Quantitative Aptitude and Reasoning

An international conference is attended by 65 people. They all speak at least one of English, French and German language. Suppose 15 speak English and French, 13 speak English and German, 12 speak French and German and 5 speak all the three languages. A total of 30 people can speak German and 30 can speak French. What is the number of people who can speak only English?

[amp_mcq option1=”17″ option2=”20″ option3=”22″ option4=”40″ correct=”option1″]

The number of people who can speak only English is 17.

– Let E, F, and G be the sets of people who speak English, French, and German, respectively.
– Total people |E U F U G| = 65 (since all speak at least one language).
– We are given: |E ∩ F| = 15, |E ∩ G| = 13, |F ∩ G| = 12, |E ∩ F ∩ G| = 5.
– We are also given: |G| = 30, |F| = 30.
– We can find the number of people speaking exactly two languages:
– |E ∩ F only| = |E ∩ F| – |E ∩ F ∩ G| = 15 – 5 = 10
– |E ∩ G only| = |E ∩ G| – |E ∩ F ∩ G| = 13 – 5 = 8
– |F ∩ G only| = |F ∩ G| – |E ∩ F ∩ G| = 12 – 5 = 7
– We can find the number of people speaking only one language using the given total for F and G:
– |G| = |G only| + |E ∩ G only| + |F ∩ G only| + |E ∩ F ∩ G|
30 = |G only| + 8 + 7 + 5 => 30 = |G only| + 20 => |G only| = 10.
– |F| = |F only| + |E ∩ F only| + |F ∩ G only| + |E ∩ F ∩ G|
30 = |F only| + 10 + 7 + 5 => 30 = |F only| + 22 => |F only| = 8.
– The total number of people is the sum of those speaking only one language, exactly two languages, and all three:
– |E U F U G| = |E only| + |F only| + |G only| + |E ∩ F only| + |E ∩ G only| + |F ∩ G only| + |E ∩ F ∩ G|
– 65 = |E only| + 8 + 10 + 10 + 8 + 7 + 5
– 65 = |E only| + 48
– |E only| = 65 – 48 = 17.

This problem can be effectively solved using a Venn diagram to visualize the different sections representing speakers of one, two, or three languages. The Principle of Inclusion-Exclusion is the formal mathematical basis for these calculations.

The angle between the hour hand and the minute hand of a clock at 10 minutes past 3 is

[amp_mcq option1=”30°” option2=”35°” option3=”37.5°” option4=”40°” correct=”option2″]

The angle between the hour hand and the minute hand of a clock at 10 minutes past 3 is 35°.

– The minute hand moves 360 degrees in 60 minutes, so its speed is 6 degrees per minute (360/60).
– The hour hand moves 360 degrees in 12 hours (720 minutes), so its speed is 0.5 degrees per minute (360/720).
– At 3:10, the time is 3 hours and 10 minutes past 12 o’clock.
– Position of the minute hand: At 10 minutes, the minute hand is at 10 * 6 = 60 degrees from the 12 o’clock mark.
– Position of the hour hand: At 3 hours and 10 minutes, the hour hand’s position relative to 12 o’clock is (3 hours * 30 degrees/hour) + (10 minutes * 0.5 degrees/minute) = 90 + 5 = 95 degrees from the 12 o’clock mark. (Note: Each hour mark is 30 degrees apart: 360/12 = 30).
– The angle between the hands is the absolute difference between their positions: |95° – 60°| = 35°.

Clock problems involve calculating the relative positions of the hour and minute hands based on their speeds. The hour hand moves continuously, not just jumping from one hour mark to the next.

What is the largest value for n (natural number) such that 6^n divides the product of the first 100 natural numbers?

[amp_mcq option1=”18″ option2=”33″ option3=”44″ option4=”48″ correct=”option4″]

The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48.

