Percentages: CAT Previous Year Questions

Ans: D

Initial quantity: 200 L with 30% acid.
Amount of acid: $0.30 \times 200 = 60$ L.

Step 1: Replacing 20% with water.
Acid remaining after this replacement: $60 \times (1 – 0.20) = 60 \times 0.8 = 48$ L.

Step 2: Replacing 10% with pure acid.
Mixture after removing 10%: The acid content becomes $48 \times 0.9 = 43.2$ L.
Adding 20 L of pure acid: Total acid content is now $43.2 + 20 = 63.2$ L.

Step 3: Replacing 15% with water.
Acid remaining after this replacement: $63.2 \times (1 – 0.15) = 63.2 \times 0.85 = 53.72$ L.

Final concentration: $\frac{53.72}{200} = 0.2686 \approx 26.86\%$.

The closest option is 27%.

Ans: C

Explanation:Let the total amount of money be T.
Pinu received 20% of the total amount, so Pinu’s share = 0.20T.
The remaining amount after Pinu’s share is T – 0.20T = 0.80T.
Meena received 40% of the remaining amount, so Meena’s share = 0.40 * (0.80T) = 0.32T.
Seema received 20% less than Pinu.
Pinu’s share = 0.20T.
Seema’s share = Pinu’s share – 0.20 * Pinu’s share
Seema’s share = 0.20T – 0.20 * (0.20T)
Seema’s share = 0.20T – 0.04T
Seema’s share = 0.16T.
The total amount distributed among Pinu, Meena, and Seema is:
0.20T (Pinu) + 0.32T (Meena) + 0.16T (Seema) = 0.68T.
The amount received by Rinu is the total amount minus the sum of amounts received by Pinu, Meena, and Seema.
Rinu’s share = T – 0.68T = 0.32T.
We need to find the ratio of the amounts received by Pinu and Rinu.
Ratio of Pinu to Rinu = Pinu’s share / Rinu’s share
Ratio = (0.20T) / (0.32T)
Ratio = 0.20 / 0.32
To simplify the ratio, we can multiply both the numerator and denominator by 100:
Ratio = 20 / 32
Divide both by their greatest common divisor, which is 4:
Ratio = 5 / 8
So, the ratio of amounts received by Pinu and Rinu is 5:8.

Correct_Option:C

Ans: B

Explanation:Let the incomes of Kamal, Amal, and Vimal in September be $8x$, $6x$, and $5x$ respectively.
The total house rent is the sum of the amounts paid by each person.
Kamal pays 15% of $8x$, which is $0.15 \times 8x = 1.2x$.
Amal pays 12% of $6x$, which is $0.12 \times 6x = 0.72x$.
Vimal pays 18% of $5x$, which is $0.18 \times 5x = 0.9x$.
The total house rent in September is $1.2x + 0.72x + 0.9x = 2.82x$.

In October, their incomes increase by 10%, 12%, and 15% respectively.
Kamal’s income in October = $8x \times (1 + 0.10) = 8x \times 1.10 = 8.8x$.
Amal’s income in October = $6x \times (1 + 0.12) = 6x \times 1.12 = 6.72x$.
Vimal’s income in October = $5x \times (1 + 0.15) = 5x \times 1.15 = 5.75x$.
The total income in October = $8.8x + 6.72x + 5.75x = 21.27x$.

The house rent remains unchanged in October. So, the total house rent in October is still $2.82x$.

We need to find the percentage of their total income that will be paid as house rent in October.
Percentage = (Total house rent / Total income) $\times$ 100
Percentage = $(2.82x / 21.27x) \times 100$
Percentage = $(2.82 / 21.27) \times 100$
Percentage $\approx 0.132581 \times 100$
Percentage $\approx 13.2581\%$

This is nearest to 13.26%.

Let’s verify the options.
Option A: 15.18
Option B: 13.26
Option C: 14.84
Option D: 12.75

The calculated value is approximately 13.2581%, which is closest to 13.26%.

Correct_Option:B

Ans: 340

Let the total initial quantity of all fruits be denoted by ‘X’.
Suppose the initial count of mangoes is represented by ‘m’ and apples by ‘p’.
The initial quantity of mangoes is 40% of X, which can be expressed as 2X/5.
The sum of all fruits sold comprises mangoes, apples, and bananas.
This total can be written as: (2X/10) + 96 + (4p/10) = X/2 (as per the problem statement).
Simplifying the equation: X/5 + 96 + 2p/5 = X/2.
Rearranging to solve for X: X = (4p + 960) / 3.
This can also be written as (4p/3) + 320.
For the term (4p/3) to yield an integer, ‘p’ must be a multiple of 3.
Additionally, for (4p/10) to represent an integer count of apples sold, ‘p’ must be a multiple of 5.
The smallest positive integer value for ‘p’ that satisfies both divisibility requirements (by 3 and 5) is 15.
Substituting this smallest value back into the expression for X:
X = (4 * 15 / 3) + 320
X = (60 / 3) + 320
X = 20 + 320
X = 340.
Hence, the value 340 is the correct solution.

Ans: C

Explanation:Let the initial salary of Gopal be S.
Let the percentage of salary increase in the first increment be x%.
Let the percentage of salary increase in the second increment be y%.

According to the problem, the percentage of salary increase in the second increment was twice of that in the first increment.
So, y = 2x.

After the first increment, Gopal’s salary became S * (1 + x/100).
After the second increment, Gopal’s salary became S * (1 + x/100) * (1 + y/100).

We are given that after two successive increments, Gopal’s salary became 187.5% of his initial salary.
So, S * (1 + x/100) * (1 + y/100) = S * (187.5/100)
(1 + x/100) * (1 + y/100) = 1.875

Substitute y = 2x into the equation:
(1 + x/100) * (1 + 2x/100) = 1.875

Let’s rewrite 1.875 as a fraction:
1.875 = 1875/1000 = 375/200 = 75/40 = 15/8.

So, (1 + x/100) * (1 + 2x/100) = 15/8.

Let’s try the options given:
Option A: x = 30
y = 2*30 = 60
(1 + 30/100) * (1 + 60/100) = (1.30) * (1.60) = 2.08. This is not 1.875.

Option B: x = 27.5
y = 2*27.5 = 55
(1 + 27.5/100) * (1 + 55/100) = (1.275) * (1.55)
1.275 * 1.55 = 1.97625. This is not 1.875.

Option C: x = 25
y = 2*25 = 50
(1 + 25/100) * (1 + 50/100) = (1.25) * (1.50) = 1.875. This matches the given condition.

