Class 6: Maths Chapter 8 solutions. Complete Class 6 Maths Chapter 8 Notes.<\/p>\n\n\n\n
RD Sharma 6th Maths Chapter 8, Class 6 Maths Chapter 8 solutions<\/p>\n\n\n\n
1. Write the following using numbers, literals and signs of basic operations. State what each letter represents:<\/strong><\/p>\n\n\n\n (i) The diameter of a circle is twice its radius.<\/strong><\/p>\n\n\n\n (ii) The area of a rectangle is the product of its length and breadth.<\/strong><\/p>\n\n\n\n (iii) The selling price equals the sum of the cost price and the profit.<\/strong><\/p>\n\n\n\n (iv) The total amount equals the sum of the principal and the interest.<\/strong><\/p>\n\n\n\n (v) The perimeter of a rectangle is two times the sum of its length and breadth.<\/strong><\/p>\n\n\n\n (vi) The perimeter of a square is four times its side.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) Consider d as the diameter and r as the radius of the circle<\/p>\n\n\n\n Hence, we get d = 2r.<\/p>\n\n\n\n (ii) Consider A as the area, l as the length and b as the breadth of a rectangle<\/p>\n\n\n\n Hence, we get A = l \u00d7 b.<\/p>\n\n\n\n (iii) Consider S.P as the selling price, C.P as the cost price and P as the profit<\/p>\n\n\n\n Hence, we get S.P = C.P + P<\/p>\n\n\n\n (iv) Consider A as the amount, P as the principal and I as the interest<\/p>\n\n\n\n Hence, we get A = P + I<\/p>\n\n\n\n (v) Consider P as the perimeter, l as the length and b as the breadth of a rectangle<\/p>\n\n\n\n Hence, P = 2 (l + b)<\/p>\n\n\n\n (vi) Consider P as the perimeter and a as the side of a square<\/p>\n\n\n\n Hence, P = 4a<\/p>\n\n\n\n 2. Write the following using numbers, literals and signs of basic operations:<\/strong><\/p>\n\n\n\n (i) The sum of 6 and x.<\/strong><\/p>\n\n\n\n (ii) 3 more than a number y.<\/strong><\/p>\n\n\n\n (iii) One-third of a number x.<\/strong><\/p>\n\n\n\n (iv) One-half of the sum of number x and y.<\/strong><\/p>\n\n\n\n (v) Number y less than a number 7.<\/strong><\/p>\n\n\n\n (vi) 7 taken away from x.<\/strong><\/p>\n\n\n\n (vii) 2 less than the quotient of x and y.<\/strong><\/p>\n\n\n\n (viii) 4 times x taken away from one-third of y.<\/strong><\/p>\n\n\n\n (ix) Quotient of x by 3 is multiplied by y.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) The sum of 6 and x can be written as 6 + x.<\/p>\n\n\n\n (ii) 3 more than a number y can be written as y + 3.<\/p>\n\n\n\n (iii) One-third of a number x can be written as x\/3.<\/p>\n\n\n\n (iv) One-half of the sum of number x and y can be written as (x + y)\/ 2.<\/p>\n\n\n\n (v) Number y less than a number 7 can be written as 7 \u2013 y.<\/p>\n\n\n\n (vi) 7 taken away from x can be written as x \u2013 7.<\/p>\n\n\n\n (vii) 2 less than the quotient of x and y can be written as x\/y \u2013 2.<\/p>\n\n\n\n (viii) 4 times x taken away from one-third of y can be written as y\/3 \u2013 4x.<\/p>\n\n\n\n (ix) Quotient of x by 3 is multiplied by y can be written as xy\/3.<\/p>\n\n\n\n 3. Think of a number. Multiply by 5. Add 6 to the result. Subtract y from this result. What is the result?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Consider x as the number.<\/p>\n\n\n\n Multiplying the number by 5 = 5x<\/p>\n\n\n\n Again add 6 to the number = 5x + 6<\/p>\n\n\n\n By subtracting y from the above equation = 5x + 6 \u2013 y.