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Ratio and Proportion Test
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Question 1 
In a students election, a applicant who gets 62% of the votes polled, is chosen by a bulk of 144 votes. The total of votes polled and the number of votes guaranteed by the applicant, who was elected a respectively
A  600: 372

B  620 : 228 
C  880 : 352 
D  700 : 128 
Question 1 Explanation:
i) If the total votes polled ="a"
Votes polled to the applicant is 62% (a)
Votes polled to other applicant is (10062) % (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
Question 2 
The price of I type milk is Rs.15/L and type II is Rs.20/L. Both are combined in the proportion 2:3, then cost/L of the combined type is :
A  Rs. 5 
B  Rs. 12 
C  Rs. 2 
D  Rs. 18 
Question 2 Explanation:
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)
dm =(20  a)
m –c (a  15)
Therefore ratio = 20a : a  15
ii) Combined type is in the proportion= 2:3
[ (20a)/(a15) ]= (2/3)
60 – 3a = 2a – 30
A= (90 / 5) = 18
Therefore the price combination = Rs. 18/L
Question 3 
What is the ratio of the current ages of Karthikh and his father?
I. The sum of the ages of Karthikh, his father and his mother is 62.
II. 5 years ago, Karthikh’s age was (1/5) of his mother’s age.
III. 2 years ago, the sum of the ages of Karthikh and his mother was 36.
A  All I, II and III are needed 
B  III only 
C  I or III only 
D  I or II only 
E  II or III only 
Question 3 Explanation:
Let X, Y and Z reffers Karthikh, His father and his mother respectively
From I , X + Y + Z = 62
From II, X5 = (1/5) (Z5)
From III, (X2) + (Z2) = 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.
Question 4 
A combination consists of metallic and copper in the proportion 5:1 and another one consists in the proportion 7:2 correspondingly. What is the weights of the both must be melted together, and combine a 5 LB size with 80% metallic?
A  2lb, 4lb 
B  3lb, 4lb 
C  1 lb, 2 lb 
D  2lb, 3lb 
Question 4 Explanation:
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
Question 5 
Three containers have the same capacity. The proportion of petrol and diesel in I , II and III containers are correspondingly 3 : 2, 7 : 3 and 11 : 4. If all the containers are combined, then locate the proportion of petrol and diesel in the combination?
A  1:2

B  70 : 7 
C  61: 29 
D  50 : 33 
Question 5 Explanation:
> 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
Question 6 
2 Pots X and Y consists of milk and water in the proportion 5:2 and 7:6 correspondingly. Track the ratio in which these combinations are combined to give a new combination containing milk and water in the ratio of 8 : 5?
A  2 : 9 
B  7 : 8 
C  2 : 8 
D  7 : 9 
Question 6 Explanation:
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
Question 7 
What is the ratio must coffee is combined with water to obtain 20% by marketing the combination at pricing?
A  1 : 3 Ratio 
B  1 : 7 Ratio 
C  1 : 5 Ratio 
D  1 : 1 Ratio 
Question 7 Explanation:
> 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
Question 8 
If a (1/4) kg of onion pricing 60 rupees, how much rupees will 200 gm costs?
A  50 rupees

B  41 rupees 
C  48 rupees 
D  30 rupees 
Question 8 Explanation:
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
Question 9 
If 12 boys and 16 girls can do a particular of job in 5 days. 13 boys and 24 girls can do the same work in 4 days then the proportion of the daytoday work done by a boy to that of a girl is
A  3 : 4 
B  2 : 1 
C  1 : 1 
D  2 : 7 
Question 9 Explanation:
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
Question 10 
The total number of students studying in an university is 35,000. If the certain years boys raises by 6% and girls by 4% the population will be 36760. What is the number of boys and girls?
A  Boys : 16,060 ; Girls: 17,200 
B  Boys : 18,000 ; Girls: 17,000 
C  None of these 
D  Boys : 19,8000 ; Girls: 18,400 
Question 10 Explanation:
Total Peoples = 35000
If the number of boys = a
& the number of girls =(35000a)
Raises in population = 6% (a) + 4% (35000a)
= 36760 – 35000
[ (6a/100) + 4(35000a)/100 ] = 1760
6a +140000 – 4a = 176000
2a = 176000 – 140000
2a = 36000
A = 18000
Number of boys = 18000
Therefore No. of girls = 35000 – 18000
= 17000
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