Quantitative Aptitude and Reasoning

81. A painter wants to paint a picture (rectangular portrait) occupying 72

A painter wants to paint a picture (rectangular portrait) occupying 72 square inches on a canvas allowing a margin of 4 inches on the top and at the bottom and 2 inches on each side. What will be the smallest dimension of the canvas?

[amp_mcq option1=”12″ x 17″” option2=”7″ x 31″” option3=”13″ x 16″” option4=”10″ x 20″” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2019
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The correct answer is (D) 10″ x 20″. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w \times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side.
– The dimensions of the canvas will be:
– Canvas Width = Picture Width + 2 * Side Margin = $w + 2 \times 2 = w + 4$
– Canvas Height = Picture Height + 2 * Top/Bottom Margin = $h + 2 \times 4 = h + 8$
– The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) \times (h+8)$ such that $w \times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options.
– The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$.
– Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$.
– To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$.
– Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 \implies 2w^2 = 72 \implies w^2 = 36$.
– Since dimensions must be positive, $w = 6$. Then $h = 72/6 = 12$.
– Check if $2w=h$: $2 \times 6 = 12$, which is true.
– The picture dimensions that minimize the canvas area are 6 inches by 12 inches.
– The corresponding canvas dimensions are:
– Canvas Width = $w + 4 = 6 + 4 = 10$ inches
– Canvas Height = $h + 8 = 12 + 8 = 20$ inches
– The calculated minimal canvas dimensions are 10″ x 20″, which is option (D).
– Let’s check if the other options correspond to valid picture dimensions and their canvas areas:
– Option A: Canvas 12×17. Picture width = 12-4=8. Picture height = 17-8=9. 8×9=72. Valid. Canvas area = 12×17 = 204.
– Option B: Canvas 7×31. Picture width = 7-4=3. Picture height = 31-8=23. 3×23=69. Not 72. Invalid option.
– Option C: Canvas 13×16. Picture width = 13-4=9. Picture height = 16-8=8. 9×8=72. Valid. Canvas area = 13×16 = 208.
– Option D: Canvas 10×20. Picture width = 10-4=6. Picture height = 20-8=12. 6×12=72. Valid. Canvas area = 10×20 = 200.
– Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10″ x 20″ canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10″x20″ option).
The problem implicitly asks for the canvas dimensions that result in the minimum possible canvas area while accommodating the picture with the specified margins. The method of minimizing the area $104 + 8w + 4h$ subject to $wh = 72$ using calculus (finding the derivative with respect to $w$ after substituting $h=72/w$) or AM-GM inequality ($8w + 4h \ge 2\sqrt{32wh} = 2\sqrt{32 \times 72} = 2\sqrt{2304} = 2 \times 48 = 96$, minimum when $8w=4h$) both confirm that the minimum area occurs when $w=6$ and $h=12$.

82. How many pairs of letters are there in the word ‘CREATIVE’ which have

How many pairs of letters are there in the word ‘CREATIVE’ which have as many letters between them in the word as in the alphabet?

[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option3″]

