PRACTICE PROBLEMS ON LINEAR EQUATIONS IN ONE VARIABLE FOR DIGITAL SAT

Problem 1 :

A) 2     B) 4       C) 9          D) 10

Solution:

So, option (D) is correct.

Problem 2 :

If 16 + 4x is 10 more than 14, what is the value of 8x?

A) 2       B) 6          C) 16          D) 80

Solution:

16 + 4x = 10 + 14

16 + 4x = 24

4x = 24 - 16

4x = 8

x = 2

8x = 8(2)

= 16

So, option (C) is correct.

Problem 3 :

A) 10       B) 5        C) 2          D) 1/2

Solution:

So, option (C) is correct.

Problem 4 :

Solution:

Problem 5 :

If 4x + 2 = 12, what is the value of 16x + 8 ?

A) 40      B) 48        C) 56            D) 60

Solution:

4x + 2 = 12

4x = 10

x = 5/2

16x + 8 = 16(5/2) + 8

= 40 + 8

16x + 8 = 48

So, option (B) is correct.

Problem 6 :

If 3r = 18, what is the value of 6r + 3 ?

A) 6        B) 27           C) 36         D) 39

Solution:

3r = 18

r = 6

By applying r = 6

6r + 3 = 6(6) + 3

= 36 + 3

= 39

So, option (D) is correct.

Problem 7 :

Solution:

So, option (D) is correct.

Problem 8 :

Solution:

So, option (D) is correct.

Problem 9 :

The number of states that joined the United States between 1776 and 1849 is twice the number of states that joined between 1850 and 1900. If 30 states joined the United States between 1776 and 1849 and x states joined between 1850 and 1900, which of the following equations is true?

A) 30x = 2      B) 2x = 30        C) x/2 = 30            D) x + 30 = 2

Solution:

To fit scenario described, 30 must be twice as large as x.

This can be written as 2x = 30.

So, option (B) is correct.

Problem 10 :

At a lunch stand, each hamburger has 50 more calories than each order of fires. If 2 hamburgers and 3 orders of fries have a total of 1700 calories, how many calories does a hamburger have?

Solution:

Let h be the calories of in a hamburger

f be the calories in an order of fries.

h = f + 50 ---> (1)

2h + 3f = 1700 ---> (2)

Substitute (1) in (2),

2(f + 50) + 3f = 1700

2f + 100 + 3f = 1700

5f + 100 = 1700

5f = 1600

f = 320

Substitute f = 320 in (1)

h = f + 50

h = 320 + 50

h = 370

Hence, the hamburger has 370 calories.

Problem 11 :

The density of an object is equal to the mass of the object divided by the volume of the object. What is the volume, in milliliters, of an object with a mass of 24 grams and a density of 3 grams per milliliters?

A) 0.125       B) 8        C) 21        D) 72

Solution:

Given, mass = 24 grams

Density = 3 grams

So, option (B) is correct.

Problem 12 :

Last week Raul worked 11 more hours than Angelica. If they worked a combined total of 59 hours, how many hours did Angelica work last week?

A) 24          B) 35          C) 40          D) 48

Solution:

If Angelica worked a hours, then Raul worked a + 11 hours.

Total hours worked = a + a + 11 = 59

2a + 11 = 59

2a = 48

a = 48/2

a = 24 hours

So, option (A) is correct.

Problem 13 :

A partially filled pool contains 600 gallons of water. A hose is turned on, and water flows into the pool at the rate of 8 gallons per minute. How many gallons of water will be in the pool after 70 minutes?

Solution:

In 70 minutes, gallons of water flowing into the pool

= 70 × 8

= 560

Total water = 600 + 560

= 1160 gallons

Problem 14 :

At a restaurant, n cups of tea are made by adding t tea bags to hot water. If t = n + 2, how many additional tea bags are needed to make each additional cup of tea?

A) None         B) One         C) Two            D) Three

Solution:

Here we need to have n +1 cups

t = n + 1 + 2

t = n + 3

When n = 1

t = 1 + 2

t = 3

When n = 2

t = 2 + 2

t = 4

When n = 3

t = 3 + 2

t = 5

For 1 unit increase in n, t is also increasing by 1.

The additional tea bags that are needed to make each additional cup of tea is one.

So, option (B) is correct.