1.
(i) Observe and complete:
1 + 3 = ?
5 + 11 = ?
21 + 47 = ?
___ + ____ = ?
From this observation, we conclude that “the sum of any two odd numbers is always an _____”
(ii) Observe and complete:
5 × 3 = ?
7 × 9 = ?
11 × 13 = ?
_____ × ____ = ?
From this observation, we conclude that “the product of any two odd numbers is always an _____”
Justify the following statements with appropriate examples:
(iii) The sum of an odd number and an even number is always an odd number.
(iv) The product of an odd and an even number is always an even number.
(v) The product of only three odd numbers is always an odd number.
Solution:
(i) 1 + 3 = 4
5 + 11 = 16
21 + 47 = 68
An odd number + another odd number = An Even number
From this observation, we conclude that the sum of any two odd numbers is always an even number.
(ii) 5 × 3 = 15
7 × 9 = 63
11 × 13 = 143
An odd number × Another odd number = An odd number
From this observation, we conclude that “the product of any two odd numbers is always an odd number.”
(iii) Take the odd number 5 and the even number 10
Their sum = 5 + 10 = 15, which is odd.
∴ Sum of an odd number and an even number is always an odd number.
(iv) Take the odd number 5 and the even number 10.
Their product = 5 × 10 = 50, which is even
Thus the product of an odd and an even number is always an even number.
(v) Consider 7 × 5 × 3
We know that the product of any two odd numbers is an odd number
7 × 5 = 35, odd number.
Also we have 35 × 3 = 105
∴ 7 × 5 × 3 = 105, an odd number.
So the product of three odd numbers is always an odd number.

Try These (Textbook Page No. 3)

Question 1.
(i) Say True or False
(a) The smallest odd natural number is 1.
(b) 2 is the smallest even whole number.
(c) 12345 + 5063 is an odd number.
(d) Every number is a factor of itself.
(e) A number which is a multiple of 6 is also a multiple of 2 and 3.
(ii) Is 7, a factor of 27?
(iii) Is 12, a factor or a multiple of 12?
(iv) Is 30, a factor or a multiple of 10?
(v) Which of the following numbers has 3 as a factor?
(a) 8
(b) 10
(c) 12
(d) 14
(vi) The factors of 24 are 1, 2, 3, __, 6, ___, 12 and 24. Find the missing ones.
(vii) Look at the following numbers carefully and find the missing multiples.

Solution:
(i) (a) True
(b) False
(c) False
(d) True
(e) True
(ii) No, 7 is not a factor of 27. Because 7 does not divide 27 exactly
(iii) 12 is both a factor and a multiple of 12
(iv) 30 is a multiple of 10
(v) (a) Factors of 8 are 1, 2, 4, 8
(b) Factors of 10 are 1, 2, 5, 10
(c) Factors of 2 are 1, 2, 3, 4, 6, 12
(d) Factors of 14 are 1, 2, 7, 14
∴ The number 12 has 3 as a factor
(vi) Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Missing Factors 4, 8.
(vii)

Try These (Textbook Page No. 6)

Question 1.
Express 68 and 128 as the sum of two consecutive primes.
Solution:
68 = 31 + 37
128 = 61 + 67

Question 2.
Express 79 and 104 as the sum of any three odd primes.
Solution:
79 = 37 + 31 + 11
79 = 41 + 31 + 7
79 = 61 + 11 + 7
79 = 59 + 13 + 7
79 = 53 + 19 + 7 and so on.
104 cannot be expressed as the sum of three odd primes.
Because we know that “ the sum of any two odd numbers is an even number”.
Also the sum of an odd and even number is always an odd number.
104 = 61 + 41 + 2
104 = 97 + 5 + 2
104 = 89 + 13 + 2 and so on.

Try These (Textbook Page No. 8)

Question 1.
Are the leap years divisible by 2?
Solution:
Leap years are divisible by 4.
Leap years are divisible by 2.

