ML Aggarwal Class 9 Maths Solutions Chapter 06 Problems on Simultaneous Linear Equations

Question 1. The sum of two numbers is 50 and their difference is 16. Find the numbers.
Answer: Suppose the two numbers are x and y. We know their sum equals 50 and their difference equals 16. Setting up the equations: x + y = 50 and x - y = 16. Adding these two equations together gives 2x = 66, so x = 33. Now substituting this value back into the first equation: 33 + y = 50, which means y = 17. The two numbers are 33 and 17.
In simple words: Add the two equations to find one number. Then use that number to find the other.

Exam Tip: When you have two equations with addition and subtraction, try adding or subtracting them to eliminate one variable - it's faster than substitution.

Question 2. The sum of two numbers is 2. If their difference is 20, find the numbers.
Answer: Let the two numbers be x and y. Given that their sum is 2 and their difference is 20, we form: x + y = 2 and x - y = 20. Adding these equations: 2x = 22, therefore x = 11. Substituting into the first equation: 11 + y = 2, so y = -9. The two numbers are 11 and -9.
In simple words: Add the equations together to remove y, then solve for x. One number turns out to be negative.

Exam Tip: Don't be surprised if one answer is negative - check by adding and subtracting your answers to verify they match the given conditions.

Question 3. The sum of two numbers is 43. If the larger is doubled and the smaller is tripled, the difference is 36. Find the two numbers.
Answer: Let x be the larger number and y be the smaller number. We know x + y = 43. According to the condition, when the larger is doubled and the smaller is tripled, their difference is 36, giving us 2x - 3y = 36. Multiplying the first equation by 3: 3x + 3y = 129. Adding this to the second equation: 5x = 165, so x = 33. Substituting back: 33 + y = 43, which gives y = 10. The two numbers are 33 and 10.
In simple words: Multiply the first equation so the y terms cancel when you add. This leaves only x to solve.

Exam Tip: When multiplying an equation to eliminate a variable, make sure the coefficients are exact opposites before adding or subtracting.

Question 4. The cost of 5 kg of sugar and 7 kg of rice is Rs 765 and the cost of 7 kg of sugar and 5 kg of rice is Rs 735. Find the cost of 6 kg of sugar and 10 kg of rice.
Answer: Let x be the cost per kg of sugar and y be the cost per kg of rice. From the given information: 5x + 7y = 765 and 7x + 5y = 735. Multiply the first equation by 7 and the second by 5 to get: 35x + 49y = 5355 and 35x + 25y = 3675. Subtracting the second from the first: 24y = 1680, so y = 70. Substituting into the first equation: 5x + 490 = 765, which gives 5x = 275, so x = 55. Therefore, the cost of 6 kg of sugar and 10 kg of rice = 6(55) + 10(70) = 330 + 700 = Rs 1,030.
In simple words: Make the x coefficients the same, subtract to find y, then find x, and finally calculate what was asked.

Exam Tip: Always make the coefficient of one variable equal in both equations before subtracting - this eliminates that variable cleanly and lets you solve for the other.

Question 5. The class IX students of a certain public School wanted to give a farewell party to the outgoing student of class X. They decided to purchase two kinds of sweets, one costing Rs 350 per kg and the other costing Rs 440 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweet was Rs 14,000, find how much sweets of each kind they purchased.
Answer: Let x kg be the quantity of the first sweet (at Rs 350 per kg) and y kg be the quantity of the second sweet (at Rs 440 per kg). We have x + y = 36, which means y = 36 - x. The total cost equation is 350x + 440y = 14,000. Substituting y: 350x + 440(36 - x) = 14,000, which simplifies to 350x + 15,840 - 440x = 14,000. This gives -90x = -1,840, so x ≈ 20.44, or approximately 20 kg. Then y = 36 - 20 = 16 kg. The sweet costing Rs 350 per kg was 20 kg and the one costing Rs 440 per kg was 16 kg.
In simple words: Use the total quantity to express one sweet in terms of the other, then substitute into the cost equation to solve.

Exam Tip: When the answer is not a whole number, round it reasonably and verify that your rounded answer still satisfies the original conditions closely enough.

Question 6. If from twice the greater of two numbers 16 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is still half the other number. What are the numbers?
Answer: Let x be the greater number and y be the smaller number. From the first condition, twice the greater number minus 16 equals half the smaller number: 2x - 16 = y/2, which gives 4x - y = 32. From the second condition, half the greater number minus 1 equals half the smaller number: x/2 - 1 = y/2, which simplifies to x - y = 2. Subtracting the second equation from the first: 3x = 30, so x = 10. Substituting into the second equation: 10 - y = 2, which gives y = 8. The two numbers are 8 and 10.
In simple words: Write both word conditions as equations, then subtract one from the other to find the first number, and use it to find the second.

Exam Tip: Convert word problems to equations very carefully - reread the condition for each equation to ensure you've captured it correctly.

Question 7. There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is Rs 8.50, how many of each are there?
Answer: Let x be the number of 20 paise coins and y be the number of 25 paise coins. Then x + y = 38. The total value in rupees is 0.2x + 0.25y = 8.50. Multiplying the second equation by 4: 0.8x + y = 34. Subtracting this from the first equation: x + y - (0.8x + y) = 38 - 34, which gives 0.2x = 4, so x = 20. Therefore, y = 38 - 20 = 18. There are 20 coins of 20 paise and 18 coins of 25 paise.
In simple words: Change the rupee value to use decimals, then multiply one equation to make y coefficients match, and subtract to find x.

Exam Tip: When dealing with money in different units (paise/rupees), convert everything to the same unit (rupees as decimals) before setting up your equations.

Question 8. A man has certain notes of denominations Rs 200 and Rs 50 which amount to Rs 3,800. If the number of notes of each kind is interchanged, they amount to Rs 600 less as before. Find the number of notes of each denomination.
Answer: Let x be the number of Rs 200 notes and y be the number of Rs 50 notes. The total value is 200x + 50y = 3,800. When the numbers are interchanged (y notes of Rs 200 and x notes of Rs 50), the new total is Rs 600 less: 200y + 50x = 3,200. Multiplying the first equation by 4: 800x + 200y = 15,200. Subtracting the second equation from this: 750x = 12,000, so x = 16. Substituting into the first equation: 200(16) + 50y = 3,800, which gives 3,200 + 50y = 3,800, so y = 12. The man has 16 notes of Rs 200 and 12 notes of Rs 50.
In simple words: Write equations for both arrangements, multiply one to match coefficients, then subtract and solve.

Exam Tip: When a problem says numbers are "interchanged," swap the variables in the second equation - this changes which denomination each variable represents.

Question 9. The ratio of two numbers is 2/3. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Answer: Let x and y be the two numbers. Given that their ratio is 2/3: x/y = 2/3, which gives x = 2y/3. When 2 is subtracted from the first and 8 from the second, the ratio becomes 3/2: (x - 2)/(y - 8) = 3/2. Substituting x = 2y/3 into this equation: (2y/3 - 2)/(y - 8) = 3/2. Simplifying the numerator: (2y - 6)/3 divided by (y - 8) equals 3/2. This gives (2y - 6)/[3(y - 8)] = 3/2. Cross-multiplying: 2(2y - 6) = 3 × 3(y - 8), which simplifies to 4y - 12 = 9y - 72. Solving: 5y = 60, so y = 12. Therefore, x = 2(12)/3 = 8. The numbers are 8 and 12.
In simple words: Use the ratio to write one number in terms of the other, then substitute into the second condition and solve.

Exam Tip: When a problem mentions ratios and their reciprocals, always double-check which ratio applies in which situation - switching them is a common mistake.

