Quantitative Aptitude and Reasoning Archives

111. During the period 2016-2020, which company’s growth rate in profit is

During the period 2016-2020, which company’s growth rate in profit is the least?

[amp_mcq option1=”A” option2=”B” option3=”C” option4=”D” correct=”option4″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

The correct answer is D) D.

Using the growth rates calculated for the period 2016-2020 from the previous question:
Growth Rate (2016-2020) = ((Profit in 2020 – Profit in 2016) / Profit in 2016) * 100%
Company A: ≈ 385.71%
Company B: ≈ 366.67%
Company C: 200%
Company D: 25%

Comparing these growth rates, the least growth rate during the period 2016-2020 is 25%, which belongs to Company D.

112. Which company’s growth rate in profit is the highest during the period

Which company’s growth rate in profit is the highest during the period 2016-2020?

[amp_mcq option1=”A” option2=”B” option3=”C” option4=”D” correct=”option1″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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The correct answer is A) A.

The growth rate during the period 2016-2020 is calculated by comparing the profit in the final year (2020) to the profit in the initial year (2016).
Growth Rate (2016-2020) = ((Profit in 2020 – Profit in 2016) / Profit in 2016) * 100%
Company A: ((34 – 7) / 7) * 100% = (27 / 7) * 100% ≈ 385.71%
Company B: ((7 – 1.5) / 1.5) * 100% = (5.5 / 1.5) * 100% ≈ 366.67%
Company C: ((12 – 4) / 4) * 100% = (8 / 4) * 100% = 200%
Company D: ((25 – 20) / 20) * 100% = (5 / 20) * 100% = 25%

Comparing the calculated growth rates for the entire period: Company A (≈385.71%), Company B (≈366.67%), Company C (200%), Company D (25%). Company A has the highest growth rate during this period.

113. The next three (3) items are based on the following data, where profit

The next three (3) items are based on the following data, where profits of four companies during the year 2016 to 2020 are displayed in rupees thousand crore :

Year	Company
Year	A	B	C	D
2016	7	1.5	4	20
2017	14	2	10	22
2018	25	6	14	23
2019	30	7	15	21
2020	34	7	12	25

Which company had maximum growth rate in profit with respect to the previous year in single year during the period 2016-2020?

[amp_mcq option1=”A” option2=”B” option3=”C” option4=”D” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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The correct answer is B) B.

To find the company with the maximum single-year growth rate, we calculate the percentage growth year-on-year for each company from 2017 to 2020 and find the maximum value among all these calculations.
Growth Rate = ((Profit in Current Year – Profit in Previous Year) / Profit in Previous Year) * 100%
Company A: Max growth = ((14 – 7) / 7) * 100% = 100% (2017 vs 2016)
Company B: Max growth = ((6 – 2) / 2) * 100% = 200% (2018 vs 2017)
Company C: Max growth = ((10 – 4) / 4) * 100% = 150% (2017 vs 2016)
Company D: Max growth = ((22 – 20) / 20) * 100% = 10% (2017 vs 2016)
Comparing the maximum single-year growth rates (100%, 200%, 150%, 10%), the highest is 200%, achieved by Company B.

Calculating other growth rates:
A: 2018 vs 2017: (25-14)/14 ≈ 78.6%; 2019 vs 2018: (30-25)/25 = 20%; 2020 vs 2019: (34-30)/30 ≈ 13.3%
B: 2019 vs 2018: (7-6)/6 ≈ 16.7%; 2020 vs 2019: (7-7)/7 = 0%
C: 2018 vs 2017: (14-10)/10 = 40%; 2019 vs 2018: (15-14)/14 ≈ 7.1%; 2020 vs 2019: (12-15)/15 = -20%
D: 2018 vs 2017: (23-22)/22 ≈ 4.5%; 2019 vs 2018: (21-23)/23 ≈ -8.7%; 2020 vs 2019: (25-21)/21 ≈ 19.0%
The maximum among all these single-year growth rates is indeed 200% for Company B.