Question 2.
Is the first 4 digit number divisible by 3?
Solution:
The first four-digit number is 1000.
Sum of the digits is 1 + 0 + 0 + 0 = 1, not divisible by 3.
1000 is not divisible by 3.

Question 3.
Is your date of birth (DDMMYYYY) divisible by 3?
Solution:
Date of birth 25.05.2007
Sum of digits = 2 + 5 + 0 + 5 + 2 + 0 + 0 + 7 = 21
Again 2 + 1 = 3, divisible by 3.
My date of birth is divisible by 3.

Question 4.
Identify the numbers in the sequence 2000, 2006, 2010, 2015, 2019, 2025 that are divisible by both 2 and 5.
Solution:
We know that a number is divisible by both 2 and 5 if it is divisible by 10. 2000 and 2010 are divisible by 10.

Question 5.
Check whether the sum of 5 consecutive numbers is divisible by 5.
Solution:
Take the first five consecutive natural numbers 1, 2, 3, 4 and 5.
Their sum 1 + 2 + 3 + 4 + 5 = 15, divisible by 5.
Also, 2 + 3 + 4 + 5 + 6 = 20, divisible by 5.
3 + 4 + 5 + 6 + 7 = 25, divisible by 5.
Generally, the sum of 5 consecutive natural numbers is divisible by 5.

Try These (Textbook Page No. 19)

A small boy went to a town to sell a basket of wood apples. On the way, some robbers grabbed the fruits from him and ate them! The small boy went to the King and complained. The King asked him, “How many wood apples did you bring?”. The boy replied, “Your Majesty! I didn’t know, but I knew that if you divided my fruits into groups of 2, one fruit would be left in the basket”. He continued saying that if the fruits were divided into groups of 3, 4, 5 and 6, the fruits left in the basket would be 2, 3, 4 and 5 respectively. Also, if the fruits were divided into groups of 7, no fruit would be there in the basket. Can you find the number of fruits, the small boy had initially?
(This problem is taken from the famous Mathematics problems collection book in Tamil called “Kanakkathikaram” under the heading of “Wood Apple Problem”)
Solution:
The total number of fruits, when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4 and 5 respectively.
Here (2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1.
∴ The required no. of fruits will be LCM (2, 3, 4, 5, 6) – 1
L CM (2, 3, 4, 5, 6) = 2 × 3 × 2 × 5 = 60

Now take the multiples of 60 and subtract 1 from it.
Also checking the conditions, multiplies of 60 are 60, 120, 180, …..
The multiple -1
59, 119, 179, ……
The required number = 119
∴ The total number of fruits = 119.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Additional Questions

Question 1.
A pair of prime numbers whose difference is 2, is called _____
Solution:
Twin primes

Question 2.
The first 4 digit number divisible by 3 ______
Solution:
1002

Question 3.
Prime triplet ______
Solution:
(3, 5, 7)

Question 4.
100 years = ______
Solution:
100 years = 1 Century

Question 5.
Loss = ____
Solution:
Loss = CP – SP

Question 6.
526 ml _____
Solution:
526 ml = 0.526 L

Question 7.
No sides are equal _____
Solution:
Scalene Triangle

Question 8.
What is the total number of primes up to 100?
Solution:
25

Question 9.
Check whether (37, 39) is a twin prime?
Solution:
No, because 39 is not a prime number.

Question 10.
Check the divisibility by 11 of 684398?
Solution:
In 684398
Sum of digits in odd places = 8 + 3 + 8 = 19
Sum of digits in even places = 6 + 4 + 9 = 19.
Difference = 19 – 19 = 0.
684398 is divisible by 11.

Question 11.
Is 53249624 is divisible by 8? How?
Solution:
In 53249624, consider the last three digits 624, which is divisible by 8.
53249624 is divisible by 8.

