3 Marks Question

Question 13 Marks

Solve algebraically the following pair of linear equations for $x$ and $y\ 31 x+29 y=33 , 29 x+31 y=27$

Answer

$31 x+29 y=33 ----------(1)$
$29 x+31 y=27----------(2)$
Multiply $(1) $ by $29$ and $(2)$ by $31 ($ Since $29,31$ are primes and $Lcm$ is $29 \times 31 )$
$(1)$ becomes $31 x \times 29+29 \times 29 y=33 \times 29 ........(3)$
$(2)$ becomes $29 x \times 31+31 \times 31 y=27 \times 31 ......(4)$
Subtracting $(3)$ from $(4),$
$(312-292) y=27 \times 31-33 \times 29=-120$
$(31-29)(31+29) y=-120$
$120 y=-120$
$y=-1$
Substituting in $(1),$
$31 x-29=33$
$31 x=62$
Hence,
$x=2 \text { and } y=-1$

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Question 23 Marks

$PQ$ is a chord of length $4.8 \ cm$ of a circle of radius $3 \ cm$. The tangents at $P$ and $Q$ intersect at a point $T$ as shown in the figure. Find the length of $TP$.

Answer

Given, Chord $PQ =4.8 \ cm$, radius of circle $=3 \ cm$
Also, The tangents at $P$ and $Q$ intersect at a point $T$ as shown in the figure.
We have to find the length of $TP$.

Let $TR = y$ and $TP = x$
We know that the perpendicular drawn from the center to the chord bisects It.
Hence, $PR = RQ \ldots ... (1)$
Since, $P Q=4.8$
or, $PR + RQ =4.8$
or, $PR + PR =4.8\ [$ from $(1)]$
So, $PR =2.4$
Now, in right triangle $\text{POR},$
Using Pythagoras theorem, we have
$PO^2=OR^2+PR^2$
$\Rightarrow 3^2=O R^2+(2.4)^2$
$\Rightarrow O R^2=3.24$
$\Rightarrow O R=1.8$
Now, in right triangle $\text{TPR},$
By Using Pythagoras theorem, we have
$T P^2=T R^2+P R^2$
$\Rightarrow x^2=y^2+(2.4)^2$
$\Rightarrow x^2=y^2+5.76 ..........(2)$
Again, In right triangle $\text{TPO}$
By Using Pythagoras theorem, we have,
$T O^2=T P^2+P O^2$
$\Rightarrow(y+1.8)^2=x^2+3^2$
$\Rightarrow y^2+3.6 y+3.24=x^2+9$
$\Rightarrow y^2+3.6 y=x^2+5.76 .......(3)$
Solving $(2)$ and $(3),$ we get
$x=4 \ cm$ and $y =3.2 \ cm$
$\therefore TP=4 \ cm$

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Question 33 Marks

Compute the median for the following cumulative frequency distribution:

Less than 20	Less than 30	Less than 40	Less than 50	Less than 60	Less than 70	Less than 80	Less than 90	Less than 100
0	4	16	30	46	66	82	92	100

Answer

We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median

Class intervals	Frequency (f)	Cumulative frequency (c.f.)
20-30	4	4
30-40	12	16
40-50	14	30
50-60	16	46
60-70	20	66
70-80	16	82
80-90	10	92
90-100	8	100
		$N =\Sigma f_i=100$

Here, $N =\Sigma f_i=100 \therefore \frac{N}{2}=50$
We observe that the cumulative frequency just greater than $\frac{N}{2}=50$ is 66 and the corresponding class is 60-70.
So, $60-70$ is the median class.
$\therefore l=60, f=20, F=46 \text { and } h=10$
Now, Median $=1+\frac{\frac{N}{2}-F}{f} \times h$
$\Rightarrow \text { Median }=60+\frac{50-46}{20} \times 10=62$

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Question 43 Marks

Prove that: $\frac{1}{\operatorname{cosec} A-\cot A}-\frac{1}{\sin A}=\frac{1}{\sin A}-\frac{1}{\operatorname{cosec} A+\cot A}$.

Answer

We have,
$\Rightarrow \frac{1}{\operatorname{cosec} A-\cot A}-\frac{1}{\sin A}=\frac{1}{\sin A}-\frac{1}{\operatorname{cosec} A+\cot A}$
$\Rightarrow \frac{1}{\operatorname{cosec} A-\cot A}+\frac{1}{\operatorname{cosec} A+\cot A}=\frac{1}{\sin A}+\frac{1}{\sin A}$
$\Rightarrow \frac{1}{\operatorname{cosec} A-\cot A}+\frac{1}{\operatorname{cosec} A+\cot A}=\frac{2}{\sin A}$
$\text { LHS }=\frac{1}{\operatorname{cosec} A-\cot A}+\frac{1}{\operatorname{cosec} A+\cot A}$
$\Rightarrow \frac{\operatorname{cosec} A+\cot A+\operatorname{cosec} A-\cot A}{(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)}$
$\Rightarrow \frac{2 \operatorname{cosec} A}{\operatorname{cosec}^2 A-\cot ^2 A}$
$\Rightarrow \frac{\frac{2}{\sin A}}{1}=\frac{2}{\sin A}=\text { RHS. }$
Hence Proved.

