Numeric

Question 1

Solve: $10 + 10 \times 100 \div 10$

110

Answer:

Explanation:

According to the BODMAS rule, division is performed first: $100 \div 10 = 10$. Next, we perform multiplication: $10 \times 10 = 100$. Finally, the addition is performed: $10 + 100 = 110$.

Question 2

Solve: $10 + 10 \div 100 \times 10$

11

Answer:

Explanation:

Applying the BODMAS rule, division is completed first: $10 \div 100 = 0.1$. Then multiplication is performed: $0.1 \times 10 = 1$. Finally, performing the addition yields: $10 + 1 = 11$.

Question 3

Simplify: $14 + (8 - 2 \times 3)$

16

Answer:

Explanation:

By BODMAS, inside the parentheses, we prioritize multiplication over subtraction: $2 \times 3 = 6$. The bracket expression simplifies to: $8 - 6 = 2$. Adding this result to 14 yields: $14 + 2 = 16$.

Question 4

Simplify: $100 \div 5 \div 5 \div 4$

1

Answer:

Explanation:

When executing consecutive divisions, the operations must be performed from left to right. First, $100 \div 5 = 20$. Then, $20 \div 5 = 4$. Finally, $4 \div 4 = 1$.

Question 5

Simplify the expression: $8 \div 8 \text{ of } 8 + 8 / 8 \div 8 \times 8 + 8$

65/128

Answer:

Explanation:

To solve this expression, we treat the forward slash (/) as the primary division bar separating the numerator and the denominator. The numerator is $8 \div 8 \text{ of } 8 + 8$. Evaluating 'of' first gives $8 \text{ of } 8 = 64$, making the numerator $8 \div 64 + 8 = \frac{1}{8} + 8 = \frac{65}{8}$. The denominator is $8 \div 8 \times 8 + 8$, which simplifies from left to right as $1 \times 8 + 8 = 16$. Dividing the numerator by the denominator gives $\frac{65}{8} \div 16 = \frac{65}{128}$.

Question 6

Simplify the expression: $18 - [6 - \{4 - (8 - 6 - 3)\}]$

17

Answer:

Explanation:

Using the BODMAS rule, we simplify the innermost parentheses first: $(8 - 6 - 3) = 2 - 3 = -1$. Next, we evaluate the expression inside the curly braces: $\{4 - (-1)\} = 4 + 1 = 5$. Now we simplify the expression inside the square brackets: $[6 - 5] = 1$. Finally, subtracting this from the initial term gives $18 - 1 = 17$.

Question 7

Which of the following is the greatest of $\frac{2}{3}$, $\frac{4}{7}$, $\frac{9}{20}$, $\frac{8}{13}$?

2/3

Answer:

Explanation:

Convert each fraction into its decimal representation to easily compare them: $\frac{2}{3} \approx 0.666$, $\frac{4}{7} \approx 0.571$, $\frac{9}{20} = 0.450$, and $\frac{8}{13} \approx 0.615$. Comparing these decimal values, $0.666$ is the largest. Therefore, $\frac{2}{3}$ is the greatest fraction.

Question 8

Which of the following are in ascending order?

2/9, 1/4, 3/8, 5/7, 10/11

Answer:

Explanation:

Convert the fractions to decimal values: $\frac{2}{9} \approx 0.222$, $\frac{1}{4} = 0.250$, $\frac{3}{8} = 0.375$, $\frac{5}{7} \approx 0.714$, and $\frac{10}{11} \approx 0.909$. Arranging them from smallest to largest yields $0.222 < 0.250 < 0.375 < 0.714 < 0.909$. This corresponds to the sequence $\frac{2}{9}, \frac{1}{4}, \frac{3}{8}, \frac{5}{7}, \frac{10}{11}$.

Question 9

Which is the largest number among the following?

(1/3)^{-4}

Answer:

Explanation:

A negative exponent means we take the reciprocal of the base and make the exponent positive. Evaluating each option, we get: $(1/2)^{-6} = 2^6 = 64$, $(1/4)^{-3} = 4^3 = 64$, $(1/3)^{-4} = 3^4 = 81$, and $(1/6)^{-2} = 6^2 = 36$. Comparing these values, 81 is the largest. Therefore, $(1/3)^{-4}$ is the greatest number among the choices.