Question 12.
Factorise 1056
Solution:

1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

Question 13.
Express 42 and 100 as the sum of two consecutive primes?
Solution:
42 = 19 + 23
100 = 47 + 53

Question 14.
A heap of stones can be made up into groups of 21. When made up into groups of 16, 20, 25 and 45 there are 3 stones left in each case. How many stones at least can there be in the heap?
Solution:
LCM of 16, 20, 25, 45 = 2 × 5 × 2 × 5 × 2 × 2 × 3 × 3 = 3600

The heap contain 3600 + 3 = 3603 stones at least.
3603 stones at least can there be in the heap.

Question 15.
Find the largest number of four digits exactly divisible by 12, 15, 18 and 27
Solution:
The largest number of four digits = 9999
lcm of 12, 15, 18, 27 is 540
Dividing 9999 by 540
We get 279 as the remainder


LCM = 2 × 3 × 3 × 2 × 5 × 3 = 540
Required number = 9999 – 279 = 9720

Question 16.
Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the same remainder 1 in each case.
Solution:
Required number = [LCM (6, 7, 8, 9, 12)] + 1
LCM (6, 7, 8, 9, 12) = 3 × 2 × 2 × 2 × 3 × 7 = 504
Required number = 504 + 1 = 505

Question 17.
Product of two co-prime numbers is 117. Then what will be their LCM?
Solution:
We know that LCM × HCF = Product of two numbers
Also, we know that HCF of two co-primes = 1
LCM × 1 = 117
LCM = 117

Question 18.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Solution:
LCM of 2, 4, 6, 8, 10 and 12 is 120
So the bell will toll together after every 120 seconds i.e 2 minutes.
In 30 minutes, they will toll together \(\frac{30}{2}\) + 1 = 16 times.
LCM = 2 × 2 × 3 × 2 × 5 = 120

Students can Download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks
(i) The variable in the expression 16x – 7 is _____
(ii) The constant term of the expression 2y – 6 is _____
(iii) In the expression 25m + 14M, the type of the terms are ______ terms
(iv) The number of terms in the expression 3ab + 4c – 9 is _____
Hint: Terms are 3ab, 4c – 9.
(v) The numerical co-efficient of the term -xy is ______
Hint: -x,y = (- 1 )xy.
Solution:
(i) x
(ii) -6
(iii) unlike
(iv) three
(v) -1

Question 2.
Say true or False
(i) x + (-x) = 0.
(ii) The co-efficient of ab in the term 15 abc is 15.
Hint: Coefficient of ab is 15c
(iii) 2pq and – 7qp are like terms.
(iv) When y = -1, the value of the expression 2y – 1 is 3.
Hint: 2(-1) – 1 = -2 – 1 = – 3
Solution:
(i) True
(ii) False
(iii) True
(iv) False

Question 3.
Fing the numerical co-efficient of each of the following terms: -3yx, 12k, y, 121bc, -x, 9pq, 2ab.
Solution:
(i) Numerical co-efficient of-3yx is – 3
(ii) Numerical co-efficient of 12k is 12
(iii) Numerical coefficient of y is 1
(iv) Numerical co-efficient of 1216c is 121
(v) Numerical co-efficient of – x is – 1
(vi) Numerical co-efficient of 9pq is 9
(vii) Numerical co-efficient of 2ab is 2

Question 4.
Write the variables, constants and terms of the following expressions,
(i) 18 + x – y
(ii) 7p – 4q + 5
(iii) 29x + 13y
(iv) b + 2
Solution:

Question 5.
Identify the like terms among the following 7x, 5y, -8x, 12y, 6z, z, -12x, -9y, 11 z
Solution:

Question 6.
If x = 2 andy = 3, then find the value of the following expressions,
(i) 2x – 3y
(ii) x + y
(iii) 4y – x
(iv) x + 1 – y
Solution:
Given x = 2; y = 3.
(i) 2x – 3y = 2 (2) – 3 (3) = 4 – 9
= 4 + (Additive inverse of 9)
= 4 +(-9) = -5
(ii) x + y = 2 + 3 = 5
(iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10
(iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0

