TITA Based Questions: Ratio & Proportion | Quantitative Aptitude (Quant) - CAT PDF Download

Ans: 42
Let the number of coins collected by A and B in one week be 3x and 4x respectfully.
The total number of coins collected by A in 5 weeks = 15x
For 15x to be a multiple of 7, x has to be a multiple of 7.
The total number of coins collected by B in 3 weeks = 12x
For 12x to be a multiple of 24, x has to be a multiple of 2.
Therefore, x has to be a multiple of 7 × 2 = 14
The minimum value that x can take is 14.
So, the minimum coins collected by A in one week = 3x = 3 × 14 = 42.

Ans: 7
Every time 10% of the mixture is replace with a pure adulterant (here water), the concentration of the mixture becomes 90% of the initial concentration.
Similarly when a proportion p of a mixture is replaced with pure adulterant, the concentration of the mixture becomes (1 - p) times the previous concentration. (0 ≤ p ≤ 1)
Here p = 4/40 = 0.1
Initially there was pure milk, so the strength of the mixture is 100% or 1 (in terms of proportion)
After repeating the process n times the concentration of Milk in the mixture will be 1 × 0.9ⁿ
For the volume of milk to be less than the volume of water, p should be less than 0.5
So we are looking for the smallest n, such that 1 × 0.9ⁿ < 0.5
This happens when n = 7

Ans: 63
Let the possible average marks of all the students combined be a.
We know that 32 < a < 60.
If the overall average is ‘a’, the ratio of the number of students in class A to the number of students in class B is (60 - a) : (a - 32).
Also when there are equal numbers of students in class A and class B, the value of a is simply . 
If there are more students in class B than class A, the value of a is greater than 46 and if there are fewer students in class B than class A, the value of a is less than 46.
Since there are 10 more students in class B than class A, 46 < a < 60.
The extreme values of the number of students depend on the extreme values of a.
When a = 47, the ratio of students in class A to class B is 13 : 15
15x - 13x = 10
2x = 10
x = 5
Total number of students in class A = 13x = 65.
When a = 56, the ratio of students in class A to class B is 1 : 6
6x - x = 10
5x = 10
x = 2
Total number of students in class A = x = 2.
The difference between the maximum and minimum possible number of students in section A is 65 - 2 = 63

Ans: 548
The common difference of the AP, 38, 55, 72, … is 17.
The smallest 3-digit number in the AP is 106 and the largest 3-digit number in it is 990
We need to find the value of,
106 + 123 + … + 973 +990
= 106 + (106 + 17) + (106 + 2 *17) + …. + (106 + 52 *17)
= 53(106) + 17(1 + 2 + 3 + … + 52)

Ans: 111

Ans: 92
The arithmetic mean of the scores of 25 students = 50
Sum of scores of these students = 25 × 50 = 1250
For the scores of the top 5 students to be as high as possible, the score of the bottom 20 students should be as low as possible.
The minimum score is 30, and the scores of the bottom 20 students are distinct integers.
In order for the bottom 20 scores to be as low as possible, they must be,
30, 31, 32, … 49
Sum of the bottom 20 scores
= 30 + 31 + 32 + … + 49
= (30 + 0) + (30 + 1) + (30 + 2) + … + (30 + 19)
= 20 × 30 + (0 + 1 + 2 + 3 + … + 19)

= 600 + 190
= 790
Therefore, the maximum sum of the scores of the top 5 students = 1250 - 790 = 460
The maximum score of the five toppers, since they all scored the same marks = 490/5 = 92

Ans: 34
Since the prices of the small and medium cups are in the ratio 2 : 5,
Let us assume the prices of the small, medium and large cups to be 2x, 5x and y.
We are given that the product of the three prices is 800.
Therefore, (2x)(5x)(y) = 800 —- (1)
If the price of the smallest and the medium cups are increased by 6, then the product becomes 3200
(2x + 6) (5x + 6) (y) = 3200 —- (2)
Dividing (2) by (1)

