RBI Grade B Officers 22 Nov 2015 Paper

(4) Let the total distance be3y km.According to the question,Speed × Time = Distance$\therefore x \times \frac{5}{2} = 2y$⇒ 5x = 4y     --- (i)Again,$(x + 2) \times \frac{50}{60} = y$$\Rightarrow (x + 2) imes 5 = 6y \quad ext{--- (ii)}$On dividing equation (ii) by (i),$\frac{(x + 2) \times 5}{5x} = \frac{6y}{4y}$$\Rightarrow \frac{x + 2}{x} = \frac{3}{2}$$\Rightarrow 3x = 2x + 4 \Rightarrow x = 4$From equation (i),$5 imes 4 = 4y \Rightarrow y = 5$∴ Total distance = 3y = 3 × 5= 15 km.Calculations (13–17) :Applications received by universityACourses of (Arts + Management+ Law)= 800 + 720 + 400 = 1920$herefore 30\% \equiv 1920$$therefore 70\% \equiv \frac{1920 \times 70}{30} = 4480$∴ Courses of Commerce +Science + Engineering⇒ 4480Engineering$\Rightarrow \frac{720 \times 100}{60} = 1200$Applications for Science = x$herefore x + x + 200 = 4480 - 1200 = 3280$$\Rightarrow 2x = 3280 - 200 = 3080$$\therefore x = \frac{3080}{2} = 1540$∴ Applications for commerce= 1740Applications received in UniversityBScience $\Rightarrow \frac{1540 \times 80}{100} = 1232$Art ⇒ 780∴ 15% = 780$\therefore 100\% = \frac{780}{15} \times 100 = 5200$Commerce $\Rightarrow \frac{780 \times 100}{60} = 1300$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 $= \frac{330}{3} = 110$∴ Engineering + Management= 1888 – 110 = 1778∴ Engineering $= \frac{1778}{2}$= 889 ⇒ Management