Option D: x = 20
y = 2*20 = 40
(1 + 20/100) * (1 + 40/100) = (1.20) * (1.40) = 1.68. This is not 1.875.

Alternatively, we can solve the quadratic equation:
(100 + x)/100 * (100 + 2x)/100 = 15/8
(100 + x)(100 + 2x) = 15/8 * 10000
10000 + 200x + 100x + 2x^2 = 15 * 1250
10000 + 300x + 2x^2 = 18750
2x^2 + 300x – 8750 = 0
Divide by 2:
x^2 + 150x – 4375 = 0

Using the quadratic formula: x = [-b ± sqrt(b^2 – 4ac)] / 2a
x = [-150 ± sqrt(150^2 – 4*1*(-4375))] / 2*1
x = [-150 ± sqrt(22500 + 17500)] / 2
x = [-150 ± sqrt(40000)] / 2
x = [-150 ± 200] / 2

Since percentage increase cannot be negative, we take the positive root:
x = (-150 + 200) / 2 = 50 / 2 = 25.

Correct_Option: C

Ans: C

The initial populace of the town in the year 2020 is reported as 100,000. From 2020 to 2021, the populace experienced a reduction of y%, and from 2021 to 2022, it saw an increase of x%, where both x and y are positive integers.

The populace in 2021 can be calculated as:
$100000\left(\frac{100-y}{100}\right)$

Consequently, the populace in 2022 is:
$100000\left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right)$

It is further stated that the populace in 2022 exceeded that of 2020, and the disparity between x and y is 10. This can be expressed as:
$100000\left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right) > 100000$
and
$(x-y) = 10$

Substituting (x = y + 10) into the inequality:
$100000\left(\frac{100-y}{100}\right)\left(\frac{100+(y+10)}{100}\right) > 100000$
$\left(\frac{100-y}{100}\right)\left(\frac{110+y}{100}\right) > 1$

To determine the smallest possible populace figure for 2021, we aim to maximize the value of y, subject to the given conditions.

Expanding the inequality:
$(100-y)(110+y) > 10000$
$11000 + 100y – 110y – y^2 > 10000$
$11000 – 10y – y^2 > 10000$
$1000 > y^2 + 10y$
$y^2 + 10y < 1000$

To find the upper limit for y, we can complete the square:
$y^2 + 10y + 25 < 1000 + 25$
$(y+5)^2 < 1025$

We look for the largest perfect square less than 1025. $32^2 = 1024$.
$(y+5)^2 \le 1024$
$y+5 \le 32$
$y \le 27$

Since y must be a natural number and we want to maximize it, the highest possible value for y is 27.

Therefore, the populace in 2021 is:
$100000 \times \left(\frac{100-27}{100}\right) = 100000 \times \left(\frac{73}{100}\right) = 73000$

The correct option is C.

Ans: D

Let the total number of eligible voters be represented by 100 units.
The quantity of votes actually cast was 80 units of the eligible votes.
One candidate secured 30% of the votes cast.
Votes obtained by this candidate = $\frac{30}{100}\times80$ units = 24 units.
The votes remaining after this candidate’s share = 80 units – 24 units = 56 units.
These remaining votes were distributed among the other three candidates in a ratio of 1:2:3.
The shares of the other three candidates are:
First other candidate: $\frac{1}{6}\times56$ units
Second other candidate: $\frac{2}{6}\times56$ units
Third other candidate: $\frac{3}{6}\times56$ units

It is stated that the difference between the votes received by the first candidate and one of the other candidates is 2512. Let’s assume this difference is between the first candidate (24 units) and the first other candidate ($\frac{1}{6}\times56$ units).
This difference can be expressed as:
24 units – $\frac{1}{6}\times56$ units = 2512
(This interpretation is flawed as it leads to negative votes for the other candidate. Re-evaluating based on the provided solution, it seems the difference is between the votes obtained by the first candidate and the votes obtained by another candidate, *not* in proportion.)

Let’s assume the difference of 2512 is between the votes received by the first candidate and the votes received by the candidate who received the smallest portion of the remaining votes (1/6th share).
Votes of the first candidate = 24x
Votes of the other three candidates in proportion 1:2:3 are $\frac{1}{6}\times56x, \frac{2}{6}\times56x, \frac{3}{6}\times56x$
The smallest share among these is $\frac{56x}{6}$
The problem statement implies a difference of 2512. From the provided calculation, it appears the difference is between the votes of the first candidate and the votes of the second other candidate (who received 2/6th of the remaining votes), *or* the difference between the first candidate and the third other candidate (who received 3/6th of the remaining votes).

Let’s follow the given calculation precisely:
Difference = 28x – 24x = 2512. This implies the first candidate received 24x votes, and another candidate received 28x votes. This contradicts the initial calculation that one candidate received 24x votes.

Let’s reinterpret the given calculation structure:
Let the total number of registered voters be 100y.
Number of votes cast = 80y.
Votes received by Candidate A = $\frac{30}{100}\times80y$ = 24y.
Remaining votes = 80y – 24y = 56y.
These remaining 56y votes are divided among three other candidates in a ratio of 1:2:3.
The portions are: $\frac{1}{6}\times56y, \frac{2}{6}\times56y, \frac{3}{6}\times56y$
The problem states that a difference of 2512 exists. The calculation provided shows:
28y – 24y = 2512.
This implies that another candidate received 28y votes. This candidate must be one of the three others.
If the third other candidate received $\frac{3}{6}\times56y$ votes, then $\frac{3}{6}\times56y = \frac{1}{2}\times56y = 28y$.
So, the difference is between Candidate A (24y) and the third other candidate (28y).
The difference in votes is 28y – 24y = 4y.
Given that this difference is 2512:
4y = 2512
y = $\frac{2512}{4}$
y = 628

The number of registered votes = 100y = 100 * 628 = 62800.
The answer corresponds to option D.

Ans: B

Explanation:Let the price of the small box be $P_s$ and the price of the large box be $P_l$.
Let the weight of chocolate in the small box be $W_s$ grams and in the large box be $W_l$ grams.
The selling price per gram of chocolate in the small box is $SP_s = \frac{P_s}{W_s}$.
The selling price per gram of chocolate in the large box is $SP_l = \frac{P_l}{W_l}$.

We are given that the large box is sold for twice the price of the small box, so $P_l = 2P_s$.

