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Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 80.
Answer (E) | deepmind/aqua_rat |
S is a set containing 10 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
A) The median of S is prime.
B) The median of T is prime
C) The median of S is equal to the median of T.
D) The sum of the terms in S is prime.
E) The sum of the terms in T is prime.
Correct Answer:E) The sum of the terms in T is prime.
Rationale: Here is my explanation. The question states:S is a set containing 9 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
(A) The median of S is prime.
Thismust be true. If there are an odd number of members of a set, then the median is a member of the set: it's the middle number, when all the numbers are ranked from smallest to biggest. Every number in S is a positive odd prime, so the median is one of them, and is prime.
(B) The median of T is prime.
Thismay or may not be true. If a set has an even number of members, the median is average of the two numbers in the middle, when ranked from smallest to biggest. The average of two odd numberscould beeven (average of 71 and 73 is 72), and hence not prime, or itcould beodd (the average of 71 and 79 is 75). For particularly well chosen odd numbers, the average can be not only odd but also prime -- for example, the average of 89 and 113 is 101, another prime number. If the two middle numbers of T were 89 and 113, the median would be 101, a prime number.
(C) The median of S is equal to the median of T. | deepmind/aqua_rat |
(C) The median of S is equal to the median of T.
Under most configurations for S and T, this wouldn't happen. If you weren't trying to make it happen, it would be unlikely to happen by chance. BUT, if the number dropped going from from S to T was the median of S (say, 101), and if the two middle numbers of T happen to have an average of that number that was dropped (for example, if the two numbers were 89 and 113), then the medians would be equal. In other words, the three middle numbers of S would have to be {. . ., 89, 101, 133, . . .}, and when 101 is dropped in going to T, the median of two would be the average of 89113, which happens to be 101. It's an exceptional case, but itcouldbe true.
(D) The sum of the terms in S is prime.
Thismay or may not be true. The sum of 9 odd numbermustbe an odd number. That odd numbercould beprime. For example, the sum of the first nine odd prime numbers {3, 5, 11, 13, 17, 19, 23, 29} is 127, which is prime. If you drop 3 and include the next prime, 31, the set {5, 11, 13, 17, 19, 23, 29, 31} has a sum of 155, which is clearly not prime.
(E) The sum of the terms in T is prime.
Thismust be false. The sum of eight odd numbers must be an even number. Only 2 is prime, and all other even numbers are not. Therefore, the sum of eight odd prime numbers will be an even number bigger than two, and absolutely cannot be prime.
E | deepmind/aqua_rat |
S is a set containing 6 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
A) The median of S is prime.
B) The median of T is prime
C) The median of S is equal to the median of T.
D) The sum of the terms in S is prime.
E) The sum of the terms in T is prime.
Correct Answer:D) The sum of the terms in S is prime.
Rationale: Here is my explanation. The question states:S is a set containing 9 different positive odd primes. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
(A) The median of S is prime.
Thismust be true. If there are an odd number of members of a set, then the median is a member of the set: it's the middle number, when all the numbers are ranked from smallest to biggest. Every number in S is a positive odd prime, so the median is one of them, and is prime.
(B) The median of T is prime.
Thismay or may not be true. If a set has an even number of members, the median is average of the two numbers in the middle, when ranked from smallest to biggest. The average of two odd numberscould beeven (average of 71 and 73 is 72), and hence not prime, or itcould beodd (the average of 71 and 79 is 75). For particularly well chosen odd numbers, the average can be not only odd but also prime -- for example, the average of 89 and 113 is 101, another prime number. If the two middle numbers of T were 89 and 113, the median would be 101, a prime number.
(C) The median of S is equal to the median of T. | deepmind/aqua_rat |
(C) The median of S is equal to the median of T.
Under most configurations for S and T, this wouldn't happen. If you weren't trying to make it happen, it would be unlikely to happen by chance. BUT, if the number dropped going from from S to T was the median of S (say, 101), and if the two middle numbers of T happen to have an average of that number that was dropped (for example, if the two numbers were 89 and 113), then the medians would be equal. In other words, the three middle numbers of S would have to be {. . ., 89, 101, 133, . . .}, and when 101 is dropped in going to T, the median of two would be the average of 89113, which happens to be 101. It's an exceptional case, but itcouldbe true.
(D) The sum of the terms in S is prime.
