Ratio, Proportion and Variation: CAT Previous Year Questions

Ans: D

Explanation:Let M be the initial number of students in the morning shift and A be the initial number of students in the afternoon shift.
Initially, the ratio of students in the morning shift to the afternoon shift was 13 : 9.
So, M/A = 13/9.
We can write M = 13x and A = 9x for some constant x.

After 21 students moved from the morning shift to the afternoon shift:
Number of students in the morning shift = M – 21 = 13x – 21
Number of students in the afternoon shift = A + 21 = 9x + 21

The new ratio became 19 : 14.
So, (13x – 21) / (9x + 21) = 19 / 14.
Cross-multiplying:
14(13x – 21) = 19(9x + 21)
182x – 294 = 171x + 399
182x – 171x = 399 + 294
11x = 693
x = 693 / 11
x = 63

Now, calculate the number of students in each shift after the transfer:
Morning shift = 13x – 21 = 13 * 63 – 21 = 819 – 21 = 798
Afternoon shift = 9x + 21 = 9 * 63 + 21 = 567 + 21 = 588

Next, some new students joined the morning and afternoon shifts in the ratio 3 : 8.
Let the number of new students who joined the morning shift be 3y and the number of new students who joined the afternoon shift be 8y.

The new number of students in the morning shift = 798 + 3y
The new number of students in the afternoon shift = 588 + 8y

The ratio of the number of students in the morning shift and the afternoon shift became 5 : 4.
So, (798 + 3y) / (588 + 8y) = 5 / 4.
Cross-multiplying:
4(798 + 3y) = 5(588 + 8y)
3192 + 12y = 2940 + 40y
3192 – 2940 = 40y – 12y
252 = 28y
y = 252 / 28
y = 9

The number of new students who joined the morning shift is 3y = 3 * 9 = 27.
The number of new students who joined the afternoon shift is 8y = 8 * 9 = 72.

The total number of new students who joined is the sum of new students in the morning and afternoon shifts:
Total new students = 27 + 72 = 99.

Correct_Option:D

Ans: A

Explanation:Let L_income and L_expenditure be the income and expenditure of Lakshmi, respectively.
Let M_income and M_expenditure be the income and expenditure of Meenakshi, respectively.

We are given the following ratios:
1. Ratio of expenditures of Lakshmi and Meenakshi:
L_expenditure : M_expenditure = 2 : 3
Let L_expenditure = 2x and M_expenditure = 3x

2. Ratio of income of Lakshmi to expenditure of Meenakshi:
L_income : M_expenditure = 6 : 7
We know M_expenditure = 3x.
So, L_income : 3x = 6 : 7
L_income = (6/7) * 3x = 18x/7

3. Excess of income over expenditure is saved.
Lakshmi’s saving (L_saving) = L_income – L_expenditure
Meenakshi’s saving (M_saving) = M_income – M_expenditure

4. Ratio of their savings is 4 : 9:
L_saving : M_saving = 4 : 9

Now, let’s express L_saving in terms of x:
L_saving = L_income – L_expenditure
L_saving = (18x/7) – 2x
L_saving = (18x – 14x) / 7
L_saving = 4x/7

From the ratio of savings, we have:
(4x/7) : M_saving = 4 : 9
M_saving = (9/4) * (4x/7)
M_saving = 9x/7

Now we can find Meenakshi’s income (M_income):
M_saving = M_income – M_expenditure
9x/7 = M_income – 3x
M_income = (9x/7) + 3x
M_income = (9x + 21x) / 7
M_income = 30x/7

We need to find the ratio of their incomes:
L_income : M_income
= (18x/7) : (30x/7)

We can cancel out x/7 from both sides:
= 18 : 30

Now, simplify the ratio by dividing both numbers by their greatest common divisor, which is 6:
= 18/6 : 30/6
= 3 : 5

Therefore, the ratio of their incomes is 3 : 5.

Comparing this with the given options:
Option A: 3:5
Option B: 5:6
Option C: 2:1
Option D: 7:8

The calculated ratio matches Option A.

Correct_Option:A

Ans: 16

Container P holds 60 litres of pure spirit, and Container Q holds 60 litres of pure water. Suppose $x$ litres are removed from container P. Container P now has $60-x$ litres of spirit. Container Q now contains 60 litres of water and $x$ litres of spirit.
After combining the contents of Q and then removing the same amount $x$, the spirit removed from Container Q would be ${\left(\dfrac{x}{60+x}\right)}$ of $x$, which equates to $\dfrac{x^2}{60+x}$ litres.
The initial total volume of 60 litres has been re-established in P after the substitution. The final quantity of spirit in Container P is:
$60 – x + \dfrac{x^2}{60+x} = \dfrac{(60+x)(60-x) + x^2}{60+x} = \dfrac{3600}{60+x}$
The total volume in container P is 60 litres, with the proportion of spirit to the total volume being $\dfrac{15}{15+4} = \dfrac{15}{19}$.
Consequently,
$\dfrac{\left(\dfrac{3600}{60+x}\right)}{60} = \dfrac{15}{19}$
$\Rightarrow \dfrac{60}{60+x} = \dfrac{15}{19}$
$\Rightarrow x = 16$
Hence, the amount transferred and replaced in both steps is 16 litres.

Ans: 272

Explanation:Let the initial number of coins in box A be $17x$ and in box B be $7x$.
The ratio of the number of coins in boxes A and B was 17:7.

After 108 coins were shifted from box A to box B:
Number of coins in box A becomes $17x – 108$.
Number of coins in box B becomes $7x + 108$.

The new ratio became 37:20.
So, we can write the equation:
$\frac{17x – 108}{7x + 108} = \frac{37}{20}$

Cross-multiply:
$20(17x – 108) = 37(7x + 108)$
$340x – 2160 = 259x + 3996$

Now, solve for x:
$340x – 259x = 3996 + 2160$
$81x = 6156$
$x = \frac{6156}{81}$
$x = \frac{684}{9}$
$x = 76$

Now, let’s find the current number of coins in boxes A and B.
Current number of coins in box A = $17x – 108 = 17(76) – 108 = 1292 – 108 = 1184$.
Current number of coins in box B = $7x + 108 = 7(76) + 108 = 532 + 108 = 640$.

We want to find the number of coins that needs to be shifted further from A to B to make the ratio 1:1. Let this number be $y$.
After shifting $y$ coins from A to B:
Number of coins in box A becomes $1184 – y$.
Number of coins in box B becomes $640 + y$.

For the ratio to be 1:1, the number of coins in both boxes must be equal:
$1184 – y = 640 + y$

Now, solve for y:
$1184 – 640 = y + y$
$544 = 2y$
$y = \frac{544}{2}$
$y = 272$

So, 272 coins need to be shifted further from A to B.

Final_Answer:272

Ans: B

Suppose the volume of the container is represented by a variable, let’s call it $V$, and it is initially full of milk.

When two-thirds of the contents are removed and then an equal amount of water is added, the proportion of milk remaining will be one-third of the original amount.

This procedure is repeated three additional times, making for a total of four operations.

Starting with an initial milk volume of $V$, after these four operations, the volume of milk will be:
$V\left(1-\frac{2}{3}\right)^4 = \frac{V}{81}$

Given that the total volume remains $V$, and the milk occupies a volume of $V/81$, the volume of water will be $V – V/81 = 80V/81$.

Therefore, the ratio of milk to water is:
$\frac{V}{81} : \frac{80V}{81}$

This simplifies to the ratio of $1:80$.

Ans: 66

Explanation:Let A be the number of apples, M be the number of mangoes, and O be the number of oranges.
We are given that the total number of fruits is 187.
So, A + M + O = 187.

The number of apples and mangoes are in the ratio 5 : 2.
So, A/M = 5/2, which means A = (5/2)M or M = (2/5)A.

Let A = 5x and M = 2x.
Then, 5x + 2x + O = 187
7x + O = 187 (Equation 1)

The fruit seller sells 75 apples, 26 mangoes, and half of the oranges.
Number of unsold apples = A – 75 = 5x – 75.
Number of unsold mangoes = M – 26 = 2x – 26.
Number of unsold oranges = O – O/2 = O/2.

The ratio of the number of unsold apples to the number of unsold oranges becomes 3 : 2.
(5x – 75) / (O/2) = 3/2
2 * (5x – 75) / O = 3/2
4 * (5x – 75) = 3 * O
20x – 300 = 3O (Equation 2)

Now we have two equations with two variables:
1) 7x + O = 187
2) 20x – 300 = 3O

From Equation 1, we can express O in terms of x:
O = 187 – 7x

Substitute this expression for O into Equation 2:
20x – 300 = 3 * (187 – 7x)
20x – 300 = 561 – 21x
20x + 21x = 561 + 300
41x = 861
x = 861 / 41
x = 21

Now we can find the initial number of each fruit:
A = 5x = 5 * 21 = 105
M = 2x = 2 * 21 = 42
O = 187 – 7x = 187 – 7 * 21 = 187 – 147 = 40

Let’s check if A + M + O = 187: 105 + 42 + 40 = 187. This is correct.

Now, let’s find the number of unsold fruits:
Number of unsold apples = A – 75 = 105 – 75 = 30.
Number of unsold mangoes = M – 26 = 42 – 26 = 16.
Number of unsold oranges = O / 2 = 40 / 2 = 20.

Let’s check the ratio of unsold apples to unsold oranges:
Unsold apples : Unsold oranges = 30 : 20 = 3 : 2. This matches the given condition.