We are also given that the selling price per gram of chocolate in the large box is 12% less than that in the small box. This means:
$SP_l = SP_s – 0.12 \times SP_s$
$SP_l = SP_s (1 – 0.12)$
$SP_l = 0.88 \times SP_s$

Now, substitute the expressions for $SP_l$ and $SP_s$:
$\frac{P_l}{W_l} = 0.88 \times \frac{P_s}{W_s}$

Substitute $P_l = 2P_s$ into the equation:
$\frac{2P_s}{W_l} = 0.88 \times \frac{P_s}{W_s}$

We can cancel $P_s$ from both sides (since $P_s > 0$):
$\frac{2}{W_l} = \frac{0.88}{W_s}$

Now, we need to find the percentage by which the weight of chocolate in the large box exceeds that in the small box. This can be calculated as:
Percentage excess $= \frac{W_l – W_s}{W_s} \times 100\%$

From the equation $\frac{2}{W_l} = \frac{0.88}{W_s}$, we can find the ratio of $W_l$ to $W_s$:
$2W_s = 0.88W_l$
$\frac{W_l}{W_s} = \frac{2}{0.88}$

Now, calculate the value of $\frac{2}{0.88}$:
$\frac{2}{0.88} = \frac{200}{88} = \frac{100}{44} = \frac{25}{11}$

So, $W_l = \frac{25}{11} W_s$.

Now, calculate the percentage excess:
Percentage excess $= \left(\frac{W_l}{W_s} – 1\right) \times 100\%$
Percentage excess $= \left(\frac{25}{11} – 1\right) \times 100\%$
Percentage excess $= \left(\frac{25 – 11}{11}\right) \times 100\%$
Percentage excess $= \left(\frac{14}{11}\right) \times 100\%$

Now, calculate the value:
$\frac{14}{11} \times 100\% \approx 1.2727 \times 100\% = 127.27\%$

The percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to 127%.

Correct_Option:B

Ans: D

Explanation:Let the initial investment be P = ₹ 10000.
Let the percentage of loss in the first year be x%.
At the end of the first year, the balance is P1 = P * (1 – x/100) = 10000 * (1 – x/100).
We are given that P1 > ₹ 5000.
10000 * (1 – x/100) > 5000
1 – x/100 > 5000/10000
1 – x/100 > 0.5
0.5 > x/100
50 > x

In the second year, the balance P1 grew by 5x%.
The balance at the end of the second year is P2 = P1 * (1 + 5x/100).
Substitute the value of P1:
P2 = [10000 * (1 – x/100)] * (1 + 5x/100)

The total gain from the initial investment over the two-year period is 35%.
This means the final amount P2 is P * (1 + 35/100).
P2 = 10000 * (1 + 0.35) = 10000 * 1.35 = 13500.

Now, we equate the two expressions for P2:
[10000 * (1 – x/100)] * (1 + 5x/100) = 13500
Divide both sides by 10000:
(1 – x/100) * (1 + 5x/100) = 13500/10000
(1 – x/100) * (1 + 5x/100) = 1.35

Let’s use the options to test.
Option A: x = 5
(1 – 5/100) * (1 + 5*5/100) = (1 – 0.05) * (1 + 25/100) = 0.95 * (1 + 0.25) = 0.95 * 1.25
0.95 * 1.25 = 1.1875. This is not 1.35.

Option D: x = 10
(1 – 10/100) * (1 + 5*10/100) = (1 – 0.10) * (1 + 50/100) = 0.90 * (1 + 0.50) = 0.90 * 1.50
0.90 * 1.50 = 1.35. This matches the required value.

Let’s verify the condition P1 > 5000 for x=10.
P1 = 10000 * (1 – 10/100) = 10000 * 0.90 = 9000.
Since 9000 > 5000, the condition is satisfied.

Therefore, the percentage of loss in the first year is 10%.

To solve algebraically:
(1 – x/100) * (1 + 5x/100) = 1.35
Let y = x/100.
(1 – y) * (1 + 5y) = 1.35
1 + 5y – y – 5y^2 = 1.35
1 + 4y – 5y^2 = 1.35
5y^2 – 4y + 0.35 = 0
Multiply by 100 to remove decimals:
500y^2 – 400y + 35 = 0
Divide by 5:
100y^2 – 80y + 7 = 0
We can use the quadratic formula y = [-b ± sqrt(b^2 – 4ac)] / 2a
Here a = 100, b = -80, c = 7.
y = [80 ± sqrt((-80)^2 – 4 * 100 * 7)] / (2 * 100)
y = [80 ± sqrt(6400 – 2800)] / 200
y = [80 ± sqrt(3600)] / 200
y = [80 ± 60] / 200

Two possible values for y:
y1 = (80 + 60) / 200 = 140 / 200 = 0.7
y2 = (80 – 60) / 200 = 20 / 200 = 0.1

Since y = x/100, then x = 100y.
If y1 = 0.7, x1 = 100 * 0.7 = 70.
If y2 = 0.1, x2 = 100 * 0.1 = 10.

We need to check the condition that the balance in the first year was more than ₹ 5000.
If x = 70%, the balance at the end of the first year would be 10000 * (1 – 70/100) = 10000 * 0.30 = 3000, which is not more than ₹ 5000.
If x = 10%, the balance at the end of the first year would be 10000 * (1 – 10/100) = 10000 * 0.90 = 9000, which is more than ₹ 5000.

Thus, the percentage of loss in the first year is 10%.

Correct_Option:D

Ans: 250

Explanation:Let W be the number of white balls and B be the number of black balls.
Total balls = W + B = 450.

Let MW be the number of metallic white balls and MB be the number of metallic black balls.
Let NW be the number of non-metallic white balls and NB be the number of non-metallic black balls.

We are given that there are as many metallic white balls as metallic black balls.
So, MW = MB.

We are also given that 40% of the white balls are metallic, which means:
MW = 0.40 * W

And 50% of the black balls are metallic, which means:
MB = 0.50 * B

Since MW = MB, we can set the two equations equal to each other:
0.40 * W = 0.50 * B

We can simplify this equation by multiplying both sides by 10:
4 * W = 5 * B

Now we have a system of two equations with two variables:
1) W + B = 450
2) 4W = 5B

From equation (2), we can express W in terms of B:
W = (5/4) * B

Substitute this expression for W into equation (1):
(5/4) * B + B = 450

Combine the terms with B:
(5/4 + 4/4) * B = 450
(9/4) * B = 450

Solve for B:
B = 450 * (4/9)
B = 50 * 4
B = 200

Now find W using equation (1):
W + 200 = 450
W = 450 – 200
W = 250

So, there are 250 white balls and 200 black balls.

Now let’s find the number of metallic and non-metallic balls.

Metallic white balls (MW):
MW = 0.40 * W = 0.40 * 250 = 100

Metallic black balls (MB):
MB = 0.50 * B = 0.50 * 200 = 100

As expected, MW = MB = 100.