Objective Type Questions

Question 1.
An algebraic statement which is equivalent to the verbal statement “Three times the sum of ‘x’ and ‘y’ is
(i) 3 (x + y)
(ii) 3 + x + y
(iii) 3x + y
(iv) 3 + xy
Solution:
(i) 3 [(x + y)]

Question 2.
The numerical co-efficient of -7mn is
(i) 7
(ii) -7
(iii) p
(iv) -p
Solution:
(ii) -7

Question 3.
Choose the pair of like terms
(i) 7p, 7x
(ii) 7r, 7x
(iii) – 4x, 4
(iv) – 4x, 7x
Solution:
(iv) -4x, 7x

Question 4.
The value of 7a – 4b when a = 3, b = 2 is
(i) 21
(ii) 13
(iii) 8
(iv) 32
Solution:
(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.3

Miscellaneous Practice Problems

Question 1.
If the cost of 7 kg of onions is ₹ 84 find the following :
(i) Weight of the onions bought for ₹ 180
(ii) The cost of 3 kg of onions
Solution:
(i) For ₹ 84 weight of onion bought
for ₹ 1 weight of onion bought
∴ For ₹ 180 weight of onion bought w
∴ For ₹ 180 weight of onion bought

(ii) Cost of 7 kg of onions = 15 kg

Question 2.
If C = kd
(i) what is the relation between C and d ?
(ii) Find k when C = 30 and d = 6
(iii) Find C, when d = 10
Solution:

As C increases d also increases
∴ It is direct proportion

Question 3.
Every 3 months Tamilselvan deposits ₹ 5000 as savings in his bank account. In how many years he can save ₹ 1,50,000.
Solution:
Let the number of years required be x.

No. of years and deposit are direct proportion as they both increases simultaneously.

He can save ₹ 1,50,000 in \(7 \frac{1}{2}\) years.

Question 4.
A printer, prints a book of 300 pages at the rate of 30 pages per minute. Then, how long will it take to print the same book if the speed of the printer is 25 pages per minute?
Solution:
Let the required time taken to print be x
As the speed increases time taken to print decreases
∴ They are in inverse proportion
Time taken to print 30 pages = 1 min

Question 5.
If the cost of 6 cans of juice in ₹ 210, then what will be the cost of 4 cans of juice?
Solution:
Let the cost required be x

As number of cans increases cost also increases.
∴ They are in direct proportion

x = 140
Cost of 4 cans of juice = 140

Question 6.
x varies inversely as twice of y. Given that when y = 6, the value of x is 4. Find the value of x when y = 8.
Solution:
Given x varies inversely as twice of y.

Question 7.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required to cover a distance of 1650 km?
Solution:
Let the required distance be x

As the distance increases fuel quantity also increases.
∴ They are direct proportion.

∴ The diesel required = 300 liters

Challenge Problems

Question 8.
If the cost of a dozen soaps is ₹ 396, what will be the cost of 35 such soaps?
Solution:
1 dozen = 12
Cost of 12 soaps = ₹ 396

Question 9.
In a school there is 7 periods a day each of 45 minutes duration. How long each period is if the school has 9 periods a day assuming the number of hours to be the same?
Solution:
Number of periods increases as duration decreases, since the number of hours is same.
Let the duration of each period be x.

Question 10.
Cost of 105 note books is ₹ 2415. How many notebooks can be bought for ₹ 1863?
Solution:
For 2415 number of notebooks bought = 105

Question 11.
10 farmers can plough a field in 21 days. Find the number of days reduced if 14 farmers ploughed the same field?
Solution:
Let the required number of days if 14 farmers ploughed = x

As number of farmers increases, number of days decreases.
∴ They are in inverse proportion

Initially the farmers worked for 21 days. Now they worked for 15 days.
∴ The number of days reduced = 21 – 15 = 6 days

Question 12.
A flood relief camp has food stock by which 80 people can be benefited for 60 days. After 10 days 20 more people have joined the camp. Calculate the number of days of food shortage due to the addition of 20 more people?
Solution:

As number of people increases food last for less number of days.

