4 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 14 Marks

Prove that the product of three consecutive positive integer is divisible by $6.$

Answer

Let, n be any positive integer. Since any positive is of the form 6q or $6q + 1 or, 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.$
If $n - 6q$, then

$n(n + 1)(n + 2) = [(6q)(6q + 1)(6q + 2)]$

$= 6[q(6q + 1)(6q + 2)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 1$, then

$n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)$

$= 6[(6q + 1)(3q + 1)(2q + 1)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 2$, then

$n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)$

$= 6[(3q + 1)(2q + 1)(6q + 4)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 3$, then

$n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)$

$= 6[(3q + 1)(3q + 2)(6q + 5)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 4$, then

$n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)$

$= 6[(6q + 4)(6q + 5)(q + 1)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 5$, then

$n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)$

$= 6[(6q + 5)(q + 1)(6q + 7)]$

$= 6m$, which is divisible by $6$

Hence, the product of three consecutive positive integer is divisible by $6.$

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Question 24 Marks

Show that the cube of a positive integer is of the form $6q + r$, where q is an integer and $r = 0, 1, 2, 3, 4, 5.$

Answer

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that.
$a=6 q+r, \text { where, } 0 \leq r<6 $
$ \Rightarrow a^3=(6 q+r)^3=216 q^3+r^3+3 \cdot 6 q \cdot r(6 q+r) $
$ {\left[\because(a+b)^3=a^3+b^3+3 a b(a+b)\right]} $
$ \Rightarrow a^3=\left(216 q^3+108 q^2 r+18 q r^2\right)+r^3 \ldots(i)$
Where, $0 \leq \mathrm{r}<6$
Case I: When $r=0$, then putting $r=0$ in
Eq. (i), we get
$a^3=216 q^3=6\left(36 q^3\right)=6 m$
Where, $m=36 q^3$ is an integer.
Case II: Where $r=1$, then putting $r=1$ in
Eq. (i), we get
$ a^3=\left(216 q^3+108 q^3+18 q\right)+1=6\left(36 q^3+18 q^3+3 q\right)+1 $
$ a^3=6 m+1$
Where $m=\left(36 q^3+18 q^3+3 q\right)$ is an integer.
Case III: When $r=2$, then putting $r=2$ in
Eq. (i), we get
$ a^3=\left(216 q^3+216 q^2+72 q\right)+8 $
$ a^3=\left(216 q^3+216 q^2+72 q+6\right)+2 $
$ \Rightarrow a^3=6\left(36 q^3+36 q^2+12 q+1\right)+2=6 m+2$
Where, $m=\left(36 q^3+36 q^2+12 q+1\right)$ is an integer.
Case IV: When $r=3$, then putting $r=3$ in
Eq. (i), we get
$a^3=\left(216 q^3+324 q^2+162 q\right)+27 $
$ =\left(216 q^3+324 q^2+162 q+24\right)+3 $
$ =6\left(36 q^3+54 q^2+27 q+4\right)+3=6 m+3$
Where, $m=\left(36 q^3+64 q^2+27 q+4\right)$ is an integer.
Case V: When $r=4$, then putting $r=4$ in
Eq. (i), we get
$a^3=\left(216 q^3+432 q^2+288 q\right)+64 $
$ a^3=6\left(36 q^3+72 q^2+48 q\right)+60+4 $
$ a^3=6\left(36 q^3+72 q^2+48 q+10\right) \text { is an integer. }$
Case VI: When $r=5$, then putting $r=5$ in
Eq. (i), we get
$ a^3=\left(216 q^3+540 q^2+450 q\right)+125 $
$ \Rightarrow a^3=\left(216 q^3+540 q^2+450 q\right)+120+5 $
$ \Rightarrow a^3=6\left(36 q^3+90 q^2+75 q+20\right)+5 $
$ \Rightarrow a^3=6 m+5$
Where, $m=\left(36 q^3+90 q^2+75 q+20\right)$ is an integer.
Hence, the cube of a positive integer of the form $6q + r, q$ is an integer and $r = 0, 1, 2, 3, 4, 5$ is also of the forms $6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.$

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Question 34 Marks

Show that the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer m.

Answer

Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that
$a=6 q+r \text {, where } 0 \leq r<6 $
$ \Rightarrow a^2=(6 q+r)^2=36 q^2+r^2+12 q r $
$ {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} $
$ \Rightarrow a^2=6\left(6 q^2+2 q r\right)+r^2 \ldots \text { (i) }$
Where, $0 \leq r<6$
Case I: Where $r=0$, then putting $r=0$ in
Eq. (i), we get
$a^2=6\left(6 q^2\right)=6 m$
Where, $m=6 q^2$ is an integer.
Case II: When $r=1$, then putting $r=1$ in
Eq. (i), we get
$a^2=6\left(6 q^2+2 q\right)+1=6 m+1$
Where, $m=\left(6 q^2+2 q\right)$ is an integer
Case III: When $r=2$, then putting $r=2$ in
Eq. (i), we get
$a^2=6\left(6 q^2+4 q\right)+4=6 m+4$
Where, $m=\left(6 q^2+4 q\right)$ is an integer.
Case IV: When $r=3$, then putting $r=3$ in
Eq. (i), we get
$ a^2=6\left(6 q^2+6 q\right)+9 $
$ =6\left(6 q^2+6 q\right)+6+3 $
$ \Rightarrow a^2=6\left(6 q^2+6 q+1\right)+3=6 m+3$
Where, $m=\left(6 q^2+6 q+1\right)$ is an integer.
Case V : When $\mathrm{r}=4$, then putting $\mathrm{r}=4$ in
Eq. (i), we get
$a^2=6\left(6 q^2+8 q\right)+16 $
$ =6\left(6 q^2+8 q\right)+12+4 $
$ \Rightarrow a^2=6\left(6 q^2+8 q+2\right)+4=6 m+4$
Where, $n=\left(6 q^2+8 q+2\right)+4=6 m+4$
Case VI: When $r=5$, then putting $r=5$ in
Eq. (i), we get
$a^2=6\left(6 q^2+10 q\right)+25 $
$ =6\left(6 q^2+10 q\right)+24+1 $
$ \Rightarrow a^2=6\left(6 q^2+10 q+4\right)+1=6 m+1$
Where, $m=\left(6 q^2+10 q+1\right)$ is an integer.
Hence, the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer m.

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Question 44 Marks

Determine the number nearest to $110000$ but greater than $100000$ which is exactly divisible by each of $8, 15$ and $21.$

Answer

To Find The number nearest to $110000$ but greater than $100000$ which is exactly divisible by each of $8, 15$ and $21.$
L.C.M of $8, 15$ and $21.$
$8 = 2^3$
$15 = 3 × 5$
$21 = 3 × 7$
L.C.M of 8, 15 and $21 = 23 × 3 × 5 × 7 = 840$
When $110000$ is divided by $840$, the remainder is obtained as $800.$
Now, $110000 - 800 = 109200$ is divisible by each of $8, 15$ and $21.$
Also, $110000 + 40 = 110040$ is divisible by each of $8, 15$ and $21.$
$109200$ and $110040$ are greater than $100000$.
Hence, $110040$ is the number nearest to $110000$ but greater than $100000$ which is exactly divisible by each of $8, 15$ and $21$.

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Question 54 Marks

In a morning walk three persons step off together, their steps measure $80\ cm, 85\ cm$ and $90\ cm$ respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

Answer

Given that three persons step off together, their steps measure $80\ cm, 85\ cm$ and $90\ cm.$
The minimum distance each should walk is equal to $L.C.M$ of $80, 85$ and $90.$
Prime factors of 80, 85 and 90 are
$ 80=2 \times 2 \times 2 \times 5=2^4 \times 5 $
$ 85=5 \times 17=5 \times 17 $
$ 90=2 \times 3 \times 3 \times 5=2 \times 3^2 \times 5 $
$ \text { L.C.M }(80,85,90)=2^4 \times 3^2 \times 5 \times 17 $
$ =16 \times 9 \times 85 $
$ =12240 \mathrm{~cm} $
$ \because 100 \mathrm{~cm}=1 \mathrm{~m} $
$ \therefore 12240 \mathrm{~cm}=122.40 \mathrm{~m} $
$ =122 \mathrm{~m} 40 \mathrm{~cm}$
So, the minimum distance each should walk is $122\ m 40\ cm.$

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Question 64 Marks

Using Euclid’s division algorithm, find the largest number that divides $1251, 9377$ and $15628$ leaving remainders $1, 2$ and $3$ respectively.

Answer

Since, $1,2$ and $3$ are the remainders of $1251, 9377$ and $15628$, respectively.
Thus, after subtracting these remainders from the numbers.
We have the numbers, $1251 - 1 = 1250, 9377 - 2 = 9375$ and $15628 - 3 = 15625$ which is divisible by the required number.
Now, required number $= HCF$ of $1250, 9375$ and $15625$ [for the largest number]
By Euclid’s division algorithm,
$a = bq + r .....(i)$
[$\because$ dividend = divisor × quotient + remainder]
For largest number, put $a = 15625$ and $b = 9375$
$15625 = 9375 \times 1 + 6250$
$\Rightarrow 9375 = 6250 \times 1 + 3125$
$\Rightarrow 6250 = 3125 \times 2 + 0$
$\therefore$ H.C.F (15625, 9375) = 3125
Now, we take c = 1250 and d = 3125, then again using Euclid's division algorithm,
$d = cq + r$ [from Eq. $(i)]$
$\Rightarrow 3125 = 1250 \times 2 + 625$
$\Rightarrow 1250 = 625 \times 2 + 0$
$\therefore$ $H.C.F (1250, 9375, 15625) = 625$
Hence, 625 is the largest number which divides $1251, 9377$ and $15628$ leaving remainder $1, 2$ and $3,$ respectively.