– To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial.
– 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3.
– The highest power of a prime p dividing n! is given by Legendre’s formula: floor(n/p) + floor(n/p^2) + floor(n/p^3) + …
– For p=3 and n=100:
Power of 3 in 100! = floor(100/3) + floor(100/9) + floor(100/27) + floor(100/81)
= 33 + 11 + 3 + 1 = 48.
– For p=2 and n=100:
Power of 2 in 100! = floor(100/2) + floor(100/4) + floor(100/8) + floor(100/16) + floor(100/32) + floor(100/64)
= 50 + 25 + 12 + 6 + 3 + 1 = 97.
– So, 100! contains 2^97 * 3^48 * …
– For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n ≤ 97 and n ≤ 48.
– The largest integer n satisfying both conditions is n = 48.

Legendre’s formula is a powerful tool for finding the exponent of a prime in the prime factorization of n!. This method works because every multiple of a prime p contributes at least one factor of p, every multiple of p^2 contributes an additional factor of p, and so on.

If a cubical container of length, breadth and height each of 10 cm can contain exactly 1 litre of water, then a spherical container of radius 10.5 cm can contain

[amp_mcq option1=”not more than 4 litres of water” option2=”more than 4 litres but less than 4.5 litres of water” option3=”more than 4.5 litres but less than 5 litres of water” option4=”more than 5 litres of water” correct=”option3″]

The correct answer is C) more than 4.5 litres but less than 5 litres of water.

A cubical container of 10 cm side length has a volume of 10 cm * 10 cm * 10 cm = 1000 cm³.
Given that this container holds exactly 1 litre, we know 1 litre = 1000 cm³.
A spherical container has a radius of 10.5 cm.
The volume of a sphere is given by the formula V = (4/3)πr³.
Using π ≈ 22/7:
V = (4/3) * (22/7) * (10.5)³
V = (4/3) * (22/7) * (10.5 * 10.5 * 10.5)
V = (4/3) * (22/7) * (1157.625)
V = (88/21) * 1157.625
V = 4.1904… * 1157.625
V ≈ 4851 cm³
To convert this volume to litres, divide by 1000:
Volume in litres = 4851 cm³ / 1000 cm³/litre = 4.851 litres.
Now compare 4.851 litres with the options:
A) not more than 4 litres (4.851 is more than 4) – Incorrect
B) more than 4 litres but less than 4.5 litres (4.851 is not less than 4.5) – Incorrect
C) more than 4.5 litres but less than 5 litres (4.5 < 4.851 < 5) - Correct D) more than 5 litres (4.851 is not more than 5) - Incorrect

The value of π used can affect the precision slightly. Using a more precise value like 3.14159 would give V = (4/3) * 3.14159 * (10.5)³ ≈ 4849.05 cm³, which is 4.84905 litres. This value also falls within the range of more than 4.5 litres but less than 5 litres.

A king ordered to make a crown from 8 kg of gold and 2 kg of silver. The goldsmith took away some amount of gold and replaced it by an equal amount of silver and the crown when made, weighed 10 kg. The king knows that under water gold loses $\frac{1}{20}$th of its weight, while silver loses $\frac{1}{10}$th. When the crown was weighed under water, it was 9.25 kg. How much gold was stolen by the goldsmith?

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

The correct answer is C) 3 kg.

Initial composition: 8 kg gold, 2 kg silver, total 10 kg.
Let ‘x’ kg of gold be stolen and replaced by ‘x’ kg of silver.
Final composition: (8-x) kg gold, (2+x) kg silver. Total weight is (8-x) + (2+x) = 10 kg.
Under water, gold loses 1/20th of its weight, so its apparent weight is (1 – 1/20) = 19/20 of its actual weight.
Under water, silver loses 1/10th of its weight, so its apparent weight is (1 – 1/10) = 9/10 of its actual weight.
The crown weighs 9.25 kg under water.
Apparent weight of gold = (19/20) * (8-x)
Apparent weight of silver = (9/10) * (2+x)
Total apparent weight = (19/20) * (8-x) + (9/10) * (2+x) = 9.25
Multiply by 20 to clear the denominators:
19 * (8-x) + 18 * (2+x) = 9.25 * 20
152 – 19x + 36 + 18x = 185
188 – x = 185
x = 188 – 185
x = 3 kg.
The goldsmith stole 3 kg of gold.

