Quantitative Aptitude and Reasoning

If the product of $n$ positive numbers is unity, then their sum is

[amp_mcq option1=”a positive integer” option2=”divisible by $n$” option3=”equal to $n + \frac{1}{n}$” option4=”never less than $n$” correct=”option4″]

The correct answer is D) never less than n.

Let the n positive numbers be $x_1, x_2, …, x_n$.
We are given that their product is unity: $x_1 \times x_2 \times \dots \times x_n = 1$.
We want to determine the property of their sum: $S = x_1 + x_2 + \dots + x_n$.
According to the Arithmetic Mean – Geometric Mean (AM-GM) inequality, for a set of non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean. Since the numbers are positive, this inequality applies:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$
Substitute the given product into the inequality:
$\frac{S}{n} \ge \sqrt[n]{1}$
$\frac{S}{n} \ge 1$
$S \ge n$
The sum of the n positive numbers whose product is unity is always greater than or equal to n.
Equality holds if and only if all the numbers are equal. If $x_1 = x_2 = \dots = x_n = x$, and their product is 1, then $x^n = 1$. Since they are positive, $x$ must be 1. In this case, the sum is $n \times 1 = n$. If the numbers are not all equal, the sum is strictly greater than n. Therefore, the sum is never less than n.

The AM-GM inequality is a fundamental concept in mathematics often used to find minimum or maximum values or to prove inequalities. It states that for non-negative numbers $a_1, a_2, \dots, a_n$, $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}$, with equality holding if and only if $a_1 = a_2 = \dots = a_n$.

5% of income of A is equal to 15% of income of B and 10% income of B is equal to 20% of income of C. If the income of C is Rs. 2,000, then what is the total income of A, B and C ?

[amp_mcq option1=”Rs. 20,000″ option2=”Rs. 18,000″ option3=”Rs. 14,000″ option4=”Rs. 6,000″ correct=”option2″]

The correct answer is B) Rs. 18,000.

We are given the following relationships between the incomes of A, B, and C:
1) 5% of A = 15% of B
(5/100) * A = (15/100) * B
0.05 A = 0.15 B
Divide both sides by 0.05: A = (0.15 / 0.05) B => A = 3B

2) 10% of B = 20% of C
(10/100) * B = (20/100) * C
0.10 B = 0.20 C
Divide both sides by 0.10: B = (0.20 / 0.10) C => B = 2C

We are given the income of C is Rs. 2,000.
Substitute the value of C into the equation for B:
B = 2 * C = 2 * 2000 = Rs. 4,000.

Substitute the value of B into the equation for A:
A = 3 * B = 3 * 4000 = Rs. 12,000.

The total income of A, B, and C is the sum of their individual incomes:
Total Income = A + B + C = 12000 + 4000 + 2000 = Rs. 18,000.

This problem requires setting up and solving a system of linear equations based on the given percentage relationships between the incomes. Starting with the known value (income of C) allows for a straightforward calculation of the other incomes.

In an examination, 25% of the candidates failed in Mathematics and 12% failed in English. If 10% of the candidates failed in both the subjects and 292 candidates passed in both the subjects, which one of the following is the number of total candidates appeared in the examination ?

[amp_mcq option1=”300″ option2=”400″ option3=”460″ option4=”500″ correct=”option2″]

The correct answer is B) 400.

Let M be the percentage of candidates who failed in Mathematics, E be the percentage who failed in English, and B be the percentage who failed in both.
Given: M = 25%, E = 12%, B = 10%.
The percentage of candidates who failed in at least one subject is given by the formula:
P(M U E) = P(M) + P(E) – P(M ∩ E)
Percentage failed in at least one subject = 25% + 12% – 10% = 37% – 10% = 27%.
The percentage of candidates who passed in both subjects is the remaining percentage:
Percentage passed in both = 100% – Percentage failed in at least one subject = 100% – 27% = 73%.
We are given that 292 candidates passed in both subjects.
Let T be the total number of candidates.
So, 73% of T = 292
(73 / 100) * T = 292
T = (292 * 100) / 73
T = 29200 / 73
Dividing 29200 by 73: 292 / 73 = 4 (since 73 * 4 = 292).
So, 29200 / 73 = 400.
The total number of candidates is 400.

This problem uses the principle of inclusion-exclusion for calculating the union of two sets (failures in Mathematics and English). The complement of the set of people failing in at least one subject is the set of people passing in both subjects.

If all students are boys and all boys are dancers, then which one of the following statements is definitely true?

