M.L Aggarwal Solutions

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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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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 Solution Class 9 Chapter 13 Rectilinear Figures Exercise 13.1

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 Exercise 13.1 Question 1 If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles. Sol : Sum of four angles of a quadrilateral=360° Sum of two given angles =40°+110°=150° ∴Sum of remaining two angles =360°-150°=210° ∴Third angle$=\frac{210^{\circ} \times 3}{3+4}$ $=\frac{210^{\circ} \times 3}{7}=90^{\circ}$ and fourth angle $=\frac{210^{\circ} \times 4}{3+4}$ $=\frac{210^{\circ} \times 4}{7}$ =120° Question 2 If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium. Sol : In trapezium ABCD ∠A : ∠B : ∠C : ∠D=1 : 2 : 3 : 4 Sum of angles of the quad ABCD=360° Sum of the ratio's=1+2+3+4=10 ∴$ \angle \mathrm{A}=\frac{360^{\circ} \times 1}{10}=36^{\circ}$ $\angle B=\frac{360^{\circ} \times 2}{10}=72^{\circ}$ $\angle C=\frac{360^{\circ} \times 3}{10}=108^{\circ}$ $\angle D=\frac{360^{\circ} \times 4}{10}=144^{\circ}$ Now , ∠A+∠D=36°+114°=180° ∵∠A+∠D=180° and there are co-interi...
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