MATH

W0RD PROBLEMS RELATED TO MENSURATION CLASS 8TH, 9TH & 10TH Formulas :- 1). Cube: Total surface area = TSA = 6a2 Lateral surface area = LSA = 4a2 Volume = V = a2 2). CUBOID: Total surface area = TSA = 2 (l*b + b*h + h*l) Lateral surface area = LSA = 2(b*h + […]

REVISE WITH ME :(FRACTION) CLASS 4TH, 5TH, 6TH, 7TH Points to remember : 1). A fraction comprises two numbers i.e. Numerator and Denominator separated by a horizontal line.2). A fraction is the simplest form when the HCF of the numerator and denominator is 1.3). Fractions having the same denominators are called like fractions and the

WORD PROBLEMS RELATED TO MENSURATION CLASS 7TH, 8TH, 9TH & 10TH Q.1) IN a bakery store, 250 cubical boxes are required. The length of the edge box measures 20 cm. How many cardboard sheets of size 50 cm*50 cm will be required to make these boxes ? SOLUTION : Cubical boxesGiven : No. of cubical

POINTS RELATED TO CHAPTER – FACTORS AND MULTIPLES CLASS 5th, 6th, 7th & 8th POINTS TO REMEMBER:- 1) A factor of a number is an exact divisor of that number and the number has a finite number of factors. 2) A multiple of a number is exactly divisible by that number and the number can

WORD PROBLEMS RELATED TO MENSURATION CLASS 8th, 9th & 10th Q.1) A cuboid has a total surface area of 160 sq. m and lateral surface area of 120 sq. m. Find the area of it’s base ? SOLUTION : CuboidGiven : Total surface area =TSA = 160 sq. m. Lateral surface area = LSA =

Word Problems on Area For Class 6th,7th,8th Q 1 ] . Split the following shapes into rectangles and squares and find their areas. (The measures are given in centimeters). SOLUTION: SOLUTION : Fig (a) Area of Fig 1 ( rectangle ) = 8cm *2cm = 16cm2 Area of Fig 2( rectangle ) = 8cm *

Word Problems On Area For Class 6th, 7th , 8th , 9th . Q 1 ] . What is the cost of tiling a rectangle plot of land 1/2Km long and 200m wide at the rate of Rs. 9 per hundred sq. m ? . SOLUTION : Rectangular Plot Given : Length = L =

WORD PROBLEMS RELATED TO AREA CLASS 6th, 7th, 8th & 9th FORMULAS: 1) Area of rectangle = L*B2) Area of square = S*S3) If rate is given as per meter – then find perimeter Hence, Cost = Perimeter * Rate4) If rate is given as per square meter – then find area Hence, Cost =

WORD PROBLEMS RELATED TO PERIMETER CLASS 6th, 7th, 8th & 9th Q.1) Aditya runs around a square park of side 50m three times. Shweta runs twice around a rectangular park of length 100m and breadth 60m. Who covers more distance? SOLUTION: Given: Aditya (Square Park)Sides = S = 50mNo. of rounds = 3 timesPerimeter of

WORD PROBLEMS ON PERIMETER : CLASS 6TH ,7TH , 8TH : Q.1) Find the perimeter of the rectangular lawn whose length is 20m 50cm and breadth is 12m 20cm. SOLUTION : Rectangular LawnGiven : Length =L = 20m 50cm =20.50 m Breadth =B = 12m 20cm = 12.20m Perimeter of rectangular lawn =? Perimeter of