[amp_mcq option1=”All dancers are boys” option2=”All boys are students” option3=”All dancers are students” option4=”All students are dancers” correct=”option4″]

The given statements are:
1. All students are boys. (If someone is a student, then that person is a boy. Student -> Boy)
2. All boys are dancers. (If someone is a boy, then that person is a dancer. Boy -> Dancer)
We can combine these two conditional statements using the principle of transitivity:
If Student -> Boy and Boy -> Dancer, then it logically follows that Student -> Dancer.
This means, “If someone is a student, then that person is a dancer,” which can be rephrased as “All students are dancers.”
Let’s examine the options based on this deduction:
A) All dancers are boys (Dancer -> Boy). This is the converse of “All boys are dancers” and is not necessarily true.
B) All boys are students (Boy -> Student). This is the converse of “All students are boys” and is not necessarily true.
C) All dancers are students (Dancer -> Student). This is the converse of the derived conclusion “All students are dancers” and is not necessarily true.
D) All students are dancers (Student -> Dancer). This is the direct logical conclusion derived from the premises.

This problem requires understanding logical implication and how to chain conditional statements (syllogism). If P implies Q, and Q implies R, then P implies R.

Using Venn diagrams can also help visualize this. Draw a circle for ‘Students’ inside a circle for ‘Boys’. Then draw the ‘Boys’ circle inside a larger circle for ‘Dancers’. It becomes clear that the ‘Students’ circle is entirely contained within the ‘Dancers’ circle, meaning all students are dancers. The reverse relationships (e.g., all dancers are boys) are not guaranteed, as there can be dancers who are not boys (and thus not students).

A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If 13Y7 is divisible by 11, then what is the value of (X+Y) ?

[amp_mcq option1=”15″ option2=”12″ option3=”11″ option4=”10″ correct=”option4″]

The addition is 4X3 + 984 = 13Y7.
Let’s perform the addition column by column:
Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place.
Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10.
Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place.
So, X + 8 = Y, where Y is a digit between 0 and 9.
The possible values for X (a digit between 0 and 9) that result in a single digit Y are:
If X=0, Y = 0+8 = 8. The sum is 1387.
If X=1, Y = 1+8 = 9. The sum is 1397.
If X=2, Y = 2+8 = 10. Y cannot be 10 as it’s a single digit.
So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397).

We are given that 13Y7 is divisible by 11.
Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11.
For 13Y7: +7 – Y + 3 – 1 = 9 – Y.
For 13Y7 to be divisible by 11, 9 – Y must be a multiple of 11 (0, 11, -11, etc.).
Since Y is a digit (0-9), 9 – Y can range from 9-0=9 to 9-9=0.
The only multiple of 11 in this range is 0.
So, 9 – Y = 0, which means Y = 9.

Now we use the relation X + 8 = Y. Substitute Y=9:
X + 8 = 9
X = 9 – 8 = 1.
So, X=1 and Y=9.

Let’s verify: 413 + 984 = 1397.
Is 1397 divisible by 11? 7 – 9 + 3 – 1 = 0. Yes, it is.

The value requested is (X + Y).
X + Y = 1 + 9 = 10.

This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of the digits at odd positions (from the right) and the sum of the digits at even positions (from the right) is divisible by 11 (i.e., 0, 11, -11, 22, etc.).

Which one of the following is the smallest number by which 2880 must be divided in order to make it a perfect square ?

[amp_mcq option1=”3″ option2=”4″ option3=”5″ option4=”6″ correct=”option3″]

To find the smallest number by which 2880 must be divided to make it a perfect square, we first find the prime factorization of 2880.
2880 = 288 * 10 = (144 * 2) * (2 * 5) = (12² * 2) * (2 * 5) = ((2² * 3)²) * 2² * 5 = (2⁴ * 3²) * 2² * 5 = 2⁶ * 3² * 5¹
For a number to be a perfect square, the exponents of all its prime factors must be even. In the factorization 2⁶ * 3² * 5¹, the exponents are 6 (even), 2 (even), and 1 (odd).
To make the exponent of 5 even (which is 1), we must divide by 5 raised to the power of its odd exponent, i.e., 5¹.
So, we must divide 2880 by 5.
2880 / 5 = 576.
Let’s check if 576 is a perfect square: 576 = 24 * 24 = 24². Yes, it is.
Therefore, the smallest number by which 2880 must be divided is 5.

A number is a perfect square if and only if the exponents of all prime factors in its prime factorization are even. To make a number a perfect square by division, divide by the product of prime factors with odd exponents, each raised to the power of their odd exponent.

If the question asked for the smallest number to *multiply* by, you would multiply by the product of the prime factors with odd exponents, each raised to the power needed to make the exponent even (which is the odd exponent itself). In this case, multiply by 5¹.