Question 10

Which number amongst $2^{40}$, $3^{21}$, $4^{18}$ and $8^{12}$ is the smallest?

3^{21}

Answer:

Explanation:

First, express the bases in prime form where possible: $4^{18} = (2^2)^{18} = 2^{36}$ and $8^{12} = (2^3)^{12} = 2^{36}$. This reduces the options to comparing $2^{40}$, $2^{36}$, and $3^{21}$. Since $2^{36}$ is smaller than $2^{40}$, we compare $2^{36}$ and $3^{21}$. We can rewrite $2^{36} = (2^{12})^3 = 4096^3$ and $3^{21} = (3^7)^3 = 2187^3$. Since $2187 < 4096$, $3^{21}$ is the smallest number.

Question 11

Which number amongst $3^{50}$, $4^{40}$, $5^{30}$ and $6^{20}$ is the greatest?

4^{40}

Answer:

Explanation:

We can express each number with a common exponent of $10$. By rewriting, we get: $3^{50} = (3^5)^{10} = 243^{10}$, $4^{40} = (4^4)^{10} = 256^{10}$, $5^{30} = (5^3)^{10} = 125^{10}$, and $6^{20} = (6^2)^{10} = 36^{10}$. Comparing the bases, $256 > 243 > 125 > 36$. Therefore, $4^{40}$ is the greatest.

Question 12

Which number amongst $25^{50}$, $4^{200}$, $30^{100}$ and $2^{567}$ is the greatest?

2^{567}

Answer:

Explanation:

First, we can write $25^{50}$ and $4^{200} = 256^{50}$ and $30^{100} = 900^{50}$. Thus, $30^{100} > 4^{200} > 25^{50}$. Next, we compare $2^{567}$ with $30^{100}$. We can express $2^{567}$ as $2^{67} \times (2^5)^{100} = 2^{67} \times 32^{100}$. Since $32^{100}$ is strictly larger than $30^{100}$ and $2^{67} > 1$, $2^{567}$ is by far the largest of all the given numbers.

Question 13

What least number should be added to 2205 to make it a perfect square?

4

Answer:

Explanation:

To find the number to be added, we find the next perfect square above 2205. The square root of 2205 is between 46 and 47 (since $46^2 = 2116$ and $47^2 = 2209$). The next perfect square is $2209$. Therefore, the least number to be added is $2209 - 2205 = 4$.

Question 14

What least number should 6300 be multiplied with to make it a perfect square?

7

Answer:

Explanation:

First, find the prime factorization of 6300: $6300 = 63 \times 100 = 7 \times 9 \times 100 = 2^2 \times 3^2 \times 5^2 \times 7^1$. For a number to be a perfect square, all the prime factors must have even exponents. Since the exponent of 7 is 1 (odd), we must multiply 6300 by 7 to make its exponent even ($7^2$).

Question 15

The sum and the product of two numbers are 25 and 144 respectively. The difference of the numbers is?

7

Answer:

Explanation:

Let the two numbers be $x$ and $y$. We are given $x + y = 25$ and $xy = 144$. We can find their difference using the algebraic identity: $(x - y)^2 = (x + y)^2 - 4xy$. Substituting the values, $(x - y)^2 = 25^2 - 4(144) = 625 - 576 = 49$. Taking the square root gives $|x - y| = 7$.

Question 16

The cost of three chairs and two tables is Rs. 2000. If one chair less and one more table are purchased it would cost Rs. 500 more. What is the cost of a chair and a table?

Rs. 900

Answer:

Explanation:

Let $C$ be the cost of a chair and $T$ be the cost of a table. From the problem, we have: $3C + 2T = 2000$ and $2C + 3T = 2500$ (one chair less, one table more). Adding both equations together, we get $5C + 5T = 4500$. Dividing this equation by 5 gives $C + T = 900$. Thus, the total cost of a chair and a table is Rs. 900.

Question 17

The total cost of five apples and six bananas is Rs. 52. If the cost of an apple is Rs. 6 more than that of a banana, what is the cost of six apples and five bananas?

