What number must be subtracted from each term of the ratio 15 23 so that the ratio becomes 7:15
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Page No 124:Answer:(i) HCF of 24 and 40 is 8. Hence, 24 : 40 in its simplest form is 3 : 5. (ii) HCF of 13.5 and 15 is 1.5. 13.515=135150The HCF of 135 and 150 is 15.=135 ÷ 15150 ÷ 15=910 Hence, 13.5 : 15 in its simplest form is 9 : 10. (iii) 203 : 152=40 : 45 ∴ 40 : 45 = 4045=40 ÷ 545 ÷ 5=89 = 8 : 9 Hence, 623 : 712 in its simplest form is 8 : 9 (iv) 9 : 6 ∴ 9 : 6 = 96=9 ÷ 36 ÷ 3 = 3 : 2 Hence, 16:19 in its simplest form is 3 : 2. (v) LCM of the denominators is 2. ∴ 4 : 5 : 92 = 8 : 10 : 9 The HCF of these 3 numbers is 1. ∴ 8 : 10 : 9 is the simplest form .(vi) 2.5 : 6.5 : 8 = 25 : 65 : 80 The HCF of 25, 65 and 80 is 5. Page No 124:Answer:(i) Converting both the quantities into the same unit, we have: = 75300
= 75 ÷ 75300 ÷ 75=14 (∵ HCF of 75 and 300 = 75) (ii) Converting both the quantities into the same unit, we have: (iii) Converting both the quantities into the same unit (iv) Converting both the quantities into the same unit, we get: (v) Converting both the quantities into the same unit, we get: 2250g : 3000 g = 22503000=2250÷7503000÷750= 34 (∵ HCF of 2250 and 3000 = 750) = 3 g : 4 g (vi) Converting both the quantities into the same unit, we get: Page No 124:Answer:AB = 75 and BC = 914 Therefore, we have: A B×BC = 75×914AC = 910 ∴ A : C = 9 : 10 Page No 124:Answer:AB=58 and BC = 1625Now, we have: AB×BC = 58×1625 ⇒AC = 25 ∴ A : C = 2 : 5 Page No 124:Answer:A : B = 3 : 5 B : C = 10 : 13 = 10÷213÷2=5 :132 Now, A : B : C = 3 : 5 : 132 ∴ A : B : C = 6 : 10 : 13 Page No 124:Answer:We have the following: A : B = 5 : 6 ∴ A : B : C = 5 : 6 : 212 = 10 : 12 : 21 Page No 124:Answer:Sum of the ratio terms = 7 + 8 = 15 Now, we have the following: Kunal's share = Rs 360 ×715= 24×7 = Rs 168 Mohit's share = Rs 360 ×815 = 24×8 = Rs 192 Page No 125:Answer:Sum of the ratio terms = 15+16=1130 Now, we have the following: Page No 125:Answer:Sum of the ratio terms is (1 + 3 + 4) = 8 We have the following: A's share = Rs 5600 ×18 =Rs 56008 = Rs 700 B's share = Rs 5600 ×38= Rs 700 × 3 = Rs 2100 C's share = Rs 5600 ×48 =Rs 700 ×4 = Rs 2800 Page No 125:Answer:Let x be the required number. ⇒9+x 16+x = 23⇒27 + 3x = 32 + 2x ⇒x =5 Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3. Page No 125:Answer:Suppose that x is the number that must be subtracted. ⇒17 - x33 - x=715⇒255 - 15x = 231 - 7x ⇒8x = 255 - 231 = 24 ⇒x = 3 Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15. Page No 125:Answer:Suppose that the numbers are 7x and 11x. Then, (7x + 7) : (11x + 7) = 2 : 3 ⇒ 21x + 21 = 22x + 14 ⇒ x = 7 Hence, the numbers are (7 × 7 =) 49 and (11 × 7 =) 77. Page No 125:Answer:Suppose that the numbers are 5x and 9x. ⇒ 5x - 39x -3=12 ⇒ 10x − 6 = 9x− 3 Hence, the numbers are (5 × 3 =) 15 and (9 × 3 =) 27. Page No 125:Answer:Let the numbers be 3x and 4x. ∴ The numbers are (3 × 15 =) 45 and (4 × 15 =) 60. Page No 125:Answer:Suppose that the present ages of A and B are 8x yrs and
