JKSSB Aptitude MCQs

Explanation

• Property: When two numbers are in the ratio $a/b$ (in simplest form) and their HCF is $H$, the numbers are $aH$ and $bH$.

• Step 1: Find the actual numbers.

  • Ratio = 7/11

  • HCF ($H$) = 5

  • First number = $7 \times 5 = 35$

  • Second number = $11 \times 5 = 55$

• Step 2: Find the LCM.

  • Method 1 (Formula): $\text{LCM} = \text{Ratio}_1 \times \text{Ratio}_2 \times \text{HCF}$

    $$\text{LCM} = 7 \times 11 \times 5 = 385$$

  • Method 2 (Product Rule): $\text{Product of numbers} = \text{HCF} \times \text{LCM}$

    $$35 \times 55 = 5 \times \text{LCM}$$

    $$1925 = 5 \times \text{LCM}$$

    $$\text{LCM} = \frac{1925}{5} = 385$$

Explanation

Finding the HCF (Highest Common Factor):

• Factors of 20: 1, 2, 4, 5, 10, 20

• Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

• The highest common factor is 10.

2. Finding the LCM (Least Common Multiple):

• Multiples of 20: 20, 40, 60, 80, ...

• Multiples of 30: 30, 60, 90, 120, ...

• The smallest common multiple is 60.

Alternative Method (Prime Factorization):

• $20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

• $30 = 2 \times 3 \times 5 = 2^1 \times 3^1 \times 5^1$

• HCF (lowest powers of common factors): $2^1 \times 5^1 = 10$

• LCM (highest powers of all factors): $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$

Explanation

• Step 1: Let the two numbers be $3x$ and $4x$ (based on the ratio 3:4).

• Step 2: Find the LCM of $3x$ and $4x$ in terms of $x$.

  • The LCM of 3 and 4 is 12.

  • Therefore, the LCM of $3x$ and $4x$ is $12x$.

• Step 3: According to the problem, the LCM is 48.

  $$12x = 48$$

• Step 4: Solve for $x$:

  $$x = \frac{48}{12} = 4$$

• Step 5: Calculate the smaller number ($3x$):

  $$\text{Smaller number} = 3 \times 4 = 12$$

Explanation

• Fundamental Rule: For any two numbers, the product of the numbers is equal to the product of their HCF and LCM.

  $$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$

• Step 1: Plug the given values into the formula.

  • $\text{HCF} = 5$

  • $\text{LCM} = 120$

  • $\text{First number} = 20$

  • Let the second number be $x$.

• Step 2: Form the equation:

  $$5 \times 120 = 20 \times x$$

  $$600 = 20x$$

• Step 3: Solve for $x$:

  $$x = \frac{600}{20} = 30$$

Explanation

• Step 1: Let the required number of years be $x$.

• Step 2: Express their ages after $x$ years:

  • Father's age = $40 + x$

  • Son's age = $10 + x$

• Step 3: According to the problem, after $x$ years, the father will be 3 times as old as the son:

  $$40 + x = 3(10 + x)$$

• Step 4: Solve for $x$:

  $$40 + x = 30 + 3x$$

  $$40 - 30 = 3x - x$$

  $$10 = 2x$$

  $$x = 5 \text{ years}$$

Explanation

• Step 1: Let the present age of the Son be $x$ and the Father be $50 - x$ (since their sum is 50).

• Step 2: Express their ages 5 years ago:

  • Son's age = $x - 5$

  • Father's age = $(50 - x) - 5 = 45 - x$

• Step 3: According to the problem, 5 years ago, the father was 4 times as old as the son:

  $$45 - x = 4(x - 5)$$

  $$45 - x = 4x - 20$$

• Step 4: Solve for $x$:

  $$45 + 20 = 4x + x$$

  $$65 = 5x$$

  $$x = \frac{65}{5} = 13$$

Explanation

• Step 1: Let the present ages of A and B be $7x$ and $9x$ (based on the ratio 7:9).

• Step 2: After 4 years, their ages will be $7x + 4$ and $9x + 4$. According to the problem, this new ratio is 9:11.

  $$\frac{7x + 4}{9x + 4} = \frac{9}{11}$$

• Step 3: Cross-multiply to solve for $x$:

  $$11(7x + 4) = 9(9x + 4)$$

  $$77x + 44 = 81x + 36$$

  $$81x - 77x = 44 - 36$$

  $$4x = 8 \implies x = 2$$

• Step 4: Calculate A's present age:

  $$\text{A's present age} = 7x = 7 \times 2 = 14 \text{ years}$$

• (Note: B's present age would be $9 \times 2 = 18$ years.)