Question 10. If 1 is added to the numerator of a fraction, it becomes 1/5; if 1 is taken from the denominator, it becomes 1/7, find the fraction.
Answer: Let x be the numerator and y be the denominator. From the first condition, adding 1 to the numerator gives 1/5: (x + 1)/y = 1/5, so y = 5(x + 1). From the second condition, subtracting 1 from the denominator gives 1/7: x/(y - 1) = 1/7, so y - 1 = 7x, which means y = 7x + 1. Equating the two expressions for y: 5(x + 1) = 7x + 1, which gives 5x + 5 = 7x + 1. Solving: 2x = 4, so x = 2. Substituting back: y = 7(2) + 1 = 15. The original fraction is 2/15.
In simple words: Each condition gives you a different expression for the denominator. Set them equal and solve for the numerator.

Exam Tip: Always verify your answer by checking that both original conditions are satisfied with the fraction you found.

Question 11. If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to 5/8 and if the numerator and denominator are each diminished by 1, the fraction becomes equal to 1/2; find the fraction.
Answer: Let x be the numerator and y be the denominator of the fraction. From the first condition, increasing the numerator by 2 and the denominator by 1 gives 5/8: (x + 2)/(y + 1) = 5/8, which gives 8(x + 2) = 5(y + 1), simplifying to 8x + 16 = 5y + 5, or 8x - 5y = -11. From the second condition, decreasing both by 1 gives 1/2: (x - 1)/(y - 1) = 1/2, which gives 2(x - 1) = y - 1, simplifying to 2x - 2 = y - 1, or 2x - y = 1. From the second equation, y = 2x - 1. Substituting into the first: 8x - 5(2x - 1) = -11, which gives 8x - 10x + 5 = -11, so -2x = -16, and x = 8. Therefore, y = 2(8) - 1 = 15. The fraction is 8/15.
In simple words: Each condition gives a linear equation. Solve the simpler one for y, substitute into the other, and find x.

Exam Tip: When two conditions modify the numerator and denominator differently, set up both equations and use substitution - pick the simpler equation to solve for one variable first.

Question 11. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac { 1 }{ 9 } \)
(ii) \( \frac { -4 }{ 3 } \)
(iii) \( \frac { 1 }{ 6 } \)
Answer: (i) \( \frac { 1 }{ 9 } = 0.\overline{1} \)
(ii) \( \frac { -4 }{ 3 } = -1.\overline{3} \)
(iii) \( \frac { 1 }{ 6 } = 0.1\overline{6} \) Only the 6 repeats, not the 1 that comes before it.
In simple words: Divide the top number by the bottom number. When a digit (or group of digits) keeps appearing over and over, draw a bar over those repeating digits. The bar tells you that part continues forever without stopping.

Exam Tip: Make sure the bar covers only the digits that repeat, not any non-repeating digits that appear first.

Question 12. Find the fraction which becomes \( \frac{1}{2} \) when the denominator is increased by 4 and is equal to \( \frac{1}{8} \) when the numerator is diminished by 5.
Answer: Suppose the numerator is x and the denominator is y. From the first condition, when the denominator increases by 4, we get \( \frac{x}{y+4} = \frac{1}{2} \), which gives us \( 2x = y + 4 \) or \( 2x - y = 4 \) ...... (i). From the second condition, when the numerator decreases by 5, we have \( \frac{x-5}{y} = \frac{1}{8} \), so \( 8(x - 5) = y \) or \( 8x - 40 = y \) ...... (ii). Substituting equation (ii) into equation (i): \( 2x - [8(x - 5)] = 4 \)

\( \implies 2x - 8x + 40 = 4 \)

\( \implies -6x = -36 \)

\( \implies x = 6 \) From equation (ii), \( y = 8(6 - 5) = 8 \). The original fraction is \( \frac{x}{y} = \frac{6}{8} \).
In simple words: Set up two equations based on the two different conditions. Solve them together by substitution to find x and y, then write the fraction.

Exam Tip: Always define your variables clearly at the start. Check your answer by substituting back into both original conditions.

Question 13. In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number, find the number.
Answer: Let the digit in the ten's place be x and the digit in the unit's place be y. From the first condition, \( x + y = 7 \) or \( x = 7 - y \) ...... (i). When the digits are reversed, the new number is \( 10y + x \). According to the second condition, this reversed number is 28 greater than twice the unit's digit of the original number: \( (10y + x) - 2y = 28 \). Substituting x from (i): \( 10y + (7 - y) - 2y = 28 \)

\( \implies 10y - y - 2y + 7 = 28 \)

\( \implies 7y = 21 \)

\( \implies y = 3 \) From equation (i), \( x = 7 - 3 = 4 \). The original number is \( 10x + y = 40 + 3 = 43 \).
In simple words: Write the original number as 10x + y. When you flip the digits, you get 10y + x. Use both conditions to make two equations, then solve.

Exam Tip: Always verify your final answer by checking it against both conditions mentioned in the problem.

Question 14. A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.
Answer: Let the ten's digit be x and the unit's digit be y. The original number is \( 10x + y \) and the reversed number is \( 10y + x \). From the first condition, the number is 6 more than four times the sum of its digits: \( 10x + y = 4(x + y) + 6 \)

\( \implies 10x + y - 4x - 4y = 6 \)

\( \implies 6x - 3y = 6 \)

\( \implies 2x - y = 2 \) ...... (i) From the second condition, reversing the digits increases the number by 9: \( 10y + x = (10x + y) + 9 \)

\( \implies 10y + x - 10x - y = 9 \)

\( \implies 9y - 9x = 9 \)

\( \implies y - x = 1 \) ...... (ii) Adding equations (i) and (ii): \( 2x - y + (y - x) = 2 + 1 \)

\( \implies x = 3 \) From equation (ii), \( y - 3 = 1 \), so \( y = 4 \). The number is \( 10(3) + 4 = 34 \).
In simple words: Create two equations from the two conditions. Add them together to remove one variable, then find both digits.

Exam Tip: When adding or subtracting equations, make sure terms cancel neatly. Always substitute back to verify your answer.

Question 15. When a two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit, the remainder is 1. What is the number?
Answer: Let the ten's digit be x and the unit's digit be y. The number is \( 10x + y \). From the first condition, when the number is divided by the sum of its digits, the quotient is 8: \( \frac{10x + y}{x + y} = 8 \)

\( \implies 10x + y = 8(x + y) \)

\( \implies 10x + y = 8x + 8y \)

\( \implies 2x - 7y = 0 \) ...... (i) From the second condition, if the ten's digit is diminished by three times the unit's digit, the remainder is 1: \( x - 3y = 1 \) ...... (ii) Multiplying equation (ii) by 2: \( 2x - 6y = 2 \) ...... (iii) Subtracting equation (i) from (iii): \( (2x - 6y) - (2x - 7y) = 2 - 0 \)

\( \implies y = 2 \) From equation (ii), \( x - 3(2) = 1 \), so \( x = 7 \). The number is \( 10(7) + 2 = 72 \).
In simple words: The division condition gives one equation, and the digit condition gives another. Solve them together by multiplying and subtracting to find x and y.

Exam Tip: Pay close attention to the wording - "quotient is 8" means division; "remainder is 1" means the result after an operation. Set up each equation carefully.

Question 16. The result of dividing a number of two digits by the number with digits reversed is \( 1\frac{3}{4} \). If the sum of digits is 12, find the number.
Answer: Let the ten's digit be x and the unit's digit be y. The original number is \( 10x + y \) and the reversed number is \( 10y + x \). From the first condition, \( \frac{10x + y}{10y + x} = 1\frac{3}{4} = \frac{7}{4} \)

\( \implies 4(10x + y) = 7(10y + x) \)

\( \implies 40x + 4y = 70y + 7x \)

\( \implies 33x - 66y = 0 \)

\( \implies 3x - 6y = 0 \) ...... (i) From the second condition, \( x + y = 12 \), so \( x = 12 - y \) ...... (ii) Substituting (ii) into (i): \( 3(12 - y) - 6y = 0 \)

\( \implies 36 - 3y - 6y = 0 \)

\( \implies 9y = 36 \)

\( \implies y = 4 \) From equation (ii), \( x = 12 - 4 = 8 \). The number is \( 10(8) + 4 = 84 \).
In simple words: Convert the mixed number to an improper fraction. Cross-multiply to get your first equation, and use the sum condition for your second equation.

