GRE Quant - Sample Questions for Practice

Are you prepared for the GRE Quantitative Reasoning section? Try some practice questions prepared by the Jamboree faculty.

• Q 1.

  A magician marks-off on a stick of length 1 yard in thirds and fifths and breaks the stick at the marked points. What is the maximum number of pieces which are equal in length?

  • 2
  • 3
  • 4
  • 5
  • 6

  Solution:

  Thirds means = 13 , 23 , 33 , 43 ,…..and so on.
  Fifth means = 15 , 25 , 35 , 45 ,….. and so on.
  A stick of length 1 yard is marked in thirds and fifths.

  Stick is marked at these points. Hence, we need to find out the length of each point.
  AB = 15 - 0
  BC = 13 -15, and so on.
  But, this method would be time-consuming, as it involves 7 fractions subtraction.
  Therefore, to reduce calculations, let's multiply the entire line by product of denominators (i.e., 3*5=15).
  Multiply all values with 15.

  Now, length of AB=3, BC=2, CD=1, DE=3, EF=1, FG=2, GH=3
  There are a maximum of 3 pieces which have the same length (AB, DE, GH)
  Key takeaway- working with integers will be more convenient than working with fractions.

• Q 2.

  Ron will walk from intersection A to intersection B along a route that is confined to the square grid of three streets and four avenues shown in the map above. How many routes from A to B can Ron take that have a minimum possible length?

  Solution:

  To go from A to B, let's represent
  1 right move ? r
  1 up move u
  On analysis, we can infer that to go from
  intersection A to intersection b along the
  grid section Ron will need to take 5 steps
  3 right and 2 up
  So, some of the different routes can be:
  RRRUU
  UURRR
  URURR
  RURRU, and so on.

  But, we won't prefer writing all the routes and rather we need a method.
  Every time when we are writing a different route, it has 2 u’s and the 3 r’s but in different arrangements. Therefore, we treat it like a question of P and C.
  [ N things can be arranged in a line in n! Ways. If there are repeating alphabets, we divide by their factorial.]
  It’s a 5 alphabetic word, with 2 u’s and the 3 r’s.
  Therefore, uuurr= 5!3!*2!
  (U represents thrice, so 3!
  R represents twice, so 2!)
  = 5*4*3*2*13*2*1*2*1
  =10

• Q 3.

  Investment of $4000 at R% per annum compounded annually will become $16000 in 8 years. If $2000 is invested at R% per annum compounded annually, in how many years will the investment become $16000?

  Solution:

  Principal, P1 = $4000
  Rate, R1 = R%
  Amount, A1 = $16000
  Time, T1 = 8 years.
  Using the formula of compound amount: A = P (1 + R100)T
  16000 = 4000 (1 + R100)8 (1 + R100)8 = 4
  Applying the property of exponent: xa = y
  x = (y)1/a
  So, (1 + R100) = (4)1/8 …………………………………………….. (1)
  Now,
  Principal, P2 = $2000
  Rate, R2 = R%
  Amount, A2 = $16000
  Time, T2 = T years.
  Using the formula of compound amount: A = P (1 + R100)T
  16000 = 2000 (1 + R100)T
  (1 + R100)T = 8
  So, (1 + R100) = (8)1/T …………………………………………….. (2)
  Equating values of (1 + R100) from equation (1) & (2);
  (4)1/8 = (8)1/T
  (22)1/8 = (23)1/T
  (2)1/4 = (2)3/T { (xa)b = (x)a b }
  14 = 3T {Xa = Xb; and x ?1 and-1 then a = b.}
  T = 12 years.

  Alternatively:
  As the investment of $4000 is becoming $16000 in 8 years so in other words we could say that the investment is becoming 4 times in 8 years or is becoming 2 times in 4 years.

  So investment of $2000 will become $4000 in 4 years
  And $4000 will become $8000 in next 4 years
  And $8000 will become $16000 in next 4 years
  or investment of $2000 will become $16000 in 12 years.