The total number of unsold fruits is the sum of unsold apples, unsold mangoes, and unsold oranges.
Total unsold fruits = 30 + 16 + 20 = 66.

Final_Answer:66

Ans: C

Explanation:Let R be the present age of Rajesh and G be the present age of Garima.
Let the age of Rajesh at that time be $R_1$ and the age of Garima at that time be $G_1$.
The problem states that when Rajesh’s age was the same as the present age of Garima, i.e., $R_1 = G$.
At that time, the ratio of their ages was 3 : 2. So, $\frac{R_1}{G_1} = \frac{3}{2}$.
Since $R_1 = G$, we can substitute this into the ratio equation: $\frac{G}{G_1} = \frac{3}{2}$.
This means $2G = 3G_1$, or $G_1 = \frac{2}{3}G$.

The difference in their ages is constant. The difference in their ages is $R – G$.
The difference in their ages at the time $R_1$ and $G_1$ was $R_1 – G_1$.
So, $R – G = R_1 – G_1$.
Substitute $R_1 = G$ and $G_1 = \frac{2}{3}G$:
$R – G = G – \frac{2}{3}G$
$R – G = \frac{1}{3}G$
$R = G + \frac{1}{3}G$
$R = \frac{4}{3}G$

Now consider the second part of the problem. When Garima’s age becomes the same as the present age of Rajesh, i.e., Garima’s age is R.
Let Garima’s age at that future time be $G_2 = R$.
The time elapsed for Garima to reach age R is $G_2 – G = R – G$.
The same amount of time will have passed for Rajesh. So, Rajesh’s age at that future time will be $R_2 = R + (R – G)$.
We know that $R = \frac{4}{3}G$. Substitute this into the expression for $R_2$:
$R_2 = \frac{4}{3}G + (\frac{4}{3}G – G)$
$R_2 = \frac{4}{3}G + \frac{1}{3}G$
$R_2 = \frac{5}{3}G$

The ratio of the ages of Rajesh and Garima at that future time will be $\frac{R_2}{G_2}$.
We have $R_2 = \frac{5}{3}G$ and $G_2 = R = \frac{4}{3}G$.
The ratio is $\frac{R_2}{G_2} = \frac{\frac{5}{3}G}{\frac{4}{3}G} = \frac{5/3}{4/3} = \frac{5}{4}$.
So, the ratio of the ages of Rajesh and Garima will become 5 : 4.

Correct_Option:C

Ans: B

Explanation:Let A be the amount of acid and W be the amount of water in the original solution.
The total volume of the original solution is A + W.

Step 1: When 2 litres of water was added.
The new amount of acid is A.
The new amount of water is W + 2.
The new total volume is A + W + 2.
The new solution had 50% acid concentration.
So, A / (A + W + 2) = 50/100 = 1/2.
This gives us the equation: 2A = A + W + 2.
Simplifying this, we get: A = W + 2. (Equation 1)

Step 2: When 15 litres of acid was further added to this new solution.
The amount of acid in the new solution from Step 1 was A.
After adding 15 litres of acid, the amount of acid becomes A + 15.
The amount of water in the new solution from Step 1 was W + 2.
The final total volume is (A + 15) + (W + 2) = A + W + 17.
The final solution had 80% acid concentration.
So, (A + 15) / (A + W + 17) = 80/100 = 4/5.
This gives us the equation: 5(A + 15) = 4(A + W + 17).
Expanding this, we get: 5A + 75 = 4A + 4W + 68.
Simplifying this, we get: A + 7 = 4W. (Equation 2)

Step 3: Solve the system of equations.
We have two equations:
1) A = W + 2
2) A + 7 = 4W

Substitute Equation 1 into Equation 2:
(W + 2) + 7 = 4W
W + 9 = 4W
9 = 3W
W = 3 litres.

Now substitute the value of W back into Equation 1 to find A:
A = W + 2
A = 3 + 2
A = 5 litres.

Step 4: Find the ratio of water and acid in the original solution.
The original amount of water was W = 3 litres.
The original amount of acid was A = 5 litres.
The ratio of water to acid in the original solution is W : A = 3 : 5.

Let’s check the options:
Option A: 5 : 3 (Acid : Water)
Option B: 3 : 5 (Water : Acid)
Option C: 5 : 4 (Acid : Water)
Option D: 4 : 5 (Water : Acid)

The question asks for the ratio of water and acid in the original solution, which is Water : Acid.
Our calculated ratio is 3 : 5.

Correct_Option:B

Ans: B

Explanation:
Let $R_w$ and $R_m$ be the cultivation area of wheat and mustard in Rajesh’s land, respectively.
Let $V_w$ and $V_m$ be the cultivation area of wheat and mustard in Vimal’s land, respectively.

Rajesh owns 20 hectares of land, so $R_w + R_m = 20$.
Vimal owns 30 hectares of land, so $V_w + V_m = 30$.

The cultivation area of wheat and mustard in Vimal’s land are in the ratio 5 : 3.
So, $V_w : V_m = 5 : 3$.
We can write $V_w = 5x$ and $V_m = 3x$.
Since $V_w + V_m = 30$, we have $5x + 3x = 30$.
$8x = 30$
$x = \frac{30}{8} = \frac{15}{4}$.
Therefore, $V_w = 5 \times \frac{15}{4} = \frac{75}{4}$ hectares and $V_m = 3 \times \frac{15}{4} = \frac{45}{4}$ hectares.

The total cultivation area of wheat and mustard are in the ratio 11 : 9.
Let the total cultivation area of wheat be $W$ and the total cultivation area of mustard be $M$.
So, $W : M = 11 : 9$.
We can write $W = 11y$ and $M = 9y$.

The total cultivation area is the sum of the areas owned by Rajesh and Vimal.
Total area = 20 + 30 = 50 hectares.
So, $W + M = 50$.
$11y + 9y = 50$
$20y = 50$
$y = \frac{50}{20} = \frac{5}{2}$.
Therefore, $W = 11 \times \frac{5}{2} = \frac{55}{2}$ hectares and $M = 9 \times \frac{5}{2} = \frac{45}{2}$ hectares.

The total cultivation area of wheat is the sum of wheat cultivation in Rajesh’s and Vimal’s land: $W = R_w + V_w$.
$\frac{55}{2} = R_w + \frac{75}{4}$
$R_w = \frac{55}{2} – \frac{75}{4} = \frac{110}{4} – \frac{75}{4} = \frac{35}{4}$ hectares.

The total cultivation area of mustard is the sum of mustard cultivation in Rajesh’s and Vimal’s land: $M = R_m + V_m$.
$\frac{45}{2} = R_m + \frac{45}{4}$
$R_m = \frac{45}{2} – \frac{45}{4} = \frac{90}{4} – \frac{45}{4} = \frac{45}{4}$ hectares.

We need to find the ratio of cultivation area of wheat and mustard in the land owned by Rajesh, which is $R_w : R_m$.
$R_w : R_m = \frac{35}{4} : \frac{45}{4}$
Multiply both sides by 4:
$R_w : R_m = 35 : 45$
Divide both sides by 5:
$R_w : R_m = 7 : 9$.

Let’s check if $R_w + R_m = 20$:
$\frac{35}{4} + \frac{45}{4} = \frac{80}{4} = 20$. This is correct.

Correct_Option:B

Ans: C

In the final blend, the proportion of coffee to cocoa stands at 16:9.
Let’s consider the coffee to be 16 parts and the cocoa to be 9 parts.
This implies that the initial quantity consisted of 25 parts coffee and no cocoa.
Suppose ‘x’ parts of this blend are extracted and then replenished with cocoa.
At this stage, the blend contains (25-x) parts of coffee and ‘x’ parts of cocoa. Let’s call this Blend P.
Subsequently, if ‘x’ parts of this blend are removed:
The quantity of coffee remaining will be (25-x) minus the coffee removed, which is:
(25-x) – [{(25-x) / 25} * x]
This simplifies to:
(25-x) * (1 – x/25) = 16
Multiplying both sides, we get:
(25-x)^2 = 16 * 25
Taking the square root of both sides:
25 – x = 20
Therefore, x = 5.
In Blend P, the amount of cocoa is x, which is 5.
In the final blend (Blend Q), the amount of cocoa is 9 parts.
The desired ratio is therefore 5:9.

Ans: C

Explanation:Let the initial salaries of Sita, Gita, and Mita be $5x$, $6x$, and $7x$, respectively.

In the first year:
Sita’s salary becomes $5x \times (1 + 0.20) = 5x \times 1.20 = 6x$.
Gita’s salary becomes $6x \times (1 + 0.25) = 6x \times 1.25 = 7.5x$.
Mita’s salary becomes $7x \times (1 + 0.20) = 7x \times 1.20 = 8.4x$.

In the second year:
Sita’s salary becomes $6x \times (1 + 0.40) = 6x \times 1.40 = 8.4x$.
Mita’s salary becomes $8.4x \times (1 + 0.25) = 8.4x \times 1.25 = 10.5x$.

Let the salary hike of Gita in the second year be $p\%$.
Gita’s salary in the second year becomes $7.5x \times (1 + p/100)$.

According to the problem, in the second year, the salary of Gita becomes equal to the mean salary of the three friends.
The salaries of the three friends in the second year are $8.4x$ (Sita), $7.5x \times (1 + p/100)$ (Gita), and $10.5x$ (Mita).