Rs. 58

Answer:

Explanation:

Let $A$ and $B$ be the costs of an apple and a banana, respectively. We have $5A + 6B = 52$ and $A = B + 6$. Substituting the value of $A$ into the first equation: $5(B + 6) + 6B = 52 \implies 11B + 30 = 52 \implies 11B = 22 \implies B = 2$. Therefore, $A = 2 + 6 = 8$. The cost of six apples and five bananas is $6(8) + 5(2) = 48 + 10 = 58$.

Question 18

Some sweets were to be distributed equally among 175 students of a school. But due to absence of 35 students, each child got 4 more sweets. How many sweets were distributed?

2800

Answer:

Explanation:

Let $S$ be the total sweets and $N$ be the original share of sweets per student. Then, $S = 175N$. Since 35 students were absent, only 140 students attended. Their new share was $N + 4$ sweets per student, so $S = 140(N + 4)$. Setting the expressions equal: $175N = 140N + 560 \implies 35N = 560 \implies N = 16$. The total sweets distributed is $175 \times 16 = 2800$.

Question 19

A general, wishing to draw his 2509 soldiers in the form of a solid square, found that he had 9 men over. Find the number of soldiers in the front row.

50

Answer:

Explanation:

Subtracting the 9 leftover men, the general forms a solid square with $2509 - 9 = 2500$ soldiers. A solid square consists of an equal number of rows and columns. To find the number of soldiers in the front row, we take the square root of 2500: $\sqrt{2500} = 50$. Therefore, there are 50 soldiers in the front row.

Question 20

Ajay and Vijay have some marbles with them. Ajay told Vijay, 'If you give me x marbles, both of us will have an equal number of marbles.' Vijay then told Ajay, 'If you give me twice as many marbles, I will have 30 more marbles than you would.' Find the value of x.

5

Answer:

Explanation:

Let Ajay have $A$ marbles and Vijay have $V$ marbles. The first statement gives $A + x = V - x \implies V - A = 2x$. The second statement gives $V + 2x = (A - 2x) + 30 \implies V - A + 4x = 30$. Substituting $V - A = 2x$ into the second equation: $2x + 4x = 30 \implies 6x = 30 \implies x = 5$.

Question 21

What is the total number of digits printed, if a book containing 150 pages is numbered from 1 to 150?

342

Answer:

Explanation:

We count the digits based on the number of digits in the page numbers. For single-digit pages (1 to 9), there are $9 \times 1 = 9$ digits. For two-digit pages (10 to 99), there are $90 \times 2 = 180$ digits. For three-digit pages (100 to 150), there are $51 \times 3 = 153$ digits. Summing these values gives the total number of digits: $9 + 180 + 153 = 342$.

Question 22

How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place?

90

Answer:

Explanation:

Numbers between 99 and 1000 are 3-digit numbers from 100 to 999 of the form $abc$. For the hundreds place ($a$), we can choose any digit from 1 to 9 (9 options). For the tens place ($b$), we can choose any digit from 0 to 9 (10 options). The units place ($c$) must be 8 (1 option). Applying the multiplication rule, there are $9 \times 10 \times 1 = 90$ such numbers.

Question 23

How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?

12

Answer:

Explanation:

First, list all integers between 1 and 100 containing the digit 4: 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, and 94 (19 numbers in total). Out of these, we remove numbers divisible by 4: 4, 24, 40, 44, 48, 64, and 84 (7 numbers). The remaining numbers not divisible by 4 are: 14, 34, 41, 42, 43, 45, 46, 47, 49, 54, 74, and 94. Counting these gives 12 integers.

Question 24

What is the number of fives used in numbering a 260-page book?

56

Answer:

Explanation:

From page 1 to 99, 5 is used 20 times (10 in units places and 10 in tens places). Similarly, from page 100 to 199, 5 is used 20 times. From page 200 to 260, 5 is used in units places: 205, 215, 225, 235, 245, 255 (6 times) and in tens places: 250 to 259 (10 times), giving 16 times. The total number of times 5 is used is $20 + 20 + 16 = 56$.

Question 25

How many numbers are there between 100 and 300 which either begin with or end with 2?

