(4) Let the total distance be3y km.
According to the question,Speed × Time = Distance
∴x×=2y⇒ 5x = 4y --- (i)
Again,
(x+2)×=y⇒(x+2)×5=6y−−−(ii)On dividing equation (ii) by (i),
=⇒=⇒3x=2x+4⇒x=4From 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%≡=4480∴ Courses of Commerce +Science + Engineering
⇒ 4480
Engineering
⇒=1200Applications for Science = x
∴x+x+200=4480–1200=3280⇒2x=3280–200=3080∴x==1540∴ Applications for commerce= 1740
Applications received in UniversityB
Science
⇒=1232Art ⇒ 780
∴ 15% = 780
∴100%=×100=5200Commerce
⇒=1300Engineering + Management +Law
⇒
5200–1232–780–1300=1888Again, Engineering + Management= 2 (Arts + Law) – 2
⇒ 1888 – Law= 2 (780 + Law) – 2
⇒ 1888 – Law = 1560 + 2 Law – 2
⇒ 3 × Law = 1888 – 1558
= 330
⇒ Law
==110∴ Engineering + Management
= 1888 – 110 = 1778
∴ Engineering
== 889 ⇒ Management