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Question 74 Marks

Prove that if a positive integer is of the form $6q + 5$, then it is of the form $3q + 2$ for some integer q, but not conversely.

Answer

To Prove: that if a positive integer is of the form $6q + 5$ then it is of the form $3q + 2$ for some integer q, but not conversely.
Proof: Let $n = 6q + 5$
Since any positive integer n is of the form of $3k$ or $3k + 1, 3k + 2$
If $q = 3k$
Then, $n = 6q + 5$

$\Rightarrow n = 18k + 5(q = 3k)$

$\Rightarrow n = 3(6k + 1) + 2$

$\Rightarrow n = 3m + 2$(where $m = (6k + 1))$

If $q = 3k + 1$
Then, $n = (6q + 5)$

$\Rightarrow n = (6(3k + 1) + 5)(q = 3k + 1)$

$\Rightarrow n = 18k + 6 + 5$

$\Rightarrow n = 18k + 11$

$\Rightarrow n = 3(6k + 3) + 2$

$\Rightarrow n = 3m + 2$(where $m = (6k + 3))$

If $q = 3k + 2$
Then, $n = (6q + 5)$

$\Rightarrow n = (6(3k + 2) + 5)(q = 3k + 2)$

$\Rightarrow n = 18k + 12 + 5$

$\Rightarrow n = 18k + 17$

$\Rightarrow n = 3(6k + 5) + 2$

$\Rightarrow n = 3m + 2$

(where $m = (6k + 5))$

Consider here 8 which is the form $3q + 2$ i.e. $3 \times 2 + 2$ but it can’t be written in the form $6q + 5$. Hence the converse is not true.

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Question 84 Marks

$105$ goats, $140$ donkeys and $175$ cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?

Answer

We are given that, $105$ goats, $140$ donkeys and $175$ cows. There is only one boat which will have to make many y trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. We need to tell the number of animals that went in each trip.
Given that
Number of goats $= 105$
Number of donkeys $= 140$
Number of cows $= 175.$
Therefore, the largest number of animals in 1 trip $= H.C.F.$ of $105, 140$ and $175.$
First we consider $105$ and $140.$
By applying Euclid’s division lemma
$140 = 105 \times 1 + 35$
$105 = 35 \times 3 + 0.$
Therefore, $H.C.F.$ of $105$ and $140 = 35$
Now, we consider $35$ and $175.$
By applying Euclid’s division lemma
$175 = 35 × 5 + 0.$
Therefore, $H.C.F.$ of $105, 140$ and $175 = 35$
Hence, the number of animals went in each trip is $35.$

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Question 94 Marks

If the $HCF$ of $408$ and $1032$ is expressible in the form $1032m - 408 \times 5$, find m.

Answer

Given that two positive integers $408$ and $1032$ and $1032 > 408.$
So, appliying Euclid’s division algorithem
$1032 = 408 \times 2 + 216 …..(i)$
$408 = 216 \times 1 + 192 …..(ii)$
$216 = 192 \times 1 + 24 …..(iii)$
$192 = 24 \times 6 + 0 …..(iv)$
So, HCF of $408$ and $1032$ is divison of eq. (iv) and remainder of eq. (iii) i.e. $24.$
$HCF (408, 1032) = 1032m - 408 \times 5$
$\Rightarrow 24 = 1032m - 2040$
$\Rightarrow 1032m = 2040 + 24$
$\Rightarrow 1032m = 2064$
$\Rightarrow\text{m} = \frac{2064}{1032}$
$\Rightarrow m = 2$
Thus, $m = 2.$

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Question 104 Marks

Show that one and only one out of n, $n + 4, n + 8, n + 12$ and $n + 16$ is divisible by $5$, where n is any positive integer.

Answer

Consider the numbers $n, (n + 4), (n + 8), (n + 12)$ and $(n + 16)$, where n is any positive integer.
Suppose $n = 5q + r$, where $0 ≤ r < 5$
$n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4$
(By Euclid's division algorithm)
Case I:
When $n = 5q$.
$n = 5q$ is divisible by $5.$
$n + 4 = 5q + 4$ is not divisible by $5.$
$n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3$ is not divisible by $5.$
$n + 12 = 5q + 10 + 2 = 5(q + 2) + 2$ is not divisible by $5.$
$n + 16 = 5q + 15 + 1 = 5(q + 3) + 1$ is not divisible by $5.$
Case II:
When $n = 5q + 1.$
$n = 5q + 1$ is not divisible by $5.$
$n + 4 = 5q + 1 + 4 = 5(q + 1)$ is not divisible by $5.$
$n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4$ is not divisible by $5.$
$n + 12 = 5q + 1 + 12 = 5(q + 2) + 3$ is not divisible by $5.$
$n + 16 = 5q + 1 + 16 = 5(q + 3) + 2 $is not divisible by $5.$
Case III:
When $n = 5q + 2.$
n = 5q + 2 is not divisible by $5.$
$n + 4 = 5q + 2 + 4 = 5(q + 1) + 1$ is not divisible by $5.$
$n + 8 = 5q + 2 + 8 = 5(q + 2)$ is not divisible by$ 5.$
$n + 12 = 5q + 2 + 12 = 5(q + 2) + 4 $is not divisible by $5.$
$n + 16 = 5q + 2 + 16 = 5(q + 3) + 3$ is not divisible by $5.$
Case IV:
When $n = 5q + 3.$
n = 5q + 3 is not divisible by $5.$
$n + 4 = 5q + 3 + 4 = 5(q + 1) + 2$ is not divisible by $5.$
$n + 8 = 5q + 3 + 8 = 5(q + 2) + 1$ is not divisible by $5.$
$n + 12 = 5q + 3 + 12 = 5(q + 3)$ is not divisible by $5.$
$n + 16 = 5q + 3 + 16 = 5(q + 3) + 4$ is not divisible by $5.$
Case V:
When $n = 5q + 4.$
$n = 5q + 4$ is not divisible by $5.$
$n + 4 = 5q + 4 + 4 = 5(q + 1) + 3$ is not divisible by $5.$
$n + 8 = 5q + 4 + 8 = 5(q + 2) + 2$ is not divisible by $5.$
$n + 12 = 5q + 4 + 12 = 5(q + 3) + 1$ is not divisible by $5.$
$n + 16 = 5q + 4 + 16 = 5(q + 4)$ is not divisible by $5.$
Hence, in each case, one and only one out of n, $n + 4, n + 8, n + 12$ and $n + 16$ is divisible by $5.$

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Question 114 Marks

Explain why $3 × 5 × 7 + 7$ is a composite number.

Answer

We have, $3 × 5 × 7 + 7 = 105 + 7 = 112$
Now, $112 = 2 × 2 × 2 × 2 × 7 = 2^4× 7$
So, it is the product of prime factors $2$ and $7$. i.e., it has more than two factors.
Hence, it is a composite number.

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Question 124 Marks

If a and b are two odd positive integers such that $a > b$, then prove that one of the two numbers $\frac{\text{a}+\text{b}}{2}$ and $\frac{\text{a}-\text{b}}{2}$ is odd and the other is even.

Answer

Let $a = 2q + 3$ and $b = 2q + 1$ be two positive odd integers such that $a > b$
Now, $\frac{\text{a}+\text{b}}{2}=\frac{2\text{q}+3+2\text{q}+1}{2}=\frac{4\text{q}+4}{2}=2\text{q}+2=$ an odd number
and $\frac{\text{a}-\text{b}}{2}=\frac{(2\text{q}+3)-(2\text{q}+1)}{2}=\frac{2\text{q}+3-2\text{q}-1}{2}=\frac{2}{2}=1=$ an odd number
Hence one of the two numbers $\frac{\text{a}+\text{b}}{2}$ and $\frac{\text{a}-\text{b}}{2}$ is odd and the other is even for any two positive odd integer.

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Question 134 Marks

Prove that the square of any positive integer is of the form $3m$ or, $3m + 1$ but not of the from $3m + 2.$

Answer

Let, n be any positive integer
We know that positive integer n is a in the form of $3q, (3q + 1)$ and $(3q + 2),$ where q is some integer
Case 1: When $n = 3q$
Squaring of both sides
$\Rightarrow n^2=(3 q)^2 $
$ \Rightarrow n^2=9 q^2 $
$ \Rightarrow n^2=3 q(3 q) $
$ \Rightarrow n^2=3 m[\text { where } m=q(3 q)]$
Case 2: When $\mathrm{n}=(3 \mathrm{q}+1)$
Squaring of both sides
$ \Rightarrow n^2=(3 q+1)^2 $
$ \Rightarrow n^2=9 q^2+6 q+1 $
$ \Rightarrow n^2=3 q(3 q+2)+1 $
$ \Rightarrow n^2=3 m+1[\text { Where } m=q(3 q+2)]$
Case 3: When $n=(3 q+2)$
Squaring of both sides
$ \Rightarrow n^2=(3 q+2)^2 $
$ \Rightarrow n 2=9 q^2+12 q+4 $
$ \Rightarrow n^2=9 q^2+12 q+3+1 $
$ \Rightarrow n^2=3\left(3 q^2+4 q+1\right)+1 $
$ \Rightarrow n^2=3 m+1\left[\text { Where } m=\left(3 q^2+4 q+1\right)\right]$
Thus, them case $1$, case $2$ and case $3$, it is proved that square of any positive odd integer is of the form $3m, (3m + 1)$ but not of the form $(3m + 2).$

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Question 144 Marks

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in yeare)	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$3$	$8$	$10$	$10$	$5$	$4$

Answer

Given;

Age (in years): $x_i$	$15$	$16$	$17$	$18$	$19$	$20$
No.of students: $f_i$	$3$	$8$	$10$	$10$	$5$	$4$

First of all prepare the frequency table in such a way that its first column consist of the values of the variate$(x_i)$ and the second column the corresponding frequencies$(f_i).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_ix_i).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$.

