M.L Aggarwal Solutions

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ML Aggarwal Solution Class 10 Chapter 1 Value Added Tax Chapter Test

Chapter Test Question 1 A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find: (i) the VAT paid by the shopkeeper. . (ii) the total amount that the consumer pays for the washing machine. Sol : (i) S.P. of washing machine $=\left(1-\frac{10}{100}\right) \times ₹ 18000$ $=\frac{90}{100} \times 18000=₹ 16200$ Sales tax=8 % of  ₹ 16200  $=\frac{8}{100} \times ₹ 16200=₹ 1296$ Printed price =₹ 18000 Price paid by shopkeeper$=\left(1-\frac{20}{100}\right) \times ₹ 18000$ $=\frac{80}{100} \times ₹ 18000=₹ 14400$ Sale tax paid by shopkeeper$=\frac{8}{100} \times ₹ 14400=₹ 1152$ (i) ∴VAT paid by shopkeeper=Tax charged - Tax paid =₹ 1296-₹ 1152=₹ 144 (ii) Price paid by customer =₹ 16200+₹ 1296=₹ 17496 Question 2 A manufacturing company sold an article to its distributor for ₹22000 includi...
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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 9 Chapter 11 Mid Point Theorem MCQS

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  MCQS Choose the correct answer from the given four options (1 to 6): Question 1 In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. IfD and E are mid-points of AB and BC respectively, then the length of DE is (a) 1.5 cm (b) 2 cm (c) 2.5 cm (d) 3.5 cm Sol : In ∆ABC, D and E are the mid-points of sides AB and BC ∴DE$=\frac{1}{2}$AC But AC=5 cm ∴DE$=\frac{1}{2}\times 5=\frac{5}{2}$ cm =2.5cm Ans (c) Question 2 In the given figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is (a) 7 cm (b) 5 cm (c) 4 cm (d) 3 cm Sol : In the given figure ABCD is a rectangle AB=6 cm , AD=8 cm P and Q are mid points of BC and CD ∴PQ$=\frac{1}{2}$BD But BD$=\sqrt{BC^2+CD^2}$ (pythagoras theorem) $=\sqrt{8^2+6^2}$ $=\sqrt{64+36}=\sqrt{100}$ =10 cm ∴PQ$=\frac{1}{2}$BD $=\frac{1}{2}\times 10=5$ cm Ans (b) Question 3 D and E are mid-points of the sides AB and AC of ∆ABC and O is any point on the side BC. O is joi...
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