The mean salary of the three friends in the second year is:
$\frac{8.4x + 7.5x \times (1 + p/100) + 10.5x}{3}$

We are given that Gita’s salary in the second year is equal to this mean salary:
$7.5x \times (1 + p/100) = \frac{8.4x + 7.5x \times (1 + p/100) + 10.5x}{3}$

Multiply both sides by 3:
$22.5x \times (1 + p/100) = 8.4x + 7.5x \times (1 + p/100) + 10.5x$

Let $G_2$ be Gita’s salary in the second year. So, $G_2 = 7.5x \times (1 + p/100)$.
The equation becomes:
$3G_2 = 8.4x + G_2 + 10.5x$
$3G_2 – G_2 = 8.4x + 10.5x$
$2G_2 = 18.9x$
$G_2 = \frac{18.9x}{2} = 9.45x$

Now, we equate Gita’s salary in the second year to this value:
$7.5x \times (1 + p/100) = 9.45x$

Divide both sides by $7.5x$:
$1 + p/100 = \frac{9.45}{7.5}$
$1 + p/100 = 1.26$

Now, solve for $p/100$:
$p/100 = 1.26 – 1$
$p/100 = 0.26$

To find the percentage hike, multiply by 100:
$p = 0.26 \times 100 = 26\%$

The salary hike of Gita in the second year is 26%.

Let’s verify the salaries in the second year:
Sita’s salary = $8.4x$
Mita’s salary = $10.5x$
Gita’s salary = $9.45x$

Mean salary = $\frac{8.4x + 9.45x + 10.5x}{3} = \frac{28.35x}{3} = 9.45x$.
This matches Gita’s salary, so the calculation is correct.

Correct_Option:C

Ans: D

It is stated that the value of a valuable gem is directly proportional to the square of its mass. Let the value be represented by V and the mass by M.
Therefore, $V ∝ M^2$ which implies $V = kM^2$ (where k is the constant of proportionality).

Sita possesses a valuable gem with a mass of 18 units.
Consequently, $V = kM^2 = k \cdot 18^2 = 324k$

If she divides it into four segments, each with a unique whole number mass, the disparity between the greatest and smallest possible total values of the four segments is 288,000.

To achieve the minimum possible total value for V, the masses of the four segments should be as closely distributed as possible (e.g., 3, 4, 5, 6). To achieve the maximum possible total value, three segments should have the smallest possible masses, and one segment should have the largest possible mass (e.g., 1, 2, 3, 12).

Thus, the maximum total value = $k(12^2 + 1^2 + 2^2 + 3^2) = k(144 + 1 + 4 + 9) = 158k$
And the minimum total value = $k(3^2 + 4^2 + 5^2 + 6^2) = k(9 + 16 + 25 + 36) = 86k$

The difference in values is $(158k – 86k) = 72k$, which is equivalent to 288,000.
So, $72k = 288000$
This gives us $k = 4000$

Therefore, the value of the original gem is $324k = 324 \times 4000 = 1296000$

The correct choice is D.

Ans: A

Explanation:
Let’s assume Anil uses 3 units of cocoa and 2 units of sugar to prepare mixture A.
The ratio of cocoa to sugar in mixture A is 3:2.
So, total parts in mixture A = 3 + 2 = 5.
Amount of sugar in mixture A = (2/5) * Total amount of A.

Let’s assume Anil uses 7 units of coffee and 3 units of sugar to prepare mixture B.
The ratio of coffee to sugar in mixture B is 7:3.
So, total parts in mixture B = 7 + 3 = 10.
Amount of sugar in mixture B = (3/10) * Total amount of B.

Anil combines mixtures A and B in the ratio 2:3 to make a new mixture C.
Let the amount of mixture A be 2x and the amount of mixture B be 3x.
Amount of sugar in 2x amount of mixture A = (2/5) * 2x = 4x/5.
Amount of sugar in 3x amount of mixture B = (3/10) * 3x = 9x/10.

Total amount of mixture C = Amount of A + Amount of B = 2x + 3x = 5x.
Total amount of sugar in mixture C = Amount of sugar in A + Amount of sugar in B
Total sugar in C = (4x/5) + (9x/10)
To add these fractions, find a common denominator, which is 10.
Total sugar in C = (8x/10) + (9x/10) = 17x/10.

The percentage of sugar in mixture C is (Total sugar in C / Total amount of C) * 100
Percentage of sugar in C = ((17x/10) / 5x) * 100
Percentage of sugar in C = (17x / 50x) * 100
Percentage of sugar in C = (17/50) * 100 = 17 * 2 = 34%.

Anil then mixes mixture C with an equal amount of milk to make a drink.
Let the amount of mixture C be M. Then the amount of milk is also M.
Total amount of the drink = Amount of C + Amount of milk = M + M = 2M.
The amount of sugar in the drink is the same as the amount of sugar in mixture C.
Amount of sugar in the drink = (34/100) * M.

The percentage of sugar in this drink is (Amount of sugar in drink / Total amount of drink) * 100
Percentage of sugar in drink = [((34/100) * M) / (2M)] * 100
Percentage of sugar in drink = [(34M / 100) / 2M] * 100
Percentage of sugar in drink = (34M / 200M) * 100
Percentage of sugar in drink = (34/200) * 100
Percentage of sugar in drink = 34/2 = 17%.

The percentage of sugar in this drink will be 17%.

Correct_Option: A

Ans: 42

The weekly coin collection rates of two individuals, A and B, are in the proportion 3:4.
Let’s represent A’s weekly collection as 3k coins and B’s as 4k coins.
The total coins A accumulates over 5 weeks is 5 * 3k = 15k. This total must be divisible by 7, implying k must be a multiple of 7.
The total coins B accumulates over 3 weeks is 3 * 4k = 12k. This total must be divisible by 24, implying k must be a multiple of 2.
Therefore, the smallest possible value for k is the least common multiple of 2 and 7, which is 14.
The number of coins A sells weekly is 3k = 3 * 14 = 42.

Ans: 43200

Explanation:Let M be the total number of males and F be the total number of females in the village.
We are given that the ratio of males to females is 5 : 4.
So, M/F = 5/4

Let ML be the number of literate males and FL be the number of literate females.
Let MI be the number of illiterate males and FI be the number of illiterate females.

We know that the total number of males is the sum of literate males and illiterate males:
M = ML + MI

And the total number of females is the sum of literate females and illiterate females:
F = FL + FI

We are given the ratio of literate males to literate females is 2 : 3.
So, ML/FL = 2/3

We are given the ratio of illiterate males to illiterate females is 4 : 3.
So, MI/FI = 4/3

We are also given that the number of literate males is 3600.
So, ML = 3600

From the ratio of literate males to literate females, we can find the number of literate females:
ML/FL = 2/3
3600/FL = 2/3
2 * FL = 3600 * 3
FL = (3600 * 3) / 2
FL = 1800 * 3
FL = 5400

Now we need to find the total number of males. We can express ML and MI in terms of a common variable.
Let ML = 2x and FL = 3x.
Since ML = 3600, we have 2x = 3600, which means x = 1800.
So, FL = 3 * 1800 = 5400.

Now consider the illiterate males and females.
Let MI = 4y and FI = 3y.

The total number of males M = ML + MI = 3600 + 4y
The total number of females F = FL + FI = 5400 + 3y

We know that the ratio of total males to total females is 5 : 4.
M/F = 5/4
(3600 + 4y) / (5400 + 3y) = 5/4

Cross-multiply:
4 * (3600 + 4y) = 5 * (5400 + 3y)
14400 + 16y = 27000 + 15y

Subtract 15y from both sides:
14400 + y = 27000

Subtract 14400 from both sides:
y = 27000 – 14400
y = 12600

Now we can find the total number of females:
F = 5400 + 3y
F = 5400 + 3 * 12600
F = 5400 + 37800
F = 43200

Alternatively, we can find the total number of males first:
M = 3600 + 4y
M = 3600 + 4 * 12600
M = 3600 + 50400
M = 54000

Then use the ratio M/F = 5/4:
54000 / F = 5/4
5 * F = 54000 * 4
F = (54000 * 4) / 5
F = 10800 * 4
F = 43200

Final_Answer:43200

Ans: D

Explanation:Let the initial mixture contain L parts lemon juice and S parts sugar syrup.
Given that the initial mixture contains lemon juice and sugar syrup in equal proportion, we can say L = S.
We can represent the initial mixture as 1 part lemon juice and 1 part sugar syrup, so the ratio of lemon juice to sugar syrup is 1:1.

A new mixture is created by adding this initial mixture and sugar syrup in the ratio 1:3.
Let’s consider taking ‘x’ units of the initial mixture and ‘3x’ units of sugar syrup.

In ‘x’ units of the initial mixture (which has a lemon juice to sugar syrup ratio of 1:1), the amount of lemon juice will be $(1/2) * x$ and the amount of sugar syrup will be $(1/2) * x$.

The amount of sugar syrup added is ‘3x’ units.

Now, let’s find the total amount of lemon juice and sugar syrup in the new mixture:
Total lemon juice = Lemon juice from the initial mixture = $(1/2) * x$
Total sugar syrup = Sugar syrup from the initial mixture + Added sugar syrup
Total sugar syrup = $(1/2) * x + 3x = (1/2) * x + (6/2) * x = (7/2) * x$

The ratio of lemon juice to sugar syrup in the new mixture is:
Ratio = Total lemon juice : Total sugar syrup
Ratio = $((1/2) * x) : ((7/2) * x)$

To simplify the ratio, we can multiply both sides by 2 and divide by x (since x is a common factor and not zero):
Ratio = $1 : 7$

Alternatively, let’s assume we have 100 ml of the initial mixture.
Since the proportion is equal, it contains 50 ml of lemon juice and 50 ml of sugar syrup.

