Question 1.
\nThree more than twice a number is equal to four less than the number. Find the number.
\nSolution:
\nLet the number = x
\nTwice the number = 2x
\nAccording to problem, 3 + 2x = x – 4
\n\u21d2 3 + 2x + 4 = x
\n\u21d2 7 = x – 2x
\n\u21d2 7 = -x
\n\u21d2 -x = 7
\n\u21d2 x = -7
\nHence, the number = -7<\/p>\n
Question 2.
\nWhen four consecutive integers are added, the sum is 46. Find the integers.
\nSolution:
\nLet x be the first integer, then the next
\nthree consecutive integers are x + 1, x + 2 and x + 3
\nAccording to problem,
\nx + (x + 1) + (x + 2) + (x + 3) = 46
\n\u21d2 x + x + 1 + x + 2 + x + 3 = 46
\n\u21d2 4x + 6 = 46
\n\u21d2 4x = 46 – 6
\n\u21d2 4x = 40
\n\u21d2 x = \\(\\frac{40}{4}\\) = 10
\nHence four consecutive integers are 10, (10 + 1), (10 + 2) and (10 + 3)
\ni.e. 10, 11, 12 and 13<\/p>\n
Question 3.
\nManjula thinks a number and subtracts \\(\\frac{7}{3}\\) from it. She multiplies the result by 6. The result now obtained is 2 less than twice the same number she thought of. What is the number?
\nSolution:
\nLet a number thought by Manjula = x
\nAccording to the condition,
\n\\(\\left(x-\\frac{7}{3}\\right)\\) \u00d7 6 = 2x – 2
\n\u21d2 6x – 14 = 2x – 2
\n\u21d2\u00a0 6x – 2x = -2 + 14 = 12
\n\u21d2\u00a0 4x = 12
\n\u21d2 x = \\(\\frac{12}{4}\\) = 3
\nFlence required number = 3<\/p>\n
Question 4.
\nA positive number is 7 times another number. If 15 is added to both the numbers, then one of the new number becomes \\(\\frac{5}{2}\\) times the other new number. What are the numbers?
\nSolution:
\nLet the required number = x
\nThen another number = \\(\\frac{x}{7}\\)
\nAccording to the condition,
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Ex-12.2-Q4.1.png\")
\nHence the numbers are 35 and 5<\/p>\n
Question 5.
\nWhen three consecutive even integers are added, the sum is zero. Find the integers.
\nSolution:
\nLet the first even integer be x,
\nthen next two consecutive even integers are (x + 2) and (x + 4)
\nAccording to given problem,
\nx + (x +2) + (x + 4) = 0
\n\u21d2 x + x + 2 + x + 4 =0
\n\u21d2 3x + 6 = 0
\n\u21d2 3x = – 6
\n\u21d2 x = \\(\\frac{-6}{3}\\)
\n\u21d2 x = \u2014 2
\nHence three consecutive integers are -2, – 2 + 2, – 2 + 4 i.e. – 2, 0, 2<\/p>\n
Question 6.
\nFind two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4.
\nSolution:
\nLet the first odd integer be x,
\nthen next consecutive odd integers is (x + 2)
\nAccording to given problem,
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Ex-12.2-Q6.1.png\")
\nHence two consecutive odd integers are
\n25 and (x + 2) = (25 + 2) = 27<\/p>\n
Question 7.
\nThe denominator of a fraction is 1 more than twice its numerator. If the numerator and denominator are both increased by 5, it becomes \\(\\frac{3}{5}\\). Find the original fraction.
\nSolution:
\nLet the numerator of the original fraction be x
\nThen, its denominator = 2x + 1
\n\u2234 The fraction = \\(\\frac{x}{2 x+1}\\)
\nAccording to given problem,
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Ex-12.2-Q7.1.png\")
\n\u21d2 5 (x + 5) = 3 (2x + 6)
\n\u21d2 5x + 25 = 6x + 18
\n\u21d2 5x – 6x = 18 – 25
\n\u21d2 -x = -7
\n\u21d2 x = 7
\nHence, the original fraction = \\(\\frac{x}{2 x+1}=\\frac{7}{2 \\times 7+1}=\\frac{7}{15}\\)<\/p>\n
Question 8.