Age (in years): $x_i$	No. of students: $f_i$	$f_ix_i$
$15$	$3$	$45$
$16$	$8$	$128$
$17$	$10$	$170$
$18$	$10$	$180$
$19$	$5$	$95$
$20$	$4$	$80$
	$\sum\text{f}_\text{i}=40$	$\sum\text{f}_\text{i}\text{x}_\text{i}=698$

We know that mean, $\overline{X}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\overline{X}=\frac{698}{40}$
$=17.45$
Hence, the mean age of the students $=17.45 \ \text{years}$

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Question 154 Marks

Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^m × 5^n$, where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.

Answer

Denominator of the rational number $\frac{257}{5000}$ is $5000.$
Now, factors of $5000=2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5=(2)^3 \times(5)^4$, which is of the type $2^m \times 5^n$, where $m = 3$ and $n = 4$ are non-negative integers.
$\therefore$ Rational number $=\frac{257}{5000}=\frac{257}{2^3\times5^4}\times\frac{2}{2}$
$[$Since, multiplying numerator and denominater by $2]$
$=\frac{514}{2^4\times5^4}=\frac{514}{(10)^4}$
$=\frac{514}{10000}=0.0514$
Hence, which is the required decimal expansion of the rational $\frac{257}{5000}$ and it is also a terminating decimal number.

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Question 164 Marks

Prove that $\sqrt{5}+\sqrt{3}$ is an irrational number.

Answer

Let us assume that $\sqrt{5}+\sqrt{3}$ is a rational number.
$\therefore\ \sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-prime numbers.
$\sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \sqrt{5}=\frac{\text{a}}{\text{b}}-\sqrt{3}$
Squaring both sides,
$\Rightarrow(\sqrt{5})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{3}\Big)^2$
$\Rightarrow5=\Big(\frac{\text{a}}{\text{b}}\Big)^2+3-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow5-3=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow2=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow\frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-2$
$\Rightarrow\ \frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)$
$\Rightarrow\ \sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)\frac{\text{b}}{2\text{a}}$
$\Rightarrow\sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{2\text{ab}}\Big)$
We know that $\sqrt{3}$ is an irrational number.
This contradicts our assumption that $\sqrt{5}+\sqrt{3}$ is a rational number.

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Question 174 Marks

Prove that the square of any positive integer is of the form $4q$ or $4q + 1$ for some integer $q.$

Answer

By Eudid's division Algorithm
$a = bm + r,$ where $0 ≤ r < b$
Put $b = 4$
$a = 4m + r,$ where $0 ≤ r < 4$
If $r = 0,$ then $a = 4m$
If $r = 1,$then $a = 4m + 1$
If $r = 2,$ then $a = 4m + 2$
If $r = 3,$ then $a = 4m + 3$
Now, $(4m)^2= 16m^2$
$= 4 × 4m^2$
$= 4q$ where q is some integer
$(4m + 1)^2 = (4m)^2 + 2(4m)(1) + (1)^2$
$= 16m^2 + 8m + 1$
$= 4(4m^2 + 2m) + 1$ where $4m^2 + 2m = q$
$= 4q + 1$ where q is some integer
$(4m + 3)^2 = (4m)^2 + 2(4m)(3) + (3)^2$
$= 16m^2 + 24m + 9$
$= 16m^2 + 24m + 8 + 1$
$= 4(4m^2 + 6m + 2) + 1$
$= 4q + 1$, where q is some integre
Hence, the square of any positive in teger is of the form $4q$ or $4q + 1$ for some integer m.

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Question 184 Marks

Show that any positive odd integer is of the form $6q + 1,$ or $6q + 3,$ or $6q + 5,$ where q is some integer.

Answer

Let a be any positive integer and $b = 6.$ Then, by Euclid’s algorithm,
$a = 6q + r$ for some integer $q ≥ 0,$ and $r = 0, 1, 2, 3, 4, 5$ because $0 ≤ r < 6.$
Therefore, $a = 6q$ or $6q + 1$ or $6q + 2$ or $6q + 3$ or $6q + 4$ or $6q + 5$
Also, $6q + 1 = 2 × 3q + 1 = 2k_1+ 1,$where $k_1$ is a positive integer
$6q + 3 = (6q + 2) + 1 = 2(3q + 1) + 1 = 2k_2 + 1,$ where $k_2$ is an integer
$6q + 5 = (6q + 4) + 1 = 2(3q + 2) + 1 = 2k_3+1,$ where $k_3$ is an integer
Clearly, $6q + 1, 6q + 3, 6q + 5$ are of the form $2k + 1$, where k is an integer.
Therefore, $6q + 1, 6q + 3, 6q + 5$ are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form $6q + 1,$ or $6q + 3$, or $6q + 5$

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Question 194 Marks

If p, q are prime positive integers, prove that $\sqrt{\text{p}}+\sqrt{\text{qp + q}}$ is an irrational number.

Answer

Let us asssume that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is rational.
Then there exist positive co-primes a and b such that
$\sqrt{\text{p}}+\sqrt{\text{q}}=\frac{\text{a}}{\text{b}}$
$\sqrt{\text{p}}=\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}$
$(\sqrt{\text{p}})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}\Big)^2$
$\text{p}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}+\text{q}$
$\text{p}-\text{q}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\Big(\frac{\text{a}}{\text{b}}\Big)^2-(\text{p}-\text{q})=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}\bigg)\Big(\frac{\text{b}}{2\text{a}}\Big)=\sqrt{\text{q}}$
$\sqrt{\text{q}}=\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{2\text{ab}}\bigg)$
Here we see that $\sqrt{\text{q}}$ is rational number which is a contradiction as we know that $\sqrt{\text{q}}$ is an irrational number
Hence $\sqrt{\text{q}}+\sqrt{\text{q}}$ is irrational

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Question 204 Marks

The following data gives the distribution of total monthly household expenditure of $200$ families of a village. .Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure $($in $₹)$	Frequency
$1000-1500$	$24$
$1500-2000$	$40$
$2000-2500$	$33$
$2500-3000$	$28$
$3000-3500$	$30$
$3500-4000$	$22$
$4000-4500$	$16$
$4500-5000$	$7$

Answer

We may observe that the given data the maximum class frequency is $40$ belonging to $1500 – 2000$ interval. So modal class $= 1500 – 2000$
$L.L (L) = 1500, f.$ of $M.C (f_1) = 40$
Frequency of class preceeding modal class $f_0 = 24$
Frequency of class succeeding modal class $f_2 = 33$
Class size $(h) = 50$
$\text{Mode}=\text{L}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=1500+\Big[\frac{40-24}{2(40)-24-33}\Big]\times500$
$=1500+\Big[\frac{16}{80-67}\times500\Big]$
$=1500+347.826$
$1847.826=1847.83$
So modal class monthly expenditure was $Rs. 1847.83$
Now we may find class mark as
$\text{Class mark}=\frac{\text{upper class limit+lower class limit}}{2}$
Class size $(h)$ of given data $= 500$
Now taking $2750$ as assumed mean $(a)$ we may calculate $d, 4$ and $f, 4$ as follows.

Expenditure in $Rs.$	No. of families $f_i$	$x_i$	$\text{d}_\text{i}=\text{x}_\text{i}-2\pi0$	$4_i$	$f_iu_i$
$1000-1500$	$24$	$1250$	$-1500$	$-3$	$-72$
$1500-2000$	$40$	$1750$	$-1000$	$-2$	$-80$
$2000-2500$	$33$	$2250$	$-500$	$-1$	$-33$
$2500-3000$	$28$	$2750$	$0$	$0$	$0$
$3000-3500$	$30$	$3250$	$500$	$1$	$30$
$3500-4000$	$22$	$3750$	$1000$	$2$	$44$
$4000-4500$	$16$	$4250$	$1500$	$3$	$28$
$4500-5000$	$7$	$4750$	$2000$	$4$	$28$
Total	$200$				$-35$

New from table we may observe that
$\sum\text{f}_\text{i}=200$
$\sum\text{f}_\text{i}\text{u}_\text{i}=-35$
$(\bar{\text{x}})\ \text{Mean}=\text{a}+\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big)\times\text{h}$
$(\bar{\text{x}})=2750+\Big(\frac{-35}{200}\Big)\times500$
$=2750-87.5$
$=2662.3$
So mean, monthly expenditure was $Rs. 2662.50$
We may observe them the given data the maximum class frequency is $10$ belonging to class
interval $30 – 35$
So modal class $30 – 35$
Class size $(h) = 5$
Lower limit $(L)$ of modal class $= 30$
Frequency $(f_1)$ of modal class $10$
Frequency $(f_0)$ of class preceeding modal class $= 9$
Frequency $(f_0)$ of class succeeding modal class $= 3$
$\text{Mode}=\text{L}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=30+\Big(\frac{10-9}{30-9-3}\Big)5$
$=30+\frac{5}{8}$
$=30.625$
$\text{Mode}=30.6$
It represents that most of states $1 \ UT$ have a teacher $-$ student ratio as $30.6$
Now we may find class marks by using the relation
$\text{Class mark}=\frac{\text{upper class limit lower class limit}}{2}$
Now taking $325$ as assumed mean $(a)$ we may calculated $d_i 4_i$ and $f_i 4_{i }$ as following.

