## Online Free Mathematics Mock Test MCQ type Questions [28.10.2019]

**Online Free Mock Test MCQ type Questions**

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**Solutions:**

**1. Answer :OptionÂ B**

Diagonal 8cm = 4cm + 4cm = legs of right triangle within rhombus.

Hence perimeter of rhombus = 4Ã—5 = 20cm

**2. Answer :OptionÂ C**

Now, length of the rectangle = (10 + 2) cm = 12 cm

We know that, 2(L + B) = Perimeter of rectangle = 40

2 Ã— 12 + 2B = 40, 2B = 40-24 = 16, B = 16/2 = 8 cm

Hence area of rectangle = (12 Ã— 8) cm

^{2}Â = 96 cm

^{2}

**3. Answer :OptionÂ B**

Sum of its non-parallel = 20 cm

Sum of parallel sides = (58 â€“ 20) = 38 cm

Area of trapezium=Â Ã—d Ã— (Sum of parallel sides)

152 =Ã—d Ã— 38

By solving this,we get

âˆ´d = 8 cm

**4. Answer :OptionÂ A**

Breadth = 11 cm

and height = 9 cm

Cubical box â€“ side = 12 cm.

Let’sÂ find the volume of the cuboidal box.

We know, volume of cuboid = length Ã— breadth Ã— height

So, volume of cuboidal box = 13 Ã— 11 Ã— 9 [as given] = 1287 cm

^{3}

We know, volume of cube = (side)

^{3}

So, volume of cubical box = (12)

^{3}Â = 1728 cm

^{3}

Total volume = 1287 + 1728 = 3015 cm

^{3}

According to the question,

Tanu wants to pack 3060 cubes of side 1 cm into this box.

Volume of one such cube of side 1 cm = (1)

^{3}Â = 1 cm

^{3}

Total volume of cubes which Tanu wants to pack = 1 Ã— 3060 = 3060 cm

^{3}

Since,Â only 3015 cm

^{3}Â volume is available, it wonâ€™t be able to accumulate 3060 cm

^{3}Â volume of total cubes.

âˆ´Â 3060Â â€“Â 3015 = 45 boxes left unpack.

Thus, Option A is correct.

**5. Answer :OptionÂ D**

Perimeter of a rectangle = Perimeter of a rectangle = 44 cm

We know that, Perimeter of square = 4 Ã— side

â‡’ 44 = 4 Ã— side

â‡’ side = 44/4 = 11 cm

Thus, area of square = (side)

^{2}Â =(11)

^{2}Â = 121 cm

^{2}â€¦(i)

According to the question,

The length of the rectangle is 5 cm more than the side of the square.

We have found that side of the square = 11 cm

â‡’ length of the rectangle = 5 cm + side of the square

â‡’ length of the rectangle = 5 cm + 11 cm = 16 cm â€¦(ii)

âˆµ, Perimeter of the rectangle = 44 cm

â‡’ 2 (length + breadth) = 44

â‡’ 2 (16 + breadth) = 44 [by (ii)] â‡’ 16 + breadth = 22

â‡’ breadth = 22 â€“ 16 = 6 cm

Length and breadth of the rectangle are 16 cm and 6 cm respectively.

Thus, Area of the rectangle = 16 Ã— 6 = 96 cm

^{2}â€¦(iii)

By equations (i) & (ii), we get

Sum of the areas of square and rectangle = 121 + 96 = 217 cm

^{2}

Hence, option D is correct.

**6. Answer :OptionÂ B**

Let h be the depth (or height) of carton.

We know that, 1 litre = 1000 cm

^{3}

Also, Volume of carton = length Ã— breadth Ã— height

â‡’Â 1000 = 8 Ã— 8 Ã— h [By substituting values in the formula] â‡’ h = 1000/64 = 15.625 â‰ˆ 16 cm

Hence, option B is correct.

**7. Answer :OptionÂ C**

= 1000 Ã— 1000 sq cm

Now, area of tile = (50)

^{2}Â = 50 Ã— 50 sq cm

âˆ´ Number of tiles needed.

{(1000*1000)/50*50}Â = 400

Hence, 400 tiles will be needed.

**8. Answer :OptionÂ A**

(2+5) = 7

(1+6) =7

(3+4) = 7

Hence they will be the opposite faces of each other.

**9.Â Option D**

If one sheet rolled out into a cylinder through its length thus, the height of the cylinder, h = 22 cm and the area of the base = 28 cm

We know that, a circumference of base = 2Ï€r

âˆ´ In that case,

If other sheet rolled out into a cylinder through its width thus, the height of the cylinder, h = 28 cm and the area of the base = 22 cm

In that case,

Therefore, required difference = (1372 â€“ 1078) cm

^{3}Â =294cm

^{3}