Thismay or may not be true. The sum of 9 odd numbermustbe an odd number. That odd numbercould beprime. For example, the sum of the first nine odd prime numbers {3, 5, 11, 13, 17, 19, 23, 29} is 127, which is prime. If you drop 3 and include the next prime, 31, the set {5, 11, 13, 17, 19, 23, 29, 31} has a sum of 155, which is clearly not prime.
(E) The sum of the terms in T is prime.
Thismust be false. The sum of eight odd numbers must be an even number. Only 2 is prime, and all other even numbers are not. Therefore, the sum of eight odd prime numbers will be an even number bigger than two, and absolutely cannot be prime.
D | deepmind/aqua_rat |
Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveler sets out from city X in the same direction as the first traveler and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveler be ahead of the first?
A)2 days
B)6 days
C)From the 2nd day after the 2nd traveler starts
D)From the 3rd day after the 2nd traveler starts
E)None
Correct Answer:B)6 days
Rationale: Explanatory Answer
Travel Pattern of Traveller 2
Day 1: 1 km; Day 2: (1 + 2) = 3 km; Day 3: (1 + 2 + 2) = 5km and so on
i.e. on each of these days he covers the following distance Day 1: 1km, Day 2: 3km, Day 3: 5km and so on.
The distances covered each day by the first traveller is in an AP with the first term being 1 and the common difference being 2
After 3 days, the first traveler would have traveled 9 kms. He is ahead of the second traveler at the end of 3 days by 9 kms.
Travel Pattern of Traveller 2
The second traveler travels as follows: On his first day he travels 12 kms, his second day he travels 13 kms and so on.
The distances covered each day by the second traveller is in an AP with the first term being 12 and the common difference being 1
Step 2
Compute total distance covered by each of the travellers
Let the number of days travelled by the second traveller be n.
Therefore, the first traveller would have travelled for (n + 3) days. Traveler 1 has covered a total distance of 1+ 3 + 5 + 7 + ..... + (n + 3) km.
The total distance covered by traveler 1 is the summation of the AP.
Sum of an AP = (n/2)(2a1 + (n-1)d)
Total distance travelled by traveller 1 in (n + 3) days = n+3/2[2(1)+(n+3−1)(2)]
The above term simplifies as (n + 3)2 | deepmind/aqua_rat |
The total distance covered by traveler 1 is the summation of the AP.
Sum of an AP = (n/2)(2a1 + (n-1)d)
Total distance travelled by traveller 1 in (n + 3) days = n+3/2[2(1)+(n+3−1)(2)]
The above term simplifies as (n + 3)2
The total distance covered by the second traveler = 12 + 13 + 14 + ... + n terms
Total distance travelled by traveller 2 in n days = n2[2(12)+(n−1)(1)] = n(n+23)2
On the days that the second traveller is ahead of the first one, the total distance covered by the second traveller till that day will be more than the total distance covered by the first traveller.
Note: All values calculated are with respect to the day from which the second traveller started travelling
Therefore, let us solve the inequality n(n+23)/2 > (n + 3)2
n2 + 23n > 2(n2 + 6n + 9)
Or n2 - 11n + 18 < 0
The quadratic factorizes as (n - 2)(n - 9) < 0
Possibility 1: (n - 2) > 0 AND (n - 9) < 0
i.e., n > 2 AND n < 9
Therefore, 2 < n < 9
Possibility 2: (n - 2) < 0 AND (n - 9) > 0
i.e., n < 2 AND n > 9
Infeasible solution
Therefore, starting the 3rd day and up to the 8th day the second traveller was ahead of the first traveller
The correct answer is Choice (2). 6 days
Answer B | deepmind/aqua_rat |
If a, q, and c are consecutive even integers and a < q < c, all of the following must be divisible by 4 EXCEPT
A)a + c
B)q + c
C)ac
D)(bc)/2
E)(abc)/4
Correct Answer:B)q + c
Rationale: I did get the correct answer within 80 seconds and it was not by luck either. I did not pick numbers but just used the concept stated by Ian.
The 3 numbers can be written as
a, (a + 2)(a + 4).
If 'a' is divisible by 4, then even 'c' or 'a + 4' is divisible by 4. However, is 'b' is divisible by 4, then both 'a' and 'a + 4' are still divisible by 2.
A - (a + c) = a + (a + 4) = 2a + 4 = 2(a + 2) = 2q. 2q will always be divisible by 4 even if 'b' is not divisible by 4. Reason: 'q' already has a prime factorization of at least a '2'. Hence '2q' has two 2s.
