# Free Number Systems 03 Practice Test - 9th Grade

34 lies between 12 and 1.

A.

True

B.

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.

A number xwas 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.

A.

11

B.

12

C.

13

D.

14

#### SOLUTION

Solution : C

Given decimal is 5.3333333......
Let, x=5.3333333......
Therefore, 10x=53.33333333....
Subtracting x from 10x, we get
10xx=53.33333..5.33333..=48
9x=48
x=489       =163=pq
So, pq=163
p  q = 13

Which of the following represents whole numbers?

A.

Positive integers

B.

Negative integers

C. π
D.

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.

The product of two irrational numbers is a/an

A.

Irrational number

B.

Rational number

C.

Real number

D.

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)×(32)=1
Both (3+2) and (32) 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.

Which of the following is an irrational number?

A.

14.287628762876.....

B.

15.2323232323......

C.

5.2731687143725186.....

D.

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.

The expression 535+3 when simplified reduces to a-b15.

Then a + b is __________.

A.

15

B.

10

C.

20

D.

5

#### SOLUTION

Solution : D

In 535+3, the denominator has to be rationalised for simplification. So ,

535+3×5353
=(53)25232
=12[(5)2+(3)2215]
=415
So, a=4 and b=1.
a + b = 4 + 1 = 5

is neither a positive nor a negative integer.

#### 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.

Find the value of (2+2)(22).

A. 0
B.

1

C. 2
D. 3

#### SOLUTION

Solution : A

A common tendency to solve this question is to apply the algebraic identity (a2b2)=(a+b)(ab). But by observation, we can see that (22), 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)(22)
=(2+2)×(0)

=0

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

A.

132863

B.

142862

C.

142876

D.

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

1417×1417 can be represented as ___________.

A.

1417+17

B.

(142)17

C.

(14×14)17

D.

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.