Quantitative Aptitude and Reasoning Archives

101. The sum of two numbers is 143. If the greater number is divided by the

The sum of two numbers is 143. If the greater number is divided by the difference of the numbers, the quotient is 7. What is the difference of the two numbers ?

[amp_mcq option1=”9″ option2=”11″ option3=”14″ option4=”15″ correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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Let the two numbers be \(x\) and \(y\), with \(x > y\).
According to the problem:
1. The sum of the two numbers is 143: \(x + y = 143\)
2. The greater number (\(x\)) divided by the difference of the numbers (\(x – y\)) is 7: \(\frac{x}{x – y} = 7\)

From the second equation, we get:
\(x = 7(x – y)\)
\(x = 7x – 7y\)
\(7y = 7x – x\)
\(7y = 6x\)
\(x = \frac{7y}{6}\)

Now substitute this expression for \(x\) into the first equation:
\(\frac{7y}{6} + y = 143\)
To combine the terms on the left side, find a common denominator:
\(\frac{7y + 6y}{6} = 143\)
\(\frac{13y}{6} = 143\)
\(13y = 143 \times 6\)
\(y = \frac{143 \times 6}{13}\)
Since \(143 = 13 \times 11\), we have:
\(y = \frac{13 \times 11 \times 6}{13}\)
\(y = 11 \times 6\)
\(y = 66\)

Now find the value of \(x\) using \(x = \frac{7y}{6}\):
\(x = \frac{7 \times 66}{6}\)
\(x = 7 \times 11\)
\(x = 77\)

The two numbers are 77 and 66.
The question asks for the difference of the two numbers, which is \(x – y\):
Difference = \(77 – 66 = 11\)

– Translate the word problem into a system of two linear equations with two variables.
– Solve the system of equations to find the values of the two numbers.
– Calculate the required difference between the numbers.

Verifying the solution:
Sum: \(77 + 66 = 143\) (Correct)
Difference: \(77 – 66 = 11\)
Greater number divided by difference: \(77 / 11 = 7\) (Correct)
The difference of the two numbers is indeed 11.

102. Given that 1st November 2019 is a Friday. Then 1st November 2022 is a

Given that 1st November 2019 is a Friday. Then 1st November 2022 is a

[amp_mcq option1=”Monday” option2=”Tuesday” option3=”Wednesday” option4=”Thursday” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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If 1st November 2019 is a Friday, then 1st November 2022 is a Tuesday.

To find the day of the week after a certain number of years, we need to calculate the number of “odd days” between the two dates. An odd day is the remainder when the total number of days is divided by 7.
A normal year (365 days) has 365 mod 7 = 1 odd day.
A leap year (366 days) has 366 mod 7 = 2 odd days.
The period from 1st November 2019 to 1st November 2022 spans three full years: the period covering 2020, the period covering 2021, and the period covering 2022 up to Nov 1.
– From Nov 1, 2019 to Nov 1, 2020: This period includes the entire year 2020, which is a leap year (divisible by 4). Thus, this period adds 2 odd days.
– From Nov 1, 2020 to Nov 1, 2021: This period includes the entire year 2021, which is a normal year. Thus, this period adds 1 odd day.
– From Nov 1, 2021 to Nov 1, 2022: This period includes the entire year 2022, which is a normal year. Thus, this period adds 1 odd day.
Total odd days from 1st November 2019 to 1st November 2022 = 2 (from 2020) + 1 (from 2021) + 1 (from 2022) = 4 odd days.
Starting from Friday, we move forward by 4 days:
Friday + 1 day = Saturday
Saturday + 1 day = Sunday
Sunday + 1 day = Monday
Monday + 1 day = Tuesday
So, 1st November 2022 is a Tuesday.

Leap years occur every 4 years, except for years divisible by 100 but not by 400. 2020 is a leap year because it is divisible by 4. Years like 1900 are not leap years, while 2000 is a leap year. This rule helps determine the number of odd days accurately over longer periods.

103. In a certain year, a school had 60% boys and 40% girls as students. In

In a certain year, a school had 60% boys and 40% girls as students. In the next five years the number of boys decreased by 10% and the number of girls increased by 10%. What is the change in total roll strength of the school in the five years ?

