Quantitative Aptitude and Reasoning Archives

31. From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran

From a pack of 5 green balls and 4 red balls, 2 balls are drawn at random. What is the probability that both the balls are of the same colour ?

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5/9

4/9

2/9

None of the above

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UPSC CISF-AC-EXE – 2022

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The correct answer is 4/9.

Total number of balls in the pack is 5 green + 4 red = 9 balls.
We are drawing 2 balls at random.
The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! / (k! * (n-k)!):
Total outcomes = C(9, 2) = 9! / (2! * 7!) = (9 × 8) / (2 × 1) = 36.
We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways:
Case 1: Both balls are green.
Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! / (2! * 3!) = (5 × 4) / (2 × 1) = 10.
Case 2: Both balls are red.
Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! / (2! * 2!) = (4 × 3) / (2 × 1) = 6.
The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16.
The probability that both balls are of the same colour is (Favourable outcomes) / (Total outcomes) = 16 / 36.
Simplifying the fraction, 16/36 = (4 × 4) / (9 × 4) = 4/9.

This problem involves basic concepts of probability and combinations. The calculation for combinations C(n, k) represents the number of ways to choose k items from a set of n distinct items without regard to the order of selection. When dealing with probabilities of multiple events, we sum the probabilities of mutually exclusive events (like drawing two green OR two red).

32. Which one of the following is the largest 3-digit number which when di

Which one of the following is the largest 3-digit number which when divided by 12, 15 and 18 respectively, gives a remainder 5 in each case ?

955

905

995

755

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UPSC CISF-AC-EXE – 2022

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The correct answer is 905.

Let the number be N. The condition states that when N is divided by 12, 15, and 18, the remainder is always 5. This means N – 5 is divisible by 12, 15, and 18. Therefore, N – 5 must be a multiple of the Least Common Multiple (LCM) of 12, 15, and 18.
The prime factorization of 12 is 2² × 3.
The prime factorization of 15 is 3 × 5.
The prime factorization of 18 is 2 × 3².
The LCM (12, 15, 18) = 2² × 3² × 5 = 4 × 9 × 5 = 180.
So, N – 5 = 180k for some integer k.
N = 180k + 5.
We are looking for the largest 3-digit number of this form. The largest 3-digit number is 999.
We need 180k + 5 ≤ 999.
180k ≤ 994.
k ≤ 994 / 180 ≈ 5.52.
The largest integer value for k is 5.
Substituting k = 5 into the formula for N:
N = 180 × 5 + 5 = 900 + 5 = 905.
905 is a 3-digit number. Let’s check if it gives a remainder of 5 when divided by 12, 15, and 18:
905 ÷ 12 = 75 with remainder 5. (900 = 12 * 75)
905 ÷ 15 = 60 with remainder 5. (900 = 15 * 60)
905 ÷ 18 = 50 with remainder 5. (900 = 18 * 50)
Since k=5 gives 905, and any larger integer k would result in a number greater than 999 (e.g., k=6 gives 180*6+5 = 1085), 905 is the largest 3-digit number satisfying the condition.

This type of problem is a classic example of finding a number that satisfies multiple congruence relations (N ≡ 5 mod 12, N ≡ 5 mod 15, N ≡ 5 mod 18), which simplifies to N ≡ 5 mod(LCM(12, 15, 18)).

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33. Suppose, ‘A’ can complete a job in 10 days, and ‘A’ and ‘B’ together c

Suppose, ‘A’ can complete a job in 10 days, and ‘A’ and ‘B’ together can complete the same job in 6 days. In how many days can ‘B’ alone complete the job ?

15 days

12 days

18 days

20 days

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UPSC CISF-AC-EXE – 2022

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Let the total amount of work be W.
A completes the job in 10 days. A’s work rate per day = W / 10.
A and B together complete the job in 6 days. Their combined work rate per day = W / 6.
Let B alone take x days to complete the job. B’s work rate per day = W / x.

The combined work rate of A and B is the sum of their individual work rates:
(A’s rate) + (B’s rate) = (A+B)’s rate
(W / 10) + (W / x) = (W / 6)

Since W represents the same job and is non-zero, we can divide the entire equation by W:
1 / 10 + 1 / x = 1 / 6

Now, solve for 1/x:
1 / x = 1 / 6 – 1 / 10

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To subtract the fractions, find a common denominator for 6 and 10. The least common multiple is 30.
1 / 6 = 5 / 30
1 / 10 = 3 / 30

So, 1 / x = 5 / 30 – 3 / 30
1 / x = (5 – 3) / 30
1 / x = 2 / 30
1 / x = 1 / 15

Therefore, x = 15.
B alone can complete the job in 15 days.