Exam Tip: Mixed numbers must be converted to improper fractions before setting up the equation. Double-check your arithmetic when dividing and cross-multiplying.

Question 17. The result of dividing a number of two digits by the number with the digits reversed is \( \frac{5}{6} \). If the difference of digits is 1, find the number.
Answer: Let the ten's digit be x and the unit's digit be y. The original number is \( 10x + y \) and the reversed number is \( 10y + x \). From the first condition, \( \frac{10x + y}{10y + x} = \frac{5}{6} \)

\( \implies 6(10x + y) = 5(10y + x) \)

\( \implies 60x + 6y = 50y + 5x \)

\( \implies 55x - 44y = 0 \)

\( \implies 5x - 4y = 0 \) ...... (i) From the second condition, the difference of digits is 1: \( y - x = 1 \), so \( x = y - 1 \) ...... (ii) Substituting (ii) into (i): \( 5(y - 1) - 4y = 0 \)

\( \implies 5y - 5 - 4y = 0 \)

\( \implies y = 5 \) From equation (ii), \( x = 5 - 1 = 4 \). The number is \( 10(4) + 5 = 45 \).
In simple words: Set up the division fraction as an equation and cross-multiply. Then use the difference condition to make a second equation, and solve both together.

Exam Tip: When you cross-multiply, simplify the resulting equation by dividing out any common factors. This makes the arithmetic cleaner.

Question 18. A number of three digits has the hundred digit 4 times the unit digit and the sum of three digits is 14. If the three digits are written in the reverse order, the value of the number is decreased by 594. Find the number.
Answer: Let the ten's digit be x and the unit's digit be y. The hundred's digit is \( 4y \). The original number is \( 100(4y) + 10x + y = 400y + 10x + y = 401y + 10x \). When digits are reversed, the number becomes \( 100y + 10x + 4y = 104y + 10x \). From the first condition, the sum of all three digits equals 14: \( 4y + x + y = 14 \)

\( \implies x + 5y = 14 \) ...... (i) From the second condition, the reversed number is 594 less than the original: \( (401y + 10x) - (104y + 10x) = 594 \)

\( \implies 297y = 594 \)

\( \implies y = 2 \) From equation (i), \( x + 5(2) = 14 \), so \( x = 4 \). The hundred's digit is \( 4y = 4(2) = 8 \). The number is \( 100(8) + 10(4) + 2 = 842 \).
In simple words: Write the three-digit number using the hundreds, tens, and units digits. Set up one equation from the sum condition and another from the difference condition, then solve for each digit.

Exam Tip: For a three-digit number, remember the place values: 100 for hundreds, 10 for tens, and 1 for units. Carefully express both the original and reversed numbers algebraically.

Question 19. Four years ago Marina was three times old as her daughter. Six years from now the mother will be twice as old as her daughter. Find their present ages.
Answer: Let Marina's present age be x years and her daughter's present age be y years. Four years ago, Marina's age was \( (x - 4) \) years and her daughter's age was \( (y - 4) \) years. From the first condition, Marina was three times as old as her daughter: \( x - 4 = 3(y - 4) \)

\( \implies x - 4 = 3y - 12 \)

\( \implies 3y - x = 8 \) ...... (i) Six years from now, Marina's age will be \( (x + 6) \) years and her daughter's age will be \( (y + 6) \) years. From the second condition, Marina will be twice as old as her daughter: \( x + 6 = 2(y + 6) \)

\( \implies x + 6 = 2y + 12 \)

\( \implies x - 2y = 6 \) ...... (ii) Adding equations (i) and (ii): \( (3y - x) + (x - 2y) = 8 + 6 \)

\( \implies y = 14 \) From equation (ii), \( x - 2(14) = 6 \), so \( x = 34 \). Marina's present age is 34 years and her daughter's present age is 14 years.
In simple words: Set up two equations - one for "four years ago" and one for "six years from now." Add them to remove x, and solve for y first, then find x.

Exam Tip: Always remember to adjust ages correctly: subtract for past time and add for future time. Verify your answer by checking both original conditions.

Question 20. On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains Rs.70. If he sells the tea set at 5% gain and lemon set at 10% gain he gains Rs.130. Find the cost price of lemon set.
Answer: Let the cost price of the tea set be Rs.x and the cost price of the lemon set be Rs.y. From the first condition, selling the tea set at 5% loss and the lemon set at 15% gain produces a total gain of Rs.70. The loss on tea is \( \frac{5x}{100} \) and the gain on lemon is \( \frac{15y}{100} \). The net gain is: \( \frac{15y}{100} - \frac{5x}{100} = 70 \)

\( \implies 15y - 5x = 7000 \)

\( \implies 3y - x = 1400 \) ...... (i) From the second condition, selling the tea set at 5% gain and the lemon set at 10% gain produces a total gain of Rs.130: \( \frac{5x}{100} + \frac{10y}{100} = 130 \)

\( \implies 5x + 10y = 13000 \)

\( \implies x + 2y = 2600 \) ...... (ii) Adding equations (i) and (ii): \( (3y - x) + (x + 2y) = 1400 + 2600 \)

\( \implies 5y = 4000 \)

\( \implies y = 800 \) The cost price of the lemon set is Rs.800.
In simple words: Profit or loss equals (selling price minus cost price). Set up two equations based on the two profit scenarios, then add them to find y.

Exam Tip: In profit-loss problems, remember: Loss = (Loss%)/(100) × Cost Price, and Gain = (Gain%)/(100) × Cost Price. Always subtract loss from gain to find net gain.

Question 21. A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of Rs.1300. If he had interchanged the amounts invested at each rate, he would have received Rs.1400 interest annually. Find the amount invested at each rate.
Answer: Let the amount invested at 12% simple interest be Rs.x and the amount invested at 10% simple interest be Rs.y. From the first condition, the total yearly interest is Rs.1300: \( \frac{12x}{100} + \frac{10y}{100} = 1300 \)

\( \implies 12x + 10y = 130000 \)

\( \implies 6x + 5y = 65000 \) ...... (i) From the second condition, if the amounts are interchanged (x is invested at 10% and y is invested at 12%), the yearly interest would be Rs.1400: \( \frac{10x}{100} + \frac{12y}{100} = 1400 \)

\( \implies 10x + 12y = 140000 \)

\( \implies 5x + 6y = 70000 \) ...... (ii) Multiplying equation (i) by 5: \( 30x + 25y = 325000 \) ...... (iii) Multiplying equation (ii) by 6: \( 30x + 36y = 420000 \) ...... (iv) Subtracting (iii) from (iv): \( (30x + 36y) - (30x + 25y) = 420000 - 325000 \)

\( \implies 11y = 95000 \)

\( \implies y = \frac{95000}{11} \) From equation (i), \( 6x + 5 \times \frac{95000}{11} = 65000 \)

\( \implies 6x = 65000 - \frac{475000}{11} \)

\( \implies 6x = \frac{715000 - 475000}{11} = \frac{240000}{11} \)

\( \implies x = \frac{40000}{11} \) The amount invested at 12% is Rs.\( \frac{40000}{11} \) (approximately Rs.3636.36) and the amount invested at 10% is Rs.\( \frac{95000}{11} \) (approximately Rs.8636.36).
In simple words: Simple interest is calculated as (Rate/100) × Principal. Write one equation for the original investment and another for the interchanged amounts, then solve both together.

Exam Tip: When dealing with simple interest problems, always convert percentages correctly into decimal form. Check that your final answer makes sense by substituting back into both original conditions.

Question 21. A man invested money at two different rates of interest. He invested Rs.x at 12% per annum and Rs.y at 10% per annum, earning Rs.1300 in yearly interest. If the amounts were swapped - Rs.y at 12% and Rs.x at 10% - he would have received Rs.40 more as yearly interest. How much did he invest at different rates?
Answer: Let the amount invested at 12% S.I. be Rs.x and at 10% S.I. be Rs.y.

