The population of city P decreased by 20% every year and population of city Q decreased by 8% every year. If after two years population of city Q was 22088 more than that of city P and initially it was 20000 more than that of population of city P, then find the population of city Q initially?

Explanation

Let population of city P and Q be 625p and 625q respectively. ATQ 625𝑞−625𝑝= 20000 𝑞−𝑝= 32 ….(i) And, 625𝑞× 23 × 23 −625𝑝× 4 × 4 = 22088 25 25 5 5 529𝑞−400𝑝= 22088…. (𝑖𝑖) From (i) and (ii) 𝑝= 40, 𝑞= 72 So, initial population of city Q = 625𝑞= 45000 5 www.bankersadda.com | www.ssca S56. Ans. (d) Sol. Let rate of interest be r%. In 8 years, the investment becomes double at simple rate of interest. So, 4800 = 4800×𝑟×8 r=12 1 % Let investment in scheme A and Scheme B be Rs. y and Rs. z respectively. ATQ, 𝑟 𝑧×𝑟×4 𝑦{(1 + 100)2 −1} − 100 =50 121 𝑧×121 2 2×4 𝑦{(1 + 100)2 −1} − 100 = 50 17 𝑧 𝑦× − = 50 64 2 17y – 32z =3200 …………. (1) And y +z = 4800 …………. (2) On solving equations, 1 and 2, We get y = Rs.3200 And z = Rs.1600 S57. Ans(c) Sol. Let the numerator and denominator of a fraction be x and y respectively. ATQ 𝑥+ 7 = 2𝑦… … . . (𝑖) And, 4𝑥 2𝑥 = 𝑦+5 𝑦 𝑦= 5 So, from (i), 𝑥= 3 Original fraction=3 Required fraction=3 × 1 = 3 5 2 10 S58. Ans. (c) Sol. Speed of train – B = 144 × 5 = 40 m/sec Length of train – B = (40 × 15) −300 = 300 m Now, let speed and length of train – A be y m/sec & x meters respectively. ATQ, 𝑥+300 = (40 + 𝑦) 𝑥+ 300 = 400 + 10𝑦 𝑥−10𝑦= 100 . …(i) And, 𝑥+300 × 3 = (40 −𝑦) 3𝑥+ 900 = 5200 −130𝑦 3𝑥+ 130𝑦= 4300 …(ii) On solving (i) & (ii): 𝑦= 25, 𝑥= 350 adda.com | www.careerpower.in | Adda247 App