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

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[wp_quiz id=”17292″]

**Solutions:**

**1. Answer :Option B**

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

Hence perimeter of rhombus = 4×5 = 20cm

**2. Answer :Option C**

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

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

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

Hence area of rectangle = (12 × 8) cm

^{2}= 96 cm

^{2}

**3. Answer :Option B**

Sum of its non-parallel = 20 cm

Sum of parallel sides = (58 – 20) = 38 cm

Area of trapezium= ×d × (Sum of parallel sides)

152 =×d × 38

By solving this,we get

∴d = 8 cm

**4. Answer :Option A**

Breadth = 11 cm

and height = 9 cm

Cubical box – side = 12 cm.

Let’s find the volume of the cuboidal box.

We know, volume of cuboid = length × breadth × height

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

^{3}

We know, volume of cube = (side)

^{3}

So, volume of cubical box = (12)

^{3}= 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}= 1 cm

^{3}

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

^{3}

Since, only 3015 cm

^{3}volume is available, it won’t be able to accumulate 3060 cm

^{3}volume of total cubes.

∴ 3060 – 3015 = 45 boxes left unpack.

Thus, Option A is correct.

**5. Answer :Option D**

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

We know that, Perimeter of square = 4 × side

⇒ 44 = 4 × side

⇒ side = 44/4 = 11 cm

Thus, area of square = (side)

^{2}=(11)

^{2}= 121 cm

^{2}…(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

⇒ length of the rectangle = 5 cm + side of the square

⇒ length of the rectangle = 5 cm + 11 cm = 16 cm …(ii)

∵, Perimeter of the rectangle = 44 cm

⇒ 2 (length + breadth) = 44

⇒ 2 (16 + breadth) = 44 [by (ii)] ⇒ 16 + breadth = 22

⇒ breadth = 22 – 16 = 6 cm

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

Thus, Area of the rectangle = 16 × 6 = 96 cm

^{2}…(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 :Option B**

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

We know that, 1 litre = 1000 cm

^{3}

Also, Volume of carton = length × breadth × height

⇒ 1000 = 8 × 8 × h [By substituting values in the formula] ⇒ h = 1000/64 = 15.625 ≈ 16 cm

Hence, option B is correct.

**7. Answer :Option C**

= 1000 × 1000 sq cm

Now, area of tile = (50)

^{2}= 50 × 50 sq cm

∴ Number of tiles needed.

{(1000*1000)/50*50} = 400

Hence, 400 tiles will be needed.

**8. Answer :Option A**

(2+5) = 7

(1+6) =7

(3+4) = 7

Hence they will be the opposite faces of each other.

**9. 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 = 2πr

∴ 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 – 1078) cm

^{3}=294cm

^{3}