Knewton Brutal GMAT Challenge – Week 10

by on May 26th, 2010

There is a Knewton shirt out in the veld, floating around all free. You can claim it if you answer this question correctly, and show your work.

The first term in a sequence is a, and each term thereafter is k/11 greater than the term before it. If a and k are both positive integers but only one of them is a prime number, what is the sum of the first 100 terms in the sequence?

  1. 2a+11k=46
  2. 3k=14-{2/3}a

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

34 comments

  • D. Each Statement Alone is Sufficient.

    A) 2A + 11K = 46; The question states that both A and K must be positive integers and that one must be a prime number. There are only three primer numbers for K – 1, 2, 3 – that will allow the equation to be less than 46 since 11*5 > 46. We can quickly eliminate 1 and 3 since K must be even because the first part of the equation cannot be odd, i.e., any number times an even number must be even.

    Therefore, K must be 2. We can now quickly solve for a, which is equal to 12. Since we know both variables from the question stem, we know we can find the 100th term (no need to solve for it!) Sufficient.

    B) 3K = 14 – 2/3A; First, multiply through by 3 to get: 9K = 42 – 2A; Second, rewrite the equation to mirror statemen 1: 2A + 9K = 42. Using the same logic from A, K must equal 2 and a must equal 12. Again, since we know both variables from the question stem, we know we can find the 100th term. Sufficient.

    • I disagree. from statement 2.
      you can not only come up with a=12 k=2, but also with a=3 k=4 , satisfying condition in question stem.
      thus answer must be A.

    • In statement 1 I don t see why A=1 and K=4 doesn t satisfy the conditions

    • Never mind. forgot 1 is not a prime

  • Answer is D

    Sum of nth terms in seqiuence : n/2( first term +( last term -1) * common difference)

    we need to find first term & common difference

    from st 1

    2a+11 k = 46

    only K =2 & a= 12 can take values

    so common difference = K/11 = 2/11
    and first term = 12

    Apply them in the formula for sum of 100 terms:
    so Sum = 100/2 (12 + 99 *2/11)
    50 * 30
    = 1500
    Sufficient..

    St 2: same thing applies

    Only for a= 12 & k = 2 ,the equation fits in.

    Sum can be found that way to be 1500

    So pick D

    • I have been trying so hard for past 9 weeks and my luck never favours me!! God just that I am so unlucky!!

      Josh r u hearing??

  • Ans D.

    Given : Current Term (a)= K/11+Previous Term (a-1)
    Between a and K only one should be prime. Both are integers
    we need to find the first term a and constant K of sequence.
    Stmt: 1
    2a+11K=46
    2a --- always even
    11K should be even inorder to satisy even+even=even
    Take K=2,4,6
    Only K=2 will be suffice to solve the problem.
    2a+22=46 -> a=12
    We know a and K, now we can substitute in orignial sequence Current Term (a)= K/11+Previous Term (a-1) and find the previous term of sequence.
    Then we find the sum of sequence using Arithmetic progression formula.
    Stmt 1 is sufficient to solve the problem
    stmt 2:
    2/3(a)+3k=14
    2a+9K=42
    EVEN+EVEN =EVEN
    so K should be even. Take K =2,4,6
    Only K=2 will be suffice to solve the problem then K =12.
    So, similar to stmt 1 we can follow the same steps to find the sum of first 100. Stmt 2 is suff.

    So Ans is D.

  • I think the answer is (a). Here's why:

    the sum of 100 terms necessarily depends on finding what is a and k. if we find a and k, we can find the sum of 100 terms.

    statement 1:
    2a + 11k = 46

    since k is a positive integer, k must be 1, 2, 3, 4. furthermore, since 2a must be even, 11k must be even as well to sum to 46. so k might = 2 or 4.

    if k = 4 (which is not prime), then a = 1 (not prime). this is impossible, so k = 4 can't be the answer.

    if k = 2 (which is prime), then a = 12 (which is not prime). great!

    statement 1 is sufficient.

    statement 2:
    3k = 14 - 2/3a
    (algebra) ==> 9k + 2a = 42

    we can have two possible solutions:
    k = 2 (prime), a = 12 (non-prime)
    k = 4 (non-prime), a = 3 (prime)

    statement 2 is insufficient. ==> A answer.

  • Good catch "mskumcp98".
    Isn't Sum of n terms in a sequence is given as S=n/2{ (2*first term) +( last term -1) * common difference}. Do correct me if I am wrong. Thanks!!!

