2018 Archives

221. 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.

222. 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.

223. 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.

224. The difference of squares of two consecutive odd numbers is always

The difference of squares of two consecutive odd numbers is always

[amp_mcq option1=”divisible by 8″ option2=”divisible by 3″ option3=”divisible by 16″ option4=”None of the above” correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The difference of squares of two consecutive odd numbers is always divisible by 8.

Let the two consecutive odd numbers be represented by (2n + 1) and (2n + 3), where n is an integer.
The difference of their squares is:
(2n + 3)^2 – (2n + 1)^2
Using the algebraic identity a^2 – b^2 = (a – b)(a + b):
= [(2n + 3) – (2n + 1)] * [(2n + 3) + (2n + 1)]
= [2n + 3 – 2n – 1] * [2n + 3 + 2n + 1]
= [2] * [4n + 4]
= 2 * 4(n + 1)
= 8(n + 1)
Since n is an integer, (n + 1) is also an integer. Therefore, the expression 8(n + 1) is always a multiple of 8.

Let’s verify with examples:
Consecutive odd numbers 1 and 3: 3^2 – 1^2 = 9 – 1 = 8. 8 is divisible by 8.
Consecutive odd numbers 3 and 5: 5^2 – 3^2 = 25 – 9 = 16. 16 is divisible by 8.
Consecutive odd numbers 5 and 7: 7^2 – 5^2 = 49 – 25 = 24. 24 is divisible by 8.
The difference is always in the form 8 * (an integer), so it is always divisible by 8.

225. If the first day of a year (other than the leap year) was Friday, then

If the first day of a year (other than the leap year) was Friday, then which one of the following was the last day of that year ?

[amp_mcq option1=”Monday” option2=”Friday” option3=”Saturday” option4=”Sunday” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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If the first day of a non-leap year was Friday, then the last day of that year was Friday.

A non-leap year has 365 days.
To find the day of the week for the last day, we need to find the number of days after the first day modulo 7.
Number of days in a non-leap year = 365.
365 days = 52 weeks + 1 day.
52 weeks contain exactly 52 * 7 = 364 days.
If the first day (Day 1) is Friday, then after 52 full weeks (on Day 365 – 1 = Day 364), the day of the week will be the same as the first day, i.e., Friday.
Day 364 is a Friday.
The last day of the year is Day 365. It will be the day after Friday, which is Saturday. This is incorrect.

Let’s rethink.
Day 1 is Friday.
After 7 days (Day 8), it’s Friday again.
After 364 days (52 weeks), the day is the same as Day 1. So, Day 364 is a Friday.
Day 365 is the day after Day 364. So, Day 365 is Saturday. This is also incorrect, as the well-known property is that the first and last days are the same.

Let’s think about the “extra” day.
The days of the year are Day 1, Day 2, …, Day 365.
Day 1 = Friday.
Day 2 = Saturday.
Day 7 = Thursday.
Day 8 = Friday (1 week after Day 1).
Day (1 + 7k) is Friday.
We want to find the day for Day 365.
365 = 1 + 364.
The number of “extra” days after Day 1 is 364.
364 modulo 7 is 0 (364 = 52 * 7).
This means Day 365 is 0 days past Friday in the weekly cycle, counting from Day 1.
So, if Day 1 is Friday, Day (1 + 364) i.e., Day 365, is Friday + 0 days = Friday.

For any year, the day of the week for the (N+1)th day is the day of the week for the 1st day plus N days (modulo 7). Here, the first day is Day 1, and the last day is Day 365. The number of days difference is 365 – 1 = 364. The day of the week for Day 365 is the day of the week for Day 1 + 364 days. Since 364 is a multiple of 7 (364 = 52 * 7), adding 364 days results in the same day of the week. Thus, the last day has the same day of the week as the first day in a non-leap year.
In a leap year (366 days), the last day would be one day later in the week than the first day.

226. A dog wants to catch a cat that is twenty-seven steps ahead of him. Th

A dog wants to catch a cat that is twenty-seven steps ahead of him. The cat takes eight steps to every five steps taken by the dog. Two steps of the dog are equal to five steps of the cat. How many steps of the dog will be required to catch the cat ?