\nFind two positive numbers in the ratio 2 : 5 such that their difference is 15.
\nSolution:
\nLet the two numbers be 2x and 5x
\n[\u2235 ratio of these two numbers = \\(\\frac{2 x}{5 x}=\\frac{2}{5}\\) = 2 : 5 ]
\nAccording to given problem,
\n5x – 2x = 15
\n\u21d2 3x = 15
\n\u21d2 x = \\(\\frac{15}{3}\\)
\n\u21d2 x = 5
\nHence the numbers are 2 \u00d7 5 and 5 \u00d7 5 i.e. 10 and 25<\/p>\n
Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Question 25. Question 26. Question 27. Question 28. Question 29.
\nWhat number should be added to each of the numbers 12, 22, 42 and 72 so that the resulting numbers may be in proportion ?
\nSolution:
\nLet the required number be x
\nThen according to given problem,
\n12 + x, 22 + x, 42 + x and 72 + x are in proportion
\n\u21d2 \\(\\frac{12+x}{22+x}=\\frac{42+x}{72+x}\\)
\n\u21d2 (12 + x) (72 + x) = (42 + x) (22 + x)
\n\u21d2 12 (72 + x) + x (72 + x) = 42 (22 + x) + x (22 + x)
\n\u21d2 864 + 12x + 72x + x2<\/sup> = 924 + 42x + 22x + x2<\/sup>
\n\u21d2 864 + 84x + x2<\/sup> = 924 + 64x + x2<\/sup>
\n\u21d2 864 + 84x + x2<\/sup> – 924 – 64x – x2<\/sup> = 0
\n\u21d2 864 + 84x – 64x – 924 = 0
\n\u21d2 84x – 64x = 924 – 864
\n\u21d2 20x = 60
\n\u21d2 x = \\(\\frac{60}{20}\\)
\n\u21d2 x = 3
\nHence, the required number is 3.<\/p>\n
\nThe digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?
\nSolution:
\nLet one’s digit of a 2-digit number = x
\nThen ten’s digit = x + 3
\n\u2234 Number = x + 10(x + 3) = x + 10x + 30 = 11x + 30
\nBy interchanging the digits,
\nOne’s digit of new number = x + 3
\nand ten’s digit = x
\n\u2234 Number = x + 3 + 10x = 11x + 3
\nAccording to the condition,
\n11x + 30+ 11x + 3 = 143
\n\u21d2 22x + 33 = 143
\n\u21d2 22x = 143-33 = 110
\n\u21d2 x = \\(\\frac{110}{22}\\) = 5
\n\u2234 Original number = 11x+ 30 = 11 \u00d7 5 + 30 = 55 + 30 = 85<\/p>\n
\nSum of the digits of a two-digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 63. Find the two-digit number.
\nSolution:
\nSum of two digits of a 2-digit number = 11
\nLet unit’s digit of a 2-digit number = x
\nThen ten’s digit = 11 – x
\n\u2234 Number = x + 10(11 – x) = x + 110 – 10x = 110 -9x
\nBy interchanging the digit,
\nOne’s digit of new number = 11 – x
\nand ten’s digit = x
\n\u2234 Number = 11 – x + 10x = 11 + 9x
\nAccording to the condition,
\n11 + 9x – (110 – 9x) = 63
\n11 + 9x – 110 + 9x = 63
\n18x = 63 – 11 + 110 = 162
\nx = \\(\\frac{162}{18}\\) = 9
\n\u2234 Original number = 110 – 9x = 110 – 9 \u00d7 9 = 110 – 81 = 29<\/p>\n
\nRitu is now four times as old as his brother Raju. In 4 years time, her age will be twice of Raju’s age. What are their present ages?
