Find the least number which when divide by 2, 3, 4, 5 and 6 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 7 leaves no remainder?
		A. 210
B. 119
C. 126
D. 154
		A. 210
B. 119
C. 126
D. 154
		A. 1/4
B. 1/2
C. 3/4
D. 3/5
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then  = 18/36 = 1/2
P(E) = 1 – 1/2 = 1/2.
		A. 12km/hr
B. 10km/hr
C. 14km/hr
D. None of these
		A. Rs.3000
B. Rs.3600
C. Rs.2400
D. Rs.4000
E. None of these
Explanation:
Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 – r) => 3r = 12000 – 2r
=> 5r = 12000 => r = 2400.
		A. 1/3
B. 7/9
C. 5/12
D. None of these
M/P = M/N * N/O * O/P = ¾ * 5/7 * 7/9 = 5/ 12
		A. 20:23
B. 34:43
C. 32:45
D. 37:45
Explanation:
A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
		A. 4 : 3
B. 8 : 7
C. 4 : 1
D. 6 : 5
Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.
Let the dimensions of the carpet in the second case be 7y, 3y units respectively.
 From the data,.
2(3x + 2x) = 2(7y + 3y)
 => 5x = 10y
=> x = 2y
Required ratio of the areas of the carpet in both the cases
= 3x * 2x : 7y : 3y
= 6×2 : 21y2
= 6 * (2y)2 : 21y2
= 6 * 4y2 : 21y2
= 8 : 7