[amp_mcq option1=”28″ option2=”30″ option3=”48″ option4=”75″ correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The dog will require 30 steps to catch the cat.

Let D_d be the distance covered by one dog step and D_c be the distance covered by one cat step.
We are given that 2 dog steps are equal to 5 cat steps in distance: 2 * D_d = 5 * D_c => D_d = (5/2) * D_c.
The cat starts 27 steps ahead of the dog. The initial distance between them is 27 * D_c.

In a certain time interval, the cat takes 8 steps and the dog takes 5 steps.
Distance covered by cat in this interval = 8 * D_c.
Distance covered by dog in this interval = 5 * D_d = 5 * (5/2) * D_c = 25/2 * D_c = 12.5 * D_c.

In this time interval, the dog reduces the distance between them by (Distance covered by dog) – (Distance covered by cat) = 12.5 * D_c – 8 * D_c = 4.5 * D_c.

The total distance the dog needs to close is the initial distance, which is 27 * D_c.
Number of intervals required to catch the cat = (Total distance to cover) / (Distance reduced per interval)
Number of intervals = (27 * D_c) / (4.5 * D_c) = 27 / 4.5 = 27 / (9/2) = 27 * (2/9) = 3 * 2 = 6 intervals.

In each interval, the dog takes 5 steps.
Total dog steps required = Number of intervals * Steps taken by dog per interval
Total dog steps = 6 * 5 = 30 steps.

Let’s check the result:
After 30 dog steps, the dog covers 30 * D_d = 30 * (5/2) * D_c = 75 * D_c.
The dog takes 5 steps per interval, so 30 dog steps means 30/5 = 6 intervals have passed.
In 6 intervals, the cat takes 6 * 8 = 48 steps.
The cat started 27 * D_c ahead. Its final position is 27 * D_c + 48 * D_c = 75 * D_c.
The dog’s final position is 75 * D_c. The dog catches the cat when their positions are the same.

227. While coming down on an escalator of a station, when a man walks down

While coming down on an escalator of a station, when a man walks down twenty-six steps of the staircase, he requires thirty seconds to reach the bottom. However, if he steps down thirty-four steps, it takes eighteen seconds to reach the bottom. How many steps are there in the escalator ?

[amp_mcq option1=”42″ option2=”44″ option3=”46″ option4=”48″ correct=”option3″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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There are 46 steps in the escalator.

Let E be the total number of steps on the escalator (when stationary).
Let V_m be the speed of the man relative to the escalator (steps/second).
Let V_e be the speed of the escalator (steps/second).
The total distance covered by the man relative to the ground is E steps. This distance is the sum of the steps the man walks relative to the escalator and the steps the escalator moves relative to the ground during the time taken.
E = (Man’s speed relative to ground) * Time
Man’s speed relative to ground = V_m + V_e.
E = (V_m + V_e) * T.

Alternatively, consider the total steps E. In time T, the man takes N steps relative to the escalator (so V_m = N/T). The escalator moves V_e * T steps.
E = N + V_e * T.

Case 1: Man takes N1 = 26 steps, T1 = 30 seconds.
E = 26 + V_e * 30 (Equation 1)

Case 2: Man takes N2 = 34 steps, T2 = 18 seconds.
E = 34 + V_e * 18 (Equation 2)

We have a system of two linear equations with two variables (E and V_e).
Equating the expressions for E:
26 + 30 * V_e = 34 + 18 * V_e
30 * V_e – 18 * V_e = 34 – 26
12 * V_e = 8
V_e = 8 / 12 = 2/3 steps/second.

Substitute the value of V_e into either equation. Using Equation 1:
E = 26 + (2/3) * 30
E = 26 + 2 * 10
E = 26 + 20
E = 46.

Using Equation 2:
E = 34 + (2/3) * 18
E = 34 + 2 * 6
E = 34 + 12
E = 46.
Both equations give the same result.

The man’s speed relative to the escalator is not constant in terms of steps per second, but the number of steps he *takes* is given. The interpretation used is that 26 steps are the steps counted on the treads by the man during his descent in 30 seconds, and 34 steps are counted in 18 seconds. The escalator adds to his progress relative to the ground.

