**8th grade math worksheet 4 answer :**

Here we are going to see 10 practice questions of mixed topics. We have been preparing this set of questions in order to test how you understood the topics.

**Question 1 :**

What is the smallest number by which 256 may be divided so that the quotient is perfect cube?

(A) 2 (B) 3 (C) 4

**Solution :**

To find the number required to divide 256 such that the quotient is a perfect cube, we have to decompose 256 into prime factors.

= ∛256 = ∛(2⋅2⋅2⋅2⋅2⋅2⋅2⋅2) When we group the prime factors inside the cube root as triples, we left over with 2⋅ 2. That is 4. Hence 4 is the smallest number required to divide 256 so that the quotient is a perfect cube. |

Let us look into the solution of next problem on "8th grade math worksheet 4 answer".

**Question 2 :**

How many digits are there in the square root of 4456321?

(A) 2 digits (B) 4 digits (C) 3 digits

**Solution :**

Square root of 1 digit and 2 digit number contains 1 digit.

Square root of 3 and 4 digit number contains 2 digits.

Square root of 5 and 6 digit number contains 3 digits.

Square root of 7 and 8 digit number contains 4 digits.

Let us look into the solution of next problem on "8th grade math worksheet 4 answer".

**Question 3 :**

Find the value of 3x^{3} - 4x^{2} + x - 5 when x = -3

(A) -54 (B) -71 (C) -32

**Solution :**

Let f(x) = 3x^{3} - 4x^{2} + x - 5

x = - 3

f(-3) = 3(-3)^{3} - 4(-3)^{2} + (-3) - 5

= 3(9) - 4(9) - 3 - 5

= 27 - 36 - 3 - 5

= 27 - 44

= -17

**Question 4 :**

The perimeter of a rectangle is 52 cm. If its width is 2 cm more than one-third of its length, find the dimensions of rectangle.

(A) 12 cm x 4 cm (B) 18 cm x 8 cm (C) 15 cm x 9 cm

**Solution :**

Perimeter of a rectangle = 52 cm

2(length + width) = 52 ==> l + w = 26

Let "x" be the length

width = (x/3) + 2

x + (x/3) + 2 = 26

[(3x + x)/3] + 2 = 26

4x / 3 = 24

4x = 24 (3)

x = 6(3) ==> 18 cm

Width = (18/3) + 2

= 6 + 2 = 8 cm

Hence the required dimension is 18 cm x 8 cm.

**Question 5 :**

What must be added to each of the numerator and the denominator of the fraction 7/11 to make it equal to 3/4

(A) 9 (B) 5 (C) 12

**Solution :**

Let "x" be the required number to be added to both numerator and denominator of the fraction.

(7 + x) / (11 + x) = 3/4

4(7 + x) = 3(11 + x)

28 + 4x = 33 + 3x

4x - 3x = 33 - 28

x = 5

Hence 5 is the required number to be added to make the fraction 7/11 as 3/4.

**Question 6 :**

A line which intersects two or more lines at a distinct points is called a ____________

(A) concurrent (B) Intersecting (C) Transversal

**Solution :**

A line which intersects two or more lines at a distinct points is called a concurrent.

**Question 7 :**

The opposite angles of a cyclic quadrilateral are _______________

(A) Supplementary (B) Complementary (C) None of these

**Solution :**

The sum of opposite angles of a cyclic quadrilateral is 180 degree.

Hence the answer is supplementary.

**Question 8 :**

In triangle ABC is inscribed in a circle with center O and BC is a diameter, if angle BAC is 50°, find angle ABC.

(A) 50° (B) 40° (C) 48°

**Solution :**

In the right triangle ABC,

<A + <B + <C = 180

50 + <B + 90 = 180

140 + <B = 180

<B = 180 - 140

<B = 40

**Question 9 :**

The circumference of a circle is 44 cm.Find its area (use π=22/7)

(A) 154 cm^{2 }(B) 130 cm^{2 }(C) 145 cm^{2}

**Solution :**

Circumference of a circle = 44 cm

2πr = 44

2 ⋅ (22/7) ⋅ r = 44

r = 7

Area of circle = πr^{2}

= (22/7) ⋅ (7)^{2}

= 154 cm^{2}

**Question 10 :**

The volume of a right circular cylinder is 1100 cm^{3} and the radius of its base is 5 cm. Calculate its curved surface area.

(A) 410 cm^{2 }(B) 125 cm^{2 }(C) 440 cm^{2}

**Solution :**

Volume of a right circular cylinder = 1100 cm^{3}

πr^{2 }h = 1100 cm^{3}

Here r = 5

(22/7) ⋅ (5)^{2} ⋅ h = 1100

h = 2/7

Curved surface area of cylinder = 2 πr^{ }h

= 2 ⋅ (22/7) ⋅ 5 ⋅ (2/7)

= 440 cm^{2}

Hence the curved surface area of cylinder is 440 cm^{2}

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