**Grade 7 math worksheet 15 answers :**

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 :**

In a school the total enrollment of class 8^{th} is 115. If the number of boys exceeds the number of girls by 33, find the number of boys in a class 8^{th}.

(A) 74 (B) 89 (C) 50

**Solution :**

Let "x" be the number of girls

Number of boys = x + 33

Total number of students in the class = 115

x + x + 33 = 115

2x + 33 = 115

Subtract 33 on both sides

2x = 115 - 33

2x = 82

Divide by 2 on both sides

x = 41

x + 33 = 74

Hence the number of boys in the class is 74.

**Question 2 :**

One of the angles of a triangles is equal to the sum of the remaining tow angles. If the ratio of these angles is 4:5, find the angles of the given triangle.

(A) 60, 60, 60 (B) 60, 80, 40 (C) 40, 50, 90

**Solution :**

Let the three angles be "9x", "4x" and "5x"

Sum of three angles = 180°

4x + 5x + 4x + 5x = 180°

8x + 10x = 180°

18x = 180° ==> x = 18°

Hence the angles are 90, 50, 40.

**Question 3 :**

Two equal sides of a triangle are each 4 m less than three times the third side.Find the dimensions of a triangle,if its perimeter is 55 m.

(A) 15, 15, 8 (B) 23, 23, 9 (C) 13, 13, 8

**Solution :**

Let "x" be the third side of the triangle

Length of two equal sides = 3x - 4

Perimeter of triangle = 55 m

3x - 4 + 3x - 4 + x = 55

7x - 8 = 55

Add both sides by 8

7x = 63

Divide both sides by 7

x = 9

3x - 4 = 3(9) - 4 ==> 27 - 4 ==> 23

Hence the required angles are 23, 23, and 9

**Question 4 :**

If a + 1/a = 2, find the value of a^{3} + (1/a)^{3}

(A) 1 (B) 2 (C) 3

**Solution :**

To find the value of a^{3} + (1/a)^{3}, we may use the formula given below.

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

a^{3} + a^{1/3} = (a + (1/a))^{3 }+ 3 a (1/a)(a + (1/a))

= 2^{3} + 3 (2)

= 8 + 6

= 14

Hence the answer is 14.

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 5 :**

A gardener bought twice the number of Lilly plants that he had in his garden but had to throw 3 bad plants. When the new Lilly plants were planted, he had in all 48 plants in the garden.Find the number of Lilly plants he had originally.

(A) 17 (B) 15 (C) 13

**Solution :**

Let "x" be the number of lilly plants in his garden

Number of plants he buys = 2x

2x + x - 3 = 48

3x - 3 = 48

Add both sides by 3

3x = 48 + 3

3x = 51

Divide both sides by 3

x = 51/3 = 17

Hence the number of Lilly plants in the garden is 17.

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 6 :**

Simplify a^{3} - b^{3} ÷ (a-b)

(A) a^{2 }+ ab + b^{2 }(B) a^{2 }- ab + b^{2 }(C) (a+b) (a+b)

**Solution :**

a^{3} - b^{3 }= (a - b) (a^{2 }+ ab + b^{2})

a^{3} - b^{3} ÷ (a - b) = (a - b) (a^{2 }+ ab + b^{2}) / (a - b)

= a^{2 }+ ab + b^{2}

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 7 :**

If a + b + c = 9 and ab + bc + ca = 23, find the value of a^{2} + b^{2} + c^{2}

(A) 15 (B) 12 (C) 35

**Solution :**

(a + b + c)^{2} = a^{2} + b^{2} + c^{2 }+ 2ab + 2bc + 2ca

(a + b + c)^{2} = a^{2} + b^{2} + c^{2 }+ 2(ab + bc + ca)

a^{2} + b^{2} + c^{2 }= (a + b + c)^{2} - 2(ab + bc + ca)

= (9)^{2} - 2 (23)

= 81 - 46 = 35

Hence the answer is 35.

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 8 :**

The difference between the compound interest and the simple interest on a certain principal for 2 years at 4% per annum is $150. Find the principal

(A) 1875 (B) 3750 (C) 1263

**Solution :**

Simple interest = PNR/100

150 = P(2)(4) / 100

150 = 8 P / 100

Multiply both sides by 100

150 (100) = 8 P

Divide both sides by 8

P = 150(100) / 8

= 1875

Hence the principal amount is 1875.

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 9 :**

The watch is marked at $1150. During off season a discount given and it is sold for $1100. Find the discount percent allowed.

(A) 2.14% (B) 4.34% (C) 3.04%

**Solution :**

Discount percentage

= (Discount amount / Marked price) ⋅ 100

Discount amount = 1150 - 1100 = 50

= (50/1150) ⋅ 100

= 0.0434 (100)

= 4.34%

Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".

**Question 10 :**

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) 2 (B) 8 (C) 5

**Solution :**

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

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

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

28 + 4x = 33 + 3x

4x - 3x = 33 - 28

x = 5

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