Quant Problems Flashcards
If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?
The tens digit of x is equal to the sum of the tens digits of y and z.
The units digit of x is equal to the sum of the units digits of y and z.
A large flower arrangement contains 3 types of flowers: carnations, lilies, and roses. Of all the flowers in the arrangement, 1/2 are carnations, 1/3 are lilies, and 1/6 are roses. The total price of which of the 3 types of flowers in the arrangement is the greatest?
(1) The prices per flower for carnations, lilies, and roses are in the ratio 1:3:4, respectively.
(2) The price of one rose is $0.75 more than the price of one carnation, and the price of one rose is $0.25 more than the price of one lily.
STATEMENT (1)–The prices per flower for carnations, lilies, and roses are in the ratio 1:3:4, respectively.
let the price of one carnation = y
one Lilly = 3y
one rose = 4y
the total price of carnation = xy2xy2 = 0.5xy
the total price of Lillies = 3xy33xy3 = xy
the total price of roses = 4xy64xy6 = 0.6xy
from here we can say the price of Lillies are the greatest that is –xy
SUFFICIENT
STATEMENT (2)–The price of one rose is $0.75 more than the price of one carnation, and the price of one rose is $0.25 more than the price of one lily.
let the price of one carnation = y
one Lilly = y+0.50
one rose = y+0.75
the total price of carnation = xy2xy2
the total price of Lillies = x(y+0.5)3x(y+0.5)3
the total price of roses = x(y+0.75)6x(y+0.75)6
sine we dont know the value of x and y we cant decide which value is the greatest
INSUFFICIENT
A is the answer
Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the preceding year. If all of the trees thrived and there were 6,250 trees in the orchard at the end of the 4-year period, how many trees were in the orchard at the beginning of the 4-year period?
Increasing the number of trees each year by of the number of trees in the orchard the preceding year is equivalent to making the number of trees increase 25% per year, compounded yearly. If there were n trees at the beginning of the 4-year period, then there will be 1.25n trees at the end of the first year, 1.25(1.25n) = (1.25)2n trees at the end of the second year, 1.25[(1.25)2n] = (1.25)3n trees at the end of the third year, and 1.25[(1.25)3n] = (1.25)4n trees at the end of the fourth year. Hence, 6,250 = (1.25)4n and n = . The arithmetic can be greatly simplified by rewriting (1.25)4 as and 6,250 as (625)(10) = (54)(10). So 2,560
If r > 0 and s > 0, is r/s < s/r?
(1) 4r = 3s
(2) s = r + 4
READ the prompt (both positive). Answer is D
If n = (33)43 + (43)33, what is the units digit of n ?
0
2
4
6
8
Answer is A (0)
When positive integer x is divided by positive integer y, the remainder is 9. If x/y = 96.12, what is the value of y?
(A) 96
(B) 75
(C) 48
(D) 25
(E) 12
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are
A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8
Step 1: Glance Read Jot
At first glance, this problem is an annoying wall of text; expect to take some time to translate this one. The answer choices are in an unusual form: two numbers per choice.
The story setup is unusual. There are 2 classes in each of 32 elementary schools (for a total of 64 classes) and there are 37 teachers. A class had exactly 1 teacher and each teacher taught from 1 to 3 classes.
32 schools @ 2 classes each = 64 classes total
37 teachers total
Each teacher = 1, 2, or 3 classes
This is shaping up to be a logic word problem. If you’re not a fan of this type, this might be a great time to choose your favorite letter and move on.
The problem asks for the least and greatest possible values for the number of teachers who taught 3 classes (n). In other words, this is also a max / min problem.
Step 2: Reflect Organize
Okay, regroup. What do you know so far?
Each teacher teaches at least 1 class, so that covers 37 classes. There are 64 classes total, so there are still 64 – 37 = 27 classes to be assigned as second or third classes among the teachers.
Glance again at the answers. They’re quite limited: for instance, the smallest possible value for n must be 0, 1, or 2. You can Work Backwards from the answers, testing the limited possible values for n.
Step 3: Work
Because the problem asks for the least possible value for n, try the smallest number, 0, first. (If you tried 1 and it worked, it could still be the case that 0 would work, too, in which case you would still have to try 0. So start there!)
Is it possible that none of the teachers taught 3 classes? Let’s assign the classes to the teachers to explore the possibilities.