114. If FATHER is coded as DCRJCT, then what is the code for UNCLE?

If FATHER is coded as DCRJCT, then what is the code for UNCLE?

[amp_mcq option1=”SPANC” option2=”SPANA” option3=”WLEJG” option4=”WLEGJ” correct=”option1″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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The correct answer is A) SPANC.

Analyzing the coding of FATHER (F-A-T-H-E-R) to DCRJCT (D-C-R-J-C-T), we observe a consistent pattern of letter shifts based on their alphabetical position:
F (6) -> D (4): Shift -2
A (1) -> C (3): Shift +2
T (20) -> R (18): Shift -2
H (8) -> J (10): Shift +2
E (5) -> C (3): Shift -2
R (18) -> T (20): Shift +2
The pattern of shifts is alternating -2, +2, -2, +2, -2, +2.

Applying this pattern to UNCLE:
U (21) -> U – 2 = S (19)
N (14) -> N + 2 = P (16)
C (3) -> C – 2 = A (1)
L (12) -> L + 2 = N (14)
E (5) -> E – 2 = C (3)
Thus, UNCLE is coded as SPANC.

115. Consider the following table : 3 8 5 2 11 9 1 -3 ?

Consider the following table :

3	8	5
2	11	9
1	-3	?

Which one of the following is the missing number in the above table?

[amp_mcq option1=”14″ option2=”4″ option3=”-2″ option4=”-4″ correct=”option4″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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The correct answer is D) -4.

Observing the relationship between the numbers in each row, it appears that the third number in each row is the difference between the second number and the first number. Let’s verify:
Row 1: 8 – 3 = 5 (Correct)
Row 2: 11 – 2 = 9 (Correct)
Following this pattern for Row 3:
-3 – 1 = -4.

Another way to express the pattern is that the second number is the sum of the first and third numbers:
Row 1: 3 + 5 = 8 (Correct)
Row 2: 2 + 9 = 11 (Correct)
For Row 3: 1 + ? = -3. Solving for ? gives ? = -3 – 1 = -4. Both patterns lead to the same answer.

116. If A, B and C are the angles of a triangle such that A: B: C=2: 3: 4,

If A, B and C are the angles of a triangle such that A: B: C=2: 3: 4, then what is the value of the following?
$$ \frac{\tan^2 B+1}{\tan^2 B-1} $$

[amp_mcq option1=”4″ option2=”2″ option3=”1″ option4=”0″ correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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Let the angles of the triangle be A, B, and C. The ratio is given as A: B: C = 2: 3: 4.
The sum of angles in a triangle is 180°.
Let the common ratio factor be $x$. Then $A = 2x$, $B = 3x$, and $C = 4x$.
$2x + 3x + 4x = 180°$
$9x = 180°$
$x = \frac{180°}{9} = 20°$.

The angles are:
A = $2x = 2 \times 20° = 40°$
B = $3x = 3 \times 20° = 60°$
C = $4x = 4 \times 20° = 80°$

We need to find the value of the expression $\frac{\tan^2 B+1}{\tan^2 B-1}$.
Substitute $B = 60°$:
$\tan B = \tan 60° = \sqrt{3}$.
$\tan^2 B = (\sqrt{3})^2 = 3$.

Now substitute this value into the expression:
$\frac{\tan^2 B+1}{\tan^2 B-1} = \frac{3+1}{3-1} = \frac{4}{2} = 2$.

The sum of angles in a triangle is 180°. Given the ratio of angles, the actual angle values can be determined. Evaluate trigonometric expressions by substituting the determined angle values and using standard trigonometric values for common angles like 60°.