Non-metallic white balls (NW):
NW = W – MW = 250 – 100 = 150

Non-metallic black balls (NB):
NB = B – MB = 200 – 100 = 100

The total number of non-metallic balls is the sum of non-metallic white balls and non-metallic black balls:
Total non-metallic balls = NW + NB = 150 + 100 = 250.

Alternatively, we can calculate the total number of metallic balls and subtract it from the total number of balls.
Total metallic balls = MW + MB = 100 + 100 = 200.
Total non-metallic balls = Total balls – Total metallic balls = 450 – 200 = 250.

Final_Answer:250

Ans: C

Explanation:Let $M$ be the total number of matches played in the tournament.
The number of matches played so far is 40.
The number of matches won so far is 30% of 40, which is $0.30 \times 40 = 12$.

Let $R$ be the number of remaining matches.
So, $M = 40 + R$.

If they win 60% of the remaining matches, their overall win percentage will be 50%.
Number of remaining matches won = $0.60 \times R$.
Total number of matches won = $12 + 0.60 \times R$.
Overall win percentage = $\frac{\text{Total matches won}}{\text{Total matches played}} \times 100$.
$50 = \frac{12 + 0.60 \times R}{40 + R} \times 100$.
$\frac{50}{100} = \frac{12 + 0.60 \times R}{40 + R}$.
$0.5 = \frac{12 + 0.60 \times R}{40 + R}$.
$0.5 \times (40 + R) = 12 + 0.60 \times R$.
$20 + 0.5R = 12 + 0.60R$.
$20 – 12 = 0.60R – 0.5R$.
$8 = 0.10R$.
$R = \frac{8}{0.10} = 80$.

So, there are 80 remaining matches.
The total number of matches in the tournament is $M = 40 + R = 40 + 80 = 120$.

Now, consider the case where they win 90% of the remaining matches.
Number of remaining matches won = 90% of 80 = $0.90 \times 80 = 72$.
Total number of matches won in this case = Matches won so far + Matches won from remaining.
Total number of matches won = $12 + 72 = 84$.

The total number of matches won by the team in the tournament will be 84.

Let’s verify the overall win percentage in the first scenario:
Total matches played = 120.
Total matches won = 12 (so far) + 0.60 * 80 (remaining) = 12 + 48 = 60.
Overall win percentage = (60 / 120) * 100 = 50%. This matches the given information.

Correct_Option:C

Ans: B

Explanation:Let M1970 and F1970 be the male and female populations in 1970, respectively.
Let T1970 = M1970 + F1970.

From 1970 to 1980:
The total population increased by 25%.
T1980 = T1970 * (1 + 0.25) = 1.25 * T1970.
The male population increased by 40%.
M1980 = M1970 * (1 + 0.40) = 1.40 * M1970.
The female population increased by 20%.
F1980 = F1970 * (1 + 0.20) = 1.20 * F1970.

We know that T1980 = M1980 + F1980.
So, 1.25 * (M1970 + F1970) = 1.40 * M1970 + 1.20 * F1970.
1.25 * M1970 + 1.25 * F1970 = 1.40 * M1970 + 1.20 * F1970.
1.25 * F1970 – 1.20 * F1970 = 1.40 * M1970 – 1.25 * M1970.
0.05 * F1970 = 0.15 * M1970.
F1970 = (0.15 / 0.05) * M1970.
F1970 = 3 * M1970.

This means the ratio of female to male population in 1970 was 3:1.
Let M1970 = x. Then F1970 = 3x.
T1970 = x + 3x = 4x.

Now we can find the populations in 1980:
M1980 = 1.40 * x.
F1980 = 1.20 * (3x) = 3.60 * x.
T1980 = M1980 + F1980 = 1.40x + 3.60x = 5.00x.
Check: T1980 = 1.25 * T1970 = 1.25 * (4x) = 5.00x. This is consistent.

From 1980 to 1990:
The female population increased by 25%.
F1990 = F1980 * (1 + 0.25) = 1.25 * F1980.
F1990 = 1.25 * (3.60 * x) = 4.50 * x.

In 1990, the female population is twice the male population.
F1990 = 2 * M1990.
4.50 * x = 2 * M1990.
M1990 = (4.50 * x) / 2 = 2.25 * x.

Now we have the populations in 1990:
M1990 = 2.25 * x.
F1990 = 4.50 * x.
T1990 = M1990 + F1990 = 2.25x + 4.50x = 6.75 * x.

We need to find the percentage increase in the total population from 1970 to 1990.
T1970 = 4x.
T1990 = 6.75x.

Percentage increase = ((T1990 – T1970) / T1970) * 100.
Percentage increase = ((6.75x – 4x) / 4x) * 100.
Percentage increase = (2.75x / 4x) * 100.
Percentage increase = (2.75 / 4) * 100.
Percentage increase = 0.6875 * 100.
Percentage increase = 68.75%.

Correct_Option:B

Ans: D

Explanation:Let the total number of people in the group be 100.

Given:
28% of the members are young.
Number of young people = 0.28 * 100 = 28
The rest are old.
Number of old people = 100 – 28 = 72

65% of the members are literates.
Number of literates = 0.65 * 100 = 65
Number of illiterates = 100 – 65 = 35

25% of the literates are young.
Number of young literates = 0.25 * 65 = 16.25

We can construct a table to represent the distribution of people:

| | Young | Old | Total |
| :———- | :—- | :—- | :—- |
| Literate | 16.25 | | 65 |
| Illiterate | | | 35 |
| Total | 28 | 72 | 100 |

Now, we can fill in the missing values:

Number of old literates = Total literates – Young literates = 65 – 16.25 = 48.75
Number of young illiterates = Total young – Young literates = 28 – 16.25 = 11.75
Number of old illiterates = Total old – Old literates = 72 – 48.75 = 23.25
Alternatively, Number of old illiterates = Total illiterates – Young illiterates = 35 – 11.75 = 23.25

We need to find the percentage of old people among the illiterates.
Percentage of old people among illiterates = (Number of old illiterates / Total number of illiterates) * 100
Percentage of old people among illiterates = (23.25 / 35) * 100
Percentage of old people among illiterates = 0.66428… * 100
Percentage of old people among illiterates ≈ 66.43%

The nearest percentage among the options is 66%.

Correct_Option:D

Ans: A

Explanation:
Let the amount of rice bought in April be $R$ kg and the amount of wheat bought in April be $W$ kg.
Let the price of rice per kg in April be $P_R$ and the price of wheat per kg in April be $P_W$.