Now, we add this mixture and sugar syrup in the ratio 1:3.
Let’s say we take 1 unit of the initial mixture and 3 units of sugar syrup.
If we consider the initial mixture as our “unit 1”, then its composition is 50% lemon juice and 50% sugar syrup.

Let the quantity of the initial mixture be Q1 and the quantity of sugar syrup be Q2.
We are given Q1 : Q2 = 1 : 3.

Let Q1 = k and Q2 = 3k for some constant k.

The amount of lemon juice in Q1 is $0.5 * k$.
The amount of sugar syrup in Q1 is $0.5 * k$.
The amount of sugar syrup added is $3k$.

Total lemon juice in the new mixture = $0.5 * k$
Total sugar syrup in the new mixture = $0.5 * k + 3k = 3.5 * k$

The ratio of lemon juice to sugar syrup in the new mixture is:
$(0.5 * k) : (3.5 * k)$

Divide both sides by k:
$0.5 : 3.5$

Multiply both sides by 2 to remove decimals:
$1 : 7$

Correct_Option:D

Ans: 111

Explanation:Let the number of persons standing ahead of Pinky be $3x$ and the number of persons standing behind her be $5x$, where $x$ is a positive integer.
The total number of persons in the queue is the sum of persons ahead of Pinky, the person (Pinky herself), and the persons behind Pinky.
Total number of persons = $3x + 1 + 5x = 8x + 1$.

We are given that the total number of persons in the queue is less than 300.
So, $8x + 1 < 300$.
Subtracting 1 from both sides, we get:
$8x < 299$.
Dividing by 8, we get:
$x < \frac{299}{8}$.
$x < 37.375$.

Since $x$ must be a positive integer, the maximum possible integer value for $x$ is 37.

The number of persons standing ahead of Pinky is $3x$.
To find the maximum possible number of persons standing ahead of Pinky, we use the maximum possible value of $x$, which is 37.
Maximum number of persons ahead of Pinky = $3 \times 37 = 111$.

Let’s check the total number of persons for $x=37$:
Total persons = $8(37) + 1 = 296 + 1 = 297$.
Since $297 < 300$, this is a valid scenario.
If we take $x=38$, then $8(38) + 1 = 304 + 1 = 305$, which is not less than 300.

Therefore, the maximum possible number of persons standing ahead of Pinky is 111.

The ratio of persons ahead of Pinky to persons behind her is 3:5.
Let the number of persons ahead be $A$ and the number of persons behind be $B$.
So, $\frac{A}{B} = \frac{3}{5}$, which means $A = 3k$ and $B = 5k$ for some positive integer $k$.
The total number of people in the queue is $A + 1 + B = 3k + 1 + 5k = 8k + 1$.
We are given that the total number of people is less than 300.
$8k + 1 < 300$
$8k < 299$
$k < \frac{299}{8}$
$k < 37.375$
Since $k$ must be an integer, the maximum possible value for $k$ is 37.
The number of persons standing ahead of Pinky is $A = 3k$.
The maximum possible number of persons ahead of Pinky is $3 \times 37 = 111$.

Final_Answer:111

Ans: 10

Explanation:Let $n_A$ be the number of sides of regular polygon A, and $n_B$ be the number of sides of regular polygon B.
Let $I_A$ be the interior angle of regular polygon A, and $I_B$ be the interior angle of regular polygon B.

The problem states that the number of sides are in the ratio 1:2. So, we can write:
$\frac{n_A}{n_B} = \frac{1}{2}$
This implies $n_B = 2n_A$.

The formula for the interior angle of a regular polygon with $n$ sides is given by:
$I = \frac{(n-2) \times 180^\circ}{n}$

So, for polygon A:
$I_A = \frac{(n_A-2) \times 180^\circ}{n_A}$

And for polygon B:
$I_B = \frac{(n_B-2) \times 180^\circ}{n_B}$

The problem states that the interior angles are in the ratio 3:4. So, we can write:
$\frac{I_A}{I_B} = \frac{3}{4}$

Substitute the formulas for $I_A$ and $I_B$:
$\frac{\frac{(n_A-2) \times 180^\circ}{n_A}}{\frac{(n_B-2) \times 180^\circ}{n_B}} = \frac{3}{4}$

Cancel out the $180^\circ$ term:
$\frac{\frac{n_A-2}{n_A}}{\frac{n_B-2}{n_B}} = \frac{3}{4}$

Rearrange the terms:
$\frac{n_A-2}{n_A} \times \frac{n_B}{n_B-2} = \frac{3}{4}$

Now, substitute $n_B = 2n_A$ into the equation:
$\frac{n_A-2}{n_A} \times \frac{2n_A}{2n_A-2} = \frac{3}{4}$

Cancel out $n_A$ from the numerator and denominator:
$\frac{n_A-2}{1} \times \frac{2}{2n_A-2} = \frac{3}{4}$

Simplify the expression:
$\frac{2(n_A-2)}{2n_A-2} = \frac{3}{4}$

Factor out 2 from the denominator:
$\frac{2(n_A-2)}{2(n_A-1)} = \frac{3}{4}$

Cancel out the 2:
$\frac{n_A-2}{n_A-1} = \frac{3}{4}$

Cross-multiply:
$4(n_A-2) = 3(n_A-1)$
$4n_A – 8 = 3n_A – 3$

Solve for $n_A$:
$4n_A – 3n_A = 8 – 3$
$n_A = 5$

Now, we need to find the number of sides of polygon B, $n_B$. We know that $n_B = 2n_A$.
$n_B = 2 \times 5$
$n_B = 10$

To verify the answer, let’s calculate the interior angles:
For polygon A with $n_A=5$: $I_A = \frac{(5-2) \times 180}{5} = \frac{3 \times 180}{5} = 3 \times 36 = 108^\circ$.
For polygon B with $n_B=10$: $I_B = \frac{(10-2) \times 180}{10} = \frac{8 \times 180}{10} = 8 \times 18 = 144^\circ$.
The ratio of interior angles is $\frac{I_A}{I_B} = \frac{108}{144}$.
Divide both by 36: $\frac{108 \div 36}{144 \div 36} = \frac{3}{4}$. This matches the given ratio.
The ratio of the number of sides is $\frac{n_A}{n_B} = \frac{5}{10} = \frac{1}{2}$. This also matches the given ratio.

Final_Answer:10

Ans: A

Explanation:
Initially, the glass has 500 cc of milk and 0 cc of water. The cup has 500 cc of water and 0 cc of milk.

Step 1: 150 cc of milk is transferred from the glass to the cup.
Glass: 500 – 150 = 350 cc milk, 0 cc water. Total volume = 350 cc.
Cup: 150 cc milk, 500 cc water. Total volume = 150 + 500 = 650 cc.

Step 2: The mixture in the cup is thoroughly mixed.
The concentration of milk in the cup is 150 cc / 650 cc = 15/65 = 3/13.
The concentration of water in the cup is 500 cc / 650 cc = 50/65 = 10/13.

Step 3: 150 cc of this mixture is transferred from the cup to the glass.
Amount of milk transferred from cup to glass = 150 cc * (3/13) = 450/13 cc.
Amount of water transferred from cup to glass = 150 cc * (10/13) = 1500/13 cc.

Now, let’s calculate the final amounts in the glass and the cup.

Glass:
Milk in glass = Initial milk in glass + Milk transferred from cup = 350 cc + 450/13 cc
Milk in glass = (350 * 13 + 450) / 13 = (4550 + 450) / 13 = 5000/13 cc.
Water in glass = Initial water in glass + Water transferred from cup = 0 cc + 1500/13 cc
Water in glass = 1500/13 cc.
Total volume in glass = 5000/13 + 1500/13 = 6500/13 = 500 cc.

Cup:
Milk in cup = Initial milk in cup – Milk transferred to glass = 150 cc – 450/13 cc
Milk in cup = (150 * 13 – 450) / 13 = (1950 – 450) / 13 = 1500/13 cc.
Water in cup = Initial water in cup – Water transferred to glass = 500 cc – 1500/13 cc
Water in cup = (500 * 13 – 1500) / 13 = (6500 – 1500) / 13 = 5000/13 cc.
Total volume in cup = 1500/13 + 5000/13 = 6500/13 = 500 cc.

We need to find the ratio of the amount of water in the glass to the amount of milk in the cup.
Amount of water in the glass = 1500/13 cc.
Amount of milk in the cup = 1500/13 cc.

The ratio of the amount of water in the glass to the amount of milk in the cup is (1500/13) : (1500/13) = 1 : 1.

Correct_Option:A

Ans: 45

Let the initial quantity of milk be $V$ and the final quantity be $F$.
The relationship is defined by the formula:
$F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$
where $n$ represents the number of replacement cycles.

In this specific scenario, $n=2$.
Thus, the equation becomes:
$F=\ V\left(1-\frac{K}{V}\right)^{2}$

We are given that $K = 9$.
Substituting this value, we have:
$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{2}$

From this, we can deduce that:
$1-\frac{9}{V}=\ \frac{4}{5}\quad\text{or}\quad 1-\frac{9}{V}=\ -\frac{4}{5}$

Considering the case where $1-\frac{9}{V}=-\frac{4}{5}$, we get $V=5$. However, this solution is invalid as it implies that 9 liters are drawn from a container with only 5 liters, which is impossible.

Therefore, we must consider the other possibility:
$1-\frac{9}{V}=\frac{4}{5}$
Solving this equation yields $V\ =\ 45$ liters.

