MCQ Questions for Class 10 Maths Real Numbers Quiz 3

State whether true or false:

$3+\sqrt{6}$ is an irrational number.

0%
True
0%
False

Explanation

AS we know 3 is a rational number and

$\sqrt{6}$ is also an irrational number

The addition of rational and irrational numbers is always an irrational number

So, 3 +

$\sqrt{6}$ is also a irrational number

Which of the following is an irrational number?

0%
$0.14$
0%
$0.14 \overline{16}$
0%
$1.1 {416}$
0%
$0.4014001400014....$

Explanation

The decimal expansion of a rational number is either terminating or non-terminating repeating.

(A)

$0.14$ is terminating, so it is a rational number

(B)

$0.14\bar{16}=0.141616....$

is also rational ( non-terminating repeating ), where digits

$16$ are repeating.

(C)

$0.1416$ is terminating, so it is a rational number

(D)

$0.4014001400014.....$

is an irrational number because it is neither terminating nor repeating.

$2-\sqrt {3}$ is an irrational number.

0%
True
0%
False

Explanation

$\sqrt{3}$ is an irrational number.

Since, the difference of a rational and irrational number is irrational, hence,

$2-\sqrt{3}$ is an irrational number.

Which of the following is always true

0%
$irrational + irrational =irrational$
0%
$\dfrac{rational }{rational }=rational$
0%
$\dfrac{integer }{integer}=integer$
0%
None of these

Explanation

Counter-example for A:

$(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C:

$\dfrac{1}{2}=0.5$

Proof for B:

Let

$q_1, q_2$ be two rational numbers such that

$q_2\neq0$ .

As they are rational, they can be written as

$a/b, c/d$ respectively for some integers

$a, b, c, d$ .

$(b,c,d\neq0)$

$\dfrac{q_1}{q_2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since,

$a,b,c,d$ were integers, even

$ad$ and

$bc(\neq0)$ are integers and therefore, the above expression is rational.

$\frac { 2 } { 2 + \sqrt { 3 } }$ is an irrational number

0%
True
0%
False

Explanation

Let if possible

$( 2 + \sqrt3 )$ is rational

then

$2 + \sqrt3 = \dfrac{a}{b} , \, b \neq = 0$

$\Rightarrow \sqrt3 = \dfrac{a}{b} - 2 \Rightarrow \sqrt3 = \dfrac{a-2b}{b}$ . . . .eq.1

$\because a \, \& \, b$ are integer.

$\because a - 2b$ is also an integer.

$\Rightarrow \dfrac{a-2b}{b}$ is rational.

LHS of the equation is the square root of a prime number.

So it is irrational and R.H.S. is rational.

Which is a contradiction. Because a rational number and an irrational number can be never equal.

So, our assumption i.e.

$2 + \sqrt 3$ is rational wrong.

Hence

$( 2 + \sqrt 3 )$ is an irrational number.

$\therefore$ This is true.

$\therefore$ option A is correct.

State whether the following statement is true or not:

$\left( 3+\sqrt { 5 } \right)$ is an irrational number.

0%
True
0%
False

Explanation

Let us suppose

$3+\sqrt 5$ is rational.

Thus

$3+\sqrt 5$ is in the form of

$\dfrac pq$ where

$p$ and

$q$ are integers and

$q\neq0$

$\Rightarrow \sqrt5=\dfrac pq-3$

$\Rightarrow \sqrt5=\dfrac{p-3q}{q}$

$p, q$ and

$3$ are integers

$\dfrac{p-3q}{q}$ is a rational number.

$\Rightarrow \sqrt 5$ is a rational number.

But we know that

$\sqrt 5$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that

$3+\sqrt 5$ is a rational number.

Hence

$3+ \sqrt5$ is an irrational number.

State the following statement is True or False

$\dfrac{987}{10500}$ will have terminating decimal expansion.