This question was previously asked in
UPSC CISF-AC-EXE – 2019
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The correct answer is (C) 3. We need to find pairs of letters in the word ‘CREATIVE’ that have the same number of letters between them in the word as they do in the English alphabet.
– List the letters in the word ‘CREATIVE’ with their positions in the word (1-indexed) and their positions in the alphabet (A=1, B=2, …):
C(1, 3), R(2, 18), E(3, 5), A(4, 1), T(5, 20), I(6, 9), V(7, 22), E(8, 5)
– A pair of letters (L1, L2) at word positions P1 and P2 forms a matching pair if the number of letters between them in the word, which is |P1 – P2| – 1, is equal to the number of letters between them in the alphabet, which is |AlphaPos(L1) – AlphaPos(L2)| – 1. This simplifies to checking if |P1 – P2| = |AlphaPos(L1) – AlphaPos(L2)|.
– Let’s check all pairs:
– C(1,3) with E(3,5): |1-3|=2, |3-5|=2. Match! (C-E)
– C(1,3) with E(8,5): |1-8|=7, |3-5|=2. No.
– E(3,5) with A(4,1): |3-4|=1, |5-1|=4. No.
– E(3,5) with I(6,9): |3-6|=3, |5-9|=4. No.
– E(3,5) with V(7,22): |3-7|=4, |5-22|=17. No.
– E(3,5) with E(8,5): |3-8|=5, |5-5|=0. No.
– A(4,1) with E(8,5): |4-8|=4, |1-5|=4. Match! (A-E)
– T(5,20) with V(7,22): |5-7|=2, |20-22|=2. Match! (T-V)
– Other pairs do not satisfy the condition. (Checked systematically in thought process).
– The three pairs are C-E (occurring forward from C to the first E), A-E (occurring forward from A to the second E), and T-V (occurring forward from T to V). The order in the alphabet doesn’t matter (|difference| is used).
– Note that C-E (positions 1 and 3 in word) have 1 letter (R) between them. In the alphabet, C and E have 1 letter (D) between them.
– A-E (positions 4 and 8 in word) have 3 letters (T, I, V) between them. In the alphabet, A and E have 3 letters (B, C, D) between them.
– T-V (positions 5 and 7 in word) have 1 letter (I) between them. In the alphabet, T and V have 1 letter (U) between them.
The question asks for the number of *pairs*, regardless of direction (forward or backward in the word). The method |P1 – P2| = |AlphaPos(L1) – AlphaPos(L2)| correctly accounts for this. We found 3 such pairs: C-E, A-E, and T-V.

83. In an artificial language, some words are translated as given below (n

In an artificial language, some words are translated as given below (not in order):
bie cie fie (good persons sing)
cie fie gie (sing good lyrics)
hie gie fie (love good lyrics)
What is the translation for ‘persons love lyrics’?

[amp_mcq option1=”cie hie gie” option2=”hie gie fie” option3=”bie hie gie” option4=”fie bie cie” correct=”option3″]

This question was previously asked in
UPSC CISF-AC-EXE – 2019
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The correct answer is (C) bie hie gie. By analyzing the provided translations and finding common words, we can decode the meaning of each artificial word.
– Given:
1. bie cie fie (good persons sing)
2. cie fie gie (sing good lyrics)
3. hie gie fie (love good lyrics)
– Comparing 1 and 2: Common words are “cie” and “fie” corresponding to “sing” and “good”.
– Comparing 2 and 3: Common words are “gie” and “fie” corresponding to “lyrics” and “good”.
– From the comparisons, “fie” is common to both sets and corresponds to the common English word “good”. So, fie = good.
– Using fie = good:
– From 2: cie fie gie becomes cie good gie (sing good lyrics). This means {cie, gie} corresponds to {sing, lyrics}.
– From 3: hie gie fie becomes hie gie good (love good lyrics). This means {hie, gie} corresponds to {love, lyrics}.
– Combining {cie, gie} = {sing, lyrics} and {hie, gie} = {love, lyrics}, the common word “gie” corresponds to “lyrics”. So, gie = lyrics.
– Using fie = good and gie = lyrics:
– From 2: cie good lyrics (sing good lyrics). This means cie = sing.
– From 3: hie lyrics good (love good lyrics). This means hie = love.
– Using fie = good, cie = sing:
– From 1: bie sing good (good persons sing). This means bie = persons.
– Final mappings: bie = persons, cie = sing, fie = good, gie = lyrics, hie = love.
– The translation for ‘persons love lyrics’ is ‘bie hie gie’.
This method of solving, known as cryptarithmetic or symbol decoding, involves identifying patterns and common elements across different statements to deduce the meaning of individual components. The order of words in the target phrase ‘persons love lyrics’ corresponds to the order of the translated words in the option ‘bie hie gie’.