\nSolution:
\nLet the age of Raju = x years
\nthen the age of Ritu = 4 \u00d7 x years = 4x years
\nIn 4 years time,
\nage of Raju = (x +4) years
\nage of Ritu = (4x + 4) years
\nAccording to given problem,
\n4x + 4 = 2 (x + 4)
\n\u21d2 4x + 4 = 2x + 8
\n\u21d2 4x – 2x = 8 – 4
\n\u21d2 2x = 4 \u21d2 x = \\(\\frac{4}{2}\\)
\n\u21d2 x = 2
\nHence, the age of Raju = 2 years
\nand the age of Ritu = 4 \u00d7 2 years = 8 years.<\/p>\n
\nA father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?
\nSolution:
\nLet the present age of son = x years
\nThen, age of his father = 7 \u00d7 x years = 7x years
\nTwo years ago age of son = (x – 2) years
\nTwo years ago age of his father = (7x – 2) years
\nAccording to given problem,
\n7x – 2 = 13 (x – 2)
\n\u21d2 7x – 2 = 13x – 26
\n\u21d2 7x – 13x = -26 + 2
\n\u21d2 -6x = -24
\n\u21d2 x = \\(\\frac{-24}{-6}\\)
\n\u21d2 x = 4
\nHence, age of son = 4 years
\nand age of his father = 7 \u00d7 4 years = 28 years.<\/p>\n
\nThe ages of Sona and Sonali are in the ratio 5 : 3. Five years hence, the ratio of their ages will be 10 : 7. Find their present ages.
\nSolution:
\nGiven ratio of ages of Sona and Sonali = 5 : 3
\nlet the present ages of Sona and Sonali is 5x and 3x years
\nfive years hence, the age of Sona = 5x + 5
\nand five years to hence the age of Sonali = 3x + 5
\nAccording to given problem,
\n\\(\\frac{5 x+5}{3 x+5}=\\frac{10}{7}\\)
\n\u21d2 7(5x + 5) = 10(3x + 5)
\n\u21d2 35x + 35 = 30x + 50
\n\u21d2 35x – 30x = 50 – 35
\n\u21d2 5x = 15 \u21d2 x = \\(\\frac{15}{5}\\)
\n\u21d2 x = 3
\nHence, the present age of Sona and Sonali is 5 \u00d7 3 and 3 \u00d7 3 years
\ni.e. 15 and 9 years.<\/p>\n
\nAn employee works in a company on a contract of 30 days on the condition that he will receive \u20b9200 for each day he works and he will be fined \u20b920 for each day he is absent. If he receives \u20b93800 in all, for how many days did he remain absent?
\nSolution:
\nPeriod of contract = 30 days
\nIf an employees works a day, he will get \u20b9200
\nIf he is absent, he will be fined \u20b920 per day
\nAt the end of contract period, he get \u20b93800
\nLet he remained absent for x days
\nThen he worked for = (30 – x) days
\nAccording to the condition,
\n(30 – x) \u00d7 200 – x \u00d7 20 = 3800
\n\u21d2 6000 – 200x – 20x = 3800
\n\u21d2\u00a0 220x = 6000 – 3800 = 2200
\n\u21d2\u00a0 x = \\(\\frac{2200}{220}\\) = 10
\nHe remained absent for 10 days.<\/p>\n
\nI have a total of \u20b9300 in coins of denomination \u20b91, \u20b92 and \u20b95. The number of coins is 3 times the number of \u20b95 coins. The total number of coins is 160. How many coins of each denomination are with me?