110

Answer:

Explanation:

Let $A$ be the set of numbers beginning with 2 (200 to 299), so $|A| = 100$. Let $B$ be the set of numbers ending with 2 between 100 and 300 (102, 112, ..., 292), so $|B| = 20$. The intersection $A \cap B$ contains numbers that start and end with 2 (202, 212, ..., 292), so $|A \cap B| = 10$. Using the Principle of Inclusion-Exclusion, the count is $|A \cup B| = 100 + 20 - 10 = 110$.

Question 26

The difference between a two-digit number and the number obtained by interchanging its digits is 54. The difference between the digits is?

6

Answer:

Explanation:

Let the two-digit number be $10x + y$. The number with interchanged digits is $10y + x$. The difference between the two numbers is $(10x + y) - (10y + x) = 9(x - y)$. We are given $9(x - y) = 54$, which simplifies to $x - y = 6$. Therefore, the difference between the digits is 6.

Question 27

If the sum of a two-digit number and the number obtained by interchanging the digits is 88, then find the sum of the digits of the number.

8

Answer:

Explanation:

Let the two-digit number be $10x + y$. The interchanged number is $10y + x$. The sum of these numbers is $(10x + y) + (10y + x) = 11(x + y)$. We are given $11(x + y) = 88$. Dividing both sides of the equation by 11 gives $x + y = 8$. Thus, the sum of the digits of the number is 8.

Question 28

There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?

None of the above

Answer:

Explanation:

Let the number be $10x + y$. The reversed number is $10y + x$, and the difference is $9|x - y| = 27 \implies |x - y| = 3$. The digit pairs can be (3,0), (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) where $x > y$ (7 numbers) and (1,4), (2,5), (3,6), (4,7), (5,8), (6,9) where $x < y$ (6 numbers). This gives a total of $7 + 6 = 13$ possible numbers. Since 13 is not among the options, 'None of the above' is correct.

Question 29

A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, then the difference between the first and last digits is:

2

Answer:

Explanation:

Let the three-digit number be $x0y$, which represents $100x + y$. We are given $x + y = 4$. Reversing the digits gives the number $y0x$, which is $100y + x$. From the problem, $(100y + x) - (100x + y) = 198 \implies 99(y - x) = 198 \implies y - x = 2$. Therefore, the difference between the first and last digits is 2.

Question 30

Let 'p' be a two-digit number and 'q' be the number consisting of same digits written in reverse order. If $p \times q = 2430$, then what is the difference between p and q?

9

Answer:

Explanation:

Let $p = 10a + b$ and $q = 10b + a$. We are given $p \times q = 2430$. Factoring 2430 into two reversed 2-digit numbers gives $45 \times 54 = 2430$. Thus, the two numbers are 45 and 54. The difference between $p$ and $q$ is $|54 - 45| = 9$.

Question 31

Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D?

3

Answer:

Explanation:

Aligning column-wise: $C+F$ must end in 2, and since $C,F > 3$, $C+F = 12$ (carry 1). For tens: $B+2+1 \text{ (carry)} = 10 \implies B = 7$ (carry 1). For hundreds: $3+E+1 \text{ (carry)} = 9 \implies E = 5$ (no carry). For thousands: $A+D = 15$. The distinct digits greater than 3 are $\{4,5,6,7,8,9\}$. Since $B=7, E=5$, remaining are $\{4,6,8,9\}$. To get $C+F=12$, we must choose $\{4,8\}$. The remaining digits are $\{6,9\}$, so $\{A,D\} = \{6,9\}$. The difference $|A - D| = |9 - 6| = 3$.

Question 32

One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers?

7, 8

Answer:

Explanation:

Let the booklet have $n$ pages. The original sum of all pages is $\frac{n(n+1)}{2}$. Let the torn consecutive page numbers be $k$ and $k+1$. The remaining sum is $\frac{n(n+1)}{2} - (2k + 1) = 195$. If $n = 20$, the sum of all pages is $\frac{20 \times 21}{2} = 210$. The sum of the torn page is $210 - 195 = 15 \implies 2k + 1 = 15 \implies k = 7$. Thus, the page contains numbers 7 and 8.