84. Suppose P and Q are distinct two-digit numbers consisting of the same

Suppose P and Q are distinct two-digit numbers consisting of the same digits. Then P-Q is

[amp_mcq option1=”a prime number” option2=”an even number” option3=”an odd number” option4=”divisible by 9″ correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2019
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The correct answer is (D) divisible by 9. Let the two distinct two-digit numbers be P and Q, formed by the digits ‘a’ and ‘b’. Since they are distinct two-digit numbers consisting of the same digits, the digits must be distinct (a ≠ b) and non-zero (otherwise one number would be a single digit). Let P = 10a + b and Q = 10b + a (where a and b are single digits from 1-9, a ≠ b).
– The difference P – Q = (10a + b) – (10b + a) = 9a – 9b = 9(a – b).
– Since ‘a’ and ‘b’ are distinct digits, (a – b) is a non-zero integer.
– Therefore, the difference P – Q is always a multiple of 9.
– Any multiple of 9 is divisible by 9.
Let’s test the other options with an example: P=72, Q=27. P-Q = 72-27 = 45.
– Is 45 a prime number? No.
– Is 45 an even number? No.
– Is 45 an odd number? Yes. But consider P=31, Q=13. P-Q = 31-13 = 18, which is even. So it’s not always odd.
– Is 45 divisible by 9? Yes (45 / 9 = 5).
The property P-Q = 9(a-b) guarantees divisibility by 9 for any pair of distinct two-digit numbers formed by swapping two distinct digits (assuming both digits are non-zero to ensure two distinct two-digit numbers, or carefully considering the case with digit 0, which would mean numbers like 20 and 02, but 02 is not a two-digit number, confirming our assumption that both digits must be non-zero).

85. 25 writers, 20 doctors, 18 dentists and 12 bank employees spent altoge

25 writers, 20 doctors, 18 dentists and 12 bank employees spent altogether ₹ 1,330 in a hotel. However, it was found that 5 writers spent as much as 4 doctors, that 12 doctors spent as much as 9 dentists, and that 6 dentists spent as much as 8 bank employees. Out of the four professional groups, which group spent the maximum amount ?

[amp_mcq option1=”Bank employees” option2=”Dentists” option3=”Doctors” option4=”Writers” correct=”option2″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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The correct answer is B) Dentists.
Let $w_s, d_s, t_s, b_s$ be the average amount spent by one writer, doctor, dentist, and bank employee, respectively. The number of people in each group is 25 writers, 20 doctors, 18 dentists, and 12 bank employees.
We are given the following relationships:
1. 5 writers spend as much as 4 doctors: $5w_s = 4d_s \implies w_s = \frac{4}{5}d_s$
2. 12 doctors spend as much as 9 dentists: $12d_s = 9t_s \implies 4d_s = 3t_s \implies t_s = \frac{4}{3}d_s$
3. 6 dentists spend as much as 8 bank employees: $6t_s = 8b_s \implies 3t_s = 4b_s$
Substitute $t_s$ from (2) into (3): $3(\frac{4}{3}d_s) = 4b_s \implies 4d_s = 4b_s \implies d_s = b_s$
So, we have $w_s = \frac{4}{5}d_s$, $t_s = \frac{4}{3}d_s$, and $b_s = d_s$.
The total amount spent is ₹ 1330: $25w_s + 20d_s + 18t_s + 12b_s = 1330$
Substitute the values in terms of $d_s$: $25(\frac{4}{5}d_s) + 20d_s + 18(\frac{4}{3}d_s) + 12d_s = 1330$
$20d_s + 20d_s + 24d_s + 12d_s = 1330$
$76d_s = 1330 \implies d_s = \frac{1330}{76} = 17.5$
Now calculate the amount spent per person for each group:
$d_s = 17.5$
$w_s = \frac{4}{5} \times 17.5 = 14$
$t_s = \frac{4}{3} \times 17.5 = \frac{70}{3} \approx 23.33$
$b_s = 17.5$
Total amount spent by each group:
Writers: $25 \times 14 = 350$
Doctors: $20 \times 17.5 = 350$
Dentists: $18 \times \frac{70}{3} = 6 \times 70 = 420$
Bank Employees: $12 \times 17.5 = 210$
Comparing the total amounts, the Dentists group spent the maximum amount (₹ 420).
This problem requires setting up equations based on the given relationships and solving for the per-person cost of one group, then using that to find the costs for others and finally the total expenditure per group.

