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Decimal Computation with [ Number Properties ] GMAT, GRE
(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)
Integer Digits [ Combinations ] GMAT, GRE
How many positive integers less than or equal to 2017 contain the digit 0?
(A) 469
(B) 471
(C) 475
(D) 478
(E) 481
(A) 469
(B) 471
(C) 475
(D) 478
(E) 481
3/8 of all students at a school are in all three of the following clubs: Albanian, Bardic, and Chess. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Chess If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 5/8
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 5/8
Simple Interest GMAT, GRE Problem Solving Guide
All interest referred to below is simple interest. The annual amount of interest paid on an investment is found by multiplying the amount invested (the principal) by the percent of interest (the rate).
Example:
Sam invested some of his money in a bank paying 4% interest annually and a second amount, $500 less than the first, in a bank paying 6% interest. If her annual income from both investments was $50, how much money did Sam invest at 6%?
Solution:
Sam invested some of his money in a bank paying 4% interest annually and a second amount, $500 less than the first, in a bank paying 6% interest. If her annual income from both investments was $50, how much money did Sam invest at 6%?
Solution:
As this is a non-ratio algebraic problem, we can handle this with a Distance = Rate x Time chart. Only we'll need an analogy for our names. Let's use:
PRINCIPAL · RATE = INTEREST INCOME
PRINCIPAL · RATE = INTEREST INCOME
Let's put it together algebraically.
x = amount invested at 4%x – 500 = amount invested at 6%
.04x = annual interest from 4% investment
.06(x – 500) = annual interest from 6% investment .04x + .06(x – 500) = 50
Multiply by 100 to remove decimals.
4x + 6(x − 500)= 5000 4x + 6x − 3000 = 5000 10x = 8000
x = 800x − 500 = 300
Sam invested $300 at 6%.
Let's try another using our matrix method
x = amount invested at 4%x – 500 = amount invested at 6%
.04x = annual interest from 4% investment
.06(x – 500) = annual interest from 6% investment .04x + .06(x – 500) = 50
Multiply by 100 to remove decimals.
4x + 6(x − 500)= 5000 4x + 6x − 3000 = 5000 10x = 8000
x = 800x − 500 = 300
Sam invested $300 at 6%.
Let's try another using our matrix method
Jill invested $7200, part at 4% and the rest at 5%. If the annual income from both investments was the same, find her total annual income from these investments.
- (A) $160
- (B) $320
- (C) $4000
- (D) $3200
- (E) $1200
7200 − x = amount invested at 5% .04x =.05(7200-x)
Multiply by 100 to eliminate decimals.
4x = 5(7200 − x)4x = 36,000 − 5x9x = 36,000
x = 4000
Her income is .04(4000) + .05(3200).
$320.
Multiply by 100 to eliminate decimals.
4x = 5(7200 − x)4x = 36,000 − 5x9x = 36,000
x = 4000
Her income is .04(4000) + .05(3200).
$320.
GMAT, GRE Factors, Primes, Functions and [ Number Properties ] Problem
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is?
A. Between 2 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
A. Between 2 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
Medium GMAT, GRE [ Rate Time Distance ] Problem
On a path, a hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20 miles per hour. the cyclist stops & waits for the hiker 5 min after passing him while the hiker continues to walk at his constant rate. how many minutes must the cyclist wait until the hiker catches up
A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3
A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3
How to handle Mixture word problems on the GMAT, GRE
Solving mixture problems follows the same method we like to do for all rate x time = Distance questions. We make a matrix for r x t = D on the top but to fill it in we'll make an analogy. What are our terms instead of r and t? Look by the numbers in the question. In the question below we have 36 and next to it is an amount of money, so one of the terms will be price. Another value is 300 and next to it is a weight, so the other term is number of pounds. Does it matter which comes first? Well, let me ask you: what's 3 x 4? And what's 4 x 3? When we multiply across our matrix to get the "D" or Total value, the order in which we do it doesn't matter. Let's look at our example.
Example:
Mr. Thomson wants to mix candy worth 36 cents a pound with candy worth 52 cents a pound to make 300 pounds of a mixture worth 40 cents a pound. How many pounds of the more expensive candy should he use?