C - ac = a(a+4). If, as stated above, one of them is divisible by 4, then the product is divisible. If both of them are not divisible by 4, then the product is still divisible by 4 because of the presence of two 2s again in the prime factorization.
D - qc/2 = (a + 2)(a + 4)/2. Either q or c is divisible by 2. Hence, if we assume that b is divisible by 2 and not divisible by 4, then it leaves us just one possibility. Is c divisible by 4? It has to be because c is the next consecutive even integer.
E - aqc/4 = a(a + 2)(a + 4)/4. One of these integers is divisible by 4 already. If we again assume 'q' to be that integer divisible by 4, then we are left with the question - Is a(a + 4) divisible by 4? This is the same as option C. | deepmind/aqua_rat |
E - aqc/4 = a(a + 2)(a + 4)/4. One of these integers is divisible by 4 already. If we again assume 'q' to be that integer divisible by 4, then we are left with the question - Is a(a + 4) divisible by 4? This is the same as option C.
B - q + c = (a + 2) + (a + 4) = 2a + 6 = 2(a + 3). (a + 3) will never be divisible by 2 because it is an odd integer. Hence, 2(a + 3), although divisible by 2, will not be divisible by 4 because it has just one 2 in its prime factorization.
As a whole, whether you choose numbers (2, 46 being the easiest) or solve conceptually, the answer is still easily obtainable within 2 minutes.B | deepmind/aqua_rat |
In how many different ways can change be provided for a one dollar bill using pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents), if at least one of each type of coin must be included?
A)39
B)42
C)60
D)65
E)66
Correct Answer:C)60
Rationale: I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we cannot have any other coins. We can't have all four coins represented if we have four quarters.
Case I does not allow any ways.
II. Case II = Three Quarters (75 cents)
This allows for one dime. Two dimes would bring us up to $0.95, and we wouldn't have room for both pennies and nickels.
Subcase 1 = 3 Q, 1 D (85 cents)
1) one nickel (N) and 10 pennies (P)
2) 2 N and 5 P
Case II allows for a total of two ways.
III. Case III = Two Quarters (50 cents)
This allows for 1-4 dimes.
Subcase 1 = 2 Q, 4 D (90 cents)
(1 way) we could have 1 nickel
Subcase 2 = 2 Q, 3 D (80 cents)
(3 ways) we could have 1-3 nickels
Subcase 3 = 2 Q, 2 D (70 cents)
(5 ways) we could have 1-5 nickels
Subcase 4 = 2 Q, 1 D (60 cents)
(7 ways) we could have 1-7 nickels
Case III allows for a total of 16 ways
IV. Case IV = One Quarter (25 cents)
This allows for 1-6 dimes
Subcase 1 = 1 Q, 6 D (85 cents)
(2 ways) we could have 1-2 nickels
Subcase 2 = 1 Q, 5 D (75 cents)
(4 ways) we could have 1-4 nickels
Subcase 3 = 1 Q, 4 D (65 cents)
(6 ways) we could have 1-6 nickels
Subcase 4 = 1 Q, 3 D (55 cents)
(8 ways) we could have 1-8 nickels
Subcase 5 = 1 Q, 2 D (45 cents)
(10 ways) we could have 1-10 nickels
Subcase 6 = 1 Q, 1 D (35 cents) | deepmind/aqua_rat |
(4 ways) we could have 1-4 nickels
Subcase 3 = 1 Q, 4 D (65 cents)
(6 ways) we could have 1-6 nickels
Subcase 4 = 1 Q, 3 D (55 cents)
(8 ways) we could have 1-8 nickels
Subcase 5 = 1 Q, 2 D (45 cents)
(10 ways) we could have 1-10 nickels
Subcase 6 = 1 Q, 1 D (35 cents)
(12 ways) we could have 1-12 nickels
Case IV allows for a total of 42 ways.
There's no other case, because we have to have at least one quarter and one dime. The total over the cases equals
Total = 0 + 2 + 16 + 42 =60 ways.
OA =(C) | deepmind/aqua_rat |
Raj is working on a set of Data Sufficiency problems for his December GMAT: a geometry problem, an algebra problem, and a data interpretation problem. He has determined that statement 1 of the geometry problem is insufficient on its own, that both statement 1 and 2 of the algebra problem are insufficient on their own, and that statement 2 of the data interpretation problem is insufficient on its own. If the probabilities are expressed as percents Q, approximately how much greater is the probability that all three answers are “C” after Raj figures out that statement 1 of the data interpretation problem is also insufficient on its own?
