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Dimensions of the matchbox (a cuboid) are l, b, h = 4 cm, 2.5 cm, 1.5 cm, respectively Formula to find the volume of the matchbox = l×b×h = (4×2.5×1.5) = 15 Volume of matchbox = 15 cm3 Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3 Therefore, the volume of the packet containing 12 matchboxes is 180 cm3. 2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l) Solution: Dimensions of the cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m Formula to find the volume of the tank, V = l×b×h Put the values, we get V = (6×5×4.5) = 135 The volume of the water tank is 135 m3 Again, The amount of water that 1 m3 volume can hold = 1000 l The amount of water that 135 m3 volume can hold = (135×1000) litres = 135000 litres Therefore, the given cuboidal water tank can hold up to 135000 litres of water. 3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? Solution: Given: Length of the cuboidal vessel, l = 10 m Width of the cuboidal vessel, b = 8m Volume of the cuboidal vessel, V = 380 m3 Let the height of the given vessel be h. Formula to find the volume of a cuboid, V = l×b×h Using the formula, we get l×b×h = 380 10×8×h = 380 Or h = 4.75 Therefore, the height of the vessels must be 4.75 m. 4. Find the cost of digging a cuboidal pit of 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3. Solution: The given pit has its length (l) as 8m, width (b) as 6m and depth (h) as 3 m. Volume of the cuboidal pit = l×b×h = (8×6×3) = 144 The required Volume is 144 m3 Now, The cost of digging per m3 volume = Rs. 30 Therefore, the cost of digging 144 m3 volume = Rs. (144×30) = Rs. 4320. 5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. Solution: The length (l) and depth (h) of the tank is 2.5 m and 10 m, respectively. To find the value of breadth, say b, The formula to find the volume of the tank = l×b×h = (2.5× b×10) m3 = 25b m3 The capacity of tank = 25b m3, which is equal to 25000b litres Also, the capacity of a cuboidal tank is 50000 litres of water (Given) Therefore, 25000 b = 50000 This implies that b = 2 Therefore, the breadth of the tank is 2 m. 6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m×15 m×6 m. For how many days will the water in this tank last? Solution: Length of the tank = l = 20 m Breadth of the tank = b = 15 m Height of the tank = h = 6 m Total population of the village = 4000 Consumption of water per head per day = 150 litres Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1) The formula to find the capacity of a tank, C = l×b×h Using the given data, we have C = (20×15×6) m3 = 1800 m3 Or C = 1800000 litres Let the water in this tank last for d days. Water consumed by all people in d days = Capacity of the tank (using equation (1)) 600000 d = 1800000 d = 3 Therefore, the water in this tank will last for 3 days. 7. A godown measures 40 m×25 m×15 m. Find the maximum number of wooden crates, each measuring 1.5m×1.25 m×0.5 m, that can be stored in the godown. Solution: From the statement, we have Length of the godown = 40 m Breadth = 25 m Height = 15 m Whereas, Length of the wooden crate = 1.5 m Breadth = 1.25 m Height = 0.5 m Since the godown and wooden crate are in cuboidal shape, we can find the volume of each using the formula, V = lbh. Now, Volume of the godown = (40×25×15) m3 = 15000 m3 Volume of the wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3 Let us consider that n wooden crates can be stored in the godown, then The volume of n wooden crates = Volume of godown 0.9375×n = 15000 Or n = 15000/0.9375 = 16000 Hence, the number of wooden crates that can be stored in the godown is 16,000. 8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. Solution: Side of the cube = 12 cm (Given) To find the volume of the cube: Volume of cube = (Side)3 = (12)3 cm3= 1728 cm3 Surface area of a cube with side 12 cm = 6a2 = 6(12) 2 cm2 …(1) The cube is cut into eight small cubes of equal volume; say the side of each cube is p. The volume of the small cube = p3 Surface area = 6p2 …(2) Volume of each small cube = (1728/8) cm3 = 216 cm3 Or (p)3 = 216 cm3 Or p = 6 cm Now, the surface areas of the cubes ratios = (Surface area of the bigger cube)/(Surface area of smaller cubes) From equations (1) and (2), we get Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4 Therefore, the required ratio is 4:1. 9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? Solution: Given: Depth of the river, h = 3 m Width of the river, b = 40 m Rate of water flow = 2 km per hour = 2000 m/60 min = 100/3 m/min So, the volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3 Therefore, 4000 m3 of water will fall into the sea in a minute. By practising problems in NCERT Solutions for Class 9 Maths Chapter 13, students can learn the easy way to solve them and can score well in the annual examination. All the 9 questions included in this exercise will help students analyse the situation and apply the formula. It includes easy-to-solve questions, which can be expected in examinations. Solve these NCERT solutions, where problems are explained in a detailed way, following each and every step. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes exercise consists of application-level questions that help to determine the volume of the cuboid. It also helps in finding the length and breadth of the cuboid by applying the formula. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.5 Solving the NCERT solutions for Class 9 Maths Chapter 13 helps students in multiple ways, as explained below: Students can self-assess their learning abilities and preparation levelStudents can analyse the type of questions that appear for the examsImprove their efficiency and speed in solving the problemsRemember the formulas in an easy way and help them apply relevantly

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