Quantitative Aptitude and Reasoning Archives

11. An amount is adequate to pay salaries to R for 20 days or to S for 30

An amount is adequate to pay salaries to R for 20 days or to S for 30 days. For how many days is the money enough to cover salaries of both of them together?

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12 days

15 days

11 days

20 days

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UPSC CISF-AC-EXE – 2024

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The correct answer is 12 days.

This is a work and time problem. We can determine the efficiency (or daily work rate) of R and S based on the number of days they take to complete the work (pay salaries).
Let the total amount of money (representing the total work) be M.
R can be paid for 20 days with M. So, R’s daily salary (work rate) = M/20 per day.
S can be paid for 30 days with M. So, S’s daily salary (work rate) = M/30 per day.
If both R and S are paid together, their combined daily salary is R’s daily salary + S’s daily salary = M/20 + M/30.
To find how many days (let’s call this D) the money M will last for both, we set M equal to the combined daily salary multiplied by D:
M = (M/20 + M/30) * D
Divide both sides by M (assuming M > 0):
1 = (1/20 + 1/30) * D
Find a common denominator for 1/20 and 1/30, which is 60:
1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12.
So, 1 = (1/12) * D
Multiply by 12 to find D:
D = 12 days.

This problem can also be viewed as calculating the time taken by two individuals to complete a task together, given the time each takes individually. If R takes ‘a’ days and S takes ‘b’ days, the time taken together is given by the formula $\frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{ab}{a+b}$. In this case, a=20 and b=30. Time together = $\frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12$ days.

12. Consider the following sum : @37 + 46@ + 5@3 = 1@@@@ In the above sum,

Consider the following sum :
@37 + 46@ + 5@3 = 1@@@@
In the above sum, @ stands for which one among the following?

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For all digits

For even digits only

For odd digits only

For no digits

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UPSC CISF-AC-EXE – 2024

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In the given sum, ‘@’ stands for any digit from 0 to 9, which means for all digits.

Let the sum be represented by the equation:
$(100 \times @ + 30 + 7) + (100 \times 4 + 60 + @) + (100 \times 5 + 10 \times @ + 3) = (1000 \times 1 + 100 \times @ + 10 \times @ + @)$
Simplifying the left side:
$100@ + 37 + 460 + @ + 500 + 10@ + 3 = (100@ + @ + 10@) + (37 + 460 + 500 + 3) = 111@ + 1000$
Simplifying the right side:
$1000 + 100@ + 10@ + @ = 1000 + 111@$
The equation becomes $111@ + 1000 = 111@ + 1000$.

This equation is an identity, which means it holds true for any value of @. Since @ represents a single digit in a number, it must be one of the digits from 0 to 9. Therefore, @ can be any digit. The structure of the column-wise addition also confirms this: in each column (units, tens, hundreds), the sum is $(10 + @)$, resulting in the units digit @ and a carry-over of 1 to the next column. The final carry-over of 1 becomes the thousands digit of the total sum, which is also 1, matching the format 1@@@@. This structure works regardless of the specific digit @, as long as it’s 0-9.

13. If $\frac{625}{0.625} = \frac{62.5}{x}$, then the value of x is

If $\frac{625}{0.625} = \frac{62.5}{x}$, then the value of x is

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0.0625

0.00625

6.25

0.000625

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This question was previously asked in

UPSC CISF-AC-EXE – 2024

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The value of x is 0.0625.

The given equation is $\frac{625}{0.625} = \frac{62.5}{x}$.
First, evaluate the left side of the equation:
$\frac{625}{0.625} = \frac{625}{\frac{625}{1000}} = 625 \times \frac{1000}{625} = 1000$.

Now substitute the value back into the equation:
$1000 = \frac{62.5}{x}$
To solve for x, multiply both sides by x and divide by 1000:
$x = \frac{62.5}{1000}$
Dividing $62.5$ by $1000$ means moving the decimal point three places to the left:
$x = 0.0625$.

14. What is the value of the following sum? $\frac{1}{(\sqrt{2}+\sqrt{1})}

What is the value of the following sum?
$\frac{1}{(\sqrt{2}+\sqrt{1})} + \frac{1}{(\sqrt{3}+\sqrt{2})} + \frac{1}{(\sqrt{4}+\sqrt{3})} + … + \frac{1}{(\sqrt{100}+\sqrt{99})}$

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This question was previously asked in

UPSC CISF-AC-EXE – 2024

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The value of the given sum is 9.

The given sum is $\sum_{k=1}^{99} \frac{1}{\sqrt{k+1}+\sqrt{k}}$. We can rationalize each term by multiplying the numerator and denominator by the conjugate of the denominator:
$\frac{1}{\sqrt{k+1}+\sqrt{k}} \times \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}} = \frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1})^2 – (\sqrt{k})^2} = \frac{\sqrt{k+1}-\sqrt{k}}{(k+1) – k} = \sqrt{k+1}-\sqrt{k}$.