Remaining food is to be used for 50 days.
But it only last for 40 days.
No. of days shortage = 50 – 40 = 10 days.
∴ 10 days of food shortage due to the addition of 20 more people.

Question 13.
Six men can complete a work in 12 days. Two days later, 6 more men joined them. How many days will they take to complete the remaining work?
Solution:

As the number of men increases number of days increases.
∴ They are inversely proportional

x = 5 days
∴ Remaining work will be complete in 5 days

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.2

Question 1.
Fill in the blanks
(i) 16 taps can fill a petrol tank in 18 minutes. The time taken for 9 taps to fill the same tank will be ___ minutes.
(ii) If 40 workers can do a project work in 8 days, then ____ workers can do it in 4 days.
Solutions:
(i) 32
(ii) 80

Question 2.
6 pumps are required to fill a water sump in 1 hr 30 minutes. What will be the time taken to fill the sump if one pump is switched off?
Solution:
Let x be the required time taken

Time taken in minutes 1 hr. 48m

Question 3.
A farmer has enough food for 144 ducks for 28 days. If he sells 32 ducks how long will the food last?
Solution:
Let the required number of days be x.

As the number of ducks decreases the food will last for more days.
∴ They are in inverse proportion. x₁y₁ = x₂y₂

The food lasts for 36 days

Question 4.
It takes 60 days for 10 machines to dig a hole. Assuming that all machines work at the same speed, how long will it take 30 machines to dig the same hole?
Solution:
Let the number of days required be x.

As the number of machines increases it takes less days to complete the work
∴ They are in inverse proportion, x₁y₁ = x₂y₂

It takes 20 days to dig the hole

Question 5.
Forty students stay in a hostel. They had food stock for 30 days. If the students are doubled then for how many days the stock will last?
Solution:
Let the required number of days be x.

As the number of students increases the food last for less number of days
∴ They are in inverse proportion.

The food stock lasts for 15 days

Question 6.
Meena had enough money to send 8 parcels each weighing 500 grams through a courier service. What would be the weight of each parcel, if she has to send 40 parcel for the same money?
Solution:
Let the required weight of the parcel be x grams.

As the number of parcels increases weight of a parcel decreases.
∴ They are in inverse proportion.

Weight of each parcel = 100 grams

Question 7.
It takes 120 minutes to weed a garden with 6 gardeners. If the same work is to be done in 30minutes, how many more gardeners are needed?
Solution:
Let the, number of gardeners needed be x.

As the number of gardeners increases the time decreases. They are in inverse proportion,
x₁y₁ = x₂y₂

∴ To complete the work in 30 min gardeners needed = 24
Already existing gardeners = 6
∴ More gardeners needed = 24 – 6 = 18
18 more gardeners are needed

Question 8.
Neelaveni goes by bicycle to her school every day. Her average speed is 12km/hr and she reaches school in 20 minutes. What is the increase in speed, If she reaches the school in 15 minutes?
Solution:
Let the speed to reach school in 15 min be x

∴ They are in inverse proportion x₁y₁ = x₂ y₂

If she reaches in 15 min the speed = 16 km/hr
Already running with 12 km / hr
∴ Increased speed = 16 – 12 = 4km / hr
Increase in speed = 4 km / hr

Question 9.
A toy company requires 36 machines to produce car toys in 54 days. How many machines would be required to produce the same number of car toys in 81 days?
Solution:
Let the required number of machines be x

As the number of machines increases number of days required decreases.

∴ 24 machines would be required

Objective Type Questions

Question 10.
12 cows can graze a field for 10 days. 20 cows can graze the same field for ____ days
(i) 15
(ii) 18
(iii) 6
(iv) 8
Solution:
(iii) 6
Hint:

Question 11.
4 typists are employed to complete a work in 12 days. If two more typists are added, they will finish the same work in days
(i) 7
(ii) 8
(iii) 9
(iv) 10
Solution:
(ii) 8

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.1

Fill in the blanks.

