Simple equations is the chapter where the maths is easy and the language is the exam. Nobody struggles to solve 3x + 5 = 20. They struggle to turn a line like the father is thrice as old as the son into 3s = f. So do the translation slowly and the solving fast, which is the opposite of what most students do. Fix one rule in your head, whatever you do to one side you must do to the other, and the equation will never mislead you. When two variables are given, look at the coefficients before you choose a method, because if one of them is already 1, substitution is quicker, and if the coefficients match, elimination is quicker.
Simple Equations Formulas
The basic rule
Whatever operation you do on one side, do the same on the other side.
Move a term across the equals sign and its sign flips. So 3x + 5 = 20 becomes 3x = 20 – 5.
Linear equation in one variable
ax + b = 0 gives x = -b ÷ a
ax + b = cx + d gives x = (d – b) ÷ (a – c)
Linear equations in two variables
For a1x + b1y = c1 and a2x + b2y = c2
x = (b2c1 – b1c2) ÷ (a1b2 – a2b1)
y = (a1c2 – a2c1) ÷ (a1b2 – a2b1)
Nature of the solution
If a1 ÷ a2 is not equal to b1 ÷ b2, there is exactly one solution
If a1 ÷ a2 = b1 ÷ b2 but not equal to c1 ÷ c2, there is no solution
If a1 ÷ a2 = b1 ÷ b2 = c1 ÷ c2, there are infinitely many solutions
The two methods
Substitution: make one variable the subject and put it into the other equation. Use this when a coefficient is already 1.
Elimination: multiply the equations to match one coefficient, then add or subtract. Use this when the coefficients are close.
Translating the words into an equation
Is, was, will be, results in, all mean equals
Of, times, product means multiply
Sum, more than, increased by means add
Less than, decreased by, difference means subtract
Per, out of, ratio means divide
A number becomes x, twice a number becomes 2x, a number more than another becomes x + y
The identities that come up
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
a² – b² = (a + b) (a – b)
a² + b² = (a + b)² – 2ab
So if x + 1/x = a, then x² + 1/x² = a² – 2, and x³ + 1/x³ = a³ – 3a
60 Simple Equations Aptitude Questions and Answers (Solved MCQs)
Question 1. What is the value of x if 5(2x−3)−3(x+4) = 2(x−5)+7?
a) 21/5
b) 24/5
c) 27/5
d) 29/5
Answer:
b) 24/5 — Expanding the equation gives 10x−15−3x−12 = 2x−10+7. Therefore, 7x−27 = 2x−3. Thus, 5x = 24 and x = 24/5.
Question 2. What is the value of x if (x−3)/4 + (x+5)/6 = 7/3?
a) 23/5
b) 24/5
c) 27/5
d) 29/5
Answer:
c) 27/5 — Multiplying the equation by 12 gives 3(x−3)+2(x+5) = 28. Expanding gives 3x−9+2x+10 = 28, so 5x+1 = 28. Hence, 5x = 27 and x = 27/5.
Question 3. If 0.4(5x−3)−0.25(4x+8) = 7, what is the value of x?
a) 49/5
b) 10
c) 51/5
d) 52/5
Answer:
c) 51/5 — Expanding the decimal terms gives 2x−1.2−x−2 = 7. Therefore, x−3.2 = 7, so x = 10.2 = 51/5.
Question 4. What is the value of x if 3−[2x−{5−(x−4)}] = −9?
a) 5
b) 6
c) 7
d) 8
Answer:
c) 7 — First, 5−(x−4) = 9−x. Therefore, 2x−(9−x) = 3x−9. The left side becomes 3−(3x−9) = 12−3x. Thus, 12−3x = −9, giving −3x = −21 and x = 7.
Question 5. What is the value of x if (2x−1)/3 − (3x+2)/5 = (2x−7)/15?
a) −7
b) −5
c) −4
d) 4
Answer:
c) −4 — Multiplying the equation by 15 gives 5(2x−1)−3(3x+2) = 2x−7. Expanding gives 10x−5−9x−6 = 2x−7, so x−11 = 2x−7. Hence, −4 = x.
Question 6. If x = 5 is the solution of (k−2)x = 3k+1, what is the value of k?
a) 9/2
b) 5
c) 11/2
d) 6
Answer:
c) 11/2 — Substituting x = 5 gives 5(k−2) = 3k+1. Therefore, 5k−10 = 3k+1, so 2k = 11. Hence, k = 11/2.
