141. The following sequence is generated by dividing 50 by natural numbers

The following sequence is generated by dividing 50 by natural numbers :
0, 0, 2, 0, 2, x, 2, 5, …
What is x?

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

This question was previously asked in
UPSC CISF-AC-EXE – 2019
The sequence is given as 0, 0, 2, 0, 2, x, 2, 5, …
The sequence is generated by “dividing 50 by natural numbers”, implying the inputs are natural numbers n=1, 2, 3, 4, 5, 6, 7, 8, …
Let f(n) be the term in the sequence corresponding to the natural number n.
f(1) = 0
f(2) = 0
f(3) = 2
f(4) = 0
f(5) = 2
f(6) = x
f(7) = 2
f(8) = 5

Let’s analyse the relationship between n and f(n).
Notice the outputs 0, 0, 2, 0, 2, ?, 2, 5.
The inputs are 1, 2, 3, 4, 5, 6, 7, 8.
f(1)=0, f(2)=0, f(4)=0. These inputs (1, 2, 4) are powers of 2 ($1=2^0$, $2=2^1$, $4=2^2$).
f(3)=2, f(5)=2, f(7)=2. These inputs (3, 5, 7) are primes greater than 2.
f(8)=5. This input (8) is $2^3$.
f(6)=x. This input (6) is not a power of 2 and not a prime.

Let’s hypothesize a rule based on the observed pattern:
– If n is a power of 2, $n=2^k$: the output depends on k? f(1)=f($2^0$)=0, f(2)=f($2^1$)=0, f(4)=f($2^2$)=0, f(8)=f($2^3$)=5.
– If n is a prime greater than 2: f(n)=2. (Matches f(3), f(5), f(7)).
– If n is composite and not a power of 2: f(n)=? (Only n=6 in the given range).

Let’s refine the rule based on inputs being powers of 2 vs not powers of 2.
– If n is a power of 2: $n=2^k$. f(1)=0, f(2)=0, f(4)=0, f(8)=5. This still doesn’t follow a simple rule like k, k+c, k^2, etc.
– If n is NOT a power of 2: f(3)=2, f(5)=2, f(7)=2. This strongly suggests f(n)=2 for n which are not powers of 2, with a potential exception for n=8.

Let’s assume the rule is:
f(n) = 2 if n is not a power of 2.
f(n) = 0 if n is a power of 2, except for n=8.
f(8) = 5.

Let’s test this rule:
f(1) = f($2^0$). Power of 2, not 8. Rule says 0. Matches.
f(2) = f($2^1$). Power of 2, not 8. Rule says 0. Matches.
f(3). Not a power of 2. Rule says 2. Matches.
f(4) = f($2^2$). Power of 2, not 8. Rule says 0. Matches.
f(5). Not a power of 2. Rule says 2. Matches.
f(6). Not a power of 2 ($6 = 2 \times 3$). Rule says 2. So x=2.
f(7). Not a power of 2. Rule says 2. Matches.
f(8) = f($2^3$). This is the special case $n=8$. Rule says 5. Matches.

This rule consistently explains the sequence values 0, 0, 2, 0, 2, ?, 2, 5, and predicts x=2.
The number 6 is not a power of 2 ($1, 2, 4, 8, 16, …$). According to the rule, f(6) should be 2.
Therefore, x = 2.

– Analyse the given sequence and the corresponding natural numbers (1, 2, 3, …).
– Look for patterns relating the input number ‘n’ to the output value in the sequence.
– Identify properties of ‘n’ that seem to determine the sequence value (e.g., being a prime, a power of 2, etc.).
This is a pattern recognition question typical in aptitude tests. The phrase “generated by dividing 50 by natural numbers” seems to be part of the context rather than implying a direct arithmetic operation like 50/n or 50 mod n. The rule identified here is based purely on the observed pattern in the sequence values corresponding to natural numbers 1 through 8. The number 50 might be irrelevant or its significance is hidden in a way not easily discernible from the small sample of the sequence.

142. How many prime numbers are there between 200 and 230?

How many prime numbers are there between 200 and 230?

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

This question was previously asked in
UPSC CISF-AC-EXE – 2019
To find the number of prime numbers between 200 and 230, we need to check each integer in this range (from 201 to 229) for primality. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. To check if a number ‘n’ is prime, we can test for divisibility by prime numbers up to the square root of ‘n’. The square root of 230 is approximately 15.17. So, we need to check for divisibility by primes up to 13 (2, 3, 5, 7, 11, 13).