According to the first condition:
\( \frac{12}{100} \times x + \frac{10}{100} \times y = 1300 \)

\( \Rightarrow 12x + 10y = 130000 \)
\( \Rightarrow 6x + 5y = 65000 \) ......(i)

Now, let the amount invested at 12% S.I. be Rs.y and at 10% S.I. be Rs.x.

According to the second condition:
\( \frac{12}{100} \times y + \frac{10}{100} \times x = 1340 \)

\( \Rightarrow 12y + 10x = 134000 \)
\( \Rightarrow 6y + 5x = 67000 \) ......(ii)

Multiplying (i) by 6 and (ii) by 5, we get:
\( 36x + 30y = 390000 \) ......(iii)
\( 30y + 25x = 335000 \) .......(iv)

Subtracting equation (iv) from (iii):
\( (36x + 30y) - (30y + 25x) = 390000 - 335000 \)
\( \Rightarrow 11x = 55000 \)
\( \Rightarrow x = 5000 \)

Substituting the value of x in (i):
\( 6(5000) + 5y = 65000 \)
\( \Rightarrow 5y = 35000 \)
\( \Rightarrow y = 7000 \)

Therefore, the investment at 12% is Rs.5000 and the investment at 10% is Rs.7000.
In simple words: When you invest money at different interest rates, you can set up two equations based on the two scenarios described. Solve them together by multiplying and subtracting to find how much was invested at each rate.

Exam Tip: Always set up both equations clearly based on each condition given, and verify your answer by substituting back into the original equations.

Question 22. A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting Rs.1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got Rs.20 more. Find the cost price of the table and the list price of the chair.
Answer: Let the cost price of the table be Rs.x and list price of the chair be Rs.y.

According to the first condition:
\( \frac{108}{100}x + \frac{90}{100}y = 1008 \)

\( \Rightarrow 108x + 90y = 100800 \)
\( \Rightarrow 18(6x + 5y) = 100800 \)
\( \Rightarrow 6x + 5y = 5600 \) ......(i)

According to the second condition:
\( \frac{110}{100}x + \frac{92}{100}y = 1028 \)

\( \Rightarrow 110x + 92y = 102800 \)
\( \Rightarrow 2(55x + 46y) = 102800 \)
\( \Rightarrow 55x + 46y = 51400 \) ......(ii)

Multiplying (i) by 55 and (ii) by 6:
\( 330x + 275y = 308000 \) ......(iii)
\( 330x + 276y = 308400 \) ......(iv)

Subtracting equation (iii) from (iv):
\( (330x + 276y) - (330x + 275y) = 308400 - 308000 \)
\( \Rightarrow y = 400 \)

Substituting the value of y in (i):
\( 6x + 5(400) = 5600 \)
\( \Rightarrow 6x = 3600 \)
\( \Rightarrow x = 600 \)

Therefore, the cost price of the table is Rs.600 and the list price of the chair is Rs.400.
In simple words: When an item is sold at a profit or discount, multiply the cost or list price by the appropriate percentage to find the selling price. Create two equations from the two scenarios and solve them together.

Exam Tip: Remember that profit is added to the cost price while discount is subtracted from the list price. Always express the conditions clearly before forming equations.

Question 23. A and B have some money with them. A said to B, 'if you give me Rs.100, my money will become 75% of the money left with you'. B said to A, 'instead if you give me Rs.100, your money will become 40% of my money. How much money did A and B have originally?
Answer: Let A have Rs.x and B have Rs.y.

According to the first condition:
\( x + 100 = \frac{75}{100}(y - 100) \)

\( \Rightarrow 100(x + 100) = 75(y - 100) \)
\( \Rightarrow 100x + 10000 = 75y - 7500 \)
\( \Rightarrow 75y - 100x = 17500 \)
\( \Rightarrow 25(3y - 4x) = 25 \times 700 \)
\( \Rightarrow 3y - 4x = 700 \) ......(i)

According to the second condition:
\( \frac{40}{100}(y + 100) = x - 100 \)

\( \Rightarrow 40(y + 100) = 100(x - 100) \)
\( \Rightarrow 40y + 4000 = 100x - 10000 \)
\( \Rightarrow 100x - 40y = 14000 \)
\( \Rightarrow 20(5x - 2y) = 20 \times 700 \)
\( \Rightarrow 5x - 2y = 700 \) ......(ii)

Multiplying (i) by 2 and (ii) by 3:
\( 6y - 8x = 1400 \) ......(iii)
\( 15x - 6y = 2100 \) ......(iv)

Adding (iii) and (iv):
\( (6y - 8x) + (15x - 6y) = 1400 + 2100 \)
\( \Rightarrow 7x = 3500 \)
\( \Rightarrow x = 500 \)

Substituting the value of x in (i):
\( 3y - 4(500) = 700 \)
\( \Rightarrow 3y = 2700 \)
\( \Rightarrow y = 900 \)

Therefore, A has Rs.500 and B has Rs.900.
In simple words: When money is transferred between two people, write down what happens to their amounts. Use the resulting equations to find how much each person originally had.

Exam Tip: Pay careful attention to what percentage is being compared - whether it's a percentage of the remaining amount or the other person's amount - and translate this into the correct mathematical statement.

Question 24. The students of a class are made to stand in (complete) rows. If one student is extra in a row, there would be 2 rows less, and if one student is less in a row, there would be 3 rows more. Find the number of students in the class.
Answer: Let the number of students in a row be x and the number of rows be y.

Total number of students = xy

According to the first condition:
\( (x + 1)(y - 2) = xy \)

\( \Rightarrow xy - 2x + y - 2 = xy \)
\( \Rightarrow y - 2x = 2 \) ......(i)

According to the second condition:
\( (x - 1)(y + 3) = xy \)

\( \Rightarrow xy + 3x - y - 3 = xy \)
\( \Rightarrow 3x - y = 3 \) ......(ii)

Adding (i) and (ii):
\( (y - 2x) + (3x - y) = 2 + 3 \)
\( \Rightarrow x = 5 \)

Substituting the value of x in (i):
\( y - 2(5) = 2 \)
\( \Rightarrow y = 12 \)

Total number of students = xy = 5 × 12 = 60

Therefore, there are 60 students in the class.
In simple words: If you change the number of students in each row, the number of rows must change to keep the total the same. Set up equations based on these changes and solve to find the total number of students.

Exam Tip: Remember that the product xy (total students) remains constant in both cases. This constraint is key to setting up the two equations correctly.

Question 25. A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams? (Pure gold is 24-carat).
Answer: Let the quantity of 18-carat gold be x gm and 12-carat gold be y gm.

Given, total weight of the new bar = 120 gm

\( x + y = 120 \) ......(i)

Pure gold is 24-carat.

Purity of 18-carat gold = \( \frac{18}{24} \times 100 = 75\% \)

Purity of 12-carat gold = \( \frac{12}{24} \times 100 = 50\% \)

Purity of 16-carat gold = \( \frac{16}{24} \times 100 = \frac{200}{3}\% \)

According to the condition (purity of mixture):
\( \frac{75}{100}x + \frac{50}{100}y = \frac{200}{3 \times 100} \times 120 \)

\( \Rightarrow \frac{3}{4}x + \frac{1}{2}y = 80 \)

\( \Rightarrow \frac{3x + 2y}{4} = 80 \)

\( \Rightarrow 3x + 2y = 320 \) ......(ii)

Multiplying (i) by 2:
\( 2x + 2y = 240 \) ......(iii)

Subtracting equation (iii) from (ii):
\( (3x + 2y) - (2x + 2y) = 320 - 240 \)
\( \Rightarrow x = 80 \)

Substituting the value of x in (i):
\( 80 + y = 120 \)
\( \Rightarrow y = 40 \)

Therefore, the jeweller requires 80 gm of 18-carat gold and 40 gm of 12-carat gold to obtain a bar of 16-carat gold weighing 120 gm.
In simple words: The purity of each type of gold tells you what fraction is pure gold. When you mix them, the total amount of pure gold in the mixture must match the purity of the final bar.