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Question 214 Marks

Compute the median for each of the following data:
$(1)$

Marks	No. of students
Less than $10$	$0$
Less than $30$	$10$
Less than $50$	$25$
Less than $70$	$43$
Less than $90$	$65$
Less than $110$	$87$
Less than $130$	$96$
Less than $150$	$100$

$(2)$

Marks	No. of students
More than $150$	$0$
More than $140$	$12$
More than $130$	$27$
More than $120$	$60$
More than $110$	$105$
More than $100$	$124$
More than $90$	$141$
More than $80$	$150$

Answer

$(1)$ We prepare the cumulative frequency table, as given below.

Marks	No. of students $(f_i)$	Cumulative frequency $(c.f.)$
$0-10$	$0$	$0$
$10-30$	$10$	$10$
$30-50$	$15$	$25$
$50-70$	$18$	$43$
$70-90$	$22$	$65$
$90-110$	$22$	$87$
$110-130$	$9$	$96$
$130-150$	$4$	$100$
	$N = 100$

Now, we have $\text{N}=100$
So, $\frac{\text{N}}{2}=50$
Now, the cumulative frequency just greater than $50$ is $65$ and the
corresponding class is $70-90$
Therefore, $70-90$ is the median class.
Here, $\text{l}=70, \text{f}=22,\text{F}=43$ and $\text{h}=20$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=70+\Big\{\frac{50-43}{22}\Big\}\times20$
$=70+\frac{7\times20}{22}$
$=70+\frac{140}{22}$
$=70+6.36$
$=76.36$
Hence, the median is $76.36.$
Note: The first class in the table can be omitted also.
$(2)$ We prepare the cumulative frequency table, as given below.

Marks	No. of students $(f_i)$	Cumulative frequency $(c.f.)$
More than $150$	$0$	$0$
$140-150$	$12$	$12$
$130-140$	$15$	$27$
$120-130$	$33$	$60$
$110-120$	$45$	$105$
$100-120$	$19$	$124$
$90-100$	$17$	$141$
$80-90$	$9$	$150$
	$N = 150$

Now, we have $\text{N}=150$
So. $\frac{\text{N}}{2}=75$
Thus, the cumulative frequency just greater than $75$ is $105$ and the corresponding class is $110-120.$
Therefore, $110-120$ is the median class.
$\text{l}=120, \text{f}=45,\text{F}=60$ and $\text{h}=-10$
$($Because class interval given in descending order$)$
We know that
$\text{Median}= \text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$= 120+\Big\{\frac{75-60}{45}\Big\}\times(-10)$
$= 120-\frac{15\times10}{45}$
$= 120-\frac{150}{45}$
$= 120-3.333$
$=116.67 ($approx$)$
Hence, the median is $116.67.$

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Question 224 Marks

The following table shows the marks scored by $140$ students in an examination of a certain paper:

Marks	$0-12$	$10-20$	$20-30$	$30-40$	$40-50$
Number of students	$20$	$24$	$40$	$36$	$20$

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Answer

We may prepare the table as shown:

Marks:	Mid value$(x_i):$	No. of students$(f_i):$	$d_i=x_i-A = x_i- 25$	$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})$ $=\frac{1}{10}(\text{d}_\text{i})$	$f_iu_i$	$f_id_i$	$f_ix_i$
$0-10$	$5$	$20$	$-20$	$-2$	$-40$	$-400$	$100$
$10-20$	$15$	$24$	$-10$	$-1$	$124$	$-240$	$-360$
$20-30$	$25$	$40$	$0$	$0$	$0$	$0$	$1000$
$30-40$	$35$	$36$	$10$	$1$	$36$	$360$	$1260$
$40-50$	$45$	$20$	$20$	$2$	$40$	$400$	$900$
		$\sum\text{f}_\text{i}=140$			$\sum\text{f}_\text{i}\text{u}_\text{i}=12$	$\sum\text{f}_\text{i}\text{d}_\text{i}=120$	$\sum\text{f}\text{i}\text{x}_\text{i}=3620$

$(1)$Direct Method:
We know that mean, $\bar{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$= 3620140 = 25.857 = 3620140 = 25.857$
Hence, the mean is $25.857.$
$(2)$Short$-$cut method:
Let the assumed mean $A = 25$
We know that mean, $\bar{\text{X}}=\text{A}+\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{d}_\text{i}\Big)$
$=25+\Big(\frac{1}{140}\times(120)\Big)$
$=25+\frac{120}{140}$
$=25+0.857$
$=25.857$
Hence, the mean is $25.857.$
$(3)$Step deviation method:
Let the assumed mean
$A = 25$ and $h = 10.$
We know that mean,
$\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=25+10\Big(\frac{1}{140}\times(12)\Big)$
$=25+\frac{120}{140}$
$=25+0.857$
$=25.857$
Hence, the mean is $25.857$

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Question 234 Marks

The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length $($in $mm)$	$118-126$	$127-135$	$136-144$	$145-153$	$154-162$	$163-171$	$172-180$
$N.$ of leaves	$3$	$5$	$9$	$12$	$5$	$4$	$2$

find the mean length of leaf.

Answer

Calculation for mean

Length $($in $mm)$	Mid$-$Values$(x_i)$	Number of lesves$(f_i)$	$f_ix_i$
$117.5-126.5$	$122$	$3$	$336$
$126.5-135.5$	$131$	$5$	$655$
$135.5-144.5$	$140$	$9$	$1260$
$144.5-153.5$	$149$	$12$	$1788$
$153.5-162.5$	$158$	$5$	$790$
$162.5-171.5$	$167$	$4$	$668$
$171.5-180.5$	$176$	$2$	$353$
		$N = 40$	$\sum\text{f}_\text{i}\text{x}_\text{i}=5880$

Mean length of the leaf $=\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{x}_\text{i}=\frac{1}{40}\times5880=147$
Calculation for median
The given series is in inclusive form. Converting it to exclusive form and
preparing the cumulative frequency table, we have

Length $($in $mm)$	Number of leaves $(f_i)$	Cumulative Frequency $(c.f.)$
$117.5-126.5$	$3$	$3$
$126.5-135.5$	$5$	$8$
$135.5-144.5$	$9$	$17$
$144.5-153.5$	$12$	$29$
$153.5-162.5$	$5$	$34$
$162.5-171.5$	$4$	$38$
$171.5-180.5$	$2$	$40$
	$N = 40$

Now, the cumulative frequency just greater than $20$ is $29$ and the corresponding class is $144.5–153.5.$
Therefore, $144.5–153.5$ is the median class.
Here, $\text{I}=145,\text{f}=12,\text{F}=17$ and $\text{h}=9$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=144.5+\Big(\frac{20-17}{12}\Big)\times9$
$=144.5+\frac{27}{12}$
$=144.5+2.25$
$=146.75$
Hence, the median length of leaf is $146.75 \ mm.$

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Question 244 Marks

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

Age in years	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Answer

Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequencies table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can definite class intervals with their respective cumulative frequencies as below

Age (in years)	No. of policy planers	Cumulative frequency (cr)
18-20	2	2
20-25	6 - 2 = 4	6
25-30	24 - 6 = 18	24
30-35	45 - 24 = 21	45
35-40	78 - 45 = 33	78
40-45	89 - 78 = 11	89
45-50	92 - 89 = 3	92
50-55	98 - 92 = 6	98
55-60	100 - 98 = 2	100

Total (n)
Now from the table we may observe that n = 100 cumulative frequencies (F) just greater than $\frac{\text{n}}{2}\Big(\text{i.e}.,\frac{100}{2}=50\Big)$ is 78 belonging to inteval 35-40
So median class = 35-40
Lower limit (1) o median class = 35
Class size (h) = 5
Frequlative (f) of median class = 33
Cumulative frequency (f) off class preceding median class = 45
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}-\text{cf}\Big)}{\text{f}}\times\text{h}$
$=35+\Big(\frac{50-45}{33}\Big)\times5$
$=35+\frac{25}{33}$
$=35.76$
So, median age is 35.76 years.