(i) If the cost of 8 apples is 56 then the cost of 12 apples is ____.
(ii) If the weight of one fruit box is \(3 \frac{1}{2}\) kg, then the weight of 6 such boxes is ____.
(iii) A car travels 60 km with 3 liters of petrol. If the car has to cover the distance of
200 km, it requires ___ liters of the petrol.
(iv) If 7 m cloth costs ₹ 294, then the cost of 5m of cloth is ____.
(v) If a machine in a cool drinks factory fills 600 bottles in 5 hrs, then it will fill _____ bottles in 3 hours.
Solutions:
(i) 84
(ii) 21 kg
(iii) 10
(iv) ₹ 210
(v) 360

Question 2.
Say True or False
(i) Distance travelled by a bus and time taken are in direct proportion.
(ii) Expenditure of a family to number of members of the family are in direct proportion.
(iii) Number of students in a hostel and consumption of food are not in direct proportion.
(iv) If Mallika walks 1km in 20 minutes, then she can convert 3km in 1 hour.
(v) If 12 men can dig a pond in 8 days, then 18 men can dig it in 6 days.
Solutions:
(i) True
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
A dozen bananas costs ₹ 20. What is the price of 48 bananas ?
Solution:
Let the required price be ₹ x. As the number of bananas increases price also increases
∴ Number of bananas and cost are in direct proportion.

Question 4.
A group of 21 students paid ₹ 840 as the entry fee for a magic show. How many students entered the magic show if the total amount paid was ₹ 1680?
Solution:
Let the required number of students be x.

As the number of students increases the entry fees also increases.
∴ They are in direct proportion .

∴ The number of students entered magic show = 42

Question 5.
A birthday party is arranged in third floor of a hotel. 120 people take 8 trips in a lift to go to the party hall. If 12 trips were made how many people would have attended the party?
Solution:
Let the number of people attended the party be x.

As the number of trips increases, number of people also increases.
∴ They are in direct proportion.


180 people attend the party in 12 trips

Question 6.
The shadow of a pole with height of 8m is 6m. If the shadow of another pole measured at the same time is 30m, find the height of the pole?
Solution:
Let the required height of the pole be ‘x’ m.

Height of the pole and its shadow are in direct proportion

∴ Height of the pole x = 40m.

Question 7.
A postman can sort out 738 letters in 6 hours. How many letters can be sorted in 9 hours?
Solution:
Let the required number of letters be x.

They are in direct proportion.

In 9 hours 1107 letters can be sorted.

Question 8.
If half a meter of cloth costs ₹ 15. Find the cost of \(8 \frac{1}{3}\) meters of the same cloth.
Solution:
Let the cost of cloth required be x.

Cost and length are in direct proportion.

Question 9.
The weight of 72 books is 9 kg. What is the weight of 40 such books (using unitary method)
Solution:
Weight of 72 books = 9 kg = 9000 g
∴ Weight of 1 book = \(\frac{9000}{72}\) = 125 g
∴ Weight of 40 books = 125 × 40 g = 5000 g = 5 kg.
Weight of 40 books = 5 kg

Question 10.
Thamarai pages ₹ 7500 as rent for 3 months. With the same rate how much does she have to pay for 1 year (using unitary method).
Solution:
Rent paid by Thamarai for 3 months = ₹ 7500
∴ Rent paid for 1 month = \(\frac{7500}{3}\) = 2500
Rent paid for 1 year or 12 moths = 2500 × 12 = ₹ 30,000
For 1 year rent to be paid = ₹ 30,000

Question 11.
If 30 men can reap a field in 15 days, then in how many days can 20 men reap the same field? (using unitary method).
Solution:

∴ 20 men can reap the field in 10 days.

Question 12.
Valli purchase 10 pens for ₹ 180 and Kamala boys 8 pens for ₹ 96. Can you say who bought the pen cheaper (using unitary method).
Solution:

∴ Kamala bought the pen cheaper.

