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Area Questions and Answers
Q) A blanket lost 20% length and 10% width when washed. Evaluate the decrease percentage of blanket area? Explanation : • Let us assume, Actual length = L and Actual width = w. • According to the question, • Decrease in area = LW - 80/100 L × 90/100 W = LW - 18/25 LW = 7/25 LW. • ∴ Decrease % = 7/25 ( LW × 1/LW × 100 ) % = 28%. Ans) 28%.
Q) A rectangular room with 20m × 15m dimensions will be layed with a carpet by leaving 2m from the walls. Calculate the cost of the carpet where 50rs is the cost per sq metre? Explanation: • According to the question,we have to leave 2m from the walls --> 2 + 2 = 4m • Area of the carpet = [(20 - 4) × (15 - 4) ]sq.ft = 176sq.ft. • ∴ Cost of the carpet = Rs. (176 × 50) =Rs. 8800. Ans) Rs. 8800
Q) Calculate the cost of gardening which has one metre broad boundary, Where the rectangular plot perimeter is 340 metres at the rate of Rs. 10 per square metre? Explanation • 2 (I + b) = 340 (given). • Area of the boundary = [ (I + 2) (b + 2) – Ib ] = 2 (I + b) + 4 = 344. • ∴ Cost of gardening =Rs. (344 × 10) = Rs. 3440 Ans) Rs. 3440
Q) Find the least number of square marbels needed to Q) Find the least number of square marbels needed to pave the floor whose dimensions are 15m 17cm × 9m 2cm? Explanation : • According to the question we need to find the least no of marbels for paving of given floor. So we have to take the largest marble. • Length of largest marble = H.C.F of 1517 cm and 902 cm = 41 cm. • Area of each marble = (41 × 41) cm². • Required number of marbles = (1517 × 902) / (41 ×41) = 814. • Ans) 814
Q) Calculate the length of the fencing of a plot, whose Q) Calculate the length of the fencing of a plot, whose area is 1360ft. Here you have to leave one side of plot open whose length is 40ft? Explanation : • According to the question, Length = 40ft and area = 1360ft • So Area = Length × width --> Width = Area/Length • ---------> Wdth = 1360/40 ---> 34 • So, Width = 34 ft. • ∴ Length of fencing = (Length + 2Width) = (20 + (2 × 34) )ft = 88 ft. Ans) 88
Q) How many poles required to lay a fence for a 90m Q) How many poles required to lay a fence for a 90m × 50m plot by leaving 10 meters apart? Explanation : • First we have to calculate perimeter of the given rectangular plot, • Perimeter of the plot = 2. (length + width) = 280 m. • ∴ Number of poles = 280/10 = 28. Ans) 28
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