Question 7. The sum of five consecutive odd positive integers is 245. What is the smallest of these integers?
a) 41
b) 43
c) 45
d) 47
Answer:
c) 45 — Let the five consecutive odd integers be x, x+2, x+4, x+6 and x+8. Their sum is 5x+20. Therefore, 5x+20 = 245, giving 5x = 225 and x = 45.
Question 8. Thirty-five per cent of a number plus 20 equals fifty per cent of the same number minus 7. What is the number?
a) 160
b) 170
c) 180
d) 190
Answer:
c) 180 — Let the number be x. The equation is 0.35x+20 = 0.50x−7. Therefore, 27 = 0.15x, giving x = 27/0.15 = 180.
Question 9. Two-thirds of a number exceeds three-fifths of the same number by 14. What is the number?
a) 180
b) 195
c) 210
d) 225
Answer:
c) 210 — Let the number be x. Then 2x/3−3x/5 = 14. Taking the LCM gives (10x−9x)/15 = 14, so x/15 = 14. Therefore, x = 210.
Question 10. What is the value of x if 4{3x−2[2x−(x−5)]} = 40?
a) 15
b) 18
c) 20
d) 25
Answer:
c) 20 — First, 2x−(x−5) = x+5. Therefore, 3x−2(x+5) = 3x−2x−10 = x−10. The equation becomes 4(x−10) = 40. Thus, x−10 = 10 and x = 20.
Question 11. For what value of k does 3(x−2)−2(2x+1) = 5−(x+k) have infinitely many solutions?
a) 11
b) 12
c) 13
d) 14
Answer:
c) 13 — Simplifying the left side gives 3x−6−4x−2 = −x−8. The right side is 5−x−k = −x+5−k. For infinitely many solutions, both sides must be identical. Therefore, −8 = 5−k, giving k = 13.
Question 12. For what value of k does the equation 4(2x−3) = 8x+k have infinitely many solutions?
a) −16
b) −12
c) 8
d) 12
Answer:
b) −12 — Expanding the left side gives 8x−12 = 8x+k. For the equation to be true for every real value of x, the constant terms must be equal. Therefore, k = −12.
Question 13. What is the value of x if (x+1)/2 − (x−2)/3 + (x−3)/4 = 5?
a) 9
b) 10
c) 11
d) 12
Answer:
c) 11 — Multiplying the equation by 12 gives 6(x+1)−4(x−2)+3(x−3) = 60. Expanding gives 6x+6−4x+8+3x−9 = 60. Therefore, 5x+5 = 60, so 5x = 55 and x = 11.
Question 14. When 7 is added to a number, the result equals five-fourths of the original number. What is the number?
a) 21
b) 24
c) 28
d) 35
Answer:
c) 28 — Let the number be x. Then x+7 = 5x/4. Multiplying by 4 gives 4x+28 = 5x. Therefore, x = 28.
Question 15. What is the value of x if (1/2)[x−(1/3){x−(x−6)/2}] = 10?
a) 118/5
b) 122/5
c) 126/5
d) 132/5
Answer:
c) 126/5 — First, x−(x−6)/2 = (2x−x+6)/2 = (x+6)/2. Therefore, x−(1/3)[(x+6)/2] = x−(x+6)/6 = (5x−6)/6. After multiplication by 1/2, the left side becomes (5x−6)/12. Thus, (5x−6)/12 = 10, giving 5x−6 = 120. Hence, 5x = 126 and x = 126/5.
Question 16. A two-digit number has a digit sum of 11. The number is 27 greater than the number obtained by reversing its digits. What is the original number?
a) 47
b) 56
c) 65
d) 74
Answer:
d) 74 — Let the tens digit be x and the units digit be y. Then x+y = 11. The original and reversed numbers are 10x+y and 10y+x. Therefore, (10x+y)−(10y+x) = 27, giving 9x−9y = 27, or x−y = 3. Solving x+y = 11 and x−y = 3 gives x = 7 and y = 4. Hence, the number is 74.