• Shortcut Method:

  • Initial Ratio ($A:B$) = $7:9$ (Difference = 2 units)

  • Ratio after 4 years = $9:11$ (Difference = 2 units)

  • Since the difference between the parts is the same (9 - 7 = 2 and 11 - 9 = 2), we can say that an increase of 2 units in the ratio corresponds to 4 years.

  • $2 \text{ units} = 4 \text{ years} \implies 1 \text{ unit} = 2 \text{ years}$.

  • A's present age = $7 \text{ units} = 7 \times 2 = 14 \text{ years}$.

Explanation

• Step 1: Let the initial number of boys be $5x$ and the initial number of girls be $6x$ (based on the ratio 5:6).

• Step 2: According to the problem, 5 more boys join the class, but the number of girls remains the same. The new ratio is 10:11.

  $$\frac{5x + 5}{6x} = \frac{10}{11}$$

• Step 3: Cross-multiply to solve for $x$:

  $$11(5x + 5) = 10(6x)$$

  $$55x + 55 = 60x$$

  $$60x - 55x = 55$$

  $$5x = 55 \implies x = 11$$

• Step 4: Calculate the number of girls:

  $$\text{Number of girls} = 6x = 6 \times 11 = 66$$

Explanation

• Step 1: To find the ratio of $a/c$ when $a/b$ and $b/c$ are given, you can multiply the two ratios together:

  $$\frac{a}{c} = \frac{a}{b} \times \frac{b}{c}$$

• Step 2: Substitute the given values ($4/7$ and $14/15$):

  $$\frac{a}{c} = \frac{4}{7} \times \frac{14}{15}$$

• Step 3: Simplify the fraction. Since 14 is divisible by 7:

  $$\frac{a}{c} = \frac{4 \times 2}{1 \times 15} = \frac{8}{15}$$

• Alternative Method (Making 'b' equal):

  • Ratio $a:b = 4:7$

  • Ratio $b:c = 14:15$

  • To combine them, make the value of '$b$' the same in both. Multiply the first ratio by 2:

    $$a:b = (4 \times 2) : (7 \times 2) = 8:14$$

  • Now, $a:b = 8:14$ and $b:c = 14:15$.

  • Therefore, $a:b:c = 8:14:15$, which gives $a:c = 8:15$.

Explanation

• Step 1: Let the two numbers be $7x$ and $11x$ (based on the ratio 7:11).

• Step 2: According to the problem, their product is 693.

  $$(7x) \times (11x) = 693$$

  $$77x^2 = 693$$

• Step 3: Solve for $x$.

  $$x^2 = \frac{693}{77} = 9$$

  $$x = \sqrt{9} = 3$$

• Step 4: Find the difference between the numbers.

  $$\text{Difference} = 11x - 7x = 4x$$

  $$\text{Difference} = 4 \times 3 = 12$$

• (Alternatively: The numbers are $7 \times 3 = 21$ and $11 \times 3 = 33$. Difference: $33 - 21 = 12$.)

Explanation

• Step 1: Calculate the daily work rates.

  • Work rate of A = $1/15$ work per day.

  • Work rate of B = $1/20$ work per day.

• Step 2: Calculate the combined work rate (A + B).

  $$\text{Combined Rate} = \frac{1}{15} + \frac{1}{20}$$

  $$\text{Using 60 as the common denominator:}$$

  $$\text{Combined Rate} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60} \text{ work per day}$$

• Step 3: Calculate the total work done in 4 days.

  $$\text{Work Done} = \text{Rate} \times \text{Time} = \frac{7}{60} \times 4 = \frac{7}{15}$$

• Step 4: Calculate the remaining fraction of work.

  $$\text{Work Left} = 1 - \text{Work Done}$$

  $$\text{Work Left} = 1 - \frac{7}{15} = \frac{8}{15}$$

• LCM Method:

  • Let Total Work = LCM(15, 20) = 60 units.

  • Efficiency of A = $60 / 15$ = 4 units/day.

  • Efficiency of B = $60 / 20$ = 3 units/day.

  • Combined Efficiency = $4 + 3$ = 7 units/day.

  • Work done in 4 days = $7 \times 4$ = 28 units.

  • Remaining Work = $60 - 28$ = 32 units.

  • Fraction Left = $32 / 60 = 8/15$.

Explanation

• Step 1: Determine the time taken by B.

  • A takes 18 days.

  • B takes half the time as A = $18 / 2$ = 9 days.

• Step 2: Calculate individual work rates.

  • Work rate of A = $1/18$ work per day.

  • Work rate of B = $1/9$ work per day.

