46
Q:

# How many cubes of 3cm edge can be cut out of a cube of 18cm edge

 A) 36 B) 232 C) 216 D) 484

Explanation:

number of cubes=(18 x 18 x 18) / (3 x 3 x 3) = 216

Q:

If a solid sphere of radius 10 cms is moulded into 8 spherical solid balls of equal radius, then surface area of each ball (in sq.cm) is ?

 A) 100 π B) 101/π C) 99 π/12 D) 54/13π

Explanation:

4/3 π x 10 x 10 x 10 = 8 x 4/3 π rxrxr
r = 5
4π x 5 x 5 = 100π

0 27
Q:

Calculate the number of bricks, each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 m x 5 m when 10% of its volume is occupied by concrete ?

 A) 6000 B) 5400 C) 3800 D) 4700

Explanation:

Let 'B' be the nuber of bricks.

=> 10 x 4/100 x 5 x 90/100 = 25/100 x 15/100 x 8/100 x B

=> 10 x 20 x 90 = 15 x 2 x B

=> B = 6000

1 18
Q:

A cuboid of dimension 24cm x 9cm x 8 cm is melted and smaller cubers are of side 3 cm is formed.find how many such cubes can be formed ?

 A) 64 B) 56 C) 48 D) 40

Explanation:

Volume of cuboid = (24 x 9 x 8) cm = 1728 cu.cm
Volume of small cube = (3 x 3 x 3) cm = 27 cu.cm
So,
No. of small cubes formed = 1728/27 = 64.

1 61
Q:

A rectangular block has the dimensions 5x6x7 cm it is dropped into a cylindrical vessel of radius 6cm and height 10 cm. If the level of the fluid in the cylinder rises by 4 cm, What portion of the block is immersed in the fluid ?

 A) 22/7 x 24/35 B) 22/7 x 36 x 4 C) 22/7 x 36/5 D) 22/7 x 37/21

Explanation:

Since level of water increased in cylinder by height 4.
This is because of the rectangular block .

Therefore , area of rectangular block immersed in water is = 22/7 x (6)^2 x 4

Thats why portion of block immersed in water is= (22/7 x 36 x 4) / total vol. of rectangle
= (22/7 x 36 x 4)/(7x5x6)
= 22/7 x 24/35.

3 101
Q:

88 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be ?

 A) 112 mts B) 84 mts C) 96 mts D) 108 mts

Explanation:

Vol of silver = $\inline \fn_jvn \small \prod R^{2}h$  = 66 C.C

given D= 1mm => R = D/2 = 1/20 cm

=> $\inline \fn_jvn \small \frac{22}{7}x(\frac{1}{20})^{2}xh$ = 66

= 66 x 400 x 7/22 = 8400 cm = 84 mtrs.