Question 17. Five years ago, Arun was three times as old as Varun. Ten years from now, Arun will be twice as old as Varun. What is Arun’s present age?
a) 45 years
b) 50 years
c) 55 years
d) 60 years
Answer:
b) 50 years — Let Arun’s and Varun’s present ages be A and V. Five years ago, A−5 = 3(V−5), giving A = 3V−10. Ten years from now, A+10 = 2(V+10), giving A = 2V+10. Therefore, 3V−10 = 2V+10, so V = 20. Hence, A = 50 years.
Question 18. A bag contains only ₹1 and ₹2 coins. If there are 70 coins with a total value of ₹110, how many ₹2 coins are there?
a) 30
b) 35
c) 40
d) 45
Answer:
c) 40 — Let the numbers of ₹1 and ₹2 coins be x and y. Then x+y = 70 and x+2y = 110. Subtracting the first equation from the second gives y = 40. Therefore, there are 40 coins of ₹2.
Question 19. A person’s monthly income is 25% more than the monthly expenditure. If the person saves ₹6,000 per month, what is the monthly income?
a) ₹24,000
b) ₹28,000
c) ₹30,000
d) ₹32,000
Answer:
c) ₹30,000 — Let the monthly expenditure be ₹x. Then the income is ₹1.25x. The saving is 1.25x−x = 0.25x. Therefore, 0.25x = 6,000, giving x = ₹24,000. The monthly income is 1.25×24,000 = ₹30,000.
Question 20. The length of a rectangle is 3 metres more than twice its breadth. If its perimeter is 66 metres, what is its length?
a) 20 metres
b) 21 metres
c) 22 metres
d) 23 metres
Answer:
d) 23 metres — Let the breadth be x metres. Then the length is 2x+3 metres. Since the perimeter is 66 metres, 2[(2x+3)+x] = 66. Therefore, 3x+3 = 33, giving x = 10. Hence, the length is 2(10)+3 = 23 metres.
Question 21. A car covers a distance of 300 km. If its speed were increased by 10 km/h, it would take one hour less to complete the journey. What is its original speed?
a) 40 km/h
b) 50 km/h
c) 60 km/h
d) 75 km/h
Answer:
b) 50 km/h — Let the original speed be x km/h. The original and reduced journey times are 300/x and 300/(x+10) hours. Therefore, 300/x−300/(x+10) = 1. Simplifying gives x²+10x−3,000 = 0, which factors as (x−50)(x+60) = 0. Since speed must be positive, x = 50 km/h.
Question 22. Twelve skilled workers and 20 unskilled workers receive a total daily wage of ₹19,000. If a skilled worker earns 50% more than an unskilled worker, what is the daily wage of one skilled worker?
a) ₹600
b) ₹650
c) ₹700
d) ₹750
Answer:
d) ₹750 — Let an unskilled worker’s daily wage be ₹x. A skilled worker earns ₹1.5x. Therefore, 12(1.5x)+20x = 19,000. This gives 18x+20x = 19,000, so 38x = 19,000 and x = ₹500. Hence, a skilled worker earns 1.5×500 = ₹750.
Question 23. Three books and five pens cost ₹370. If the price of one book is ₹30 more than three times the price of one pen, what is the price of one book?
a) ₹80
b) ₹90
c) ₹100
d) ₹110
Answer:
b) ₹90 — Let the price of one pen be ₹x. Then one book costs ₹(3x+30). Therefore, 3(3x+30)+5x = 370. This gives 9x+90+5x = 370, so 14x = 280 and x = ₹20. Hence, the price of one book is 3(20)+30 = ₹90.
Question 24. When a positive integer is divided by 7, the remainder is 3. If the quotient is 5 more than the remainder, what is the integer?
a) 52
b) 56
c) 59
d) 63
Answer:
c) 59 — The remainder is 3, so the quotient is 3+5 = 8. Using Dividend = Divisor×Quotient+Remainder, the integer is 7×8+3 = 56+3 = 59.
Question 25. The numerator and denominator of a fraction have a sum of 29. If 3 is added to both, the fraction becomes 3/4. What is the original fraction?
a) 11/18
b) 12/17
c) 13/16
d) 14/15
Answer:
b) 12/17 — Let the numerator be x, so the denominator is 29−x. According to the condition, (x+3)/(32−x) = 3/4. Cross-multiplying gives 4x+12 = 96−3x. Therefore, 7x = 84 and x = 12. The denominator is 29−12 = 17, so the original fraction is 12/17.
