# SAT MATH QUESTIONS WITH ANSWERS - 4

Question 1 :

In three plays, a football team loses 5 yards and then gains 32 yards by completing a pass. Then  a penalty was called and the team lost 10 yards. How many yards did the team actually gain?

= -5 + 32 - 10

= 17 yards

Question 2 :

On the number line above, if BC = 2AB, find the value x.

Length of BC :

BC = C - B

BC = x - (-0.4)

BC = x + 0.4

Length of AB :

AB = B - A

AB = -0.4 - (-1.25)

AB = -0.4 + 1.25

AB = 0.85

Given : BC = 2AB.

x + 0.4 = 2(0.85)

x + 0.4 = 1.7

Subtract 0.4 from both sides.

x = 1.3

Question 3 :

If n > 0, how much is (n + 3) greater than (n - 11)?

(A)  8
(B)  10
(C)  12
(D)  14

To know know how much (n + 3) is greater than (n - 11), subtract the smaller value from the greater value.

= (n + 3) - (n - 11)

= n + 3 - n + 11

= 14

Therefore, (n + 3) is greater than (n - 11) by 14.

The correct answer is option (D).

Question 4 :

What number is halfway between - between ?

(A)  -¼
(B)  -
(C)  -½
(D)  -⁵⁄₁₂

To find a number which is halfway between - between  is to find the average of the given two numbers.

Average of - and ⅓ :

The correct answer is option (A).

Question 5 :

How many minutes are there in 2h hours and 6m minutes?

(A)  60h + 12m

(B)  120h + 6m

(C)  60h + 6m

(D)  120h + 60m

To find the number of minutes in 2h hours and 6m minutes, convert 2h hours to minutes.

Since 1 hour = 60 minutes, we have to multiply 2h hours by 60 to convert it to minutes.

2h hours and 6m minutes :

2h hours + 6m minutes

= 2⋅ 60 minutes + 6m minutes

= 120h minutes + 6m minutes

= (120h + 6m) minutes

The correct answer is option (B).

Question 6 :

If x = 10, what is the value of ˣ⁄₂ ˣ⁄₂₀ ˣ⁄₂₀₀?

= ˣ⁄₂ ˣ⁄₂₀ ˣ⁄₂₀₀

Substitute x = 10.

= ¹⁰⁄₂ ¹⁰⁄₂₀ ¹⁰⁄₂₀₀

= 5 ⁵⁰⁄₁₀₀ ⁵⁰⁄₁₀₀₀

= 5 + 0.5 + 0.05

= 5.55

Question 7 :

If x and y are positive integers and 2x + 5y = 18, what is the value of x?

Since both x and y are posoitve integers,

x = 1, 2, 3,.......

y = 1, 2, 3,.......

We can test the values 1, 2, 3,....... one by one for x.

When x = 1,

2(1) + 5y = 18

2 + 5y = 18

5y = 16

y = ¹⁶⁄₅ (not an integer)

When x = 2,

2(2) + 5y = 18

4 + 5y = 18

5y = 14

y = ¹⁴⁄₅ (not an integer)

When x = 3,

2(3) + 5y = 18

6 + 5y = 18

5y = 12

y = ¹²⁄₅ (not an integer)

When x = 4,

2(4) + 5y = 18

8 + 5y = 18

5y = 10

y = 2 (positive integer)

Therefore,

x = 4

Question 8 :

The sum of four consecutive odd integers is 296. What is the greatest of the four consecutive odd integers?

For example, if 5 is the first odd integer, then the four consecutive odd integers are

5, 7, 9, 11

But, we do not the first odd integer. So, we can assume x as the first odd integer. Then the four consecutive integers are

x, x + 2, x + 4, x + 6

Given :  The sum of four consecutive odd integers is 296.

x + (x + 2) + (x + 4) + (x + 6) = 296

x + x + 2 + x + 4 + x + 6 = 296

4x + 12 = 296

Subtract 12 from both sides.

4x = 284

Divide both sides by 4.

x = 71

The greatest of four consecutive odd integers :

= x + 6

Substitute x = 71

= 71 + 6

= 77

Verify :

When x = 71, the four consecutive odd integers,

71, 73, 75, 77

Sum of the four consecutive odd integers :

= 71 + 73 + 75 + 77

= 296 (verified)

Question 9 :

The product of four consecutive positive integers is 120. Find the smallest of the four consecutive positive integers.

Let us assume the following are the four positive consecutive positive integers.

y, y + 1, y + 2, y + 3

Given : The product of four consecutive positive integers is 120.

y(y + 1)(y + 2)(y + 3) = 120

Solving the above equation for y is a difficult process. Because,  if you multiply the terms on the left side, the exponent of y is 4. Try to write 120 as a product of four consecutive positive integers in ascending order (smallest to greatest).

y(y + 1)(y + 2)(y + 3) = 2 ⋅ 60

y(y + 1)(y + 2)(y + 3) = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 5

y(y + 1)(y + 2)(y + 3) = 2 ⋅ 3 ⋅ (2 ⋅ 2) ⋅ 5

y(y + 1)(y + 2)(y + 3) = 2 ⋅ 3 ⋅ 4 ⋅ 5

Comparing the smallest values on both sides,

y = 2

The smallest of the four consecutive positive integers is 2.

Question 10 :

Find the sum of all three digit numbers which can be formed with the digits 3, 5 and 7, if each digit is used only once in each arrangement.

The list of three digit numbers which can be formed with the digits 3, 5 and 7 (each digit is used only once in each arrangement).

357

375

537

573

735

753

Sum of the above three digit numbers is 3330.

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