{"id":46919,"date":"2023-12-20T10:25:42","date_gmt":"2023-12-20T04:55:42","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=46919"},"modified":"2023-12-20T17:21:54","modified_gmt":"2023-12-20T11:51:54","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-6-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-6-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 6 Problems on Simultaneous Linear Equations Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 6 Problems on Simultaneous Linear Equations Chapter Test<\/h2>\n

Question 1.
\nA 700 gm dry fruit packcosts Rs. 216. It contains some almonds and the rest cashew kernel. If almonds cost Rs. 288 ‘per kg and cashew kernel cost Rs. 336 per kg, what are the quantities of the two dry fruits separately ?
\nAnswer:
\nCost of 700 gm of fruit = Rs. 216
\nCost of almonds = Rs. 288 per kg
\nand cost of cashew = Rs. 336 per kg
\nLet weight of almond = x gm
\nThen weigh of cashew = (700 – x) gm
\nAccording to the sum,
\n\"ML
\n\u2234 weight of almond = 400 gm
\nand weight of cashew = 700 – 400
\n= 300 gm<\/p>\n

Question 2.
\nDrawing pencils cost 80 paise each and coloured pencils cost Rs. 1.10 each. If altogether two dozen pencils cost Rs. 21.60, how many coloured pencils are there ?
\nAnswer:
\nLet the number of drawing pencils = x
\nand let the number of coloured pencils = y
\nThen, according to first condition of given problem,
\nx + y = 2 \u00d7 12 (2 dozen = 2 \u00d7 12)
\n\u21d2 x + y = 24 ….(1)
\nAccording to second condition of given problem,
\n\"ML
\nSubstituting the value ofy in equation (1), we get
\nx + 8 = 24 \u21d2 x = 24 – 8 \u21d2 x = 16
\nHence, Number of coloured pencils y = 8<\/p>\n

Question 3.
\nShikha works in a factory. In one week she earned Rs. 390 for working 47 hours, of which 7 hours were overtime. The next week she earned Rs. 416 for working 50 hours, of which 8 hours were overtime. What is Shikha\u2019s hourly earning rate?
\nAnswer:
\nLet Shikha\u2019s earning be Rs. x per regular hour and Rs. y per hour overtime,
\nthen according to first condition of given problem, 40x + 7y = 390 ….(1)
\nAccording to second condition of given problem, 42x + 8y = 416 ….(2)
\nMultiplying equation (1) by 8 and equation (2) by 7
\n\"ML
\nSubstituting the value of x in equation (1), we get
\n40 \u00d7 8 + 7y = 390 \u21d2 320 + 7y = 390
\n\u21d2 7y = 390 – 320 \u21d2 7y = 70
\n\u21d2 y = 10
\nHence, Shika\u2019s earning be Rs. 8 per regular hours and Rs. 10 per hour overtime.<\/p>\n

Question 4.
\nThe sum of the 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.
\nAnswer:
\nLet the digit at tens place = x
\nand let the digit at unit place = y
\nthen number = 10 \u00d7 x + 1 \u00d7 y = 10x +y
\nReversing the number =10 \u00d7 y + x + 1 = 10y + x
\nAccording to first condition of given problem
\nx + y = 7 ….(1)
\nAccording to second condition of given problem,
\n10y + x = 4(10x + y) – 3 \u21d2 10y + x = 40x + 4y – 3
\n\u21d2 10y + x – 40x – 4y = -3 \u21d2 -39x + 6y = -3
\n\u21d2 +3(-13x + 2y) = -3 \u21d2 -3 (13 – 2y) = -3
\n\u21d2 13x – 2y = \\(\\frac{-3}{-3}\\) \u21d2 13x – 2y = 1 ….(2)
\nMultiplying equation (1) by 2, we get
\n\"ML
\nSubstituting the value of x in equation (1), we get
\n1 + y = 7 \u21d2 y = 7 – 1 \u21d2 y = 6
\nHence, Number = 10x + y = 10 \u00d7 1 + 6
\n= 10 + 6 = 16<\/p>\n

Question 5.
\nThree years hence a man\u2019s age will be three times his son\u2019s age and 7 years ago he was seven times as old as his son. How old are they now ?
\nAnswer:
\nLet the man\u2019s age now = x years
\nand his son\u2019s age now = y years
\nthree years hence man\u2019s age = (x + 3) years
\nthree years hence his son\u2019s age = (y + 3) years
\n7 years ago man\u2019s age = (x – 7) years
\n7 years ago his son\u2019s age = (y – 7) years
\nAccording to first condition of given problem,
\n(x + 3) = 3 (y + 3) \u21d2 x + 3 – 3y + 9
\n\u21d2 x – 3y = 9 – 3 \u21d2 x – 3y = 6 ….(1)
\nAccording to second condition of given problem,
\n(x – 7) = 7 (y – 7) \u21d2 x – 7 = 7y – 49
\n\u21d2 x – 7y = -49 + 7 \u21d2 x – 7y = -42 ….(2)
\nFrom equation (1) and equation (2), we get
\n\"ML
\nSubstituting the value ofy in equation (1), we get
\nx – 3 \u00d7 12 = 6 \u21d2 x – 36 = 6 \u21d2 x = 6 + 36
\n\u21d2 x = 42
\nHence, Man\u2019s age = 42 years
\nHis son\u2019s age y years = 12 years<\/p>\n

