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Grade 7 Math Word Problems

The document provides detailed solutions and explanations to 7 math word problems involving fractions, percentages, ratios, and other concepts. The problems cover topics like finding the number of balls of a certain color in a bag, determining the number of students in different age groups at a school, calculating ratios of areas when dimensions are changed, and more. Step-by-step workings are shown to systematically arrive at the final answers for each problem.

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Aya Tanquis
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0% found this document useful (0 votes)
1K views13 pages

Grade 7 Math Word Problems

The document provides detailed solutions and explanations to 7 math word problems involving fractions, percentages, ratios, and other concepts. The problems cover topics like finding the number of balls of a certain color in a bag, determining the number of students in different age groups at a school, calculating ratios of areas when dimensions are changed, and more. Step-by-step workings are shown to systematically arrive at the final answers for each problem.

Uploaded by

Aya Tanquis
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Grade 7 Math word Problems

With Solutions and Explanations


Detailed solutions and full explanations to grade 7 math word problems are
presented.

1. In a bag of small balls 1/4 are green, 1/8 are blue, 1/12 are yellow and the remaining 26
white. How many balls are blue?

Solution

Let us first find the fraction of green, blue and yellow balls

1/4 + 1/8 + 1/12 = 6 / 24 + 3 / 24 + 2 / 24 , common denominator

= 11 / 24 , add numerators

The fraction of white balls is given by

24 / 24 - 11 / 24 = 13 / 24

So the fraction 13 / 24 corresponds to 26 balls. If x is the total number of balls then

(13 / 24) of x = 26 balls

or (13 / 24) × x = 26

x = 26 × (24 / 13) = 48 , total number of balls

The fraction of blue balls is 1 / 8 of x. The number of blue balls is given by

(1 / 8) of 48 = 6 balls.

2. In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are
older than 10 but younger than 12, the remaining 70 students are 12 years or older. How
many students are 10 years old?

Solution

Let us write the fraction for each group of students

Group A: younger than 10: 50% = 50/100 = 1/2


Group B: 10 years old: 1/20

Group C: older that 10 but younger than 12: 1/10

Group D: 12 years or older: 70 students

The fraction for group A, B and C together is given by

1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator

= 13 / 20 , add numerators

The fraction for group D is given by

20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students

If X is the total number of students, then

7 / 20 of X = 70

or (7 / 20) × X = 70

Solve for X

X = 70 × (20 / 7) = 200

Students who are 10 years old have a fraction of 1 / 20 of the total X and their number is
equal to

(1 / 20) of 200 = 10 students

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3. If the length of the side of a square is doubled, what is the ratio of the areas of the original
square to the area of the new square?

Solution

If x be the side of the original square, then its area is equal to

x2

If x is doubled to 2x, then the new area is equal to

(2x)2 = 4 x2

The ratio of the areas of the original square to the area of the new square
x2 / (4 x2) = 1 / 4 or 1:4

4. The division of a whole number N by 13 gives a quotient of 15 and a remainder of 2. Find


N.

Solution

According to the division process of whole numbers, N can be written, using multiplication,
as follows

N = quotient × divisor + remainder = 15 × 13 + 2 = 197

5. In the rectangle below, the line MN cuts the rectangle into two regions. Find x the length of
segment NB so that the area of the quadrilateral MNBC is 40% of the total area of the
rectangle.

Solution

We first note that

MC = 20 - 5 = 15

The quadrilateral MNBC is a trapezoid and its area A is given by

A = (1/2) × 10 × (x + MC) = 5 (x + 15)


40% of the area of the rectangle is equal to

40% × (20 × 10) = (40 / 100) × 200 = 80

Since the area of MNBC is equal to 40% the area of the rectangle, we can write

5(x + 15) = 80

5x + 75 = 80

5x = 5

x = 1 meter

6. A person jogged 10 times along the perimeter of a rectangular field at the rate of 12
kilometers per hour for 30 minutes. If field has a length that is twice its width, find the area
of the field in square meters.

Solution

Let us first find the distance d jogged

distance = rate × time = (12 km / hr) × 30 minutes

= (12 km/hr) × 0.5 hr = 6 km

The distance of 6 km corresponds to 10 perimeters and therefore 1 perimeter is equal to

6 km / 10 = 0.6 km = 0.6 × 1000 meters = 600 meters

Let L and W be the length and width of the field. The length is twice the width. Hence

L=2W

The perimeter is 600 meters and is given by

2 (L + W) = 600

Substitute L by 2 W

2 (2 W + W) = 600

Simplify and solve for W

4 W + 2 W = 600

6 W = 600
W = 100

Find L

L = 2 W = 200

Find the area A of the rectangle

A = L * W = 200 * 100 = 20,000 square meters

7. Four congruent isosceles right triangles are cut from the 4 corners of a square with a side
of 20 units. The length of one leg of the triangles is equal to 4 units. What is the area of the
remaining octagon?

