HW8

1. Given a Venn-diagram.
   1. Define improper subset.
   2. Write the improper subset of set A.
   3. If x and y are only the members of set B, then what type of sets are A and B. Write with reason.
1. An improper subset of a set is the set itself.
2. The improper subset of set A is A itself.
   \(A=\{m,n,x,y\}\)
3. If x and y are only the members of set B, then
   A and B are disjoint set, because there will be no common elements
2. 1. Convert 216 into binary number system.
   2. If \(\blacksquare\) and \(\square\) denote 1 and 0 respectively, which of the following rectangle should be shaded to denote 35 in binary number system? Show by shading.
      \(\square \quad \square \quad \square \quad \square \quad \square \quad \square\)
      \(a \qquad b \quad c \quad d \quad e \qquad f\)
   3. Identify rational and irrational number between 3 and \(\sqrt{10}\). Also compare these two real numbers.
1. The solution is

   2	216		0↑
   2	108		0↑
   2	54		0↑
   2	27		1↑
   2	13		1↑
   2	6		0↑
   2	3		1↑
   	1		1↑

   Therefore
   \( 216 = 11011000_2 \)
2. The conversion of 35 in binary system is

   2	35		1↑
   2	17		1↑
   2	8		0↑
   2	4		0↑
   2	2		0↑
   	1		1↑

   Therefore
   \( 35 = 100011_2 \)
   Now, the shading is
   \(\blacksquare \quad \square \quad \square \quad \square \quad \blacksquare \quad \blacksquare\)
   \(a \qquad b \quad c \quad d \quad e \qquad f\)
3. The solution is
   3 is rational number.
   \(\sqrt{10}=3.16227766 \cdots\) is irrational number.
   The comparision between 3 and \(\sqrt{10}\) is
   \(3< \sqrt{10}\)
3. If the numerator is divided by denominator the quotient is \(0.454545 \cdots\), Answer the following questions.
   1. Write the quotient in the symbol of recurring decimal.
   2. Convert the quotient into fraction?
   3. If the fraction is multiplied by 1100 then write the product in scientific notation.
1. The recurring decimal \(0.454545\ldots\) has the digits "45" repeating. It is written as:
   \(0.\overline{45}\)
2. Converting \(0.\overline{45}\) into a fraction
   Let
   \(x = 0.\overline{45}\)
   \(x = 0.454545\ldots\)(1)
   Multiply both sides by 100 (since two digits repeat)
   \(100x = 45.454545\ldots\)(2)
   Subtract (1) from (2), we get
   \(100x - x = 45.4545\ldots - 0.4545\ldots\)
   or\(99x = 45\)
   or\(x = \dfrac{45}{99} = \dfrac{5}{11}\)
   So, the fraction is \(\dfrac{5}{11}\).
3. Multiplying the fraction by 1100 and writing in scientific notation:
   \(\dfrac{5}{11} \times 1100 = 5 \times 100 = 500\)
   In scientific notation:
   \(500 = 5 \times 10^2\)
   Hence, the product in scientific notation is \(5 \times 10^2\).
4. The ratio of the number of boys and girls in a school is 2:3.
   1. Who are smaller in number, boys or girls? Give reason.
   2. If 360 are girls, find the number of boys.
   3. If 9 workers can complete a piece of work in 20 days, how many workers should be added to complete the same work in 15 days?
1. Boys are smaller in number.
   Because, in the ratio 2:3, the part for boys (2) is less than the part for girls (3).
2. Given ratio of boys : girls = 2 : 3
   Let number of boys = 2x, number of girls = 3x
   Given that
   girls = 360
   or3x = 360
   orx = 360 ÷ 3 = 120
   Thus, number of boys is
   boys = 2x = 2 × 120 = 240
3. Given that
   

   Workers \(\uparrow\)	Days \(\downarrow\)
   9	20
   9 + x	15

   We know that "number of workers" and "number of days" have an indirect relation, therefore, using indirect proportion, we get
   \(\dfrac{9 + x}{9} = \dfrac{20}{15}\)
   
   or\(9 + x = 9 \times \dfrac{4}{3} \)
   
   or\(9 + x = 12\)
   or\(x = 12 - 9 \)
   or\(x = 3 \)
   So, 3 workers should be added to complete the same work in 15 days
5. Ms. Nehita has a cuboidal box of 12 cm, 10 cm high and its volume is 1800 cm3.
   1. Find the length of the box.
   2. Find the cost of painting the box at Rs 5 per sq.cm.
   3. If she has an another cubical box of 1cm3, then find 2000 number of such cubical boxes can easily fit in cuboidal box or not.