Let’s check the numbers between 200 and 230:
– 201: Divisible by 3 (sum of digits 2+0+1=3). Not prime.
– 202: Divisible by 2. Not prime.
– 203: $203 = 7 \times 29$. Not prime.
– 204: Divisible by 2. Not prime.
– 205: Divisible by 5. Not prime.
– 206: Divisible by 2. Not prime.
– 207: Divisible by 3 (sum of digits 2+0+7=9). Not prime.
– 208: Divisible by 2. Not prime.
– 209: $209 = 11 \times 19$. Not prime.
– 210: Divisible by 10. Not prime.
– 211: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $14^2 = 196$, $15^2 = 225$. $\sqrt{211} \approx 14.5$. Primes to check up to 13. 211 is prime.
– 212: Divisible by 2. Not prime.
– 213: Divisible by 3 (sum of digits 2+1+3=6). Not prime.
– 214: Divisible by 2. Not prime.
– 215: Divisible by 5. Not prime.
– 216: Divisible by 2. Not prime.
– 217: $217 = 7 \times 31$. Not prime.
– 218: Divisible by 2. Not prime.
– 219: Divisible by 3 (sum of digits 2+1+9=12). Not prime.
– 220: Divisible by 10. Not prime.
– 221: $221 = 13 \times 17$. Not prime.
– 222: Divisible by 2. Not prime.
– 223: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{223} \approx 14.9$. Primes to check up to 13. 223 is prime.
– 224: Divisible by 2. Not prime.
– 225: Divisible by 5. Not prime.
– 226: Divisible by 2. Not prime.
– 227: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{227} \approx 15.07$. Primes to check up to 13. 227 is prime.
– 228: Divisible by 2. Not prime.
– 229: Check divisibility by primes 2, 3, 5, 7, 11, 13. Not divisible by any of these. $\sqrt{229} \approx 15.13$. Primes to check up to 13. 229 is prime.

The prime numbers between 200 and 230 are 211, 223, 227, and 229.
There are 4 prime numbers in this range.

– A prime number is a natural number greater than 1 with no positive divisors other than 1 and itself.
– To check for primality of a number ‘n’, one needs to test divisibility only by prime numbers up to $\sqrt{n}$.
Checking primality efficiently requires testing only prime divisors. For numbers up to 230, the largest prime divisor we need to check is 13, since the next prime is 17, and $17^2 = 289 > 230$.
The prime numbers less than or equal to 13 are 2, 3, 5, 7, 11, 13.

143. 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
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 $h=12$.

144. 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
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.

145. 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
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’.

146. 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
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).

147. Which one of the following statements is not correct?

Which one of the following statements is not correct?

[amp_mcq option1=”Hepatitis B virus is transmitted much like HIV.” option2=”Hepatitis B, unlike Hepatitis C, does not have a vaccine.” option3=”Globally, the number of people infected with Hepatitis B and C viruses are several times more than those infected with HIV.” option4=”Some of those infected with Hepatitis B and C viruses do not show the symptoms for many years.” correct=”option2″]

This question was previously asked in
UPSC IAS – 2019
The correct option is B because the statement “Hepatitis B, unlike Hepatitis C, does not have a vaccine” is incorrect.
Vaccination is a highly effective way to prevent Hepatitis B infection. While there is no widely available prophylactic vaccine for Hepatitis C, research is ongoing.
A) Hepatitis B and HIV share similar transmission routes, including sexual contact, exposure to infected blood (e.g., through sharing needles), and mother-to-child transmission. This statement is correct.
B) There is a safe and effective vaccine available to prevent Hepatitis B infection. In contrast, there is currently no effective preventive vaccine for Hepatitis C, although treatments for Hepatitis C have significantly advanced. Thus, the statement that Hepatitis B *does not* have a vaccine is false.
C) According to WHO estimates, significantly more people globally live with chronic Hepatitis B and C infections compared to HIV. This statement is correct.
D) Both Hepatitis B and C infections can be asymptomatic for many years, with chronic liver damage occurring silently before symptoms appear. This statement is correct.

148. What is Cas9 protein that is often mentioned in news?

What is Cas9 protein that is often mentioned in news?