Question 26. At an event, adult tickets cost ₹150 and children’s tickets cost ₹80. If 120 tickets are sold for a total of ₹13,100, how many adult tickets are sold?
a) 40
b) 45
c) 50
d) 55
Answer:
c) 50 — Let the number of adult tickets be x. Then the number of children’s tickets is 120−x. Therefore, 150x+80(120−x) = 13,100. Expanding gives 150x+9,600−80x = 13,100. Thus, 70x = 3,500 and x = 50.
Question 27. The total cost of manufacturing a product consists of a fixed cost and a cost per unit. Manufacturing 400 units costs ₹9,200, while manufacturing 650 units costs ₹14,200. What will it cost to manufacture 800 units?
a) ₹16,400
b) ₹17,000
c) ₹17,200
d) ₹18,000
Answer:
c) ₹17,200 — Let the fixed cost be ₹F and the cost per unit be ₹x. Then F+400x = 9,200 and F+650x = 14,200. Subtracting gives 250x = 5,000, so x = ₹20. Hence, F = 9,200−400(20) = ₹1,200. The cost of 800 units is 1,200+800(20) = ₹17,200.
Question 28. A shop offers a 20% discount on the marked price of an item and then charges 5% GST on the discounted price. If the customer pays ₹1,680, what is the marked price?
a) ₹1,800
b) ₹1,900
c) ₹2,000
d) ₹2,100
Answer:
c) ₹2,000 — Let the marked price be ₹x. After a 20% discount, the price becomes 0.8x. Adding 5% GST gives 1.05×0.8x = 0.84x. Therefore, 0.84x = 1,680, so x = ₹2,000.
Question 29. A container has 40 litres of a milk-and-water mixture in the ratio 3:2. How much water must be added to make the ratio 6:5?
a) 3 litres
b) 4 litres
c) 5 litres
d) 6 litres
Answer:
b) 4 litres — The original quantities of milk and water are 24 litres and 16 litres respectively. Let x litres of water be added. Then 24/(16+x) = 6/5. Cross-multiplying gives 120 = 96+6x. Therefore, 6x = 24 and x = 4 litres.
Question 30. A group consisting of 8 adults and 12 children pays ₹4,200 for entry tickets. If each children’s ticket costs half as much as an adult ticket, what is the price of one adult ticket?
a) ₹250
b) ₹280
c) ₹300
d) ₹350
Answer:
c) ₹300 — Let the price of one adult ticket be ₹x. Then a children’s ticket costs ₹x/2. Therefore, 8x+12(x/2) = 4,200. This gives 8x+6x = 4,200, so 14x = 4,200 and x = ₹300.
Question 31. If 3x+2y = 23 and 2x−3y = −2, what is the value of x+y?
a) 7
b) 8
c) 9
d) 10
Answer:
c) 9 — Multiplying the first equation by 3 gives 9x+6y = 69. Multiplying the second equation by 2 gives 4x−6y = −4. Adding gives 13x = 65, so x = 5. Substituting into 3x+2y = 23 gives 15+2y = 23, so y = 4. Therefore, x+y = 9.
Question 32. Three notebooks and two pens cost ₹130, while two notebooks and five pens cost ₹160. What is the cost of one notebook?
a) ₹20
b) ₹25
c) ₹30
d) ₹35
Answer:
c) ₹30 — Let the costs of a notebook and a pen be ₹x and ₹y respectively. Then 3x+2y = 130 and 2x+5y = 160. Multiplying the first equation by 5 and the second by 2 gives 15x+10y = 650 and 4x+10y = 320. Subtracting gives 11x = 330, so x = ₹30.
Question 33. The sum of the digits of a two-digit number is 12. When its digits are reversed, the resulting number is 18 greater than the original number. What is the original number?
a) 48
b) 57
c) 66
d) 75
Answer:
b) 57 — Let the tens digit be x and the units digit be y. Then x+y = 12. The reversed number exceeds the original by 18, so (10y+x)−(10x+y) = 18. Hence, 9(y−x) = 18, giving y−x = 2. Solving x+y = 12 and y−x = 2 gives x = 5 and y = 7. Therefore, the original number is 57.