• Step 3: Calculate the combined work rate (A + B).

  $$\text{Combined Rate} = \frac{1}{18} + \frac{1}{9}$$

  $$\text{Using 18 as the common denominator:}$$

  $$\text{Combined Rate} = \frac{1}{18} + \frac{2}{18} = \frac{3}{18} = \frac{1}{6} \text{ work per day}$$

• Step 4: Find the total days taken together.

  $$\text{Total Time} = \frac{1}{\text{Combined Rate}} = 6 \text{ days}$$

• Shortcut Formula:

  $$\text{Total Time} = \frac{xy}{x + y} = \frac{18 \times 9}{18 + 9} = \frac{162}{27} = 6 \text{ days}$$

Explanation

• Step 1: Calculate the work rates (work done per day).

  • Combined rate (A + B) = $1/10$ work per day.

  • Individual rate of A = $1/30$ work per day.

• Step 2: Find B's work rate by subtracting A's rate from the combined rate.

  $$\text{B's rate} = \frac{1}{10} - \frac{1}{30}$$

  $$\text{Using 30 as the common denominator:}$$

  $$\text{B's rate} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \text{ work per day}$$

• Step 3: Convert the rate back into time.

  $$\text{Time taken by B alone} = \frac{1}{\text{Rate}} = 15 \text{ days}$$

• Alternative LCM Method:

  • Let Total Work = LCM of 10 and 30 = 30 units.

  • Efficiency of (A + B) = $30 / 10$ = 3 units/day.

  • Efficiency of A = $30 / 30$ = 1 unit/day.

  • Efficiency of B = $(A + B) - A = 3 - 1$ = 2 units/day.

  • Time for B = $\text{Total Work} / \text{Efficiency} = 30 / 2$ = 15 days.

Explanation

• Step 1: Calculate the combined work rate of the man and woman.

  $$\text{Work rate (Man + Woman)} = \frac{1}{8} \text{ work per day}$$

• Step 2: Calculate the work rate of the man alone.

  $$\text{Work rate (Man)} = \frac{1}{10} \text{ work per day}$$

• Step 3: Subtract the man's rate from the combined rate to find the woman's work rate.

  $$\text{Work rate (Woman)} = \frac{1}{8} - \frac{1}{10}$$

  $$\text{To subtract, find a common denominator (40):}$$

  $$\text{Work rate (Woman)} = \frac{5}{40} - \frac{4}{40} = \frac{1}{40} \text{ work per day}$$

• Step 4: Convert the rate back into time.

  $$\text{Time taken by woman alone} = \frac{1}{\text{Work rate}} = 40 \text{ days}$$

Explanation

• The sum of the first $n$ natural numbers is given by the formula:

  $$\text{Sum} = \frac{n(n + 1)}{2}$$

• The average is defined as the $\frac{\text{Sum}}{\text{Total Count}}$:

  $$\text{Average} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2}$$

Explanation

• Method 1 (Algebraic):

  Let the six consecutive even numbers be $x, x+2, x+4, x+6, x+8,$ and $x+10$.

  The average is the sum divided by the count:

  $$\frac{x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10)}{6} = 25$$

  $$\frac{6x + 30}{6} = 25$$

  $$x + 5 = 25 \implies x = 20$$

  • Smallest number ($x$) = 20

  • Largest number ($x+10$) = $20 + 10$ = 30

• Method 2 (Logic):

  For a set of consecutive numbers, the average is always the middle value. For 6 numbers, the average lies exactly between the $3^{rd}$ and $4^{th}$ numbers.

  Middle value = 25.

  The two even numbers surrounding 25 are 24 ($3^{rd}$) and 26 ($4^{th}$).

  The sequence is: $20, 22, 24, \text{[25]}, 26, 28, 30$.

Explanation

• Step 1: Find the total sum of the first group of numbers.

  $$\text{Sum}_1 = \text{Average}_1 \times \text{Count}_1 = 20 \times 30 = 600$$

• Step 2: Find the total sum of the second group of numbers.

  $$\text{Sum}_2 = \text{Average}_2 \times \text{Count}_2 = 30 \times 20 = 600$$

• Step 3: Calculate the total sum of all numbers combined.

  $$\text{Total Sum} = 600 + 600 = 1200$$

• Step 4: Calculate the total count of all numbers.

  $$\text{Total Count} = 30 + 20 = 50$$

• Step 5: Find the combined average.

  $$\text{Combined Average} = \frac{\text{Total Sum}}{\text{Total Count}} = \frac{1200}{50} = 24$$

Explanation

• Step 1: Find the incorrect total sum of the 10 numbers.

