Quant Boosters  CatOMania  Set 1

what is the probability of always having x + y > 0 ,where y < 0
a)1/2
b)1/8
c)3/4
d)1/44 cases
X negative
X zero
X positive and greater than y
X positive and less than y
We want only 1 case so 1/4In base 7, 65324 + 50231  133 × P = 130462. What is the value of 'P'?
a 110
b 101
c 1001
d 111Going 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.Aseem writes three singledigit numbers, starting from 1, on a black board. The numbers written by him form an arithmetic progression. Ruchika, who is a friend of Aseem, erases one of the numbers written on the black board. If the average of the remaining numbers is 5.5. The number erased by Ruchika is
a 7
b 1
c 4
d Cannot be determineda  d, a, a + D
a + a + D = 11
2a + D = 11
a  d = 1
3a = 12
A = 4Sum of (1/101)+ (1/102)+ (1/103) + ... +(1/200) is
a. 1
b. 0.5
c. 1.85
d. None of theseMethod 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 < 0.831 + 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 2^100Method 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 problemConsider 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) 8Let 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.What is the remainder when 12345678987654321 is divided by 1001 ?
111111111^2mod 1001
111111mod 1001=0
(111111 * 1000+111)^2mod 1001
111^2mod1001
12321
12 × 1000 + 321 mod 1001
12 × 1 + 321 = 309The least number which must be added to 3775 * 3789 to make the result the result a perfect square is?
a)64
b)49
c)121
d) 9Method 1
Take y = 3782
(y  7)(y + 7) = y^2  49
So if we add 49 it will be a perfect square.Method  2
Don't take x,
Take 1
1×15=15
1
Minimum number from options  15 + 9 = 24, not a square
Next 15 + 49 = 64 => doneSimilar questions came this year in CAT
See, if you generalize above like may be it is possible for 1 not for the number given
Taking the question first
2 × 16 = 32 + 49 = 81
3 × 17 = 51 + 49 = 100
So you can see it is a perfect square for every number.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 7/92xy  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 7/9Two joggers run around an oval track in opposite directions. One jogger runs around the track in 56 seconds. The joggers meet every 24 seconds. How many seconds does it take for the second jogger to run around the track?
As they run in opposite direction and they meet in 24 seconds 1st runner would run as much as 2nd runner would do in 32 seconds ( 56  24 = 32)
So ratio of time= 32:24 = 8:6 = 4:3
So total time will also be, 56:42