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Question 254 Marks

Find the mean, median and mode of the following data:

Classes	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$	$250-300$	$300-350$
Frequency	$2$	$3$	$5$	$6$	$5$	$3$	$1$

Answer

Classes	Frequency $(f_i)$	$x_i$	$f_ix_i$	$c.f.$
$0-50$	$2$	$25$	$50$	$2$
$50-100$	$3$	$75$	$225$	$5$
$100-150$	$5$	$125$	$625$	$10$
$150-200$	$6$	$175$	$1050$	$16$
$200-250$	$5$	$225$	$1125$	$21$
$250-.300$	$3$	$275$	$825$	$24$
$300-350$	$1$	$325$	$325$	$25$
	$\text{N}=\sum\text{f}=25$		$\sum\text{f}_\text{i}\text{x}_\text{i}=4225$

Here, the maximum frequency is $6$ so the modal class $150-200.$
Therefore,
$l = 150 \ h = 50 \ f = 6 \ f_1 = 5 \ f_2 = 5 \ F = 10$
$\text{Mean}=\frac{\sum\text{f}_i\text{x}_i}{\sum\text{f}}$
$=\frac{4225}{25}$
$\text{Mean}=169$
Thus, the mean of the data is $169.$
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=150+\frac{12.5-10}{6}\times50$
$=150+\frac{2.5}{6}\times50$
$=150+\frac{125}{6}$
$\text{Median}=170.83$
Thus, the median of the data is $170.83$
$\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=150+\frac{6-5}{12-5-5}\times50$
$=150+\frac{1}{2}\times50$
$=150+25$
$=175$
Thus, the mode of the data is $175.$

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Question 264 Marks

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is $50.$

$x$	$10$	$30$	$50$	$70$	$90$
$f$	$17$	$f_1$	$32$	$f_2$	$19$	Total $120$

Answer

Mean $= 50$

$x$	$f$	$fx$
$10$	$17$	$170$
$30$	$f_1$	$30f_1$
$50$	$32$	$1600$
$70$	$f_2$	$70f_2$
$90$	$19$	$1710$
Total	$68 + f_1 + f_2 = 120$	$3480 + 30f_1 +70f_2$

$68+\text{f}_1+\text{f}_2=120$
$\Rightarrow\text{f}_1+\text{f}_2=120-68=52$
$\Rightarrow\text{f}_2=52-\text{f}_1$
And $\overline{\text{x}}=\frac{\sum\text{f}\times\text{x}}{\sum\text{f}}$
$\Rightarrow50=\frac{3480+30\text{f}_1+70\text{f}_2}{120}$
$\Rightarrow3480+30\text{f}_1+70\text{f}_2=120\times50=6000$
$\Rightarrow30\text{f}_1+70\text{f}_2=6000-3480=2520$
$\Rightarrow3\text{f}_1+7\text{f}_2=252$
$\Rightarrow3\text{f}_1+7(52-\text{f}_1)=252$
$\Rightarrow3\text{f}_1+364-7\text{f}_1=252$
$\Rightarrow-4\text{f}_1=252-364=-112$
$\Rightarrow\text{f}_1=\frac{-112}{-4}=28$
And $\text{f}_2=52-28=24$
Hence $\text{f}_1=28,\ \text{f}_2=24$

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Question 274 Marks

Find the mean of each of the following frequency distributions:

Classes	$25-29$	$30-34$	$35-39$	$40-44$	$45-49$	$50-54$	$55-59$
Frequency	$14$	$22$	$16$	$6$	$5$	$3$	$4$

Answer

The given series is an inclusive series. Firstly, make it an exclusive series.

Classes interval	Mid$-$value$(x_i)$	Frequency$(f_i)$	$\text{d}_\text{i}=\text{x}_\text{i}-\text{A}$ $=\text{x}_\text{i}-42$	$\text{u}_\text{i}=\frac{\text{d}_\text{i}}{\text{h}_\text{i}}=\frac{\text{d}_\text{i}}{5}$	$f_iu_i$
$24.5-29.5$	$27$	$14$	$-15$	$-3$	$-42$
$29.5-34.5$	$32$	$22$	$-10$	$-2$	$-44$
$34.5-39.5$	$37$	$16$	$-5$	$-1$	$-16$
$39.5-44.5$	$A = 42$	$6$	$0$	$0$	$0$
$44.5-49.5$	$47$	$5$	$5$	$1$	$5$
$49.5-54.5$	$52$	$3$	$10$	$2$	$6$
$54.5-59.5$	$57$	$4$	$15$	$3$	$12$
		$\sum\text{f}_\text{i}=70$			$\sum\text{f}_\text{i}\text{u}_\text{i}=-79$

Let the assumed mean be $A = 42$ and $h = 5$
We know that mean, $\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}= \sum \text{f}_\text{i}=70, \sum \text{f}_\text{i} \text{u}_\text{i}=-79,\text{h}=5$ and $A = 42$
Putting the values in the above formula, we have
$\overline{\text{X}} =\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=42+5\Big(\frac{1}{70}\times(-79)\Big)$
$=42-\frac{395}{70}$
$=42-5.6438$
$=36.3571$
Hence, the mean is $36.357.$

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Question 284 Marks

$100$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	$1-4$	$4-7$	$7-10$	$10-13$	$13-16$	$16-19$
Number surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, and the modal size of the surnames.

Answer

onsider the following table.

Number of letters	No. of surnames	$x_i$	$f_ix_i$	$c.f.$
$1-4$	$6$	$2.5$	$15$	$6$
$4-7$	$30$	$5.5$	$165$	$36$
$7-10$	$40$	$8.5$	$340$	$76$
$10-13$	$16$	$11.5$	$184$	$92$
$13-16$	$4$	$14.5$	$58$	$96$
$16-19$	$4$	$17.5$	$70$	$100$
	$\text{N}=\sum\text{f}=100$		$\sum\text{f}_\text{i}\text{x}_\text{i}=832$

Here, the maximum frequency is $40$ So the modal class is $7-10.$
Therefore,
$l = 7$
$h = 3$
$f = 40$
$f_{1 }= 30$
$f_{2 }= 16$
$\Rightarrow\text{Modal}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=7+\frac{10}{34}\times3$
$=7+\frac{30}{34}$
$\text{Mode}=7.88$
Thus, the modal sizes of the surnames is $7.88$
$\text{Mean}=\frac{\sum\text{f}_i\ \text{x}_i}{\sum\text{f}}$
$=\frac{832}{100}$
$\text{Mean}=8.32$
Thus, the mean number of letters in the surnames is $8.32.$
Median
$=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=7+\frac{50-36}{40}\times3$
$=7+\frac{14}{40}\times3$
$=7+\frac{21}{20}$
$\text{Mean}=8.05$
Thus, the median number of letters in the surnames is $8.05.$

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Question 294 Marks

The mean of the following frequency distribution is $62.8$ and the sum of all the frequencies is $50.$ Compute the missing frequency.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	$5$	$f_1$	$10$	$f_2$	$7$	$8$

Answer

Class	Class Marks $(x)$	Frequency $(y)$	$fx$
$0-20$	$10$	$5$	$50$
$20-40$	$30$	$f_1$	$30f_1$
$40-60$	$50$	$10$	$500$
$60-80$	$70$	$f_2$	$70f_2$
$80-100$	$90$	$7$	$630$
$100-120$	$110$	$8$	$880$
Total		$30 + f_1 + f_2$	$2060 + 30f_{1 }+ 70f_2$

Mean $=62.8$ and $\sum\text{f}=50$
$\therefore 30+\text{f}_1+\text{f}_2=50$
$\Rightarrow\text{f}_1+\text{f}_2=50-30$
$\Rightarrow\text{f}_1+\text{f}_2=20$
$\Rightarrow\text{f}_2=20-\text{f}_1$
Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2060+30\ \text{f}_1+70\ \text{f}_2}{50}$
$\Rightarrow62.8=\frac{2060+30\ \text{f}_1+70\ \text{f}_2}{50}$
$\Rightarrow 30\text{f}_1+70\ (20-\text{f}_1)=1080$
$\Rightarrow 30\text{f}_1+1400-70\text{f}_1=1080$
$\Rightarrow -40\text{f}_1=1080-1400=-320$
$\Rightarrow \text{f}_1=\frac{-320}{-40}=8$
and $ \text{f}_2=20- \text{f}_1=20-8=12$
Hence $ \text{f}_1=8,\text{f}_2=12$
$2060+30 \text{f}_1+70\text{f}_2=50\times62.8=3140$
$\Rightarrow30 \text{f}_1+70\text{f}_2=3140-2060=1080$

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Question 304 Marks

The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000	14 56 60 86 74 62 48

Find the median life.

Answer

Life time	f	c.f.
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400
	N = 400

$\frac{\text{N}}{2}=\frac{400}{2}=200$
f = 86, C.F. = 130
l = 3000
h = 500
$\text{M}=\text{l}+\bigg(\frac{\frac{\text{N}}{2}-\text{C.F.}}{\text{f}}\bigg)\text{h}$
$=3000+\Big(\frac{200-130}{86}\Big)500$
$=3000+\Big(\frac{70}{86}\Big)500$
$=3000+\frac{35000}{86}$
$=3000+406.97=3406.97$

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Question 314 Marks

The median of the following data is $525.$ Find the missing frequency, if it is given that there are $100$ observations in the data:

Class interval	Frequency
$0-100$	$2$
$100-200$	$5$
$200-300$	$f_1$
$300-400$	$12$
$400-500$	$17$
$500-600$	$20$
$600-700$	$f_2$
$700-800$	$9$
$800-900$	$7$
$900-1000$	$4$

Answer

Median $= 525, N = 100$

Class interval	Frequency $(f)$	$c.f.$
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$f_1$	$7 + f_1$
$300-400$	$12$	$19 + f_1$
$400-500$	$17$	$36 + f_1$
$500-600$	$20$	$56 + e_1$
$600-700$	$f_2$	$56 + f_1 + f_2$
$700-800$	$9$	$65 + f_1 + f_2$
$800-900$	$7$	$72 + f_1 + f_2$
$900-1000$	$4$	$76 + f_1 +f_2$
Total	100

$\therefore 76+\text{f}_1+\text{f}_2=100$
$\Rightarrow\text{f}_1+\text{f}_2=100-76=24$
$\text{f}_2=24-\text{f}_1\ ......(1)$
$\therefore$ Median $= 525$ which lies in the alss $500-600$
$\text{l}=500,\text{F}=36+\text{f}_1,\text{f}=20,\text{h=100}$
$\therefore\text{Median}= \text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=500+\frac{\frac{100}{2}-(36+\text{f}_1)}{20}\times100$
$\Rightarrow525=500+\frac{50-36-\text{f}_1}{20}\times100$
$\Rightarrow 525=500+(14-\text{f}_1)\times5$
$\Rightarrow 25=70-5\text{f}_1$
$\Rightarrow5\text{f}_1=70-25=45$
$\Rightarrow\text{f}_1=\frac{45}{5}=9$
and $\text{f}_2=24-\text{f}_1=24-9=15$
Hence $\text{f}_1=9,\text{f}_2=15$