[amp_mcq option1=”3% increase” option2=”2% decrease” option3=”No change” option4=”5% decrease” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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The change in total roll strength of the school in the five years is a 2% decrease.

Assume the initial total roll strength is 100 units.
Initial Boys = 60% of 100 = 60 units.
Initial Girls = 40% of 100 = 40 units.
In the next five years:
Number of boys decreased by 10%. New number of boys = 60 * (1 – 0.10) = 60 * 0.90 = 54 units.
Number of girls increased by 10%. New number of girls = 40 * (1 + 0.10) = 40 * 1.10 = 44 units.
New total roll strength = New Boys + New Girls = 54 + 44 = 98 units.
The change in total roll strength = New Total – Initial Total = 98 – 100 = -2 units.
The percentage change = (Change / Initial Total) * 100 = (-2 / 100) * 100 = -2%.
A -2% change represents a 2% decrease.

This calculation shows how percentage changes applied to different subgroups affect the overall total, depending on the initial proportions of the subgroups.

104. X paid ₹ 47 for certain cups of tea and coffee. If tea costs ₹ 5 per c

X paid ₹ 47 for certain cups of tea and coffee. If tea costs ₹ 5 per cup and coffee costs ₹ 8 per cup, which one of the following statements is correct ?

[amp_mcq option1=”He drank 8 cups of tea and coffee.” option2=”He drank the same number of cups of tea and coffee.” option3=”He drank more tea than coffee.” option4=”He drank more coffee than tea.” correct=”option4″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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He drank more coffee than tea.

Let ‘t’ be the number of cups of tea and ‘c’ be the number of cups of coffee. The cost equation is 5t + 8c = 47. We need to find non-negative integer solutions for t and c.
We can try possible values for c:
If c=0, 5t = 47 (not possible for integer t)
If c=1, 5t = 47 – 8 = 39 (not possible for integer t)
If c=2, 5t = 47 – 16 = 31 (not possible for integer t)
If c=3, 5t = 47 – 24 = 23 (not possible for integer t)
If c=4, 5t = 47 – 32 = 15 => t = 3. This is a valid integer solution (t=3, c=4).
If c=5, 5t = 47 – 40 = 7 (not possible for integer t)
If c=6, 5t = 47 – 48 = -1 (not possible for non-negative t)
The only valid solution is t=3 cups of tea and c=4 cups of coffee.
Based on this solution:
A) Total cups = 3 + 4 = 7, not 8. (Incorrect)
B) Number of tea cups (3) is not the same as coffee cups (4). (Incorrect)
C) He drank 3 cups of tea and 4 cups of coffee. 3 is not more than 4. (Incorrect)
D) He drank 4 cups of coffee and 3 cups of tea. 4 is more than 3. (Correct)

This is a simple linear Diophantine equation where we are looking for non-negative integer solutions. Since the coefficients are relatively small, trial and error is an efficient method to find the solution.

105. How many times will the digit 5 come in counting from 1 to 99, excludi

How many times will the digit 5 come in counting from 1 to 99, excluding those numbers which are divisible by 3 ?

[amp_mcq option1=”16″ option2=”15″ option3=”14″ option4=”13″ correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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The digit 5 comes 16 times in counting from 1 to 99, excluding those numbers which are divisible by 3.

First, count the total occurrences of the digit 5 from 1 to 99. The digit 5 appears in the units place in 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 (10 times). It appears in the tens place in 50, 51, 52, 53, 54, 55, 56, 57, 58, 59 (10 times). The number 55 contains the digit 5 twice. So, the digit 5 appears a total of 10 + 10 = 20 times.
Next, identify the numbers between 1 and 99 that contain the digit 5 and are divisible by 3. These numbers are:
– 15 (1+5=6, divisible by 3) – contains one 5
– 45 (4+5=9, divisible by 3) – contains one 5
– 51 (5+1=6, divisible by 3) – contains one 5
– 54 (5+4=9, divisible by 3) – contains one 5
– 57 (5+7=12, divisible by 3) – contains one 5
– 75 (7+5=12, divisible by 3) – contains one 5
These are 6 numbers. Let’s count the occurrences of the digit 5 *within* these numbers: 15 (one 5), 45 (one 5), 51 (one 5), 54 (one 5), 57 (one 5), 75 (one 5). Total occurrences of 5 in these numbers are 6.
The numbers containing 5 are: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85, 95.
The numbers from this list that are divisible by 3 are: 15, 45, 51, 54, 57, 75.
We need to count the 5s in the numbers *not* divisible by 3: 5, 25, 35, 50, 52, 53, 55, 56, 58, 59, 65, 85, 95.
Counting the 5s in this remaining list:
5 (one 5), 25 (one 5), 35 (one 5), 50 (one 5), 52 (one 5), 53 (one 5), 55 (two 5s), 56 (one 5), 58 (one 5), 59 (one 5), 65 (one 5), 85 (one 5), 95 (one 5).
Total count = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 16.