Ans: A

Let the amounts invested by Anil, Bobby, and Chintu be represented by $A$, $B$, and $C$ respectively.
Assume the total investment is $T$.
So, $A + B + C = T$.

According to the problem, Anil’s investment constitutes 70% of the total investment.
Anil’s investment $= 0.70 \times T$.

The problem states that Anil’s profit share reduces by ₹ 420 when the overall profit rate declines from 18% to 15%.
Profits are distributed proportionally to the investments.
A 3% decrease in the overall profit rate leads to a ₹ 420 reduction in Anil’s profit. Anil’s profit is 70% of the total profit.
Therefore, a 3% decrease in profit corresponds to a ₹ 420 loss for Anil.
This means that a 3% profit is equivalent to a total profit of ₹ $\frac{420 \times 100}{3 \times 70} = ₹ 2000$.

Alternatively, if a 3% decrease in profit translates to a ₹ 420 decrease in Anil’s share (which is 70% of the total), then the total loss across all investors for that 3% decrease would be:
Total loss $= ₹ 420 \times \frac{100}{70} = ₹ 600$.
Thus, a 3% profit is equivalent to ₹ 600.

Given that the total profit rate was 18%, and 3% of this profit is ₹ 600.
This implies that 1% of the profit is ₹ 200.
Therefore, 18% of the profit is $18 \times ₹ 200 = ₹ 3600$.
This total profit amount of ₹ 3600 is distributed among Anil, Bobby, and Chintu.
The total investment that generated this profit is ₹ 20000.

Chintu’s profit share increased by ₹ 80 when the profit rate increased by 2%.
A 2% increase in profit, based on a total investment of ₹ 20000, amounts to $₹ 20000 \times \frac{2}{100} = ₹ 400$.
Out of this ₹ 400 total profit increase, Chintu earned ₹ 80.
This means Chintu’s share of the profit is $\frac{80}{400} = \frac{1}{5} = 20\%$ of the total profit.
Since profits are distributed in proportion to investment, Chintu invested 20% of the total amount.
Chintu’s investment $= 0.20 \times ₹ 20000 = ₹ 4000$.

Bobby invested the remaining 10 percent of the total amount.
Bobby’s investment $= 0.10 \times ₹ 20000 = ₹ 2000$.

Anil’s investment $= 70\%$ of ₹ 20000 $= ₹ 14000$.
Check: Anil (₹ 14000) + Bobby (₹ 2000) + Chintu (₹ 4000) = ₹ 20000.

Ans: 34

Let the cost of the smallest container be $2x$, the medium container be $5x$, and the large container be $y$.

According to the first given information:
$(2x) \times (5x) \times y = 800$
This simplifies to:
$10x^2y = 800$
$x^2y = 80 \quad (1)$

Now, considering the second condition:
$(2x + 6) \times (5x + 6) \times y = 3200 \quad (2)$

Divide equation (2) by equation (1):
$\frac{(2x + 6)(5x + 6)y}{x^2y} = \frac{3200}{80}$
$\frac{(2x + 6)(5x + 6)}{x^2} = 40$
Expand the numerator:
$(10x^2 + 12x + 30x + 36) = 40x^2$
$10x^2 + 42x + 36 = 40x^2$
Rearrange the terms:
$30x^2 – 42x – 36 = 0$
Divide by 6 to simplify:
$5x^2 – 7x – 6 = 0$

Solving this quadratic equation for $x$, we find $x = 2$.

Therefore, the cost of the smallest container is $2x = 2 \times 2 = 4$.
The cost of the medium container is $5x = 5 \times 2 = 10$.

Substitute $x=2$ into equation (1) to find $y$:
$(2)^2 \times y = 80$
$4y = 80$
$y = 20$

The total sum of the costs is the sum of the costs of the smallest, medium, and large containers:
$4 + 10 + 20 = 34$

Ans: 8

Explanation:
The initial volume of the solution is 40 litres.
The initial proportion of dye to water is 2:3.
This means the total parts in the ratio are 2 + 3 = 5 parts.

Initial amount of dye = (2/5) * 40 litres = 16 litres.
Initial amount of water = (3/5) * 40 litres = 24 litres.

Water is added to change the proportion to 2:5. Let ‘x’ litres of water be added.
The amount of dye remains 16 litres.
The new amount of water will be 24 + x litres.
The new proportion is dye : water = 2 : 5.
So, 16 / (24 + x) = 2 / 5.

Cross-multiply:
16 * 5 = 2 * (24 + x)
80 = 48 + 2x
80 – 48 = 2x
32 = 2x
x = 16 litres.

So, 16 litres of water were added.
The new volume of the diluted solution is 40 litres + 16 litres = 56 litres.
The amount of dye in the diluted solution is still 16 litres.
The amount of water in the diluted solution is 24 litres + 16 litres = 40 litres.
Check the proportion: 16 : 40 = 2 : 5 (dividing by 8).

One fourth of this diluted solution is taken out.
The volume of solution taken out = (1/4) * 56 litres = 14 litres.
When a part of the solution is taken out, the proportion of dye and water in that part is the same as in the original diluted solution.

Amount of dye taken out = (1/4) * 16 litres = 4 litres.
Amount of water taken out = (1/4) * 40 litres = 10 litres.

Remaining amount of dye = 16 litres – 4 litres = 12 litres.
Remaining amount of water = 40 litres – 10 litres = 30 litres.

Now, dye is added to the remaining solution to bring the proportion back to 2:3. Let ‘y’ litres of dye be added.
The new amount of dye will be 12 + y litres.
The amount of water remains 30 litres.
The target proportion is dye : water = 2 : 3.
So, (12 + y) / 30 = 2 / 3.

Cross-multiply:
3 * (12 + y) = 2 * 30
36 + 3y = 60
3y = 60 – 36
3y = 24
y = 8 litres.

Therefore, 8 litres of dye must be added.

Final_Answer: 8

Ans: B

Explanation:Let the volumes of metals A, B, and C in the alloy be 3x, 4x, and 7x respectively.
The ratio of the weights of the same volume of metals A, B, and C is 5 : 2 : 6.
This means if we consider a unit volume, the weight of A is 5 units, the weight of B is 2 units, and the weight of C is 6 units.

Let the weight per unit volume for A, B, and C be $w_A, w_B, w_C$.
We are given $w_A : w_B : w_C = 5 : 2 : 6$.
So, we can say $w_A = 5k, w_B = 2k, w_C = 6k$ for some constant k.

The weight of metal A in the alloy is (volume of A) * (weight per unit volume of A) = (3x) * (5k) = 15xk.
The weight of metal B in the alloy is (volume of B) * (weight per unit volume of B) = (4x) * (2k) = 8xk.
The weight of metal C in the alloy is (volume of C) * (weight per unit volume of C) = (7x) * (6k) = 42xk.

The total weight of the alloy is the sum of the weights of A, B, and C:
Total weight = 15xk + 8xk + 42xk = (15 + 8 + 42)xk = 65xk.

We are given that the total weight of the alloy is 130 kg.
So, 65xk = 130 kg.
This implies xk = 130 / 65 = 2 kg.

We need to find the weight of metal C in the alloy.
Weight of metal C = 42xk.
Substitute the value of xk:
Weight of metal C = 42 * 2 kg = 84 kg.

Therefore, in 130 kg of the alloy, the weight of the metal C is 84 kg.

Correct_Option:B

Ans: 800

Explanation:Let the shares of Amal, Sunil, and Mita be A, S, and M respectively.
We are given the ratio of the shares of Amal and Sunil as A:S = 3:2.
We are also given the ratio of the shares of Sunil and Mita as S:M = 4:5.

To combine these ratios, we need to make the share of Sunil (S) common in both ratios.
The ratio A:S is 3:2.
The ratio S:M is 4:5.

To make S common, we can multiply the first ratio by 2 (since 2 * 2 = 4, which is the value of S in the second ratio).
So, A:S = (3*2) : (2*2) = 6:4.
Now we have A:S = 6:4 and S:M = 4:5.
The combined ratio of Amal, Sunil, and Mita is A:S:M = 6:4:5.

Let the common factor for these shares be ‘x’.
Then, Amal’s share (A) = 6x
Sunil’s share (S) = 4x
Mita’s share (M) = 5x

The shares are 6x, 4x, and 5x.
The largest share is 6x (Amal’s share).
The smallest share is 4x (Sunil’s share).

The difference between the largest and the smallest of these three shares is given as Rs.400.
So, Largest Share – Smallest Share = 400
6x – 4x = 400
2x = 400
x = 400 / 2
x = 200

We need to find Sunil’s share.
Sunil’s share (S) = 4x
S = 4 * 200
S = 800

Therefore, Sunil’s share is Rs.800.

Final_Answer:800

Ans: C

Let the two numbers be represented by variables ‘a’ and ‘b’. We are given that their product is 616, so ab = 616.

