3 Marks Question

Question 13 Marks

If both $( x -2)$ and $\left(x-\frac{1}{2}\right)$ are factors of $p x^2+5 x+r$, Show that $p = r$.

Answer

Suppose, $p(x)=p x^2+5 x+r$
As $(x-2)$ is a factor of $p(x)$
$\therefore p(2)=0$
$\Rightarrow p(2)^2+5(2)+r=0$
$\Rightarrow 4 p+10+r=0 \ldots(1)$
Again, $\left(x-\frac{1}{2}\right)$ is factor of $p(x)$.
$\therefore p\left(\frac{1}{2}\right)=0$
Now, $p\left(\frac{1}{2}\right)=p\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+r$
$=\frac{1}{4} p+\frac{5}{2}+r$
$\therefore p\left(\frac{1}{2}\right)=0$
$\Rightarrow \frac{1}{4} p+\frac{5}{2}+r=0 \ldots(2)$
From equation $(1),$ we have $4 p+r=-10$
From equation $(2),$ we have $p+10+4 r=0$
$\Rightarrow p+4 r=-10$
$\therefore 4 p+r=p+4 r[\because \text { Each }=-10]$
$\therefore 3 p=3 r$
$\Rightarrow p=r$
Hence, proved.

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Question 23 Marks

Read the bar graph given in Figure and answer the following questions:

i. What information is given by the bar graph?
ii. In which years the areas under the sugarcane crop were the maximum and the minimum?
iii. State whether true or false:
The area under the sugarcane crop in the year 1982-83 is three times that of the year 1950-51.

Answer

i. It gives the information about the areas (in lakh hectors) under sugarcane crop during different years in India.
ii. The areas under the sugarcane crops were the maximum and minimum in 1982-83 and 1950-51 respectively.
iii. The area under sugarcane crop in the year 1982-83= 34 lakh hectares.
The area under sugarcane crop in the year 1950-51= 17 lakh hectares.
Clearly, the area under sugarcane crop in the year 1982-83 is not 3 times that of the year 1950-51 So, the given statement is false.

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Question 33 Marks

The marks scored by 750 students in an examination are given in the form of a frequency distribution table.

Marks:	600-640	640-680	680-720	720-760	760-800	800-840	840-880
No. of Students:	16	45	156	284	172	59	18

Represent this data in the form of a histogram and construct a frequency polygon.

Answer

In Figure, a histogram and a frequency table of the above frequency distribution are drawn on the same scale.

To construct a frequency polygon without using the histogram of a given frequency distribution, we use the following algorithm.
STEP-I: Obtain the frequency distribution.
STEP-II: Compute the mid-points of class intervals i.e. class marks.
STEP-III: Represent class marks on X-axis on a suitable scale.
STEP-IV: Represent frequencies on Y-axis on a suitable scale.
STEP-V: Plot the points, where x denotes class mark and f corresponding frequency.
STEP-VI: Join the points plotted in step V by line segments.
STEP-VII: Take two class intervals of zero frequency, one at the beginning and the other at the end. Obtain their mid-points. These classes are known as imagined classes.
STEP-VIII: Complete the frequency polygon by joining the mid-points of first and last class intervals to the mid-points of the imagined classes adjacent to them.

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Question 43 Marks

Find the solution of the linear equation $x + 2y = 8$ which represents a point on
$i.$ The $x-$axis
$ii.$ The $y-$axis

Answer

$i.$ On $x-$axis $y = 0$
$\Rightarrow x+2 \times 0=8$
$\Rightarrow x=8$
Therefore, the required point is $(8, 0).$
$ii.$ On $y-$axis $x = 0$
$\Rightarrow 0+2 y=8$
$\Rightarrow y=\frac{8}{2}$
$\Rightarrow y=4$
Thus, the required point is $(0, 4).$

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Question 53 Marks

In Fig. X and Y are respectively the mid-points of the opposite sides AD and BC of a parallelogram ABCD. Also, BX and DY intersect AC at P and Q, respectively. Show that AP = PQ = QC.

Answer

$AD = BC$ (Opposite sides of a parallelogram)
Therefore, $DX = BY \left(\frac{1}{2} AD =\frac{1}{2} B C\right)$
Also, $DX \| BY ( As AD \| BC )$
So, XBYD is a parallelogram (A pair of opposite sides equal and parallel) i.e., PX || QD
Therefore, $A P=P Q$ (From $\triangle A Q D$ where $X$ is mid-point of $A D\ldots(1)$
Similarly, from $\triangle C P B, C Q=P Q\ldots(2)$
Thus, AP = PQ = CQ [From (1) and (2)]

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Question 63 Marks

A random survey of the number of children of various age groups playing in a park was found as follows :

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.

Answer

Age (in years)	Number of children(frequency)	Width of the class	Length of the rectangle
1-2	5	1	$\frac{5}{1} \times 1=5$
2-3	3	1	$\frac{3}{1} \times 1=5$
3-5	6	2	$\frac{6}{2} \times 1=3$
5-7	12	2	$\frac{12}{2} \times 1=6$
7-10	9	3	$\frac{9}{3} \times 1=3$
10-15	10	5	$\frac{10}{5} \times 1=2$
15-17	4	2	$\frac{4}{2} \times 1=2$

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Question 73 Marks

Locate $\sqrt{10}$ on the number line.

Answer

We can write $10$ as
$10=9+1=3^2+1^2$
Draw $OA =3$ units, on the number line
Draw $BA =1$ unit, perpendicular to $OA .$
Join $OB$
Figure:

Now, by Pythagoras theorem,
$\ce{OB^2=AB^2 + OA^2}$
$OB^2=1^2+3^2=10$
$\Rightarrow OB=\sqrt{10}$
Taking $O$ as centre and $OB$ as a radius, draw an arc which intersects the number line at point $C.$
Clearly, $OC$ corresponds to $\sqrt{10}$ on the number line.

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