<\/p>\n\n\n\n Hence, the result is 5x + 6 \u2013 y.<\/p>\n\n\n\n 4. The number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor. If the first floor has x rooms, how many rooms does the ground floor has?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Consider y as the number of rooms on the ground floor<\/p>\n\n\n\n We know that<\/p>\n\n\n\n The number of rooms on the first floor = x<\/p>\n\n\n\n It is given that number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor<\/p>\n\n\n\n So we get<\/p>\n\n\n\n y = 2x \u2013 12<\/p>\n\n\n\n Hence, the rooms on the ground floor is y = 2x \u2013 12.<\/p>\n\n\n\n 5. Binny spend Rs a daily and saves Rs b per week. What is her income for two weeks?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Amount spent by Binny = Rs a<\/p>\n\n\n\n Amount saved by Binny = Rs b<\/p>\n\n\n\n Amount spent by Binny in one week = 7a<\/p>\n\n\n\n So the total income for one week = Amount spent by Binny in one week + Amount saved by Binny<\/p>\n\n\n\n Substituting the values<\/p>\n\n\n\n Total income for one week = 7a + b<\/p>\n\n\n\n We get Binny\u2019s income for 2 weeks = 2 (7a + b) = Rs 14a + 2b<\/p>\n\n\n\n Hence, the income of Binny for two weeks is Rs 14a + 2b.<\/p>\n\n\n\n 6. Rahul scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Marks scored by Rahul in English = 80<\/p>\n\n\n\n Marks scored by Rahul in Hindi = x<\/p>\n\n\n\n So the total scores in the two subjects = x + 80<\/p>\n\n\n\n Hence, the total score of Rahul in two subjects is x + 80.<\/p>\n\n\n\n 7. Rohit covers x centimetres in one step. How much distance does he cover in y steps?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Distance covered by Rohit in one step = x cm<\/p>\n\n\n\n So the distance covered by Rohit in y steps = xy cm<\/p>\n\n\n\n Hence, Rohit covers xy cm in y steps.<\/p>\n\n\n\n 8. One apple weighs 75 grams and one orange weighs 40 grams. Determine the weight of x apples and y oranges.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Weight of one apple = 75 g<\/p>\n\n\n\n Weight of one orange = 40 g<\/p>\n\n\n\n So the weight of x apples = 75x g<\/p>\n\n\n\n So the weight of y oranges = 40y g<\/p>\n\n\n\n We get the weight of x apples and y oranges = (75x + 40y) g<\/p>\n\n\n\n Hence, the weight of x apples and y oranges is (75x + 40y) g.<\/p>\n\n\n\n 9. One pencil costs Rs 2 and one fountain pen costs Rs 15. What is the cost of x pencils and y fountain pens?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Cost of one pencil = Rs 2<\/p>\n\n\n\n Cost of one fountain pen = Rs 15<\/p>\n\n\n\n Cost of x pencils = 2x<\/p>\n\n\n\n Cost of y fountain pens = 15y<\/p>\n\n\n\n So the cost of x pencils and y fountain pens = Rs (2x + 15y)<\/p>\n\n\n\n Hence, the cost of x pencils and y fountain pens is Rs (2x + 15y).<\/p>\n\n\n\n 1. Write each of the following products in exponential form:<\/strong><\/p>\n\n\n\n (i) a \u00d7 a \u00d7 a \u00d7 a \u00d7 \u2026\u2026.. 15 times<\/strong><\/p>\n\n\n\n (ii) 8 \u00d7 b \u00d7 b \u00d7 b \u00d7 a \u00d7 a \u00d7 a \u00d7 a<\/strong><\/p>\n\n\n\n (iii) 5 \u00d7 a \u00d7 a \u00d7 a \u00d7 b \u00d7 b \u00d7 c \u00d7 c \u00d7 c<\/strong><\/p>\n\n\n\n (iv) 7 \u00d7 a \u00d7 a \u00d7 a \u2026\u2026.. 