The given equation is:
$\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$ = $\ \frac{\ 157}{3}$

Cross-multiplying yields:
$ 3(a^3 – b^3) = 157(a-b)^3 $

We know the expansion of $(a-b)^3$ is $a^3 – b^3 – 3ab(a-b)$. Substituting this into the equation:
$ 3(a^3 – b^3) = 157(a^3 – b^3 – 3ab(a-b)) $
$ 3(a^3 – b^3) = 157(a^3 – b^3) – 157 \cdot 3ab(a-b) $

Rearranging the terms to group $(a^3 – b^3)$:
$ 157 \cdot 3ab(a-b) = 157(a^3 – b^3) – 3(a^3 – b^3) $
$ 471ab(a-b) = 154(a^3 – b^3) $

Using the identity $a^3 – b^3 = (a-b)(a^2 + ab + b^2)$:
$ 471ab(a-b) = 154(a-b)(a^2 + ab + b^2) $

Assuming $a \neq b$ (otherwise $a^3-b^3$ would be 0, which contradicts the given ratio unless $a-b=0$), we can divide both sides by $(a-b)$:
$ 471ab = 154(a^2 + ab + b^2) $

Now, substitute the given value of ab = 616:
$ 471 \cdot 616 = 154(a^2 + 616 + b^2) $

Divide both sides by 154. Note that $616/154 = 4$:
$ 471 \cdot 4 = a^2 + 616 + b^2 $
$ 1884 = a^2 + b^2 + 616 $

Isolate $a^2 + b^2$:
$ a^2 + b^2 = 1884 – 616 $
$ a^2 + b^2 = 1268 $

We are looking for $(a+b)^2$. We know that $(a+b)^2 = a^2 + b^2 + 2ab$.
Substitute the values of $a^2 + b^2$ and $ab$:
$ (a+b)^2 = 1268 + 2(616) $
$ (a+b)^2 = 1268 + 1232 $
$ (a+b)^2 = 2500 $

Taking the square root of both sides:
$ a+b = \sqrt{2500} $
$ a+b = 50 $

Ans: B

Explanation:Let the salaries of Ramesh, Ganesh, and Rajesh in 2010 be $6x$, $5x$, and $7x$ respectively, based on the given ratio 6:5:7.
Let the salaries of Ramesh, Ganesh, and Rajesh in 2015 be $3y$, $4y$, and $3y$ respectively, based on the given ratio 3:4:3.

We are given that Ramesh’s salary increased by 25% from 2010 to 2015.
So, Ramesh’s salary in 2015 is Ramesh’s salary in 2010 + 25% of Ramesh’s salary in 2010.
$3y = 6x + 0.25 \times 6x$
$3y = 6x + 1.5x$
$3y = 7.5x$
Now we can express $y$ in terms of $x$:
$y = \frac{7.5x}{3}$
$y = 2.5x$

We need to find the percentage increase in Rajesh’s salary during this period.
Rajesh’s salary in 2010 was $7x$.
Rajesh’s salary in 2015 was $3y$.
Substitute the value of $y$ in terms of $x$ into Rajesh’s salary in 2015:
Rajesh’s salary in 2015 = $3(2.5x) = 7.5x$.

The increase in Rajesh’s salary is:
Increase = Rajesh’s salary in 2015 – Rajesh’s salary in 2010
Increase = $7.5x – 7x$
Increase = $0.5x$

The percentage increase in Rajesh’s salary is calculated as:
Percentage Increase = $\frac{\text{Increase}}{\text{Rajesh’s salary in 2010}} \times 100$
Percentage Increase = $\frac{0.5x}{7x} \times 100$
Percentage Increase = $\frac{0.5}{7} \times 100$
Percentage Increase = $\frac{50}{7}$
Percentage Increase $\approx 7.14\%$

Now, let’s compare this with the given options:
Option A: 10
Option B: 7
Option C: 9
Option D: 8

The calculated percentage increase (approximately 7.14%) is closest to 7%.

Correct_Option:B

Ans: B

Explanation:Let R, M, and A be the original scores of Rama, Mohan, and Anjali, respectively.
According to the problem statement, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali:
R = (1/12)(M + A)
12R = M + A (Equation 1)

After a review, the score of each of them increased by 6.
The revised scores are:
Anjali’s revised score = A + 6
Mohan’s revised score = M + 6
Rama’s revised score = R + 6

The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3.
So, we can write:
(A + 6) : (M + 6) : (R + 6) = 11 : 10 : 3

From this ratio, we can set up the following equations:
(A + 6) / (R + 6) = 11 / 3
3(A + 6) = 11(R + 6)
3A + 18 = 11R + 66
3A – 11R = 66 – 18
3A – 11R = 48 (Equation 2)

(M + 6) / (R + 6) = 10 / 3
3(M + 6) = 10(R + 6)
3M + 18 = 10R + 60
3M – 10R = 60 – 18
3M – 10R = 42 (Equation 3)

Now we have a system of three equations with three variables:
1) 12R = M + A => A = 12R – M
2) 3A – 11R = 48
3) 3M – 10R = 42

Substitute Equation 1 into Equation 2:
3(12R – M) – 11R = 48
36R – 3M – 11R = 48
25R – 3M = 48 (Equation 4)

Now we have a system of two equations with two variables (R and M):
3) 3M – 10R = 42
4) 25R – 3M = 48

Add Equation 3 and Equation 4 to eliminate M:
(3M – 10R) + (25R – 3M) = 42 + 48
-10R + 25R = 90
15R = 90
R = 90 / 15
R = 6

Now substitute the value of R back into Equation 3 to find M:
3M – 10(6) = 42
3M – 60 = 42
3M = 42 + 60
3M = 102
M = 102 / 3
M = 34

Now substitute the values of R and M back into Equation 1 to find A:
12(6) = 34 + A
72 = 34 + A
A = 72 – 34
A = 38

So, the original scores were:
Rama (R) = 6
Mohan (M) = 34
Anjali (A) = 38

Let’s check the original condition: R = (1/12)(M + A)
6 = (1/12)(34 + 38)
6 = (1/12)(72)
6 = 6 (This is correct)

Now, let’s find the revised scores:
Anjali’s revised score = A + 6 = 38 + 6 = 44
Mohan’s revised score = M + 6 = 34 + 6 = 40
Rama’s revised score = R + 6 = 6 + 6 = 12

Let’s check the ratio of revised scores: 44 : 40 : 12
Divide by the greatest common divisor, which is 4:
44/4 : 40/4 : 12/4 = 11 : 10 : 3
This matches the given ratio, so our calculations for the scores are correct.

The question asks: “Then Anjali’s score exceeded Rama’s score by”
This refers to the difference in their revised scores.
Difference = Anjali’s revised score – Rama’s revised score
Difference = 44 – 12
Difference = 32

Correct_Option:B

Ans: D

Explanation:Let the original number of marbles with Raju be 4x and with Lalitha be 9x, where x is a positive integer.
The original ratio is Raju:Lalitha = 4x:9x.

Let Lalitha give y marbles to Raju.
After the transfer, the number of marbles with Raju becomes 4x + y.
The number of marbles with Lalitha becomes 9x – y.

The new ratio of marbles with Raju to Lalitha is given as 5:6.
So, we have the equation:
(4x + y) / (9x – y) = 5/6

Cross-multiply the equation:
6 * (4x + y) = 5 * (9x – y)
24x + 6y = 45x – 5y

Now, we need to solve for y in terms of x.
Combine the y terms on one side and the x terms on the other side:
6y + 5y = 45x – 24x
11y = 21x

This equation implies a relationship between y and x. If we want to find the fraction of her original marbles given by Lalitha, we need to express y as a fraction of Lalitha’s original marbles (9x).

From 11y = 21x, we can write y = (21/11)x.
This suggests that y is not an integer if x is an integer, which might seem problematic. However, the question asks for a fraction of her original marbles, so we are looking for the value of y / (9x).

We have 11y = 21x.
We want to find the fraction y / (9x).
We can rearrange the equation 11y = 21x to get y/x = 21/11.

Now, let’s calculate the fraction of her original marbles given by Lalitha:
Fraction = y / (original number of marbles with Lalitha)
Fraction = y / (9x)

We can rewrite this as:
Fraction = (y/x) / 9

Substitute the value of y/x = 21/11 into the expression:
Fraction = (21/11) / 9
Fraction = (21/11) * (1/9)
Fraction = 21 / 99

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
Fraction = 21 ÷ 3 / 99 ÷ 3
Fraction = 7 / 33

Let’s check if this result leads to integer values for marbles.
If we choose x such that 11y = 21x, the smallest integer solution for x and y would occur if x is a multiple of 11 and y is a multiple of 21.
Let x = 11k. Then 11y = 21 * (11k), which means y = 21k.
Original marbles for Raju = 4x = 4 * 11k = 44k
Original marbles for Lalitha = 9x = 9 * 11k = 99k
Lalitha gives y = 21k marbles to Raju.
New marbles for Raju = 44k + 21k = 65k
New marbles for Lalitha = 99k – 21k = 78k
The new ratio is Raju:Lalitha = 65k:78k.
Divide both by 13: 65k/13 = 5k, 78k/13 = 6k.
The new ratio is 5:6, which matches the problem statement.

The fraction of her original number of marbles given by Lalitha is y / (9x).
y = 21k
9x = 99k
Fraction = 21k / 99k = 21/99 = 7/33.

Correct_Option:D

Ans: A

Explanation:Let the original scores of Amal and Bimal be $11x$ and $14x$ respectively.
Let the amount by which their scores increased be $y$.
So, their new scores are $(11x + y)$ and $(14x + y)$.
The new scores are in the ratio 47 : 56.
Therefore, we can write the equation:
$\frac{11x + y}{14x + y} = \frac{47}{56}$

Cross-multiply to solve for the relationship between $x$ and $y$:
$56(11x + y) = 47(14x + y)$
$616x + 56y = 658x + 47y$
$56y – 47y = 658x – 616x$
$9y = 42x$
Divide both sides by 3:
$3y = 14x$
$y = \frac{14x}{3}$

We need to find the ratio of Bimal’s new score to his original score.
Bimal’s original score = $14x$
Bimal’s new score = $14x + y$
Substitute the value of $y$ into Bimal’s new score:
Bimal’s new score = $14x + \frac{14x}{3}$
Bimal’s new score = $\frac{3 \times 14x + 14x}{3}$
Bimal’s new score = $\frac{42x + 14x}{3}$
Bimal’s new score = $\frac{56x}{3}$

Now, find the ratio of Bimal’s new score to his original score:
Ratio = $\frac{\text{Bimal’s new score}}{\text{Bimal’s original score}}$
Ratio = $\frac{\frac{56x}{3}}{14x}$
Ratio = $\frac{56x}{3 \times 14x}$
Ratio = $\frac{56}{42}$
Divide both numerator and denominator by 14:
Ratio = $\frac{4}{3}$

Thus, the ratio of Bimal’s new score to that of his original score is 4 : 3.