One year ago, a father was four times as old as his son. After six years his age exceeds twice his son’s age by 9 years. The ratio of their present age is

[amp_mcq option1=”9 : 2″ option2=”11 : 3″ option3=”12 : 5″ option4=”13 : 4″ correct=”option2″]

The ratio of their present age is 11 : 3.

– Let the present age of the son be $S$ years and the present age of the father be $F$ years.
– One year ago, the son’s age was $S-1$ and the father’s age was $F-1$.
– According to the first condition: $F-1 = 4(S-1)$. This simplifies to $F-1 = 4S – 4$, so $F = 4S – 3$ (Equation 1).
– After six years, the son’s age will be $S+6$ and the father’s age will be $F+6$.
– According to the second condition: The father’s age ($F+6$) exceeds twice his son’s age ($2(S+6)$) by 9 years. So, $F+6 = 2(S+6) + 9$.
– This simplifies to $F+6 = 2S + 12 + 9$, which is $F+6 = 2S + 21$. So, $F = 2S + 15$ (Equation 2).
– Now we have a system of two linear equations for $F$ and $S$:
1) $F = 4S – 3$
2) $F = 2S + 15$
– Equating the expressions for $F$: $4S – 3 = 2S + 15$.
– Subtract $2S$ from both sides: $2S – 3 = 15$.
– Add 3 to both sides: $2S = 18$.
– Divide by 2: $S = 9$. The son’s present age is 9 years.
– Substitute $S=9$ into Equation 1 (or Equation 2) to find $F$:
$F = 4(9) – 3 = 36 – 3 = 33$. The father’s present age is 33 years.
– The ratio of their present age (Father : Son) is $F : S = 33 : 9$.
– This ratio can be simplified by dividing both numbers by their greatest common divisor, which is 3.
$33 \div 3 = 11$
$9 \div 3 = 3$
– The simplified ratio is 11 : 3.

It’s always a good idea to check the answer with the original conditions.
Present ages: Father = 33, Son = 9.
One year ago: Father = 32, Son = 8. Is 32 four times 8? Yes, $32 = 4 \times 8$. (Condition 1 satisfied).
After six years: Father = $33+6 = 39$, Son = $9+6 = 15$. Is 39 equal to twice the son’s age plus 9? $2 \times 15 + 9 = 30 + 9 = 39$. Yes. (Condition 2 satisfied).
The calculated ages satisfy both conditions.

Two pipes A and B can fill a tank in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time, B should be closed so that the tank is full in 9 minutes ?

[amp_mcq option1=”3 ½ min” option2=”4 min” option3=”4 ½ min” option4=”4 ¾ min” correct=”option2″]

Pipe B should be closed after 4 minutes so that the tank is full in 9 minutes.

– Pipe A fills the tank in 12 minutes, so its filling rate is $1/12$ of the tank per minute.
– Pipe B fills the tank in 16 minutes, so its filling rate is $1/16$ of the tank per minute.
– Let $t$ be the time (in minutes) for which both pipes are open together.
– After time $t$, pipe B is closed, and only pipe A continues to fill the tank for the remaining time.
– The total time to fill the tank is 9 minutes. So, pipe A works for the entire 9 minutes. Pipe B works only for the first $t$ minutes.
– Amount filled by A in 9 minutes = Rate of A $\times$ Time A worked = $(1/12) \times 9 = 9/12 = 3/4$ of the tank.
– Amount filled by B in $t$ minutes = Rate of B $\times$ Time B worked = $(1/16) \times t = t/16$ of the tank.
– The total amount filled is the sum of the amounts filled by A and B, which is 1 full tank.
– So, $(3/4) + (t/16) = 1$.
– To solve for $t$, multiply the entire equation by the least common multiple of 4 and 16, which is 16:
$16 \times (3/4) + 16 \times (t/16) = 16 \times 1$
$4 \times 3 + t = 16$
$12 + t = 16$
$t = 16 – 12 = 4$.
– Therefore, pipe B should be closed after 4 minutes.

Alternatively, we can set up the equation based on the duration both pipes work together and the duration only A works. Let B be closed after $t$ minutes. Both A and B work for $t$ minutes, and A works alone for $(9-t)$ minutes.
Work done by A and B together in $t$ minutes = $(1/12 + 1/16) \times t = (4/48 + 3/48) \times t = (7/48)t$.
Work done by A alone in $(9-t)$ minutes = $(1/12) \times (9-t)$.
Total work = $(7/48)t + (1/12)(9-t) = 1$.
Multiply by 48: $7t + 4(9-t) = 48 \Rightarrow 7t + 36 – 4t = 48 \Rightarrow 3t + 36 = 48 \Rightarrow 3t = 12 \Rightarrow t = 4$.
Both methods yield the same result.