86. Find the missing letter. – The question contains a table which is not

Find the missing letter.
– The question contains a table which is not rendered here.

[amp_mcq option1=”A” option2=”D” option3=”O” option4=”N” correct=”option1″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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The question refers to a missing table which is essential to identify the pattern and find the missing letter. Without the table or the context of the pattern, it is impossible to determine the correct answer from the given options. Therefore, a definitive correct option cannot be provided based solely on the text of the question.
Questions involving missing letters or numbers typically require identifying a sequence, pattern, or relationship within a given set of data, often presented in a table, matrix, series, or diagram. The pattern could be based on alphabetical position, arithmetic operations, visual arrangement, or a combination thereof.
To solve such problems, one would need to carefully observe the provided data (the table, in this case) and test potential patterns (e.g., row-wise, column-wise, diagonal, sum, difference, product, letter values, etc.) to find the rule governing the arrangement. Once the rule is found, it is applied to determine the missing element. As the table is unavailable, this step cannot be performed.

87. Find the odd one out of the following items : EF22, JK42, GH24, VW90,

Find the odd one out of the following items :
EF22, JK42, GH24, VW90, IJ38

[amp_mcq option1=”EF22″ option2=”GH24″ option3=”IJ38″ option4=”VW90″ correct=”option2″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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The correct answer is B) GH24.
The pattern for most items is that the sum of the alphabetical positions of the two letters, when multiplied by 2, equals the number.
– EF: E is the 5th letter, F is the 6th. 5 + 6 = 11. 11 * 2 = 22. (Matches EF22)
– JK: J is the 10th letter, K is the 11th. 10 + 11 = 21. 21 * 2 = 42. (Matches JK42)
– GH: G is the 7th letter, H is the 8th. 7 + 8 = 15. 15 * 2 = 30. The number given is 24. (Does not match GH24)
– VW: V is the 22nd letter, W is the 23rd. 22 + 23 = 45. 45 * 2 = 90. (Matches VW90)
– IJ: I is the 9th letter, J is the 10th. 9 + 10 = 19. 19 * 2 = 38. (Matches IJ38)
All items except GH24 follow this pattern.
This is a common type of reasoning question that tests pattern recognition based on alphabetical order and arithmetic operations. Always check for simple relationships between letter positions and the given numbers, such as sums, differences, products, or sequences.

88. Delhi is bigger than Pune, Kanpur is bigger than Raipur. Jhansi is not

Delhi is bigger than Pune, Kanpur is bigger than Raipur. Jhansi is not as big as Pune, but is bigger than Kanpur. Which one of the following is the smallest city ?

[amp_mcq option1=”Delhi” option2=”Pune” option3=”Kanpur” option4=”Raipur” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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The smallest city is Raipur.
We are given the following comparisons regarding size:
1. Delhi is bigger than Pune: Delhi > Pune
2. Kanpur is bigger than Raipur: Kanpur > Raipur
3. Jhansi is not as big as Pune (i.e., smaller than Pune): Jhansi < Pune 4. Jhansi is bigger than Kanpur: Jhansi > Kanpur

Let’s combine these inequalities:
From (3) and (4): Pune > Jhansi and Jhansi > Kanpur. Combining these gives Pune > Jhansi > Kanpur.
From (2) and the result above (Jhansi > Kanpur): Jhansi > Kanpur > Raipur.
From (1) and the result Pune > Jhansi: Delhi > Pune > Jhansi.

Combining all the relationships, we get a clear order of size:
Delhi > Pune > Jhansi > Kanpur > Raipur.

The city at the smaller end of this chain is Raipur.

The comparisons allow us to establish a strict linear ordering of the four cities by size: Delhi is the largest, followed by Pune, then Jhansi, then Kanpur, and finally Raipur is the smallest.
Delhi > Pune
Pune > Jhansi (from Jhansi < Pune) Jhansi > Kanpur
Kanpur > Raipur
Combining these gives Delhi > Pune > Jhansi > Kanpur > Raipur.

