Chapter 6 - Formulas

Question 1

4 major types of Formula problems

Answer 1

1) Plug in Formula
2) Strange Symbol Formulas
3) Formulas with unspecified amounts
4) Sequence formulas

Question 2

Plug in Formula

Ex: C= QL/J

Answer 2

Trick: unfamiliarity of the given formula bc it will be out of the blue. Simply write the equation down, plug in #’s carefully and solve for required unknown.

Look for language such as “is defined as” to identify what equals what.

Question 3

Strange Symbols Formulas:

Symbols are arbitrary symbol which defines a certain procedure. The symbol is irrelevant. Just carefully follow each step in the procedure that the symbol indicates.

Watch out for symbols that invert the order of an operation. It is easy to automatically translate the function in a “left to right” manner even when that is NOT what the function specfies.

Lastly, be careful and look that the order of operation is not **reversed **

Answer 3

x  y= (√x)^yfor all integers x and y. What is 4  9?

(√4)⁹= 2⁹= 512

x ⌘ y= (√y)^x for all integers x and y. What is 4⌘9? (INVERT)

(√9)⁴ = 3⁴= 81

x ⌘ y= (√y)^xfor all integers x and y. What is 4⌘ (3⌘16)?

(3⌘16) = (√16)³= 4³ = 64 ⇒ Perform procedure inside ( ) 1st.

4 ⌘ 64 = (√64)⁴ = 8⁴ = 4,096

Question 4

Formula with Unspecified Amounts:

Some formula problems are tricky because they never give you real values; they only tell you how the value of a variable has changed.

Answer 4

Ex: Cost is expressed by the formula tb⁴. If b is doubled, the new cost is what percent of the original cost?

You need to express the new cost in terms of the **original cost. **The only diff bw the cost is that b is doubled.

Original cost = tb⁴

New cost = t(2b)⁴= 16tb⁴

So, the new cost is 16 times the org cost. 16 is equiv to 1600%

Question 5

Sequence formulas: sequence is a collection of numbers in a set order.

Answer 5

Ex: If S_n = 15n-7, what is the value of S₇- S₅?

S₇= 15(7)-7 = 105 -7 =98

S₅ = 15(5)-7 = 75-7 = 68

S₇- S₅ = 98-68 = 30

Question 6

Recursive Sequences: A recursive sequence defines each term relative to other terms. When a sequence is defined recursively, the question will have to give you the value of at least one of the terms. Those values can be used to find the value of the desired term.

Answer 6

If a_n = 2a_n-1-4, and a₆ = - 4, what is the value of a₄?

a_n represents the nth term, then a_n-1 is the previous term.

— — -4

a₄ a₅ a₆

a₆ = 2a(_6-1)-4 ⇒ Use a₆ to find the value of a₅

(-4) = 2a₅ -4

0= 2a₅

0= a₅

— 0 -4

a₄ a₅ a₆

a₅ = 2a₄ - 4 ⇒ Use a₅ to find a₄

0= 2a₄ -4

4 = 2a₄

2=a₄

Question 7

Sequence Problems: Alternate Method

Answer 7

If each # in a sequence is 3 more than the previous #, and the 6th # is 32 what is the 100th #?

From the 6th to the 100th term, there are 94 jumps of 3. Since 94 * 3 = 282, there is an inc of 282 from the 6th term to the 100th term.

32 + 282 = 314

Question 8

Sequence and Patterns:

Note: most sequences on GMAT are defined for integer n ≥ 1. That is the sequene S_n almost always starts at S₁. Occassionaly, a sequence might start at S₀ but in that case, you are told that n ≥ 0. So the 1st term in the sequene would be S_0, the 2nd term would be S₁, the 3rd term would be S_2….

Answer 8

If S_{n =}3ⁿ, what is the units digit of S₆₅?

HINT: GMAT doesn’t expect to multiply out 3⁶⁵ on exam. List out power of three.

3¹=3 Note: The pattern of unit digit in the power of 3:

3² =9 3, 9, 7, 1, 3 (repeating). So you know the unit

3³ = 27 digit up to 3⁵. Since the pattern repeats every

3⁴ = 8**1 ** 4 #’s, we can use 4 as an “anchor point” to find

3⁵ = 243 mutliple of 4. Since 65 goes into 4 16 times with a remainder of 1, 65 is 1 more than 64 ( the closest multiple of 4). (Imagine there are 16 buckets of 4 #’s that have a pattern of 3,9,7,1). So th 64th bucket will end at 1 since 1 is the last # in the pattern of 3,9,7,1. So the unit digit of S₆₅ will be 3 which is the 65 #.

Question 9

If S_n = (4ⁿ) + 3, what is the units digit of S₁₀₀?

Answer 9

S₁ = 4¹ + 3 = 4 + 3= 7 ** S₂ = 4² + 3 = 16 + 3 = 19**

S₃ = 4³ + 3 = 64 + 3 = 67 ** S₄ = 4⁴ + 3 = 256 + 3 = 259**

The unit digit of all odd numbered terms is 7. The unit digit of all even numbered terms is 9. Bc S₁₀₀ is an even numbered, its unit digit will be 9. (Look out for odd/even exponents and its affect on the changing value)

Alternative Method:

4¹= 4 ⇒ Find a pattern in the sequence. Here, the #’s

4² = 1**6 **have sequence of 4,6,4,6 so the pattern

4³ = 64 repeats every 2nd #**. **Find the multiple of n

4⁴= 25**6 ** **using the pattern. N is 100 so divide by 2 = 50 with no remainders. Therefore, there are 50 buckets of patterns of 4,6,4,6. 50th bucket will end with unit digit of 6 since 6 is the last # in the bucket of 4,6. 6 + 3 = 9

Question 10

The first term in an arithmetic sequence is -5 and the 2nd term is -3. What is the 50th term. ( Recall that in an arithmetic sequence, the diff bw successive terms is constant.

Answer 10

The first term is -5 and the 2nd term is -3, so you are adding +2 to each successive term. How many times do you have to add 2? There are 50-1 =49 additional “steps” after the 1st term, so you have to add +2 a total of 49 times, beginning with your starting point of -5. -5+ 2(49)= 93