Quant Boosters  Quant Mixed Bag  Set 1

[Credits : Tanveer Iqbal]
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 days

Q21) 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

[Credits : Tanveer Iqbal]
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 answer

Q22) 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

[Credits : Tanveer Iqbal]
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

Q23) 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

[Credits : Taveer Iqbal]
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.

Q24) 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 = ?

[Credits : Tanveer Iqbal]
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 : 9

Q25) What digit does “a” represent, if 35! = 10333147966386144929a66651337523200000000?
(a) 4
(b) 6
(c) 2
(d) 1

[Credits : Tanveer Iqbal]
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

Q26) In base 7, 65324 + 50231  133 × P = 130462. What is the value of 'P'?
a) 110
b) 101
c) 1001
d) 111

[Credits : Ashutosh Mishra]
Going by last digits
4 + 1  3 = 2
Now second last digit
2 + 3  x = 6
5  x = 6
Only possible when you would have carried over other wise 6 won't be the result
12  6 = 6
So x = 6
Now only multiplying 133 by 111 gives second last digit as 6.

Q27) Sum of (1/101)+ (1/102)+ (1/103) + ... +(1/200) is
a. 1
b. 0.5
c. 1.85
d. None of these

[Credits : Ashutosh Mishra]
Method 1:
Sum of series of 1/n is ln(n). Ln is log to the base e.
Ln 200ln100
Ln(200/100)=ln(2)
E=2.3
Ln e base e=1
So answer will be less than 1 but can it be 0.5??
Rt 2=1.414
So rt e=1.7 approx
But 2 is greater than 1.7
So answer will lie between 0.5 to 1 so dMethod 2:
S = 1/200 + 1/2000... = 100 / 200 = 0.5
but we have terms 1/101 , 1/102..likewise which will increase the sum
and if we substitute S = 1/101 + 1/101 ... = 100 / 101 = 0.99.
S = 0.5 < S < 1Method 3
S = {(1/101) + (1/102) + .. + (1/150)} + {(1/151) + (1/152) + ... + (1/200)}
1/100 > 1/101 so if we find the sum of below it wil be greater than the asked one
S < {(1/100) + (1/100) + .. + (1/100)} + (1/150) + (1/150) + ... + (1/150)
= 50/100+ 50/150 = (1/2) + (1/3) = 5/6 ~ 0.83
also S > {(1/150) + (1/150) + .. + (1/150)} + {(1/200) + (1/200) + ... + (1/200)} = {50/150} + {50/200} = (1/3) + (1/4) ~ 0.58
So 0.58 < S < 0.83

Q28) 1 + 2 x 2 + 3 x 2^2 + 4 x 2^3 + ... + 100 x 2^99 = ?
a. 99 x 2^100
b. 99 x 2^100 + 1
c. 99 x 2^100  1
d. 100 x 2^100

[Credits : Ashutosh Mishra]
Method 1: (that I wouldn't recommend)
S = 1 + 2.2 + 3.2² + 4.2³ .... 100.2^99
2S = 2 + 2.2² + 3.2³ .... 99.2^99 + 100.2^100
S  2S = (10) + (2.22) +( 3.2²  2.2²) ..... (100.2^99  99.2^99)  100.2^100
S = 1 + 2 + 4 + 8 ..... 2^99  2^100.
S = 100.2^100  ( 1 + 2 + 4 + 8+...... 2^99)
S = 100.2^100  1(2^100  1)/(21).
S = 100.2^100  2^100 + 1
S = 99.2^100 + 1Method 2 : Generalize
Take first 2 terms
1 + 2 × 2 = 5
So if answer would have been for only 2 terms then answers would have been
1 × 2^2
1 × 2^2 + 1
1 × 2^2  1
2 × 2^2
Now you have the answer and this is a CAT problem :)

Q29) Consider the set S = (1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?
(a) 3
(b) 4
(c) 6
(d) 7
(e) 8

[Credits : Ashutosh Mishra]
Let number of terms in the arithmetic progression be n, then
1000 = 1 + (n–1) d
=>(n–1) d = 999
=> n – 1 = 999/d
Since n is an integer, so n – 1 is also an integer. This means that ‘d’ is a factor of 999.
Now 999 = 33 × 37. so the total factors of 999 are 4 × 2 = 8.
Out of these 8 factors one factor is 999 and we will reject it as in that case there will be only two terms in the
A.P. i.e. 1 and 1000, which is not possible.
Hence, ‘d’ can take 7 different values.
So, in total, 7 APs are possible

Q30) If 3x + y + 4 = 2xy, where x and y are natural numbers, then find the ratio of the sum of all possible values of x to the sum of all possible values of y.
a) 2/3
b) 15/19
c) 17/21
d) 7/9