SAT PERCENTAGE WORD PROBLEMS

Problem 1 :

Jane is playing a board game in which she must collect as many cards as many cards as possible. On her first turn, she loses 18 percent of her cards. On the second turn, she increases her card count by 36 percent. If her final card count after these two turns is n, which of the following represents her starting card count in terms of n ?

Solution :

Let n be the number of cards initially.

Number of cards she has in the first turn =  82% of n

Number of cards she has in the second turn

= 136% of (82% of n)

= 1.36 (0.82 n)

So, option d is correct.

Problem 2 :

Due to deforestation, researchers expect the deer population to decline by 6 percent every year. If the current deer population is 12000, what is the approximate expected population size 10 years from now ?

a)  4800    b)  6460      c)  7240       d)  7980

Solution :

Initial population of deer = 12000

P(t) = 12000(1 - 6%)^x

P(t) = 12000(1 - 0.06)^x

P(t) = 12000(0.94)^x

When x = 10

P(10) = 12000(0.94)¹⁰

= 6463.38

Accordingly options given, option b is correct.

Problem 3 :

Kyle bought a $2000 government bond that yields 6% in simple interest each year. Which of the following equations gives the total amount A, in dollars. Kyle will receive when he sells the bond after t years ?

a)  A = 2000 (1 + 0.06)t          b)  A = 2000(1 + 0.06t)

c)  A = 2000 (1 + 0.06)^t          d)  A = 2000(1 + 0.06^t)

Solution :

The given situation comes under the concept of exponential growth.

Exponential growth function :

P(x) = a(1 + r%)^x

a = 2000, r = 6% and x = t

A = 2000(1 + 6%)^t

 = 2000(1 + 0.06)^t

 = 2000(1.06)^t

So, option c is correct.

Problem 4 :

A small clothing store sells 3 different types of accessories. 20% are scarves, 60% are ties and the other 40 accessories are belts. If half of the ties are replaced with scarves, how many scarves will the store have ?

Solution :

20% + 60% = 80%

Let x be the total number of accessories.

20% of x = 40 belts

0.20x = 40

x = 40/0.20

x = 200

Number of ties = 60% of 200

= 0.60 (200)

= 120

Half of the ties = 60

Number of scarves already = 20% of 200

= 0.20(200)

= 40

Total number of scarves at present = 60 + 40

= 100

Problem 5 :

Daniel has $1000 in a checking account and $3000 in a savings account. The checking account earns him 1 percent interest compounded annually. The savings account earns him 6 percent interest compounded annually. Assuming he leaves both these accounts alone, how much more interest Daniel will have earned from the savings account than from the checking account after 5 years ?

Solution :

Formula to find amount in compound interest :

A = a(1 + r%)^x

Formula to find interest in compound interest :

I = a(1 + r%)^{x -} A

x = period of investing = 5

Difference : 

= 3000(1 + 0.06)⁵ - 3000 - 1000(1 + 0.01)⁵ - 1000

= 3000(1.06)⁵ - 1000(1.01)⁵ - 4000

Problem 6 :

Kristen opens a bank account that earns 4% interest each year, compounded once every two years. If she opened the account with k dollars, which of the following expression represents the total amount in the account after t years ?

a)  A = k(1.04)^2t          b)  A = k(1.04)^t/2

c)  A = k (1.08)^t          d)  A = k(1.08)^t/2

Solution :

Rate of interest = 4%

Interest calculating once in every two years.

A = a(1 + r%)^t

A = k(1 + 2(4)%)^2t

= k(1 + 8%)^2t

= k(1 + 0.08)^2t

= k(1.08)^2t

Problem 7 :

Before a big test, James memorized 10 percent more words than Zach. Zach memorized 30 percent more words than Amy, If Amy memorized a words, how many words did James memorize in terms of a ?

a)  1.10 (1.30a)      b)  a/1.10 (1.30)       c)  a/(0.9)(0.7)

d)  1.40a

Solution :

Let a be number of words memorized by Amy.

Number of words memorized by Zach = 130% of a

Number of words memorized by James = 110% of 130% of a

= 1.10 (1.30a)

So, option a is correct.

Problem 8 :

If X is 20% of Y and Y is 30% of Z, then what percent of Z is X ?

a)  5%      b) 6%      c)  8%     d) 12%

Solution :

X = 20% of Y

Y = 30% of Z

X = 0.20Y ==> Y = X/0.20  -----(1)

Y = 0.30Z -----(2)

(1) = (2)

X/0.20 = 0.30Z

X = 0.20(0.30Z)

X = 0.06Z

X = 6% of Z

Problem 9 :

The number of subscribers, S to a magazine increases by 21 percent each year. If the current number of subscribers to the magazine is 3000. Which of the following equations models the number of subscribers to the magazine h half year from now ?

a)  S = 3000(1.1)^h      b)  S = 3000(1.21)^h

c)  S = 3000(1.105)^h      b)  S = 3000(1.4641)^h

Solution :

Let t be the number of years that they pass.

S = 3000(1 + 21%)^h/2

= 3000(1 + 0.21)^h/2

= 3000[ (1.21)^1/2 ]^h

= 3000 [√1.21]^h

= 3000 (1.1)^h

Problem 10 :

When a certain gas tank is 70% empty, it contains 12 gallons. How many gallons can a full tank hold ?

a)  30    b)  36     c)  40    d) 42

Solution :

Let x be the capacity of tank.

30% of capacity of tank = 12 gallons

0.30 x = 12

x = 12/0.30

x = 40

Capacity of tank is 40 gallons.