  $$\text{Incorrect Sum} = \text{Average} \times \text{Number of items} = 15 \times 10 = 150$$

• Step 2: Calculate the difference between the correct value and the incorrect value.

  $$\text{Difference} = \text{Correct value} - \text{Incorrect value} = 36 - 26 = +10$$

• Step 3: Find the correct sum by adding the difference.

  $$\text{Correct Sum} = 150 + 10 = 160$$

• Step 4: Calculate the correct average.

  $$\text{Correct Average} = \frac{\text{Correct Sum}}{\text{Number of items}} = \frac{160}{10} = 16$$

• Short-cut Method:

  $$\text{Change in Average} = \frac{\text{Difference in values}}{\text{Total items}} = \frac{36 - 26}{10} = \frac{10}{10} = +1$$

  $$\text{New Average} = \text{Old Average} + 1 = 15 + 1 = 16$$

Explanation

• Step 1: Calculate the amount of decrease.

  $$15\% \text{ of } 400 = \frac{15}{100} \times 400 = 15 \times 4 = 60$$

• Step 2: Subtract the decrease from the original number to find the new number.

  $$\text{New Number} = 400 - 60 = 340$$

• Alternative Method: If a number is decreased by 15%, the remaining value is $100\% - 15\% = 85\%$ of the original number.

  $$\text{New Number} = 85\% \text{ of } 400 = 0.85 \times 400 = 340$$

Explanation

• Method 1 (Assumption): Let the original population be 100.

  1. Increase by 10% in the first year: $100 + 10 = 110$.

  2. Increase the new population (110) by 20% in the second year: $110 + (20\% \text{ of } 110) = 110 + 22 = 132$.

  3. Net Change: The population went from 100 to 132, which is a total increase of 32%.

• Method 2 (Formula): For successive increases of $x\%$ and $y\%$, the effective percentage increase is:

  $$\text{Total Increase} = x + y + \frac{xy}{100}$$

  Substituting $x = 10$ and $y = 20$:

  $$\text{Total Increase} = 10 + 20 + \frac{10 \times 20}{100} = 30 + \frac{200}{100} = 30 + 2 = 32\%$$

Explanation

• Method 1 (Assumption): Let the original number be 100.

  1. Increase by 20%: $100 + 20 = 120$.

  2. Decrease the result (120) by 20%: $120 - (20\% \text{ of } 120) = 120 - 24 = 96$.

  3. Net Change: The number went from 100 to 96, which is a decrease of 4%.

• Method 2 (Formula): For successive changes of $a\%$ and $b\%$, the net change is given by:

  $$\text{Net Change} = a + b + \frac{ab}{100}$$

  Substituting $a = +20$ and $b = -20$:

  $$\text{Net Change} = 20 - 20 + \frac{20 \times (-20)}{100} = 0 - \frac{400}{100} = -4\%$$

Explanation

1. Breakdown of the Purchase

• Total Cost Price (CP): Rs. 56,000

• Part 1 (Sold): $1/3$ of the total

• Part 2 (Remaining): $2/3$ of the total

2. Calculate the Loss on the First Part

Since we are looking for a percentage to "cover the loss" (meaning reaching a break-even point where total profit/loss is zero), the actual rupee value of 56,000 doesn't change the final percentage. We can use the fractions directly:

• Loss on 1/3 of the stock: $40\%$

• Total Loss (relative to the whole): $\frac{1}{3} \times 40\% = 13.33\%$

3. Calculate the Required Profit on the Remaining Part

To cover a total loss of $13.33\%$, the remaining $2/3$ of the stock must generate an equivalent amount of profit.

Let $x$ be the required profit percentage on the remaining $2/3$:

$$\frac{2}{3} \times x = 13.33\%$$

$$\frac{2}{3}x = \frac{40}{3}$$

Now, solve for $x$:

$$2x = 40$$

$$x = 20\%$$

Explanation

1. Set Up the Relationship

Let the Cost Price (CP) of one book be $x$.

• The CP of 18 books is $18x$.

• The problem states that the Selling Price (SP) of 12 books is equal to the CP of 18 books.

• So, SP of 12 books = $18x$.