In April, John spent ₹ 450 on rice. So, $R \times P_R = 450$.

In May, John bought the same amount of rice and wheat as in April, so he bought $R$ kg of rice and $W$ kg of wheat.
The price of rice increased by 20%, so the new price of rice is $P_R’ = P_R + 0.20 P_R = 1.20 P_R$.
The price of wheat increased by 12%, so the new price of wheat is $P_W’ = P_W + 0.12 P_W = 1.12 P_W$.

The total amount spent in April is $450 + W \times P_W$.
The total amount spent in May is $R \times P_R’ + W \times P_W’ = R \times (1.20 P_R) + W \times (1.12 P_W)$.

John spent ₹ 150 more in May than in April.
Total spent in May – Total spent in April = ₹ 150
$(1.20 R P_R + 1.12 W P_W) – (R P_R + W P_W) = 150$
$1.20 R P_R – R P_R + 1.12 W P_W – W P_W = 150$
$0.20 R P_R + 0.12 W P_W = 150$

We know that $R P_R = 450$. Substitute this value into the equation:
$0.20 \times 450 + 0.12 W P_W = 150$
$90 + 0.12 W P_W = 150$
$0.12 W P_W = 150 – 90$
$0.12 W P_W = 60$
$W P_W = \frac{60}{0.12} = \frac{6000}{12} = 500$.

The amount John spent on wheat in April is $W P_W = ₹ 500$.

We need to find how much he spent on wheat in May.
Amount spent on wheat in May = $W \times P_W’ = W \times (1.12 P_W) = 1.12 \times (W P_W)$.
Amount spent on wheat in May = $1.12 \times 500$.
Amount spent on wheat in May = $1.12 \times 500 = 560$.

So, John spent ₹ 560 on wheat in May.

The final answer is $\boxed{560}$.
Correct_Option:A

Ans: D

Explanation:Let R be the marks obtained by Ramesh.
Let M be the maximum marks of the examination.

According to the problem statement:
Bishnu scored 52% of the total marks.
Bishnu’s marks = 0.52 * M

Asha scored 64% of the total marks.
Asha’s marks = 0.64 * M

The marks obtained by Bishnu is 23 less than Ramesh’s marks.
Bishnu’s marks = R – 23
0.52 * M = R – 23 (Equation 1)

The marks obtained by Asha is 34 more than Ramesh’s marks.
Asha’s marks = R + 34
0.64 * M = R + 34 (Equation 2)

We have a system of two linear equations with two variables (M and R). We can solve this system to find the values of M and R.

Subtract Equation 1 from Equation 2:
(0.64 * M) – (0.52 * M) = (R + 34) – (R – 23)
0.12 * M = R + 34 – R + 23
0.12 * M = 57

Now, solve for M:
M = 57 / 0.12
M = 57 / (12/100)
M = 57 * (100/12)
M = 5700 / 12
M = 1900 / 4
M = 475

So, the maximum marks of the examination is 475.

Now we can find Ramesh’s marks (R) using either Equation 1 or Equation 2. Let’s use Equation 2:
0.64 * M = R + 34
0.64 * 475 = R + 34
304 = R + 34
R = 304 – 34
R = 270

Ramesh obtained 270 marks.

Geeta scored 84% of the total marks.
Geeta’s marks = 0.84 * M
Geeta’s marks = 0.84 * 475

Calculate Geeta’s marks:
0.84 * 475 = (84/100) * 475
= (21/25) * 475
= 21 * (475/25)
= 21 * 19
= 399

Geeta obtained 399 marks.

Let’s check our values with Bishnu’s marks.
Bishnu’s marks = 0.52 * 475 = 247
According to the problem, Bishnu’s marks = R – 23 = 270 – 23 = 247. This matches.

The marks obtained by Geeta is 399.

Correct_Option:D

Ans: A

Explanation:Let A be Amala’s income, B be Bimala’s income, and K be Kamala’s income.

According to the problem statement:
1. Amala’s income is 20% more than that of Bimala.
A = B + 0.20B = 1.20B

2. Amala’s income is 20% less than that of Kamala.
A = K – 0.20K = 0.80K

From these two equations, we can relate B and K:
1.20B = 0.80K
B = (0.80 / 1.20)K
B = (8/12)K
B = (2/3)K

Let’s assume Kamala’s income is 100 units for simplicity (K=100).
Then, B = (2/3) * 100 = 200/3.
And A = 0.80 * 100 = 80.
We can check if A = 1.20B: 80 = 1.20 * (200/3) = (12/10) * (200/3) = (4/10) * 200 = 4 * 20 = 80. This is consistent.

Now, let’s consider the changes in income:
Kamala’s income goes down by 4%.
New Kamala’s income (K’) = K – 0.04K = 0.96K
If K = 100, then K’ = 0.96 * 100 = 96.

Bimala’s income goes up by 10%.
New Bimala’s income (B’) = B + 0.10B = 1.10B
If B = 200/3, then B’ = 1.10 * (200/3) = (11/10) * (200/3) = 11 * (20/3) = 220/3.

We need to find the percentage by which Kamala’s income would exceed Bimala’s.
This means we need to calculate ((K’ – B’) / B’) * 100.

K’ = 96
B’ = 220/3

K’ – B’ = 96 – 220/3 = (96*3 – 220)/3 = (288 – 220)/3 = 68/3

Percentage difference = ((68/3) / (220/3)) * 100
= (68/220) * 100
= (68/22) * 10
= (34/11) * 10
= 340/11

Now, let’s perform the division:
340 / 11 ≈ 30.9090…

We need to find the nearest percentage.
30.9090… is closest to 31.

Let’s recheck calculations with a common base for K and B.
Let K = 300 (to avoid fractions with 100 and 2/3).
A = 0.80 * K = 0.80 * 300 = 240.
B = (2/3) * K = (2/3) * 300 = 200.
Check: A is 20% more than B. 200 * 1.20 = 240. This is correct.

New Kamala’s income (K’): 4% decrease from 300.
K’ = 300 * (1 – 0.04) = 300 * 0.96 = 288.

New Bimala’s income (B’): 10% increase from 200.
B’ = 200 * (1 + 0.10) = 200 * 1.10 = 220.

Kamala’s income exceeds Bimala’s by: K’ – B’ = 288 – 220 = 68.
Percentage by which Kamala’s income exceeds Bimala’s:
((K’ – B’) / B’) * 100
= (68 / 220) * 100
= (68 / 22) * 10
= (34 / 11) * 10
= 340 / 11

340 divided by 11:
340 = 11 * 30 + 10
So, 340/11 = 30 and 10/11.
10/11 is approximately 0.9090…
So, 340/11 ≈ 30.9090…

Rounding to the nearest integer, we get 31.