A)Q=2.3%
B)Q=2.8%
C)Q=3.3%
D)Q=5.6%
E)Q=8.3%
Correct Answer:B)Q=2.8%
Rationale: Step 1: Figure out what we need to know.We have two probabilities that we need to figure out: the one before Raj figures out that statement 1 of the data interpretation problem is also insufficient, and the one after. The answer is the difference between them, in the form of a percent.
Step 2: The first probability.
If statement 1 of the geometry problem is insufficient, then the only valid answer choices are B, C, and E; the probability that it is C is 1/3.
If both statements of the algebra problem are insufficient on their own, then the only remaining valid answers are C and E; the probability that it is C is 1/2
If statement 2 of the data interpretation problem is insufficient, then the remaining answers are A, C, and E, and the probability that C is correct is 1/3.
The probability of all three occurring is the product of the probability fractions: (1/3)*(1/2)*(1/3) = 1/18.
Step 3: The second probability.
Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.
The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12. | deepmind/aqua_rat |
Step 3: The second probability.
Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.
The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12.
Step 4: The answer.Note that here, as will sometimes happen on the GMAT, values that you compute “along the way” appear as wrong answer choices. This problem calls for the difference between our two numbers,Q (1/12)-(1/18) = 1/36, which is a 2.77777% chance, rounded to 2.8% —the correct answer is Choice B. | deepmind/aqua_rat |
Raj is working on a set of Data Sufficiency problems for his December GMAT: a geometry problem, an algebra problem, and a data interpretation problem. He has determined that statement 1 of the geometry problem is insufficient on its own, that both statement 1 and 2 of the algebra problem are insufficient on their own, and that statement 2 of the data interpretation problem is insufficient on its own. If the probabilities are expressed as percents E, approximately how much greater is the probability that all three answers are “C” after Raj figures out that statement 1 of the data interpretation problem is also insufficient on its own?
A)E=2.3%
B)E=2.8%
C)E=3.3%
D)5.6%
E)8.3%
Correct Answer:B)E=2.8%
Rationale: Step 1: Figure out what we need to know.We have two probabilities that we need to figure out: the one before Raj figures out that statement 1 of the data interpretation problem is also insufficient, and the one after. The answer is the difference between them, in the form of a percent.
Step 2: The first probability.
If statement 1 of the geometry problem is insufficient, then the only valid answer choices are B, C, and E; the probability that it is C is 1/3.
If both statements of the algebra problem are insufficient on their own, then the only remaining valid answers are C and E; the probability that it is C is 1/2
If statement 2 of the data interpretation problem is insufficient, then the remaining answers are A, C, and E, and the probability that C is correct is 1/3.
The probability of all three occurring is the product of the probability fractions: (1/3)*(1/2)*(1/3) = 1/18.
Step 3: The second probability.
Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.
The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12. | deepmind/aqua_rat |
Step 3: The second probability.
Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.
The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12.
Step 4: The answer.Note that here, as will sometimes happen on the GMAT, values that you compute “along the way” appear as wrong answer choices. This problem calls for the difference between our two numbers, E(1/12)-(1/18) = 1/36, which is a 2.77777% chance, rounded to 2.8% —the correct answer is Choice B. | deepmind/aqua_rat |
If n is a positive integer and n^2 is divisible by 264, then the largest positive integer that must divide n is
A)6
B)12
C)24
D)36
E)48
Correct Answer:E)48
Rationale: The question asks aboutthe largest positive integer that MUST divide n, not COULD divide n. Since the least value of n for which n^2 is a multiple of 72 is 12 then the largest positive integer that MUST divide n is 12.
Complete solution of this question is given above. Please ask if anything remains unclear.
I spent a few hours on this one alone and I'm still not clear. I chose 12 at first, but then changed to 48.
I'm not a native speaker, so here is how I interpreted this question:the largest positive integer that must divide n=the largest positive factor of n. Since n is a variable (i.e. n is moving), so is its largest factor. Please correct if I'm wrong here.
I know that if n = 12, n^2 = 144 = 2 * 72 (satisfy the condition). When n = 12, the largest factor of n is n itself, which is 12. Check: 12 is the largest positive number that must divide 12 --> true
However if n = 48, n^2 = 48 * 48 = 32 * 72 (satisfy the condition too). When n = 48, the largest factor of n is n itself, which is 48. Check: 48 is the largest positive number that must divide 48 --> true
So, I also notice that the keyword isMUST, notCOULD. The question is, why is 48 notMUST divide 48, but instead onlyCOULD divide 48? I'm not clear right here. Why is 12MUST divide 12? What's the difference between them?