0%
True
0%
False

Explanation

Terminating decimal expansion, because

$\dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}$

H.C.F. of

$26$ and

$91$ is:

0%
$13$
0%
$2366$
0%
$91$
0%
$182$

Explanation

Prime factors of

$26=2\times 13$
Prime factors of

$91=7\times 13$
As

$13$ is the only common factor.

Therefore,

$HCF(26,91)=13$

The product of a non-zero rational and an irrational number is

0%
always irrational
0%
always rational
0%
rational or irrational
0%
one

Explanation

Let

$x$ be a rational number and

$y$ be an irrational number.
Let

$xy = a$ .

Let us assume that

$a$ is rational.

Since,

$a$ is rational it can be expressed as

$\dfrac{p}{q}$ , where

$p$ and

$q$ are integers.

Let

$x =\dfrac{m}{n}$ , where

$m$ and

$n$ are integers.
Now,

$xy = a$

$\dfrac{my}{n} = \dfrac{p}{q}$ .
On cross multipliying we get,

$\Rightarrow y =\dfrac{pn}{qm}$ .

Now,

$pn$ and

$qm$ are integers.

Hence,

$\dfrac{pn}{qm}$ is a rational number.

However,

$y$ is irrational.

Hence, our assumption is incorrect.
Hence, the product of a non-zero rational and an irrational number is always an Irrational number.

So, option

$A$ .

According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers

$a$ and

$b$ with

$a > b$ enables us to find the

0%
HCF
0%
LCM
0%
Decimal expansion
0%
Quotient

Explanation

According to Euclid's division lemma,

For each pair of positive integers

$a$ and

$b$ , we can find unique integers

$q$ and

$r$ satisfying the relation

$a = bq + r , 0 ≤ r < b$

$r = 0$ then q will be HCF of

$a$ and

$b$

The basis of the Euclidean division algorithm is Euclid’s division lemma.

To calculate the Highest Common Factor (HCF) of two positive integers

$a$ and

$b$ we use Euclid’s division algorithm.

HCF is the largest number which exactly divides two or more positive integers.

By exactly we mean that on dividing both the integers

$a$ and

$b$ by HCF, the remainder is zero.

So correct answer is option A

State True or False:

$6+\sqrt{2}$ is a rational number.

0%
True
0%
False

Explanation

Let's assume that

$6+\sqrt2$ is rational.....

then

$6+\sqrt2 = p/q$

$\sqrt2 =( p-6q)/(q)$

now take

$p-6q$ to be P and

$q$ to be Q........where P and Q are integers

which means,

$\sqrt2= P/Q$ ......

But this contradicts the fact that

$\sqrt2$ is rational

So our assumption is wrong and

$6+\sqrt2$ is irrational.

Euclids division lemma states that if a and b are any two

$+$ ve integers, then there exists unique integers q and r such that :

0%
$a=bq+r, 0 < r < b$
0%
$a=bq+r, 0 \leq r \leq b$
0%
$a=bq+r, 0 \leq r < b$
0%
$a=bq+r, 0 < b < r$

Explanation

Division Algorithm: Euclids division lemme states that if a and b are any two positive integers then there exist unique integers q and r such that,

$a = q b + r \, \, 0 \leq r < b$

$\therefore$ option C is correct.

Proof: We begin by proving that the set s = {a - xb | x is an integer: a-xb

$\geq$ 0 }

is non-empty. To do this, if suffices to exhibit a value of x making a-xb nonnegative a value of x making a-xb nonnegative. Because the integer b

$\geq$ 1 , we have |a| b

$\geq$ |a| and so -

$a - (-|a|) b = a + |a| b \geq a + |a| \geq 0$

For the choice x = -|a|, then a - xb lies in s. This paves the way for an application of well ordering principle, from which we infer that the set S contains a smallest integer : call it r. By the definition S,

$\exists$ an integer q satisfying , r = a - qb ,

$0\leq r$

we argue that r < b. If this were not this case, then r

$\geq$ 0 and a - ( q + 1 ) b = ( a - q b ) - b = r - b

$\geq$ 0

This implication is that the integer a - ( q + 1 ) b has the proper form to belong to the set S. But a - ( q + 1 ) b = r - b < r , leading to a contradiction of the choice of r as the smallest member of S. Hence r < b .