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Question 324 Marks

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them:

Monthly consumption $($in units$)$	$65-85$	$85-105$	$105-125$	$125-145$	$145-165$	$165-185$	$185-205$
No. of consumers	$4$	$5$	$13$	$20$	$14$	$8$	$4$

Answer

Let assumed mean $(A) = 135$ Here $\text{N}=68,\frac{\text{N}}{2}=\frac{68}{2}=34$

Monthly consumption $($in units$)$	Class Marks $(x)$	No.of consumers $(f)$	$c.f.$	$d = x - A$	$f.d.$
$65-85$	$75$	$4$	$4$	$-60$	$-240$
$85-105$	$95$	$5$	$9$	$-40$	$-200$
$105-125$	$115$	$13$	$22$	$-20$	$-260$
$125-145$	$135 - A$	$20$	$42$	$0$	$0$
$145-165$	$155$	$14$	$56$	$20$	$280$
$165-185$	$175$	$8$	$64$	$40$	$320$
$185-205$	$195$	$4$	$68$	$60$	$240$
Total		$68$			$140$

Medim: $\frac{\text{N}}{2}=34$ which lies in the class $125-145$
Here $l = 125, F = 22, f = 20, h = 20$
$\therefore\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{f}}{\text{f}}\times\text{h}$
$=125 +\frac{34-22}{22}\times20=125+12=137\text{ units}$
$\text{Mean}=\text{A}+\frac{\sum\text{f}_i\text{d}_i}{\sum\text{f}_i}=135+\frac{140}{68}$
$=135+2.05=137.05\text{ units}$
Mode; We see that the class $125 - 145$ has maximum frequency, therefore it a modal class
Here $l = 125, f = 20, f_1= 13, f_2 = 14, h = 20$
$\text{Model}=\text{I}+\frac{\text{f}_-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times \text{h}$
$=125+\frac{20-13}{20\times2-14-13}\times20$
$=125+\frac{7\times20}{40-27}=125+\frac{140}{13}$
$=125+10.76=135.76\text{ units}$

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Question 334 Marks

Find the mean, median and mode of the following data,

Classes	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$	$120-140$
Frequency	$6$	$8$	$10$	$12$	$6$	$5$	$3$

Answer

Class interval	Mid value $x$	Frequency $f$	$fx$	Cumulative frequency
$0-20$	$10$	$6$	$60$	$6$
$20-40$	$30$	$8$	$240$	$14$
$40-60$	$50$	$10$	$500$	$24$
$60-80$	$70$	$12$	$840$	$36$
$80-100$	$90$	$06$	$540$	$42$
$100-120$	$110$	$5$	$550$	$47$
$120-140$	$130$	$3$	$390$	$50$
		N = 50	$\sum\text{fx}=3120$

$\text{Mean}=\frac{\sum\text{f}\ \text{x}}{\text{N}}=\frac{320}{50}=62.4$
We have, $N = 50$
Then $\frac{\text{N}}{2}=\frac{50}{2}=25$
$\text{c},>\frac{\text{N}}{2}$ is $36$ then median class $60-80$ such that
$l = 60, h = 20, f = 12, F = 24$
$\text{Median}=\text{I}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=60+\frac{25.24}{12}\times20=60+1.67$
Modal class $ l = 60, h = 20, f = 10, f_2 = 6$
$\text{Mode}=\text{l}+\Big[\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\Big]\text{h}$
$=60+\Big[\frac{12-10}{24-10-6}\Big]20$
$=60+\frac{40}{8}=65$
$\text{Mode}=65$

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Question 344 Marks

An incomplete distribution is given below:

Variable	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$12$	$30$	$-$	$65$	$-$	$25$	$18$

You are given that the median value is $46$ and the total number of items is $230.$
$(1)$ Using the median formula fill up missing frequencies.
$(2)$ Calculate the $AM$ of the completed distribution.

Answer

$(1)$ The median formula fill up missing frequencies.

Class interval	Frequency $(f_i)$	Cumulative frequency $(c.f.)$
$10-20$	$12$	$12$
$20-30$	$30$	$42$
$30-40$	$x$	$42 + f_1$
$40-50$	$65$	$107 + f_1$
$50-60$	$y$	$107 + f_1 + f_2$
$60-70$	$25$	$132 + f_1 + f_2$
$70-80$	$18$	$150 + x + y$
	$N = 230$

Given,
Median $= 46$
The median class $= 40-50$
$l = 40, h = 50 - 40 = 10, f = 65, F = 42 + x$
Median $=\text{l}+\frac{\big(\frac{\text{N}}{2}\big)-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow46=40+\frac{115-(42+x)}{65}\times10$
$\Rightarrow46-40=\frac{115-42-\text{x}}{65}\times10$
$\Rightarrow6=\frac{73-\text{x}}{65}=10$
$\Rightarrow\frac{6\times65}{10}=73-\text{x}$
$\Rightarrow\frac{390}{10}=73-\text{x}$
$39=73-\text{x}$
$\text{x}=34$
Given, $N = 230$
$\Rightarrow12+30+\text{x}+65+\text{y}+25+18=230$
$\Rightarrow12+30+34+65+\text{y}+25+18=230$
$\Rightarrow184+\text{y}=230$
$\Rightarrow\text{y}=230-184$
$\Rightarrow\text{y}=46$
Hence, the missing frequencies are $34$ and $46.$
$(2)$ The $AM$ of the completed distribution.

Class interval	Mid value $(x)$	Frequency $(f)$	$fx$
$10-20$	$15$	$12$	$180$
$20-30$	$25$	$30$	$750$
$30-40$	$35$	$34$	$1190$
$40-50$	$45$	$65$	$2925$
$50-60$	$55$	$46$	$2530$
$60-70$	$65$	$25$	$1625$
$70-80$	$75$	$18$	$1350$
		$N = 230$	$\sum\text{fx}=10550$

Mean $=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{10550}{230}=45.87$
Hence, the mean is $45.87.$

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Question 354 Marks

Calculate the missing frequency from the following distribution, it being given that the median of the distribution

Age in years	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
No. of persons	$5$	$25$	$?$	$18$	$7$

Answer

Let the frequency of the class $20-30$ be $f_1.$ It is given that median is $35$ which lies
in the class $20-30.$ So $20-30$ is the median class.
Now, lower limit of median class $(l) = 20$
Height of the class $(h) = 10$
Frequency of median class $= (f) = f_1$
Cumulative frequency of preceding median class $(F) = 5 + 25$
Total frequency $(N) =55 + f_1$
Formula to be used to calculate median,
$=\text{l}+\bigg(\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg)\text{(h)}$
Where,
$l-$ Lower limit of median class
$h-$ Height of the class
$f-$ Frequency of median class
$F-$ Cumulative frequency of preceding median class
$N-$ Total frequency
Put the values in the above,
$=24=20+\Bigg(\frac{\frac{\text{(55}+\text{f}_1)}{2}-30}{\text{f}_1}\Bigg)(10)$
$\frac{4}{10}=\frac{55+\text{f}_1-60}{2\text{f}_1}$
$2\ \text{f}_1=50$
$\text{f}_1=25$
Hence, the required frequency is $25. $

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Question 364 Marks

If the median of the following data is $32.5,$ find the missing frequencies.

Class intertval	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	Total
Frequency	$f_1$	$5$	$9$	$12$	$f_2$	$3$	$2$	$40$

Answer

Class interval	Frequency	Cumulative frequency
$0-10$	$f_1$	$f_1$
$10-20$	$5$	$5 + f_1$
$20-30$	$9$	$14 + f_1(f)$
$30-40$	$12$	$26 + f_1$
$40-50$	$f_2$	$26 + f_1 + f_2$
$50-60$	$3$	$29 + f_1 + f_2$
$60-70$	$2$	$31 + f_1 + f_2$
	$N = 40$

Given
Median $= 32.5$
The median class $= 90-40$
$\text{l}=30 \therefore 40-32=10,$
$\text{f}=12,\text{F}= 14+\text{f}_1$
$\text{Median} = 1+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow32.5=30+\frac{20-(14+\text{f}_1)}{12}\times10$
$\Rightarrow2.5=\frac{6-\text{f}_1}{6}\times5$
$\Rightarrow15=(6-\text{f}_1)5$
$\Rightarrow3=6-\text{f}_1$
$\Rightarrow\frac{15}{5}=6-\text{f}_1$
$\Rightarrow\text{f}_1=3$
Given sum of frequencies $= 40$
$\Rightarrow3+5+9+12+\text{f}_2+3+2=40$
$\Rightarrow34+\text{f}_2=40$
$\Rightarrow\text{f}_2=6$
$\therefore \text{f}_1=3; \text{f}_2=6$

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Question 374 Marks

The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Class interval:	0-6	6-16	12-18	18-24	24+30
Frequency:	4	x	5	y	1

Answer

Class interval	Frequency	Cumulative frequency
0-6	4	4
6-12	x	4 + x
12-18	5	9 + x
18-24	y	9 + x + y
24-30	1	10 + x + y

It is given that n = 20
So, $10+\text{x}+\text{y}=20,\text{i.e.}\ \text{x}+\text{y}=10\ ......(1)$
It is also given that median = 14.4
Which lies in the class interval 12-18
So, $\text{I}=12,\text{f}=5,\text{cf}=4+\text{x},\text{h}=6$
Using the formula,
$\text{Median} = \text{l}+\bigg(\frac{\frac{\text{n}}{2}-\text{cf}}{\text{f}}\bigg)\text{h}$
We get, $14.4=12+\Big(\frac{10-(4+{\text{x})}}{5}\Big)6$
or $14.4=12+\Big(\frac{6-{\text{x}}}{5}\Big)6$
or $\text{x}=4\ ....(2)$
Now,
$10+\text{x}+\text{y}=20$
$4+\text{y}=20-10$
$\text{y}=10-4$
$\text{y} = 6.$

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Question 384 Marks

The marks in science of $80$ students of class $X$ are given below: Find the mode of the marks obtained by the students in science.

