9th Grade > Mathematics
NUMBER SYSTEMS MCQs
Total Questions : 197
| Page 1 of 20 pages
Answer: Option D. -> 8
:
D
Let x= 1.6666...... ----- (i)
then, 10x= 16.666666 ----- (ii)
Subtracting (i) from (ii), we get
10x - x = (16.666666...) - (1.666666...)
⇒ 9x = 15
Hence x = 159=15393=53
i.e., p = 5 and q = 3
So, p + q = 8.
:
D
Let x= 1.6666...... ----- (i)
then, 10x= 16.666666 ----- (ii)
Subtracting (i) from (ii), we get
10x - x = (16.666666...) - (1.666666...)
⇒ 9x = 15
Hence x = 159=15393=53
i.e., p = 5 and q = 3
So, p + q = 8.
Answer: Option D. -> All of the above
:
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 andifa=b, the expression becomes (a×a)m=(a2)m=a2m.
So, using first rule, we have
1417+17=1427
Now, using secondrule,
(142)17= 1427
Finally, using thirdrule,
(14×14)17=1427
Hence, we see that all the three results are same.
:
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 andifa=b, the expression becomes (a×a)m=(a2)m=a2m.
So, using first rule, we have
1417+17=1427
Now, using secondrule,
(142)17= 1427
Finally, using thirdrule,
(14×14)17=1427
Hence, we see that all the three results are same.
Answer: Option A. -> 0
:
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
:
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
Answer: Option A. -> 132863
:
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
:
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
Answer: Option C. -> 13
:
C
Given decimal is5.3333333......
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
:
C
Given decimal is5.3333333......
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
Answer: Option A. -> Positive integers
:
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.
:
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.
Answer: Option D. -> 5
:
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
:
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
:
The integers include positive integers i.e. 1, 2, 3, ... , negative integers i.e. -1, -2, -3, ... and zero. Hence, zero isneither a positive nor a negative integer.
Answer: Option A. -> True
:
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 that34 lies between 12 and 1.
:
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 that34 lies between 12 and 1.
Answer: Option C. -> 5.2731687143725186.....
:
C
Irrational numbers are defined as those numbers which havea non-terminating and non-repeating decimal expansion and hence cannot be represented in the pqform. 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.
:
C
Irrational numbers are defined as those numbers which havea non-terminating and non-repeating decimal expansion and hence cannot be represented in the pqform. 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.
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