Quantitative Aptitude and Reasoning Archives

91. 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

Download PDF Attempt Online

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.

92. 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

Download PDF Attempt Online

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.

93. 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

Download PDF Attempt Online

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.

94. 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

Download PDF Attempt Online

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.

95. 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

Download PDF Attempt Online

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.

96. 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

Download PDF Attempt Online

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.

97. What is the number of all possible positive integer values of ‘n’ for

What is the number of all possible positive integer values of ‘n’ for which n² + 96 is a perfect square ?

[amp_mcq option1=”2″ option2=”4″ option3=”5″ option4=”Infinite” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

Download PDF Attempt Online

The correct answer is B.

We are looking for positive integer values of ‘n’ such that n² + 96 is a perfect square.
Let n² + 96 = k², where k is an integer.
Since n is a positive integer, n² is positive, so k² must be greater than n². This implies k > n (assuming k is positive, which it must be since k² = n² + 96).
Rearranging the equation, we get k² – n² = 96.
This is a difference of squares, which can be factored as (k – n)(k + n) = 96.
Since k and n are integers, (k-n) and (k+n) must be integer factors of 96.
Also, (k+n) – (k-n) = 2n. Since n is an integer, 2n is an even integer. This means (k-n) and (k+n) must have the same parity. Since their product (96) is even, both factors must be even.
Furthermore, since n is positive, k+n > k-n. Also, k+n > 0 (as n>0 and k>n).

We need to find pairs of even factors (a, b) of 96 such that a * b = 96 and b > a.
The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.
The even factors are 2, 4, 6, 8, 12, 16, 24, 32, 48, 96.
Pairs (a, b) where a and b are even, a*b=96, and b > a:
1. (2, 48) -> k – n = 2, k + n = 48. Adding gives 2k=50, k=25. Subtracting gives 2n=46, n=23. (n=23 is a positive integer)
2. (4, 24) -> k – n = 4, k + n = 24. Adding gives 2k=28, k=14. Subtracting gives 2n=20, n=10. (n=10 is a positive integer)
3. (6, 16) -> k – n = 6, k + n = 16. Adding gives 2k=22, k=11. Subtracting gives 2n=10, n=5. (n=5 is a positive integer)
4. (8, 12) -> k – n = 8, k + n = 12. Adding gives 2k=20, k=10. Subtracting gives 2n=4, n=2. (n=2 is a positive integer)
We found 4 distinct positive integer values for n: 23, 10, 5, and 2.
Therefore, there are 4 possible positive integer values of ‘n’ for which n² + 96 is a perfect square.

98. The sum of two positive integers is 52 and their LCM is 168. What is t

The sum of two positive integers is 52 and their LCM is 168. What is the ratio between the numbers?

[amp_mcq option1=”2:3″ option2=”5:4″ option3=”7:6″ option4=”7:8″ correct=”option3″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

Download PDF Attempt Online

The correct answer is C.

Let the two positive integers be x and y.
Given: x + y = 52 and LCM(x, y) = 168.
Let g be the Greatest Common Divisor (GCD) of x and y. So, x = ga and y = gb, where a and b are coprime positive integers (gcd(a, b) = 1).
Using the given information:
1. x + y = ga + gb = g(a + b) = 52.
2. LCM(x, y) = g * a * b = 168.

From g(a + b) = 52, we know that g must be a divisor of 52. The divisors of 52 are 1, 2, 4, 13, 26, 52.
From gab = 168, we know that g must be a divisor of 168.
So, g must be a common divisor of 52 and 168.
Divisors of 52: 1, 2, 4, 13, 26, 52
Divisors of 168: 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168
Common divisors of 52 and 168 are 1, 2, 4. So g can be 1, 2, or 4.
We also have the constraints: a and b are positive integers and gcd(a, b) = 1.

Case 1: g = 1
a + b = 52/1 = 52
ab = 168/1 = 168
We look for two coprime numbers a and b such that a+b=52 and ab=168. This is equivalent to solving the quadratic equation t² – 52t + 168 = 0. The discriminant is 52² – 4*168 = 2704 – 672 = 2032, which is not a perfect square, so integer solutions for t (a, b) do not exist. Alternatively, list factor pairs of 168 and check sums: (1, 168) sum 169; (2, 84) sum 86; (3, 56) sum 59; (4, 42) sum 46; (6, 28) sum 34; (7, 24) sum 31; (8, 21) sum 29; (12, 14) sum 26. None sum to 52.

Case 2: g = 2
a + b = 52/2 = 26
ab = 168/2 = 84
We look for two coprime numbers a and b such that a+b=26 and ab=84. Factors of 84: (1, 84) sum 85, gcd 1 (valid); (2, 42) sum 44, gcd 2 (invalid); (3, 28) sum 31, gcd 1 (valid); (4, 21) sum 25, gcd 1 (valid); (6, 14) sum 20, gcd 2 (invalid); (7, 12) sum 19, gcd 1 (valid). None sum to 26.