Question 34. The present ages of a mother and her daughter add up to 57 years. Six years ago, the mother was four times as old as her daughter. What is the mother’s present age?
a) 39 years
b) 40 years
c) 42 years
d) 45 years
Answer:
c) 42 years — Let the present ages of the mother and daughter be M and D. Then M+D = 57. Six years ago, M−6 = 4(D−6), giving M = 4D−18. Substituting into the first equation gives 4D−18+D = 57. Thus, 5D = 75 and D = 15. Hence, M = 42 years.
Question 35. The sum of two positive numbers is 49. The larger number exceeds the smaller number by one-third of the smaller number. What is the larger number?
a) 21
b) 24
c) 28
d) 30
Answer:
c) 28 — Let the smaller number be x and the larger number be y. Then x+y = 49 and y−x = x/3. Therefore, y = 4x/3. Substituting gives x+4x/3 = 49, so 7x/3 = 49 and x = 21. Hence, y = 28.
Question 36. A box contains ₹5 and ₹10 coins. If there are 80 coins worth ₹610 in total, how many ₹10 coins are there?
a) 38
b) 40
c) 42
d) 44
Answer:
c) 42 — Let the numbers of ₹5 and ₹10 coins be x and y. Then x+y = 80 and 5x+10y = 610. Multiplying the first equation by 5 gives 5x+5y = 400. Subtracting this from the value equation gives 5y = 210, so y = 42.
Question 37. For what value of k do the equations kx+3y = 7 and 4x+6y = 10 have no solution?
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2 — Two linear equations have no solution when the coefficients of x and y are proportional but the constants are not. Therefore, k/4 = 3/6, giving k/4 = 1/2 and k = 2. However, 7/10 ≠ 1/2, so the lines are parallel and distinct.
Question 38. For what value of k do the equations 3x+ky = 12 and 6x+8y = 24 have infinitely many solutions?
a) 2
b) 3
c) 4
d) 6
Answer:
c) 4 — For infinitely many solutions, the corresponding coefficients and constants must be proportional. Since 3/6 = 12/24 = 1/2, we require k/8 = 1/2. Therefore, k = 4.
Question 39. Two shirts and three pairs of trousers cost ₹5,100, while three shirts and two pairs of trousers cost ₹4,900. What is the price of one pair of trousers?
a) ₹900
b) ₹1,000
c) ₹1,100
d) ₹1,200
Answer:
c) ₹1,100 — Let the prices of one shirt and one pair of trousers be ₹x and ₹y. Then 2x+3y = 5,100 and 3x+2y = 4,900. Adding gives 5x+5y = 10,000, so x+y = 2,000. Subtracting the second original equation from the first gives y−x = 200. Solving the two equations gives 2y = 2,200, so y = ₹1,100.
Question 40. At a museum, 25 adult tickets and 40 student tickets cost ₹7,300. Fifteen adult tickets and 20 student tickets cost ₹4,100. What is the price of one adult ticket?
a) ₹140
b) ₹160
c) ₹180
d) ₹200
Answer:
c) ₹180 — Let the adult and student ticket prices be ₹x and ₹y. Then 25x+40y = 7,300 and 15x+20y = 4,100. Multiplying the second equation by 2 gives 30x+40y = 8,200. Subtracting the first equation gives 5x = 900, so x = ₹180.
Question 41. A boat travels at 18 km/h downstream and 10 km/h upstream. What is the speed of the boat in still water?
a) 12 km/h
b) 13 km/h
c) 14 km/h
d) 15 km/h
Answer:
c) 14 km/h — Let the speed of the boat in still water be x km/h and the speed of the current be y km/h. Then x+y = 18 and x−y = 10. Adding the equations gives 2x = 28, so x = 14 km/h.
Question 42. How many kilograms of rice costing ₹70 per kg must be mixed with rice costing ₹40 per kg to obtain 30 kg of a mixture worth ₹58 per kg?
a) 16 kg
b) 18 kg
c) 20 kg
d) 22 kg
Answer:
b) 18 kg — Let x kg of the ₹40 rice and y kg of the ₹70 rice be mixed. Then x+y = 30 and 40x+70y = 58×30 = 1,740. Substituting x = 30−y gives 40(30−y)+70y = 1,740. Thus, 1,200+30y = 1,740, so y = 18 kg.