Next we turn to the task of showing the uniqueness of q and r. Suppose that a has two representation of the desired form say,

$a = q b + r = q^{'}b + r^{'}$

Where

$0 \leq r < b, \, 0 \leq r^{'} < b .$ Then

$r^{'}- r = b (q - q^{'})$ and, owing to the fact that

$| r^{'} - r | = b | q - q^{'} |$ upon adding two inequalities

$-b < -r \leq 0$ and

$0 \leq r^{'} < b$ we obtain

$-b < r^{'}-r < b \Rightarrow |r^{'} - r | < b$

Thus,

$b|q-q^{'}| < b$ which yields

$0 \leq |q-q^{'} | < 1$

As,

$|q-q^{'} |$ is positive, the only possibility is that

$|q-q^{'} | = 0$ where

$q = q^{'} :$ this is turns give

$r = r^{'}$ and ends the proof.

Write whether every positive integer can be of the form

$4q + 2$ , where

$q$ is an integer.

0%
Yes
0%
No
0%
Ambiguous
0%
Data Insufficient

Explanation

No, all positive integers cannot be written in the form of

$4q+2$
Because,

$4q+2=2\left( 2q+1 \right)$
Therefore,

$4q+2$ is an even number.
So, we can't write odd numbers in the form

$4q+2$ (where

$q$ is an integer)
Also

$2q+1$ is an odd number, hence the maximum power of

$2$ that divides

$4q+1$ is

$1$ . Therefore, we can't represent the numbers divisible by

$4$ this form.

$d$ is the

$HCF$ of 45 and 27, then

$x, y$ satisfying

$d=27x+45y$ are :

0%
$x=2, y=1$
0%
$x=2, y=-1$
0%
$x=-1, y=2$
0%
$x=-1, y=-2$

Explanation

Applying Euclid's division lemma to

$27$ and

$45$ , we get

$45 = 27 \times 1 + 18$ ...(1)

$27 = 18 \times 1 + 9$ ...(2)

$18 = 9 \times 2 + 0$ ...(3)
Since the remainder is zero, therefore, last divisor

$9$ is the

$HCF$ of

$27$ and

$45$ .

From (2), we get

$9 = 27 - 18 \times 1$

$= 27 - \left(45 - 27 \times 1\right) \times 1 \ \ \ \ \ \ \left[using (1)\right]$

$= 27 - 45 \times 1 + 27 \times 1 \times 1$

$= 27 - 45 + 27$

$= 54 - 45$

$\Rightarrow 9 = 27 \times 2 - 45 \times 1$ ...(4)

Comparing (4) with

$d = 27x + 45y$ , we get

$d = 9, x = 2 \ and\ y = -1$

Say true or false:

A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

0%
True
0%
False

Explanation

Let the positive integer

$n$ is of the form

$3q, 3q+1,$ and

$3q+2$
If

$n=3q$
Squaring both sides, we get,

$=>{ n }^{ 2 }=9{ q }^{ 2 }$

$=>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right)$

$=>{ n }^{ 2 }=3m$ , where

$m=3{ q }^{ 2 }$
Now, if

$n=3q+1$

$=>{ n }^{ 2 }={ \left( 3q+1 \right) }^{ 2 }$

$=>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1$

$=>{ n }^{ 2 }={ 3q\left( 3q+2 \right) }+1$

$=>{ n }^{ 2 }=3m+1 ,$ where

$m=q\left( 3q+2 \right)$
Now, if

$n=3q+2$

$=>{ n }^{ 2 }={ \left( 3q+2 \right) }^{ 2 }$

$=>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4$

$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+4$

$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+3+1$

$=>{ n }^{ 2 }=3m+1$ where

$m=\left( 3{ q }^{ 2 }+4q+1 \right)$
Hence,

${ n }^{ 2 }$ integer is of the form

$3m$ and

$3m+1$ not

$3m+2$

H.C.F. of

$6, 72$ and

$120$ is:

0%
$6$
0%
$2$
0%
$3$
0%
$1$

Explanation

Factors of 6 are

$1,2,3,6$
Factors of 72 are

$1,2,3,4,6,8,9,12,18,24,36,72$
Factors of 120 are

$1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$
The highest factor in all the three is

$6$ .