\nSolution:
\nAmount of coins = \u20b9300
\nand total coins = 160
\nLet number of coins of \u20b95 = x
\nThen number of coins of \u20b92 = 3x
\nand number of coins of \u20b91 = 150 – (x + 3x) = 150 – 4x
\nAccording to the condition,
\n(160 – 4x) \u00d7 1 + 3x \u00d7 2 + x \u00d7 5 = 300
\n\u21d2 160 – 4x + 6x + 5x = 300 \u21d2 160 + 7x = 300
\n\u21d2 7x = 300 – 160 = 140
\n\u21d2 x = \\(\\frac{140}{7}\\) = 20
\n\u2234 5 rupee coins = 20
\n2 rupee coins = 3 \u00d7 20 = 60
\nand 1 rupee coins = 160 – 60 – 20 = 80<\/p>\n
\nA local bus is carrying 40 passengers, some with \u20b95 tickets and the remaining with \u20b97.50 tickets. If the total receipts from these passengers is \u20b9230, find the number of passengers with \u20b95 tickets.
\nSolution:
\nLet the number of passengers with \u20b95 tickets = x
\nThen, the number of passengers with \u20b97.50 tickets = (40 – x)
\nAccording to given problem,
\n5 \u00d7 x + (40 – x) \u00d7 7.50 = 230
\n\u21d2 5x + 300 – 7.5x = 230
\n\u21d2 5x – 7.5x = 230 – 300
\n\u21d2 -2.5x = -70
\n\u21d2 x = \\(\\frac{70}{2.5}\\) = 28
\nHence, the number of passengers with \u20b95 tickets = 28.<\/p>\n
\nOn a school picnic, a group of students agree to pay equally for the use of a full boat and pay \u20b910 each. If there had been 3 more students in the group, each would have paid \u20b92 less. How many students were there in the group ?
\nSolution:
\nLet, the number of students in a group = x
\nwhen 3 students are more then,
\nthe total number of students in the group = x + 3
\nAccording to given problem,
\n\u21d2 10 \u00d7 x = (x + 3) \u00d7 (10 – 2)
\n\u21d2 10x = (x + 3) \u00d7 8
\n\u21d2 10x = 8 (x + 3)
\n\u21d2 10x = 8x + 24
\n\u21d2 10x – 8x = 24
\n\u21d2 2x = 24
\n\u21d2 x = \\(\\frac{24}{2}\\)
\n\u21d2 x = 12
\nHence, the number of students in the group = 12<\/p>\n
\nHalf of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
\nSolution:
\nLet total deer in the herd = x
\nNumber of deer grazing in the field = \\(\\frac{x}{2}\\)
\nRemaining = x – \\(\\frac{x}{2}=\\frac{x}{2}\\)
\n\\(\\frac{3}{4}\\) of the remaining deer playing
\n= \\(\\frac{3}{4} \\times \\frac{1}{2} x=\\frac{3}{8} x\\)
\nRest of deer = \\(\\frac{x}{2}-\\frac{3}{8} x=\\frac{1}{8} x\\)
\n\u2234 \\(\\frac{1}{8}\\)x = 9
\n\u21d2 x = 9 \u00d7 8 = 72
\n\u2234 Total number of deer = 72<\/p>\n
\nSakshi takes some flowers in a basket and visits three temples one by one. At each temple, she offers one half of the flowers from the basket. If she is left with 6 flowers at the end, find the number of flowers she had in the beginning.
\nSolution:
\nLet total flowers in a basket = x
\nFlowers offered in the first temple = \\(\\frac{x}{2}\\)
\nRemaining flowers = x – \\(x-\\frac{x}{2}=\\frac{x}{2}\\)
\nFlowers offered in the second temple
\n\\(\\frac{x}{2} \\times \\frac{1}{2}=\\frac{x}{4}\\)
\nRemaining flower = \\(\\frac{x}{2}-\\frac{x}{4}=\\frac{x}{4}\\)
\nFlowers offered in the third temple = \\(\\frac{x}{4} \\times\\) \\(\\frac{1}{2}=\\frac{x}{8}\\)
\nRemaining flowers = \\(\\frac{x}{4}-\\frac{x}{8}=\\frac{x}{8}\\)
\n\u2234 Total , number of flowers = 48<\/p>\n
\nTwo supplementary angles differ by 50\u00b0. Find the measure of each angle.