Correct_Option:A

Ans: A

Explanation:Let the length of the rectangle be $l$ and the width be $w$.
The area of the rectangle is $A = l \times w$.
The perimeter of the rectangle is $P = 2(l+w)$.
The square of its perimeter is $P^2 = (2(l+w))^2 = 4(l+w)^2$.

The problem states that the ratio of the area of the rectangle to the square of its perimeter is 1:25.
So, $\frac{A}{P^2} = \frac{l \times w}{4(l+w)^2} = \frac{1}{25}$.

We can cross-multiply to get:
$25 \times (l \times w) = 1 \times 4(l+w)^2$
$25lw = 4(l^2 + 2lw + w^2)$
$25lw = 4l^2 + 8lw + 4w^2$

Now, rearrange the terms to form a quadratic equation. Assume $l \ge w$ without loss of generality. Divide the entire equation by $w^2$:
$25\frac{l}{w} = 4(\frac{l}{w})^2 + 8\frac{l}{w} + 4$

Let $x = \frac{l}{w}$. The equation becomes:
$25x = 4x^2 + 8x + 4$
$4x^2 + 8x – 25x + 4 = 0$
$4x^2 – 17x + 4 = 0$

We can solve this quadratic equation for $x$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$.
Here, $a=4$, $b=-17$, and $c=4$.

$x = \frac{-(-17) \pm \sqrt{(-17)^2 – 4(4)(4)}}{2(4)}$
$x = \frac{17 \pm \sqrt{289 – 64}}{8}$
$x = \frac{17 \pm \sqrt{225}}{8}$
$x = \frac{17 \pm 15}{8}$

We have two possible values for $x$:
$x_1 = \frac{17 + 15}{8} = \frac{32}{8} = 4$
$x_2 = \frac{17 – 15}{8} = \frac{2}{8} = \frac{1}{4}$

Since we assumed $l \ge w$, the ratio $\frac{l}{w}$ should be greater than or equal to 1. Therefore, we take $x = 4$.
So, $\frac{l}{w} = 4$.
This means the ratio of the longer side to the shorter side is 4:1.
The question asks for the ratio of the shorter and longer sides, which is $\frac{w}{l} = \frac{1}{4}$.

Let’s check the options given:
Option A: 1:4 (This matches our result for shorter to longer sides)
Option B: 2:9
Option C: 1:3
Option D: 3:8

If the ratio of the shorter side to the longer side is 1:4, let the shorter side be $k$ and the longer side be $4k$.
Area $A = k \times 4k = 4k^2$.
Perimeter $P = 2(k+4k) = 2(5k) = 10k$.
Square of perimeter $P^2 = (10k)^2 = 100k^2$.
Ratio of Area to Square of Perimeter = $\frac{4k^2}{100k^2} = \frac{4}{100} = \frac{1}{25}$.
This confirms that the ratio of the shorter side to the longer side is 1:4.

Correct_Option:A

Ans: B

It is stated that in the first container, the proportion of A to B is 18 to 7.
Let’s consider that in the second container, the proportion of A to B is represented as x to 1.
We are informed that the contents of the first and second containers are combined in a 3:4 ratio, resulting in a final blend where the proportion of A to B is 13 to 7.
By evaluating the proportion of A in the mixture:
$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$
$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$
$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$
$\Rightarrow$ $x = \dfrac{239}{161}$
Consequently, it can be concluded that in the second container, the proportion of A to B is $\dfrac{239}{161}$ to 1, or equivalently, 239 to 161.

Ans: B

Let ‘a’, ‘b’, and ‘c’ represent the salt concentrations in solutions A, B, and C, respectively.

When solutions A, B, and C are combined in a ratio of 1:2:3, the resulting mixture has a salt concentration of 20%.
This can be expressed as:
$ \frac{a \times 1 + b \times 2 + c \times 3}{1 + 2 + 3} = 20 $
$ \frac{a + 2b + 3c}{6} = 20 $
$ a + 2b + 3c = 120 $ … (Equation 1)

Alternatively, if the solutions are mixed in a ratio of 3:2:1, the resulting mixture has a salt concentration of 30%.
This can be expressed as:
$ \frac{a \times 3 + b \times 2 + c \times 1}{3 + 2 + 1} = 30 $
$ \frac{3a + 2b + c}{6} = 30 $
$ 3a + 2b + c = 180 $ … (Equation 2)

By manipulating Equations 1 and 2, we can derive further relationships:
Subtracting Equation 1 from Equation 2:
$ (3a + 2b + c) – (a + 2b + 3c) = 180 – 120 $
$ 2a – 2c = 60 $
$ a – c = 30 $
$ a = 30 + c $

We can also find a relationship involving ‘b’. From Equation 1:
$ 2b = 120 – a – 3c $
Substitute $a = 30 + c$:
$ 2b = 120 – (30 + c) – 3c $
$ 2b = 120 – 30 – c – 3c $
$ 2b = 90 – 4c $
$ b = 45 – 2c $

Now, consider solution D, formed by mixing solutions B and C in a ratio of 2:7. The salt concentration in D is:
$ \text{Concentration in D} = \frac{b \times 2 + c \times 7}{2 + 7} $
$ \text{Concentration in D} = \frac{2b + 7c}{9} $
Substitute $b = 45 – 2c$:
$ \text{Concentration in D} = \frac{2(45 – 2c) + 7c}{9} $
$ \text{Concentration in D} = \frac{90 – 4c + 7c}{9} $
$ \text{Concentration in D} = \frac{90 + 3c}{9} $

The question asks for the ratio of the concentration of salt in solution D to the concentration of salt in solution A.
$ \text{Required Ratio} = \frac{\text{Concentration in D}}{\text{Concentration in A}} $
$ \text{Required Ratio} = \frac{\frac{90 + 3c}{9}}{a} $
Substitute $a = 30 + c$:
$ \text{Required Ratio} = \frac{\frac{90 + 3c}{9}}{30 + c} $
$ \text{Required Ratio} = \frac{90 + 3c}{9(30 + c)} $
$ \text{Required Ratio} = \frac{3(30 + c)}{9(30 + c)} $
$ \text{Required Ratio} = \frac{3}{9} $
$ \text{Required Ratio} = 1 : 3 $

Therefore, the correct option is B.

Ans: A

Explanation:Let the profits of C1, C2, and C3 be $P_1, P_2, P_3$ respectively.
Given that $P_1 : P_2 : P_3 = 9 : 10 : 8$.
We can write $P_1 = 9k$, $P_2 = 10k$, $P_3 = 8k$ for some constant $k$.

Let the profits of C2, C4, and C5 be $P_2, P_4, P_5$ respectively.
Given that $P_2 : P_4 : P_5 = 18 : 19 : 20$.
We can write $P_2 = 18m$, $P_4 = 19m$, $P_5 = 20m$ for some constant $m$.

We have two expressions for $P_2$: $P_2 = 10k$ and $P_2 = 18m$.
To equate these, we find a common multiple for 10 and 18, which is 90.
Let $P_2 = 90x$.
Then, from the first ratio:
$P_1 : P_2 = 9 : 10$
$P_1 : 90x = 9 : 10$
$10 P_1 = 9 \times 90x$
$P_1 = 81x$

$P_2 : P_3 = 10 : 8$
$90x : P_3 = 10 : 8$
$10 P_3 = 8 \times 90x$
$P_3 = 72x$

From the second ratio:
$P_2 : P_4 = 18 : 19$
$90x : P_4 = 18 : 19$
$18 P_4 = 19 \times 90x$
$P_4 = \frac{19 \times 90x}{18} = 19 \times 5x = 95x$

$P_2 : P_5 = 18 : 20$
$90x : P_5 = 18 : 20$
$18 P_5 = 20 \times 90x$
$P_5 = \frac{20 \times 90x}{18} = 20 \times 5x = 100x$

So, the profits of the five companies are:
$P_1 = 81x$
$P_2 = 90x$
$P_3 = 72x$
$P_4 = 95x$
$P_5 = 100x$

We are given that C5 has made a profit of Rs 19 crore more than C1.
$P_5 – P_1 = 19$
$100x – 81x = 19$
$19x = 19$
$x = 1$

Now we can find the profits of each company:
$P_1 = 81 \times 1 = 81$ crore
$P_2 = 90 \times 1 = 90$ crore
$P_3 = 72 \times 1 = 72$ crore
$P_4 = 95 \times 1 = 95$ crore
$P_5 = 100 \times 1 = 100$ crore

The total profit made by all five companies is $P_1 + P_2 + P_3 + P_4 + P_5$.
Total Profit = $81 + 90 + 72 + 95 + 100 = 438$ crore.

Correct_Option:A

Ans: A

Explanation:Let the number of large, super, and jumbo popcorn packets be $7x$, $17x$, and $16x$ respectively.
Let the number of large, super, and jumbo chips packets be $6y$, $15y$, and $14y$ respectively.