Question 33

How many numbers will be left out of the numbers 1 to 40, if all numbers exactly divisible by 4 and the ones having 4 as one of the digits are removed?

28

Answer:

Explanation:

Numbers divisible by 4 from 1 to 40 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 (10 numbers). Numbers containing the digit 4 are: 4, 14, 24, 34, 40 (5 numbers). The union of these two sets is: {4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40}, which contains 12 unique numbers. Removing these 12 numbers from 40 leaves $40 - 12 = 28$ numbers.

Question 34

What is the difference between the largest number and the least number, which are of five-digit numbers, written with the figures 3, 4, 7, 0, 3?

43983

Answer:

Explanation:

The digits given are $3, 4, 7, 0, 3$. The largest 5-digit number is formed by arranging the digits in descending order: $74330$. The smallest 5-digit number cannot begin with 0, so we start with the smallest non-zero digit 3 and arrange the remaining digits in ascending order: $30347$. The difference between them is $74330 - 30347 = 43983$.

Question 35

The sum of the digits of a two-digit number is 12. The difference of the digits is 6. Find the number.

Either (a) or (b)

Answer:

Explanation:

Let the digits of the number be $x$ and $y$. We have $x + y = 12$ and $|x - y| = 6$. Case 1: If $x - y = 6$, adding the equations gives $2x = 18 \implies x = 9$, which means $y = 3$. The number is 93. Case 2: If $y - x = 6$, adding gives $2y = 18 \implies y = 9$, which means $x = 3$. The number is 39. Since both numbers satisfy the conditions, the correct option is 'Either (a) or (b)'.

Question 36

The ratio of a two-digit natural number to a number formed by reversing its digits is 4: 7. The number of such pairs is:

4

Answer:

Explanation:

Let the number be $10x + y$ and the reversed number be $10y + x$. We are given $\frac{10x + y}{10y + x} = \frac{4}{7} \implies 7(10x + y) = 4(10y + x) \implies 70x + 7y = 40y + 4x \implies 66x = 33y \implies 2x = y$. Since $x$ and $y$ are non-zero single digits, the possible pairs $(x,y)$ are $(1,2)$, $(2,4)$, $(3,6)$, and $(4,8)$. This gives exactly 4 such pairs.

Question 37

If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is:

always greater than the original fraction

Answer:

Explanation:

Let the original fraction be $x/y$ with $0 < x < y$. Let the positive quantity added be $k > 0$. The new fraction is $\frac{x + k}{y + k}$. Comparing the two, we compute $\frac{x + k}{y + k} - \frac{x}{y} = \frac{xy + ky - xy - kx}{y(y + k)} = \frac{k(y - x)}{y(y + k)}$. Since $y > x$, the numerator $k(y - x)$ is positive. Thus, the resulting fraction is always greater than the original fraction.

Question 38

For what value of n, the sum of digits in the number $(10^n+1)$ is 2?

For any whole number n

Answer:

Explanation:

Let's evaluate $10^n + 1$ for non-negative integers. For $n = 0$, $10^0 + 1 = 2$ (sum of digits is 2). For $n = 1$, $10^1 + 1 = 11$ (sum of digits is $1+1=2$). For any whole number $n \ge 1$, $10^n + 1$ has the digits $1$ at the start and $1$ at the end with $n-1$ zeros in between, which always sums to 2. Thus, the condition holds true for any whole number $n$.

Question 39

In the expression $5 * 4 * 3 * 2 * 1$, * is chosen from +, -, x each at most two times. What is the smallest nonnegative value of the expression?

0

Answer:

Explanation:

We want to find if the expression can evaluate to 0. Let's arrange the operations as $5 - 4 - 3 + 2 \times 1$. Using the standard order of operations (multiplication first), we get: $5 - 4 - 3 + 2 = 1 - 3 + 2 = 0$. Since we used two '-' operations, one '+' operation, and one '$\times$' operation, we respected the constraint of using each operation at most two times. Thus, the smallest non-negative value is 0.

Question 40

On January 1st, 2023, a person saved Rs. 1. On January 2nd 2023, he saved Rs. 2 more than that on the previous day. On January 3rd, 2023, he saved Rs 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube?