This problem is an application of Archimedes’ principle. The weight loss under water is equal to the weight of the water displaced, which is proportional to the volume of the object. Different densities of gold and silver cause different weight losses for the same weight. Pure gold is much denser than silver, so 1 kg of gold occupies less volume than 1 kg of silver and displaces less water, thus losing less weight. Replacing gold with silver increases the overall volume for the same weight, leading to a greater weight loss under water. The initial weight loss for the pure crown would have been (1/20)*8 + (1/10)*2 = 0.4 + 0.2 = 0.6 kg, making its underwater weight 10 – 0.6 = 9.4 kg. The actual underwater weight is 9.25 kg, a loss of 0.75 kg (10 – 9.25). The increased loss of 0.15 kg (0.75 – 0.6) is due to the replacement of ‘x’ kg of gold with ‘x’ kg of silver. The difference in weight loss per kg is (1/10) – (1/20) = 1/20. So, the increased loss is (1/20)*x = 0.15 kg. x = 0.15 * 20 = 3 kg.

Suppose a, b, c, d and e are five consecutive odd numbers in ascending order. Consider the following statements:

• 1. Their average is (a+4).
• 2. Their average is (b+2).
• 3. Their average is (e-4).

Which of the statements given above is/are correct?

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

The correct answer is D) 1, 2 and 3.

Let the five consecutive odd numbers in ascending order be a, a+2, a+4, a+6, and a+8. Thus, b=a+2, c=a+4, d=a+6, e=a+8.
The average of these five numbers is (a + (a+2) + (a+4) + (a+6) + (a+8)) / 5 = (5a + 20) / 5 = a + 4.
Now let’s check the statements:
1. Their average is (a+4). This is correct.
2. Their average is (b+2). Since b = a+2, b+2 = (a+2)+2 = a+4. This is correct.
3. Their average is (e-4). Since e = a+8, e-4 = (a+8)-4 = a+4. This is correct.
All three statements are correct.

For any sequence of consecutive numbers (arithmetic progression), the average is equal to the median. In this case, c is the middle number, and c = a+4. The average is also the average of the first and last term: (a+e)/2 = (a + a+8)/2 = (2a+8)/2 = a+4. Similarly, the average of the second and fourth term: (b+d)/2 = (a+2 + a+6)/2 = (2a+8)/2 = a+4.

If the number 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰ is divisible by 10ⁿ, then which one of the following is the maximum value of n?

[amp_mcq option1=”78″ option2=”85″ option3=”89″ option4=”98″ correct=”option4″]

The correct answer is D) 98.

A number is divisible by 10ⁿ if its prime factorization contains at least n factors of 2 and n factors of 5. The maximum value of n is determined by the minimum exponent of 2 and 5 in the prime factorization of the given number.
The given number is N = 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰.
Let’s express all terms in terms of prime factors 2, 3, 5:
N = 2² * 5⁴ * (2²)⁶ * (2*5)⁸ * (2*3)¹⁰ * (3*5)¹² * (2³)¹⁴ * (2²*5)¹⁶ * (2*5)¹⁸ * (5²)²⁰
N = 2² * 5⁴ * 2¹² * 2⁸ * 5⁸ * 2¹⁰ * 3¹⁰ * 3¹² * 5¹² * 2⁴² * 2³² * 5¹⁶ * 2¹⁸ * 5¹⁸ * 5⁴⁰

Total power of 2: 2 + 12 + 8 + 10 + 42 + 32 + 18 = 124. So, 2¹²⁴.
Total power of 5: 4 + 8 + 12 + 16 + 18 + 40 = 98. So, 5⁹⁸.
Total power of 3: 10 + 12 = 22. So, 3²².

The number is 2¹²⁴ * 3²² * 5⁹⁸.
For this number to be divisible by 10ⁿ = 2ⁿ * 5ⁿ, the maximum possible value of n is min(124, 98) = 98.

Divisibility by 10ⁿ is equivalent to divisibility by (2*5)ⁿ, which means divisibility by 2ⁿ and 5ⁿ simultaneously. The number of trailing zeros in an integer is equal to the exponent of 10 in its prime factorization, which is the minimum of the exponents of 2 and 5.

