## ML Aggarwal CBSE Solutions Class 6 Math Fourteenth Chapter Practical Geometry Exercise 14.3

ML Aggarwal CBSE Solutions Class 6 Math 14th Chapter Practical Geometry Exercise 14.3 (1) Draw an angle of 80o and make a copy of it using ruler and compass. (2) Draw an angle of measure 127o and construct its bisector. (3) Draw <POQ = 64o. Also draw its line of symmetry. Solution: (4) Draw a […]

## ML Aggarwal CBSE Solutions Class 6 Math Fourteenth Chapter Practical Geometry Exercise 14.1

ML Aggarwal CBSE Solutions Class 6 Math 14th Chapter Practical Geometry Exercise 14.1 (1) Construct a circle of radius: (i) 2 cm (ii) 3.5 cm (2) With the same centre O, draw two circles of radii 2.6 cm 4.1 cm. Solution: (3) Draw any circle and mark points A, B and C such that – […]

## New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Fractions Chapter 5H Solution

New Learning Composite Mathematics Class 6 SK Gupta Anubhuti Gangal Fractions Chapter 5H Solution (1) (a) 2/5, 9/10 LCD of 5, 10 = 10 = 2 x 2 / 5 x 2 , 9/10 = 4/10, 9/10 (b) 3/8 , 7/12 = LCD of 8, 12 = 24 = 9/24 , 14/24 (c) 5/6 , […]

## ML Aggarwal CBSE Solutions Class 8 Math Seventh Chapter Compound Interest Exercise 7.3

ML Aggarwal CBSE Solutions Class 8 Math 7th Chapter Compound Interest Exercise 7.3 (6) V0 = Present value = 16000 V = Value after 2 years = 14440 N = 2 years R =? V = V0 [1- r/100]r Or, 14440 = 16000 (1 – r/100)2 Or, 14440/16000 = (1 – r/100)2 Or, (38/40)2 = […]

## ML Aggarwal CBSE Solutions Class 8 Math Seventh Chapter Compound Interest Exercise 7.2

ML Aggarwal CBSE Solutions Class 8 Math 7th Chapter Compound Interest Exercise 7.2 (2) For SI – I – PRT/100 = 2500 X 4 X 2/100 = 200/- For CI – P = 2500, R = 4/2 = 2% [∵ semi annually] Years = 2 = 24 month. ∴ n = 24/6 = 4 ∴ […]

## ML Aggarwal CBSE Solutions Class 8 Math Seventh Chapter Compound Interest Exercise 7.1

ML Aggarwal CBSE Solutions Class 8 Math 7th Chapter Compound Interest Exercise 7.1 (1) A = P (1+ R/100)n Given P = 6000/- R = 10% N = 2 years = 6000 (1+ 10/100)2 = 6000 X (100+10/100)2 = 6000 X (110/100)2 = 6000 X 110/100 X 110/100 = 7260 ∴ C.I. = A – […]

## ML Aggarwal CBSE Solutions Class 8 Math Sixth Chapter Percentage and its Applications Exercise 6.4

ML Aggarwal CBSE Solutions Class 8 Math 6th Chapter Percentage and its Applications Exercise 6.4 (1) (i) CP = 50/- GST = 5% of 50 ∴ = 5/100 X50 = 5/2 /- ∴ SP = 50 + 5/2 = 100+5/2 = 105/2 = 50.50/- (Ans) (ii) 1kg = 150/- ∴ CP of 5kg = 150X5 […]

## ML Aggarwal CBSE Solutions Class 8 Math Sixth Chapter Percentage and its Applications Exercise 6.3

ML Aggarwal CBSE Solutions Class 8 Math 6th Chapter Percentage and its Applications Exercise 6.3 (1) (i) MP = 575 /- Dis = 12% SP = 575 X 88/100 = 506/- (Ans) Dis = 575 – 506 = 69/- (Ans) (ii) MP = 12750/- dis = 8 1/3 % = 25/3 % SP = 12750 […]

## ML Aggarwal CBSE Solutions Class 8 Math Sixth Chapter Percentage and its Applications Exercise 6.2

ML Aggarwal CBSE Solutions Class 8 Math 6th Chapter Percentage and its Applications Exercise 6.2 (1) (i) CP = 400/- SP = 468/- Profit = 468 – 400 = 68 Profit % = Profit/CP X 100 = 68/400 X 100 = 17 % (Ans) (ii) C.P. = 13600/- S.P. = 12104/- Loss = 13600 – […]

## ML Aggarwal CBSE Solutions Class 8 Math Sixth Chapter Percentage and its Applications Exercise 6.1

ML Aggarwal CBSE Solutions Class 8 Math 6th Chapter Percentage and its Applications Exercise 6.1 (15) Let, total number of vote polled = x ∴ Loser polled = x X 42/100 = 42x/100 ∴ Winner poled = x – 42x/100 = 100x – 42x/100 = 58x/100 ∴ The difference between winner & loser polled = […]