8 times \u00d7 b \u00d7 b \u00d7 b \u00d7 \u2026\u2026 5 times<\/strong><\/p>\n\n\n\n (v) 4 \u00d7 a \u00d7 a \u00d7 \u2026\u2026 5 times \u00d7 b \u00d7 b \u00d7 \u2026\u2026. 12 times \u00d7 c \u00d7 c \u2026\u2026 15 times<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) a \u00d7 a \u00d7 a \u00d7 a \u00d7 \u2026\u2026.. 15 times is written in exponential form as a15<\/sup>.<\/p>\n\n\n\n (ii) 8 \u00d7 b \u00d7 b \u00d7 b \u00d7 a \u00d7 a \u00d7 a \u00d7 a is written in exponential form as 8a4<\/sup>b3<\/sup>.<\/p>\n\n\n\n (iii) 5 \u00d7 a \u00d7 a \u00d7 a \u00d7 b \u00d7 b \u00d7 c \u00d7 c \u00d7 c is written in exponential form as 5a3<\/sup>b2<\/sup>c3<\/sup>.<\/p>\n\n\n\n (iv) 7 \u00d7 a \u00d7 a \u00d7 a \u2026\u2026.. 8 times \u00d7 b \u00d7 b \u00d7 b \u00d7 \u2026\u2026 5 times is written in exponential form as 7a8<\/sup>b5<\/sup>.<\/p>\n\n\n\n (v) 4 \u00d7 a \u00d7 a \u00d7 \u2026\u2026 5 times \u00d7 b \u00d7 b \u00d7 \u2026\u2026. 12 times \u00d7 c \u00d7 c \u2026\u2026 15 times is written in exponential form as 4a5<\/sup>b12<\/sup>c15<\/sup>.<\/p>\n\n\n\n 2. Write each of the following in the product form:<\/strong><\/p>\n\n\n\n (i) a2<\/sup> b5<\/sup><\/strong><\/p>\n\n\n\n (ii) 8x3<\/sup><\/strong><\/p>\n\n\n\n (iii) 7a3<\/sup>b4<\/sup><\/strong><\/p>\n\n\n\n (iv) 15 a9<\/sup>b8<\/sup>c6<\/sup><\/strong><\/p>\n\n\n\n (v) 30x4<\/sup>y4<\/sup>z5<\/sup><\/strong><\/p>\n\n\n\n (vi) 43p10<\/sup>q5<\/sup>r15<\/sup><\/strong><\/p>\n\n\n\n (vii) 17p12<\/sup>q20<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) a2<\/sup> b5<\/sup> is written in the product form as a \u00d7 a \u00d7 b \u00d7 b \u00d7 b \u00d7 b \u00d7 b.<\/p>\n\n\n\n (ii) 8x3<\/sup> is written in the product form as 8 \u00d7 x \u00d7 x \u00d7 x.<\/p>\n\n\n\n (iii) 7a3<\/sup>b4<\/sup> is written in the product form as 7 \u00d7 a \u00d7 a \u00d7 a \u00d7 b \u00d7 b \u00d7 b \u00d7 b.<\/p>\n\n\n\n (iv) 15 a9<\/sup>b8<\/sup>c6<\/sup> is written in the product form as 15 \u00d7 a \u00d7 a \u2026\u2026 9 times \u00d7 b \u00d7 b \u00d7 \u2026 8 times \u00d7 c \u00d7 c \u00d7 \u2026.. 6 times.<\/p>\n\n\n\n (v) 30x4<\/sup>y4<\/sup>z5<\/sup> is written in the product form as 30 \u00d7 x \u00d7 x \u00d7 x \u00d7 x \u00d7 y \u00d7 y \u00d7 y \u00d7 y \u00d7 z \u00d7 z \u00d7 z \u00d7 z \u00d7 z.<\/p>\n\n\n\n (vi) 43p10<\/sup>q5<\/sup>r15<\/sup> is written in the product form as 43 \u00d7 p \u00d7 p \u2026. 10 times \u00d7 q \u00d7 q \u2026. 5 times \u00d7 r \u00d7 r \u00d7 \u2026. 15 times.<\/p>\n\n\n\n (vii) 17p12<\/sup>q20<\/sup> is written in the product form as 17 \u00d7 p \u00d7 p \u2026. 12 times \u00d7 q \u00d7 q \u00d7 \u2026.. 20 times.<\/p>\n\n\n\n 3. Write down each of the following in exponential form:<\/strong><\/p>\n\n\n\n (i) 4a3 <\/sup>\u00d7 6ab2<\/sup> \u00d7 c2<\/sup><\/strong><\/p>\n\n\n\n (ii) 5xy \u00d7 3x2<\/sup>y \u00d7 7y2<\/sup><\/strong><\/p>\n\n\n\n (iii) a3<\/sup> \u00d7 3ab2<\/sup> \u00d7 2a2<\/sup>b2<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) 4a3 <\/sup>\u00d7 6ab2<\/sup> \u00d7 c2<\/sup> is written in exponential form as 24a4<\/sup>b2<\/sup>c2<\/sup>.<\/p>\n\n\n\n (ii) 5xy \u00d7 3x2<\/sup>y \u00d7 7y2<\/sup> is written in exponential form as 105x3<\/sup>y4<\/sup>.