Quantitative Aptitude and Reasoning

What is the value of the missing number ?
[Image of a circle with sectors containing numbers: 3,4; 1,5; 3,2; 3,1; 2,3; 2,7; 5,6; 4,5. Inner circle sectors: 5, 4, 5, 6, 7, 8, 3, ?]

[amp_mcq option1=”9″ option2=”7″ option3=”5″ option4=”3″ correct=”option4″]

The puzzle involves pairs of outer numbers and a single inner number in sectors of a circle. Let the outer numbers be `a` and `b`, and the inner number be `c`. We need to find a relationship between `a`, `b`, and `c`.
Observing the first sector with outer numbers (3, 4) and inner number 5, we notice that 3, 4, and 5 form a Pythagorean triple (3² + 4² = 5²). Here, the outer numbers appear to be the legs of a right triangle, and the inner number is the hypotenuse.
Let’s examine the last sector with outer numbers (4, 5) and a missing inner number ?. These numbers (4 and 5) are also part of the 3-4-5 Pythagorean triple. If 4 and 5 were legs, the hypotenuse would be sqrt(4² + 5²) = sqrt(16 + 25) = sqrt(41), which is not an integer option. However, if 5 is the hypotenuse and 4 is one leg, the other leg is calculated by x² + 4² = 5², which gives x² = 25 – 16 = 9, so x = 3.
This suggests a pattern related to the 3-4-5 Pythagorean triple:
– Sector 1 (3,4) -> 5: (leg, leg) -> hypotenuse (3² + 4² = 5²)
– Sector 8 (4,5) -> ?: If 5 is the hypotenuse and 4 is a leg, the missing number (the other leg) is 3 (3² + 4² = 5²).
While this specific pattern might not apply uniformly to *all* sectors using only standard leg/hypotenuse relations (e.g., (1,5)->4, (3,2)->5 don’t form 3-4-5 triples this way), the presence of the (3,4,5) triple in the first sector and the possibility of completing it in the last sector (4,5,?) using one of the options (3) provides the most plausible rule among typical puzzle patterns, especially when other arithmetic operations don’t yield a consistent rule across all sectors. Thus, the missing number is likely 3.

– Examine the relationship between the outer and inner numbers in each sector.
– Look for simple arithmetic operations (addition, subtraction, multiplication, division).
– Look for sequences or patterns across different sectors.
– Consider common mathematical concepts like Pythagorean triples if the numbers suggest it.
– The numbers 3, 4, and 5 in the first sector and 4 and 5 in the last sector strongly suggest a connection to the 3-4-5 Pythagorean triple.

Pythagorean triples are sets of three positive integers a, b, and c, such that a² + b² = c². The most well-known triple is (3, 4, 5). Number puzzles often incorporate basic mathematical properties or sequences. While a perfect, simple pattern across all sectors might not be immediately obvious, the prominent 3-4-5 relationship in the context of the numbers involved makes option 3 the most likely correct answer based on typical puzzle design principles.

If the day before yesterday was Tuesday, when will Saturday be ?

[amp_mcq option1=”Today” option2=”Tomorrow” option3=”Day after tomorrow” option4=”Two days after tomorrow” correct=”option3″]

The statement “the day before yesterday was Tuesday” sets the timeline.
If today is represented as ‘T’, then ‘yesterday’ is T-1 day, and ‘the day before yesterday’ is T-2 days.
So, T-2 days = Tuesday.
To find ‘Today’ (T), we add 2 days to Tuesday: Tuesday + 1 day = Wednesday, Wednesday + 1 day = Thursday.
Therefore, Today is Thursday.
We are asked when Saturday will be.
If Today is Thursday, then Tomorrow (T+1 day) is Friday, and the Day after tomorrow (T+2 days) is Saturday.

– Establish ‘Today’ as the reference point.
– Work backward from the given information (“the day before yesterday was Tuesday”) to determine ‘Today’.
– Work forward from ‘Today’ to find when the target day (Saturday) occurs.

This is a simple logic puzzle involving days of the week. It relies on understanding relative time references like ‘yesterday’ and ‘tomorrow’.

If South-east is called East, North-west is called West, South-west is called South and so-on, what will North be called ?

[amp_mcq option1=”East” option2=”North-east” option3=”North-west” option4=”South” correct=”option1″]

The given changes in direction indicate a consistent rotation. If South-east (SE) is called East (E), North-west (NW) is called West (W), and South-west (SW) is called South (S), let’s determine the type and magnitude of the rotation.
– South-east (SE) to East (E) is a 90-degree clockwise rotation.
– North-west (NW) to West (W) is a 90-degree clockwise rotation.
– South-west (SW) to South (S) is a 90-degree clockwise rotation.
The rule is a 90-degree clockwise rotation of the compass points.
Applying this rule to North (N), a 90-degree clockwise rotation from North results in East (E).

– Identify the pattern of change in the given directions.
– Recognize the rotation (clockwise or counter-clockwise) and its magnitude (degrees).
– Apply the established rotation rule to the direction asked (North).

