M.L Aggarwal Solutions

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ML AGGARWAL CLASS 8 CHAPTER 7 Percentage Exercise 7.3

     Exercise 7.3 Question 1 Sol : (i) Marked price = 575, discount =12 % Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$ $\begin{array}{l}12=\frac{\text { Discont }}{575} \times 100 \\ \text { Discount }=\frac{12 \times 575}{100}\end{array}$ Discount =₹69 Selling price = Marked price - Discount =575-69 =₹506 (ii)  Printed price = 12,750, discount $=8 \frac{1}{3} \%$ Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$ $\frac{25}{3}=\frac{D \text { discount }}{12,750} \times 100$ Discount $=\frac{25}{3} \times \frac{12,750}{1004}=1062.5$ Selling price = Marked price or printed price - discount =12,750-1062.5 =11,687.5 Question 2 Sol : (i) Marked price=780,  Selling price =2721.5 Selling price = Marked price - Discount Discount = Marked price - Selling price =780-721.5 =58.5 Discount percentage $=\frac{\text { Discount }}{\text { Marked price }} \times 100 \%$ $=\fr...
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ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

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 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
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ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.1

 Exercise 3.1 Question 1 (i) (2x+7y) 2 It is in the form of (a+b) 2 =a 2 +2ab+b 2 ∴a=2x, b=7y ∴(2x+7y) 2 =(2x) 2 +2.2x.7y+(7y) 2 =4x 2 +28xy+49y 2 (ii) $\left(\frac{1}{2}x+\frac{2}{3}y\right)^2$ Sol : $\left(\frac{1}{2}x \right)^2+2\times \frac{1}{2} \times x \times \frac{2}{3}\times y+\left(\frac{2}{3} y\right)^2$ $\frac{x^2}{4}+\frac{2xy}{3}+\frac{4}{9}y^2$ Question 2 (i) $\left(3x+\frac{1}{2x}\right)^2$ It is in the form of $(a+b)^2=a^2+2ab+b^2$ ∴$(3x)^2+2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$ $9x^3+3+\frac{1}{4x^2}$ (ii) $(3x^2y+5z)^2$ It is in the form of $(a+b)^2=a^2+2ab+b^2$ Here $a=3x^2y$ , b=5x $(3x^2y)^2+2.3x^2y.5z+(5z)^2$ $9x^4y^2+30x^2yz+25z^2$ Question 3 (i) $\left(3x-\frac{1}{2x}\right)^2$ It is in the form of $(a-b)^2=a^2-2ab+b^2$ Here, a=3x ; $b=\frac{1}{2x}$ $(3x)^2-2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$ $3^2-x^2-3+\frac{1}{2^2x^2}$ $9x^2-3+\frac{1}{4x^2}$ (ii) $\left(\frac{1}{2}x-\frac{3}{2}y\right)^2$ It is in the form of $(a-b)^2=a^2-2ab+b^2$ Here...
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