(2x + 6) (5x + 6) = 40 x²
10 x² + 42x + 36 = 40 x²
30 x² - 42x - 36 = 0
10 x² - 14x - 12 = 0
10 x² - 20x + 6x - 12 = 0
10 x(x - 2) + 6(x - 2) = 0
(10x + 6) (x - 2) = 0
x = 2 or x = -0.6
x can’t be negative and hence x = 2
WKT, (2x)(5x)(y) = 800
(4)(10)(y) = 800
y = 20
Therefore, the sum of the prices = 2x + 5x + y
= 2(2) + 5(2) + 20
= 4 + 10 + 20
= 34

Ans: 45
Let’s assume the capacity of the container is ‘c’
Let’s track down the milk which is retained.
The milk retained after taking first 9 litres from the container is
The milk retained after taking next 9 litres from the container is
The volumes of milk and water in the container are now in the ratio of 16 : 9.
The milk in the container is 16/25

c = 45

Ans: 10
Let the number of matches already played be x, then over the next ten matches if one more goal is scored, the average goals become 0.15. Further if the player scores two goals the average becomes 0.2.
From this we can see that one extra goal is increasing the average by 0.05 (0.2 - 0.15). 
Hence it can be said that the total number of matches played till the end is 20.
Therefore, x + 10 = 20
Hence, x = 10

Ans: 200
Given that, there are two 800cc bottles of indigo solution with strengths of 33% and 17% respectively.
Some amount of the solution from bottle 1 is replaced by the same amount of solution from bottle 2.
In 800cc of a new solution of strength 21% has two solutions in the ratio 1:3.
200cc of 33% solution and 600cc of 17% solution combined to give 21% solution of indigo
Hence, the remaining solution in the bottle 2 is 200cc.

Ans: 35
Let's consider the number of patients admitted in hospital A and hospital B to be 'x' and ‘x + 21' 
Given that, the sum of recovery days for patients in hospitals A and B were 200 and 152, respectively.
The average recovery days for patients admitted in hospital A was 3 more than the average in-hospital B.

By solving the above equation we get x = 35.
Hence, the number of patients admitted to hospital A is 35.

Ans: 18
Let's consider Tom's age to be 'x' years.
Rick is thrice as old as Tom,
Rick's age = 3 × x = 3x
Harry is twice as old as Rick,
Harry's age = 2 × 3x = 6x
Rick's age is 1 year less than the average age of all three,
3x = Average(x, 3x, 6x) - 1
1 = x/3
x = 3
Harry's age = 6x = 6 × 3 = 18 years.

Ans: 800
A : S : M
3 : 2
4 : 5
6 : 4 : 5
Given difference 6k – 4k = 2k = 400
k = 800

Ans: 100
Ratio of pen : pencil is 2:3 and he spends Rs 86
Next time The ratio of pen and pencil is 4:1 and he spends Rs 112
Second time Ashok purchased 4 pens, So the ratio of Pen and Pencil is 4:1
Hence he purchased 4 pens and 1 pencil during his second visit
Let price of pen be 'x' and price of pencil be 'y'
Then,
4x + y = 112. -----(2)
2x + 3y = 86 -----(1)----(From first visit)
(1) Is in Ratios, so it can be 4x + 6y = 86 (or)
6x + 9y = 86 (or)
8x + 12y = 86
Since, From (2), 4 pen and a pencil costs Rs 112
4 pens and 6 pencils has too cost more than 112 (Given that Cost of a pen and pencil is a positive integer)
In that case 4x + 6y = 86 (or)
6x + 9y = 86 (or)
8x + 12y = 86
These possibilities are ruled out
Only possibility is 2x + 3y = 86
Solving (1) and (2) by Elimination Method
We get y = 12 ---> Cost of a pen
On second visit, he pays a total of Rs 112, One pencil and 4 pen.
1 Pencil costs Rs 12, 112 - 12 = Rs 100
He paid Rs 100 during his second visit for the pens.
The answer is 100