Ratio and proportion Online Test [Quantitative Ability Test]

-> If C.P. of 1 cup of coffee = Rs.1 S.Price of 1 cup coffee = Rs.1 Gain is 20% Then CP of 1cup of combination ={[ 100/(100+20)]x1}=(100/120)=Rs.(5/6) Mean cost = Rs.(5/6) -> Using the allegation rule we get, => CP of 1 cup water is (c) = 0 => CP of 1 cup coffee(d)= 1 => Mean cost(m) = Rs.(5/6) => d – m = [1 – (5/6) = (1/6) => m – c = [(5/6) – 0]= (5/6) Therefore Ratio of coffee and water = 1/6 : 5/6 = 1: 5

i) Milk in 1 litre combination of X = (5/7)L Milk in 1 litre combination of Y is (7/13)L Milk in 1 litre combination of last combination= (8/13) L Mean amount =( 8/13)L ii) Using the allegation rule, -> Some of milk in X (c) = 5/7 -> Some of milk in Y (d)= 7/13 -> Mean price(m) = 8/13 -> d – m = (5/7) – (8/13) = (9/19) -> m – c = (8/13) – (7/13) = (1/13) Therefore needed ratio = [1/13] : [9/91] = 7 : 9

i) If the total votes polled ="a" Votes polled to the applicant is 62% (a) Votes polled to other applicant is (100-62) % (a) = 38% (a) ii) 62% (a) – 38% (a) = 144 24% (a) = 144 [ (24a)/100 ] = 144 A =( 14400/24 )/24= 600 Number of votes polled = 600 Number of votes guaranteed by the applicant = 62% (600) =[ (62/100)] * 600 = 372

Total Peoples = 35000 If the number of boys = a & the number of girls =(35000-a) Raises in population = 6% (a) + 4% (35000-a) = 36760 – 35000 [ (6a/100) + 4(35000-a)/100 ] = 1760 6a +140000 – 4a = 176000 2a = 176000 – 140000 2a = 36000 A = 18000 Number of boys = 18000 Therefore No. of girls = 35000 – 18000 = 17000

i) If the cost of combined type is "a" rupees Mean cost= Rs."a" Price of 1 L of Type I (c) = Rs. 15 Price of 1 L of Type II (d) =Rs.20 Mean weight (m) d-m =(20 - a) m –c (a - 15) Therefore ratio = 20-a : a - 15 ii) Combined type is in the proportion= 2:3 [ (20-a)/(a-15) ]= (2/3) 60 – 3a = 2a – 30 A= (90 / 5) = 18 Therefore the price combination = Rs. 18/L

Let X, Y and Z reffers Karthikh, His father and his mother respectively From I , X + Y + Z = 62 From II, X-5 = (1/5) (Z-5) From III, (X-2) + (Z-2) = 36 From II and III, we will get X and Z Substituting X and Z in (I) we get Y So, All I, II and III are needed.

Let the needed price= "a" rupees Weight Cost 250gm 60 rupees 200 gm (a)rupees Less weight, less price = Direct proportion Therefore price of 200 gm onions = 48 rupees

i) Silver in I kind = 5/6 Silver in II kind = 7/9 Silver in Final mix = 80/100 = 4/5 (mean wt) Step (ii) By the rule of allegation, Silver in I kind (%) (d) 5/6 Silver in II kind (c) 7/9 Mean wt (m) Needed ratio = 1/45 : 1/30 = 45 :30 = 3 : 2 The weights are = 2lb +3lb = 5lb

-> Three containers have the same capacity, Let "a" litres of combination has. Ratio of combination in I container = 3:2 Then petrol in the I container = (a)(3/5)= (3a) / 5L Diesel in I container= (2a)/5L Ratio of combination in II container= 7:3 Then petrol in the II container =(7a)/10 L. Diesel in the II container= (4a)/10 L Ratio of combination in III container=11:4 Then petrol in the III container = (11a) / 15 L, Diesel in the III container =(4a)/15 L, -> Total petrol in last combination =[(3a/5)+(7a/10)+(11a/15)]=[(61a)/30]L Total diesel in last combination =[ (2a/5)+(3a/10)+(4a/15)]=[(29a)/30] L Needed ratio of petrol and diesel in the last combination =(61a / 30) : (29a/30) = 61 : 29

If one boy’s single day work = "a" and one girl’s single day work ="b" Then , 12a + 16b = (1/5) ---Eqn(1) 13a + 24 b = (1/4) ----Eqn( 2) Solving Eqn(1) and (2) , we will get a =(1/100) and b = (1/200) Therefore needed ratio = (1/100) : (1/200) = 2:1