RBI Grade B Officers 22 Nov 2015 Paper

(4) Let the total distance be3y km.
According to the question,Speed × Time = Distance
∴x×

5

2

=2y
⇒ 5x = 4y     --- (i)
Again,
(x+2)×

50

60

=y
⇒(x+2)×5=6y−−−(ii)
On dividing equation (ii) by (i),

(x+2)x5

5x

=

6y

4y

⇒

x+2

x

=

3

2

⇒3x=2x+4⇒x=4
From equation (i),
5×4=4y⇒y=5
∴ Total distance = 3y = 3 × 5
= 15 km.
Calculations (13–17) :
Applications received by universityA
Courses of (Arts + Management+ Law)
= 800 + 720 + 400 = 1920
∴30%≡1920
∴70%≡

1920×70

30

=4480
∴ Courses of Commerce +Science + Engineering
⇒ 4480
Engineering⇒

720×100

60

=1200Applications for Science = x
∴x+x+200=4480–1200=3280
⇒2x=3280–200=3080
∴x=

3080

2

=1540
∴ Applications for commerce= 1740
Applications received in UniversityB
Science ⇒

1540×80

100

=1232
Art ⇒ 780
∴ 15% = 780
∴100%=

780

15

×100=5200
Commerce ⇒

780×100

60

=1300
Engineering + Management +Law
⇒5200–1232–780–1300=1888
Again, Engineering + Management= 2 (Arts + Law) – 2
⇒ 1888 – Law= 2 (780 + Law) – 2
⇒ 1888 – Law = 1560 + 2 Law – 2
⇒ 3 × Law = 1888 – 1558
= 330
⇒ Law =

330

3

=110
∴ Engineering + Management
= 1888 – 110 = 1778
∴ Engineering =

1778

2

= 889 ⇒ Management