Real Numbers Test

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Real Numbers Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

A number when divided by 61 gives 27 as quotient and 32 as remainder, then the number is:

A
1679

B
1967

C
1569

D
1796

Solution

Dividend = Divisor × Quotient + Remainder

→ Number(dividend) = D × Q + R

Therefore the number (Dividend) = 61 × 27 + 32

= 1647 + 32

= 1679
Question 2
1 / -0

Every positive even integer is of the form ____ for some integer ‘q’.

A
2q

B
2q – 1

C
2q + 1

D
none of these

Solution

Let a be any positive integer and b=2

Then by applying Euclid’s Division Lemma, we have,

a = 2q + r where 0 ⩽ r < 2 r = 0 or 1

Therefore, a = 2q or 2q + 1

Thus, it is clear that a = 2q

i.e., a is an even integer in the form of 2q
Question 3
1 / -0

Every positive odd integer is of the form ________ where ‘q’ is some integer.

A
3q + 1

B
2q + 2

C
5q + 1

D
2q + 1

Solution

Let aa be any positive integer and b = 2

Then by applying Euclid’s Division Lemma,

we have, a = 2q +r,

where 0 ⩽ r < 20 ⩽ r < 2 ⇒ r = 0 or 1 ∴∴ a = 2q or 2q + 1 .

Therefore, it is clear that a = 2q i.e., a is an even integer.

Also, 2q and are two 2q + 1 consecutive integers, therefore, 2q + 1 is an odd integer.
Question 4
1 / -0

For any two positive integers a and b, such that a > b. there exist (unique) whole numbers q and r such that

A
b = aq + r, 0 ⩽ r < b

B
q = ar + b, 0 ⩽ r < b

C
a = bq + r, 0 ⩽ r < b

D
a = qbr

Solution

Euclid’s Division Lemma states that for given positive integer a and b,

there exist unique integers q and r satisfying a = bq + r ; 0 ⩽ r < b.
Question 5
1 / -0

For any positive integer ‘a’ and 3, there exist unique integers ‘q’ and ‘r’ such that a = 3q + r where ‘r’ must satisfy

A
1 < r < 3

B
0 < r < 3

C
0 ⩽ r < 3

D
0 < r ⩽ 3

Solution

Since a is a positive integer, therefore, r = 0, 1, 2 only.
So, that a = 3q, 3q + 1, 3q + 2.
Question 6
1 / -0

If 112 = q × 6 + r,112 = q × 6 + r, then the possible values of r are:

A
0, 1, 2, 3

B
2, 3, 5

C
1, 2, 3, 4

D
0, 1, 2, 3, 4, 5

Solution

For the relation x = qy + r, 0 ⩽ r < y

So, here r lies between 0 ⩽ r < 6.

Hence, r = 0, 1, 2, 3, 4, 5.
Question 7
1 / -0

By Euclid’ division lemma x = qy + r, x > y the value of q and r for x = 27 and y = 5 are:

A
cannot be determined

B
q = 5, r = 2

C
q = 6, r = 3

D
q = 5, r = 3

Solution

x = qy + r

⇒ 27 = 5 × 5 + 2

⇒ q = 5, r = 2
Question 8
1 / -0

Every positive odd integer is of the form 2q + 1, where ‘q’ is some

A
whole number

B
natural number

C
None of these

D
integer

Solution

Euclid’s Division Lemma states that

for given positive integer a and b, there exist unique integers q and r

satisfying a = bq + r;

0 ⩽ r < b

let b=6 then possible values of r will be 0,1,2,3,4,5

when b =6 , r = 0 then a = 6q +0

r = 1 a = 6q +1

r = 2 a = 6q +2

r = 3 a = 6q + 3

r = 4 a = 6q +4

r = 5 a = 6q + 5

but 6q , 6q + 2, 6q + 4 cannot be because they are all positive even integers while a is odd integer

thus we can say that a can be 6q + 1 or 6q + 3 or 6q + 5

Hence, general form is bq + 1.
Question 9
1 / -0

Any ____________ is of the form 4q + 1 or 4q + 3 for some integer ‘q’.

A
prime number

B
positive odd integer

C
positive even integer

D
composite number

Solution

Let a be a given positive odd integer.

Applying Euclid’s Division Lemma to a and b=4,

We have, a = 4q + r where 0 ⩽ r < 4

⇒ r = 0, 1, 2, 3

⇒ a = 4q or 4q + 1 or 4q + 2 or 4q + 3

But a=4q and 4q + 2 = 2 ( 2 q + 1 ) are clearly even.

Also a = 4q, 4q + 1, 4q + 2, 4q + 3 are consecutive integers,

therefore any positive odd integer is of the form 4q+1 and 4q+3

where q is some integer.
Question 10
1 / -0

The product of three consecutive positive integers is divisible by

A
4

B
6

C
5

D
10

Solution

Let n be a positive integer,

then three consecutive positive integers are (n+1) (n+2) (n+3) = n(n+1) (n+2)+3 (n+1) (n+2)

Here, the first term is divisible by 6 and the second term is also divisible by 6

Because it contains a factor 3 and one of the two consecutive integers (n+1) or (n+2) is even and thus is divisible by 2

Therefore, the sum of multiple of 6 is also a multiple of 6.

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r = 1	a = 6q +1
r = 2	a = 6q +2
r = 3	a = 6q + 3
r = 4	a = 6q +4
r = 5	a = 6q + 5

Real Numbers Test - 8

Result

Real Numbers Test - 8

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Question 1

1679

1967

1569

1796

Solution

Question 2

2q

2q – 1

2q + 1

none of these

Solution

Question 3

3q + 1

2q + 2

5q + 1

2q + 1

Solution

Question 4

b = aq + r, 0 ⩽ r < b

q = ar + b, 0 ⩽ r < b

a = bq + r, 0 ⩽ r < b

a = qbr

Solution

Question 5

1 < r < 3

0 < r < 3

0 ⩽ r < 3

0 < r ⩽ 3

Solution

Question 6

0, 1, 2, 3

2, 3, 5

1, 2, 3, 4

0, 1, 2, 3, 4, 5

Solution

Question 7

cannot be determined

q = 5, r = 2

q = 6, r = 3​​​​

q = 5, r = 3

Solution

Question 8

whole number

natural number

None of these

integer

Solution

Question 9

prime number

positive odd integer

positive even integer

composite number

Solution

Question 10

4

6

5

10

Solution

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q = 6, r = 3