State true or false:

$\sqrt{2}$ is not a rational number.

0%
True
0%
False

Explanation

Let us assume that

$\sqrt{2}$ is a rational number

$\Rightarrow \sqrt{2}=\dfrac{p}{q}\left [ \dfrac{p}{q}\ is\ in\ simplest\ form\ \right ]$

$\Rightarrow 2q^{2}=p^{2}$

$p^{2}$ is even

$\Rightarrow p$ is even

$\Rightarrow p=2k$

$2q^{2}=4k^{2}\Rightarrow q^{2}=2k^{2}\Rightarrow q^{2}$ is even

$\Rightarrow q$ is even.

$\Rightarrow p,q$ have 2 as a common factor which is contradiction to assumption.

$\therefore \sqrt{2}=\dfrac{p}{q}$ is a false

$\Rightarrow \sqrt{2}$ is not a rational.

Use Euclid's division lemma to find the HCF of the following

16 and 176

0%
$16$
0%
$14$
0%
$10$
0%
$12$

Explanation

by using Euclid's division lemma,

$a=bq+r$ ;

$r$ greater than or equal to 0, and

$r$ less than

$b$

$176=(16)(11)+0, H.C.F=16$

Euclid's division lemma states that for two positive integers a and b, there exist unique integers q and r such that

$a = bq + r$ , where r must satisfy

0%
1 < r < b
0%
$0 < r \leq b$
0%
$0\leq r < b$
0%
0 < r < b

Explanation

$r$ must satisfy

$0 \le r < b$
Proof,

$..,a-3b,a-2b,a-b,a,a+b,a+2b,a+3b,..$
clearly it is an arithmetic progression with common difference

$b$ and it extends infinitely in both directions.
Let

$r$ be the smallest non-negative term of this arithmetic progression.Then,there exists a non-negative integer

$q$ such that,

$a-bq=r$

$=>a=bq+r$
As,

$r$ is the smallest non-negative integer satisfying the result.Therefore,

$0\quad \le \quad r\quad \le \quad b$
Thus, we have

$a=bq_1+r_1$ ,

$0 \le r_1 \le b$

If HCF of

$210$ and

$55$ is of the form

$(210) (5) + 55 y$ , then the value of

$y$ is :

0%
$-19$
0%
$-18$
0%
$5$
0%
$55$

Explanation

HCF of 210 and 55 is 5

Given HCF =

$(210)(5)+55y$

$5 =(210)(5)+55y$

$55y$ =

$-1045$

On solving this equation we get

$y=-19$

So, correct answer will be option A

Use Euclid's division lemma to find the HCF of

$40$ and

$248$ .

0%
$4$
0%
$10$
0%
$6$
0%
$8$

Explanation

Here,

$a=248$ and

$b=40$

by using Euclid's division lemma,

$a=bq+r$ ;

$0\le r<b$

$248=40\times 6 +8$

$40=8\times 5+0$

$\therefore H.C.F (40,248)=8$

hence option

$D$ is correct.

The decimal expansion of the rational number

$\displaystyle\frac{23}{2^{3}5^{2}}$ , will terminate after how many places of decimal?

0%
1
0%
3
0%
4
0%
5

Explanation

$\displaystyle\frac{23}{2^{3}5^{2}}$

=

$\displaystyle\frac{23}{2(2)^{2}(5)^{2}}$

=

$\displaystyle\frac{23}{2(2\times5)^{2}}$

=

$\displaystyle\frac{23}{2(10)^{2}}$

=

$\displaystyle\frac{11.5}{100} = 0.115$

Therefore, the decimal expansion of the rational number

$\displaystyle\frac{23}{2^{3}5^{2}}$
will terminate after three places of decimal.