8th January, 2023

Answer:

Explanation:

The daily savings form an Arithmetic Progression of consecutive odd numbers: $1, 3, 5, 7, \dots$. The sum of the first $n$ odd numbers is always $n^2$. Thus, at the end of the $n$-th day, the total savings is $n^2$. For the savings to be both a perfect square and a perfect cube, $n^2$ must be of the form $k^6 \implies n = k^3$ (a perfect cube). The first such integer $n > 1$ is $n = 2^3 = 8$. This corresponds to 8th January, 2023.

Question 41

Consider the following statements: 1. 1000 litres = $1 \text{ m}^3$ 2. 1 metric ton = 1000 kg 3. 1 hectare = $10000 \text{ m}^2$ Which of the statements given above are correct?

1, 2 and 3

Answer:

Explanation:

Statement 1 is correct because 1 cubic meter ($1\text{ m}^3$) is exactly equal to 1000 liters of water volume. Statement 2 is correct because 1 metric ton is defined as 1000 kilograms. Statement 3 is correct because 1 hectare is a unit of area equal to a square with 100-meter sides, which translates to $100 \times 100 = 10000\text{ m}^2$. Thus, all three statements are correct.

Question 42

Let X be a two-digit number and Y be another two-digit number formed by interchanging the digits of X. If $(X+Y)$ is the greatest two-digit number, then what is the number of possible values of X?

8

Answer:

Explanation:

Let $X = 10a + b$ and $Y = 10b + a$. The sum is $X + Y = 11(a + b)$. The greatest two-digit number that is a multiple of 11 is 99. Therefore, $11(a + b) = 99 \implies a + b = 9$. Since $X$ and $Y$ are both two-digit numbers, neither $a$ nor $b$ can be 0. The valid single-digit integer pairs $(a,b)$ are: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), and (8,1), giving 8 possible values of $X$.

Question 43

What is the sum of the first 28 terms in the sequence: 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, ...?

84

Answer:

Explanation:

The sequence can be grouped as: G1 = [1], G2 = [1, 2], G3 = [1, 3, 2], G4 = [1, 4, 3, 2], up to G7 = [1, 7, 6, 5, 4, 3, 2]. The number of terms up to Group $k$ is $\frac{k(k+1)}{2}$. For $k=7$, the number of terms is $\frac{7 \times 8}{2} = 28$. The sum of elements in G_k is also $\frac{k(k+1)}{2}$. The sum of the first 28 terms is the sum of G1 through G7: $1 + 3 + 6 + 10 + 15 + 21 + 28 = 84$.

Question 44

If the sum of the two-digit numbers AB and CD is the three-digit number 1CE, where the letters A, B, C, D, E denote distinct digits, then what is the value of A?

9

Answer:

Explanation:

We set up the column addition: $AB + CD = 1CE$. In the tens column, we have $A + C + \text{carry} = 10 + C \implies A + \text{carry} = 10$. Since the carry from the units column ($B+D$) can only be 0 or 1, and $A$ is a single digit, the carry must be 1. Thus, $A + 1 = 10 \implies A = 9$. This satisfies all conditions of distinct digits.

Question 45

The total cost of 4 oranges, 6 mangoes and 8 apples is equal to twice the total cost of 1 orange, 2 mangoes and 5 apples. Consider the following statements: 1. The total cost of 3 oranges, 5 mangoes and 9 apples is equal to the total cost of 4 oranges, 6 mangoes and 8 apples. 2. The total cost of one orange and one mango is equal to the cost of one apple. Which of the statements given above is/are correct?

Both 1 and 2

Answer:

Explanation:

Let $O$, $M$, and $A$ represent the costs. The given equation is $4O + 6M + 8A = 2(O + 2M + 5A) \implies 4O + 6M + 8A = 2O + 4M + 10A \implies 2O + 2M = 2A \implies O + M = A$. This confirms Statement 2. Substituting $A = O + M$ into Statement 1, both sides simplify to $12O + 14M$, confirming Statement 1 is also correct. Thus, both statements are correct.

