ADURO Quant Lessons  Set 1

Q1) With a given amount X .. Mr. Y can buy 20 E or 12 S or 5 P. If he purchased an equal number of E and S and P with an amount Z and given that Z = 3X, Find the number of Ps bought by him ?
Let X =100
Price of E/S/P  5/8.33/20
Price of E+S+P= 33.33
Number of sets for X(100 ) = 100/33.33 =3
Number of sets fo 3X (300 ) = 3X3 = 9Other Methods:
#1
Price ratio of P:S:E=12:5:3.
Consider X=60 = > Z=180 = > Total share of P=108. Number of P=108/12=9
#2
One can think of an analogous problem in Time work Domain:
P, S, E can finish a job in 5, 12, 20 days respectively. They are assigned to complete a work together which thrice as large as the previous one. In how many days the work will get completed if each one works with the same efficiency.
#3
Treat the first set of equations as amount of work done and time taken by each person respectively to complete the task
In the next set of equations, they all work together to complete a task working equal no.of days
We have to find that "equal" no .of daysQ2) There are z digits in the decimal expression of the natural number N,while there are y digits in the decimal expression of N^3. Then which of the following cannot be equal to y+z?
(a) 20 (b) 26 (c) 35 (d) 45 (e) none of these
Solution:
As per the question,
10^(z−1) < = N < 10^z = > 10^(3z−3) < = N^3 < 10^3z.
So, y can be {3z − 2, 3z − 1, 3z} and y+z can be {4z2 ,4z1,4z)
Clearly all the natural numbers except those of type 4k+1 satisfy the value for y+z.
So, 35 cannot be the value.
= > Choice (d) is the right answerQ3) A real number x is chosen at random in the interval [21/2, 21/2]. What is the probability that the closest integer to x is odd?
a) 1/21 (b) 5/11 (c) 10/21 (d) 11/21 (e) 1/2
From [10.5,10.5] There can be 21 intervals [10.5,9.5],[9.5,8.5].....[9.5,10.5]..and there are 10 intervals [9.5,8.5]....[8.5,9.5]..in which any x selected in those intervals is closest to an odd integer.
So the probability is 10/21
= > Choice (c) is the right answer
Q4) A positive integer p is called "CAT number" if p^3 + 7p = q^3 + 133 for some positive integer q. The sum of all such "CAT numbers" is
(a) 24 (b) 26 (c) 30 (d) 34 (e) 38
We consider the size of p relative to q. If p = q, then q is a CAT number if and only if
7q = 133. In this case, we get 19 is a good number. Now, suppose that p > = q + 1. Then
q^3 + 7q − 133 = p^3 > = (q + 1)^3 = q^3 + 3q^2 + 3q + 1
so that 3q^2 − 4q + 134 < = 0. For positive integers q, it is easy to see that this is impossible. It remains to consider p < = q − 1. For such p, we have q^3 + 7q − 133 = p^3 < = (q − 1)^3 = q^3 − 3q^2 + 3q − 1 which implies 3q^2 + 4q − 132 < 0. It follows here that q < 6. One checks that 6^3 + 7 · 6 − 133 = 5^3 and 5^3 + 7 · 5 − 133 = 3^3,
so 6 and 5 are CAT numbers. For q < = 4, we see that q^3 +7q−133 < 0 and, hence, q^3 +7q−133 cannot equal p^3 for a positive integer p. Therefore, the sum of all CAT numbers is 19+6+5 = 30.Q5) If P : Q = 1 / 2 : 3 /8 , Q : R = 1 / 3 : 5 / 9 and R : S = 5 / 6 : 3 / 4 , Then P : Q : R : S = ?
P : Q = 1 / 2 : 3 / 8 = 4 : 3 , Q : R = 1 / 3 : 5 / 9 = 3 : 5 and R : S = 5 / 6 : 3 / 4 = 10 : 9
P : Q = 4 : 3 , Q : R = 3 : 5 , R : S = 10 / 2 : 9 / 2 = 5 : 9 / 2
P : Q : R : S = 4 : 3 : 5 : 9 / 2 = 8 : 6 : 10 : 9Q6) What digit does “a” represent, if 35! = 10333147966386144929a66651337523200000000?
(a) 4
(b) 6
(c) 2
(d) 1What digit does “a” represent, if
35! = 10333147966386144929a66651337523200000000?
(a) 4 (b) 6 (c) 2 (d) 1
35!= 1 * 2 * 3 * 4 * 5 *... * 35
so the above number will also be divisible by 3,9,11( such number which uses the sum of number to check the divisibility)
Check by 3: you will get the different values of a which also have to satisfy for 9,11.
Check by 9: (Sum of digits +a) has to be divisible by 9.
Check by 11 directly,
Sum of odd places sum of even places= 11 n or 0.
(66+a)72=0, as a cant be greater than 9. to give the difference multiple of 11.
so a=6