Exam Tip: Always convert carat values to percentages of purity before setting up the equation. The sum of pure gold from both bars must equal the pure gold in the final mixture.

Question 26. A and B together can do a piece of work in 15 days. If A's one day work is 1\( \frac{1}{2} \) times the one day's work of B, find in how many days can each do the work.
Answer: Let A's one day work be x and B's one day work be y.

According to the first condition given in the problem:
\( x = \frac{3}{2}y \)

\( \Rightarrow 2x = 3y \)
\( \Rightarrow 2x - 3y = 0 \) ......(i)

Also given, A and B together can do the work in 15 days:
\( x + y = \frac{1}{15} \)

\( \Rightarrow 15(x + y) = 1 \)
\( \Rightarrow 15x + 15y = 1 \) ......(ii)

Multiplying (i) by 5:
\( 10x - 15y = 0 \) ......(iii)

Adding (ii) and (iii):
\( 15x + 15y + 10x - 15y = 1 + 0 \)
\( \Rightarrow 25x = 1 \)
\( \Rightarrow x = \frac{1}{25} \)

Substituting x in (i):
\( 2 \times \frac{1}{25} - 3y = 0 \)
\( \Rightarrow 3y = \frac{2}{25} \)
\( \Rightarrow y = \frac{2}{75} \)

Since A's one day work is x and B's one day work is y:
\( \frac{1}{x} = \frac{1}{\frac{1}{25}} = 25 \) days (A can complete the work in 25 days)

\( \frac{1}{y} = \frac{1}{\frac{2}{75}} = \frac{75}{2} = 37\frac{1}{2} \) days (B can complete the work in 37\( \frac{1}{2} \) days)

Therefore, A will do the work in 25 days and B will do the work in 37\( \frac{1}{2} \) days.
In simple words: If A works 1.5 times faster than B, then A takes less time to finish the job alone. The one day work rates add up to give the combined work rate.

Exam Tip: Express work rates as fractions of the job completed per day. The reciprocal of the work rate gives the number of days needed to complete the work.

Question 27. 2 men and 5 women can do a piece of work in 4 days, while one man and one woman can finish it in 12 days. How long would it take for 1 man to do the work?
Answer: Let 1 man take x days to do the work and 1 woman take y days.

Work done by 1 man in 1 day = \( \frac{1}{x} \)

Work done by 1 woman in 1 day = \( \frac{1}{y} \)

Work done by 2 men in 1 day = \( \frac{2}{x} \)

Work done by 5 women in 1 day = \( \frac{5}{y} \)

According to the first condition:
\( \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \) ......(i)

According to the second condition:
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \) ......(ii)

Multiplying (ii) by 5:
\( \frac{5}{x} + \frac{5}{y} = \frac{5}{12} \) ......(iii)

Subtracting (i) from (iii):
\( \left(\frac{5}{x} + \frac{5}{y}\right) - \left(\frac{2}{x} + \frac{5}{y}\right) = \frac{5}{12} - \frac{1}{4} \)

\( \Rightarrow \frac{5}{x} - \frac{2}{x} = \frac{5 - 3}{12} \)

\( \Rightarrow \frac{3}{x} = \frac{2}{12} \)

\( \Rightarrow \frac{3}{x} = \frac{1}{6} \)

\( \Rightarrow x = 18 \)

Therefore, one man can do the work in 18 days.
In simple words: When different numbers of people work together, their individual work rates add up. Set up equations for the combined work and solve to find each person's individual rate.

Exam Tip: Express the work rates as reciprocals of time. Multiply an equation by a suitable number to eliminate one variable when subtracting equations.

Question 28. A train covered a certain distance at a uniform speed. If the train had been 30 km/h faster, it would have taken 2 hours less than scheduled time. If the train were slower by 15 km/h, it would have taken 2 hours more than the scheduled time. Find the length of the journey.
Answer: Let the actual speed of the train be x km/h and the scheduled time be y hours.

Distance = Speed × Time = xy

According to the first condition:
\( (x + 30)(y - 2) = xy \)

\( \Rightarrow xy - 2x + 30y - 60 = xy \)
\( \Rightarrow -2x + 30y = 60 \) ......(i)

According to the second condition:
\( (x - 15)(y + 2) = xy \)

\( \Rightarrow xy + 2x - 15y - 30 = xy \)
\( \Rightarrow 2x - 15y = 30 \) ......(ii)

Adding (i) and (ii):
\( (-2x + 30y) + (2x - 15y) = 60 + 30 \)
\( \Rightarrow 15y = 90 \)
\( \Rightarrow y = 6 \)

Substituting the value of y in (i):
\( -2x + 30(6) = 60 \)
\( \Rightarrow -2x + 180 = 60 \)
\( \Rightarrow -2x = -120 \)
\( \Rightarrow x = 60 \)

Distance = xy = 60 × 6 = 360 km

Therefore, the distance of the journey is 360 km.
In simple words: When speed changes, time changes in the opposite direction while distance stays the same. Use this fact to create two equations and solve for speed and time.

Exam Tip: Always verify that the distance remains constant by checking both conditions. The distance formula (Speed × Time = Distance) is key to setting up correct equations.

Question 29. A boat takes 2 hours to go 40 km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.

Speed of boat downstream = (x + y) km/h

Speed of boat upstream = (x - y) km/h

Given, the boat takes 2 hours to go 40 km down the stream:
\( \frac{40}{x + y} = 2 \)

\( \Rightarrow 2(x + y) = 40 \)
\( \Rightarrow x + y = 20 \) ......(i)

Given, the boat returns in 4 hours:
\( \frac{40}{x - y} = 4 \)

\( \Rightarrow 4(x - y) = 40 \)
\( \Rightarrow x - y = 10 \) ......(ii)

Adding (i) and (ii):
\( (x + y) + (x - y) = 20 + 10 \)
\( \Rightarrow 2x = 30 \)
\( \Rightarrow x = 15 \)

Subtracting (ii) from (i):
\( (x + y) - (x - y) = 20 - 10 \)
\( \Rightarrow 2y = 10 \)
\( \Rightarrow y = 5 \)

Therefore, the speed of the boat in still water is 15 km/h and the speed of the stream is 5 km/h.
In simple words: When a boat moves downstream, the stream helps it move faster. When it moves upstream, the stream works against it. The combined and opposing speeds lead to the two equations.

Exam Tip: Downstream speed = boat speed + stream speed. Upstream speed = boat speed - stream speed. Use distance = speed × time to set up equations, then add and subtract them to find each speed separately.

Question 30. A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48 minutes longer to cover the same distance against the current. Find the speed of the boat in still water and the speed of the current.
Answer: Let the boat's speed in still water be x km/h and the current's speed be y km/h.

When moving with the current, the boat's speed becomes (x + y) km/h. When moving against the current, it becomes (x - y) km/h.

Since the boat covers 44 km in 4 hours with the current:
\[ \frac{44}{x + y} = 4 \]
\[ \Rightarrow 4(x + y) = 44 \]
\[ \Rightarrow x + y = 11 \text{ .......(i)} \]

The boat takes 4 hours 48 minutes (which is 4 + 48/60 = 4 + 4/5 = 24/5 hours) against the current:
\[ \frac{44}{x - y} = 4 + \frac{4}{5} \]
\[ \Rightarrow \frac{44}{x - y} = \frac{20 + 20 + 4}{5} \]
\[ \Rightarrow \frac{44}{x - y} = \frac{44}{5} \]
\[ \Rightarrow x - y = 5 \text{ .......(ii)} \]

Adding equations (i) and (ii):
\[ (x + y) + (x - y) = 11 + 5 \]
\[ \Rightarrow 2x = 16 \]
\[ \Rightarrow x = 8 \]

Substituting x = 8 in equation (i):
\[ 8 + y = 11 \]
\[ \Rightarrow y = 3 \]

Therefore, the boat's speed in still water is 8 km/h and the current's speed is 3 km/h.
In simple words: Set up two equations - one for going with the current and one for going against it. Add the equations to find the boat's speed, then subtract to find the current's speed.