228. Consider the following sequence : 5, 7, 10, 15, 22, 33, 46, ,

Consider the following sequence :
5, 7, 10, 15, 22, 33, 46, ____, ____
Which one of the alternatives will come at the end of the above sequence ?

[amp_mcq option1=”61, 79″ option2=”63, 82″ option3=”65, 85″ option4=”59, 71″ correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The next two numbers in the sequence are 63 and 82.

Let’s find the difference between consecutive terms:
7 – 5 = 2
10 – 7 = 3
15 – 10 = 5
22 – 15 = 7
33 – 22 = 11
46 – 33 = 13
The differences are 2, 3, 5, 7, 11, 13. This sequence of differences is the sequence of prime numbers in ascending order.
The next prime number after 13 is 17.
The term after 46 is 46 + 17 = 63.
The next prime number after 17 is 19.
The term after 63 is 63 + 19 = 82.

The sequence is generated by adding successive prime numbers starting from the first prime number (2) to the previous term.
Term 1: 5
Term 2: 5 + 2 = 7
Term 3: 7 + 3 = 10
Term 4: 10 + 5 = 15
Term 5: 15 + 7 = 22
Term 6: 22 + 11 = 33
Term 7: 33 + 13 = 46
Term 8: 46 + 17 = 63
Term 9: 63 + 19 = 82
The sequence continues: 5, 7, 10, 15, 22, 33, 46, 63, 82.

229. What is the greatest number less than 1000 which when divided respecti

What is the greatest number less than 1000 which when divided respectively by 5, 7 and 9 leaves the remainders 3, 5 and 7 respectively ?

[amp_mcq option1=”943″ option2=”963″ option3=”953″ option4=”989″ correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The greatest number less than 1000 which satisfies the conditions is 943.

Let the number be N. The conditions are:
N ≡ 3 (mod 5)
N ≡ 5 (mod 7)
N ≡ 7 (mod 9)
Notice that in each case, the remainder is 2 less than the divisor. This means N + 2 is divisible by 5, 7, and 9.
Thus, N + 2 must be a multiple of the Least Common Multiple (LCM) of 5, 7, and 9.
Since 5, 7, and 9 are pairwise coprime, LCM(5, 7, 9) = 5 * 7 * 9 = 315.
So, N + 2 = 315k for some integer k.
N = 315k – 2.
We are looking for the greatest number less than 1000.
For k=1, N = 315(1) – 2 = 313.
For k=2, N = 315(2) – 2 = 630 – 2 = 628.
For k=3, N = 315(3) – 2 = 945 – 2 = 943.
For k=4, N = 315(4) – 2 = 1260 – 2 = 1258, which is greater than 1000.
The greatest number less than 1000 is 943.

This problem is an application of the Chinese Remainder Theorem, but can be solved more directly by observing the pattern in the remainders. Checking the answer: 943 divided by 5 gives 188 with remainder 3. 943 divided by 7 gives 134 with remainder 5 (134 * 7 = 938). 943 divided by 9 gives 104 with remainder 7 (104 * 9 = 936). The conditions are satisfied.

230. Which one of the following elements has the oxidation state as +7 ?

Which one of the following elements has the oxidation state as +7 ?

[amp_mcq option1=”Nitrogen” option2=”Phosphorus” option3=”Manganese” option4=”Magnesium” correct=”option3″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

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The correct answer is Manganese.

Manganese (Mn) is a transition metal that can exhibit various oxidation states, including +2, +3, +4, +6, and +7. In compounds like potassium permanganate (KMnO4), Manganese has an oxidation state of +7.

Nitrogen typically exhibits oxidation states ranging from -3 to +5 (e.g., +5 in HNO3 or N2O5). Phosphorus typically exhibits oxidation states of -3, +3, and +5 (e.g., +5 in H3PO4 or P4O10). Magnesium is an alkaline earth metal and almost exclusively exhibits an oxidation state of +2 in its compounds. Therefore, among the given options, only Manganese can have an oxidation state of +7.