Question 46

A Question is given followed by two Statements 1 and 2. Consider the Question and the Statements. Question: What is the time required to download the software? Statement-1: The size of the software is 12 megabytes. Statement-2: The transfer rate is 2.4 kilobytes per second. Which of the following is correct in respect of the above Question and the Statements?

The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

Answer:

Explanation:

To calculate the time required to download, we need both the total size of the file and the transfer rate since $\text{Time} = \text{Size} \div \text{Rate}$. Statement-1 gives only the size (12 MB), and Statement-2 gives only the speed (2.4 KB/s). Neither statement is sufficient on its own. Using both statements together, we can compute the downloading time. Thus, they are sufficient only when taken together.

Question 47

A Question is given followed by two Statements 1 and 2. Consider the Question and the Statements. P, Q, R and S appeared in a test. Question: Has P scored more marks than Q? Statement-1: The sum of the marks scored by P and Q is equal to the sum of the marks scored by R and S. Statement-2: The sum of the marks scored by P and S is more than the sum of the marks scored by Q and R. Which of the following is correct in respect of the above Question and the Statements?

The Question cannot be answered even by using both the Statements together.

Answer:

Explanation:

We are given: (1) $p + q = r + s$ and (2) $p + s > q + r$. Adding the equations yields $2p + s + q > 2r + s + q \implies p > r$. Subtracting the equations yields $s - q > q - s \implies s > q$. However, we cannot establish a definitive relationship between $p$ and $q$. For example, the values $(10, 8, 9, 9)$ yield $p > q$, whereas the values $(5, 6, 3, 8)$ satisfy both statements but yield $p < q$. Thus, the question cannot be answered even by combining both statements.

Question 48

A Question is given followed by two Statements 1 and 2. Consider the Question and the Statements. Age of each of P and Q is less than 100 years but more than 10 years. If you interchange the digits of the age of P, the number represents the age of Q. Question: What is the difference of their ages? Statement-1: The age of P is greater than the age of Q. Statement-2: The sum of their ages is 11/6 times their difference. Which of the following is correct in respect of the above Question and the Statements?

The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

Answer:

Explanation:

Let P's age be $10x + y$ and Q's age be $10y + x$. The difference is $9|x - y|$. Statement-1 alone does not give a unique difference. From Statement-2, the sum $11(x + y)$ is equal to $\frac{11}{6} \times 9|x - y| \implies 2(x + y) = 3|x - y|$. Solving this yields $x = 5y$ (if $x > y$) or $y = 5x$ (if $x < y$). Since $x$ and $y$ are non-zero single digits, the only possible digit combinations are (5,1) or (1,5). In both cases, the ages are 15 and 51, and the difference is exactly 36. Thus, Statement-2 alone is sufficient.

Question 49

If a+b means a-b; a-b means a x b; a x b means a/b; a/b means a+b; then what is the value of $10+30-100\times50\div25$? (Operations are to be replaced simultaneously)

-25

Answer:

Explanation:

Replacing the operators simultaneously as defined: '+' becomes '-', '-' becomes '$\times$', '$\times$' becomes '$\div$', and '$\div$' becomes '+'. The expression becomes $10 - 30 \times 100 \div 50 + 25$. According to BODMAS, we first perform division: $100 \div 50 = 2$. Next, perform multiplication: $30 \times 2 = 60$. Finally, evaluating left-to-right: $10 - 60 + 25 = -50 + 25 = -25$.

Question 50

In some code, letters P, Q, R, S, T represent numbers 4, 5, 10, 12, 15. It is not known which letter represents which number. If $Q-S=2S$ and $T=R+S+3$, then what is the value of $P+R-T$?

2

Answer:

Explanation:

From $Q - S = 2S \implies Q = 3S$. Out of $\{4, 5, 10, 12, 15\}$, the only matching pairs are $Q=15, S=5$ or $Q=12, S=4$. Case 1: If $S=4$ and $Q=12$, then $T = R + 7$. No remaining digits $\{5, 10, 15\}$ satisfy this. Case 2: If $S=5$ and $Q=15$, then $T = R + 8$. From the remaining digits $\{4, 10, 12\}$, only $R=4, T=12$ satisfies this. This leaves $P=10$. Substituting into the target expression: $P + R - T = 10 + 4 - 12 = 2$.