The key is to correctly identify which numbers containing 5 are excluded (those divisible by 3) and then count the 5s in the remaining set of numbers that contain the digit 5.

106. At 8:30 am, the angle between the minute and hour hands is

At 8:30 am, the angle between the minute and hour hands is

[amp_mcq option1=”70°” option2=”75°” option3=”80°” option4=”85°” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

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At 8:30 am, the angle between the minute and hour hands is 75°.

The relative speed of the hour hand and minute hand needs to be considered.
– The hour hand moves 360° in 12 hours, or 30° per hour, or 0.5° per minute.
– The minute hand moves 360° in 60 minutes, or 6° per minute.
– At 8:30:
– The minute hand is exactly on the number 6, which corresponds to 30 minutes past 12. Its position from the 12 o’clock mark is 30 minutes * 6°/minute = 180°.
– The hour hand is past the number 8, halfway between 8 and 9. Its position from the 12 o’clock mark is calculated based on the hours (8) and the minutes (30). The position is (8 hours * 30°/hour) + (30 minutes * 0.5°/minute) = 240° + 15° = 255°.
– The angle between the hands is the absolute difference between their positions: |255° – 180°| = 75°.

A useful formula for finding the angle between hour and minute hands is |30H – (11/2)M|, where H is the hour and M is the minute. For 8:30 am, H=8, M=30. Angle = |30*8 – (11/2)*30| = |240 – 11*15| = |240 – 165| = 75°.

107. Aneesh borrowed some money at the rate of 5% per annum for the first 3

Aneesh borrowed some money at the rate of 5% per annum for the first 3 years, 8% per annum for the next 7 years and 12% per annum for the period beyond 10 years. If the total interest paid by him at the end of 13 years is ₹ 5,350, how much money did Aneesh borrow, if the interest is charged as simple interest?

[amp_mcq option1=”₹ 5,000″ option2=”₹ 5,500″ option3=”₹ 5,100″ option4=”₹ 5,800″ correct=”option1″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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Let the principal amount borrowed be P. The total time period is 13 years. The simple interest rate changes over periods:
Period 1: Rate (R1) = 5% p.a., Time (T1) = 3 years.
Simple Interest for Period 1 (SI1) = (P * R1 * T1) / 100 = (P * 5 * 3) / 100 = 15P / 100.
Period 2: Rate (R2) = 8% p.a., Time (T2) = next 7 years. Total time elapsed = 3 + 7 = 10 years.
Simple Interest for Period 2 (SI2) = (P * R2 * T2) / 100 = (P * 8 * 7) / 100 = 56P / 100.
Period 3: Rate (R3) = 12% p.a., Time (T3) = period beyond 10 years until 13 years. T3 = 13 – 10 = 3 years.
Simple Interest for Period 3 (SI3) = (P * R3 * T3) / 100 = (P * 12 * 3) / 100 = 36P / 100.
The total interest paid is the sum of simple interests for each period.
Total SI = SI1 + SI2 + SI3 = (15P / 100) + (56P / 100) + (36P / 100) = (15P + 56P + 36P) / 100 = 107P / 100.
Given that the total interest paid is ₹ 5,350.
107P / 100 = 5350.
107P = 5350 * 100.
P = (5350 * 100) / 107.
Since 5350 = 107 * 50, we have:
P = (107 * 50 * 100) / 107 = 50 * 100 = ₹ 5,000.
The amount borrowed was ₹ 5,000.

In simple interest, the interest for each period is calculated on the original principal amount. When the rate changes over different periods, the total simple interest is the sum of the interest accumulated during each period with its respective rate and duration.