Question 6.
\nRectangles are drawn on line segments of fixed lengths. When the breadths are 6 m and 5 m respectively the sum of the areas of the rectangles is 83 m2<\/sup>. But if the breadths are 5 m and 4 m respectively the sum of the areas is 68 m2<\/sup>. Find the sum of the areas of the squares drawn on the line segments.
\nAnswer:
\nLet the length of first fixed line segment =x
\nand length of second line segment = y
\nthen in first case sum of areas = 83 m2<\/sup>
\nand breadths are 6 m and 5 m respectively.
\n\u2234 6x + 5y = 83 …(i)
\nIn second case,
\n5x + 4y = 68 …(ii)
\nMultiply (i) by 4 and (ii) by 5,
\n\"ML
\nHence first line segment = 8 m
\nand second line segment = 7 m
\nNow sum of areas of the square on these two line segments = (8)2<\/sup> + (7)2<\/sup>
\n= 64 + 49 = 113 m2<\/sup><\/p>\n

Question 7.
\nIf the length and the breadth of a room are increased by 1 metre each, the area is increased by 21 square metres. If the length is decreased by 1 metre and the breadth is increased by 2 metres, the area is increased by 14 square metres. Find the perimeter of the room.
\nAnswer:
\nLet the length of room = x metre
\nand let the breadth of the room = y metres
\nArea of room = length \u00d7 breadth = x \u00d7 y sq. metre = xy sq. metre.
\nwhen the length is increased by 1 metre, then new length = (x + 1) m
\nwhen the breadth is increased by 1 metre, then new breadth = (y + 1) m
\nNew Area = new length \u00d7 new breadth
\n= (x + 1) (y + 1) sq. metre
\nwhen the length is decreased by 1 metre, then new length = (x – 1) m
\nwhen the breadth is increased by 2 metre, new breadth = (y + 2) m
\nNew Area = new length \u00d7 New breadth
\n= (x – 1) (y + 2) sq. metre
\nAccording to first condition of given problem,
\nxy = (x + 1)(y + 1) – 21
\n\u21d2 xy = x(y + 1) + 1 (y + 1) – 21
\n\u21d2 xy = xy + x + y + 1 – 21 \u21d2 0 = x + y – 20
\n\u21d2 20 = x + y \u21d2 x + y = 20 …(1)
\nAccording to second condition of given problem,
\nxy = (x – 1)(y + 2) – 14
\n\u21d2 xy = x(y + 2) – 1(y + 2) – 14
\n\u21d2 xy = xy + 2 – y – 2 – 14
\n\u21d2 0 = 2x – y – 2 – 14 \u21d2 0 = 2x – y – 16
\n\u21d2 16 = 2x – y \u21d2 2x – y = 16 …(2)
\nFrom equation (1) and (2), we get
\n\"ML
\nSubstituting the value of x in equation (1), we get
\n12 + y = 20 \u21d2 y = 20 – 12 \u21d2 y = 8
\n\u2234 Length of the room = x m = 12m
\n\u2234 Breadth of the room = y m = 8 m
\nHence perimeter of the room = 2 (length + breadth)
\n= 2 (12 + 8) metre = 2 \u00d7 20 metre = 40 metre<\/p>\n

Question 8.
\nThe lenghts (in metres) of the sides of a triangle are \\(2 x+\\frac{y}{2}, \\frac{5 x}{3}+y+\\frac{1}{2} \\text { and } \\frac{2}{3} x+\\)\\(2 y+\\frac{5}{2}\\). If the triangle is equilateral, find its perimeter.
\nAnswer:
\nLength of the sides of an equilateral triangle are
\n\"ML
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 9.
\nOn 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 the 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 ?
\nAnswer:
\nLet the longer candle shorten at the rate of x cm\/hr in burning case and the smaller candle shorter at the rate of y cm\/hr.
\nIn burning case the longer candle bums out completely in 6 hours and that the smaller candle in 4 hours,
\n\u2234 Their lengths are 6x cm and 4y cm respectively
\nAccording to first condition of given problem,
\n6x = 4y + 3 \u21d2 6x – 4y = 3 ….(1)
\nAt 9.30 p.m the length of longer candle = (6x- 4x) cm = 2x cm
\n\"ML
\nHence, lengths of longer candle = 6x = 6 \u00d7 4.5 cm = 27 cm
\nlength of smaller candle = 4y cm = 4 \u00d7 6 cm = 24 cm<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 6 Problems on Simultaneous Linear Equations Chapter Test Question 1. A 700 gm dry fruit packcosts Rs. 216. It contains some almonds and the rest cashew kernel. 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A 700 gm dry fruit packcosts Rs. 216. It contains some almonds and the rest cashew kernel. 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