A piece of string is 50cm long. It is cut into three pieces such that the longest piece is three times as long as the shortest piece and the third piece is 20cm shorter than the longest piece. If the pieces are joined to form a triangular region, find the area (in sq. cm) of the region formed.

Answer: D Let the length of the longest, the shortest and the third pieces be a,b and c respectively. 50= a+b+c a=3c and b=a-20 => 50 = c+3c+3c-20=> c=10 a=30, b= 10, c= 10. Thus, Triangle cannot be formed as (b+c)<a

Q. No. 2:

Sachin had a triangle in his mind, its longest side has a length of 10 and one of its sides has a length of 5.What is the exact length of its third side, if the area of the triangle is 20?

Answer: A Let assume a triangle point A is on the top and the base is BC, a perpendicular is drwan on BC which is point D. Let AB=5 and BC =10 Now, 1/2 BC*AD = Area of triangle ABC = 20 * 1/2 * 10* AD => AD=4. BD =√(AB^{2}-AD^{2}) = √5^{2}-4^{2} = 3 DC= BC-BD =7 In a triangle ADC, AC= √65

Q. No. 3:

A thin rectangular strip of cloth which is 80cm long is cut into 3 pieces along its length. The longest piece is 3 times as long as the middle sized piece and the shortest piece is 46cm shorter than the longest piece. Find the length of the shortest piece.

Answer: C Let X be the length of the middle strip of cloth Length of the longest strip= 3X Shortest strip = 3X-46 Total length = X+3X+3X-46 = 7X-46 =80 => X =18 cm. The length of the shortest piece is 3X-46 = 54-46 = 8cm.

Q. No. 4:

A rectangular sheet of paper was folded along the line joining the mid points of its longer sides. The rectangle which resulted had the same ratio of its longer and the shorter sides as that of the original rectangle. If the breadth of the original rectangle is √2 cm, find the area (in sq cm) of the smaller rectangle.

Answer: C Let the length and the breadth of the original rectangle be l and b respectively. l/b = b/(l/2) Thus, √2 b =l As b=√2, l=2 Area of the smaller rectangle = lb/2 = √2 sq cm.

Q. No. 5:

A cuboid has a volume of 64 cubic units. Find the minimum possible vale (in units) of the sum of the lengths of the edges of the cuboid.

Answer: D Let the length, breadth and the height of the cuboid be l,b and h lbh=64 As the product of l,b and h is constant their sum will be minimum is l=b=h so, l=b=h=4 Hence, the sum of the edges of the cuboid = 4(l+b+h) =48 units.

Q. No. 6:

A small cylinder container is of height 7cm. the outer radius of the container is 8cm and inner radius is 6cm. Water whose density is 1gm/cm^{3} is filled to the brim in the container. If the density of the material of the container is 16gm/cm^{3}, find the weight of the container filled completely with water.(The weight of the bottom is negligible).

Answer: D Volume of the container = pie(8^{2}-6^{2})*7 = 22*28 cm^{2} Weight of the container = 22*28*16 = 9856 gm. Volume of water = pie*r^{2}*h = 22/7 *6^{2}*7 = 792 cm^{3} Weight of water = 792*1 gm/cm^{3} =792 gm Total weight = 9856+792 = 10648 gm.