This method applies specifically to simple interest. For compound interest, the principal would change after each period as accumulated interest is added to the principal.

108. A shopkeeper fixes the marked price of an item 25% above its cost pric

A shopkeeper fixes the marked price of an item 25% above its cost price. What percentage of discount may be allowed to gain 6%?

[amp_mcq option1=”15.5%” option2=”15.2%” option3=”15.0%” option4=”15.8%” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

Download PDF Attempt Online

Let the Cost Price (CP) of the item be ₹ 100.
The shopkeeper fixes the Marked Price (MP) 25% above the cost price.
MP = CP + 25% of CP = 100 + (25/100) * 100 = 100 + 25 = ₹ 125.
The desired gain is 6%. The Selling Price (SP) must be CP plus the gain.
SP = CP + 6% of CP = 100 + (6/100) * 100 = 100 + 6 = ₹ 106.
The discount is the difference between the Marked Price and the Selling Price.
Discount = MP – SP = 125 – 106 = ₹ 19.
The discount percentage is calculated on the Marked Price.
Discount Percentage = (Discount / MP) * 100% = (19 / 125) * 100%.
(19 / 125) * 100 = (19 * 4 / 125 * 4) * 100 = (76 / 500) * 100 = 76 / 5 = 15.2%.

Profit percentage is calculated on Cost Price, while Discount percentage is calculated on Marked Price.

This type of problem involves the relationship between Cost Price, Marked Price, Selling Price, Profit/Loss percentage, and Discount percentage. The Marked Price is typically set above the Cost Price, and the discount is offered on the Marked Price to arrive at the Selling Price.

109. A man buys a cycle for ₹ 1,400 and sells it at a loss of 15%. What is

A man buys a cycle for ₹ 1,400 and sells it at a loss of 15%. What is the selling price of the cycle?

[amp_mcq option1=”₹ 1,190″ option2=”₹ 1,160″ option3=”₹ 210″ option4=”₹ 1,180″ correct=”option1″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

Download PDF Attempt Online

The cost price (CP) of the cycle is ₹ 1,400.
The cycle is sold at a loss of 15%.
The amount of loss is 15% of the cost price.
Loss amount = 15% of ₹ 1,400 = (15/100) * 1400 = 15 * 14 = ₹ 210.
The selling price (SP) is the cost price minus the loss.
SP = CP – Loss amount = ₹ 1,400 – ₹ 210 = ₹ 1,190.
Alternatively, the selling price can be calculated directly as CP * (1 – Loss%/100).
SP = 1400 * (1 – 15/100) = 1400 * (85/100) = 14 * 85 = ₹ 1,190.

Loss is calculated as a percentage of the cost price. Selling price in case of loss is Cost Price minus Loss amount.

Profit or loss percentages are typically calculated on the cost price unless otherwise stated.

110. In 2020, the medical expense of a person increased by 80%. If his medi

In 2020, the medical expense of a person increased by 80%. If his medical expense in 2019 was 10% of his salary, then what percent of his salary was the medical expense in 2020?

[amp_mcq option1=”20%” option2=”18%” option3=”16%” option4=”15%” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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Let the person’s salary in 2019 be S.
His medical expense in 2019 was 10% of his salary, which is 0.10 * S.
In 2020, the medical expense increased by 80%. The increase amount is 80% of the 2019 expense.
Increase = 80% of (0.10 * S) = (80/100) * (0.10 * S) = 0.80 * 0.10 * S = 0.08 * S.
The medical expense in 2020 is the 2019 expense plus the increase:
Expense in 2020 = (0.10 * S) + (0.08 * S) = 0.18 * S.
To express this as a percentage of his salary in 2020 (assuming salary remained the same, as implied by the question structure), we have:
Percentage = (Expense in 2020 / Salary) * 100%
Percentage = (0.18 * S / S) * 100% = 0.18 * 100% = 18%.

The key is to calculate the increase amount based on the original expense and then add it to the original expense to find the new expense.

This is a straightforward percentage calculation problem. If the base for the percentage (salary) changes, the question would specify the change in salary as well. Since it doesn’t, we assume the salary remains constant relative to which the expense percentage is calculated.