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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  700 : 128 
B  620 : 228 
C  880 : 352 
D  600: 372

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 
If a (1/4) kg of onion pricing 60 rupees, how much rupees will 200 gm costs?
A  48 rupees 
B  30 rupees 
C  50 rupees

D  41 rupees 
Question 2 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 3 
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  1 lb, 2 lb 
B  2lb, 3lb 
C  2lb, 4lb 
D  3lb, 4lb 
Question 3 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 4 
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  61: 29 
B  1:2

C  70 : 7 
D  50 : 33 
Question 4 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 5 
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  7 : 8 
B  7 : 9 
C  2 : 9 
D  2 : 8 
Question 5 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 6 
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  1 : 1 
C  2 : 7 
D  2 : 1 
Question 6 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 7 
What is the ratio must coffee is combined with water to obtain 20% by marketing the combination at pricing?
A  1 : 1 Ratio 
B  1 : 3 Ratio 
C  1 : 7 Ratio 
D  1 : 5 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 
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. 2 
C  Rs. 18 
D  Rs. 12 
Question 8 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 9 
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 : 18,000 ; Girls: 17,000 
B  None of these 
C  Boys : 16,060 ; Girls: 17,200 
D  Boys : 19,8000 ; Girls: 18,400 
Question 9 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
Question 10 
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  I or III only 
B  II or III only 
C  III only 
D  I or II only 
E  All I, II and III are needed 
Question 10 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.
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