<\/p>\n\n\n\n (iii) a3<\/sup> \u00d7 3ab2<\/sup> \u00d7 2a2<\/sup>b2<\/sup> is written in exponential form as 6a6<\/sup>b4<\/sup>.<\/p>\n\n\n\n 4. The number of bacteria in a culture is x now. It becomes square of itself after one week. What will be its number after two weeks?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Number of bacteria in a culture = x<\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Number of bacteria becomes square of itself in one week = x2<\/sup><\/p>\n\n\n\n So the number of bacteria after two weeks = (x2<\/sup>)2<\/sup> = x4<\/sup><\/p>\n\n\n\n Hence, the number of bacteria after two weeks is x4<\/sup>.<\/p>\n\n\n\n 5. The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area if its breadth is x cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Area of rectangle = l \u00d7 b<\/p>\n\n\n\n Breadth = x cm<\/p>\n\n\n\n Length = (2\/3) x cm<\/p>\n\n\n\n So the area of the rectangle = (2\/3) x \u00d7 x = 2\/3 x2<\/sup> cm2<\/sup><\/p>\n\n\n\n Hence, the area of rectangle is (2\/3) x2<\/sup> cm2<\/sup>.<\/p>\n\n\n\n 6. If there are x rows of chairs and each row contains x2<\/sup> chairs. Determine the total number of chairs.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Number of rows of chairs = x<\/p>\n\n\n\n Each row contains = x2<\/sup> chairs<\/p>\n\n\n\n So the total number of chairs = number of rows of chairs \u00d7 chairs in each row<\/p>\n\n\n\n We get<\/p>\n\n\n\n Total number of chairs = x \u00d7 x2<\/sup> = x3<\/sup><\/p>\n\n\n\n Hence, the total number of chairs is x3<\/sup>.<\/p>\n\n\n\n Mark the correct alternative in each of the following:<\/strong><\/p>\n\n\n\n 1. 5 more than twice a number x is written as Solution:<\/strong><\/p>\n\n\n\n The option (b) is correct answer.<\/p>\n\n\n\n 5 more than twice a number x is written as 2x + 5.<\/p>\n\n\n\n 2. The quotient of x by 2 is added to 5 is written as Solution:<\/strong><\/p>\n\n\n\n The option (a) is correct answer.<\/p>\n\n\n\n The quotient of x by 2 is added to 5 is written as x\/2 + 5.<\/p>\n\n\n\n 3. The quotient of x by 3 is multiplied by y is written as Solution:<\/strong><\/p>\n\n\n\n The option (d) is correct answer.<\/p>\n\n\n\n It can be written as<\/p>\n\n\n\n x\/3 \u00d7 y = xy\/3<\/p>\n\n\n\n 4. 9 taken away from the sum of x and y is Solution:<\/strong><\/p>\n\n\n\n The option (a) is correct answer.<\/p>\n\n\n\n 9 taken away from the sum of x and y is x + y \u2013 9.<\/p>\n\n\n\n 5. The quotient of x by y added to the product of x and y is written as Solution:<\/strong><\/p>\n\n\n\n The option (a) is correct answer.<\/p>\n\n\n\n The quotient of x by y added to the product of x and y is written as x\/y + xy.<\/p>\n\n\n\n 6. a2<\/sup><\/em>b3<\/sup><\/em> \u00d7 2ab
\n\n\n\nExercise 8.2 page: 8.11<\/h4>\n\n\n\n
\n\n\n\nObjective Type Questions PAGE: 8.13<\/h4>\n\n\n\n
(a) 5 + x + 2
(b) 2x + 5
(c) 2x \u2212 5
(d) 5x + 2<\/strong><\/p>\n\n\n\n
(a) x\/2 + 5
(b) 2\/x+5
(c) (x+2)\/ 5
(d) x\/ (2+5)<\/strong><\/p>\n\n\n\n
(a) x\/3y
(b) 3x\/y
(c) 3y\/x
(d) xy\/3<\/strong><\/p>\n\n\n\n
(a) x + y \u2212 9
(b) 9 \u2212 (x+y)
(c) x+y\/ 9
(d) 9\/ x+y<\/strong><\/p>\n\n\n\n
(a) x\/y + xy
(b) y\/x + xy
(c) xy+x\/ y
(d) xy+y\/ x<\/strong><\/p>\n\n\n\n