The total number of popcorn packets is $7x + 17x + 16x = 40x$.
The total number of chips packets is $6y + 15y + 14y = 35y$.

Given that the total number of popcorn packets in its stock is the same as that of chips packets, we have:
$40x = 35y$

We can simplify this ratio by dividing both sides by 5:
$8x = 7y$

This implies that the ratio of $x$ to $y$ is $x/y = 7/8$. We can assume $x = 7k$ and $y = 8k$ for some constant $k$.

We need to find the ratio of jumbo popcorn packets to jumbo chips packets.
Number of jumbo popcorn packets = $16x$
Number of jumbo chips packets = $14y$

The ratio is $\frac{16x}{14y}$.

Substitute the values of $x$ and $y$ in terms of $k$:
Ratio = $\frac{16(7k)}{14(8k)}$

Cancel out $k$ from the numerator and denominator:
Ratio = $\frac{16 \times 7}{14 \times 8}$

Now, simplify the fraction:
Ratio = $\frac{(2 \times 8) \times 7}{(2 \times 7) \times 8}$
Ratio = $\frac{112}{112}$
Ratio = $1 : 1$

Alternatively, we can use the relation $8x = 7y$, which means $\frac{x}{y} = \frac{7}{8}$.
The ratio of jumbo popcorn packets to jumbo chips packets is $\frac{16x}{14y} = \frac{16}{14} \times \frac{x}{y}$.
$\frac{16}{14} = \frac{8}{7}$
So, the ratio is $\frac{8}{7} \times \frac{7}{8} = 1$.
Thus, the ratio is $1 : 1$.

Correct_Option:A

Ans: D

Explanation:
We are given two equations with integers a and b of opposite signs:
1. $(a + 3)^{2} : b^{2} = 9 : 1$
2. $(a – 1)^{2} : (b – 1)^{2} = 4 : 1$

From equation 1, we have:
$\frac{(a + 3)^{2}}{b^{2}} = \frac{9}{1}$
Taking the square root of both sides:
$\frac{a + 3}{b} = \pm 3$

This gives us two possibilities:
Case 1.1: $a + 3 = 3b \Rightarrow a – 3b = -3$
Case 1.2: $a + 3 = -3b \Rightarrow a + 3b = -3$

From equation 2, we have:
$\frac{(a – 1)^{2}}{(b – 1)^{2}} = \frac{4}{1}$
Taking the square root of both sides:
$\frac{a – 1}{b – 1} = \pm 2$

This gives us two possibilities:
Case 2.1: $a – 1 = 2(b – 1) \Rightarrow a – 1 = 2b – 2 \Rightarrow a – 2b = -1$
Case 2.2: $a – 1 = -2(b – 1) \Rightarrow a – 1 = -2b + 2 \Rightarrow a + 2b = 3$

Now we need to solve these systems of linear equations and check if a and b are integers of opposite signs.

Let’s combine Case 1.1 with the possibilities from Case 2.

Combination 1: Case 1.1 and Case 2.1
$a – 3b = -3$
$a – 2b = -1$
Subtracting the second equation from the first:
$(a – 3b) – (a – 2b) = -3 – (-1)$
$-b = -2 \Rightarrow b = 2$
Substitute b = 2 into $a – 2b = -1$:
$a – 2(2) = -1$
$a – 4 = -1 \Rightarrow a = 3$
Here, a = 3 and b = 2. They are both positive, so they are not of opposite signs.

Combination 2: Case 1.1 and Case 2.2
$a – 3b = -3$
$a + 2b = 3$
Subtracting the first equation from the second:
$(a + 2b) – (a – 3b) = 3 – (-3)$
$5b = 6 \Rightarrow b = 6/5$
Since b is not an integer, this combination is invalid.

Now let’s combine Case 1.2 with the possibilities from Case 2.

Combination 3: Case 1.2 and Case 2.1
$a + 3b = -3$
$a – 2b = -1$
Subtracting the second equation from the first:
$(a + 3b) – (a – 2b) = -3 – (-1)$
$5b = -2 \Rightarrow b = -2/5$
Since b is not an integer, this combination is invalid.

Combination 4: Case 1.2 and Case 2.2
$a + 3b = -3$
$a + 2b = 3$
Subtracting the second equation from the first:
$(a + 3b) – (a + 2b) = -3 – 3$
$b = -6$
Substitute b = -6 into $a + 2b = 3$:
$a + 2(-6) = 3$
$a – 12 = 3 \Rightarrow a = 15$
Here, a = 15 and b = -6. They are integers and have opposite signs. This is a valid solution.

We need to find the ratio $a^{2} : b^{2}$.
$a^{2} = (15)^{2} = 225$
$b^{2} = (-6)^{2} = 36$
The ratio $a^{2} : b^{2} = 225 : 36$.
To simplify this ratio, we can divide both numbers by their greatest common divisor, which is 9.
$225 \div 9 = 25$
$36 \div 9 = 4$
So, $a^{2} : b^{2} = 25 : 4$.

Let’s double check the original conditions with a = 15 and b = -6.
Opposite signs: Yes.
$(a + 3)^{2} : b^{2} = (15 + 3)^{2} : (-6)^{2} = (18)^{2} : 36 = 324 : 36 = 9 : 1$. Correct.
$(a – 1)^{2} : (b – 1)^{2} = (15 – 1)^{2} : (-6 – 1)^{2} = (14)^{2} : (-7)^{2} = 196 : 49 = 4 : 1$. Correct.

The ratio $a^{2} : b^{2}$ is $25:4$.

Correct_Option: D

Ans: C

Explanation:We are given that a, b, and c are three positive integers.
The ratio of a to b is 3:4, which can be written as a/b = 3/4.
This means that a = 3k and b = 4k for some positive integer k.

The ratio of b to c is 2:1, which can be written as b/c = 2/1.
This means that b = 2m and c = m for some positive integer m.

Since b appears in both ratios, we need to make the value of b consistent. We can find a common multiple for the values of b in the two ratios (4 and 2). The least common multiple of 4 and 2 is 4.

To make the ratio a:b = 3:4 consistent with the ratio b:c = 2:1, we can rewrite the second ratio such that the value corresponding to b is 4.
If b:c = 2:1, then to make b equal to 4, we multiply both parts of the ratio by 2.
So, b:c = (2*2):(1*2) = 4:2.

Now we have the ratios:
a:b = 3:4
b:c = 4:2

Since the value of b is the same in both adjusted ratios, we can combine them to get the ratio a:b:c = 3:4:2.

This means that a, b, and c can be expressed as:
a = 3x
b = 4x
c = 2x
where x is a positive integer.

We need to find a possible value of (a + b + c).
a + b + c = 3x + 4x + 2x = 9x.

So, the sum (a + b + c) must be a multiple of 9. Let’s check the given options:
Option A: 201. 201 / 9 = 22 with a remainder of 3. So, 201 is not a multiple of 9.
Option B: 205. 205 / 9 = 22 with a remainder of 7. So, 205 is not a multiple of 9.
Option C: 207. 207 / 9 = 23. So, 207 is a multiple of 9. If 9x = 207, then x = 23.
With x = 23, we have:
a = 3 * 23 = 69
b = 4 * 23 = 92
c = 2 * 23 = 46
These are positive integers. Let’s check the ratios:
a:b = 69:92 = (3*23):(4*23) = 3:4 (Correct)
b:c = 92:46 = (2*23):(1*23) = 2:1 (Correct)
So, 207 is a possible value for (a + b + c).

Option D: 210. 210 / 9 = 23 with a remainder of 3. So, 210 is not a multiple of 9.

Therefore, the only possible value among the options is 207.

Correct_Option:C

Ans: C

Explanation:Let’s assume we take 4 units of the first mixture, 3 units of the second mixture, and 2 units of the third mixture.

Mixture 1 (Water : A = 1:2):
In 4 units of mixture 1, the amounts are:
Water = (1/3) * 4 = 4/3 units
Liquid A = (2/3) * 4 = 8/3 units

Mixture 2 (Water : B = 1:3):
In 3 units of mixture 2, the amounts are:
Water = (1/4) * 3 = 3/4 units
Liquid B = (3/4) * 3 = 9/4 units

Mixture 3 (Water : C = 1:4):
In 2 units of mixture 3, the amounts are:
Water = (1/5) * 2 = 2/5 units
Liquid C = (4/5) * 2 = 8/5 units

Now, let’s find the total amount of water and each liquid in the resulting mixture:

Total Water = (4/3) + (3/4) + (2/5)
To add these fractions, find a common denominator, which is 60.
Total Water = (4*20)/60 + (3*15)/60 + (2*12)/60 = 80/60 + 45/60 + 24/60 = (80 + 45 + 24)/60 = 149/60 units

Total Liquid A = 8/3 units = (8*20)/60 = 160/60 units

Total Liquid B = 9/4 units = (9*15)/60 = 135/60 units

Total Liquid C = 8/5 units = (8*12)/60 = 96/60 units

Now let’s compare the amounts:

* Water vs Liquid A: 149/60 vs 160/60. Liquid A is more than water. (Option D is incorrect)
* Water vs Liquid B: 149/60 vs 135/60. Water is more than liquid B. (Option A is incorrect, Option C is correct)
* Liquid B vs Liquid C: 135/60 vs 96/60. Liquid B is more than liquid C. (Option B is incorrect)

Therefore, the resulting mixture has more water than liquid B.

Correct_Option:C