Exam Tip: Always convert time to a single unit (hours as fractions) before setting up equations. Checking: with current gives 44/11 = 4 hours; against current gives 44/5 = 8.8 hours = 8 hours 48 minutes, which is 4 hours 48 minutes more than 4 hours.

Question 31. An aeroplane flies 1680 km with a head wind in 3.5 hours. On the return trip with same wind blowing, the plane takes 3 hours. Find the plane's air speed and the wind speed.
Answer: Let the plane's air speed be x km/h and the wind speed be y km/h.

When flying against the wind, the plane's effective speed becomes (x - y) km/h. When flying with the wind, it becomes (x + y) km/h.

The plane covers 1680 km against the wind in 3.5 hours:
\[ \frac{1680}{x - y} = 3.5 \]
\[ \Rightarrow 3.5(x - y) = 1680 \]
\[ \Rightarrow x - y = 480 \text{ .......(i)} \]

The plane covers 1680 km with the wind in 3 hours:
\[ \frac{1680}{x + y} = 3 \]
\[ \Rightarrow 1680 = 3(x + y) \]
\[ \Rightarrow x + y = 560 \text{ .......(ii)} \]

Adding equations (i) and (ii):
\[ (x - y) + (x + y) = 480 + 560 \]
\[ \Rightarrow 2x = 1040 \]
\[ \Rightarrow x = 520 \]

Substituting x = 520 in equation (i):
\[ 520 - y = 480 \]
\[ \Rightarrow y = 40 \]

Therefore, the plane's air speed is 520 km/h and the wind speed is 40 km/h.
In simple words: The plane moves faster with the wind and slower against it. Use the two time equations to set up a system, then solve for the plane's actual speed and the wind's speed separately.

Exam Tip: Remember that against the wind, time increases; with the wind, time decreases. Always define variables clearly as the object's own speed plus/minus the external speed (wind or current).

Question 32. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When Bhawana takes food for 20 days, she has to pay Rs.2600 as hostel charges; whereas when Divya takes food for 26 days, she pays Rs.3020 as hostel charges. Find the fixed charges and the cost of food per day.
Answer: Let the fixed charge be Rs. x and the per-day food charge be Rs. y.

Bhawana pays Rs. 2600 for 20 days of food:
\[ x + 20y = 2600 \text{ .......(i)} \]

Divya pays Rs. 3020 for 26 days of food:
\[ x + 26y = 3020 \text{ .......(ii)} \]

Subtracting equation (i) from equation (ii):
\[ (x + 26y) - (x + 20y) = 3020 - 2600 \]
\[ \Rightarrow 6y = 420 \]
\[ \Rightarrow y = 70 \]

Substituting y = 70 in equation (i):
\[ x + 20(70) = 2600 \]
\[ \Rightarrow x + 1400 = 2600 \]
\[ \Rightarrow x = 1200 \]

Therefore, the fixed charge is Rs. 1200 and the cost of food per day is Rs. 70.
In simple words: The total cost for each person is the fixed part plus a daily food charge multiplied by the number of days. Subtract one equation from the other to find the daily charge, then find the fixed charge.

Exam Tip: When working with word problems involving a fixed part and a variable part, subtraction of equations is most efficient to eliminate one variable in a single step.

Multiple Choice Questions

Question 1. Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is
(1) 35
(2) 53
(3) 26
(4) 62
Answer: (1) 35
In simple words: Set the tens digit as x and the ones digit as y. The original number is 10x + y. When you reverse the digits, you get 10y + x. Use the two conditions (sum of digits and the reversal difference) to form two equations and solve.

Exam Tip: Always express a two-digit number in the form 10 × (tens digit) + (ones digit). This makes it easy to set up the reversal equation correctly.

Question 2. The sum of two natural numbers is 25 and their difference is 7. The numbers are
(1) 17 and 8
(2) 16 and 9
(3) 18 and 7
(4) 15 and 10
Answer: (2) 16 and 9
In simple words: When you have the sum and the difference of two numbers, add the two equations to find double the larger number, then subtract to find double the smaller number.

Exam Tip: For sum and difference problems, adding the equations always gives you 2 times the larger number; subtracting gives you 2 times the smaller number. This avoids fractions and works quickly.

Question 3. The sum of two natural numbers is 240 and their ratio is 3 : 5. Then the greater number is
(1) 180
(2) 160
(3) 150
(4) 90
Answer: (3) 150
In simple words: If two numbers are in the ratio 3 : 5, you can write them as 3k and 5k for some value k. Their sum is 3k + 5k = 8k. Use this to find k, then multiply to get the actual numbers.

Exam Tip: When dealing with ratio problems, always express the numbers in terms of a common multiplier (like 3k and 5k). This converts a ratio problem into a single-variable problem.

Question 4. The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is
(1) 27
(2) 72
(3) 63
(4) 36
Answer: (4) 36
In simple words: Express the number as 10x + y where x is the tens digit and y is the ones digit. When you add 27, you get the reversed number 10y + x. Form two equations from these conditions and solve for x and y.

Exam Tip: After forming the equation from "adding 27 gives the reversed number," simplify carefully to avoid sign errors. A common mistake is reversing the signs when moving terms across the equals sign.

Question 5. The sum of the digits of a two digit number is 12. If the number is decreased by 18, its digits get reversed. The number is
(1) 48
(2) 84
(3) 57
(4) 75
Answer: (4) 75
In simple words: Write the original number as 10x + y. When you subtract 18, you get 10x + y - 18, which equals the reversed number 10y + x. Rearrange to get x and y, then calculate the number.

Exam Tip: When digits are reversed after a decrease, the tens digit becomes smaller and the ones digit becomes larger. Check your answer by verifying both conditions: the digit sum and the reversal after the decrease.

Question 6. Aruna has only Rs.1 and Rs.2 coins with her. If the total number of coins that she has is 50 and the amount of the money with her is Rs.75, then the number of Rs.1 and Rs.2 coins are respectively
(1) 35 and 15
(2) 35 and 20
(3) 15 and 75
(4) 25 and 25
Answer: (4) 25 and 25
In simple words: Set up one equation for the total number of coins and another for the total money. The first equation counts all coins; the second equation multiplies each coin type by its value and adds them.

Exam Tip: Always set up two distinct equations: one for counting (number of items) and one for value (amount of money). Subtracting the count equation from the value equation directly gives you the difference in coin counts.

Question 7. The age of a woman is four times the age of her daughter. Five years hence, the age of the woman will be three times the age of her daughter. The present age of the daughter is
(1) 40 years
(2) 20 years
(3) 15 years
(4) 10 years
Answer: (4) 10 years
In simple words: Let the daughter's present age be x years. Then the woman's age is 4x years. In 5 years, the daughter will be x + 5 and the woman will be 4x + 5. The second condition says the woman's future age will be 3 times the daughter's future age.

Exam Tip: For age problems involving future time ("five years hence"), always add the same number of years to both people's current ages. The relationship between their ages may change over time, so be careful to apply the new condition to the future ages, not the present ones.

Question 8. Father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present age in years of the son and the father are, respectively,
(1) 4 and 24
(2) 5 and 30
(3) 6 and 36
(4) 3 and 24
Answer: (4) 3 and 24
In simple words: Let the son's present age be x years, so the father's age is 6x years. After 4 years, the son will be x + 4 and the father will be 6x + 4. Use the condition that the father's future age is 4 times the son's future age to form an equation and solve.

Exam Tip: In age problems, the ratio between two people's ages changes as time passes. Always apply each given ratio or relationship to the correct time period - either "now" (present age) or "then" (future or past age). Double-check by substituting back into both original conditions.