The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area ?

[amp_mcq option1=”37.5 %” option2=”60%” option3=”75%” option4=”120%” correct=”option1″]

The width would have to be decreased by 37.5% to maintain the same area.

– Let the original length be $L$ and original width be $W$. The original area is $A = L \times W$.
– The length is increased by 60%, so the new length $L’$ is $L + 0.60L = 1.60L$.
– Let the new width be $W’$. The new area $A’ = L’ \times W’ = 1.60L \times W’$.
– To maintain the same area, $A’ = A$, so $1.60L \times W’ = L \times W$.
– We can solve for $W’$: $W’ = \frac{L \times W}{1.60L} = \frac{W}{1.60} = \frac{W}{8/5} = \frac{5}{8}W$.
– The decrease in width is $W – W’ = W – \frac{5}{8}W = \frac{3}{8}W$.
– The percentage decrease in width is $\frac{\text{Decrease in width}}{\text{Original width}} \times 100\% = \frac{(3/8)W}{W} \times 100\% = \frac{3}{8} \times 100\%$.
– $\frac{3}{8} = 0.375$, so the percentage decrease is $0.375 \times 100\% = 37.5\%$.

This problem demonstrates the inverse relationship between dimensions when the area is kept constant. If one dimension is increased, the other must be decreased proportionally to maintain the same area. The percentage change in one dimension results in a different percentage change in the other dimension for a constant area, especially when the changes are expressed relative to the original values.

Two men set out at the same time to walk towards each other from points A and B, 72 km apart. The first man walks at the speed of 4 kmph while the second walks 2 km in the first hour, 2½ km in the second hour, 3 km in the third hour, and so on. The two men will meet

[amp_mcq option1=”in 8 hours” option2=”nearer to A than B” option3=”nearer to B than A” option4=”midway between A and B” correct=”option4″]

The two men will meet midway between A and B.

Let the time taken for them to meet be T hours. The distance between A and B is 72 km.
Man 1 starts from A at a constant speed of 4 kmph. In T hours, Man 1 covers a distance of 4T km.
Man 2 starts from B. In the first hour, he covers 2 km. In the second hour, 2.5 km. In the third, 3 km, and so on. This is an arithmetic progression of distances covered per hour with first term a = 2 and common difference d = 0.5.
The distance covered by Man 2 in T hours is the sum of the first T terms of this AP:
Sum = (T/2) * [2a + (T-1)d] = (T/2) * [2*2 + (T-1)*0.5] = (T/2) * [4 + 0.5T – 0.5] = (T/2) * [3.5 + 0.5T].
When they meet, the sum of the distances covered by both men equals the total distance:
Distance by Man 1 + Distance by Man 2 = 72
4T + (T/2) * (3.5 + 0.5T) = 72
4T + 1.75T + 0.25T^2 = 72
0.25T^2 + 5.75T – 72 = 0
Multiplying by 4 to clear decimals:
T^2 + 23T – 288 = 0
Using the quadratic formula T = [-b ± sqrt(b^2 – 4ac)] / 2a:
T = [-23 ± sqrt(23^2 – 4*1*(-288))] / 2*1
T = [-23 ± sqrt(529 + 1152)] / 2
T = [-23 ± sqrt(1681)] / 2
Since sqrt(1681) = 41 (as 40^2=1600, 41^2=1681), and time must be positive:
T = (-23 + 41) / 2 = 18 / 2 = 9 hours.
They meet after 9 hours.
Distance covered by Man 1 in 9 hours = 4 kmph * 9 hours = 36 km from A.
Distance covered by Man 2 in 9 hours = Sum of 9 terms of AP (a=2, d=0.5) = (9/2) * [2*2 + (9-1)*0.5] = 4.5 * [4 + 8*0.5] = 4.5 * [4 + 4] = 4.5 * 8 = 36 km from B.
Since they meet after covering 36 km from A and 36 km from B, which adds up to the total distance of 72 km, they meet exactly midway between A and B.

Option A states they meet in 8 hours, which was disproven by the calculation (at 8 hours, the total distance covered is 32km + 30km = 62km, less than 72km). Options B and C are incorrect because meeting midway means they are equidistant from A and B. Therefore, only option D accurately describes the meeting point based on the given data. The problem likely includes option A as a distractor based on a potential miscalculation or a similar problem with different parameters.