At what time between 2 o’clock and 3 o’clock will the hour and minute hands of a clock be 12 minutes division apart?

[amp_mcq option1=”12 minutes past 2 o’clock” option2=”18 minutes past 2 o’clock” option3=”24 minutes past 2 o’clock” option4=”30 minutes past 2 o’clock” correct=”option3″]

The correct answer is C) 24 minutes past 2 o’clock.

At 2 o’clock, the hour hand is at the ‘2’ mark and the minute hand is at the ’12’ mark. The distance between them is 10 minute divisions (each minute mark represents 1 minute or 6 degrees; 5 minute divisions between 12 and 1, 1 and 2, so 10 divisions). We want the hands to be 12 minute divisions (12 * 6 = 72 degrees) apart.
The minute hand moves 6 degrees per minute. The hour hand moves 0.5 degrees per minute. The relative speed of the minute hand with respect to the hour hand is 5.5 degrees per minute.
At 2:00, the hour hand is at 60 degrees from 12 (2 * 30), and the minute hand is at 0 degrees. The hour hand is 60 degrees ahead of the minute hand.
We want the difference in angles to be 72 degrees. This can be when the minute hand is 72 degrees behind the hour hand, or 72 degrees ahead.
Let ‘t’ be the time in minutes past 2.
Angle of minute hand = 6t.
Angle of hour hand = 60 + 0.5t.
We want |6t – (60 + 0.5t)| = 72.
|5.5t – 60| = 72.
Case 1: 5.5t – 60 = 72 => 5.5t = 132 => t = 132 / 5.5 = 1320 / 55 = 24.
Case 2: 5.5t – 60 = -72 => 5.5t = -12 => t = -12 / 5.5, which is not possible for time after 2:00.
So, the time is 24 minutes past 2 o’clock. At 2:24, the minute hand is at 24 * 6 = 144 degrees. The hour hand is at 60 + 24 * 0.5 = 60 + 12 = 72 degrees. The difference is 144 – 72 = 72 degrees, which is 12 minute divisions (72/6).

The relative speed of the minute hand w.r.t. hour hand is 5.5 degrees per minute, or 11/12 minute divisions per minute. At 2 o’clock, the hour hand is 10 minute divisions ahead of the minute hand. To be 12 divisions apart, the minute hand must gain (10 + 12) = 22 divisions if it’s ahead, or gain (10 – (-12)) = 22 if behind (but this means passing the hour hand, which happens at 2:24 anyway). The minute hand needs to gain 22 minute divisions relative to the hour hand from the 12 o’clock position where both hypothetically start together. The hour hand is effectively at 10 minute divisions at 2:00. The minute hand starts at 0. The relative gain required is the initial gap plus the desired gap. Initial gap is 10 min divisions. Desired gap is 12 min divisions. Total gain required = 10 + 12 = 22 minute divisions (for the minute hand to be ahead). Time = Gain / Relative speed = 22 / (11/12) = 22 * (12/11) = 2 * 12 = 24 minutes.

On simplification the product (x + y)(x² + y²)(x⁴ + y⁴)(x⁸ + y⁸) … how many such terms are there which will have only single x and rest ys?

[amp_mcq option1=”21″ option2=”10″ option3=”20″ option4=”1″ correct=”option4″]