2. Calculate the Selling Price (SP) of One Book

To find the selling price of a single book:

$$\text{SP of 1 book} = \frac{18x}{12}$$

$$\text{SP of 1 book} = 1.5x$$

3. Calculate the Profit per Book

$$\text{Profit} = \text{SP} - \text{CP}$$

$$\text{Profit} = 1.5x - x = 0.5x$$

4. Calculate the Profit Percentage

$$\text{Profit\%} = \left( \frac{\text{Profit}}{\text{CP}} \right) \times 100$$

$$\text{Profit\%} = \left( \frac{0.5x}{x} \right) \times 100$$

$$\text{Profit\%} = 0.5 \times 100 = 50\%$$

Alternatively, using the Ratio Method:

$$\frac{\text{SP of 12 books}}{\text{CP of 18 books}} = 1 \implies 12 \times \text{SP}_1 = 18 \times \text{CP}_1$$

$$\frac{\text{SP}_1}{\text{CP}_1} = \frac{18}{12} = \frac{3}{2}$$

If CP = 2 and SP = 3, then:

$$\text{Profit} = 3 - 2 = 1$$

$$\text{Profit\%} = \frac{1}{2} \times 100 = 50\%$$

Explanation

1. Assume a Cost Price (CP)

Let the Cost Price (CP) = $100$.

2. Find the Marked Price (MP)

The shopkeeper marks his goods 30% above the cost price:

$$\text{Marked Price (MP)} = CP + 30\% \text{ of } CP$$

$$\text{MP} = 100 + 30 = 130$$

3. Find the Selling Price (SP)

He allows a discount of 10% on the marked price:

$$\text{Discount} = 10\% \text{ of } 130 = 13$$

$$\text{Selling Price (SP)} = \text{MP} - \text{Discount}$$

$$\text{SP} = 130 - 13 = 117$$

4. Calculate the Gain Percentage

$$\text{Gain} = \text{SP} - \text{CP} = 117 - 100 = 17$$

$$\text{Gain\%} = \left( \frac{\text{Gain}}{\text{CP}} \right) \times 100$$

$$\text{Gain\%} = \left( \frac{17}{100} \right) \times 100 = 17\%$$

Alternatively, using the successive percentage formula:

$\text{Net Change} = x + y + \frac{xy}{100}$

Where $x = +30$ (markup) and $y = -10$ (discount):

$$\text{Net Gain} = 30 - 10 + \frac{30 \times (-10)}{100}$$

$$\text{Net Gain} = 20 - 3 = 17\%$$

Explanation

1. Find the Cost Price (CP)

The original selling price is Rs. 2,850 with a gain of 14%. The formula for Cost Price is:

$$CP = \frac{SP \times 100}{100 + \text{Profit\%}}$$

Substituting the values:

$$CP = \frac{2,850 \times 100}{100 + 14}$$

$$CP = \frac{285,000}{114}$$

$$CP = 2,500$$

2. Find the New Selling Price

Now, we want to find the selling price if the profit is reduced to 8%. The formula for Selling Price is:

$$SP = \frac{CP \times (100 + \text{Profit\%})}{100}$$

Substituting the new values:

$$SP = \frac{2,500 \times (100 + 8)}{100}$$

$$SP = \frac{2,500 \times 108}{100}$$

$$SP = 25 \times 108$$

$$SP = 2,700$$

Explanation

The probability of getting a number greater than 6 on a standard six-sided die is 0.

Explanation

The probability of a sure event (or certain event) is 1 (or 100%).

Explanation

1. Identify the number of cards for each event:

• Total cards in the deck = $52$

• Red cards ($A$): There are two red suits (Hearts and Diamonds), each with 13 cards.

  $|A| = 13 + 13 = 26$

• Face cards ($B$): Each suit has 3 face cards (Jack, Queen, King). There are 4 suits in total.

  $|B| = 3 \times 4 = 12$

• Red face cards ($A \cap B$): These are the face cards from the red suits (Hearts and Diamonds).

  $|A \cap B| = 3 (\text{Hearts}) + 3 (\text{Diamonds}) = 6$

2. Calculate the probability:

$$P(A \cup B) = \frac{26}{52} + \frac{12}{52} - \frac{6}{52}$$

$$P(A \cup B) = \frac{26 + 12 - 6}{52}$$

$$P(A \cup B) = \frac{32}{52}$$

3. Simplify the fraction:

To simplify $\frac{32}{52}$, divide both the numerator and the denominator by their greatest common divisor, which is $4$:

$$\frac{32 \div 4}{52 \div 4} = \frac{8}{13}$$

Explanation

-42/21

Explanation

$$480 \div \{ 16 - 8 \times (3/4) \} + 35\% \text{ of } 200 - 18$$

The calculation would be:

1. Simplify the braces:

   $$16 - 8 \times (3/4) = 16 - 6 = 10$$

2. Divide the first term:

   $$480 \div 10 = 48$$

3. Calculate the percentage:

   $$35\% \text{ of } 200 = 70$$

4. Combine the results:

   $$48 + 70 - 18$$

   $$118 - 18 = 100$$