– Represent work rates as the reciprocal of the time taken (assuming the total work is 1 unit or W).
– The combined work rate is the sum of individual work rates.
– Set up an equation based on work rates and solve for the unknown time.

This type of problem can also be approached by considering “units of work”. If the LCM of 10 and 6 is 30, assume the total work is 30 units.
A does 30 units in 10 days, so A’s rate = 30/10 = 3 units/day.
A and B together do 30 units in 6 days, so their combined rate = 30/6 = 5 units/day.
B’s rate = (A+B)’s rate – A’s rate = 5 units/day – 3 units/day = 2 units/day.
Time taken by B alone = Total Work / B’s rate = 30 units / (2 units/day) = 15 days.
Both methods yield the same result.

34. ‘B’ is 28 years older than ‘A’. After 12 years from now, the age of ‘B

‘B’ is 28 years older than ‘A’. After 12 years from now, the age of ‘B’ will be twice the age of ‘A’. The present age of ‘A’ is :

10 years

12 years

30 years

16 years

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UPSC CISF-AC-EXE – 2022

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Let the present age of A be A years.
Let the present age of B be B years.
According to the first statement, ‘B’ is 28 years older than ‘A’:
B = A + 28 (Equation 1)

After 12 years from now:
Age of A will be A + 12 years.
Age of B will be B + 12 years.

According to the second statement, the age of ‘B’ will be twice the age of ‘A’ after 12 years:
B + 12 = 2 * (A + 12) (Equation 2)

Now, we substitute Equation 1 into Equation 2:
(A + 28) + 12 = 2 * (A + 12)
A + 40 = 2A + 24

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Now, we solve for A:
40 – 24 = 2A – A
16 = A

So, the present age of A is 16 years.
Let’s verify:
If A = 16, then B = 16 + 28 = 44.
After 12 years, A’s age = 16 + 12 = 28.
After 12 years, B’s age = 44 + 12 = 56.
Is B’s age (56) twice A’s age (28)? Yes, 56 = 2 * 28. The condition is satisfied.

– Represent the unknown ages with variables.
– Translate the given information into algebraic equations.
– Solve the system of equations to find the unknown age.

This is a typical age-based problem that can be solved using simultaneous linear equations. Setting up the equations correctly based on the time references (present, after 12 years) is crucial.

35. A solid metallic sphere of 2 cm radius is melted and converted to a cu

A solid metallic sphere of 2 cm radius is melted and converted to a cube. The side of the cube is approximately equal to :

2.4 cm

2.8 cm

3.2 cm

3.6 cm

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UPSC CISF-AC-EXE – 2022

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When a solid metallic sphere is melted and converted into a cube, the volume of the material remains constant.
Radius of the sphere (r) = 2 cm.
Volume of the sphere (V_sphere) = (4/3)πr³.
V_sphere = (4/3) * π * (2 cm)³ = (4/3) * π * 8 cm³ = (32/3)π cm³.
Let the side of the cube be ‘s’ cm.
Volume of the cube (V_cube) = s³.
Since the volume is conserved, V_cube = V_sphere.
s³ = (32/3)π.
To find ‘s’, we take the cube root: s = ³√[(32/3)π].
Using the approximate value of π ≈ 3.14159:
(32/3)π ≈ (10.6667) * 3.14159 ≈ 33.51 cm³.
s ≈ ³√33.51 cm.
Let’s evaluate the cube of the given options:
A) (2.4)³ ≈ 13.824
B) (2.8)³ ≈ 21.952
C) (3.2)³ ≈ 32.768
D) (3.6)³ ≈ 46.656
The value 3.2³ = 32.768 is closest to 33.51.

– The volume of the material is conserved during melting and recasting.
– Formula for the volume of a sphere: V = (4/3)πr³.
– Formula for the volume of a cube: V = s³.
– Solve for the side ‘s’ by equating the volumes and taking the cube root.