Question 43. The perimeter of a rectangle is 74 metres, and its length is 9 metres greater than its breadth. What is its area?
a) 294 m²
b) 308 m²
c) 322 m²
d) 336 m²
Answer:
c) 322 m² — Let the length and breadth be x and y metres. Since the perimeter is 74 metres, x+y = 37. Also, x−y = 9. Adding gives 2x = 46, so x = 23. Therefore, y = 14. The area is 23×14 = 322 m².
Question 44. A farm has chickens and goats. There are 50 heads and 140 legs altogether. How many goats are there?
a) 15
b) 18
c) 20
d) 25
Answer:
c) 20 — Let the numbers of chickens and goats be x and y. Then x+y = 50. Since chickens have two legs and goats have four, 2x+4y = 140. Multiplying the first equation by 2 gives 2x+2y = 100. Subtracting gives 2y = 40, so y = 20.
Question 45. A and B together can complete a job in 12 days. A alone can complete it in 20 days. In how many days can B alone complete the job?
a) 24 days
b) 28 days
c) 30 days
d) 36 days
Answer:
c) 30 days — Let A’s and B’s one-day work rates be x and y. Together, x+y = 1/12, while A’s rate is x = 1/20. Therefore, y = 1/12−1/20 = (5−3)/60 = 1/30. Hence, B alone can complete the job in 30 days.
Question 46. What is the value of x if (2x−3)/5 − (3x+1)/7 = (x−19)/35?
a) −9/2
b) −7/2
c) −5/2
d) 7/2
Answer:
b) −7/2 — Multiplying the equation by 35 gives 7(2x−3)−5(3x+1) = x−19. Expanding gives 14x−21−15x−5 = x−19. Therefore, −x−26 = x−19, so −7 = 2x and x = −7/2.
Question 47. If α and β are the real solutions of 1/(x−1) + 1/(x+1) = 3/4, what is the value of α+β?
a) 2
b) 7/3
c) 8/3
d) 3
Answer:
c) 8/3 — The domain excludes x = ±1. Combining the fractions gives 2x/(x²−1) = 3/4. Cross-multiplying gives 8x = 3x²−3, or 3x²−8x−3 = 0. The roots are 3 and −1/3, both of which are valid. Therefore, α+β = 3−1/3 = 8/3.
Question 48. If α and β are the real solutions of |3x−7| = 2x+1, what is the value of α+β?
a) 41/5
b) 44/5
c) 46/5
d) 49/5
Answer:
c) 46/5 — For 3x−7 ≥ 0, the equation becomes 3x−7 = 2x+1, giving x = 8. For 3x−7 < 0, it becomes 7−3x = 2x+1, giving 5x = 6 and x = 6/5. Both values satisfy the original equation. Therefore, α+β = 8+6/5 = 46/5.
Question 49. For what value of k does the equation (k−1)x+2 = 3x+k have no solution?
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4 — Rearranging gives (k−4)x = k−2. An equation has no solution when the coefficient of x is zero but the constant on the other side is non-zero. Thus, k−4 = 0, giving k = 4. The equation then becomes 2 = 4, which is impossible.
Question 50. What is the value of x if (x−2)/3 + (x−3)/4 = (x−4)/5?
a) 31/23
b) 35/23
c) 37/23
d) 41/23
Answer:
c) 37/23 — Multiplying the equation by 60 gives 20(x−2)+15(x−3) = 12(x−4). Expanding gives 20x−40+15x−45 = 12x−48. Therefore, 35x−85 = 12x−48, so 23x = 37 and x = 37/23.
Question 51. Four years ago, a father was five times as old as his son. Eight years from now, he will be three times as old as his son. What is the sum of their present ages?
a) 72 years
b) 76 years
c) 80 years
d) 84 years
Answer:
c) 80 years — Let their present ages be F and S. Four years ago, F−4 = 5(S−4), giving F = 5S−16. Eight years from now, F+8 = 3(S+8), giving F = 3S+16. Therefore, 5S−16 = 3S+16, so 2S = 32 and S = 16. Hence, F = 64 and their total present age is 64+16 = 80 years.
Question 52. The denominator of a fraction is 5 greater than its numerator. If 2 is subtracted from the numerator and 1 is added to the denominator, the fraction becomes 1/2. What is the original fraction in its simplest form?
a) 1/2
b) 2/3
c) 3/4
d) 4/5
Answer:
b) 2/3 — Let the numerator be x, so the denominator is x+5. According to the condition, (x−2)/(x+6) = 1/2. Cross-multiplying gives 2x−4 = x+6, so x = 10. The original fraction is 10/15, which simplifies to 2/3.