1. Given that
   Volume of cuboid = 1800 cm³
   breadth (b) = 12 cm
   Height (h) = 10 cm
   Now, volume is
   volume = length × breadth × height
   or\(18000=l \times 12 \times 10\)
   or\( l = \dfrac{1800}{120} = 15 \)
   Therefore
   Length of the box = 15 cm
2. Total surface area of cuboid is
   \(A= 2(lb + bh + hl) \)
   or\( A=2(15 \times 12 + 12 \times 10 + 10 \times 15) \)
   or\(A= 900 \text{ cm}^2 \)
   Now
   Cost of painting is
   Cost= A x rate
   orCost=900 × Rs 5 = Rs 4,500
3. 2000 cubical boxes of \(1cm^3\) cannot fit in the cuboidal box, because the volume of cuboidal box is \(1800cm^3\)
6. Two algebraic expressions are, \(x^2−7x+10\) and \(x^2−8x+15\).
   1. Factorize the first expression.
   2. Find the HCF of the two expressions.
   3. Also, calculate their LCM.
   4. For what values of x, the expression \(x^2−7x+10\) equals to zero?
1. Factorize \(x^2 - 7x + 10\)
   We look for two numbers that multiply to 10 and add to –7, which are –2 and –5, thus
   \(x^2 - 7x + 10\)
   or\((x - 2)(x - 5)\)
2. Factorize the second expression \(x^2 - 8x + 15\)
   We look for two numbers that multiply to 15 and add to –8, which are –3 and –5, thus
   \(x^2 - 8x + 15\)
   or\((x - 3)(x - 5)\)
   Now, HCF is
   HCF = \(x - 5\)
3. Next, the LCM is
   LCM = HCF x remaining factors
   orLCM=\((x - 2)(x - 3)(x - 5)\)
4. Given expression is
   \(x^2 - 7x + 10=(x - 2)(x - 5)\)
   So, for \(x=2\) or \(x=5\), the expression \(x^2−7x+10\) equals to zero.
7. 1. Simplify: \(\dfrac{6^{n+2} - 6^n}{6^{n+1} + 6^n}\)
   2. Simplify: \(\left(\dfrac{x^m}{x^{-n}}\right)^{m-n} \times \left(\dfrac{x^n}{x^{-p}}\right)^{n-p} \div \left(\dfrac{x^m}{x^p}\right)^{m+p}\)
1. Simplify
   \(\dfrac{6^{n+2} - 6^n}{6^{n+1} + 6^n}\)
   or\(\dfrac{6^n 6^2 - 6^n}{6^n6^1 + 6^n}\)
   or\(\dfrac{6^n(6^2 - 1)}{6^n(6^1 + 1)}\)
   or\(\dfrac{36 - 1}{6 + 1}\)
   or\(\dfrac{35}{7} = 5\)
2. Simplify
   \(\left(\dfrac{x^m}{x^{-n}}\right)^{m-n} \times \left(\dfrac{x^n}{x^{-p}}\right)^{n-p} \div \left(\dfrac{x^m}{x^p}\right)^{m+p}\)
   or\(\left(x^{m+n}\right )^{m-n} \times \left(x^{n+p}\right)^{n-p} \div \left(x^{m-p}\right)^{m+p}\)
   or\(x^{m^2-n^2} \times x^{n^2-p^2} \div x^{m^2-p^2}\)
   or\(x^{m^2-n^2} \times x^{n^2-p^2} \times x^{p^2-m^2}\)
   or\(x^{m^2-n^2+n^2-p^2+m^2-p^2}\)
   or\(x^0=1\)
   
8. From the given figure, find the area of shaded region. If \(x = 5\) cm and \(y = 3\) cm, then find the actual area of shaded region.



From the figure, area of shaded region is
A=Area of outer square – Area of inner square
or\(A=(4x)(4x) - (3y)(3y)\)
or\(A=16x^2 - 9y^2\)
Given: \(x = 5\) cm, \(y = 3\) cm, thus
\(A=16(5)^2 - 9(3)^2 \)
or\(A=16 \times 25 - 9 \times 9\)
or\(A=400 - 81 = 319\) cm²
9. In the given figure, \(AB || EF || CD\). If \(\measuredangle ABG = 110^0\) and \(\measuredangle DCG = 140^o\), then answer the following questions.
   
   
   
   1. Define transversal line. Identify transversal line from a given figure.
   2. Write an angle co-interior to \(\measuredangle ABG\)
   3. Find the value of the angle \(a\) formed at point G.
   4. If DC is produced where DC and BG intersect at H where \(AB||HD\), then what will be the measure of \(\measuredangle BHC\)?
1. A transversal line is a line that intersects two or more parallel lines.
   In the given figure, line BG and line CG are transversals lines. Line BG is transversal line because it is intersecting the parallel lines \(AB || EF\).
   Also, line CG is transversal line because it is intersecting the parallel lines \(CD || EF\)
2. Co-interior angle to \(\measuredangle ABG\) is \(\measuredangle BGE\) (on the same side of transversal BG.
3. To find angle \(a = \measuredangle BGC\), we do
   b+110=180
   orb=70°
   Again
   c+ 140° = 180°
   orc=40°
   Now
   a+b+c= 180°
   ora+70+40=180°
   ora+110=180°
   ora=70°
   
4. If DC is produced to meet BG at H, and \(AB || HD\), then:
   
   Here
   \(\measuredangle BHC = 110°\)
10. In the given graph.
    1. Write the co-ordinates of the object.
    2. Also, Reflect ΔABC in x-axis. Plot the image on graph paper.
1. The co-ordinates of the object are
   \(A=(2,4)\)
   \(B=(-2,-2)\)
   \(C=(1,-4)\)
   
2. Also, Reflect ΔABC in x-axis. Plot the image on graph paper. The reflection of ΔABC in x-axis is as follows.
   \(A'=(2,-4)\)
   \(B'=(-2,2)\)
   \(C'=(1,4)\)
   The object and image of ΔABC are as below.