# Free Number Systems 03 Practice Test - 9th Grade

### Question 1

34 lies between 12 and 1.

True

False

#### SOLUTION

Solution :A

To check if a number lies between any two numbers, we first convert the numbers into the decimal form. 34=0.75 ,12=0.5 and we have to check if 0.75 lies between 0.5 and 1 and plot the numbers on the number line as follows :

Therefore, we can say that 34 lies between 12 and 1.

### Question 2

A number ′x′was found to be represented in the simplest pq form. The decimal expansion of this number was found to be 5.3333333333..... Find the value of p - q.

11

12

13

14

#### SOLUTION

Solution :C

Given decimal is 5.3333333......

Let, x=5.3333333......

Therefore, 10x=53.33333333....

Subtracting x from 10x, we get

10x−x=53.33333..−5.33333..=48

⇒9x=48

⇒x=489 =163=pq

So, p−q=16−3

∴ p − q = 13

### Question 3

Which of the following represents whole numbers?

Positive integers

Negative integers

Integers

#### SOLUTION

Solution :A

Integers are constituted by natural numbers, their negatives and 0. Removing the negative numbers from integers would leave us with the whole numbers. Therefore, whole numbers include 0 as well as all the positive integers.

### Question 4

The product of two irrational numbers is a/an

Irrational number

Rational number

Real number

Integer

#### SOLUTION

Solution :A, B, C, and D

The product of two irrational numbers will be either a rational or an irrational number.

Consider the following example: (√3+√2)×(√3−√2)=1.

Both (√3+√2) and (√3−√2) are irrational numbers and their product is 1.

1 is an integer and since, all integers are rational numbers as well, we can infer that product of two irrational numbers may be an integer, a rational number or an irrational number and they are all real numbers.

When we multiply the irrational numbers √2 and √3, we get √6, which is also an irrational number.

### Question 5

Which of the following is an irrational number?

14.287628762876.....

15.2323232323......

5.2731687143725186.....

1.33333333...

#### SOLUTION

Solution :C

Irrational numbers are defined as those numbers which have a non-terminating and non-repeating decimal expansion and hence cannot be represented in the pq form. So, here we see that out of the given options the number 5.2731687143725186..... has a non-terminating and non-repeating decimal expansion as clearly one can interpret that there are no patterns formed in the decimal expansion and the number goes on expanding arbitrarily.

### Question 6

The expression √5−√3√5+√3 when simplified reduces to a-b√15.

Then a + b is __________.

15

10

20

5

#### SOLUTION

Solution :D

In √5−√3√5+√3, the denominator has to be rationalised for simplification. So ,

√5−√3√5+√3×√5−√3√5−√3

=(√5−√3)2√52−√32

=12[(√5)2+(√3)2−2√15]

=4−√15

So, a=4 and b=1.

∴a + b = 4 + 1 = 5

### Question 7

#### SOLUTION

Solution :The integers include positive integers i.e. 1, 2, 3, ... , negative integers i.e. -1, -2, -3, ... and zero. Hence, zero is neither a positive nor a negative integer.

### Question 8

Find the value of (√2+√2)(√2−√2).

1

#### SOLUTION

Solution :A

A common tendency to solve this question is to apply the algebraic identity (a2−b2)=(a+b)(a−b). But by observation, we can see that (√2−√2), which is one of the factors, will result in 0, thus making the final value of the whole expression as 0 i.e.

(√2+√2)(√2−√2)

=(√2+√2)×(0)

=0

### Question 9

If 14.287628762876........ can be represented in pq form, then 'p - q ' is ________.

132863

142862

142876

142863

#### SOLUTION

Solution :A

Let 'x' = 14.287628762876 ------- (i)

then 10000x = 142876.287628762876 --------(ii)

Subtracting (i) from (ii) we get

10000x - x = 142876.287628762876 - 14.287628762876

9999x = 142862

then 'x' = 1428629999

p = 142862, q = 9999

So, p-q = 132863

### Question 10

1417×1417 can be represented as ___________.

1417+17

(142)17

(14×14)17

All of the above

#### SOLUTION

Solution :D

For any positive real number 'a' and integers 'm' and 'n', we define

1. am×an=am+n

2. (am)n=amn

3. (ab)m=ambm and if a=b, the expression becomes (a×a)m=(a2)m=a2m.

So, using first rule, we have

1417+17=1427

Now, using second rule,

(142)17= 1427

Finally, using third rule,

(14×14)17 =1427

Hence, we see that all the three results are same.