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Question 53 Marks

In a given figure $, AB$ is a chord of length $8 \ cm$ of a circle of radius $5 \ cm$. The tangents to the circle at $A$ and $B$ intersect at $P$. Find the length of $AP$.

Answer

According to the question, radius of circle is $=5 \ cm$.
Also $, AB =8 \ cm$
Now $, AM =\frac{A B}{2}=4 \ cm$

$\text { In } \triangle OMA \text {, }$
By using pythagoras theorem, we get
$ OA ^2= OM ^2+ AM ^2$
$\therefore OM =\sqrt{5^2-4^2}=3 \ cm$
Let $ AP = y \ cm , PM = x \ cm $
In $ \triangle OAP \text {, }$
By using pythagoras theorem, we get
$OP^2=OA^2=AP^2$
$(x+3)^2=y^2+25$
$\Rightarrow x^2+9+6 x=y^2+25 ......(i)$
In $\triangle AMP$,
By using pythagoras theorem, we get
$x^2+4^2=y^2.........(ii)$
Substituting eq.$(ii)$ in eq.$(i),$ we get
$\Rightarrow x^2+6 x+9$
$=x^2+16+25$
$\Rightarrow 6 x=32$
$\Rightarrow x=\frac{32}{6}$
$=\frac{16}{3} cm$
Now, $y ^2= x ^2+16$
$=\frac{256}{9}+16$
$=\frac{400}{9}$
$\Rightarrow y =\frac{20}{3} \ cm$.
Therefore, length of $AP = y =\frac{20}{3} \ cm$.

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Question 63 Marks

Half of the difference between two numbers is $2$ . The sum of the greater number and twice the smaller number is $13$. Find the numbers.

Answer

Let the greater numbers be $'x\ '$ and the smaller number $'y\ '.$
It is said that half the difference between the two numbers is $2$ .
So, we can write it as;
$\frac{1}{2} \times(x-y)=2$
$\Rightarrow(x-y)=4 \ldots$
It is also said that the sum of the greater number and twice the smaller number is $13$ .
So, we can write it as; $x+2 y=13 \ldots(2)$
Subtracting $(2)$ from $(1),$ we get;
$(x-y)-(x+2 y)=4-13$
$\Rightarrow x-y-x-2 y=-9$
$\Rightarrow-3 y=-9$
$\Rightarrow y=\frac{9}{3}$
$\Rightarrow y=3$
Putting $y=3$ in $(1), $ we get;
$x-y=4$
$\Rightarrow x-3=4$
$\Rightarrow x=4+3$
$\Rightarrow x=7$
Hence, the greater number is $7$ .
The smaller number is $3$ .

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Question 73 Marks

Find the zeroes of the quadratic polynomial $6 x^2-3-7 x$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Answer

The given polynomial is
$p(x)=6 x^2-7 x-3$
Factorize the above quadratic polynomial, we have
$6 x^2-7 x-3$
$=6 x^2-9 x+2 x-3$
$=3 x(2 x-3)+1(2 x-3)$
$=(3 x+1)(2 x-3)$
For $p ( x )=0$, either $3 x +1=0$ or $2 x -3=0$
$\Rightarrow x=\frac{-1}{3} $ or $ x=\frac{3}{2}$
Verification : we have $a =6, b=-7, c =-3$
Sum of zeroes $=\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}$
Also, $\frac{-b}{a}=\frac{-(-7)}{6}=\frac{7}{6}$
Now, product of zeroes $=\left(-\frac{1}{3}\right) \times \frac{3}{2}=\frac{-1}{2}$
Also, $\frac{c}{a}=\frac{-3}{6}=\frac{-1}{2}$

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Question 83 Marks

For a morning walk, three persons steps off together. The measure of their steps is $80,85$ and $90 \ cm$ respectively.What is the minimum distance each should walk so that all can cover the same distance in complete steps?
Which value is preferred in this situation?

Answer

Since, the three persons start walking together.
$\therefore$ The minimum distance each should walk so that all can cover the same distance in complete steps will be equal to $\text{LCM}$ of $80,85$ and $90$ .
Prime factors of $80,85$ and are as following
$80=16 \times 5=2^4 \times 5$
$85=5 \times 17$
$90=2 \times 9 \times 5=2 \times 3^2 \times 5$
So $\text{LCM}$ of $80,85$ and $90$
$=2^4 \times 3^2 \times 5 \times 17$
$=16 \times 9 \times 5 \times 17=12240$
$\therefore$ Each person should walk the minimum distance
$=12240 \ cm=122 $ meter $ 40 \ cm$
Value of morning walk :
"An early morning walk is a blessing for the whole day."

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