Correct_Option:A

Ans: C

Explanation:Let the maximum marks of the examination be M.
Let the passing score be P.

Meena’s initial score = 40% of M = 0.40M.

After review, her score is increased by 50%.
Meena’s post-review score = 0.40M + 50% of 0.40M = 0.40M + 0.50 * 0.40M = 0.40M + 0.20M = 0.60M.

She fails by 35 marks after the review. This means her post-review score is 35 marks less than the passing score.
So, 0.60M = P – 35 —(1)

If her post-review score is increased by 20%, she will have 7 marks more than the passing score.
New score = 0.60M + 20% of 0.60M = 0.60M + 0.20 * 0.60M = 0.60M + 0.12M = 0.72M.

This new score is 7 marks more than the passing score.
So, 0.72M = P + 7 —(2)

Now we have two equations:
1) 0.60M = P – 35
2) 0.72M = P + 7

We can solve these two equations for M and P.
From equation (1), P = 0.60M + 35.
Substitute this value of P into equation (2):
0.72M = (0.60M + 35) + 7
0.72M = 0.60M + 42
0.72M – 0.60M = 42
0.12M = 42
M = 42 / 0.12
M = 42 / (12/100)
M = 42 * (100/12)
M = (42/12) * 100
M = 3.5 * 100
M = 350

Now find the passing score P using equation (1):
P = 0.60M + 35
P = 0.60 * 350 + 35
P = 210 + 35
P = 245

The question asks for the percentage score needed for passing the examination.
Percentage passing score = (P / M) * 100
Percentage passing score = (245 / 350) * 100
Percentage passing score = (245/350) * 100 = (7 * 35 / 10 * 35) * 100 = (7/10) * 100 = 70%.

Correct_Option:C

Ans: 20

Let the total student count be represented by 100k.
Consequently, the count of female students is 60k, and the count of male students is 40k.
The difference between the number of female and male students is given as 30:
60k – 40k = 30
This implies 20k = 30, so k = 1.5.

The total number of female students is 60 * 1.5 = 90.

The count of female students who achieved a passing grade is 68k – 30, which translates to 68 * 1.5 – 30 = 102 – 30 = 72.

The count of female students who did not achieve a passing grade is the total number of female students minus those who passed: 90 – 72 = 18.

Therefore, the proportion of female students who did not pass is (18 / 90) * 100, which equals 20%.

Ans: D

Explanation:Initially, we have the following:
Vessel A: 500 ml solution, strength 10%. Amount of salt in A = (10/100) * 500 = 50 grams.
Vessel B: 500 ml solution, strength 22%. Amount of salt in B = (22/100) * 500 = 110 grams.
Vessel C: 500 ml solution, strength 32%. Amount of salt in C = (32/100) * 500 = 160 grams.

Step 1: 100 ml of solution from A is transferred to B.
Amount of salt in 100 ml of A = (10/100) * 100 = 10 grams.
After transfer:
Vessel A: 500 – 100 = 400 ml solution, salt = 50 – 10 = 40 grams.
Vessel B: 500 + 100 = 600 ml solution. Salt in B = 110 (initial) + 10 (from A) = 120 grams.
Strength of solution in B = (120 / 600) * 100 = 20%.

Step 2: 100 ml of solution from B is transferred to C.
The strength of solution in B is now 20%.
Amount of salt in 100 ml of B = (20/100) * 100 = 20 grams.
After transfer:
Vessel B: 600 – 100 = 500 ml solution, salt = 120 – 20 = 100 grams.
Strength of solution in B = (100 / 500) * 100 = 20%.
Vessel C: 500 + 100 = 600 ml solution. Salt in C = 160 (initial) + 20 (from B) = 180 grams.
Strength of solution in C = (180 / 600) * 100 = 30%.

Step 3: 100 ml of solution from C is transferred to A.
The strength of solution in C is now 30%.
Amount of salt in 100 ml of C = (30/100) * 100 = 30 grams.
After transfer:
Vessel C: 600 – 100 = 500 ml solution, salt = 180 – 30 = 150 grams.
Strength of solution in C = (150 / 500) * 100 = 30%.
Vessel A: 400 (from previous step) + 100 = 500 ml solution. Salt in A = 40 (from previous step) + 30 (from C) = 70 grams.

The strength, in percentage, of the resulting solution in vessel A is (70 / 500) * 100 = 14%.

Correct_Option:D

Ans: B

Explanation:Let F be the number of fiction books in 2010 and N be the number of nonfiction books in 2010.
According to the problem statement:
In 2010, the total number of books was 11500.
So, F + N = 11500 (Equation 1)

In 2015, there was a 10% increase in the fiction category and a 12% increase in the nonfiction category.
Number of fiction books in 2015 = F + 0.10F = 1.10F
Number of nonfiction books in 2015 = N + 0.12N = 1.12N

In 2015, the total number of books was 12760.
So, 1.10F + 1.12N = 12760 (Equation 2)

Now we have a system of two linear equations with two variables:
1) F + N = 11500
2) 1.10F + 1.12N = 12760

From Equation 1, we can express N in terms of F:
N = 11500 – F

Substitute this expression for N into Equation 2:
1.10F + 1.12(11500 – F) = 12760
1.10F + 1.12 * 11500 – 1.12F = 12760
1.10F + 12880 – 1.12F = 12760

Combine the F terms:
(1.10 – 1.12)F + 12880 = 12760
-0.02F + 12880 = 12760

Subtract 12880 from both sides:
-0.02F = 12760 – 12880
-0.02F = -120

Divide by -0.02 to solve for F:
F = -120 / -0.02
F = 120 / 0.02
F = 120 * (100 / 2)
F = 120 * 50
F = 6000

So, there were 6000 fiction books in 2010.

The question asks for the number of fiction books in the library in 2015.
Number of fiction books in 2015 = 1.10F
Number of fiction books in 2015 = 1.10 * 6000
Number of fiction books in 2015 = 6600

Let’s verify this with nonfiction books.
N = 11500 – F = 11500 – 6000 = 5500 (nonfiction books in 2010)
Nonfiction books in 2015 = 1.12N = 1.12 * 5500 = 6160
Total books in 2015 = 6600 (fiction) + 6160 (nonfiction) = 12760. This matches the given total.

The number of fiction books in the library in 2015 is 6600.