Only restriction we have on positive integer n is that n^2 is divisible by 72. The least value of n for which n^2 is divisible by 72 is 12, thus nmustbe divisible by 12 (n is in any case divisible by 12). For all other values of n, for which n^2 is divisible by 72, n will still be divisible by 12. This means that n is always divisible by 12 if n^2 is divisible by 72. | deepmind/aqua_rat |
Now, ask yourself: if n=48, is n divisible by 48? No. So, n is not always divisible by 48.
E | deepmind/aqua_rat |
The difference between the squares of two consecutive odd integers
is a square<1000 whose root equals the combined sum of the digits
of the two integers. What is the sum of the digits M of the larger integer?
A)M=2
B)M=5
C)M=8
D)10
E)11
Correct Answer:A)M=2
Rationale: Really hard. But let's start with few concepts:
1. Square of an odd integer will be odd.
2. Difference of odd and odd will be even.
3. Any odd integer can be expressed as difference of two squares.
4. An even integer can be expressed as difference of two squares only if that even integer is a multiple of 4.
Assume two odd integers to be (2x-1) and (2x+1).
Difference of their squares = (2x+1)^2 - (2x-1)^2
= 4x^2 + 1 + 4x - (4x^2 + 1 - 4x)
= 4x^2 + 1 + 4x - 4x^2 - 1 + 4x
= 8x
Now root of 8x needs to be an integer such that it is equal to the sum of the digits of the two odd integers.
8 = 2^3, so x needs to be such that itcompletesa perfect square.
If we find x, we can find the value of 2x+1 (larger integer) and then sum of its digits.
Let's check the options, starting with c.
For sum to be 8, few possibilities are: 17, 35, 53, 71
If we take 17, the pair is 15 and 17, meaning x = 8.
8x = 64
root of 64 = 4 but 4 is not equal to 1+5+1+7.
Reject.
If we take 35, the pair is 33 and 35, meaning x = 17.
8x = 8*17
Reject since perfect square is not possible.
If we take 53, the pair is 51 and 53, meaning x = 26.
8x = 8*26
Reject since perfect square is not possible.
If we take 71, the pair is 69 and 71, meaning x = 35.
8x = 8*35
Reject since perfect square is not possible.
I tried each option and the possibilities and then got lost.
Then, it occurred to me that I was only checking 2-digit integers. What about 3-digits? | deepmind/aqua_rat |
If we take 53, the pair is 51 and 53, meaning x = 26.
8x = 8*26
Reject since perfect square is not possible.
If we take 71, the pair is 69 and 71, meaning x = 35.
8x = 8*35
Reject since perfect square is not possible.
I tried each option and the possibilities and then got lost.
Then, it occurred to me that I was only checking 2-digit integers. What about 3-digits?
Starting with option a, only 2-digit odd integer whose sum is 2 is 101.
If we take 101, the pair is 99 and 101, meaning x = 50.
8x = 8*50 = 400
root of 400 = 20 which is indeed equal to 9+9+1+1.
Answer (A). | deepmind/aqua_rat |
Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum V in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is
A)90%
B)80%
C)70%
D)60%
E)40%
Correct Answer:A)90%
Rationale: The problem asks for the approximate chance that no more than 1/3 of the original investment is lost. We can apply the “1 – x” technique: what’s the chance that more than 1/3 of the original investment is lost? There are two outcomes we have to separately measure:
(a) All 3 investments become worthless.
(b) 2 of the 3 investments become worthless, while 1 doesn’t.
Outcome (a): The probability is (0.2)(0.2)(0.2) = 0.008, or a little less than 1%.
Outcome (b): Call the investments X, Y, and Z. The probability that X retains value, while Y and Z become worthless, is (0.8)(0.2)(0.2) = 0.032. Now, we have to do the same thing for the specific scenarios in which Y retains value (while X and Z don’t) and in which Z retains value (while X and Y don’t). Each of those scenarios results in the same math: 0.032. Thus, we can simply multiply 0.032 by 3 to get 0.096, or a little less than 10%.
The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.
The correct answer is A. | deepmind/aqua_rat |
The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.
The correct answer is A.
