GMAT Problem Solving Questions
August 27th, 2012 | 
Author: 1. How many integers less than 1000 have no factors (other than 1) in common with 1000?
(1) 400
(2) 410
(3) 411
(4) 412
(5) None of the above
Soln: 1000 – multiples of 2 and/or 5
multiples of 2 = 500 (all even #)
multiples of 5 = (995 -5)/10 + 1 [ Using AP formula]
= 100
Answer = 1000 – (500 + 100)
= 400
You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The difference in the AP series should be 10 instead of 5 because you’re looking for the integers that have 5 as a unit’s digit. Therefore we divide by 10 and not 5.
2. Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers’ sum was divided by the same divisor, the remainder was 4. What was the divisor?
36, 28, 12, 9 or none
Soln: Let the divisor be a.
x = a*n + 11 —- (1)
y = a*m + 21 —– (2)
also given, (x+y) = a*p + 4 —— (3)
adding the first 2 equations. (x+y) = a*(n+m) + 32 —– (4)
equate 3 and 4.
a*p + 4 = a*(n+m) + 32
or
a*p + 4 = [a*(n+m) + 28] + 4
cancel 4 on both sides.
u will end up with.
a*p = a*(n+m) + 28.
which implies that 28 should be divisible by a. or in short a = 28 works.
Another method:
I think the easiest (not necessarily the shortest), way to solve this is to use given answer choices. Since the remainders are given as 11 and 21, therefore the divisor has to be greater than 21 which leaves with two choices 28 and 36. Try 28 first; let the two numbers be 28+11= 39 and 28+21= 49. Summing them up and dividing by 28 gives (49+39=88), 88/28 remainder is 4, satisfies the given conditions. Check for 36 with same approach, does not work, answer is 28
3. There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?
Soln: let’s assume Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.
so the total is (1c1.6c3)/8c4 which is 2/7
4. How many 5-digit positive integers exist where no two consecutive digits are the same?
A.) 9*9*8*7*6
B.) 9*9*8*8*8
C.) 9^5
D.) 9*8^4
E.) 10*9^4
Soln: C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5
5. How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
(A) 15
(B) 96
(C) 216
(D) 120
(E) 180
Soln: The sum of digits of a multiple of 3 should be div by 3.
for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15.
For a sum of 12, the digits that can be used : 0,1,2,4,5
for a sum of 15: 1,2,3,4,5
Number of numbers from the first set = 4.4! (0 cannot be the first digit in the numbers)
for the second set : 5!
total = 5! +4.4! = 4!(5+4) = 24*9 = 216
Since 0 cannot be the first digit of a number, for the first position, you have 4 choices (all digits except zero). No such constraints exist for the rest of the positions; hence the next choices are 4,3,2,1 – all multiplying up to give a 4!. Had there been no 0 involved, the choices would’ve been 5! Instead of 4.4!
6. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420
B. 2520
C. 168
D. 90
E. 105
Soln: out of 8 people one team can be formed in 8c2 ways.
8c2*6c2*4c2*2c2= 2520.
The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )
7. My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
Soln: there are actually 20 ways to interchange the letters, namely, the first letter could be one of 5, and the other letter could be one of 4 left. So total pairs by product rule = 20.
Now, there are two cases when it wouldn’t change the name. First, keeping them all the same. Second, interchanging the two EEs together. Thus 2 options would leave the name intact.
Prob = 2/20 = 0.1, or 10%.
8. A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?
A. y > ROOT2
B. ROOT3/2 < y < ROOT2
C. ROOT2/3 < y < ROOT3/2
D. ROOT3/4 < y < ROOT2/3
E. y < ROOT3/4
Soln: right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1
i.e xy = 2
If x were equal to y, we would have had xy = y^2 = 2. And y = root2
But, x<y and so y>root2.
9. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16
Soln: Chance of drawing a blue on the first draw = 2/8, so chance of not drawing a blue on the first draw is 6/8
similarly chance of not drawing blue on second draw = 5/7
Therefore probability of not drawing blue in 2 draws = 6/8*5/7 = 15/28
10. How many integers between 100 and 150, inclusive can be evenly divided by neither 3 nor 5?
Soln: Number of integers that divide 3:
the range is 100-150
Relevant to this case, we take 102 – 150 (since 102 is the first to div 3)
102 = 34*3
150= 50*3, so we have 50-34+1 = 17 multiples of 3
For multiples of 5,
100=5*20
150=5*30
30-20+1 =11
Now we have a total of 27 integers, but we double counted the ones that divide BOTH 3 AND 5, ie 15.
105 is the first to divide 15.
105=15*7
150=15*10
10-7+1 = 4 integers
So our total is 17+11-4 = 24 integers that can be divided by either 3 or 5 or both.
51 integers – 24 integers = 27 that cannot be evenly divided.
11. Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%
Soln: (x+y)30/100 = x*40/100 + y*25/100
30x + 30y = 40x + 25y
y = 2x or y/x = 2/1 or y:x = 2:1 hence x = 33 1/3%
12.
Sequence A and B. a1=1, b1=k. an=b(n-1)-a(n-1) bn=b(n-1)+a(n-1). What is a4=?
Soln: a2 = k-1 ; b2 = k+1
a3= (k+1)-(k-1) = 2 ; b3 = (k+1)+(k-1) = 2k
a4 = 2k – 2 = 2(k-1) = 2(b1-a1)
13. A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum?
A) 38% B) 42% C) 19% D) 40%
Soln: assume, interest = r
so after 3 years total money = 1000*(1+r)^3 = 1000*1.19
(1+r)^3 = 1.19
so after 6 years total money = 1000*(1+r)^6 = 1000*1.19^2 = 1000*1.42
so percentage of interest is 42%
14. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they
meet for the first time after they start the race?
A. 750 seconds
B. 50 seconds
C. 250 seconds
D. 375 seconds
E. 75 seconds
Soln: When two people are running in the same direction the relative speed is a difference in speeds of the two people.
In this case A=3 B=6 C=18
So relative speed of B wrt A is 6-3 = 3m/s
Relative speed of A wrt to C is 18-3 =15m/s
Therefore relative distances will be:
B wrt A is 750/3 =250
C wrt to A 750/15 = 50
So they have to bridge this distance of 250 and 50 between them which is the LCM of 250 and 50 which is 250.
Another Method: Simply put, Runner A’s time take to run one lap is 250
Runner B’s time is 125s
and Runner C’s time is 41.67s
We can notice that A=2B
and thet B=3C
So when 250 s elapse, they will be at their starting point. A will have completed one lap, B 2 laps, and C 6 laps
15. X percents of the rooms are suits, Y percent of the rooms are painted light blue. Which of the following best represents the least percentage of the light blue painted suits?
1) X-Y
2)Y-X +100
3)100X-Y
4)X+Y-100
e)100-XY
Soln: Equation from set theory:
n(AUB)=n(A)+n(B)-n(A^B)
where,
A= % of rooms which are suites
B= % of rooms painted blue
A^B means the intersection of the two sets
Now in this case, what we need to find is n(A^B), therefore
n(A^B)=n(A)+n(B)-n(AUB)
= X + Y – n(AUB)
Now this would be least when n(AUB) is maximum, which would happen if these two kinds of rooms are only two kinds available, making n(AUB)=100
Therefore the answer should be X+Y-100
Posted in Management