[amp_mcq option1=”A molecular scissors used in targeted gene editing” option2=”A biosensor used in the accurate detection of pathogens in patients” option3=”A gene that makes plants pest-resistant” option4=”A herbicidal substance synthesized in genetically modified crops” correct=”option1″]

This question was previously asked in
UPSC IAS – 2019
The correct option is A. Cas9 protein is a key component used in targeted gene editing.
Cas9 (CRISPR-associated protein 9) is an enzyme that acts as a molecular scissors. It is guided by an RNA molecule to a specific location on a DNA strand and makes a precise cut. This makes it a fundamental tool in the CRISPR-Cas9 gene editing system.
The CRISPR-Cas9 system has revolutionized molecular biology and genetic engineering by providing a relatively simple, fast, and inexpensive way to modify genes. It allows scientists to precisely edit DNA sequences, enabling potential applications in treating genetic diseases, developing new crops, and fundamental biological research.
B) While gene editing tools *could* potentially be involved in developing biosensors, Cas9 itself is not primarily a biosensor.
C) Cas9 is a tool for editing genes, not a gene that confers pest resistance directly, although it could be used to insert or modify genes to make plants pest-resistant.
D) Cas9 is a protein used for DNA modification, not a herbicidal substance.

149. Recently, scientists observed the merger of giant ‘blackholes’ billion

Recently, scientists observed the merger of giant ‘blackholes’ billions of lightyears away from the Earth. What is the significance of this observation?

[amp_mcq option1=”Higgs boson particles’ were detected.” option2=”‘Gravitational waves’ were detected.” option3=”Possibility of inter-galactic space travel through ‘wormhole’ was confirmed.” option4=”It enabled the scientists to understand ‘singularity’.” correct=”option2″]

This question was previously asked in
UPSC IAS – 2019
The correct option is B. The observation of the merger of giant black holes billions of lightyears away from Earth led to the first direct detection of gravitational waves.
Gravitational waves are ripples in the fabric of spacetime caused by accelerating massive objects. Highly energetic events like the merger of black holes or neutron stars are strong sources of gravitational waves.
The LIGO and Virgo collaborations detected the first direct evidence of gravitational waves in 2015 (event GW150914), which originated from the merger of two black holes. This observation was a landmark discovery, confirming a major prediction of Einstein’s general theory of relativity and opening a new window for observing the universe through gravitational wave astronomy.
A) Higgs boson particles were detected at the Large Hadron Collider (LHC) at CERN, which is unrelated to black hole mergers.
C) While black holes and general relativity are related to the concept of wormholes, the detection of gravitational waves from a merger does not confirm the possibility of inter-galactic space travel through wormholes.
D) Observing black hole mergers and the resulting gravitational waves provides insights into the nature of gravity in extreme conditions, but it does not directly enable scientists to understand the singularity at the heart of a black hole, which remains a theoretical concept under current physics.

150. RNA interference (RNAi)’ technology has gained popularity in the last

RNA interference (RNAi)’ technology has gained popularity in the last few years. Why?

  • It is used in developing gene silencing therapies.
  • It can be used in developing therapies for the treatment of cancer.
  • It can be used to develop hormone replacement therapies.
  • It can be used to produce crop plants that are resistant to viral pathogens.

Select the correct answer using the code given below.

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

This question was previously asked in
UPSC IAS – 2019
The correct option is A, as statements 1, 2, and 4 accurately describe applications of RNA interference technology, while statement 3 does not.
RNA interference (RNAi) is a biological process in which RNA molecules inhibit gene expression or translation, by neutralizing targeted mRNA molecules. This mechanism is utilized to “silence” specific genes.
1. **Gene silencing therapies:** RNAi is a primary mechanism used in gene silencing. This property is being explored to develop therapies for various genetic disorders and diseases by turning off specific problematic genes.
2. **Cancer treatment:** RNAi can be used to silence genes that promote cancer cell growth, survival, or metastasis, making it a promising area of research for developing new cancer therapies.
3. **Hormone replacement therapies:** Hormone replacement therapy involves administering exogenous hormones to correct a deficiency or imbalance. RNAi deals with regulating gene expression and is not directly used in hormone replacement therapy.
4. **Crop resistance to viral pathogens:** RNAi can be engineered in plants to target and silence viral genes, thereby making the plants resistant to specific viral infections. This is a recognized application in agricultural biotechnology.