Question 13.
A motorbike requires 2 liters of petrol to cover 100 kilometres. How many liters of petrol will be required to cover 250 kilometers? (using unitary method).
Solution:
To cover 100 km quantity of petrol required = 2 litres

5 litres of petrol required to cover 250 km

Objective Type Questions

Question 14.
If the cost of 3 books is ₹ 90, then find the cost of 12 books.
(i) ₹ 300
(ii) ₹ 320
(iii) ₹ 360
(iv) ₹ 400

Solution:
(iii) ₹ 360

Question 15.
If Mani buys 5 kg of potatoes for ₹ 75 then he can buy ₹ 105.
(i) 6
(ii) 7
(iii) 8
(iv) 5

Solution:
(ii) 7

Question 16.
35 cycles were produced in 5 days by a company then ___ cycles will be produced in 21 days.
(i) 150
(ii) 70
(iii) 100
(iv) 147

Solution:
(iv) 147

Question 17.
An aircraft can accommodate 280 people in 2 trips. It can take ______ trips to take 1400 people.
(i) 8
(ii) 10
(iii) 9
(iv) 12
Solution:
(ii) 10

Question 18.
Suppose 3 kg of sugar is used to prepare sweets for 50 members, then ___ kg of sugar is required for 150 members.
(i) 9
(ii) 10
(iii) 15
(iv) 6

Solution:
(i) 9

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Exercise 2.1

Parallelogram

(Try These Text book Page No. 33)

Question 1.
Find the missing values for the following:

Solution:
(i) Given length l = 12 m; Breadth b = 8 cm
∴ Area of rectangle = l × b sq. units = 12 × 8 m² = 96 m²
Perimeter of the rectangle = 2 × (1 + b) units = 2 × (12 + 8)m = 2 × 20 = 40m

(ii) Given Length l = 15 cm ; Area of the rectangle = 90 sq. cm
l × b = 90; 15 × 6 = 90; b = \(\frac{90}{15}\) = 6 cm
Perimeter of the rectangle = 2 × (l+ b) units = 2 × (15 + 6) cm = 2 × 21 cm = 42 cm

(iii) Given Breadth of rectangle = 50 mm ; Perimeter of the rectangle = 300 mm
2 × (l + b) = 300
2 × (l + 50) = 300
l + 50 = \(\frac{300}{2}\) = 150
l = 150 – 50
l = 100
Area = l × b sq. untis = 100 × 50 mm² = 5000 mm²

(iv) Length of the rectangle = 12 cm ; Perimeter = 44 cm
2(l + b) = 44
2(12 + b) = 44
12 + b = \(\frac{44}{2}\)
12 + b = 22 ; b = 22 – 12; b = 10 cm
Area = l × b sq. units
= 12 × 10 cm² = 120 cm²

Question 2.

Solution:
(i) Given side a = 60 cm
Area of the square = a × a sq.units = 60 × 60 cm² = 3600 cm²
Perimeter of the square = 4 × a units = 4 × 60 cm = 240 cm

(ii) Given area of a square = 64 sq. m
a × a = 64
a × a = 8 × 8
a = 8m
Perimeter = 4 × a
= 4 × 8
= 32 m

(iii) Given perimeter of the square = 100 mm
4 × a = 100
a = \(\frac{100}{4}\) mm
a = 25 mm
Area = a × a sq. units
= 25 × 25 mm²
= 625 mm²

Question 3.

Solution:
(i) Given base of the right angled triangle = 13 m ; height = 5 m

(iii) Given height h = 6 mm ; Area = 84 sq. mm
\(\frac{1}{2}\) × b × h = 84 ; \(\frac{1}{2}\) × b × 6 = 84
b = \(\frac{{84} \times 2}{6}\); b = 28 mm

(Try This Textbook Page No. 35)

Question 1.
Explain the area of the parallelogram as sum of the areas of the two triangles.
Solution:
ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.

