RS Aggarwal Class 8 Chapter 1 Assertion Reason Solutions
In this post we have provided assertion reason questions and solution of RS Aggarwal class 8 chapter 1. The chapter 1 is based on the basic concepts rational numbers.
Complete RS Aggarwal solution for class 8 is also available in our website.
Rational Numbers
Assertion – Reason Questions
Directions (Questions 1 to 10): Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
a.) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
b.) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
c.) Assertion (A) is true but Reason (R) is false.
d.) Assertion (A) is false but Reason (R) is true.
1.) Assertion (A): If p/q is a rational number then q cannot be equal to 0.
Reason (R): Division by zero is not defined.
Ans:
Option (a) is the correct option as both the assertion (A) and the reason (R) are true and Reason (R) is the correct explanation for the Assertion.
2.) Assertion (A): Every whole number is a rational number.
Reason (R): 0 is a whole number which is not a rational number.
Ans:
Option (c) is correct as the assertion A is true and but the reason R is false.
0 is both a whole numbers as well as a rational numbers. That’s why, A is true R is false.
3.) Assertion (A): The additive inverse of a rational number is always a negative rational number.
Reason (R): If ‘a’ is a rational number then its additive inverse is (-a).
Ans:
Option (d) is correct as the assertion is false but the reason is true.
We know that both positive and negative numbers can be rational numbers. So the additive inverse of a positive rational number will be a negative rational number while the additive inverse of a negative rational number will be a positive rational number. So the A is wrong.
The additive inverse of a rational number ‘a’ is (-a). So the reason is true.
Thus, A is false but R is true.
4.) Assertion (A): Between every two rational numbers there exists a unique rational number.
Reason (R): If x and y be two rational numbers such that x<y then ½ (x + y) is a rational number between x and y.
Ans:
Option (d) is correct as the assertion A is wrong and the reason R is correct.
Between every two rational numbers there are infinite numbers of rational numbers. So, A is false.
However when we divide the sum of two rational numbers by two the answer is always between the two given numbers. So R is true.
5.) Assertion (A): There are three rational numbers which are their own reciprocals namely -1, 0 and 1.
Reason (R): b/a is called the reciprocal of a/b
Ans:
Option (d) is correct as the assertion A is wrong and the reason R is correct.
We know that – 1 and 1 are their own reciprocals, but reciprocal of 0 cannot be defined. So, A is false.
a/b and b/a are reciprocal to each other so R is true.
6.) Assertion (A): 2/3 × (4/5+6/7) = (2/3×4/5) + (2/3×6/7)
Reason (R): Multiplication is distributive over addition for rational numbers.
Ans:
Option (a) is correct as both the assertion and the reason are correct and the given reason is the correct explanation for the assertion.
According to distributive law of multiplication over addition of rational numbers, we have
a/b× (c/d + e/f)=(a/b × c/d)+(a/b × e/f)
So, both A and R are true and R correctly explains A.
7.) Assertion (A): Nonzero rational numbers are closed under addition, subtraction, multiplication and division.
Reason (R): The sum, difference and product of two rational numbers is a rational number but division by 0 is not defined.
Ans:
Option (a) is correct as the given assertion A and the reason R is correct.
8.) Assertion (A): The product of the additive inverse and multiplicative inverse of a rational number is – 1.
Reason (R): If ‘a’ is a rational number then its additive inverse is (-a) and its multiplicative inverse is 1/a.
Ans:
Option (a) is correct as the given the assertion is true
As we can see (-a) × 1/a=-1.
The Reason given is also correct
9.) Assertion (A): Addition and multiplication of rational numbers is both commutative and associative.
Reason (R): The rational numbers may be added or multiplied in any order or by grouping in any order. The sum or product remains the same.
Ans: Option (a) is correct as the assertion and reason are both true.
If a, b and c are rational numbers then by commutative law, we have
a + b=b + a and a × b= b × a
And, by associative law, we have
a+(b + c)= (a +b)+c and (a × b) ×c= a×(b × c).
10.) Assertion (A): There exists a unique rational number whose additive inverse and multiplicative inverse do not exist.
Reason (R): The additive inverse of 1 is -1 and its multiplicative inverse is 1.
Ans:
Option (d) is correct as the assertion is wrong and the reason is correct.
0 is a unique rational number whose multiplicative inverse does not exist. But every rational number including 0 has an additive inverse. So, A is false.
The additive inverse of 1 is -1 [1 + (-1) = 0 ] and multiplicative inverse of 1 is 1 [1 × (1/1) = 1]. So the reason is true.
Case- Based Questions
Directions (Questions 11 to 14): Examine the following case study carefully and answer the questions that follow.
A train starts from station A with a certain number of passengers. At station B, the train drops one third of the passengers and takes in 96 more. At the next station C one half of the passengers on board get down while 12 new passengers get on board. At station D, a quarter of the passengers get down and 20 new passengers board the train. It then reaches its final destination – station E- with 200 passengers.
11.) How many passengers were on board when the train left station A?
(a.) 460
(b.) 520
(c.) 540
(d.) 560
11.) (c): let us assume there were x passengers on board when the train left station A.
x/4+65=200
=> x/4 = 135
=>x= 135×4 = 540
∴ A total of 540 students were on board when the train left station A.
12.) How many passengers were on board when the train left station B?
(a.) 428
(b.) 436
(c.) 444
(d.) 456
Ans: (d): Let the number of passengers in train when it left In station
Required number=
x = (2x/3+96)=(2×540/3+96)=360+96=456.
So the number of passengers in train when it left station B was 456.
13.) How many passengers were on board when the train left station C?
(a.) 228
(b.) 240
(c.) 256
(d.) 264
Ans: The numbers of passengers who are were in the train when it left station C is 240.
Option (b) is the Required number = (x/3+60) = (540/3+60)=180+60=240.
14.) Had the train started with 720 passengers from station A, how many passengers would have got down at station E?
(a.) 228
(b.) 236
(c.) 245
(d.) 256
Ans:
If there were 720 people in the train when it started from station A then the number of passengers would have gotten down in station E is = (720/4+65)=180+65=245
So, the option c 245 is the correct answer.