The expression can also be simplified using trigonometric identities:
$\frac{\tan^2 B+1}{\tan^2 B-1} = \frac{\sec^2 B}{\frac{\sin^2 B}{\cos^2 B}-1} = \frac{\frac{1}{\cos^2 B}}{\frac{\sin^2 B – \cos^2 B}{\cos^2 B}} = \frac{1}{\sin^2 B – \cos^2 B}$.
Using the identity $\cos(2B) = \cos^2 B – \sin^2 B = -(\sin^2 B – \cos^2 B)$, the expression is $\frac{1}{-\cos(2B)}$.
For $B=60°$, $2B=120°$. $\cos(120°) = -\frac{1}{2}$.
The value is $\frac{1}{-(-\frac{1}{2})} = \frac{1}{\frac{1}{2}} = 2$.

117. What is the angle in degree between the hour hand and the minute hand

What is the angle in degree between the hour hand and the minute hand of a clock when the time it shows is 5:20 PM?

[amp_mcq option1=”35°” option2=”40°” option3=”42°” option4=”45°” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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At 5:20 PM, the time is 5 hours and 20 minutes.
First, calculate the angle of the hour hand from the 12 o’clock position.
The hour hand moves 360° in 12 hours, or 30° per hour (360°/12).
It also moves due to the minutes past the hour. In 60 minutes, the hour hand moves 30°. So, in 1 minute, it moves $30°/60 = 0.5°$.
At 5:20, the time is 5 hours and 20 minutes past 12. The total time in minutes past 12 is $(5 \times 60) + 20 = 300 + 20 = 320$ minutes.
Angle covered by hour hand = $320 \times 0.5° = 160°$.
Alternatively, angle = (5 hours $\times$ 30°/hour) + (20 minutes $\times$ 0.5°/minute) = 150° + 10° = 160°.

Second, calculate the angle of the minute hand from the 12 o’clock position.
The minute hand moves 360° in 60 minutes, or 6° per minute (360°/60).
At 20 minutes past the hour, the minute hand is at the 20 minute mark.
Angle covered by minute hand = $20 \times 6° = 120°$.

The angle between the hour hand and the minute hand is the absolute difference between their positions.
Angle = $|160° – 120°| = 40°$.

The hour hand moves 0.5° per minute. The minute hand moves 6° per minute. The angle between the hands at H hours and M minutes past 12 is $|30H – 5.5M|$ degrees (where $5.5M = (6M – 0.5M)$, the difference in speeds). Using the position method: Hour hand angle = $30H + 0.5M$. Minute hand angle = $6M$. Angle = $|(30H + 0.5M) – 6M| = |30H – 5.5M|$. At 5:20, H=5, M=20. Angle = $|30 \times 5 – 5.5 \times 20| = |150 – 110| = 40°$.

Clock problems require understanding the relative speeds of the hour and minute hands. The 12-hour mark is usually taken as the reference point (0 degrees). Angles are measured clockwise from 12.

118. A wheel of circumference 2 m rolls on a circular path of radius 80 m.

A wheel of circumference 2 m rolls on a circular path of radius 80 m. What is the angle made by the wheel about the centre of the path, if it rotates 64 times on its own axis?

[amp_mcq option1=”1.6 rad” option2=”1.4 rad” option3=”1.2 rad” option4=”0.8 rad” correct=”option1″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference.
The wheel rotates 64 times on its own axis.
Total distance covered by the wheel rolling on the circular path = Number of rotations $\times$ Circumference of the wheel
Total distance = $64 \times 2$ m = 128 m.

This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($\theta$) subtended at the center (in radians) is $s = R \times \theta$.
Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m.
So, $128 = 80 \times \theta$.
Solve for $\theta$:
$\theta = \frac{128}{80}$ radians.

Simplify the fraction:
$\theta = \frac{128 \div 16}{80 \div 16} = \frac{8}{5}$ radians.
$\frac{8}{5} = 1.6$ radians.

When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = R\theta$, where $R$ is the radius and $\theta$ is the angle subtended at the center *in radians*.

The radius of the wheel itself ($\frac{1}{\pi}$ m) is not directly needed to solve this problem, as the distance covered is given by the circumference and the number of rotations. The problem asks for the angle subtended at the *centre of the path*, not the centre of the wheel.