The standard eight compass points in clockwise order are N, NE, E, SE, S, SW, W, NW. A 90-degree clockwise rotation maps each direction to the one two positions ahead in this sequence (e.g., N -> E, NE -> SE, E -> S, etc.).

A is the smallest positive integer which when divided by 9 and 12 leaves remainder 8. B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5. Which one of the following is the value of A – B ?

[amp_mcq option1=”3″ option2=”2″ option3=”1″ option4=”0″ correct=”option1″]

3

This problem involves finding integers based on remainders when divided by multiple numbers, using the concept of the Least Common Multiple (LCM).

A is the smallest positive integer which when divided by 9 and 12 leaves remainder 8.
This means A $\equiv$ 8 (mod 9) and A $\equiv$ 8 (mod 12).
Equivalently, A – 8 is divisible by both 9 and 12. Therefore, A – 8 must be a multiple of the LCM of 9 and 12.
The prime factorization of 9 is $3^2$. The prime factorization of 12 is $2^2 \times 3$.
LCM(9, 12) = $2^2 \times 3^2 = 4 \times 9 = 36$.
So, A – 8 must be a multiple of 36. A – 8 = 36k for some integer k. A = 36k + 8.
We are looking for the smallest positive integer A.
If k=0, A = 36(0) + 8 = 8. Check: $8 \div 9$ gives remainder 8, $8 \div 12$ gives remainder 8. 8 is a positive integer.
So, A = 8.

B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5.
This means B $\equiv$ 5 (mod 9) and B $\equiv$ 5 (mod 12).
Equivalently, B – 5 is divisible by both 9 and 12. Therefore, B – 5 must be a multiple of LCM(9, 12) = 36.
So, B – 5 = 36m for some integer m. B = 36m + 5.
We are looking for the smallest positive integer B.
If m=0, B = 36(0) + 5 = 5. Check: $5 \div 9$ gives remainder 5, $5 \div 12$ gives remainder 5. 5 is a positive integer.
So, B = 5.

Finally, we need to find the value of A – B:
A – B = 8 – 5 = 3.

The alphabets from A to J are numbered from 0 to 9 respectively. Which one of the following is the value of AGJ – CEG + EDB ?

[amp_mcq option1=”CFE” option2=”DGF” option3=”GFD” option4=”FCE” correct=”option1″]

CFE

Interpret the three-letter codes as numerical values in base 10 based on the provided alphabet-to-number mapping (A=0, B=1, …, J=9), perform the arithmetic operation, and then convert the result back to the letter code.

The mapping is A=0, B=1, C=2, D=3, E=4, F=5, G=6, H=7, I=8, J=9.
AGJ corresponds to the digits 0, 6, 9. As a number, this is 069 = 69.
CEG corresponds to the digits 2, 4, 6. As a number, this is 246.
EDB corresponds to the digits 4, 3, 1. As a number, this is 431.
The expression is AGJ – CEG + EDB, which translates to the arithmetic operation:
$69 – 246 + 431$
$69 – 246 = -177$
$-177 + 431 = 254$
Now, convert the result 254 back into a three-letter code using the mapping:
The digit 2 corresponds to C.
The digit 5 corresponds to F.
The digit 4 corresponds to E.
The resulting code is CFE.

A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length?

[amp_mcq option1=”50%” option2=”56.25%” option3=”62.25%” option4=”75%” correct=”option2″]

56.25%

Assuming the wire is a cylinder and its material is incompressible, the volume of the wire remains constant when it is stretched. The volume of a cylinder is given by $V = \pi R^2 L$.

Let the original length be $L_1 = 6$ m and the original radius be $R_1$. The original volume is $V_1 = \pi R_1^2 L_1 = 6\pi R_1^2$.
The wire is stretched such that its radius is reduced by 20%. The new radius is $R_2 = R_1 – 0.20R_1 = 0.80R_1$.
Let the new length be $L_2$. The new volume is $V_2 = \pi R_2^2 L_2 = \pi (0.80R_1)^2 L_2 = \pi (0.64R_1^2) L_2$.
Since the volume remains constant, $V_1 = V_2$:
$6\pi R_1^2 = \pi (0.64R_1^2) L_2$
Assuming $R_1 > 0$, we can cancel $\pi R_1^2$:
$6 = 0.64 L_2$
$L_2 = \frac{6}{0.64} = \frac{600}{64} = \frac{75}{8} = 9.375$ m.
The increase in length is $L_2 – L_1 = 9.375 – 6 = 3.375$ m.
The percentage increase in length is $\frac{\text{Increase in length}}{\text{Original length}} \times 100\% = \frac{3.375}{6} \times 100\%$.
$\frac{3.375}{6} = \frac{3375}{6000} = \frac{675}{1200} = \frac{135}{240} = \frac{27}{48} = \frac{9}{16}$.
Percentage increase $= \frac{9}{16} \times 100\% = 0.5625 \times 100\% = 56.25\%$.