The sum becomes a telescoping series:
$(\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4}-\sqrt{3}) + \dots + (\sqrt{100}-\sqrt{99})$
The intermediate terms cancel out: $(-\sqrt{1} + \sqrt{2}) + (-\sqrt{2} + \sqrt{3}) + (-\sqrt{3} + \sqrt{4}) + \dots + (-\sqrt{99} + \sqrt{100})$.
The sum simplifies to $\sqrt{100} – \sqrt{1}$.
$\sqrt{100} = 10$ and $\sqrt{1} = 1$.
The sum is $10 – 1 = 9$.

15. A company calculates its profit or loss quarterly. The company makes 2

A company calculates its profit or loss quarterly. The company makes 25% profit, 20% loss and 20% loss in three consecutive quarters. What minimum profit or loss should the company make in fourth quarter to have no loss overall?

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20% profit

20% loss

25% profit

25% loss

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This question was previously asked in

UPSC CISF-AC-EXE – 2024

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To have no loss overall, the company should make a minimum profit of 25% in the fourth quarter.

Let the initial capital be $C$.
Quarter 1: Profit of 25%. Capital becomes $C \times (1 + 0.25) = 1.25C$.
Quarter 2: Loss of 20%. Capital becomes $1.25C \times (1 – 0.20) = 1.25C \times 0.80 = C$.
Quarter 3: Loss of 20%. Capital becomes $C \times (1 – 0.20) = C \times 0.80 = 0.8C$.
After three quarters, the capital is $0.8C$.

To have no loss overall after four quarters, the final capital must be at least the initial capital, $C$. Let the profit percentage in the fourth quarter be $p$ (as a decimal). The capital after the fourth quarter will be $0.8C \times (1+p)$.
We need $0.8C \times (1+p) \ge C$.
Dividing by $0.8C$ (assuming $C>0$): $1+p \ge \frac{C}{0.8C} = \frac{1}{0.8} = \frac{1}{4/5} = 1.25$.
$p \ge 1.25 – 1 = 0.25$.
So, the minimum profit percentage in the fourth quarter must be $0.25 \times 100\% = 25\%$.

16. A man purchases 5 shirts in a ‘buy two, get five’ scheme. How much dis

A man purchases 5 shirts in a ‘buy two, get five’ scheme. How much discount percentage did he get ?

50%

60%

40%

45%

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This question was previously asked in

UPSC CISF-AC-EXE – 2023

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The scheme is “buy two, get five”. In the context of retail promotions, this is usually interpreted as “pay for 2 items and receive 5 items in total”. Thus, the man receives 5 shirts by paying the price of only 2 shirts.

– Total number of shirts received = 5
– Number of shirts paid for = 2
– Number of free shirts received = 5 – 2 = 3
– The discount percentage is calculated based on the total number of items received at the discounted price compared to the original price of all items.
– Discount percentage = (Number of free items / Total number of items received) * 100

Let the price of one shirt be P.
The actual price of 5 shirts would be 5 * P.
Under the scheme, the man pays the price of 2 shirts, which is 2 * P, to get 5 shirts.
The amount of discount received is the price of the free shirts, which is 3 * P.
Discount percentage = (Discount amount / Original price of items received) * 100
Discount percentage = (3P / 5P) * 100 = (3/5) * 100 = 60%.

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17. In a game between X and Y, X has to give ₹ 10 each time he loses to Y.

In a game between X and Y, X has to give ₹ 10 each time he loses to Y. If he wins, then he gets ₹ 50 from Y. If they play 15 times and X earns ₹ 450, how many times does X win ?

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This question was previously asked in

UPSC CISF-AC-EXE – 2023

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Let W be the number of times X wins and L be the number of times X loses. The total number of games played is W + L = 15. When X wins, he gains ₹ 50, so the total gain from wins is 50W. When X loses, he gives ₹ 10, so the total loss from losses is 10L. X’s net earning is the total gain minus the total loss, which is 50W – 10L. We are given that X earns ₹ 450, so 50W – 10L = 450. We now have a system of two linear equations: 1) W + L = 15 and 2) 50W – 10L = 450. From equation 1, L = 15 – W. Substituting this into equation 2: 50W – 10(15 – W) = 450. 50W – 150 + 10W = 450. 60W – 150 = 450. 60W = 600. W = 600 / 60 = 10. X wins 10 times.

– Define variables for the number of wins and losses.
– Set up one equation based on the total number of games played.
– Set up a second equation based on the total net earning (total gain from wins minus total loss from losses).
– Solve the system of linear equations for the number of wins.

This is an algebra word problem that requires setting up and solving a system of equations. It’s important to correctly represent the gain and loss per game and relate the total gain/loss to the net earning. Checking the answer (10 wins, 5 losses: 10*50 – 5*10 = 500 – 50 = 450) confirms the solution.