119. A sum of money invested at compound interest amounts to ₹ 400 in 2 yea

A sum of money invested at compound interest amounts to ₹ 400 in 2 years and to ₹ 420 in 3 years. The rate of interest per annum is

[amp_mcq option1=”2.5%” option2=”5.5%” option3=”4%” option4=”5%” correct=”option4″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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Let the principal amount be P and the annual rate of interest be r (as a decimal).
The formula for the amount A after t years at compound interest is $A = P(1+r)^t$.
Given:
Amount after 2 years ($A_2$) = ₹ 400
Amount after 3 years ($A_3$) = ₹ 420

So, we have two equations:
1) $400 = P(1+r)^2$
2) $420 = P(1+r)^3$

To find the rate r, divide equation (2) by equation (1):
$\frac{420}{400} = \frac{P(1+r)^3}{P(1+r)^2}$
$\frac{42}{40} = (1+r)^{3-2}$
$\frac{21}{20} = 1+r$

Now, solve for r:
$r = \frac{21}{20} – 1$
$r = \frac{21 – 20}{20}$
$r = \frac{1}{20}$

To express the rate as a percentage, multiply by 100:
Rate = $\frac{1}{20} \times 100\% = 5\%$.

In compound interest, the amount for any year becomes the principal for the next year. The ratio of the amount at the end of year (t+1) to the amount at the end of year (t) is equal to $(1+r)$, where r is the annual interest rate.
$\frac{A_{t+1}}{A_t} = 1+r$

In this problem, the amount after 2 years (₹400) acts as the principal for the 3rd year. The interest earned during the 3rd year is $420 – 400 = ₹20$.
The interest rate can be calculated as the interest earned in the 3rd year divided by the principal at the beginning of the 3rd year (which is the amount after 2 years).
Interest rate = $\frac{\text{Interest earned in 3rd year}}{\text{Amount after 2 years}} = \frac{20}{400} = \frac{1}{20}$.
As a percentage, this is $\frac{1}{20} \times 100\% = 5\%$. This method is valid only because the time difference is exactly one year.

120. Kumar completes a work in 8 days and Raj completes the same work in 16

Kumar completes a work in 8 days and Raj completes the same work in 16 days. In how many days can Kumar and Raj together complete the work?

[amp_mcq option1=”$3\frac{1}{3}$ days” option2=”$5\frac{1}{3}$ days” option3=”$1\frac{1}{3}$ days” option4=”$\frac{1}{3}$ day” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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Kumar completes the work in 8 days, so his work rate is $\frac{1}{8}$ of the work per day.
Raj completes the work in 16 days, so his work rate is $\frac{1}{16}$ of the work per day.
When working together, their work rates add up.
Combined work rate = Work rate of Kumar + Work rate of Raj
Combined work rate = $\frac{1}{8} + \frac{1}{16}$ per day.
To add the fractions, find a common denominator, which is 16.
Combined work rate = $\frac{2}{16} + \frac{1}{16} = \frac{3}{16}$ of the work per day.
The time taken to complete the work together is the reciprocal of the combined work rate.
Time taken together = $\frac{1}{\text{Combined work rate}} = \frac{1}{\frac{3}{16}} = \frac{16}{3}$ days.
Converting the improper fraction to a mixed number: $\frac{16}{3} = 5$ with a remainder of 1, so $5\frac{1}{3}$ days.

If a person completes a work in $n$ days, their work rate is $\frac{1}{n}$ of the work per day. When multiple people work together, their individual work rates add up to find the combined work rate. The total time taken is the reciprocal of the combined work rate.

This type of problem is a classic “work and time” problem. The formula for two people is $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$, where $T_1$ and $T_2$ are the times taken by individuals and $T$ is the time taken together.
$\frac{1}{T} = \frac{1}{8} + \frac{1}{16} = \frac{2+1}{16} = \frac{3}{16}$.
$T = \frac{16}{3} = 5\frac{1}{3}$ days.