89. How many times in a day are the hour hand and the minute hand of a wal

How many times in a day are the hour hand and the minute hand of a wall clock straight (i.e., the angle between them is 180°)?

[amp_mcq option1=”20″ option2=”21″ option3=”22″ option4=”24″ correct=”option3″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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The hour hand and the minute hand of a wall clock are straight (180° apart) 22 times in a day (24 hours).
In a 12-hour period, the minute hand completes 12 revolutions while the hour hand completes 1 revolution. The minute hand gains 11 full revolutions over the hour hand. During this process, the hands coincide (0° apart) 11 times and are opposite (180° apart) 11 times.
The hands are 180° apart once in every hour interval, except for the interval between 6 o’clock and 7 o’clock, where they are 180° apart exactly at 6 o’clock. The 6 o’clock position is counted in both the 5-6 interval and the 6-7 interval if we consider specific time points, but considering distinct occurrences in a 12-hour cycle, it happens 11 times.
For example, between 12 pm and 12 am, the hands are opposite at approximately 12:33, 1:38, 2:44, 3:49, 4:55, 6:00, 7:05, 8:11, 9:16, 10:22, 11:27. (These are approximate times, the exact times are fractions). This is 11 distinct times in a 12-hour period.
A full day is 24 hours, which consists of two 12-hour periods.
Therefore, in 24 hours, the hands will be straight (180° apart) 11 times + 11 times = 22 times.
The general formula for the time t (in minutes past H o’clock) when the hands are at an angle $\theta$ is $t = \frac{2}{11} (30H \pm \theta)$.
For the hands to be straight, $\theta = 180^\circ$.
$t = \frac{2}{11} (30H \pm 180)$.
Let’s check for H from 1 to 12.
For H=6, $t = \frac{2}{11} (180 \pm 180)$. $t = \frac{2}{11} (360) \approx 65.45$ min (past 6) or $t = \frac{2}{11} (0) = 0$ min (past 6). This confirms 6:00 is one time.
For H=5, $t = \frac{2}{11} (150 \pm 180)$. $t = \frac{2}{11} (330) = 60$ min (past 5, which is 6:00) or $t = \frac{2}{11} (-30)$ (not valid in this hour).
The hands are exactly opposite at 6:00. This instance is the boundary point that is counted only once in a 12-hour period when counting the intervals between hours.
The number of times the hands are 180° apart is 11 in 12 hours, and 22 in 24 hours.

90. In a bag, there are notes of ₹ 10, ₹ 20 and ₹ 50 in the ratio of 1 : 2

In a bag, there are notes of ₹ 10, ₹ 20 and ₹ 50 in the ratio of 1 : 2 : 3. If the total money is ₹ 1,000, how many notes of ₹ 10 are there ?

[amp_mcq option1=”5″ option2=”10″ option3=”15″ option4=”30″ correct=”option1″]

This question was previously asked in
UPSC CISF-AC-EXE – 2018
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There are 5 notes of ₹10 in the bag.
The ratio of the number of notes of ₹10, ₹20, and ₹50 is 1 : 2 : 3.
Let the number of ₹10 notes be x.
Then, the number of ₹20 notes is 2x.
And the number of ₹50 notes is 3x.

The total value of the money in the bag is the sum of the values of each type of note:
Value from ₹10 notes = x * ₹10 = 10x
Value from ₹20 notes = 2x * ₹20 = 40x
Value from ₹50 notes = 3x * ₹50 = 150x

Total money = 10x + 40x + 150x = 200x.
We are given that the total money is ₹1,000.
So, 200x = 1000.
Solving for x:
x = 1000 / 200
x = 5.

The number of ₹10 notes is x, which is 5.

Number of ₹10 notes = 5 (Value = 5 * ₹10 = ₹50)
Number of ₹20 notes = 2 * 5 = 10 (Value = 10 * ₹20 = ₹200)
Number of ₹50 notes = 3 * 5 = 15 (Value = 15 * ₹50 = ₹750)
Total value = ₹50 + ₹200 + ₹750 = ₹1000. This confirms the calculation.
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