\nSolution:
\nLet the angle be x
\nThen, its supplementary angle = 180\u00b0 – x
\nAccording to given problem,
\nx – (180\u00b0 – x) = 50\u00b0
\n\u21d2 x – 180\u00b0 + x = 50\u00b0
\n\u21d2 2x – 180\u00b0 = 50\u00b0
\n\u21d2 2x = 180\u00b0 + 50\u00b0
\n\u21d2 2x = 230\u00b0
\n\u21d2 x = \\(\\frac{230^{\\circ}}{2}\\)
\n\u21d2 x = 115\u00b0
\nHence, the measurement of each angle be 115\u00b0 and (180\u00b0 – 115\u00b0)
\ni.e. 115\u00b0 and 65\u00b0.<\/p>\n
\nIf the angles of a triangle are in the ratio 5 : 6 : 7, find the angles.
\nSolution:
\nLet the angles of a triangle are 5x, 6x, and 7x
\nThen, we know that,
\n5x + 6x + 7x = 180\u00b0
\n\u21d2 18x = 180\u00b0
\n\u21d2 x = \\(\\frac{180^{\\circ}}{18}\\)
\n\u21d2 x= 10\u00b0
\nHence, the angle of a triangle are 5 \u00d7 10\u00b0, 6 \u00d7 10\u00b0,
\nand 7 \u00d7 10\u00b0 i.e. 50\u00b0, 60\u00b0 and 70\u00b0.<\/p>\n
\nTwo equal sides of an isosceles triangle are 3x – 1 and 2x + 2 units. The third side is 2x units. Find x and the perimeter of the triangle.
\nSolution:
\nTwo equal sides of an isosceles triangle are 3x – 1 and 2x + 2
\ni. e. 3x – 1 = 2x + 2
\n\u21d2 3x – 2x = 2 + 1
\n\u21d2 x = 3
\nThird side of triangle = 2x = 2 \u00d7 3 = 6 units
\nequal sides of an triangle = 3 \u00d7 3 – 1= 9 – 1 = 8 units
\n\u2234 Perimeter of the triangle = (8 + 8 + 6) units = 22 units<\/p>\n
\nIf each side of a triangle is increased by 4 cm, the ratio of the perimeters of the new triangle and the given triangle is 7 : 5. Find the perimeter of the given triangle.
\nSolution:
\nLet the perimeter of original triangle = x cm
\nBy increasing each side by 4 cm The perimeter will be
\n= x + 4 \u00d7 3 = (x + 12) cm
\nNow Ratio of perimeter of new triangle and given triangle = 7 : 5
\n\u21d2 \\(\\frac{x+12}{x}=\\frac{7}{5}\\)
\n\u21d2 5x + 60 = 7x (By corss multiplication)
\n\u21d2 7x – 5x = 60
\n\u21d2 2x = 60
\n\u21d2 x = \\(\\frac{60}{2}\\) = 30
\n\u2234 Perimeter of given triangle = 30 cm<\/p>\n
\nThe length of a rectangle is 5 cm less than twice its breadth. If the length is decreased by 3 cm and breadth increased by 2 cm, the perimeter of the resulting rectangle is 72 cm. Find the area of the original rectangle.
\nSolution:
\nLet, the breath of the original rectangle = x cm
\nThen, length of the original rectangle = (2x – 5) cm
\nWhen, length is decreased by 3 cm then new length
\n= [(2x – 5) – 3)] cm = (2x – 8) cm
\nWhen breadth is increased by 2 cm,
\nthen new length = (x + 2) cm
\nNew perimeter=2(new length + new breadth)
\n= 2 [(2x – 8) + (x + 2)]
\n= 2 [2x – 8 + x + 2]
\n= 2 (3x – 6) = 6x – 12
\nAccording to the given problem,
\n6x – 12 = 72
\n\u21d2 6x = 72 + 12
\n\u21d2 6x = 84
\n\u21d2 x = \\(\\frac{84}{6}\\)
\n\u21d2 x = 14
\nBreadth of the original rectangle = 14 cm
\nand length of the original rectangle = (2 \u00d7 14 – 5) cm = 23 cm
\nArea of the original rectangle = Length \u00d7 Breadth
\n= 23 \u00d7 14 cm2<\/sup> = 322 cm2<\/sup><\/p>\n
\nA rectangle is 10 cm long and 8 cm wide. When each side of the rectangle is increased by x cm, its perimeter is doubled. Find the equation in x and hence find the area of the new rectangle.
