CWC ACCOUNTANT OTHERS Q.A. TIME DIST. AGE -PL GPAY RS. 499 TO 9003037557...
Qn.11 let the distance between A and B be x
x/48 = x/(48+24) + 5/2
x/48 = x/72 + 5/2
x/48 = (2x+360)/ 144
144x = 96x + 360*48
144x – 96x = 360*48
48x = 360* 48
x= 360 km
12. A/B = 5/4 also A /C = 5/2
4A = 5 B 2A = 5C
= 10C = 5B or B = 2C
(2C -6) / (C+2) = 3/2
4C -12 = 3C +6
4C – 3C = 6 + 12 = 18 or C = 18
2A = 5C or 2 A = 18*5 = 90
A = 45.
A ‘s age 3 years ago = 45 -3 = 42
13. Answer: C
Let the speed of train A and train B be (x – 10)
m/sec and x m/sec respectively.
Length of Train A = 3a
Length of Train B = 4a
3a + 90 = (x – 10) * 16 ---(1)
x = 4a/8 = a/2
Substitute x value in equation (1),
3a + 90 = 16(a/2) – 160
a = 250/5 = 50
x = 50/2 = 25 m/sec
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