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency	$3$	$5$	$16$	$12$	$13$	$20$	$5$	$4$	$1$	$1$

Answer

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$602-70$	$70-80$	$80-90$	$90-100$
Frequency	$3$	$5$	$16$	$12$	$13$	$20$	$5$	$4$	$1$	$1$

Here, the maximum frequency is $20$ so the model class is $50-60.$
Therefore.
$i = 50$
$h = 10$
$f = 20$
$f_{1 }= 13$
$f_{2 }= 5$
Now,
$\text{Model}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=50+\frac{20-13}{40-13-5}\times10$
$=50+\frac{7}{22}\times10$
$=50+\frac{70}{22}$
$=50+3.17$
$\text{Model}=53.17$
Thus, the mode of the marks obtained by the students in science is $53.17.$

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Question 394 Marks

Find the mean of each of the following frequency distributions:

Class interval	$0-6$	$6-12$	$12-18$	$18-24$	$24-30$
Frequency	$7$	$5$	$10$	$12$	$6$

Answer

Let the assumed mean be $A = 15$ and $h = 6.$

Class inteval$:$	Mid value$(x_i):$	Frequency$(f_i):$	$d_i = x_i - A = x_i - 15$	$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i}) $$=\frac{1}{6}(\text{d}_\text{i})$	$f_iu_i$
$0-6$	$3$	$7$	$-12$	$-2$	$-14$
$6-12$	$9$	$5$	$-6$	$-1$	$-5$
$12-18$	$15$	$10$	$0$	$0$	$0$
$18-24$	$21$	$12$	$6$	$1$	$12$
$24-30$	$27$	$6$	$12$	$2$	$12$
		$\sum\text{f}_\text{i}=40$			$\sum\text{f}_\text{i}\text{u}_\text{i}=5$

We know that mean, $\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N} =\sum\text{f}_\text{i}=40,\sum\text{f}_\text{i}\text{u}_\text{i}=5,\text{h}=6$ and $A = 15$
Putting the values in the above formula, we get
$\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=15+6\Big(\frac{1}{40}\times(5)\Big)$
$=15+\frac{30}{40}$
$=15+0.75$
$=15.75$
Hence, the mean is $15.75.$

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Question 404 Marks

Class interval	$50-70$	$70-90$	$90-110$	$110-130$	$130-150$	$150-180$
Frequency	$18$	$12$	$13$	$27$	$8$	$22$

Answer

Let the assumed mean be $A = 100$ and $h = 20.$

Class interval	Mid value$(x_i):$	Frequency$(f_i)$	$d_i = x_i - A = x_i - 100$	$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})=\frac{1}{20}(\text{d}_\text{i})$	$f_iu_i$
$50-70$	$60$	$18$	$-40$	$-2$	$-36$
$70-90$	$80$	$12$	$-20$	$-1$	$-12$
$90-110$	$100$	$13$	$0$	$0$	$0$
$110-130$	$120$	$27$	$20$	$1$	$27$
$130-150$	$140$	$8$	$40$	$2$	$16$
$150-170$	$160$	$22$	$60$	$3$	$66$
		$\sum\text{f}_\text{i}=100$			$\sum\text{f}_\text{i}\text{u}_\text{i}=61$

We know that mean $\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big) $
Now we have ${\text{N}}= \sum\text{f}_\text{i}=100, \sum\text{f}_\text{i}\text{u}_\text{i},=61,=\text{h}=20$ and $A = 100$
Putting the values in the above formula, we get
$\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=100+20\Big(\frac{1}{100}\times(61)\Big)$
$=100+\frac{1220}{100}$
$=100+12.20$
$=112.20$
Hence, the mean is $112.20.$

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Question 414 Marks

Find the value of $p,$ if the mean of the following distribution is $20.$

$x$	$15$	$17$	$17$	$20 + p$	$23$
$f$	$2$	$3$	$4$	$5p$	$6$

Answer

Given:

$x_i$	$15$	$17$	$19$	$20 + p$	$23$
$f_i$	$2$	$3$	$4$	$5Pp$	$6$

Mean $= 20$ First of all prepare the frequency table in such a way that its first column consist of the values of the variate $(x_i)$ and the second column the corresponding frequencies $(f_i).$ Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing $(f_ix_i).$ Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$.

$x_i$	$f_i$	$f_ix_i$
$15$	$2$	$30$
$17$	$3$	$51$
$19$	$4$	$76$
$20 + p$	$5p$	$5p(20 + p)$
$23$	$6$	$138$
	$\sum\text{f}_\text{i}=15+5\text{p}$	$\sum\text{f}_\text{i}\text{x}_\text{i}=295+5\text{p}(20+\text{p})$

We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$20=\frac{295+5\text{p}(20+\text{p})}{15+5\text{p}}$By using cross miltiplication method,
$5\text{p}^2+100\text{p}+295=300+100\text{p}$
$\Rightarrow5\text{p}^2=300-295=5$
$\Rightarrow\text{p}^2=1$
$\Rightarrow\text{p}=1$ Hence, $\text{p}=1$

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Question 424 Marks

The frequency distribution table of agriculture holdings in a village is given below:

Area of land $($in hactares$):$	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$	$11-13$
Number of families:	$20$	$45$	$80$	$55$	$40$	$12$

Find the modal agriculture holdings of the village.

Answer

Here the maximum class frequency is $80,$
and the class corresponding to this frequency is $5-7.$
So, the modal class is $5-7.$
$l \ ($lower limit of modal class$) = 5$
$f_1 \ ($frequency of the modal class$) = 80$
$f_0 \ ($frequency of the class preceding the modal class$) = 45$
$f_2 \ ($frequency of the class succeeding the modal class$) = 55$
$h \ ($class size$) = 2$
$\text{Modal}=\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{\text{2f}_1-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=5+\Big(\frac{80-45}{2(80)-45-55}\Big)\times2$
$=5+\frac{35}{60}\times2=5+\frac{35}{30}$
$=5+1.2=6.2$
Hence, the modal agricultural holdings of the village is $6.2$ hectares

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Question 434 Marks

Compare the modal ages of two groups of students appearing for an entrance test:

Age (in years)	16-18	18-20	20-22	22-24	24-26
Group A	50	78	46	28	23
Group B	54	89	40	25	17

Answer

age (in years)	16-18	18-20	20-22	22-24	24-26
Group A	50	78	46	28	23
Group B	54	89	40	25	17

For Group A
Here the maximum frequency is 78, then the correspondingb class 18-20 is modal class
$\text{L}=18,\text{h}=20-18=2,\text{f}=48,\text{f}_1=50,\text{f}_2=46$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=18+\frac{78-50}{156-50-46}\times2$
$=18+\frac{56}{60}=18+0.93$
$=18.9\ \text{years}$
Here the maximum frequency is 89, then the corresponding class 18-20 is modal class
$\text{L}=18,\text{h}=20-18=2,\text{f}=89,\text{f}_1=54,\text{f}_2=40$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=18+\frac{78-54}{156-54-40}\times5$
$=18+0.83$
$=18.83$
Here the mode of age for the group A is higher than group B.

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Question 444 Marks

The following table shows the ages of the patients admitted in a hospital during a year :

Age $($in years$)$	$5-15$	$15-25$	$25-35$	$35-45$	$45-55$	$55-65$
No. of students	$6$	$11$	$21$	$23$	$14$	$5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer

We may observe compute class marks $(xi)$ as per the relation
$\text{x}_\text{i}=\frac{\text{Uppre class limit+ lower claas limit}}{2}$
Now taking $30$ as assumed mean $(a) $ we may calculated and $f_1d_1$ as follows

Age$($in years$)$	No. of patients $(f_1)$	Class Marks $x_1$	$d_1 = x_1 - 30$	$f_1d_1$
$5-15$	$6$	$10$	$-20$	$-120$
$15-25$	$11$	$20$	$-10$	$-110$
$25-35$	$21$	$30$	$0$	$0$
$35-45$	$23$	$40$	$10$	$230$
$45-55$	$14$	$50$	$20$	$150$
$55-65$	$5$	$60$	$30$	$150$
Total	$80$			$430$

From the table we may observe that $\sum\text{f}_\text{i}=80$
$\sum\text{f}_\text{i}\text{d}_\text{i}=430$
$\text{Mean}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=30+\big(\frac{430}{80}\big)$
$=30+5.375$
$=35.38$
Clearly mean of this data is $35.38$ It represents that on an average the age of patient
admitted to hospital was $35.58$ years.As we may observe that maximum class frequency $23$
belonging to class interval $35-45$
So, modal class $= 35-45$
Lower limit $(L)$ of modal class $= 35$
Frequency $(f_1)$ of modal class $= 23$
Class size $(h) 10$
Frequency $(f_0)$ of class Preceding the Modal $= 21$
Frequency $(f_2)$ of class succeeding the modal $= 14$
Now, $\text{ mode}=\text{L}+\Big(\frac{\text{f}-\text{f}_0} {2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=35+\Big[\frac{23-21}{2(23)-21-14}\Big]\times6$
$=35+\frac{12}{11}$
$=36.8$

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Question 454 Marks

Following is the distribution of $I.Q.$ of $100$ students. Find the median $I.Q.$

$I.Q.$	$55-64$	$65-74$	$75-84$	$85-94$	$95-104$	$105-114$	$115-124$	$125-134$	$135-144$
No. of students	$1$	$2$	$9$	$22$	$33$	$22$	$8$	$2$	$1$

Answer

Here, the frequency table is given in inclusive form. Transforming the given table into exclusive form and prepare the cumulative frequency table.