Case 3: g = 4
a + b = 52/4 = 13
ab = 168/4 = 42
We look for two coprime numbers a and b such that a+b=13 and ab=42. Factors of 42: (1, 42) sum 43, gcd 1 (valid); (2, 21) sum 23, gcd 1 (valid); (3, 14) sum 17, gcd 1 (valid); (6, 7) sum 13, gcd 1 (valid).
The pair (6, 7) satisfies a+b=13 and gcd(6, 7)=1.
So, possible values for (a, b) are (6, 7) or (7, 6).
If (a, b) = (6, 7), then x = g*a = 4*6 = 24 and y = g*b = 4*7 = 28.
Check: 24 + 28 = 52. LCM(24, 28) = LCM(2³*3, 2²*7) = 2³*3*7 = 8*21 = 168. This is correct.
The two numbers are 24 and 28.
The ratio between the numbers is 24:28 or 28:24.
24:28 simplifies to 6:7 (dividing by 4).
28:24 simplifies to 7:6 (dividing by 4).
Option C is 7:6.

Therefore, the ratio between the numbers is 7:6 or 6:7. Option C provides 7:6.

99. The rabbit population in community A increases at 25% per year while t

The rabbit population in community A increases at 25% per year while that in community B increases at 50% per year. If the present populations of A and B are equal, what will be the ratio of the number of rabbits in community B to that in community A after 2 years ?

[amp_mcq option1=”1.44″ option2=”1.72″ option3=”1.90″ option4=”1.25″ correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

Download PDF Attempt Online

The correct answer is A.

Let the present population of rabbits in community A and B be P.
Population growth in community A is 25% per year. After ‘n’ years, the population will be P * (1 + 0.25)^n.
Population growth in community B is 50% per year. After ‘n’ years, the population will be P * (1 + 0.50)^n.
We need to find the ratio of the number of rabbits in community B to that in community A after 2 years (n=2).

Population of A after 2 years = P_A(2) = P * (1 + 0.25)² = P * (1.25)² = P * 1.5625.
Population of B after 2 years = P_B(2) = P * (1 + 0.50)² = P * (1.50)² = P * 2.25.
The ratio of the number of rabbits in community B to that in community A after 2 years is P_B(2) / P_A(2).
Ratio = (P * 2.25) / (P * 1.5625) = 2.25 / 1.5625.
To calculate 2.25 / 1.5625, we can write it as 22500 / 15625.
Divide both by 25: 900 / 625.
Divide both by 25 again: 36 / 25.
36 / 25 = 1.44.
The ratio is 1.44.

100. ‘A’ and ‘B’ can complete a work in 2 hours and 55 minutes. ‘A’ alone c

‘A’ and ‘B’ can complete a work in 2 hours and 55 minutes. ‘A’ alone can do the same work two hours faster than ‘B’ alone can. How long will ‘B’ take to do the work alone ?

[amp_mcq option1=”2 hours and 10 minutes” option2=”4 hours and 10 minutes” option3=”5 hours” option4=”7 hours” correct=”option4″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

Download PDF Attempt Online

The correct answer is D.

Let the time taken by B to complete the work alone be `x` hours.
Since A can do the same work two hours faster than B, the time taken by A alone is `x – 2` hours.
The work rate of B is 1/x work per hour.
The work rate of A is 1/(x-2) work per hour.
A and B together complete the work in 2 hours and 55 minutes.
2 hours 55 minutes = 2 + 55/60 hours = 2 + 11/12 hours = (24+11)/12 hours = 35/12 hours.
The combined work rate of A and B is 1 / (35/12) = 12/35 work per hour.
The combined work rate is also the sum of their individual rates: 1/(x-2) + 1/x = 12/35.

To solve for x:
[x + (x-2)] / [x(x-2)] = 12/35
(2x – 2) / (x² – 2x) = 12/35
35(2x – 2) = 12(x² – 2x)
70x – 70 = 12x² – 24x
12x² – 94x + 70 = 0
Divide by 2: 6x² – 47x + 35 = 0
Using the quadratic formula x = [-b ± sqrt(b² – 4ac)] / 2a:
x = [47 ± sqrt((-47)² – 4 * 6 * 35)] / (2 * 6)
x = [47 ± sqrt(2209 – 840)] / 12
x = [47 ± sqrt(1369)] / 12
x = [47 ± 37] / 12
Two possible solutions for x:
x1 = (47 + 37) / 12 = 84 / 12 = 7
x2 = (47 – 37) / 12 = 10 / 12 = 5/6
If x = 5/6 hours, then A takes x – 2 = 5/6 – 2 = -7/6 hours, which is not possible as time cannot be negative.
So, x = 7 hours is the valid solution.
B takes 7 hours to do the work alone.