Question 9. Consider the following two statements: Statement 1: A husband is 2 years older than his wife, and sum of their ages is 52 years. Then the wife is 25 years old. Statement 2: A father is twice as old as his daughter, and difference of their ages is 26 years. Then the father is 50 years old. Which of the following is valid?
(1) Both the statements are true.
(2) Both the statements are false.
(3) Statement 1 is true, and Statement 2 is false.
(4) Statement 1 is false, and Statement 2 is true.
Answer: (1) Both the statements are true
In simple words: When you check both statements by solving the equations, you find that they both give you the correct ages. The wife turns out to be 25, and the father turns out to be 52.

Exam Tip: Always verify each statement separately using the given conditions - set up two equations, solve them, and then check if the answer matches what the statement claims.

Question. Assertion (A): Difference between ages of two brothers is 5 years, while sum of their ages is 25 years. Then the younger is 10 years old. Reason (R): The difference between age of two brothers remains constant, even when they grow older.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A)
In simple words: The assertion is correct - the younger brother is indeed 10 years old. The reason is also true - the age gap between brothers never changes. However, the reason doesn't actually explain why the younger brother is 10; we get that age from the two given equations, not from knowing the gap stays constant.

Exam Tip: In assertion-reason questions, always check both parts separately first. A statement can be true but not be the actual reason that makes the assertion true - this is a key distinction examiners test.

Question. Assertion (A): Perimeter of a garden is 24 cm, while the difference between its length and width is 2 units. Then its area is 35 sq. cm. Reason (R): If length of a rectangle is doubled, while width remains the same, then the perimeter also gets doubled.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (1) Assertion (A) is true, Reason (R) is false
In simple words: When you solve for the garden's dimensions using the perimeter and length-width difference, you get length = 7 cm and width = 5 cm, so the area really is 35 sq. cm. But the reason is wrong - if you double just the length, the perimeter does not double; it increases by twice the original length.

Exam Tip: Always test the reason independently. For the rectangle perimeter claim, try a simple example: if a 2 × 3 rectangle has perimeter 10, doubling the length gives 4 × 3 with perimeter 14, not 20.

Chapter Test

Question 1. A 700 g dry fruit pack costs Rs. 432. It contains some almonds and the rest cashew kernel. If almonds cost Rs. 576 per kg and cashew kernel cost Rs. 672 per kg, what are the quantities of the two dry fruits separately?
Answer: Set the quantity of almonds as x gm and cashew kernel as y gm. The total weight gives us x + y = 700, so x = 700 - y. For the cost, almonds cost \( \frac{576x}{1000} \) and cashew costs \( \frac{672y}{1000} \). The total cost is Rs. 432, so \( \frac{576x + 672y}{1000} = 432 \), which simplifies to 576x + 672y = 432000. Substituting x = 700 - y into this equation:
\( 576(700 - y) + 672y = 432000 \)
\( \implies 403200 - 576y + 672y = 432000 \)
\( \implies 96y = 28800 \)
\( \implies y = 300 \)
Therefore x = 700 - 300 = 400. The pack contains 400 gm of almonds and 300 gm of cashew kernel.
In simple words: We write two equations - one for the total weight and one for the total cost. Then we solve by replacing one variable in the second equation using the first equation.

Exam Tip: Remember to convert costs to the same unit (paise or rupees per gram) before setting up the equation. Always verify your answer by checking both the total weight and total cost.

Question 2. Drawing pencils cost Rs. 4 each and coloured pencils cost Rs. 5.50 each. If altogether two dozen pencils cost Rs. 108, how many coloured pencils are there?
Answer: Let the number of drawing pencils be x and coloured pencils be y. Two dozen means 24 total pencils, so x + y = 24, giving us x = 24 - y. The total cost equation is 4x + 5.50y = 108. Substituting x = 24 - y:
\( 4(24 - y) + 5.50y = 108 \)
\( \implies 96 - 4y + 5.50y = 108 \)
\( \implies 96 + 1.5y = 108 \)
\( \implies 1.5y = 12 \)
\( \implies y = 8 \)
Therefore, there are 8 coloured pencils in the purchase.
In simple words: We use the total number of pencils to create one equation and the total money spent to create another. By combining these two equations, we can find how many of each type.

Exam Tip: When dealing with two different items and costs, always set up both a quantity equation and a cost equation. Check your answer: 16 drawing pencils at Rs. 4 = Rs. 64, and 8 coloured pencils at Rs. 5.50 = Rs. 44, totalling Rs. 108.

Question 3. Shikha works in a factory. In one week she earned Rs. 3,900 for working 47 hours, of which 7 hours were overtime. The next week she earned Rs. 4,160 for working 50 hours, of which 8 hours were overtime. What is Shikha's hourly earning rate?
Answer: Let Shikha earn Rs. x per hour for regular work and Rs. y per hour for overtime. In week 1, she worked 40 regular hours (47 - 7) and 7 overtime hours, earning Rs. 3,900:
\( 40x + 7y = 3900 \) ... (1)
In week 2, she worked 42 regular hours (50 - 8) and 8 overtime hours, earning Rs. 4,160:
\( 42x + 8y = 4160 \) ... (2)
Multiply equation (1) by 8: \( 320x + 56y = 31200 \) ... (3)
Multiply equation (2) by 7: \( 294x + 56y = 29120 \) ... (4)
Subtract (4) from (3):
\( 26x = 2080 \)
\( \implies x = 80 \)
Substitute back into equation (1):
\( 40(80) + 7y = 3900 \)
\( \implies 3200 + 7y = 3900 \)
\( \implies 7y = 700 \)
\( \implies y = 100 \)
Shikha earns Rs. 80 per hour for regular work and Rs. 100 per hour for overtime work.
In simple words: We set up two equations from two different weeks, making sure to separate regular hours from overtime hours. Then we solve by eliminating one variable.

Exam Tip: Always identify which hours are regular and which are overtime before writing equations. Use the elimination method when the coefficients allow it - multiply the equations strategically so one variable cancels out.

Question 4. The sum of digits of a two digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the number.
Answer: Let x be the tens digit and y be the units digit. The original number is 10x + y, and the reversed number is 10y + x. The sum of digits is 7, so:
\( x + y = 7 \)
\( \implies x = 7 - y \) ... (i)
When the digits are reversed and the new number is increased by 3, it equals 4 times the original:
\( 10y + x + 3 = 4(10x + y) \)
\( \implies 10y + x + 3 = 40x + 4y \)
\( \implies 6y - 39x = -3 \)
\( \implies 13x - 2y = 1 \) ... (ii)
Substitute x = 7 - y into equation (ii):
\( 13(7 - y) - 2y = 1 \)
\( \implies 91 - 13y - 2y = 1 \)
\( \implies 91 - 15y = 1 \)
\( \implies 15y = 90 \)
\( \implies y = 6 \)
Therefore x = 7 - 6 = 1, and the number is 10(1) + 6 = 16.
In simple words: For a two-digit number, the tens place and units place are different positions. When we reverse them, we get a completely different number. We use this fact along with the given conditions to set up equations.

Exam Tip: Always check your answer: 16 reversed is 61. Is 61 + 3 = 64? Is 64 = 4 × 16? Yes. Do 1 and 6 add to 7? Yes. Your answer is confirmed.

Question 5. Three years hence a man's age will be three times his son's age, and 7 years ago he was seven times as old as his son. How old are they now?
Answer: Let the son's current age be x and the man's current age be y. Three years from now, the man's age will be three times the son's age:
\( y + 3 = 3(x + 3) \)
\( \implies y + 3 = 3x + 9 \)
\( \implies y - 3x = 6 \) ... (i)
Seven years ago, the man was seven times as old as his son:
\( y - 7 = 7(x - 7) \)
\( \implies y - 7 = 7x - 49 \)
\( \implies y - 7x = -42 \) ... (ii)
Subtract equation (i) from equation (ii):
\( (y - 7x) - (y - 3x) = -42 - 6 \)
\( \implies -4x = -48 \)
\( \implies x = 12 \)
Substitute into equation (i):
\( y - 3(12) = 6 \)
\( \implies y - 36 = 6 \)
\( \implies y = 42 \)
The son is currently 12 years old and the man is currently 42 years old.
In simple words: We set up two equations using future and past conditions. One equation talks about "three years from now" and the other about "seven years ago." We solve both together to find their ages today.