$IQ$	Frequency $(f_i)$	Cummulative Frequency $(c.f.)$
$54.5-64.5$	$1$	$1$
$64.5-74.5$	$2$	$3$
$74.5-84.5$	$9$	$12$
$84.5-94.5$	$22$	$34$
$94.5-104.5$	$33$	$67$
$104.5-114.5$	$22$	$89$
$114.5-124.5$	$8$	$97$
$124.5-134.5$	$2$	$99$
$134.5-144.5$	$1$	$100$
	$N = 100$

Here, $N = 100$
$\frac{\text{N}}{2}=50$
So,
Thus, the cumulative frequency just greater than $50$ is $67$ and
the corresponding class is $94.5-104.5.$
Therefore, $94.5-104.5$ is the median class.
Here, $l = 94.5, f = 33, F = 34$ and $h = 9$
We know that,
$\text{Median} \ =\text{l} +\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=94.5+\Big(\frac{50-34}{33}\Big)\times10$
$=94.5+\frac{160}{33}$
$=94.5+4.85$
$=99.35$
Hence, the median is $99.35.$

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Question 464 Marks

If the mean of the following frequency distribution is $18,$ find the missing frequency.

Class interval:	$11-13$	$13-15$	$15-17$	$17-19$	$19-21$	$21-23$	$23-25$
Frequency	$3$	$6$	$9$	$13$	$f$	$5$	$4$

Answer

Class interval	Mid$-$point $(x_i)$	Frequency $(f_i)$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-18}{2}$	$f_iu_i$
$11-13$	$12$	$3$	$-3$	$-9$
$13-15$	$14$	$6$	$-2$	$-12$
$15-17$	$16$	$9$	$-1$	$-9$
$17-19$	$18$	$13$	$0$	$0$
$16-21$	$20$	$f$	$1$	$f$
$21-23$	$22$	$5$	$2$	$10$
$23-25$	$24$	$4$	$3$	$12$
		$\sum\text{f}_\text{i}=40+\text{f}$		$\sum\text{f}_\text{i}\text{u}_\text{i}=\text{f}-8$

Let us take assumed mean $(a) = 18.$
Here $h = 2$
$\text{Mean}=\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{{{\sum\text{f}_\text{i}\text{u}_\text{i}}}}{{\sum\text{f}_\text{i}}}\Big)$
$= 18+2 \Big(\frac{\text{f}-8}{40+\text{f}}\Big)$
$\bar{\text{x}}=18 \ ($given$)$
So, $18=18+\frac{2(\text{f}-8)}{40+\text{f}}$
or $\text{f}=8$
Hence, the frequency of the class interval $19-21$ is $8.$

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Question 474 Marks

If the mean of the following distribution is $27,$ find the value of $p.$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$8$	$p$	$12$	$13$	$10$

Answer

Given: Mean $= 27$
Let the assumed mean $A = 25$ and $h = 10.$

Class	Mid$-$value$(x_i):$	Frequency$(f_i):$	$d_i=x_i-A = x_i - 25$	$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})$ $=\frac{1}{10}(\text{d}_\text{i})$	$f_iu_i$
$0-10$	$5$	$8$	$-20$	$-2$	$-16$
$10-20$	$15$	$p$	$-10$	$-1$	$-p$
$20-30$	$25$	$12$	$0$	$0$	$0$
$30-40$	$35$	$13$	$10$	$1$	$13$
$40-50$	$45$	$10$	$20$	$2$	$20$
		$\sum\text{f}_\text{i}=43+\text{p}$			$\sum\text{f}_\text{i}\text{u}_\text{i}=17-\text{p}$

We know that mean,$\bar{\text{X}}=\text{A}+\text{h}\bigg(\frac{1}{\text{N}}{\sum\text{f}_\text{i}\text{u}_\text{i}}\bigg)$
Now, we have ${\sum\text{f}_\text{i}}=43+\text{p}$
${\sum\text{f}_\text{i}\text{u}_\text{i}}=17-\text{p}$
$\text{h}=10$ and $\text{A}=25$
Putting the values in the above formula, we have
$27=25+10\Big(\frac{1}{43+\text{p}}\times(17-\text{p})\Big)$
$\frac{2}{10}=\Big(\frac{(17-\text{p)}}{43+\text{p}}\Big)$
$43+\text{p=85-5}\text{p}$
$6\text{p}=42$
$\text{p}=7$
Thus, the value of $p$ is $7.$

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Question 484 Marks

To find out the concentration of $S0_2$ in the air $($in parts per million, i.e., ppm$),$ the data was collected for $30$ localities in a certain city and is presented below:

Concentration of $SO_2 ($in ppm$)$	Frequency
$0.00-0.04$	$4$
$0.04-0.08$	$9$
$0.08-0.12$	$9$
$0.12-0.16$	$2$
$0.16-0.20$	$4$
$0.20-0.24$	$2$

Find the mean concentration of $SO_2$ in the air.

Answer

Let the assumed mean $A = 0.1$ and $h = 0.04.$

Concentration of $SO_2 ($in ppm$):$	Midvalue$(x_i):$	Frequency$(f_i)$	$d_i = x_i - A = x_i - 0.10$	$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})$ $=\frac{1}{0.04}(\text{d}_\text{i})$	$f_iu_i$
$0.00-0.04$	$0.02$	$4$	$-0.08$	$-2$	$-8$
$0.04-0.08$	$0.06$	$9$	$-0.04$	$-1$	$-9$
$0.08-0.12$	$0.10$	$9$	$0$	$0$	$0$
$0.12-0.16$	$0.14$	$2$	$0.04$	$1$	$2$
$0.16-0.20$	$0.18$	$4$	$0.08$	$2$	$8$
$0.20-0.24$	$0.22$	$2$	$0.12$	$3$	$6$
		$\sum\text{f}_\text{i}=30$			$\sum\text{f}_\text{i}\text{u}_\text{i}=-1$

We know that mean, $\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}\Big)$
Now, we have $\text{N}=\sum\text{f}_\text{i}=30,\sum\text{f}_\text{i}\text{u}_\text{i}=-1$
$h = 0.04$ and $A = 0.10$
Putting the values in the above formula, we have
$\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}{\sum\text{f}_\text{i}\text{u}_\text{i}}\Big)$
$=0.10+0.04\Big[\frac{1}{30}\times(-1)\Big]$
$=0.10-\frac{0.04}{30}$
$=0.10-0.001$
$=0.099$
Hence, the mean concentration of $SO_2$ in the air is $1.099$ ppm.

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Question 494 Marks

If the mean of the following data is 20.6. Find the value of p.

x	10	15	p	25	35
y	3	10	25	7	5

Answer

x	f	fx
5	6	30
10	p	10p
15	6	90
20	10	200
25	5	125
	N = P127	$\sum \text{fx} = 10\text{P} + 445$

Given
$\Rightarrow \text{Mean} = 15$
$\Rightarrow\frac{\sum\text{Px}}{\text{N}}=5$
$\Rightarrow\frac{109+445}{\text{P}+127}=15$
$\Rightarrow10\text{P}+445=15\text{P}+405$
$\Rightarrow15\text{P}-10\text{P}=445-405$
$\Rightarrow5\text{P}=40$
$\Rightarrow\text{P}=\frac{40}{5}$
$\Rightarrow\text{P}=8$

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Question 504 Marks

The given distribution shows the number of runs scored by some top batsmen of the world in one$-$day international cricket matches.

Runs scored	Number of batsman	Runs scored	Number of batsman
$3000-4000$ $4000-5000$ $5000-6000$ $6000-7000$	$4$ $18$ $9$ $7$	$7000-8000$ $8000-9000$ $9000-10000$ $10000-11000$	$6$ $3$ $1$ $1$

Find the mode of the data.

Answer

The given data is shown below.

Runs scored	Number of batsmen
$3000-4000$	$4$
$4000-5000$	$18$
$5000-6000$	$9$
$6000-7000$	$7$
$7000-8000$	$6$
$8000-9000$	$3$
$9000-10000$	$1$
$10000-11000$	$1$

Here, the maximum frequency is $18$ so the modal class is $4000-5000.$
Therefore
$l = 4000$
$h = 1000$
$f_1 = 18$
$f_0 = 4$
$f_2 = 9$
$\therefore\text{Modal}=\text{l}+\frac{\text{f}_1-\text{f}_0}{2\text{f}_1-\text{f}_0-\text{f}_2}\times\text{h}$
$=4000+\frac{18-4}{2\times18-4-9}\times1000$
$=4000+\frac{14}{23}\times1000$
$=4000+608.7$
$=4608.7$
Thus, the mode of the data$($or runs scored$)$ is $4608.7.$

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4 Marks Questions - Maths STD 10 Questions - Vidyadip