Aladin_Baniel_GE-1B
EXERCISE VI. A
43. The length of a rectangle is 3 inches greater than its width. Its area is 70 square inches.
What are
its dimensions?
Step 1: Given
a rectangle with a length 3 inches greater than its width
it has an area of 70 square inches
Required: dimensions of the rectangle
Step 2: Assign variables
Let x = width of the rectangle
x + 3 = length of the rectangle
x+3
x
Step 3: Write an equation
Area of rectangle = (length)(width)
70 = (x + 3)(x)
Step 4: Solve
70 = (𝑥 + 3)(𝑥)
70 = 𝑥 2 + 3𝑥
𝑥 2 + 3𝑥 − 70 = 0
(𝑥 + 10)(𝑥 − 7) = 0
𝑥 + 10 = 0 𝑥 − 7 = 0
𝑥 = −10 𝑥 = 7
Step 5: State the answer
In this equation the answer cannot be negative therefore the length of the rectangle is 10
inches
and the width is 7 inches.
Step 6: Check
(7)(7 + 3) = 70
(7)(10) = 70
70 = 70 
Aladin_Baniel_GE-1B
44. The diagonal of a rectangle is 2 inches greater than its length and 9 inches greater than its
width.
Find the dimensions of the rectangle.
Step 1: Given
a rectangle with a diagonal 2 inches greater than length and 9 inches greater than width
Required: dimensions of the rectangle
Step 2: Assign variables
Let x = diagonal of the rectangle
x – 2 = length of the rectangle
x – 9 = width of the rectangle
x- 2
x
x- 9
Step 3: write an equation
𝑥 2 = (𝑥 − 2)2 + (𝑥 − 9)2
Derived from Pythagorean Theorem
Step 4: Solve
𝑥 2 = (𝑥 − 2)2 + (𝑥 − 9)2
𝑥 2 = 𝑥 2 − 4𝑥 + 4 + 𝑥 2 − 18𝑥 + 81
𝑥 2 = 2𝑥 2 − 22𝑥 + 85
𝑥 2 − 22𝑥 + 85 = 0
(𝑥 − 17)(𝑥 − 5) = 0
𝑥 − 17 = 0
𝑥−5=0
𝑥 = 17
𝑥=5
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 𝑙𝑤
(15)(8) = 90
Step 5:State the answer
In this equation, the sides of the rectangle should not be negative, therefore, the
length of the
rectangle is 15 inches and the width is 8
inches of which the diagonal is 17 inches.
Step 6:Check
172 = 152 + 82
289 = 225 + 64
289 = 289 
Aladin_Baniel_GE-1B
45. Find two consecutive positive integers whose product is 1332.
Step: Given
two consecutive integers
product is 1332
Required: the two integers
Step 2: Assign variables
Let x = first integer
x + 1 = second integer
Step 3: Write an equation
(x)(x + 1) = 1332
Step 4: Solve
(𝑥)(𝑥 + 1) = 1332
𝑥 2 + 𝑥 = 1332
2
𝑥 + 𝑥 − 1332 = 0
(𝑥 + 37)(𝑥 − 36) = 0
𝑥 + 37 = 0 𝑥 − 36 = 0
𝑥 = −37
𝑥 = 36
Step 5: State the answer
In this equation the answer should not be negative, therefore the first integer is 36 and
the second integer is 37.
Step 6: Check
(36)(36 + 1) = 1332
(36)(37)=1332 
Aladin_Baniel_GE-1B
47. Find two consecutive positive positive odd integers whose product is 783.
Step 1: Given
two consecutive odd integers
their product is 783
Required: the integers
Step 2: Assign variables
Let x = first integer
x + 2 = second integer
Step 3: Write an equation
(x)(x+2) = 783
Step 4: Solve
(𝑥)(𝑥 + 2) = 783
𝑥 2 + 2𝑥 = 783
2
𝑥 + 2𝑥 − 783 = 0
(𝑥 − 27)(𝑥 + 29) = 0
𝑥 − 27 = 0
𝑥 + 29 = 0
𝑥 = 27
𝑥 = −29
Step 5: State the answer
In this equation the answer cannot be negative, therefore the first integer is 27 and the
second integer is 29.