Exam Tip: Verify both conditions: In 3 years, the man will be 45 and the son will be 15, and 45 = 3 × 15. Seven years ago, the man was 35 and the son was 5, and 35 = 7 × 5. Both check out.

Question 6. Rectangles are drawn on line segments of fixed lengths. When the breadths are 6m and 5m respectively the sum of the areas of the rectangles is 83 m2. But if the breadths are 5m and 4m respectively the sum of the areas is 68 m2. Find the sum of the areas of squares drawn on the line segments.
Answer: Let the length of the first fixed line segment be x and the second line segment be y. In the first case, when the breadths measure 6m and 5m, the total area equals 83 m2, giving us the equation \( 6x + 5y = 83 \) ......(i). In the second case, with breadths of 5m and 4m, the total area is 68 m2, so \( 5x + 4y = 68 \) ......(ii). To solve, multiply equation (i) by 4 and equation (ii) by 5, yielding \( 24x + 20y = 332 \) ......(iii) and \( 25x + 20y = 340 \) ......(iv). Subtract (iii) from (iv): \( 25x + 20y - (24x + 20y) = 340 - 332 \) giving \( x = 8 \). Substitute x = 8 into (i): \( 6(8) + 5y = 83 \) \( \implies 48 + 5y = 83 \) \( \implies 5y = 35 \) \( \implies y = 7 \). The sum of the areas of squares on these two line segments is \( x^2 + y^2 = 8^2 + 7^2 = 64 + 49 = 113 \) m2.
In simple words: Set up two equations using the rectangle areas for different breadths. Solve them to find the line segment lengths (8 and 7), then add their squares to get 113 m2.

Exam Tip: Always form equations from the given conditions carefully, then use elimination or substitution to find the unknowns. The final answer requires you to find x2 + y2, not just x and y.

Question 7. If the length and breadth of a room are increased by 1 metre each, the area is increased by 21 square meters. If the length is decreased by 1 meter and the breadth is increased by 2 meters, the area is increased by 14 square meters. Find the perimeter of the room.
Answer: Let the length be x metres and breadth be y metres, so the original area is xy m2. From the first condition, when both dimensions increase by 1 metre: \( (x + 1)(y + 1) = xy + 21 \) \( \implies xy + x + y + 1 = xy + 21 \) \( \implies x + y = 20 \) ......(i). From the second condition, the length decreases by 1 metre while breadth increases by 2 metres: \( (x - 1)(y + 2) = xy + 14 \) \( \implies xy + 2x - y - 2 = xy + 14 \) \( \implies 2x - y = 16 \) ......(ii). Add equations (i) and (ii): \( x + y + (2x - y) = 20 + 16 \) \( \implies 3x = 36 \) \( \implies x = 12 \). Substitute x = 12 into (i): \( 12 + y = 20 \) \( \implies y = 8 \). Therefore, the perimeter is \( 2(x + y) = 2(12 + 8) = 2(20) = 40 \) metres.
In simple words: Write two equations from the area changes, solve for length and breadth, then multiply their sum by 2 to find the perimeter.

Exam Tip: Always expand the product form of the new dimensions carefully and simplify to get the equations. Verify your answer by checking both original conditions with x = 12 and y = 8.

Question 8. The lengths (in meters) of the sides of a triangle are \( 2x + \frac{y}{2} \), \( \frac{5x}{3} + y + \frac{1}{2} \), and \( \frac{2x}{3} + 2y + \frac{5}{2} \). If the triangle is equilateral, find its perimeter.
Answer: Since the triangle is equilateral, all three sides must be equal. Setting the first side equal to the second side:
\( 2x + \frac{y}{2} = \frac{5x}{3} + y + \frac{1}{2} \)
\( \implies 2x - \frac{5x}{3} + \frac{y}{2} - y = \frac{1}{2} \)
\( \implies \frac{6x - 5x}{3} - \frac{y}{2} = \frac{1}{2} \)
\( \implies \frac{x}{3} - \frac{y}{2} = \frac{1}{2} \)
\( \implies \frac{2x - 3y}{6} = \frac{1}{2} \)
\( \implies 2x - 3y = 3 \) ......(i) Setting the second side equal to the third side:
\( \frac{5x}{3} + y + \frac{1}{2} = \frac{2x}{3} + 2y + \frac{5}{2} \)
\( \implies \frac{5x}{3} - \frac{2x}{3} + y - 2y = \frac{5}{2} - \frac{1}{2} \)
\( \implies \frac{3x}{3} - y = 2 \)
\( \implies x - y = 2 \) ......(ii) Multiply equation (ii) by 2: \( 2x - 2y = 4 \) ......(iii). Subtract (i) from (iii): \( 2x - 2y - (2x - 3y) = 4 - 3 \) \( \implies y = 1 \). Substitute y = 1 into (i): \( 2x - 3(1) = 3 \) \( \implies 2x = 6 \) \( \implies x = 3 \). The length of one side is \( 2x + \frac{y}{2} = 2(3) + \frac{1}{2} = 6 + 0.5 = 6.5 \) m. Since the triangle is equilateral, all three sides measure 6.5 m each. Therefore, the perimeter is \( 3 \times 6.5 = 19.5 \) m.
In simple words: For an equilateral triangle, all sides are equal. Set them equal to each other and solve for x and y, then calculate one side length and multiply by 3.

Exam Tip: Always simplify the equations carefully before solving. Double-check by substituting your values back into all three original side expressions to confirm they are all equal.

Question 9. On Diwali eve, two candles, one of which is 3 cm longer than the other, are lighted. The longer one is lighted at 5.30 p.m. and the shorter at 7 p.m. At 9.30 p.m. they both are of same length. The longer one burns out at 11.30 p.m. and the shorter one at 11 p.m. How long was each candle originally?
Answer: Assume the longer candle burns at a rate of x cm/hr and the shorter candle at a rate of y cm/hr. Given that the longer candle burns for 6 hours (5.30 p.m. to 11.30 p.m.) and the shorter burns for 4 hours (7 p.m. to 11 p.m.), their original lengths are 6x cm and 4y cm respectively. From the first condition, the longer candle is 3 cm longer than the shorter: \( 6x = 4y + 3 \) \( \implies 6x - 4y = 3 \) ......(i). At 9.30 p.m., the longer candle has burned for 4 hours, so its remaining length is \( 6x - 4x = 2x \) cm. The shorter candle has burned for 2.5 hours (7 p.m. to 9.30 p.m.), so its remaining length is \( 4y - 2.5y = 1.5y = \frac{3y}{2} \) cm. Since both candles have the same length at 9.30 p.m.:
\( 2x = \frac{3y}{2} \)
\( \implies 4x = 3y \)
\( \implies 4x - 3y = 0 \) ......(ii) Multiply (i) by 3 and (ii) by 4: \( 18x - 12y = 9 \) ......(iii) and \( 16x - 12y = 0 \) ......(iv). Subtract (iv) from (iii): \( 18x - 12y - (16x - 12y) = 9 - 0 \) \( \implies 2x = 9 \) \( \implies x = 4.5 \) cm/hr. Substitute into (ii): \( 4(4.5) - 3y = 0 \) \( \implies 18 - 3y = 0 \) \( \implies y = 6 \) cm/hr. Therefore, the longer candle's original length is \( 6x = 6(4.5) = 27 \) cm, and the shorter candle's original length is \( 4y = 4(6) = 24 \) cm.
In simple words: Let x and y be the burning rates. Use the total burn times and initial length difference to set up two equations, then solve to find that the longer candle was 27 cm and the shorter was 24 cm.

Exam Tip: Carefully track the burn times for each candle - the longer one burns for 6 hours total while the shorter burns for 4 hours. At the 9.30 p.m. checkpoint, they have burned for different durations from their respective start times.