Use Euclid's division lemma to find the HCF of the following

27727 and 53124

0%
$233$
0%
$200$
0%
$154$
0%
$212$

Explanation

Here,

$a=53124$ and

$b=27727$

by using Euclid's division lemma,

$a=bq+r$ ;

$0<r<b$

$53124=(27727)(1)+25397$

$27727=(25397)(1)+2330$

$25397=(2330)(10)+2097$

$2330=(2097)(1)+233$

$2097=(233)(9)+0$
H.C.F is

$233$

Use Euclid's division lemma to find the HCF :

65 and 495.

0%
$5$
0%
$10$
0%
$15$
0%
$0$

Explanation

According to Euclid’s Division Lemma if we have two positive integers

$a$ and

$b$ , then there exists unique integers

$q$ and

$r$ which satisfies the condition

$a = bq + r$ where

$0 ≤ r < b$ .

$HCF$ is the largest number which exactly divides two or more positive integers.

By Euclid's division lemma, we mean that on dividing both the integers

$a$ and

$b$ , the remainder is zero.

The given integers are

$a=65$ and

$b=495$ .

Clearly

$495 > 65$ .

So, we will apply Euclid’s division lemma to

$65$ and

$495$ , we get,

$495 = (65\times 7) + 40$

Since the remainder

$40≠ 0$ . So we again apply the division lemma to the divisor

$65$ and remainder

$40$ . We get,

$65 = (40\times 1) + 25$

Again the remainder

$25≠ 0$ , so applying the division lemma to the new divisor

$40$ and remainder

$25$ . We get,

$40 = (25\times 1) + 15$

Now, again the remainder

$15≠ 0$ , so applying the division lemma to the new divisor

$25$ and remainder

$15$ . We get,

$25 = (15\times 1) + 10$

Again the remainder

$10≠ 0$ , so applying the division lemma to the new divisor

$15$ and remainder

$10$ . We get,

$15 = (10\times 1) + 5$

Again the remainder

$5≠ 0$ , so applying the division lemma to the new divisor

$10$ and remainder

$5$ . We get,

$10 = (5\times 1) + 0$

Finally we get the remainder

$0$ and the divisor is

$5$ .

Hence, the

$HCF$ of

$65$ and

$495$ is

$5$ .

Use Euclid's division lemma to find the

$HCF$ of the following numbers:

$8068$ and

$12464$

0%
$4$
0%
$2$
0%
$0$
0%
$1$

Explanation

Here,

$a=12464$ and

$b=8068$

So, by using Euclid's division lemma,

$a=bq+r$ ;

$0<r<b$

$\begin{aligned}{}12464 &= (8068)(1) + 4396\\8068 &= (4396)(1) + 3672\\4396& = (3672)(1) + 724\\3672 &= (724)(5) + 52\\724& = (52)(13) + 48\\52 &= (48)(1) + 4\\48 &= (4)(12) + 0\end{aligned}$

Hence,

$HCF$ of

$8068$ and

$12464$ is equal to

$4.$

Without doing any actual division, find which of the following rational numbers have terminating decimal representation :
(i)

$\displaystyle \dfrac{7}{16}$ (ii)

$\displaystyle \dfrac{23}{125}$
(iii)

$\displaystyle \dfrac{9}{14}$ (iv)

$\displaystyle \dfrac{32}{45}$
(v)

$\displaystyle \dfrac{43}{50}$ (vi)

$\displaystyle \dfrac{17}{40}$
(vii)

$\displaystyle \dfrac{61}{75}$ (viii)

$\displaystyle \dfrac{123}{250}$

0%
(i), (iii), (v), (vi) and (vii)
0%
(i), (ii), (v), (vi) and (viii)
0%
(i), (iii), (v), (vi) and (viii)
0%
(i), (ii), (v), (vi) and (vii)

Explanation

The rational not having denominator as multiple of

$2^{m}\times 5^{n}$ will be non terminating.
(1)

$\dfrac{7}{16}=\dfrac{7}{2\times2\times2\times2}-$ denominator is multiple of

$2^{m}\times 5^{n}$ hence terminating.
(2)