Step 6: Check
(27)(27+2) = 783
(27)(29) = 783
783 = 783 
Aladin_Baniel_GE-1B
48. One number is ten more than another. Their product is 299. What are the numbers?
Step 1: Given
two numbers
the second number is ten more than the other
their product is 299
Required: the numbers
Step 2: Assign variables
Let x = first number
x + 10 = second number
Step 3: Write an equation
(x)(x+10) = 299
Step 4: Solve
(𝑥)(𝑥 + 10) = 299
𝑥 2 + 10𝑥 = 299
𝑥 2 + 10𝑥 − 299 = 0
(𝑥 − 13)(𝑥 + 23) = 0
𝑥 − 13 = 0 𝑥 + 23 = 0
𝑥 = 13
𝑥 = −23
Step 5: State the answer
In this equation the answer cannot be negative, therefore the first number 13 and the
second number is23
Step 6: Check
(13)(13+10) = 299
(13)(23) = 299
299 = 299 
Aladin_Baniel_GE-1B
50. A tray with square base and a volume of 648 cubic inches is to be constructed from a square
sheet of tin by cutting 2-inch squares from the corners and bending up the sides. What size of
sheet of tin should be used?
Step 1: Given
a tray with a volume of 648
is made from a square sheet that is cut 2-inch square from its edge
Required: size of the square sheet
Step 2: Assign variables
Let x = side of the square base of the tray
x + 4 = side of the square sheet
2
x
2
2
2
x
x
2
2
2
x
X
Volume = 648cubic inch
2 inch
2
Step 3: Write an equation
(x)(x)(2) = 648
Step 4: Solve
(𝑥)(𝑥)(2) = 648
2𝑥 2 = 648
2𝑥 2
2
=
Side of square sheet = 18 + 4
= 22 inch
648
2
𝑥 = √324
𝑥 = 18
Step 5: State the answer
Therefore the size of the square sheet should be 22x22 suare inch.
Step 6: Check
(18)(18)(2) = 648
648 = 648 
Aladin_Baniel_GE-1B
51. A rectangular sheet of cardboard is twice as long as it is wide. An open box having a capacity
of 168 cubic inches is made by cutting 3 inch squares from the cornes and banding up the sides
and ends. Find the dimensions of the cardboard.
Step 1: Given
a rectangular sheet of card board having a length of twice as long as its width
its cut 3 inch squares from its corners
making an open box that has a volume of 168 cubic inch
Required: dimensions of the rectangular sheet of cardboard
Step 2: Assign variables
Let x = width of the rectangular sheet of cardboard
2x = length of the rectangular sheet of cardboard
x – 6 = width of the box
2x – 6= length of the box
3
2x
3
3
3
x
3
2x-6
3
3
x-6
3inch
3
Volume=168 cubic inch
Step 3: Write an equation
(2x – 6)(x – 6)(3) = 168
Step 4: Solve
(2𝑥 − 6)(𝑥 − 6)(3) = 168
(2𝑥 − 6)(3𝑥 − 18) = 168
2
6𝑥 − 36𝑥 − 18𝑥 + 108 = 168
6𝑥 2 − 54𝑥 + 108 − 168 = 0
6𝑥 2 − 54𝑥 − 60 = 0
(6𝑥 − 60)(𝑥 + 1) = 0
6𝑥 − 60 = 0
𝑥+1=0
6𝑥
60
=6
𝑥 = −1
6
𝑥 = 10
Step 5: State the answer
In this equation the answer cannot be negative, therefore the width of the rectangular
sheet is 10inches and the length is 20inches and its area is 200-square inch.
Step 6: Check
(10 – 6)(20 – 6)(3)=168
(4)(14)(3)=168
168 = 168 
Aladin_Baniel_GE-1B
52. A party of fishermen rented a boat for $240. Two of the men had to withdraw from the
party and, as a result, the share of each of the others was increased by $10. How many were in
the original party?