3x yrs. Now, present age of A = 8 ×
6 yrs = 48 yrs Page No 125:Answer:Suppose that the weight of zinc is x g. Then, 48.6 : x = 9 : 5 ⇒ x = 48.6×59=2439 = 27 Hence, the weight of zinc in the alloy is 27 g. Page No 125:Answer:Suppose that the number of boys is x. ⇒ x = 8×3753=8×125 = 1000 Hence, the number of girls in the school is 1000. Page No 125:Answer:Suppose that the monthly income of the family is Rs x. Then, x : 2500 = 11 : 2 ⇒ x = 11×25002=11×1250 Hence, the income is Rs 13,750. Page No 125:Answer:Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively. Total value of these coins = (5 x ×100100+ 8x×50100 + 4x ×25100) ⇒5x + 8x2 + 4x4= 20x + 16x + 4x4=40x4=10x However, the total value is Rs 750. Hence, number of one rupee coins = 5 × 75 = 375 Page No 125:Answer:(4x + 5) : (3x + 11) = 13 : 17 ⇒4x+ 53x + 11=1317⇒68x + 85 = 39x + 143⇒29x = 58 ⇒x = 2 Page No 125:Answer:x y = 34⇒x=3y4 Now, we have (3x + 4y) : (5x + 6y) = 25 : 39 Page No 125:Answer:
xy = 611⇒x = 6y11 Now, we have: 8x -3y3 x + 2y = 8×6y11 -3y3×6y11+2y =48y-33y18y + 22y =15y40y=38 ∴ (8x − 3y) : (3x + 2y) = 3 : 8 Page No 125:Answer:Suppose that the numbers are 5x and 7x. Hence, the numbers are (5 × 60 =) 300 and (7 × 60 =) 420. Page No 125:Answer:(i) The LCM of 6 and 9 is 18. 56=5×36×3=151879 =7×29×2=1418Clearly, 1418<1518 ∴ (7 : 9) < (5 : 6) (ii) The LCM of 3 and 7 is 21. 23=2×73×7=1421 47=4×37×3=1221Clearly, 1221< 1421 ∴ (4 : 7) < (2 : 3) (iii) The LCM of 2 and 7 is 14. 1×72×7=714 4×27×2= 814 Clearly, 714<814 ∴ (1 : 2) < (4 : 7) (iv) The LCM of 5 and 13 is 65. 35=3×135×13= 3965 813= 8×5 13×5= 4065Clearly, 3965<4065 ∴ (3 : 5) < (8 : 13) Page No 125:Answer:(i) We have 56, 89 and 1118. The LCM of 6, 9 and 18 is 18. Therefore, we have: 56= 5×36 ×3=151889= 8×29×2=1618 1118=1118Clearly, 1118< 1518<1618 Hence, (11 : 18) < (5 : 6) < (8 : 9) (ii) We
have 1114, 1721, 57 and 23. The LCM of 14, 21, 7 and 3 is 42. 1114=11×314×3=33281721=17×221×2=344257=5×67×6=304223=2×143×14=2842Clearly , 2842<3042<3328<3442Hence, (2 : 3) < (5 : 7) < (11 : 14) < (17 : 21) Page No 128:Answer:We have: Product of the extremes = 30 × 60 = 1800 Hence, 30 : 40 :: 45 : 60 Page No 128:Answer:We have: Hence, 36, 49, 6 and 7 are not in proportion. Page No 128:Answer:Product of the extremes = 2 × 27 = 54 Since 2 : 9 ::
x : 27, we have: Page No 128:Answer:Product of the extremes = 8 × 35 = 280 Since 8 : x :: 16 : 35, we
have: Page No 128:Answer:Product of the extremes = x × 60 = 60x Since x : 35 :: 48 : 60, we have: Page No 128:Answer:(i) Let the fourth proportional be x. 8 × x = 36 × 6