The value of π is an irrational number, so the exact value of ‘s’ is ³√[(32/3)π]. The question asks for an approximate value, so we use the numerical value of π. The calculation shows that 3.2 cm is the closest approximation among the given options.

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36. The sum of the numbers 3, 6, 9, 12, … up to the 20 th term is :

The sum of the numbers 3, 6, 9, 12, … up to the 20^th term is :

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600

1260

630

960

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This question was previously asked in

UPSC CISF-AC-EXE – 2022

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The given sequence is 3, 6, 9, 12, … which is an arithmetic progression (AP).
The first term (a) is 3.
The common difference (d) is 6 – 3 = 3 (or 9 – 6 = 3, etc.).
We need to find the sum of the first 20 terms (n = 20).
The formula for the sum of the first n terms of an AP is S_n = n/2 * [2a + (n-1)d].
Substituting the values:
S₂₀ = 20/2 * [2 * 3 + (20 – 1) * 3]
S₂₀ = 10 * [6 + 19 * 3]
S₂₀ = 10 * [6 + 57]
S₂₀ = 10 * 63
S₂₀ = 630

– Recognize the sequence as an Arithmetic Progression (AP).
– Identify the first term (a) and the common difference (d).
– Use the formula for the sum of the first n terms of an AP: S_n = n/2 * [2a + (n-1)d].

Alternatively, the sequence is 3 * 1, 3 * 2, 3 * 3, …, 3 * 20.
The sum is 3 * (1 + 2 + 3 + … + 20).
The sum of the first n natural numbers is n(n+1)/2.
So, the sum of 1 to 20 is 20(20+1)/2 = 20 * 21 / 2 = 10 * 21 = 210.
The sum of the sequence is 3 * 210 = 630.
This confirms the result obtained using the AP sum formula.

37. In a group of 30 students, each student has opted for at least one of

In a group of 30 students, each student has opted for at least one of the two subjects, Hindi and English. Twelve of them have opted for Hindi and twenty-two have opted for English. The number of students who have opted for only English is :

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Let H be the set of students who opted for Hindi and E be the set of students who opted for English. We are given the total number of students, $|H \cup E| = 30$ (since each student opted for at least one subject). We are given $|H| = 12$ and $|E| = 22$. The number of students who opted for both subjects is the intersection of H and E, denoted by $|H \cap E|$. Using the principle of inclusion-exclusion for two sets, we have $|H \cup E| = |H| + |E| – |H \cap E|$. Substituting the given values, $30 = 12 + 22 – |H \cap E|$. This gives $30 = 34 – |H \cap E|$, so $|H \cap E| = 34 – 30 = 4$. The number of students who opted for only English is the number of students in E minus the number of students in both H and E. So, number of students who opted for only English = $|E| – |H \cap E| = 22 – 4 = 18$.

– Total students = Students who opted for Hindi only + Students who opted for English only + Students who opted for both.
– $|H \cup E| = |H \text{ only}| + |E \text{ only}| + |H \cap E|$.
– $|H| = |H \text{ only}| + |H \cap E|$.
– $|E| = |E \text{ only}| + |H \cap E|$.

From $|H \cap E|=4$, we can also find the number of students who opted for only Hindi: $|H \text{ only}| = |H| – |H \cap E| = 12 – 4 = 8$.
Check: Total students = (Only Hindi) + (Only English) + (Both) = 8 + 18 + 4 = 30, which matches the given information.

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38. If the sum of five consecutive even numbers is equal to the product of

If the sum of five consecutive even numbers is equal to the product of first five natural numbers, then which one of the following is the largest of those even numbers ?

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The product of the first five natural numbers (1, 2, 3, 4, 5) is $1 \times 2 \times 3 \times 4 \times 5 = 120$. Let the five consecutive even numbers be represented as $x-4, x-2, x, x+2, x+4$, where $x$ is the middle number. The sum of these five consecutive even numbers is $(x-4) + (x-2) + x + (x+2) + (x+4) = 5x$. We are given that this sum is equal to the product of the first five natural numbers, so $5x = 120$. Solving for $x$, we get $x = 120 / 5 = 24$. The five consecutive even numbers are $24-4=20$, $24-2=22$, $24$, $24+2=26$, and $24+4=28$. The largest of these even numbers is 28.