    • (D) EACH statement ALONE is sufficient.
      Series got numbers {a, (a+k/11), (a+2k/11),..(a+99k/11)}
      Sum=n/2(first term+last term)=100/2(a+a+99k/11)=50(2a+9K)
      => Sum=50(2a+9k)

      1) 2a+11k=46 (Even+Even=Even
      If K is prime then k=2,3,5,7,....
      k=2 gives a=12 (K prime and a non-prime) satisfies the given condition. Substitute values of k and a in Sum=50(2a+9k)
      Sufficient

      2) 2a+9k=42
      Substitute 42 in Sum=50(2a+9k).
      Sufficient

  • ANSWER IS D):
    using 1) only we get (a=1 , k=4 { not correct as no one is prime }) and (a=12 and k=2 { correct as only k = prime} ).
    thus a= 12 and k= 2 sufficient.
    1st term = 12 and 100th term = 12 + 99 * (2/11) = 30
    thus sum = 100/2 (12+30) = 2100 :: NOTE : { Using n/2 (first + last term) }

    using 2) only:
    we get ( a= 3, k=4 { correct as only a = prime} ) and but Using a= 3 and k = 4
    => we get 1st term = 3 and 100th term = 3 + 99 * (4/11) = 39
    => thus sum = 100/2 (3+39) = 2100 :: NOTE : { Using n/2 (first + last term) }

    also we get (a=12 and k=2 { correct as only k = prime} )
    1st term = 12 and 100th term = 12 + 99 * (2/11) = 30
    thus sum = 100/2 (12+30) = 2100 :: NOTE : { Using n/2 (first + last term) }

    So both the cases give us sum 2100 thus sufficient.

    • Nice work, Dinesh! The only thing I'd point out is that as Vibhavender (and others) have shown, statement 2 actually gives you the sum for ANY values of a and k (that satisfy 3k = 14 – 2/3a), so you don't even have to calculate that the sum for both pairs of a and k (3, 4 and 12, 2) is equal to 2,100.

      Great job though: you certainly got to a solid proof of the right answer, D, so you and Vibhavender (for a more direct understanding of Statement 2) each win a Knewton T-shirt! If you send an email to abby AT knewton.com, telling her your physical address, we'll get your shirt to you as soon as possible!

    • Why are we not considering for statement 1:
      a=23 and k=0 (23*2 + 11*0 = 46)

      this gives the sum of 100 terms as:
      Sum = (100/2)(2*23 + 99*0) = 50*46 = 2300
      ???
      please justify John

    • Anirudh,
      That's because its mentioned that a and k are both positive integers and u cannot consider zero for any of the variables...

    • @Vibhavendra

      But isn't zero both non -ve and non +ve??
      Kindly explain further.

  • Answer D.
    I will go with anthony's explanation for statement 1 sufficiency. (off course somebody has mentioned that prime numbers are 1,2 and 3). I wouldn't mind that as long as first statement is sufficient.

    Second statement is also sufficient because....
    As everyone got it it is simplified version 2a+9k = 42
    now...
    sum of 100 terms will be...
    a + a+(k/11) + a+(2k/11) +......+ a +(99k/11)
    = 100a + [(k/11)(1+2+...+99)]
    = 100a + [(k/11)(99(99+1)/2)..since sum of first n numbers is (n+1)/2

    = 100a + (k/11)(100*99/2)

    =100(a+99k/22)

    =100(a+9k/2)

    = 50(2a+9k)

    where statement 2 gives u the value of 2a+9k...
    so sufficient....

    would like to add one more thing...
    for all those who consider statement 2 sufficient JUST BECAUSE a=12 and k=2 satisfies it..
    THINK AGAIN....
    a=3,k=4 also satisfies the equation and every other condition mentioned..both positive..both integers and one prime and other non prime...
    please let me know in case I have overlooked any thing here....

    • Vibhavender, great job! Statement 1 is indeed sufficient, as shown above, and your approach to statement 2 shows that finding the sum allows you to factor out 50(2a + 9k), so knowing that 2a + 9k = 42 lets you find the sum immediately, without spending time determining all the possible values for a and k.

      For your more complete explanation of Statement 2, you and Dinesh Signh each win a Knewton T-shirt! If you send an email to abby AT knewton.com, telling her your physical address, we'll get your shirt to you as soon as possible! Congratulations!

  • sorry sum of first n terms is n*(n+1)/2...typo there....

  • Sum of integers in arithmetic sequence is Sn = n/2 (2a+(n-1)d)
    from this we get equation for sum of 100 terms as S100 = 100/2 [2a+99*k/11]
    which is equal to 50(2a+9k)
    for knowing the sum we should know values of a and k, it is given that one of them is prime, which one is not known if it is a, a could be any number 2/3/5 , if it is k, k could be any number divisible by 11 (i am considering k not to be prime as k/11 should be integer value, hope we are dealing with sequences of integer numbers) if you go by that logic then k can never be prime, which leaves us a, as stated a can be any prime number.

