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