This problem illustrates the power of diversification in financial investments. All else being equal, it’s less risky to hold a third of your money in three uncorrelated (independent) but otherwise equivalent investments than to put all your eggs in one of the baskets. That said, be wary of historical correlations! Housing price changes in different US cities were not so correlated—and then they became highly correlated during the recent housing crisis (they all fell together), fatally undermining spreadsheet models that assumed that these price changes were independent. | deepmind/aqua_rat |
A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight, in kg, of the heaviest box ?
A)60
B)61
C)64
D)Can't be determined
E)None
Correct Answer:B)61
Rationale: Explanation :
Two boxes are selected at at time and weighed together. Thus, the selection order does not matter and selecting box 1 and then box 2 is the same as selecting box 2 and then box 1. So this is a combination problem.
2 boxes can be selected from 5 boxes in 5C2 ways, i.e. 10 ways. The given list of weighings is 10 in number. So we can assume that all the possible combinations of boxes have been weighed. Lets assume that all boxes are lined as per increasing weights and named as b1, b2, b3, b4, b5. From the given list (Arranges in increasing order of weights) ,
what we have is b1 + b2 = 110, as b1 is the lightest and b2 is the next lightest box heavier than b1. So the sum of these two should be the minimum value in the list.
The second lightest sum of weights, the second value in the list, will be the sum of the b1 + b3 = 112.
Similarly, the maximum value in the list should be the sum of weights of b4 and b5, giving b4 + b5 = 121.
The second last heaviest will be b3 + b5 = 120.
Adding the 4 equations, we get, 2 b1 + 2 b3 + 2 b5 + b2 + b4 = 463 - Equation 1.
Sum of all the weights in the given list is 1160. This sum is the sum of each of the cases b1+b2, b1+b3, b1+b4, b1+b5, b2+b3, b2+b4, b2+b5, b3+b4, b3+b5, b4+b5. Thus sum of all this will be 4(b1 + b2 + b3 + b4 + b5) = 1160. Thus, b1 + b2 + b3 + b4 + b5 = 290 - Equation 2. | deepmind/aqua_rat |
Subtracting equation 2 from equation 1, we get, b1 + b3 + b5 = 173.
We know that b1 + b3 = 112, So b5 = 173 - 112 = 61 = Desired answer.
Answer : B | deepmind/aqua_rat |
A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% sunflower. If a customer purchases a mix of the two types of birdseed that is 50% sunflower, what percent of the mix is Brand A?
A)40%
B)45%
C)50 %
D)55 %
E)60 %
Correct Answer:E)60 %
Rationale: Yes there is a simple method :
Consider the following method
Brand A : 40% millet and 60% sunflower
Brand B : 65% millet and 35% sunflower
Mix : 50% sunflower
Here the weighted average is 50%,
Now Brand A has 60 % sunflower, which is 10% more than the weighted average of mix = + 0.10 A --------------- I
Similarly, Brand B has 35 % sunflower, which is 15 % less than the weighted average of mix = - 0.15 B ------------ II
Now, both Brand A and Brand B are combined to give a 50% mix containing millet, so equate I and II
implies, 0.10 A = 0.15 B
Therefore A/B = 0.15/0.10 = 3/2
A : B : (A + B) = 3 : 2 : (3+2) = 3 : 2 : 5
We have to find, percent of the mix is Brand A i.e. A : (A + B) = 3 : 5 = (3 / 5) * 100 = 60 %
Here is a pictorial representation :
Brand A= 60%------------------------10% or 0.10 above average, A times-----------------Total below = + 0.10 A | deepmind/aqua_rat |
Here is a pictorial representation :
Brand A= 60%------------------------10% or 0.10 above average, A times-----------------Total below = + 0.10 A
----------------------------------------------------------------------------------------Average = 50% or 0.50
Brand B = 35 %--------------------------15% or 0.15 below average, B times-----------------Total above = - 0.15 B
Since the amount below the average has to equal the average above the average; therefore,
0.10 A = 0.15 B
A/B = 3/2
A:B: Total = 3:2:5
Therefore
A/Total = 3:5 = 60 %
Answer: E | deepmind/aqua_rat |
The difference between the squares of two consecutive odd integers
is a square<1000 whose root equals the combined sum of the digits
of the two integers. What is the sum of the digits Q of the larger integer?
A)Q=2
B)Q=5
C)Q=8
D)Q=10
E)11
Correct Answer:A)Q=2
Rationale: Really hard. But let's start with few concepts:
1. Square of an odd integer will be odd.
2. Difference of odd and odd will be even.
3. Any odd integer can be expressed as difference of two squares.
4. An even integer can be expressed as difference of two squares only if that even integer is a multiple of 4.
Assume two odd integers to be (2x-1) and (2x+1).