Question 53. A railway counter sells 60 tickets consisting of first-class tickets priced at ₹120 each and second-class tickets priced at ₹80 each. If the total collection is ₹6,000, how many first-class tickets are sold?
a) 25
b) 30
c) 35
d) 40
Answer:
b) 30 — Let the number of first-class tickets be x. Then the number of second-class tickets is 60−x. Therefore, 120x+80(60−x) = 6,000. Expanding gives 120x+4,800−80x = 6,000. Hence, 40x = 1,200 and x = 30.
Question 54. Forty per cent of the larger of two positive numbers equals two-thirds of the smaller number. If their sum is 192, what is the larger number?
a) 108
b) 112
c) 120
d) 128
Answer:
c) 120 — Let the larger and smaller numbers be x and y. The condition gives 2x/5 = 2y/3. Cancelling 2 and cross-multiplying gives 3x = 5y, so x:y = 5:3. Since their sum is 192, eight parts equal 192 and one part equals 24. Therefore, the larger number is 5×24 = 120.
Question 55. The average marks of eight students increase by 2 when one student’s marks are replaced by those of a new student. If the replaced student scored 56 marks, how many marks did the new student score?
a) 68
b) 70
c) 72
d) 74
Answer:
c) 72 — An increase of 2 in the average of eight students means the total marks increase by 8×2 = 16. Therefore, the new student’s marks are 16 more than the replaced student’s marks. Hence, the new student scored 56+16 = 72 marks.
Question 56. One pipe can fill a tank in 6 hours, while a second pipe can empty the full tank in 9 hours. If both pipes are opened together when the tank is empty, how long will they take to fill it?
a) 12 hours
b) 15 hours
c) 18 hours
d) 24 hours
Answer:
c) 18 hours — The filling pipe completes 1/6 of the tank per hour, while the emptying pipe removes 1/9 per hour. Their net rate is 1/6−1/9 = (3−2)/18 = 1/18 of the tank per hour. Therefore, the tank will be filled in 18 hours.
Question 57. For which values of k does the equation |2x−5| = k have two distinct positive real solutions?
a) k < 0
b) 0 < k < 5
c) k ≥ 5
d) k > 0
Answer:
b) 0 < k < 5 — The two solutions are x = (5+k)/2 and x = (5−k)/2. They are distinct when k > 0. For both solutions to be positive, we also require 5−k > 0, giving k < 5. Therefore, the required condition is 0 < k < 5.
Question 58. A salesperson’s monthly salary consists of a fixed amount and an 8% commission on sales. When the monthly sales are ₹1,50,000, the salary is ₹22,000. What will the salary be when the sales are ₹2,50,000?
a) ₹28,000
b) ₹30,000
c) ₹32,000
d) ₹34,000
Answer:
b) ₹30,000 — Let the fixed salary be ₹x. For sales of ₹1,50,000, the commission is 8% of ₹1,50,000 = ₹12,000. Therefore, x+12,000 = 22,000, giving x = ₹10,000. For sales of ₹2,50,000, the commission is ₹20,000. Hence, the total salary is ₹10,000+₹20,000 = ₹30,000.
Question 59. The sum of four consecutive positive multiples of 3 is 114. What is the largest of these numbers?
a) 30
b) 33
c) 36
d) 39
Answer:
b) 33 — Let the four consecutive multiples of 3 be x, x+3, x+6 and x+9. Their sum is 4x+18. Therefore, 4x+18 = 114, giving 4x = 96 and x = 24. The largest number is x+9 = 33.
Question 60. What is the value of x if (1/2)[3x−(2/3){x−(3/4)(2x−8)}] = 19?
a) 57/5
b) 61/5
c) 63/5
d) 67/5
Answer:
c) 63/5 — First, (3/4)(2x−8) = 3x/2−6. Therefore, x−(3x/2−6) = 6−x/2. Multiplying by 2/3 gives 4−x/3. The expression inside the square brackets becomes 3x−(4−x/3) = 10x/3−4. Thus, half of this expression is 5x/3−2. The equation becomes 5x/3−2 = 19, so 5x/3 = 21 and x = 63/5.
