1. The surface area of a cube is 726 sq cm. Find the volume of the cube.

1) 1331 cm3 2) 1232 cm3 3) 1626 cm3 4) 1836 cm3 5) None of these

Solution: (1)

According to the question,

6a2 = 726 [ a = edge of the cube]

=a2 = 726/6 = 121

Therefore a = 121 = 11 cm

Therefore required volume = a3 = 113 = 1331 cm3

2. The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm X 6cm X

2cm, is

1) 213 cm 2) 214 cm 3) 226 cm 4) 102 5) None of these

Solution: (3)

Length of largest pencil that can be kept in a box

= Diagonal of box = l2 + b2 + h2

Where, I = 8 cm, b = 6 cm, h = 2 cm

= 64 + 36 + 4

= 104 = 2 26 cm

3. Internal length , breadth and height of a rectangular box are 10 cm, 8 cm and 6 cm, respecti vely.

How many boxes are needed which can be packed in a cube whose volume is 6240 cu.cm.?

1) 12 2) 13 3) 15 4) 17 5) None of these

Solution: (2)

Volume of rectangular box = 10 x 8 x 6 = 480cm3

Volume of cubes = 6240 cm3

Therefore required boxes = Volume of cubes / Volume of rectangular box = 6240 / 480 = 13

Hence, 13 boxes are needed.

4. A cube has each edge 2 cm and a cuboid is 1 cm long, 2 cm wide and 3 cm high. The paint in a

certain container is sufficient to paint an area equal to 54 cm2. Which one of the following is correct?

1) Both cube and cuboid can be painted

2) Only cube can be painted

3) Only cuboid can be painted

4) Neither cube nor cuboid can be painted

5) None of these

Solution: (1)

Surface area of cube which can be painted = 6(side)2 = 6 (2)2 = 24 cm2

Now, surface area of cuboid which can be painted

= 2 ( lb + bh + 1h)

= 2 (2 + 6+ 3) = 22cm

Total surface area of both cube and cuboid

= 22 + 24 = 46 cm2 < 54 cm2

Therefore , both cube and cuboid can be painted.

5. The whole surface area of a rectangular block is 8788 sq.cm. If length, breadth and height are in the

ratio of 4:3:2, then find the length.

1) 26 cm 2) 52 cm 3) 104 cm 4) 13 cm 5) None of these

Solution: (2)

Let length, breadth and height be 4x, 3x and 2x, respectively .

Whole surface area = 2 (lb + bh + lh)

= (lb + bh + lh) = 8788/2 = 4394

= ( 4x 3+3x2+2x4)x2 = 4394

26x2 = 4394

X2 = 169

X = 13

Therefore length = 4x = 4 x 13 = 52cm.

Also read

IBPS PO CWE V Notification out

IBPS RRB CWE IV Notification released

IBPS PO Preliminary Practice Set Download Here

**Advertisements**

## 0 comments :

## Post a Comment