M.L Aggarwal Solutions

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ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
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ML Aggarwal Solution Class 10 Chapter 1 Value Added Tax Exercise 1

Exercise 1 Question 1 A manufacturing company sells a T.V. to a trader A for ₹ 18000. Trader A sells it to a trader B at a point of ₹ 750 and trader B sells it to a consumer at a profit of ₹ 900. If the rate of sales tax (under VAT) is 10%, find (i) the amount of tax received by the Government. (ii) the amount paid by the consumer for the T.V. Sol : Sale price of a T.V. to trader A = ₹ 18000 Rate of VAT tax = 10% ∴Amount of VAT$=18000 \times \frac{10}{100}$ =₹1800 Gain of A=₹750 ∴VAT on gain $=₹ \frac{750 \times 10}{100}=₹ 75$ Gain of trader B=₹ 900 VAT tax paid $=\frac{900 \times 10}{100}=₹ 90$ (i) Total tax paid to Govt. = ₹ (1800+75+90)  =₹ 1965 (ii) Amount paid by the consumer to trader B =₹ 18000 + 750 + 900 + Tax 1965  = ₹ 21615 Question 2 A manufacturer sells a washing machine to a wholesaler for ₹ 15000. The wholesaler sells it to a trader at a profit of ₹ 1200 and the trader sells it to a consumer at a profit of ₹ 1800. If the rate of VAT is 8%, find : (i) The amount of VAT re
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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^{\c
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