The product is (x + y)(x² + y²)(x⁴ + y⁴)(x⁸ + y⁸)… Let’s consider a product of n terms: P_n = (x^2^0 + y^2^0)(x^2^1 + y^2^1)…(x^2^(n-1) + y^2^(n-1)). When expanding this product, each term is formed by choosing either the first term (x raised to a power of 2) or the second term (y raised to the same power of 2) from each bracket. For example, from the i-th bracket (x^2^i + y^2^i), we pick either x^2^i or y^2^i. A term in the expansion is the product of one chosen term from each of the n brackets. We are looking for terms with a “single x”, which means the power of x in the term must be 1 (x¹). Let’s say we pick x^2^i from one bracket and y^2^j from all other brackets (j ≠ i). The resulting term is (x^2^i) * (product of y^2^j for all j ≠ i). The power of x in this term is 2^i. For this power to be 1 (x¹), we must have 2^i = 1. This occurs only when i = 0. So, the only way to get a term with x¹ is to choose x^2^0 = x from the first bracket (x+y) and y^2^j from all subsequent brackets (j = 1, 2, …, n-1). The resulting term is x * y² * y⁴ * y⁸ * … * y^(2^(n-1)) = x * y^(2¹ + 2² + … + 2^(n-1)). The power of y is the sum of a geometric series: 2(2^(n-1) – 1)/(2-1) = 2^n – 2. The term is x¹y^(2^n – 2). This is the only term in the expansion that has x raised to the power of 1. Any other combination will result in x raised to a power other than 1 (e.g., choosing x from multiple brackets, or choosing x^2^i where i > 0 results in a power 2^i > 1). Thus, there is only exactly one such term. The number of such terms is 1.

In the expansion of a product of factors of the form (x^a + y^a), terms are formed by selecting one element from each factor and multiplying them. To find terms with a specific power of x, analyze the possible combinations of selections from each factor.

This type of product (x+y)(x²+y²)(x⁴+y⁴)… is related to the identity (x-y)(x+y) = x²-y², (x²-y²)(x²+y²) = x⁴-y⁴, and generally (x^2^n – y^2^n)(x^2^n + y^2^n) = x^2^(n+1) – y^2^(n+1). If the product were multiplied by (x-y), it would telescope nicely to x^(2^n) – y^(2^n). However, the question asks about the terms in the expansion of the product itself.

In an examination, 53% students passed in Mathematics, 61% passed in Physics, 60% passed in Chemistry, 24% passed in Mathematics and Physics, 35% in Physics and Chemistry, 27% in Mathematics and Chemistry and 5% in none. The ratio of percentage of passes in Mathematics and Chemistry but not in Physics in relation to the percentage of passes in Physics and Chemistry but not in Mathematics is:

[amp_mcq option1=”7:5″ option2=”5:7″ option3=”4:5″ option4=”5:4″ correct=”option2″]

Let M, P, and C represent the sets of students who passed in Mathematics, Physics, and Chemistry, respectively. We are given the following percentages: |M|=53, |P|=61, |C|=60, |M∩P|=24, |P∩C|=35, |M∩C|=27, |None|=5. The percentage of students who passed in at least one subject is |M∪P∪C| = 100 – |None| = 100 – 5 = 95%. Using the principle of inclusion-exclusion: |M∪P∪C| = |M| + |P| + |C| – |M∩P| – |M∩C| – |P∩C| + |M∩P∩C|. So, 95 = 53 + 61 + 60 – 24 – 27 – 35 + |M∩P∩C|. 95 = 174 – 86 + |M∩P∩C|. 95 = 88 + |M∩P∩C|. Thus, |M∩P∩C| = 95 – 88 = 7%.
Percentage of passes in Mathematics and Chemistry but not in Physics is the region (M∩C) excluding the triple intersection (M∩C∩P). This is |M∩C| – |M∩C∩P| = 27% – 7% = 20%.
Percentage of passes in Physics and Chemistry but not in Mathematics is the region (P∩C) excluding the triple intersection (M∩P∩C). This is |P∩C| – |M∩P∩C| = 35% – 7% = 28%.
The ratio of the percentage of passes in Mathematics and Chemistry but not in Physics to the percentage of passes in Physics and Chemistry but not in Mathematics is 20 : 28. Simplifying the ratio by dividing by 4, we get 5 : 7.

This problem requires using the principle of inclusion-exclusion for three sets and calculating the percentages corresponding to specific regions in a Venn diagram (intersections of two sets excluding the third).

Using a Venn diagram helps visualize the different regions. The percentage of students in M only is |M| – (|M∩P| + |M∩C|) + |M∩P∩C|. Similar calculations can be done for P only, C only, M∩P only, P∩C only, and M∩C only. The sum of percentages in all 7 regions plus the ‘none’ percentage should equal 100%.