Step 1: Given
a group of fishermen rent a boat for $240
if two of them leave the group each payment will increase by $10
Required: number of fishermen in the original group
Step 2: Assign variables
Let x = number of original group of fishermen
Step 3: Write an equation
240
240
=
− 10
𝑥
𝑥−2
Step 4: Solve
240
𝑥
240
𝑥
240
𝑥
240
= 𝑥−2 − 10
=
=
240−10𝑥+20
𝑥−2
260−10𝑥
𝑥−2
240𝑥 − 480 = −10𝑥 2
10𝑥 2 + 240𝑥 − 260𝑥 − 480 = 0
10𝑥 2 + 20𝑥 − 480 = 0
(5𝑥 + 30)(2𝑥 − 16) = 0
5𝑥 + 30 = 0
2𝑥 − 16 = 0
5𝑥
−30
2𝑥
16
= 5
= 2
5
2
𝑥 = −6
𝑥=8
Step 5: State the answer
In this equation the answer cannot be negative, therefore the number of the
original group of fishermen is 8.
Step 6: Check
240
Payment of original group = 8
= 30
240
When two of the men left = 6
= 40
40 – 30 = 10 
Aladin_Baniel_GE-1B
53. A man drove 156 miles at a constant rate. If he had driven 5 miles per hour faster he would
have made the trip in 12 minutes less time. How fast did he drive?
Step 1: Given
A man drove 156 miles at a constant rate
He had driven 5 miles per hour faster
He would have made the trip in 12 minutes less time
Required: How fast did he drive?
Step 2: Assign values
Let 156 miles=distance
x mph=rate
156/x hours=time
Let 156 miles=distance
x+5 mph=rate
156 miles/x+5 mph=time
Step 3: Write an equation
Note that 12 minutes is equal to 1/5 hour…
156/x-(156/(x+5))=1/5 hour
Step 4:Solve
156
𝑥
156
1
− (𝑥+5) = 5 ℎ𝑜𝑢𝑟
𝑥(𝑥+5)
156(𝑥 + 5) − 156 = 5
156 ∗ 5 ∗ 5 = 𝑥 2 + 5𝑥
𝑥 2 + 5𝑥 − 3900 = 0
(𝑥 − 60)(𝑥 + 65) = 0
𝑥 − 60 = 0 𝑥 + 65 = 0
𝑥 = 60
𝑥 = −65
Step 5:State the answer
In this equation the answer cannot be negative, therefore the man drove at the speed
of 60 mph
Step 6:Check
156
𝑥
156
60
156
1
− (𝑥+5) = 5 ℎ𝑜𝑢𝑟
−
156
60+5
=
1
5
156(65)−156(60)
(60)(65)
1
5
1
=5
1
=5
Aladin_Baniel_GE-1B
54. An airplane flying north at the rate of 240 miles per hour passed over a flying field at noon.
A second plane flying east at 200 miles per hour passed over the same field 5 minutes later.
When were they 250 miles apart?
Step 1: Given
an airplane flying north at the rate of 240mph passed a over a flying field
a second airplane flying east at the rate of 200mph passed over the same field 5 mins
later
Required: time in which they were 250 miles apart
Step 2: Assign variables and values
Let t= no. of minutes after 12 noon when the two planes were 250 miles apart
240/60 t=4t= no. of miles that the first airplane travel
200/60 t=3.33(t-5) = no. of miles that the second airplane travel
Step 3: Write an equation
200
(4𝑡)2 + ( 60 (𝑡 − 5))2 = 2502
Step 4: Solve
200
2
(4𝑡)2 + ( 60 (𝑡 − 5)) = 2502
16𝑡 2 + 11.1𝑡 2 − 111.1𝑡 + 277.79 = 2502
27.1𝑡 2 − 111.1𝑡 − 62 222.2 = 0
Solve for t using the quadratic equation
𝑡=
𝑡=
−𝑏±√𝑏 2 −4𝑎𝑐
2𝑎
−(−111.1)±√(−111.1)2 −4(27.1)(62 222.2)
𝑡 = 50
2(27.1)
𝑡 = −45.9
Step 5: State the answer
Therefore the planes were 250 miles apart at 12:50 pm and 11:14 am.
Step 6: Check
2
(200)2 + (3.3333333(45)) = (250)2
40000 + 22500 = 62500
62500 = 62500 