$\dfrac{23}{125}=\dfrac{23}{5\times5\times5}$ -- denominator is multiple of

$2^{m}\times 5^{n}$ hence terminating.
(3)

$\dfrac{9}{14}=\dfrac{9}{7\times2}$ -- denominator is not multiple of

$2^{m}\times 5^{n}$ hence non-terminating
(4)

$\dfrac{32}{45}=\dfrac{32}{3\times3\times5}$ -- denominator is not multiple of

$2^{m}\times 5^{n}$ hence non-terminating.
(5)

$\dfrac{43}{50}=\dfrac{43}{5\times5\times2}$ -- denominator is multiple of

$2^{m}\times 5^{n}$ hence terminating.
(6)

$\dfrac{17}{40}=\dfrac{17}{2\times2\times2\times5}$ -- denominator is multiple of

$2^{m}\times 5^{n}$ hence terminating.
(7)

$\dfrac{61}{75}$ -- denominator is not multiple of

$2^{m}\times 5^{n}$ hence non-terminating.
(8)

$\dfrac{123}{250}=\dfrac{123}{5\times5\times5\times2}$ -- denominator is multiple of

$2^{m}\times 5^{n}$ hence terminating.
(i), (ii), (v), (vi) and (viii) will have terminating decimal.

The given pair of numbers

$231, 396$ are __________ .

0%
multiple of each other
0%
factors of each other
0%
co-prime to each other
0%
having a difference of $120$ among each other

Explanation

Euclid's division lemma,

$a=bq+r$ ;

$0\le r<b$

Applying Euclid's Algorithim to

$231 , 395$

$395 = 231\times 1+164$

$231 = 164 \times 1+67$

$164 = 67 \times 2+30$

$67 = 30 \times 2 +7$

$30 = 7 \times 4 +2$

$7 = 2 \times 3+1$

$2 = 1 \times 2+0$

The given numbers have only

$1$ as common factor.

Hence, the given numbers are co-prime.

State: True or False

$3+2\sqrt 5$ is an irrational number.

0%
True
0%
False

Explanation

Let us assume, to the contrary, that

$3+2\sqrt{5}$ is rational.

That is, we can find coprime integers

$a$ and

$b$ (b does not equal to 0) such that

$3+2\sqrt{5}=\dfrac{a}{b}$ .

Therefore,

$\dfrac{a}{b} - 3=2\sqrt{5}$

$\dfrac{a-3b}{b}=2\sqrt{5}$

$\dfrac{a-3b}{2b}=\sqrt{5}$

$\dfrac{a}{2b}-\dfrac{3}{2}=\sqrt{5}$

Since

$a$ and

$b$ are integers and rational, we get

$\dfrac{a}{2b}-\dfrac{3}{2}$ is rational, and so

$\dfrac{a-3b}{2b}=\sqrt{5}$ is rational.

But this contradicts the fact that

$\sqrt{5}$ is irrational.

This contradiction has arisen because of our incorrect assumption that

$3+2\sqrt{5}$ is rational.

So, we conclude that

$3+2\sqrt{5}$ is irrational.

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

$\dfrac {29}{343}$

0%
Terminating
0%
Non-terminating
0%
Ambiguous
0%
Data insufficient

Explanation

Given,

$\displaystyle \frac {29}{343}= \frac {29}{7^{3}}$
As it is not in the form of

${ 2 }^{ m }\times { 5 }^{ n }$ .
So, the rational number

$\displaystyle \frac {29}{343}$ has a non terminating decimal expansion

If these numbers form positive odd integer 6q+1, or 6q+3 or 6q+5 for some q then q belongs to:

0%
integers
0%
irrational
0%
real
0%
none of these

Explanation

We know that product and sum of integers will give integer.

Given

$6q+1, 6q+3 , 6q+5$ are integers so

$6q$ will be integer and then q will be integer.

So correct answer will be option A.

CBSE Questions for Class 10 Maths Real Numbers Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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