Solution:
Let's set up our r x t = D matrix using our analogie values of pounds, price and total.
The value of the more expensive candy plus the value of the less expensive candy must be equal to the value of the mixture. Almost all mixture problems derive their equation from adding the final column (The "D" or "Distance" value) in the chart. In this case, we changed our Distance to read Total Value.
52x + 36(300 – x) = 12000
Notice that all values were computed in cents to avoid decimals. As long as you multiply all the values of an equation you can manipulate them however you want.
52x+10,800−36x =12,000 16x = 1200
x = 75
He should use 75 pounds of the more expensive candy.
700 Level GMAT [ Work Rate Ratio ] Problem
Twenty-four men can complete a work in sixteen days.Thirty-two women can complete the same work in twenty-four days.Sixteen men and sixteen women started working for twelve days.How many more men are to be added to complete the work remaining work in 2 days?
A .16
B. 24
C. 36
D. 48
E. 54
Distance = rate x time
For our purposes: Work = Rate (of work for each man person) * (total time)
1 Man 1 Dday work=1/(16*24)
1 Woman 1 D work=1/(24*32)
16 M 1 D work=1/24=2/48
16 W 1 D work=1/48
16 M&W 1 D work=3/48=1/16
16 M&W 12 D work=12/16
The remaining work is 4/16 or 1/4
this work is to be completed by the additional men
1 M 1 D work = 1/(16*24)
1 M 2 D work = 1/(16*12)
x M 2 D work = x/(16*12)
this is set to be equal to 1/4
x/(16*12) =1 /4
x = 48
D
A .16
B. 24
C. 36
D. 48
E. 54
Distance = rate x time
For our purposes: Work = Rate (of work for each man person) * (total time)
1 Man 1 Dday work=1/(16*24)
1 Woman 1 D work=1/(24*32)
16 M 1 D work=1/24=2/48
16 W 1 D work=1/48
16 M&W 1 D work=3/48=1/16
16 M&W 12 D work=12/16
The remaining work is 4/16 or 1/4
this work is to be completed by the additional men
1 M 1 D work = 1/(16*24)
1 M 2 D work = 1/(16*12)
x M 2 D work = x/(16*12)
this is set to be equal to 1/4
x/(16*12) =1 /4
x = 48
D
GMAT, GRE [ Remainders ]
Let's look at the cycle of remainders:
What is the remainder of 7 divided by 4? Ans = 3
What is the remainder of 6 divided by 4? Ans = 2
What is the remainder of 5 divided by 4? Ans = 1
What is the remainder of 4 divided by 4? Ans = 0
What is the remainder of 3 divided by 4? Ans = -1
(11^1)/12 gives a remainder of 11 or -1
(Remainder of 11/12)^12gives us (-1)^12
which is 1
GMAT, GRE [Inequalities] With Squares
Which of the following describes all values of x for which 1–x^2 ≥ 0?
(A) x ≥ 1
(B) x ≤ –1
(C) 0 ≤ x ≤ 1
(D) x ≤ –1 or x ≥ 1
(E) –1 ≤ x ≤ 1
(A) x ≥ 1
(B) x ≤ –1
(C) 0 ≤ x ≤ 1
(D) x ≤ –1 or x ≥ 1
(E) –1 ≤ x ≤ 1
Range With [Rate] Gmat, Gre Quantitative Problem
On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between
A. 290/12.5 and 290/11.5
B. 295/12 and 285/11.5
C. 285/12 and 295/12
D. 285/12.5 and 295/11.5
E. 295/12.5 and 285/11.5
A. 290/12.5 and 290/11.5
B. 295/12 and 285/11.5
C. 285/12 and 295/12
D. 285/12.5 and 295/11.5
E. 295/12.5 and 285/11.5
Cindy drove her car 290 miles, rounded to the nearest 10 miles --> 285≤m<295 ;
Used 12 gallons of gasoline, rounded to the nearest gallon -->11.5≤g<12.5 ;
Minimum Miles per gallon, m/g -->28512.5<mg<29511.5
Used 12 gallons of gasoline, rounded to the nearest gallon -->
Minimum Miles per gallon, m/g -->
D
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