[Product of extremes = Product of means] (ii) Let the fourth proportional be x. Hence, the fourth proportional is 42. (iii) Let the fourth proportional be x. Page No 128:Answer:36, 54 and x are in continued proportion. Page No 128:Answer:27, 36 and x are in continued proportion. Hence, the value of x is 48. Page No 128:Answer:(i) Suppose that x is the third
proportional to 8 and 12. Hence, the required third proportional is 18. (ii) Suppose that x is the third proportional to 12 and 18. Hence, the third proportional is 27. (iii) Suppose that x is the third proportional to 4.5 and 6. Hence, the third proportional is 8. Page No 128:Answer:The third proportional to 7 and x is 28. Page No 128:Answer:(i) Supposethat x is the mean proportional. Then, 6 : x :: x : 24 ⇒ 6 × 24 = x × x
(Product of extremes = Product of means) Hence, the mean proportional to 6 and 24 is 12. (ii) Suppose that x is the mean proportional. Then, 3 : x :: x : 27 Hence, the mean proportional to 3 and 27 is 9. (iii) Suppose that x is the mean proportional. Then, 0.4 : x :: x : 0.9 ⇒0.4 × 0.9 = x × x
⇒x2 = 0.36 (Product of extremes =Product of means) Hence, the mean proportional to 0.4 and 0.9 is 0.6. Page No 128:Answer:Suppose that the number is x. Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x) ⇒(5 + x) ×(12 + x) = (9 + x) × (7 + x) (Product of extremes = Product of means)⇒60 +5 x + 12 x + x2 = 63 + 9x + 7x + x2⇒60 + 17x = 63 + 16x⇒x = 3 Hence, 3 must be added to each of the numbers: 5, 9, 7 and 12, to get the numbers which are in proportion. Page No 128:Answer:Suppose that x is the number that is to be subtracted. Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x) ⇒(10- x) ×(24 - x)
=(12 - x) ×(19 - x) (Product
of extremes =Product of means)⇒240 - 10x -24x + x2 = 228 -
12x -19x + x2⇒240 - 34x = 228 - 31x⇒3x = 12⇒x = 4 Page No 128:Answer:Distance represented by 1 cm on the map = 5000000 cm = 50 km Distance represented by 3 cm on the map = 50 × 4 km = 200 km ∴ The actual distance is 200 km. Page No 128:Answer:(Height of tree) : (height of its shadow) = (height of the pole) : (height of its shadow) Suppose that the height of pole is x cm. Then, 6 : 8 = x : 20 ⇒ x = 6×208 = 15
Page No 128:Answer:The correct option is (d). ac= ab×bc = 34×89 = 23 Hence, a : c = 2 : 3 Page No 128:Answer:(a) 15 : 8 AB= 23 BC= 45Then, AB×BC = 23×45= 815 Hence, C : A = 15 : 8 Page No 128:Answer:The correct option is (d). A =
3B2C = 4B5∴ A : C = AC = 3B
