5 Marks Questions

Question 15 Marks

The following table shows the average daily earnings of 40 general stores in a market, during a certain week:

Daily earning (in rupees)	700-750	750-800	800-850	850-900	900-950	950-1000
Number of stores	6	9	2	7	11	5

Draw a histogram to represent the above data.

Answer

The following table shows the average daily earnings of 40 general stores in a market, during a certain week:

Daily earning (in ₹)	700-750	750-800	800-850	850-900	900-950	950-1000
Number of stores	6	9	2	7	11	5

Clearly, the given frequency distribution is in the exclusive form. We take class intervals, i.e. daily earnings (in ₹) along x-axis and frequencies i.e. number of stores along y-axis. So , we get the required histogram.
Since the scale on X-axis starts at 700, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 700.

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Question 25 Marks

In each of the figures given below, $AB \| CD$. Find the value of $x^{\circ}$ in each case.

Answer

Draw $E F\|A B\| C D$
Now, $AB \| EF$ and $BE$ is the transversal.
Then,
$\angle A B E=\angle B E F [$Alternate Interior Angles$]$
$\Rightarrow \angle B E F=35^{\circ}$
Again, $EF \| CD$ and $DE$ is the transversal
then,
$\angle D E F=\angle F E D$
$\Rightarrow \angle F E D=65^{\circ}$
$\therefore x^{\circ}=\angle B E F+\angle F E D$
$x^{\circ}=35^{\circ}+65^{\circ}$
$x ^{\circ}=100^{\circ}$

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Question 35 Marks

In the given figure, $AB \| CD \| EF , \angle D B G=x, \angle E D H=y, \angle A E B=z, \angle E A B=90^{\circ}$ and $\angle B E F=65^{\circ}$. Find the values of $x , y$ and z .

Answer

$EF \| CD$ and $ED$ is the transversal.
$\therefore \angle F E D+\angle E D H=180^{\circ}$ [co-interior angles]
$\Rightarrow 65^{\circ}+y=180^{\circ}$
$\Rightarrow y=\left(180^{\circ}-65^{\circ}\right)=115^{\circ}$
Now $C H \| A G$ and $D B$ is the transversal
$\therefore x = y =115^{\circ} [$corresponding angles$]$
Now, ABG is a straight line.
$\therefore \angle A B E+\angle E B G=180^{\circ}[$ sum of linear pair of angles is $180^{\circ}]$
$\Rightarrow \angle A B E+x=180^{\circ}$
$\Rightarrow \angle A B E+115^{\circ}=180^{\circ}$
$\Rightarrow \angle A B E=\left(180^{\circ}-115^{\circ}\right)=65^{\circ}$
We know that the sum of the angles of a triangle is $180^{\circ}$.
From $\triangle E A B$, we get
$\angle E A B+\angle A B E+\angle B E A=180^{\circ}$
$\Rightarrow 90^{\circ}+65^{\circ}+ z =180^{\circ}$
$\Rightarrow z =\left(180^{\circ}-155^{\circ}\right)=25^{\circ}$
$\therefore x =115^{\circ}, y =115^{\circ}$ and $z =25^{\circ}$

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Question 45 Marks

i. AB = BC, M is the mid-point of AB and N is the mid-point of BC. Show that AM = NC.
ii. BM = BN, M is the mid-point of AB and N is the mid-point of BC. Show that AB = BC.

Answer

i.From the above figure, We have AB = BC…(1) [Given]
Now, A, M, B are the three points on a line, and M lies between A and B such that M is the mid point of AB [Given], then AM + MB = AB …(2) Also B, N, C are three points on a line such that N is the mid point of BC [Given]
Similarly, BN + NC = BC $\ldots$ (3)
So, we get AM + MB = BN + NC
From (1), (2), (3) and Euclid’s first axiom
Since M is the mid-point of AB and N is the mid-point of BC, therefore
2AM = 2NC i.e. AM = NC
Hence, AM = NC. Proved
Using axiom 6, things which are double of the same thing are equal to one another.
ii. From the above figure, We have BM = BN $\ldots$(1) [ Given ]
As M is the mid-point of AB [Given] , so that
BM = AM $\ldots$(2)
And N is the mid-point of BC [Given]
BN = NC $\ldots$(3)
From (1), (2) and (3) and Euclid’s first axiom, we get
AM = NC $\ldots$(4)
Adding (4) and (1), we get
AM + BM = NC + BN
Hence, AB = BC Proved
[By axiom 2 if equals are added to equals, the wholes are equal]

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Question 55 Marks

If x is a positive real number and exponents are rational numbers, simplify $\left(\frac{x^b}{x^c}\right)^{b+c-a} \cdot\left(\frac{x^c}{x^a}\right)^{c+a-b} \cdot\left(\frac{x^a}{x^b}\right)^{a+b-c}$.

Answer

Given. $\left(\frac{x^b}{x^c}\right)^{b+c-a} \cdot\left(\frac{x^c}{x^a}\right)^{c+a-b} \cdot\left(\frac{x^a}{x^b}\right)^{a+b-c}$
$=\left(\frac{x^{b^2+b c-a b}}{x^{b c+c^2-a c}}\right) \cdot\left(\frac{x^{c^2+a c-b c}}{x^{a c+a^2-a b}}\right) \cdot\left(\frac{x^{a^2+a b-a c}}{x^{a b+b^2-b c}}\right)$
$=\left(x^{b^2+b c-a b-b c-c^2+a c}\right)\left(x^{c^2+a c-b c-a c-a^2+a b}\right)\left(x^{a^2+a b-a c-a b-b^2+b c}\right)$
$=\left(x^{b^2-a b-c^2+a c}\right)\left(x^{c^2-b c-a^2+a b}\right)\left(x^{a^2-a c-b^2+b c}\right)$
$=x^{b^2-a b-c^2+a c+c^2-b c-a^2+a b+a^2-a c-b^2+b c}$
$=x^0$
$=1$

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Question 65 Marks

If $a=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $b=\frac{\sqrt{2}-1}{\sqrt{2}+1}$, then find the value of $a ^2+ b ^2-4 ab$.

Answer

Given, $a=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $b=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
Here, $a=\frac{\sqrt{2}+1}{\sqrt{2}-1}$
$=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$
$=\frac{(\sqrt{2}+1)^2}{(\sqrt{2})^2-1^2}$
$=\frac{(\sqrt{2})^2+1+2 \sqrt{2}}{2-1}$
$=\frac{2+1+2 \sqrt{2}}{1}$
$=3+2 \sqrt{2}$
$\therefore a=3+2 \sqrt{2} \ldots( i )$
$b=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$=\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}$
$=\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-1}$
$=\frac{(\sqrt{2})^2+1^2-2 \sqrt{2}}{2-1}$
$=\frac{2+1-2 \sqrt{2}}{1}$
$=3-2 \sqrt{2}$
$\therefore b=3-2 \sqrt{2} \ldots \text { (ii) }$
From equation $(i)$ and $(ii)$
$a+b=3+2 \sqrt{2}+3-2 \sqrt{2}=6$
$a b=(3+2 \sqrt{2})(3-2 \sqrt{2})=3^2-(2 \sqrt{2})^2$
$=9-4 \times 2=9-8=1$
$\therefore a^2+b^2-4 a b$
$=a^2+b^2+2 a b-6 a b$
$=(a+b)^2-6 a b$
$=6^2-6$
$=36-6$
$=30$

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