Correct_Option:B

Ans: 80

Explanation:Let A, B, C, and D represent the scores of the respective individuals.
We are given the following information:
1. The score of A was 10% less than that of B.
This can be written as: A = B – 0.10B = 0.90B
2. The score of B was 25% more than that of C.
This can be written as: B = C + 0.25C = 1.25C
3. The score of C was 20% less than that of D.
This can be written as: C = D – 0.20D = 0.80D
4. A scored 72.
So, A = 72.

We need to find the score of D. We can work backward from A’s score.

From statement 1, we know A = 0.90B.
Substitute the value of A: 72 = 0.90B
To find B, divide 72 by 0.90:
B = 72 / 0.90 = 72 / (9/10) = 72 * (10/9) = 8 * 10 = 80.
So, B scored 80.

From statement 2, we know B = 1.25C.
Substitute the value of B: 80 = 1.25C
To find C, divide 80 by 1.25:
C = 80 / 1.25 = 80 / (5/4) = 80 * (4/5) = 16 * 4 = 64.
So, C scored 64.

From statement 3, we know C = 0.80D.
Substitute the value of C: 64 = 0.80D
To find D, divide 64 by 0.80:
D = 64 / 0.80 = 64 / (8/10) = 64 * (10/8) = 8 * 10 = 80.
So, D scored 80.

Alternatively, we can express A in terms of D:
A = 0.90B
B = 1.25C
C = 0.80D

Substitute C into B:
B = 1.25 * (0.80D) = 1.00D = D

Substitute B into A:
A = 0.90 * (1.00D) = 0.90D

We are given A = 72.
So, 72 = 0.90D
D = 72 / 0.90 = 80.

Final_Answer:80

Ans: B

Explanation:Let N be the maximum possible score in the examination.
The pass mark is 45% of N, which can be written as 0.45N.
A candidate obtains 36 marks.
The candidate falls short of the pass mark by 68%. This means that the difference between the pass mark and the candidate’s score is 68% of the pass mark.

We can write this relationship as:
Pass Mark – Candidate’s Score = 68% of Pass Mark
0.45N – 36 = 0.68 * (0.45N)

Now, let’s solve for N:
0.45N – 36 = 0.68 * 0.45N
0.45N – 36 = 0.306N

Now, rearrange the equation to group the N terms:
0.45N – 0.306N = 36
0.144N = 36

Now, solve for N:
N = 36 / 0.144
N = 36 / (144/1000)
N = 36 * (1000/144)
N = (36 * 1000) / 144

We can simplify this calculation. Notice that 144 = 4 * 36.
N = (36 * 1000) / (4 * 36)
N = 1000 / 4
N = 250

Now let’s check which option is correct for N = 250.
Option A: $N \leq 200$ (250 is not less than or equal to 200) – False
Option B: $243 \leq N \leq 252$ (243 <= 250 <= 252) - True
Option C: $201 \leq N \leq 242$ (250 is not within this range) – False
Option D: $N \geq 253$ (250 is not greater than or equal to 253) – False

Therefore, the correct option is B.

Correct_Option:B

Ans: B

Explanation:
Initially, the total volume of the mixture is 175 ml (water) + 700 ml (alcohol) = 875 ml.

Step 1: Gopal takes out 10% of the mixture.
Amount removed = 10% of 875 ml = 0.10 * 875 = 87.5 ml.
Amount of water removed = 10% of 175 ml = 0.10 * 175 = 17.5 ml.
Amount of alcohol removed = 10% of 700 ml = 0.10 * 700 = 70 ml.

After removing 87.5 ml:
Remaining water = 175 ml – 17.5 ml = 157.5 ml.
Remaining alcohol = 700 ml – 70 ml = 630 ml.

He substitutes the removed amount with water.
Amount of water added = 87.5 ml.

After the first substitution:
New amount of water = 157.5 ml + 87.5 ml = 245 ml.
New amount of alcohol = 630 ml.
New total volume = 245 ml + 630 ml = 875 ml.

The percentage of water after the first operation is (245 / 875) * 100 = 28.57%.

Step 2: The process is repeated once again.
Gopal takes out 10% of the new mixture.
Amount removed = 10% of 875 ml = 87.5 ml.
Amount of water removed = 10% of 245 ml = 0.10 * 245 = 24.5 ml.
Amount of alcohol removed = 10% of 630 ml = 0.10 * 630 = 63 ml.

After removing 87.5 ml:
Remaining water = 245 ml – 24.5 ml = 220.5 ml.
Remaining alcohol = 630 ml – 63 ml = 567 ml.

He substitutes the removed amount with water.
Amount of water added = 87.5 ml.

After the second substitution:
New amount of water = 220.5 ml + 87.5 ml = 308 ml.
New amount of alcohol = 567 ml.
New total volume = 308 ml + 567 ml = 875 ml.

The percentage of water in the mixture now is (308 / 875) * 100.
(308 / 875) * 100 = 0.352 * 100 = 35.2%.

Alternatively, we can use the formula for the amount of a component remaining after repeated dilution:
Final amount of component = Initial amount * (1 – (fraction removed))^n
where n is the number of operations.

In this case, we are interested in the amount of alcohol, as the water is being added.
Initial amount of alcohol = 700 ml.
Fraction removed = 10% = 0.1.
Number of operations = 2.

Final amount of alcohol = 700 * (1 – 0.1)^2 = 700 * (0.9)^2 = 700 * 0.81 = 567 ml.
The total volume of the mixture remains 875 ml.
The final amount of water = Total volume – Final amount of alcohol = 875 ml – 567 ml = 308 ml.
Percentage of water = (308 / 875) * 100 = 35.2%.

Correct_Option: B

Ans: 20

Explanation:Let Arun’s present age be $A$ and Barun’s present age be $B$.
According to the first statement, Arun’s present age is 40% of Barun’s.
So, $A = 0.40B$ (Equation 1)

Let the number of years after which Arun’s age will be half of Barun’s be $x$.
In $x$ years, Arun’s age will be $A+x$.
In $x$ years, Barun’s age will be $B+x$.
According to the second statement, Arun’s age will be half of Barun’s after $x$ years.
So, $A+x = \frac{1}{2}(B+x)$ (Equation 2)

Substitute the value of $A$ from Equation 1 into Equation 2:
$0.40B + x = \frac{1}{2}(B+x)$
Multiply both sides by 2 to eliminate the fraction:
$2(0.40B + x) = B+x$
$0.80B + 2x = B+x$
Rearrange the terms to solve for $x$:
$2x – x = B – 0.80B$
$x = 0.20B$

This means that after $0.20B$ years, Arun’s age will be half of Barun’s.