– First five natural numbers are 1, 2, 3, 4, 5. Their product is 120.
– Consecutive even numbers differ by 2. Representing them around the middle term simplifies the sum calculation.
– If the middle term is $x$, the sum of five consecutive even numbers is $5x$.

Using $x-4, x-2, x, x+2, x+4$ simplifies the sum calculation as the constant terms cancel out. Alternatively, if the first even number is $a$, the numbers are $a, a+2, a+4, a+6, a+8$. Their sum is $5a+20$. So $5a+20 = 120$, $5a = 100$, $a=20$. The numbers are 20, 22, 24, 26, 28. The largest is $a+8 = 20+8 = 28$.

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39. When the square of the sum of two numbers are added to the square of t

When the square of the sum of two numbers are added to the square of their difference, we get 416. The difference between the square of the sum and square of the difference is 384. What are the numbers ?

18 and 24

12 and 16

8 and 12

10 and 12

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UPSC CISF-AC-EXE – 2022

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Let the two numbers be ‘a’ and ‘b’. The sum of the numbers is (a + b) and the difference is (a – b). The problem gives two conditions based on the squares of the sum and difference.
Condition 1: The square of the sum added to the square of the difference is 416.
(a + b)² + (a – b)² = 416.
Using the identities (a+b)² = a² + 2ab + b² and (a-b)² = a² – 2ab + b², we get:
(a² + 2ab + b²) + (a² – 2ab + b²) = 416.
2a² + 2b² = 416.
Dividing by 2, we get: a² + b² = 208. (Equation 1)

Condition 2: The difference between the square of the sum and the square of the difference is 384.
(a + b)² – (a – b)² = 384.
Using the identities again:
(a² + 2ab + b²) – (a² – 2ab + b²) = 384.
a² + 2ab + b² – a² + 2ab – b² = 384.
4ab = 384.
Dividing by 4, we get: ab = 96. (Equation 2)

We now have a system of two equations:
1) a² + b² = 208
2) ab = 96
We can check the options given to see which pair of numbers satisfies both equations.
Option C gives the numbers 8 and 12.
Let’s test if a=8 and b=12 satisfy the equations:
Equation 1: 8² + 12² = 64 + 144 = 208. (Satisfied)
Equation 2: 8 * 12 = 96. (Satisfied)
Since the numbers 8 and 12 satisfy both conditions derived from the problem statement, they are the correct numbers. Alternatively, we can solve the system using the identities (a+b)² = a²+b²+2ab and (a-b)² = a²+b²-2ab.
(a+b)² = 208 + 2(96) = 208 + 192 = 400 => a+b = √400 = 20 (assuming positive numbers).
(a-b)² = 208 – 2(96) = 208 – 192 = 16 => a-b = √16 = 4 (assuming a>b).
Solving a+b=20 and a-b=4: Adding gives 2a=24 => a=12. Substituting gives 12+b=20 => b=8. The numbers are 12 and 8.

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40. In order to convert a loss of 5% to a profit of 5%, a shopkeeper raise

In order to convert a loss of 5% to a profit of 5%, a shopkeeper raises the price of an item by ₹ 500. What is the cost price of the item ?

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₹ 1,000

₹ 5,000

₹ 10,000

₹ 1,200

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UPSC CISF-AC-EXE – 2022

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Let the cost price (CP) of the item be ₹ x. Initially, the shopkeeper incurs a loss of 5%. The selling price (SP1) in this case is CP – 5% of CP = x – 0.05x = 0.95x. To convert this to a profit of 5%, the new selling price (SP2) must be CP + 5% of CP = x + 0.05x = 1.05x.

The problem states that the shopkeeper raises the price of the item by ₹ 500 to achieve the desired change. This means the difference between the new selling price (SP2) and the original selling price (SP1) is ₹ 500.
SP2 – SP1 = ₹ 500.
(1.05x) – (0.95x) = 500.

Subtracting the two expressions for the selling prices in terms of CP (x):
(1.05 – 0.95) * x = 500.
0.10 * x = 500.
To find x (the cost price), divide 500 by 0.10:
x = 500 / 0.10 = 500 / (1/10) = 500 * 10 = ₹ 5,000.
The price increase of ₹ 500 represents the change from a 5% loss margin to a 5% profit margin, which is a total change of 10% of the cost price (5% below CP to 5% above CP). So, 10% of CP = 500, which directly leads to CP = 5000.