Quant Boosters  Quant Mixed Bag  Set 1

[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

2xy  3x  y = 4
2x(y  3/2)  y = 4
add 3/2on both sides
2x(y  3/2)  y + 3/2 = 4 + 3/2
2x(y  3/2)  1(y  3/2) = 11/2
(2x  1)(y  3/2) = 11/2
(2x  1)(2y  3) = 11
11 = 1 * 11
Case1) 2x  1 = 1 => x=1
2y  3 = 11, y = 7
Case2) 2x  1 = 11
x = 6
2y  3 = 1, y = 2
so 7/9