First class: all 37 teachers are assigned a first class, leaving 64 – 37 = 27 more classes to assign
Second class: assign a second class to 27 of the 37 teachers, leaving no more classes
Third class: all 64 classes have already been assigned, so no teacher teaches a third class in this case
Yes, it is possible that n = 0. Cross off answers (C), (D), and (E).
Now, test out the largest possible number remaining, 14. Can n = 14?
If n = 14, then 14 teachers teach 3 classes each, covering 14 ´ 3 = 42 classes. There are 64 classes total, so there are still 64 – 42 = 22 more classes to assign.
There are 37 teachers total, but 14 have already been given their classes, so there are 37 – 14 = 23 teachers that still need to be given classes.
If 14 teachers each taught three classes, then there would be only 22 more classes to assign to the 23 remaining teachers. One teacher wouldn’t have a class to teach. Since the problem indicates that each teacher does teach at least 1 class, n cannot be 14. Eliminate answer (B).
Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
3 members at one table and 4 members at each of the other tables, means that the total number of members is 3 more than a multiple of 4: x=4m+3.
3 members at one table and 5 members at each of the other tables, means that the total number of members is 3 more than a multiple of 5: x=5n+3.
Thus x-3 is a multiple of both 4 and 5, so a multiple of 20. Therefore x is 3 more than a multiple of 20. Since 10
The remainder when 23 is divided by 6 is 5.
Answer: E.
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?
A) (m+4)/[m(m-4)]
B) (2m-4)/[m(m-4)]
C) (4m-4)/[m(m-4)]
D) (4m-8)/[m(m-4)]
E) (4m-12)/[m(m-4)]
Step 1: Glance Read Jot
This story problem has variables in the answer choices; this is an indication that you can use Smart Numbers.
Jot down what you are given: m = people, m > 4. The information about the specific sandwiches is fairly complicated so reflect on the information before writing anything else down. Finally, make a note that you are solving for the amount that Carol eats.
Step 2: Reflect Organize
The variable, m, is a “simple” input into the problem and therefore is a good candidate for smart numbers; if m had represented a complex value such as an average or difference, then smart numbers might be harder to apply. The complexity of the answer choices also indicates that smart numbers is potentially easier than an algebraic approach. If you would like to see the algebraic solution, please see the Official Guide.
Step 3: Work
To apply smart numbers to a problem you need to decide two things: (1)which unknown you will replace with a number, and (2) what value you will assign to that unknown.
The only unknown in the problem is the number of students, m, so assign the smart number here. The only restriction on m is that it is a number of students (and integer) and it is greater than 4; choose the next largest integer, 5.
You are told that Carol eats a piece from each sandwich. If all 5 share the first sandwich, Carol eats 1515. The same is true for the second and the third sandwich, for a total so far of 15+15+15=3515+15+15=35 of a sandwich. The final sandwich is a bit different; 4 of the students did not want any, so Carol ate the entire thing. In total she eats 35+1=35+55=8535+1=35+55=85 of a sandwich.
The final step is to plug the smart number, m = 5, in for m in the answer choices. For every choice the denominator is the same: 5(5-4) = 5; therefore you only need to find the answer with a numerator of 8.
(A) 5+4 = 9
(B) 2(5) – 4 = 10 – 4 = 6
(C) 4(5) – 4 = 20 – 4 = 16
(D) 4(5) – 8 = 20 – 8 = 12
(E) 4(5) – 12 = 20 – 12 = 8 CORRECT.
The correct answer is (E).
If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?
(1) The tens digit of x is equal to the sum of the tens digits of y and z.
(2) The units digit of x is equal to the sum of the units digits of y and z.
A
A school administrator will assign each student in a group of N students to one of M classrooms. If 3
(1) It is possible to assign each of 3N students to one of M classrooms so that each classroom has the same number of students assigned to it.
(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it.
Basically the question asks whether nn (# of students) is a multiple of mm (# of classrooms), or whether nm=integernm=integer, because if it is then we would be able to assign students to classrooms so that each classroom has the same number of students assigned to it.
Given: 3
(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it –> 3nm=integer3nm=integer, from this we cannot say whether nm=integernm=integer. For example nn indeed might be a multiple of mm (n=14n=14 and m=7m=7) but also it as well might not be (n=14n=14 and m=6m=6). Not sufficient.
(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it –> 13nm=integer13nm=integer, now as given that 3
Answer: B.