\nSolution:
\nLength of rectangle (l) = 10 cm
\nand width (b) = 8 cm
\nPerimeter = 2(l + b) = 2(10 + 8) cm = 2 \u00d7 18 = 36 cm
\nBy increasing each side by x cm
\nThen perimeter = 2[10 + x + 8 + x]
\n= 2(18 + 2x) = (36 + 4x) cm
\nAccording to the condition,
\n36 + 4x = 2(36)
\n\u21d2 36 + 4x = 72
\n\u21d2 4x = 72 – 36 = 36
\n\u21d2 x = \\(\\frac{36}{4}\\)
\n\u21d2 x = 9
\nLength of new rectanlge = l + x = 10 + 9 = 19 cm
\nand breadth = b + x = 8 + 9 = 17 cm
\nArea = Length x Breadth = 19 \u00d7 17 cm2\u00a0<\/sup>= 323 cm2<\/sup><\/p>\n
\nA steamer travels 90 km downstream in the same time as it takes to travel 60 km upstream. If the speed of the stream is 5 km\/hr, find the speed of the streamer in still water.
\nSolution:
\nLet, the speed of the streamer in still water be x km\/hr,
\nThen, the speed downstream = (x + 5) km\/hr
\nAnd, the speed upstream = (x – 5) km\/r.
\nAccording to given problem
\n\\(\\frac{90}{x+5}=\\frac{60}{x-5}\\)
\n\u21d2 90(x – 5) = 60(x + 5)
\n\u21d2 90x – 450 = 60x + 300
\n\u21d2 90x – 60x = 300 + 450
\n\u21d2 30x = 750
\n\u21d2 x = \\(\\frac{750}{30}\\)
\n\u21d2 x = 25
\nHence, the speed of the streamer in still water be 25 km\/hr.<\/p>\n
\nA steamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km\/h, find the speed of the streamer in still water and the distance between two ports.
\nSolution:
\nSpeed of the stream in still water = 1 km\/h
\nLet speed of streamer = x km\/h
\n\u2234 It down speed = (x + 1) km\/h
\nand up speed = (x – 1) km\/h
\nAccording to the condition,
\n(x + 1) \u00d7 5 = (x – 1) \u00d7 6
\n\u21d2 5x + 5 = 6x – 6
\n\u21d2 6x – 5x = 5 + 6
\n\u21d2 x = 11
\n\u2234 Speed of streamer in still water = 11 km\/h
\nand distance between two points = (11 + 1) \u00d7 5 = 60 km\/h<\/p>\n
\nDistance between two places A and B is 350 km. Two cars start simultaneously from A and B towards each other and the distance between them after 4 hours is 62 km. If the speed of one car is 8 km\/h less than the speed of other cars, find the speed of each car.
\nSolution:
\nDistance between two places A and B = 350 km
\nLet speed of car C1<\/sub> = x km\/h
\nThe speed of car C2<\/sub> = (x – 8) km\/h
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-12-Linear-Equations-and-Inequalities-in-one-Variable-Ex-12.2-Q29.1.png\")
\nIn 4 hours, there will be 62 km distance between these two cars.
\n\u2234 x \u00d7 4 + (x – 8) \u00d7 4 = 350 – 62
\n\u21d2 4x + 4x – 32 = 288
\n\u21d2 8x = 288 + 32 = 320
\n\u21d2 x = \\(\\frac{320}{8}\\) = 40
\nSpeed of one car C1<\/sub> = 40 km\/h
\nand speed of car C2<\/sub> = 40 – 8 = 32 km\/h<\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"