Difference of their squares = (2x+1)^2 - (2x-1)^2
= 4x^2 + 1 + 4x - (4x^2 + 1 - 4x)
= 4x^2 + 1 + 4x - 4x^2 - 1 + 4x
= 8x
Now root of 8x needs to be an integer such that it is equal to the sum of the digits of the two odd integers.
8 = 2^3, so x needs to be such that itcompletesa perfect square.
If we find x, we can find the value of 2x+1 (larger integer) and then sum of its digits.
Let's check the options, starting with c.
For sum to be 8, few possibilities are: 17, 35, 53, 71
If we take 17, the pair is 15 and 17, meaning x = 8.
8x = 64
root of 64 = 4 but 4 is not equal to 1+5+1+7.
Reject.
If we take 35, the pair is 33 and 35, meaning x = 17.
8x = 8*17
Reject since perfect square is not possible.
If we take 53, the pair is 51 and 53, meaning x = 26.
8x = 8*26
Reject since perfect square is not possible.
If we take 71, the pair is 69 and 71, meaning x = 35.
8x = 8*35
Reject since perfect square is not possible.
I tried each option and the possibilities and then got lost.
Then, it occurred to me that I was only checking 2-digit integers. What about 3-digits? | deepmind/aqua_rat |
If we take 53, the pair is 51 and 53, meaning x = 26.
8x = 8*26
Reject since perfect square is not possible.
If we take 71, the pair is 69 and 71, meaning x = 35.
8x = 8*35
Reject since perfect square is not possible.
I tried each option and the possibilities and then got lost.
Then, it occurred to me that I was only checking 2-digit integers. What about 3-digits?
Starting with option a, only 2-digit odd integer whose sum is 2 is 101.
If we take 101, the pair is 99 and 101, meaning x = 50.
8x = 8*50 = 400
root of 400 = 20 which is indeed equal to 9+9+1+1.
Answer (A). | deepmind/aqua_rat |
Tanya prepared 4 different letters to 3 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
A)1/24
B)1/8
C)1/4
D)1/3
E)3/8
Correct Answer:C)1/4
Rationale: I hope you are familiar with basic probability fundas -
Let's say you have just ONE letter and TWO envelopes ONE of which is correctly addressed and the other addressed incorrectly.
What's the probability of putting the Letter in the correctly addressed envelope -
To answer this question - we see IN HOW MANY WAYS can the letter be put into the envelope - you could put it (assuming you don't know which envelope is which) in either of the two - so in total you have TWO ways of shoving the letter in. However, there's only ONE way in which it can go into the correctly addressed envelope -
so 1/2 is the prob of putting in correct envelope. This is easy.
Now in our current problem - let's say we have just ONE letter but FOUR envelopes. Only one of these envelopes has the address corresponding to the letter. The remaining three envelopes are incorrectly addressed.
So the probability that you will put the letter correctly is 1/4. Right?
What happens if i ask you the reverse question? what is the prob for putting it in the incorrect envelope. Suddenly you have three envs that are incorrect so you can put the letter incorrectly with a prob of 3/4. Right?
The whole problem can be broken down into Four Events that will fulfill the requirement of the question
Event 1 - E1
We know that prob of putting ONE Letter correctly is 1/4. Now once ONE letter has been put CORRECTLY, what are you LEFT with? You are left with THREE ENVELOPES and the remaining THREE letters. Since the one letter has been put correctly (though technically we have just calculated the PROBABILITY that the first letter goes into the correct envelope) we have the remaining THREE Letters and THREE envelopes.
Event 2 - E2 | deepmind/aqua_rat |
Event 1 - E1
We know that prob of putting ONE Letter correctly is 1/4. Now once ONE letter has been put CORRECTLY, what are you LEFT with? You are left with THREE ENVELOPES and the remaining THREE letters. Since the one letter has been put correctly (though technically we have just calculated the PROBABILITY that the first letter goes into the correct envelope) we have the remaining THREE Letters and THREE envelopes.
Event 2 - E2
Let's take letter number 2 now - what is the probability that it LANDS in the INCORRECT envelope. Again by the same logic as above - there are 3 envelopes remaining out of which ONLY ONE has the correct address for LETTER number 2. The remaining 2 have INCORRECT address and LETTER NUMBER 2 could go in either of these 2 to meet our condition.