Now we need to find the percentage increase in Barun’s age during this period.
Barun’s present age is $B$.
Barun’s age after $x$ years is $B+x$.
We found that $x = 0.20B$.
So, Barun’s age after $x$ years is $B + 0.20B = 1.20B$.

The increase in Barun’s age is $(B+x) – B = x = 0.20B$.
The percentage increase in Barun’s age is calculated as:
Percentage Increase = $\frac{\text{Increase in Age}}{\text{Original Age}} \times 100$
Percentage Increase = $\frac{0.20B}{B} \times 100$
Percentage Increase = $0.20 \times 100$
Percentage Increase = $20\%$

Final_Answer:20

Ans: 70000

Explanation:Let Ravi’s total monthly savings be S.
He invests 50% of his monthly savings in fixed deposits.
Amount invested in fixed deposits = 0.50 * S

The rest of his savings = S – 0.50 * S = 0.50 * S

Thirty percent of the rest of his savings is invested in stocks.
Amount invested in stocks = 0.30 * (0.50 * S) = 0.15 * S

The rest goes into Ravi’s savings bank account.
Amount in savings bank account = (0.50 * S) – (0.15 * S) = 0.35 * S

The total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500.
Total amount deposited = Amount in fixed deposits + Amount in savings bank account
59500 = (0.50 * S) + (0.35 * S)
59500 = 0.85 * S

To find Ravi’s total monthly savings (S), we can rearrange the equation:
S = 59500 / 0.85
S = 59500 / (85/100)
S = 59500 * (100/85)
S = (59500 * 100) / 85

We can simplify the division:
59500 / 85 = 700
So, S = 700 * 100
S = 70000

Ravi’s total monthly savings is Rs 70000.

Check:
Fixed deposits = 0.50 * 70000 = 35000
Rest of savings = 70000 – 35000 = 35000
Stocks = 0.30 * 35000 = 10500
Savings bank account = 35000 – 10500 = 24500
Total deposited in bank (Fixed Deposits + Savings Account) = 35000 + 24500 = 59500. This matches the given information.

Final_Answer:70000

Ans: D

Explanation:Let the number of boys appearing for the test be $B$.
According to the question, the number of girls appearing for the test is twice the number of boys, so the number of girls is $2B$.
The total number of candidates appearing for the test is $B + 2B = 3B$.

The percentage of girls who get admission is 30%.
Number of girls who get admission = $30\%$ of $2B = 0.30 \times 2B = 0.6B$.
The percentage of girls who do not get admission = $100\% – 30\% = 70\%$.
Number of girls who do not get admission = $70\%$ of $2B = 0.70 \times 2B = 1.4B$.

The percentage of boys who get admission is 45%.
Number of boys who get admission = $45\%$ of $B = 0.45 \times B = 0.45B$.
The percentage of boys who do not get admission = $100\% – 45\% = 55\%$.
Number of boys who do not get admission = $55\%$ of $B = 0.55 \times B = 0.55B$.

The total number of candidates who do not get admission is the sum of the number of girls and boys who do not get admission:
Total not admitted = Number of girls not admitted + Number of boys not admitted
Total not admitted = $1.4B + 0.55B = 1.95B$.

The percentage of candidates who do not get admission is calculated as:
Percentage not admitted = (Total not admitted / Total candidates) $\times 100$
Percentage not admitted = $(1.95B / 3B) \times 100$
Percentage not admitted = $(1.95 / 3) \times 100$
Percentage not admitted = $0.65 \times 100$
Percentage not admitted = $65\%$.

Correct_Option:D

Ans: B

Explanation:Let the total number of shirts produced in the factory be $T$.
The number of defective shirts is 15% of $T$, which is $0.15T$.
The number of non-defective shirts is $T – 0.15T = 0.85T$.

20% of the rest (non-defective shirts) are sold in the domestic market.
So, the number of shirts sold in the domestic market is 20% of $0.85T$, which is $0.20 \times 0.85T = 0.17T$.

The remaining shirts are left for export.
The number of shirts left for export is the total non-defective shirts minus the shirts sold in the domestic market.
Number of shirts for export = $0.85T – 0.17T = 0.68T$.

We are given that the remaining 8840 shirts are left for export.
So, we can set up the equation:
$0.68T = 8840$

To find the total number of shirts produced ($T$), we can divide 8840 by 0.68:
$T = \frac{8840}{0.68}$
$T = \frac{8840}{\frac{68}{100}}$
$T = 8840 \times \frac{100}{68}$

Now, let’s simplify the fraction.
$T = \frac{884000}{68}$

We can perform the division:
$884000 \div 68$

Let’s divide 884 by 68:
$884 \div 68 = 13$

So, $884000 \div 68 = 13000$.

Therefore, the number of shirts produced in the factory is 13000.

Let’s verify:
Total shirts = 13000
Defective shirts = 15% of 13000 = $0.15 \times 13000 = 1950$
Non-defective shirts = $13000 – 1950 = 11050$
Domestic sales = 20% of non-defective shirts = 20% of 11050 = $0.20 \times 11050 = 2210$
Export shirts = Non-defective shirts – Domestic sales = $11050 – 2210 = 8840$
This matches the given information.

Correct_Option:B

Ans: C

Explanation:Let P be the population of the village and F be the total production of food grains.
Let pc be the per capita production of food grains.
The relationship between these variables is given by:
pc = F / P

We are given that the production of food grains increased by 40%.
So, the new production of food grains, F’, is F * (1 + 40/100) = 1.40F.

We are also given that the per capita production of food grains increased by 27%.
So, the new per capita production of food grains, pc’, is pc * (1 + 27/100) = 1.27pc.

Let the percentage increase in the population be x%.
So, the new population, P’, is P * (1 + x/100).

Now, we can write the new per capita production in terms of the new total production and new population:
pc’ = F’ / P’

Substitute the expressions for F’, P’, and pc’:
1.27pc = 1.40F / (P * (1 + x/100))

We know that pc = F / P. Substitute this into the equation:
1.27 * (F / P) = 1.40F / (P * (1 + x/100))

We can cancel out F/P from both sides of the equation (assuming F and P are not zero):
1.27 = 1.40 / (1 + x/100)

Now, we need to solve for x:
1 + x/100 = 1.40 / 1.27
1 + x/100 = 1.10236…

Subtract 1 from both sides:
x/100 = 1.10236… – 1
x/100 = 0.10236…

Multiply by 100 to find x:
x = 0.10236… * 100
x = 10.236…

The percentage by which the population of the village increased is approximately 10.24%.
We need to find the nearest option.
Option A: 16
Option B: 13
Option C: 10
Option D: 7

The value 10.24 is nearest to 10.

Correct_Option:C