The number of ways by which 6 distinct balls can be put in 5 distinct boxes are

[amp_mcq option1=”7776″ option2=”15625″ option3=”720″ option4=”120″ correct=”option2″]

15625

This is a problem of placing distinct objects into distinct containers. For each distinct object, there are multiple independent choices of container.

We have 6 distinct balls and 5 distinct boxes.
Consider the balls one by one:
The first ball can be put into any of the 5 distinct boxes (5 choices).
The second ball can be put into any of the 5 distinct boxes (5 choices).
…
The sixth ball can be put into any of the 5 distinct boxes (5 choices).
Since the choice for each ball is independent, the total number of ways is the product of the number of choices for each ball.
Total ways = $5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^6$.
$5^6 = 15625$.
This is the general formula for placing $n$ distinct items into $m$ distinct bins: $m^n$. Here $n=6$ and $m=5$, so $5^6$.

A sum triples in ten years under compound interest at a certain rate of interest, the interest is being compounded annually. In how many years, it would become nine times ?

[amp_mcq option1=”20 years” option2=”30 years” option3=”40 years” option4=”50 years” correct=”option1″]

20 years

Under compound interest, if a principal amount becomes ‘x’ times in ‘T’ years, it will become ‘$x^n$’ times in ‘nT’ years, provided the interest rate and compounding frequency remain constant.

Let the principal be P and the annual interest rate be r. The amount A after t years is given by $A = P(1+r)^t$.
We are given that the sum triples in ten years:
$3P = P(1+r)^{10} \implies (1+r)^{10} = 3$.
We want to find the number of years ‘t’ it takes for the sum to become nine times the principal:
$9P = P(1+r)^t \implies (1+r)^t = 9$.
Since $9 = 3^2$, we can substitute the value of 3 from the first equation:
$(1+r)^t = ( (1+r)^{10} )^2$
$(1+r)^t = (1+r)^{10 \times 2}$
$(1+r)^t = (1+r)^{20}$
Equating the exponents, we get $t = 20$.
It would take 20 years for the sum to become nine times.

The average age of the boys in a class is 12 years. The average age of the girls in the class is 11 years. There are 50% more girls than boys in the class. Which one of the following is the average age of the class (in years)?

[amp_mcq option1=”11.2 years” option2=”11.4 years” option3=”11.6 years” option4=”11.8 years” correct=”option2″]

11.4 years

The average age of the class is the weighted average of the average ages of boys and girls, where the weights are the number of boys and girls.

Let $N_B$ be the number of boys and $N_G$ be the number of girls.
Average age of boys ($A_B$) = 12 years.
Average age of girls ($A_G$) = 11 years.
$N_G$ is 50% more than $N_B$, so $N_G = N_B + 0.50 N_B = 1.5 N_B$.
The total sum of ages of boys is $N_B \times A_B = 12N_B$.
The total sum of ages of girls is $N_G \times A_G = 11N_G = 11(1.5N_B) = 16.5N_B$.
The total sum of ages in the class is $12N_B + 16.5N_B = 28.5N_B$.
The total number of students is $N_B + N_G = N_B + 1.5N_B = 2.5N_B$.
The average age of the class is $\frac{\text{Total sum of ages}}{\text{Total number of students}} = \frac{28.5N_B}{2.5N_B} = \frac{28.5}{2.5} = \frac{285}{25}$.
$\frac{285}{25} = \frac{57 \times 5}{5 \times 5} = \frac{57}{5} = 11.4$.
The average age of the class is 11.4 years.

The cost of gold varies directly as the cube of its weight. A gold piece weighing 20 decigram costs ₹1,000. If it is broken into two pieces whose weights are in the ratio 2 : 3, then what is the profit or loss incurred?

[amp_mcq option1=”₹280 profit” option2=”₹280 loss” option3=”₹720 profit” option4=”₹720 loss” correct=”option4″]

₹720 loss

The cost of gold is directly proportional to the cube of its weight ($C \propto W^3$). When the piece is broken, its total weight remains the same, but the combined cost of the pieces is calculated based on the cubed weights of the individual pieces.

Let the cost be C and the weight be W. Given $C = kW^3$ for some constant k.
For the original piece: $W_1 = 20$ dg, $C_1 = ₹1000$.
$1000 = k \times (20)^3 = k \times 8000$
$k = \frac{1000}{8000} = \frac{1}{8}$.
The piece is broken into two pieces with weights in the ratio 2:3. The total weight is 20 dg. The weights of the two pieces are $\frac{2}{2+3} \times 20 = \frac{2}{5} \times 20 = 8$ dg and $\frac{3}{2+3} \times 20 = \frac{3}{5} \times 20 = 12$ dg.
Let the costs of the two pieces be $C_2$ and $C_3$.
$C_2 = k \times (8)^3 = \frac{1}{8} \times 512 = 64$.
$C_3 = k \times (12)^3 = \frac{1}{8} \times 1728 = 216$.
The total value of the broken pieces is $C_2 + C_3 = 64 + 216 = ₹280$.
The original cost was ₹1,000. The value after breaking is ₹280.
The profit or loss is $280 – 1000 = -720$. This is a loss of ₹720.