Thus the probability of this event is 2/3
So till now what we have done is -
we have calculated the prob of shoving Letter number 1 in correct env -- 1/4
we have calculated the prob of shoving Letter number 2 in INcorrect env --- 2/3
Event 3 - E3
Now let's take letter number 3 - again according to question we want to shove this in the WRONG envelope. There are 2 remaining envelopes and hence the prob of shoving this in the wrong env (or equally in the RIght env) is 1/2.
Finally we come to event E4 - the Letter number 4. This has only one way of going in so its probability of being put into the WRONG envelope is 1.
ok so we can see that our grand event is actually a combination of FOUR EVENTS happening - each with a probability of its own. So to calculate the total probability of the Grand Event itself we just multiply the individual probabilities since each event happens INDEPENDENTLY of each other
Egrand = 1/4 * 2/3 * 1/2 * 1/1 = 1/12
However at this point - I must introduce one last element in this question -since there are FOUR Letters - what we saw above was JUST ONE SEQUENCE of events leading to the desired result.
If we arbitrarily call the letters L1 thru L4, and let's say the above was an example in which we started by Picking up Letter L1 and worked thru the remaining letters, we could have equally well started out with letter L2 or L3 or L4. | deepmind/aqua_rat |
However at this point - I must introduce one last element in this question -since there are FOUR Letters - what we saw above was JUST ONE SEQUENCE of events leading to the desired result.
If we arbitrarily call the letters L1 thru L4, and let's say the above was an example in which we started by Picking up Letter L1 and worked thru the remaining letters, we could have equally well started out with letter L2 or L3 or L4.
Thus since each of these events ARE MUTUALLY EXCLUSIVE, meaning THEY CAN NEVER HAPPEN ALL THE SAME TIME BUT ONLY ONE LETTER AT A TIME, to calculate the TOTAL PROBABILITY of we will add the individual probabilities 1/12 + 1/12 + 1/12 + 1/12 which works out to 1/4.
C | deepmind/aqua_rat |
Of the 14,210 employees of the anvil factory, 3/7 are journeymen. If half of the journeymen were laid off, what percentage of the total remaining employees would be journeymen?
A) 14.3%
B) 16.67%
C) 33%
D) 28.6%
E) 49.67%
Correct Answer:A) 14.3%
Rationale: The exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers (also because otherwise the ratio would be incorrect). Since the question is about percentages, the actual numbers will be meaningless, as only the ratio of that number versus others will be meaningful. Nonetheless, for those who are curious, each 1/7 portion represents (14210/7) 2,030 employees. This in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers.
If half the journeymen were laid off, that would mean 1/7 of the total current workforce would be removed. This statistic is what leads many students to think that since half the journeymen are left, the remaining journeymen would represent half of what they used to be, which means 1/7 of the total workforce. If 1/7 of the workforce is journeymen, and 1/7 is roughly 14.3%, then answer choice A should be the right answer. In this case, though, it is merely the tempting trap answer choice.
What changed between the initial statement and the final tally? Well, you let go of 1/7 of the workforce, so the total number of workers went down. The remaining workers are still 1/7 of the initial workers, but the group has changed. The new workforce is smaller than the original group, specifically 6/7 of it because 1/7 was eliminated. The remaining workers now account for 1/7 out of 6/7 of the force, which if we multiply by 7 gives us 1 out of 6. This number as a percentage is answer choice B, 14.3%. | deepmind/aqua_rat |
Using the absolute numbers we calculated before, there were 4,060 journeymen employees out of 14,210 total. If 2,030 of them are laid off, then there are 2,030 journeyman employees left, but now out of a total of (14,210-2,030) 12,180 employees. 2,030/12,180 is exactly 1/6, or 16.67%. The answer will work with either percentages or absolute numbers, but the percentage calculation will be significantly faster and applicable to any similar situation.
The underlying principle of percentages (and, on a related note, ratios) can be summed up in the brainteaser I like to ask my students: If you’re running a race and you overtake the 2nd place runner just before the end, what position do you end up in?
The correct answer is 2nd place.
Percentages, like ratios and other concepts of relative math, depend entirely on the context. Whether 100% more of something is better than 50% more of something else depends on the context much more than the percentages quoted. When it comes to percentages on the GMAT, the goal is to understand them enough to instinctively not fall into the traps laid out for you.
A | deepmind/aqua_rat |
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