CSAT 2nd Second Paper 4 solution
1. Age of left = previous age - name of person present * increase in average age
=15-(10-1)*(18-15)
=15-9*1
=15-9= 6 years
2. ( 5 +1)(2 *5+1) / 6 = 6*11/6=66/6=11
3) (10 +1) (10+2)/3
=11*12/3= 132/3
=44
4) n+1/2 = 9+1/2=10/2 =5
5) average speed = 2xy/x+y
=2*15*12 all divided by15 +12
= 360/27= 131/3
6) 20= 200000
12*12500=150000
remains8(200000-150000)
=50000
= 50000/8 = 6250
7)10*26=260
lets tax passed and failed student is x
Average failed student = (10 -x)
passed = 80*x= 80x
failed student (10 - x )* 20
= 200 - 20 x
80x +200 - 20x = 260
60 x = 60
x = 60/60 = 1
8. E D C A B
9.b
11.5% increase = 16000 *5/100= 800
16000+800=16800 salary
reduce transport allowance = 800
16800 - 800 = 16000.
12. lets tax passed and failed student is =x
average failed = (10-x)
passed = 80 * x = 80x
failed = (10-x)*20
200-20x
80x+200-20x= 260
60x = 260 -200
60x=60
x=60/60=1
14. Dice = ( 1,2,3,4,5,6)
n(s) = 6( 1,2,3,4,5,6)
n(E) = 3( 2, 4, 6)=3/6=1/2
15) Multiple of 2 = 3 , 6
n(E) = 2( 3,6)
n(s) = 6( 1,2,3,4,5,6)
Comments
Post a Comment