24B5= 158 Page No 128:Answer:The correct option is (b). 15100A = 20100B⇒AB =43 Hence, A : B = 4 : 3 Page No 128:Answer:(a) 1 : 3 : 6 A = 1 3BC = 2B∴ A : B : C = 13B : B : 2B = 1 : 3 : 6 Page No 129:Answer:(b) 30 : 42 : 77 AB = 57⇒ A = 5B7BC = 611⇒C =11B6∴ A : B : C = 5B 7 : B : 11B6 = 30 : 42 : 77 Page No 129:Answer:(c) 6 : 4 : 3 2A=3B = 4CThen, A = 3B2 and C = 3B4∴ A : B : C = 3 B2 : B : 3B4 = 6 : 4 : 3 Page No 129:Answer:(a) 3 : 4 : 5 A =3B4C = 5B4∴ A : B : C = 3B4 : B : 5B4 = 3 : 4 : 5 Page No 129:Answer:(b) 15 : 10 : 6 1x :1y=2 : 3Then, y : x = 2 : 3 and y = 23x1y:1z = 3 : 5Then, z : y =3 : 5 and z = 35y∴ x : y : z = x : 23x : 35y = x : 23x: 35×23x =x : 23x : 25x = 15 : 10 : 6 Page No 129:Answer:xy= 34 x = 3y4∴ 7x + 3y7x - 3y = 73y4+3y73y4 - 3y=21y + 12y 21y - 12y = 33y9y= 113 Hence, (7x + 3y) : (7x − 3y) = 11 : 3 The correct option is (c). Page No 129:Answer:(c) 5 : 2 3a + 5b3a - 5b=513a + 5b = 15a - 25b12a = 30bab= 3012=52 ∴ a : b = 5 : 2 Page No 129:Answer:(c) 9 7 × 45 = x × 35 (Product of extremes = Product of means )⇒35x = 315⇒x = 9 Page No 129:Answer:(b) 7 Suppose that x is the number that is to be added. Then, (3 + x) : (5 + x) = 5 : 6 ⇒3 +x5 + x= 56⇒18 + 6x = 25+ 5 x⇒x =7 Page No 129:Answer:(d) 40 Suppose that the numbers are x and y. Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7 xy=35x = 3y5=>x + 10y + 10= 57=>7x+70 = 5y + 50=>7×3y5 + 70 =5y + 50=>5y -21y5 = 20=>4y5= 20=>y = 25Therefore, x =3 ×255= 15 Hence, sum of numbers = 15 + 25 = 40 Page No 129:Answer:(a) 3 ⇒15 - x19 - x=34Cross multiplying, we get:60 - 4x = 57 - 3x⇒x = 3 Page No 129:Answer:(a) Rs 180 A's share = 37 × 420 = 180 Page No 129:Answer:(d) 416 Let x be the number of boys. ⇒85= x160 ⇒x=8×1605= 256 ∴ Total strength of the school =256 + 160 = 416 Page No 129:Answer:(a) (2 :3) LCM of 3 and 7 = 7×3= 21 2×73×7 = 1421 and 4×37×3= 1221Clearly, 1221<1421Hence, (4 : 7) < (2 : 3) Page No 129:Answer:(c) 16 Suppose that the third proportional is x. ⇒9 × x = 12 × 12 (Product of extremes= Product of means)⇒9x = 144⇒x = 16 Page No 129:Answer:(b) 12 Suppose that the mean proportional is x. Then, 9 : x :: x : 16 9 ×16 = x × x (Product of extremes =Product of means)⇒x2 = 144⇒x = 12 Page No 129:Answer:(a) 18 years Suppose that the
present ages of A and B are 3x yrs and 8x yrs, respectively. ⇒3x + 68x+6= 4 9⇒27x + 54 = 32x + 24⇒5x = 30⇒x = 6Hence, the present ages of A and B are 18 yrs and 48 yrs, respectively. Page No 131:Answer:The given fractions are 45 and 79. Now, we have:4×95×9=3645 and 7×59×5= 3545 Clearly, 3545<3645Hence, (7 : 9 ) <(4 : 5) Page No 131:Answer:The sum of ratio terms is 10. Then, we have: A's share = Rs 210×1100 = Rs 220 B's share =Rs 310× 1100 =Rs 330 C's share =Rs 510× 1100 = Rs 550 Page No 131:Answer:Product of the extremes = 25 × 6 = 150 The product of the extremes is not equal to that of the means. Hence, 25, 36, 5 and 6 are not in proportion. Page No 131:Answer:x : 18 :: 18 : 108 ⇒x × 108 = 18×18 (Product of extremes =Product of means)⇒108x = 324⇒x =3Hence, the value of x is 3. Page No 131:Answer:Suppose that the numbers are 5x and 7x. Hence, the numbers are (5 × 7 =) 35 and (7 × 7 =) 49. Page No 131:Answer:Suppose that the present ages of A and B are 4x yrs and 3xyrs, respectively. ⇒4x-8 3x - 8=107⇒28x - 56 = 30x - 80⇒2x = 24⇒x= 12Hence, present age of A = 4×12 yrs = 48 yrs and present age of B = 3×12 yrs = 36 yrs Page No 131:Answer:Distance covered in 60 min = 54 km ∴ Distance covered in 40 min = 5460×40 = 36 km Page No 131:Answer:Suppose that the third proportional to 8 and
