A rectangular plot of land is 25 3 4 m long and 15 2 3 m wide. Find the perimeter and area of the plot.

Length of the plot is 25 3 4 m Breadth of the plot is 15 2 3 m Perimeter of the rectangle is = 2 l + b = 2 25 3 4 + 15 2 3 = 2 103 4 + 47 3 = 2 309 + 188 12 = 2 × 497 12 = 497 6 Area of rectangle = length × breadth = 25 3 4 × 15 2 3 = 103 4 × 47 3 = 4841 12 Thus, perimeter of rectangle is 497 6 m and area of rectangle is 4841 12 m .

Represent 5 2 on a number line.

We need to represent 5 2 on the number line. In a rational number, the numeral below the bar (denominator), tells the number of equal parts into which the first unit has been divided. The numeral above the bar (numerator) tells about the number of parts that has to be considered to denote the given point on the number line. Let us represent point O to represent 0 , point B to the right of O to represent 1 . Here the denominator is 2 , the line segment OB between 0 and 1 is divided into two equal parts. Now since the numerator is 5 so five parts are taken to get the point that represents the rational number 5 2 denoted by the point A . Therefore, the number line is shown below:

Verify and name the following property: 4 7 + 2 3 + 3 5 = 4 7 + 2 3 + 3 5

We know the associative property for the addition of rational numbers say that if a , b and c are three rational numbers, then a + b + c = a + b + c is true. In the given mathematical statement 4 7 + 2 3 + 3 5 = 4 7 + 2 3 + 3 5 , the associative property for the addition of rational numbers is used. Now, we shall verify the property. We shall convert all the three given rational numbers into equivalent rational numbers with denominators equal to their LCM, which is 105 . Thus, we have, 4 7 = 4 × 15 7 × 15 = 60 105 2 3 = 2 × 35 3 × 35 = 70 105 3 5 = 3 × 21 5 × 21 = 63 105 LHS = 4 7 + 2 3 + 3 5 = 60 105 + 70 105 + 63 105 = 60 + 70 105 + 63 105 = 130 105 + 63 105 = 130 + 63 105 = 193 105 RHS = 4 7 + 2 3 + 3 5 = 60 105 + 70 105 + 63 105 = 60 105 + 70 + 63 105 = 60 105 + 133 105 = 60 + 133 105 = 193 105 So, LHS = RHS. Hence, proved.

Find three rational numbers between 2 3 and 1 .

We are given to find three rational numbers between 2 3 and 1 . We shall write 2 3 = 2 × 4 3 × 4 = 8 12 and 1 = 1 1 = 1 × 12 1 × 12 = 12 12 . We observe that the denominators of 8 12 and 12 12 are he same. There are three integers 9 , 10 and 11 between the integers 8 and 12 in the numerators. Hence, the three rational numbers between 2 3 and 1 will be 9 12 , 10 12 and 11 12 OR 3 4 , 5 6 and 11 12 .

E M B I B E

Math Sight book>Chapter 1 - Rational Numbers>Test Paper 1

Jharne De and Ranjana Kumar Solutions for Chapter: Rational Numbers, Exercise 5: Test Paper 1

Author:Jharne De & Ranjana Kumar

Jharne De Mathematics Solutions for Exercise - Jharne De and Ranjana Kumar Solutions for Chapter: Rational Numbers, Exercise 5: Test Paper 1

Attempt the free practice questions on Chapter 1: Rational Numbers, Exercise 5: Test Paper 1 with hints and solutions to strengthen your understanding. Math Sight book solutions are prepared by Experienced Embibe Experts.

Questions from Jharne De and Ranjana Kumar Solutions for Chapter: Rational Numbers, Exercise 5: Test Paper 1 with Hints & Solutions

HARD

8th CBSE

IMPORTANT

Math Sight book>Chapter 1 - Rational Numbers>EXERCISE 1.4>Q 10

A rectangular plot of land is $25 \frac{3}{4} m$ long and $15 \frac{2}{3} m$ wide. Find the perimeter and area of the plot.

HARD

8th CBSE

IMPORTANT

Math Sight book>Chapter 1 - Rational Numbers>Test Paper 1>Q 1

Represent $\frac{5}{2}$ on a number line.

MEDIUM

8th CBSE

IMPORTANT

Math Sight book>Chapter 1 - Rational Numbers>Test Paper 1>Q 3

Write the additive and multiplicative inverse of $\frac{3}{7}$ .

MEDIUM

8th CBSE

IMPORTANT

Math Sight book>Chapter 1 - Rational Numbers>Test Paper 1>Q 6

The number of girls in section $A$ of class $V I I I$ is $3 \frac{1}{2}$ times the number of girls in section $B$ of class $V I I I$ . If there are $10$ girls in section $B$ , how many girls are there in section $A$ of class $V I I I$ ?