A plot has a concrete path within its borders on all sides having uniform width of 4m. The plot is rectangular with sides 20m and 15m. Charge of removing concrete is Rs. 6 per sq.m. How much is spent in removing all the concrete?
Answer: B
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Ramesh runs along a circular park every day at a speed of 60m/min. One day he runs along the diameter with same speed and crosses the park. He needed 60 minutes less to cross the park. What is the radius of the park?
Answer: A
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A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?
Answer: B
Length * Breadth = Area
⇒20 * Breadth=680
⇒Breadth = 34 feet
Area to be fenced=2B+L=2*34+20
=88 feet
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The area of a square field 3136 sq m, if the length of cost of drawing barbed wire 3 m around the field at the rate of Rs.1.50 per meter. Two gates of 1 m width each are to be left for entrance. What is the total cost?
Answer: C
a⊃2; = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.5 = 999
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A square farm of side 240 meters has a small rectangular area for barn construction. This barn area has its length 10 meters more than its breadth and its area measures one-eighth the square farm. How much the owner has to spend to build a wall around the barn area at rate of Rs. 1.5 per meter?
Answer: A
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The area of a triangle will be when a = 1m, b = 2m, c = 3m, a, b, c being lengths of respective sides.
Answer: A
S = (1 + 2 + 3)/2 = 3
=> No triangle exists
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A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be?
Answer: D
lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
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The ratio of the areas of a square and rhombus whose base is same is:
Answer: C
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50 square stone slabs of equal size were needed to cover a floor area of 72 sq.m. Find the length of each stone slab.
Answer: C
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A big rectangular plot of area 4320 Sq.m is divided into 3 square shaped smaller plots by fencing parallel to the smaller side of the plot. However, some area of land was still left as a square plot could not be formed. So 3 more square shaped plots were formed by fencing parallel to the longer side of the original plot. Such that no area of the plot was left a surplus. What are the dimensions of the original plot?
Answer: C
Let the side of smaller square at the extreme right be a
Then side of the larger square is 3a
Total area of the field = a⊃2; + a⊃2; +a⊃2; + (3a)⊃2; +(3a)⊃2; = 30a⊃2;
Therefore, 30 a⊃2; = 4320 => a⊃2; = 432 / 3 = 144 => a = 12
Smaller side of the field = 36
Larger side of the field = (3a) × 3 + a = 10 a = 10 × 12 = 120
Original Dimensions of the field = 120m × 36m
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