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18. A man takes a loan at 5% simple interest for a period of 2 years. He i

A man takes a loan at 5% simple interest for a period of 2 years. He immediately gives this money on loan at 5% compound interest for 2 years. What is the amount of loan he has taken if he makes a profit of ₹ 2,100 ?

₹ 8,00,000

₹ 4,20,000

₹ 1,00,000

₹ 8,40,000

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This question was previously asked in

UPSC CISF-AC-EXE – 2023

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Let the principal amount be P. The simple interest (SI) paid by the man for 2 years at 5% is SI = (P * 5 * 2) / 100 = 0.10P. The compound interest (CI) received by the man for 2 years at 5% is CI = P(1 + 5/100)^2 – P = P(1.05)^2 – P = P(1.1025) – P = 0.1025P. The man’s profit is the difference between the CI received and the SI paid: Profit = CI – SI = 0.1025P – 0.10P = 0.0025P. Given the profit is ₹ 2,100, we have 0.0025P = 2100. Solving for P: P = 2100 / 0.0025 = 2100 / (1/400) = 2100 * 400 = ₹ 8,40,000.

– Simple Interest formula: SI = PRT/100.
– Compound Interest formula: CI = Amount – Principal = P(1 + R/100)^T – P.
– The profit in this scenario is the difference between the interest earned (CI) and the interest paid (SI).
– Set up an equation with the profit difference equal to the given profit amount and solve for the principal P.

This question tests the understanding of simple and compound interest calculations and how to find the profit earned by an intermediary who borrows at one rate/type and lends at another rate/type. The difference in interest earned versus interest paid constitutes the profit.

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19. A businessman lost 20% of his goods due to rain. The remaining items w

A businessman lost 20% of his goods due to rain. The remaining items were sold at a profit of 50%. The businessman earned an overall profit of:

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10%

30%

20%

40%

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This question was previously asked in

UPSC CISF-AC-EXE – 2023

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Assume the total cost price of the goods was ₹ 100. Let the total quantity be 100 units, so the cost per unit is ₹ 1. 20% of goods were lost, meaning 20 units were lost. Remaining goods are 100 – 20 = 80 units. These 80 units are sold at a profit of 50% on their cost. The cost of the 80 units is ₹ 80 (since cost per unit was ₹ 1). Selling price of these 80 units = Cost * (1 + Profit%) = ₹ 80 * (1 + 0.50) = ₹ 80 * 1.50 = ₹ 120. The businessman invested ₹ 100 initially (the cost of the total goods) and received ₹ 120 from the sale. The overall profit is ₹ 120 – ₹ 100 = ₹ 20. Overall profit percentage = (Overall Profit / Total Cost) * 100 = (₹ 20 / ₹ 100) * 100 = 20%.

– The profit/loss is calculated on the original total cost of the goods, even if a portion is lost.
– The selling price is calculated based on the goods that *are* sold, applying the given profit margin to their cost.
– Overall profit = Total Revenue – Total Cost.
– Overall profit percentage = (Overall Profit / Total Cost) * 100.

Setting the initial cost price to a convenient value like ₹ 100 simplifies calculations for percentage problems. The loss of goods means the cost invested in those goods is not recovered, and this must be accounted for in the overall profit calculation.

20. ‘A’ and ‘B’ have pocket money in the ratio of 3 : 4. After the day’s w

‘A’ and ‘B’ have pocket money in the ratio of 3 : 4. After the day’s work, ‘A’ earned ₹ 600 while ‘B’ earned ₹ 500. However, ‘A’ spent ₹ 150 and ‘B’ spent ₹ 100 during the day. If they have equal amount of money at the end of the day, then the pocket money ‘A’ had in the morning is:

₹ 150

₹ 200

₹ 250

₹ 100

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This question was previously asked in

UPSC CISF-AC-EXE – 2023

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Let A’s initial pocket money be 3x and B’s initial pocket money be 4x. At the end of the day, A’s total money is 3x + ₹ 600 – ₹ 150 = 3x + ₹ 450. B’s total money is 4x + ₹ 500 – ₹ 100 = 4x + ₹ 400. Since they have equal amounts at the end, 3x + 450 = 4x + 400. Solving for x, we get x = 50. A’s initial pocket money was 3x, which is 3 * 50 = ₹ 150.

– Represent the initial amounts using the given ratio with a variable (e.g., 3x and 4x).
– Formulate expressions for the final amounts for A and B by adding earnings and subtracting spending.
– Set the final amounts equal to each other based on the problem statement.
– Solve the resulting linear equation for the variable x.
– Calculate the initial pocket money for A using the value of x.

This is a typical word problem involving ratios and linear equations. Careful tracking of money earned and spent for each person is crucial. The phrase “equal amount of money at the end of the day” provides the equation needed to solve for the unknown variable.

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