12 is x. ⇒ 8x = 144 (Product of extremes = Product of means) Hence, the third proportional is 18 . Page No 131:Answer:40 men can finish the work in 60 days. Hence, 75 men will finish the same work in 32 days. Page No 131:Answer:(d) 6 : 4 : 3 A = 32 BC = 34B∴ A : B : C = 32B : B : 34B = 6 : 4 : 3 Page No 131:Answer:(a) 2 : 3 : 4 A = 23BC = 43B∴ A : B : C = 23B : B : 43 B = 2 : 3 : 4 Page No 131:Answer:(c) 11 : 3 We have x = 34yNow, 7x + 3y 7x - 3y = 7×34y +3y7×34y - 3y = 21y + 12y21y -12y=33y9y=113 Page No 131:Answer:(a) 3 Suppose that the number to be subtracted is x. ⇒15 - x19 - x =34⇒ 60 - 4x = 57 - 3x⇒x = 3 Page No 131:Answer:(b) 360 Sum of the ratio terms = 4 + 3 = 7 ∴ B's share = Rs 840 ×37 = Rs 360 Page No 131:Answer:(c) 40 years Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively. Then, (5x+ 5) : (2x + 5) = 15 : 7 ⇒ 5x+52x +5= 157 Cross multiplying, we get: 35x+ 35 = 30x + 75 Hence, the present age of A is 5 × 8 = 40 yrs. Page No 131:Answer:(b) 896 Suppose that the number of boys in the school is x. Hence, total strength of the school = 576 + 320 = 896 Page No 131:Answer:(i) 15 : 8 AC = AB×BC = 23× 45= 815 ∴ C : A=15 : 8 (ii) 5 : 4 16100A = 20100B⇒AB= 2016 = 54 (iii) 1 : 3 : 6 A : B : C = 13 B : B : 2B = 1 : 3 : 6 (iv) 30 : 42 : 77 AB= 5×67×6=3042⇒BC= 6×711×7= 4277⇒ A : B : C = 30 : 42 : 77 Page No 131:Answer:(i) F (ii) F (iii) T (iv) T 3a + 5b3a-5b=51
⇒3a + 5b = 15a - 25 b ⇒ a : b = 5 : 2 View NCERT Solutions for all chapters of Class 7 What must be subtracted from each term of ratio 15 to 23 so that the ratio becomes 7 by 15?7(23-x)=15(15-x).
What must be subtracted from each term of the ratio 3/7 so that the ratio become 2 5?Therefore, if 1/3 is subtracted from each term of the ratio 3:7, the ratio becomes 2:5.
What number must be subtracted from each term of the ratio 19/21 to make 7 8?This is Expert Verified Answer
It means that we have to find out a number which should be subtracted from 19 : 21 to make it in the ratio 7 : 8. Ratio that make 7 : 8 are :- 14 : 16, 21 : 24, 28 : 32 etc. Condition is that if both the results are equal then only we can subtract them as told in question. = 5.
What number should be subtracted from each of the numbers 18/24 28 38 so that the remainders may be in proportion?Answer: The required number is 3. To find : What number should be subtracted from each of the number 18,24,28,38 so that remainders may be in proportion?
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