Area of the parallelogram ABCD = base × height
= AB × DE

Question 2.
A rectangle is a parallelogram but a parallelogram is not a rectangle. Why?
Solution:
(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to 90°. But for parallelogram opposite angles are equal.
∴ All rectangles are parallelograms. But all parallelograms are not rectan¬gles as their angles need not be equal to 90°.

(Try These Textbook Page No. 36)

Question 1.
Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).

(a) ______ sq. units
(b) ______ sq. units
(c) ______ sq. units
(d) ______ sq. units
Solution:
Converting the given parallelograms into rectangles we get.

(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Question 2.
Draw the heights for the given parallelograms and mark the measure of their bases and find the area. Analyze your result.

Solution:
(a) Area of the parallelogram = b × h sq. units
= 4 × 2 sq. units = 8 sq. units

By counting the small squares also we get number of full
squares + number of square more than half = 6 + 2 = 8 sq. units.

(b) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

(c) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

Also area = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) Number of half squares = 4 + 4 = 8 sq. units

(d) Area of the parallelogram = (base × height) sq. units = 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + \(\frac{1}{2}\) [Number of half squares] + Number of squares more than half = 4 + 0 + 4 = 8sq. units

(e) Area of parallelogram = (base × height) sq. units
= 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) [Number of half squares] = 2 + 6 = 8sq. units

Question 3.
Find the area o the following parallelograms by measuring their base and height, using formula.

(a) _____ sq. units
(b) _____ sq. units
(c) _____ sq. units
(d) _____ sq. units
(e) _____ sq. units
Solution:
(a) Area of the rectangle = (base × height) sq. units
base = 5 units
height = 5 units
∴ Area = (5 × 5 ) = sq. units = 25 sq. units

(b) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 1 units
∴ Area = (4 × 1 ) = sq. units = 4 sq. units

(c) Area of the rectangle = (base × height) sq. units
base = 2 units
height = 3 units
∴ Area = (2 × 3 ) = sq. units = 6 sq. units

(d) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 4 units
∴ Area = (4 × 4 ) = sq. units = 16 sq. units

(e) Area of the parallelogram = (base × height) sq. units
base = 7 units
height = 5 units
= 7 × 5 = 35 sq. units

Question 4.
Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.
Solution:

Area of parallelogram (a), (b) or (c) = 20 sq. units

Exercise 2.2

Rhombus

(Try These Textbook Page No. 41)

Question 1.
Observe the figure and answer the following questions.
(i) Name two pairs of opposite sides.
(ii) Name two pairs of adjacent sides.
(iii) Name the two diagonals.

Solution:
(i) (a) PQ and RS (b) QR and PS
(ii) (a) PQ and QR (b) PS and RS
(iii) (a) PR and Question are diagonals.

Question 2.
Find the area of the rhombus given in (i) and (ii).

Solution:
(i) Area of the rhombus = \(\frac{1}{2}\) (d₁ + d₂) sq. units = \(\frac{1}{2}\)(11 + 13) sq. units
= \(\frac{1}{2}\) × (24) cm² = 12 cm²

(ii) Base = 10 cm ; Height = 7 cm
Area of the rhombus = b × h sq. units = 10 × 7 cm² = 70 cm²

Question 3.
Can you find the perimeter of the rhombus?
Solution:
If we know the length of one side we can find the perimeter using 4 × side units.

Question 4.
Can diagonals of a rhombus be of the same length?
Solution:
When the diagonals of a rhombus become equal it become a square.

Question 5.
A square is a rhombus but a rhombus is not a square. Why?
Solution:
In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to 90°.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.

Question 6.
Can you draw a rhombus in such a way that the side is equal to the diagonal.
Solution:
Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.

Exercise 2.3

(Try These Textbook Page No. 46)

Question 1.
Can you find the perimeter of the trapezium? Discuss.
Solution:
If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.

Question 2.
In which case a trapezium can be divided into two equal triangles?
Solution:
If two parallel sides are equal in length. Then it can be divided into two equal triangles.

Question 3.
Mention any three life situations where the isosceles trapeziums are used?
Solution:
(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.