    Statement 1: 2a + 11k = 46
    not sufficient as our equ needs value of 2a+9k
    Statement 2: resolving 3k=14-2/3a
    2/3a=14-3k
    2a=3(14-3k)
    2a=42-9k
    2a+9k=42
    hence we have sum of 100 in the sequence as 50(2a+9k)=50(42)
    Hence B alone is sufficient.

    • Hey Nikhil, The question never mentioned that K/11 shud be an integer, it says only K shud be integer....so 2a + 11k = 46 has only one set of answer that is a= 12 and K=2 satisfying all the conditions..i.e both positive, both integers and one prime and other non-prime...

  • I think the correct answer is (b) :

    Here the sequence is a, a+k/11,a+2(k/11),a+3(k/11).....
    Now the sum of the first 100 terms a+k/11+a+2(k/11)+a+3(k/11)+.....+ a+99(k/11) is:
    100a+ k/11(1+2+...+99)
    100a+k*(4950)/11 (sum of n terms= N*(N+1)/2)
    so, 100a+450k is the sum of first 100 terms.

    1) 2a+11k=46
    From this you cannot get the value of 100a+450k.
    So, this is insufficient.

    2) 3k=14-2*a/3
    from this equation...
    9k=42-2a => 2a+9k=42 , now if you multiply this equation with 50
    you can get 100a+450k=2100.
    so, this is sufficient.

    Statement 2 alone is sufficient.
    so, the correct answer is B.

    • Correct answer is D:

      form 1) you can get a=12 and k=2,
      From this you can get the value of 100a+450K.

  • solving for the sum is exactly what GMAC wants you to do in this problem.
    all you need to know is if you can come up with consistent values of A and K, satisfying the condition in question stem.
    st 1 gives you a=12 and k =2 so SUFF
    st 2 gives you a=12 k=2 AND a=3 k=4 so INSUFF

    • But irrespective of the values of a and K in 2a+9k = 42, we can always get the exact value of the sum of the first 100 terms in the sequence which is 50(2a+9k) ......which I have deduced in my explanation above..so statement 2 is sufficient...

    • you're right.
      I should've checked to see if the sum would be the same regardless when I found values were inconsistent.
      my mistake.
      D

  • Ans .D

    Sum = n/2 (2a + (n-1)d)
    n = 100
    d = k/11

    Upon simplification
    Sum = 50 (2a + 9k)

    Stmt 1 : Only one pair of value satisfies
    a = 12 k = 2 (prime)
    Sum can be computed by plugging the values of a & k

    Stmt 2 : 3k = 14 - 2/3a
    The eqn. can be reduced to
    9k = 52 -2a
    2a + 9k = 52
    Again sum can be computed by plugging the value of 2a + 9k

    • Hey priya ..typo there...shud be 42 rather than 52

  • Answer is A.
    1st equation has 1 soln. (12,2)

    2nd eq has two soln (12,2) and (4,3)

    • Sorry 2 soln are (12,2) and (3,4)

    • yea but
      the sum is the same.

    • sry CALC mistake Both soln do give similar sum..

  • Answer is D.

    Here is why

    sum of first 100 terms is

    a+a+k/11+a+2k/11+...+a+99k/11
    = 100a+ k/11((1+99)*99/2)=100a+450k
    =50(2a+9k)

    Statement one is 2a+11k=46, and we know that a and k are both positive integers and only one of them is a prime number.
    the only possible solution to this statement is (k=2, a=12) and (k=4, a=1). So obviously from statement 1, we can deduct that k=2 and a=12, so it's sufficient to solve 50(2a+9k).

    statement two is 3k=14-2a/3 which can be rearranged to 2a+9k=42, then it's sufficient to solve 50(2a+9k).

    • Ans D.

      The sum = 50(2a + 9k)

      Now option A says 2a + 11k = 46. => k is even number because the sum is even.
      Now k can be 2 (only even prime) in that case a = 12
      Also k can be 4 a = 1. considering 1 as not a prime number, there is only one solution for 2a+k. So SUFF.

      From option B we can calculate 2a+9k SUFF.

  • Hi ,can any one of you tell me what triggered you to start with finding prime values for 'k' for the equation 2a+11k =46.

    I started with finding prime values for 'a' and by the time I reached to correct combination it was already more than 2 minutes.

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