Solution

Let us first find area A of the square

A = 20 × 20 = 400 square units

The area B of a small triangle is

B = (1/2) × 4 × 4 = 8 square units

The area of the octagon obtained by subtracting the ares of the 4 triangles from the area of
the large square

A - 4 B = 200 - 4 × 8 = 168 square units


8. A car is traveling 75 kilometers per hour. How many meters does the car travel in one
minute?

Solution

Convert hour into minutes ( 1 hour = 60 minutes) and kilometers into meters (1 km = 1000
m) and simplify

75 kilometers per hour = 75 km/hr

= (75 × 1000 meters) / (60 minutes) = 1,250 meters / minute

9. Linda spent 3/4 of her savings on furniture and the rest on a TV. If the TV cost her $200,
what were her original savings?

Solution

If Linda spent 3/4 of her savings on furnitute, the rest

4 / 4 - 3 / 4 = 1 / 4 on a TV

But the TV cost her $200. So 1 / 4 of her savings is $200. So her original savings are 4
times $200 = $800

10. Stuart bought a sweater on sale for 30% off the original price and another 25% off the
discounted price. If the original price of the sweater was $30, what was the final price of
the sweater?

Solution

The price with 30% off

30 - 30% of 30 = 30 - (30 / 100) × 30 = 30 - 9 = 21

The price with another 25% off


21 - 25% of 21 = 21 - (25/100) × 21

= 21 - (525 / 100) = 21 - 5.25 = $15.75

11. 15 cm is the height of water in a cylindrical container of radius r. What is the height of this
quantity of water if it is poured into a cylindrical container of radius 2r?

Solution

The volume V of water in the container of radius r

V1 = 15*(πr2)

The volume V of water in the container of radius 2r

V2 = H * (2π (2r)2) (H is the height to be found)

Since is it is the same quantity of water, V1 = V2

15*(πr2) = H * (2π (2r)2)

Solve the above for H to obtain

H = 15/4 = 3.75 cm

12. John bought a shirt on sale for 25% off the original price and another 25 % off the
discounted price. If the final price was $16, what was the price before the first discount?

Solution

let x be the price before the first discount. The price after the first discount is

x - 25%x (price after first discount)

A second discount of 25% of the discounted price after which the final price is 16

(x - 25%x) - 25%(x - 25%x) = 16

Solve for x

x = $28.44

13. How many inches are in 2000 millimeters? (round your answer to the nearest hundredth of
of an inch).
Solution

One inch is the same as 25.4 mm. Let x inches be the same as 1000 mm

x = 1 inch * 2000 mm / 25.4 mm = 78.74 inches

14. The rectangular playground in Tim's school is three times as long as it is wide. The area of
the playground is 75 square meters. What is the perimeter of the playground?

Solution

Let L be the length and W be the width of the playground. "The rectangular playground in
Tim's school is three times as long as it is wide means":

L=3W

The area A = L * W. Hence

75 = L * W = (3W) * W = 3 W 2

Solve for W

3 W 2 = 75 give W = 75/3 = 25 gives W = &sqrt;(25) = 5 m

L = 3 W = 3 * 5 = 15 m

Perimeter = 2L + 2W = 2(15) + 2(5) = 40 m

15. John had a stock of 1200 books in his bookshop. He sold 75 on Monday, 50 on Tuesday,
64 on Wednesday, 78 on Thursday and 135 on Friday. What percentage of the books were
not sold?

Solution

Let N be the total number of books sold. Hence

N = 75 + 50 + 64 + 78 + 135 = 402

Let M be the books NOT sold

M = 1200 - N = 1200 - 402 = 798

Percentage

Books not sold / total number of books = 798/1200 = 0.665 = 66.5%


16. N is one of the numbers below. N is such that when multiplied by 0.75 gives 1. Which
number is equal to N?

Solution

"N is such that when multiplied by 0.75 gives 1" is written mathematically as

N * 0.75 = 1

Solve for N

N = 1/0.75 = 100/75 = (75 + 25) / 75 = 75/75 + 25/75 = 1+1/3 Answer: B

A) 1 1/2
B) 1 1/3
C) 5/3
D) 3/2

17. In 2008, the world population was about 6,760,000,000. Write the 2008 world population in
scientific notation.

Solution

a number in scientific notation is written as

m * 10n , such that |m| is greater than or equal to 1 and smaller than 10.

6,760,000,000 = 6.76 * 109

18. Calculate the circumference of a circular field whose radius is 5 centimeters.

Solution

Circumference C is given by

C = 2πr = 2π*5 = 10 π cm

Answers to the Above Questions

2. 6 balls are blue


3. 10 students are 10 years old
4. 1:4
5. N = 197
6. x = 1 meter
7. 20,000 square meters
8. 368 square units
9. 1250 meters per minute
10. $800
11. $15.75
12. 3.75 cm
13. $28.44
14. 78.74 inches
15. 40 meters
16. 66.5%
17. B
18. 6.76 109
19. 10π centimeters

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