0
609

# CAT Averages, Ratio and Proportion Questions PDF [Most important]

Ratio and Proportion is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Ratio and Proportion questions in the CAT Previous year’s papers. If you want to learn the basics, you can watch these videos on Ratio and Proportion basics. This article will look into some important Ratio and Proportion Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Ratio and Proportion Most Important Questions PDF below, which is completely Free.

Question 1:Â Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?

a)Â 0

b)Â 1

c)Â (1/2)*n

d)Â (n+1)/2n

e)Â 2008

Solution:

The odd numbers in the set are 3, 5, 7, …2n+1

Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$

Average of odd numbers =Â $n^2 + 2n$/n = n+2

Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Question 2:Â A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a)Â 2

b)Â 3

c)Â 4

d)Â 5

Solution:

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

Question 3:Â A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kgs. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kgs. What is the weight, in kgs, of the heaviest box?

a)Â 60

b)Â 62

c)Â 64

d)Â cannot be determined

Solution:

Let the individual weights be a,b,c,d,e in increasing order such that e is max and a is min. Adding all the addition of weight together we get 4*(a+b+c+d+e) = 1156 so a+b+c+d+e = 289 . Out of these a+b will be lowest sum and d+e will be the max . so a+b=110 and d+e=121 so we get value of c as 58 . now c have the 3rd highest weight so addition of c and e must give the second largest total i.e 120 . hence e = 120-58 = 62

Question 4:Â The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

a)Â 55

b)Â 60

c)Â 62

d)Â Cannot be determined

Solution:

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 – 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Question 5:Â My son adores chocolates. He likes biscuits. But he hates apples. I told him that he can buy as many chocolates he wishes. But then he must have biscuits twice the number of chocolates and should have apples more than biscuits and chocolates together. Each chocolate cost Re 1. The cost of apple is twice the chocolate and four biscuits are worth one apple. Then which of the following can be the amount that I spent on that evening on my son if number of chocolates, biscuits and apples brought were all integers?

a)Â Rs. 34

b)Â Rs. 33

c)Â Rs. 8

d)Â None of these

Solution:

When numbers of chocolates, biscuits and apples are integers.
Now let’s say number of chocolates taken 1 , then biscuits will be 2 and apples can be 4,5,6,7
Hence minimum money that should be spent = 1+1+8 = 10 (Hence option C is cancelled)
Now when number of chocolates are 4
Biscuits will be 8
And apples can be 13,14,15….
Now total money spent can be 4+4+26 = 34 and more of it.

Question 6:Â Arun’s present age in years is 40% of Barun’s. In another few years, Arun’s age will be half of Barun’s. By what percentage will Barun’s age increase during this period?

Solution:

Let Arun’s current age be A. Hence, Barun’s current age is 2.5A
Let Arun’s age be half of Barun’s age after X years.
Therefore, 2*(X+A) = 2.5A + X
Or, X = 0.5A
Hence, Barun’s age increased by 0.5A/2.5A = 20%

Question 7:Â The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

a)Â 30

b)Â 28

c)Â 32

d)Â 26

Solution:

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x – 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

Question 8:Â Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

a)Â $\frac{1}{5}$

b)Â $\frac{6}{19}$

c)Â $\frac{1}{4}$

d)Â $\frac{7}{33}$

Solution:

Let the number of marbles with Raju and Lalitha initially be 4x and 9x.
Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6
24x +Â 6a = 45x – 5a
11a = 21x
a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).
a/9x = 21/99
= 7/33.

Therefore, option D is the right answer.

Question 9:Â The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimalâ€™s new score to that of his original score is

a)Â 4 : 3

b)Â 8 : 5

c)Â 5 : 4

d)Â 3 : 2

Solution:

Let the score of Amal and Bimal be 11k and 14k
Let the scores be increased by x
So, after increment, Amal’s score =Â  11k + x and Bimal’s score = 14k + x
According to the question,
$\dfrac{\text{11k + x}}{\text{14k + x}}$ = $\dfrac{47}{56}$
On solving, we get x = $\dfrac{42}{9}$k
Ratio of Bimal’s new score to his original score

=Â $\dfrac{\text{14k + x}}{\text{14k}}$

=$\dfrac{14k +\frac{42k}{9}}{14k}$

=$\dfrac{\text{168k}}{\text{14*9k}}$

=$\dfrac{4}{3}$

Hence, option A is the correct answer.

Question 10:Â The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a)Â 3 : 10

b)Â 1 : 3

c)Â 1 : 4

d)Â 2 : 5

Solution:

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given thatÂ three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$Â $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

Question 11:Â John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

Solution:

It is given that John works altogether 172 hours i.e including regular and overtime hours.

Let a be the regular hours, 172-a will be the overtime hours

John’s income from regular hours = 57*a

John’s income for working overtime hours = (172-a)*144

It is given thatÂ his income from overtime hours is 15% of his income from regular hours

a*57*0.15 = (172-a)*114

a=160

The number of hours for which he worked overtime = 172-160=12 hrs

Question 12:Â Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?

a)Â 6350

b)Â 6000

c)Â 7000

d)Â 7250

Solution:

Assuming the investment ofÂ Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina’s interest income exceeds Amala by 20x-18x=2x=250Â  => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

Question 13:Â An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a)Â 48

b)Â 84

c)Â 70

d)Â 96

Solution:

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

Question 14:Â The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Solution:

Let Bottle A haveÂ an indigo solution of strength 33%Â whileÂ Bottle B have an indigo solution of strength 17%.

The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% :Â $\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$

Hence, three parts of the solution from BottleÂ B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace 600 cc of solution from Bottle A and replace it with 600 cc of solution from Bottle B {since both bottles have 800 cc, three parts of this volume = 600cc}.As a result, 200 cc of the solution remains in Bottle B.

Hence, the correct answer is 200 cc.

Question 15:Â In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

a)Â 80

b)Â 78

c)Â 84

d)Â 86

Solution:

Initially number of matches = 40
Now matches won = 12
Now let remaining matches be x
Now number of matches won = 0.6x
Now as per the condition :
$\frac{\left(12+0.6x\right)}{40+x}=\frac{1}{2}$
24 +1.2x=40+x
0.2x=16
x=80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84