NCERT Solutions for Class 7 Maths Chapter 8 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among class 7 students for Maths Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 7 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 7 Maths are prepared by experts and are 100% accurate.
Page No 124:
Question 1:
Answer:
(i) HCF of 24 and 40 is 8.
∴ 24 : 40 = = 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
(ii) HCF of 13.5 and 15 is 1.5.
Hence, 13.5 : 15 in its simplest form is 9 : 10.
(iii)
The HCF of 40 and 45 is 5.
∴ 40 : 45 = = 8 : 9
Hence, in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
Hence, in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
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.
∴ 25 : 65 : 80 = = 5 : 13 : 16
Page No 124:
Question 2:
(i) HCF of 24 and 40 is 8.
∴ 24 : 40 = = 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
(ii) HCF of 13.5 and 15 is 1.5.
Hence, 13.5 : 15 in its simplest form is 9 : 10.
(iii)
The HCF of 40 and 45 is 5.
∴ 40 : 45 = = 8 : 9
Hence, in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
Hence, in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
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.
∴ 25 : 65 : 80 = = 5 : 13 : 16
Answer:
(i) Converting both the quantities into the same unit, we have:
75 paise : (3 100) paise = 75 : 300
= (âµ HCF of 75 and 300 = 75)
= 1 paise : 4 paise
(ii) Converting both the quantities into the same unit, we have:
105 cm : 63 cm = (âµ HCF of 105 and 63 = 21)
= 5 cm : 3 cm
(iii) Converting both the quantities into the same unit
65 min : 45 min = (âµ HCF of 65 and 45 = 5)
= 13 min : 9 min
(iv) Converting both the quantities into the same unit, we get:
8 months : 12 months = (âµ HCF of 8 and 12 = 4)
= 2 months : 3 months
(v) Converting both the quantities into the same unit, we get:
2250g : 3000 g = (âµ HCF of 2250 and 3000 = 750)
= 3 g : 4 g
(vi) Converting both the quantities into the same unit, we get:
1000 m : 750 m = (âµ HCF of 1000 and 750 = 250)
= 4 m : 3 m
Page No 124:
Question 3:
(i) Converting both the quantities into the same unit, we have:
75 paise : (3 100) paise = 75 : 300
= (âµ HCF of 75 and 300 = 75)
= 1 paise : 4 paise
(ii) Converting both the quantities into the same unit, we have:
105 cm : 63 cm = (âµ HCF of 105 and 63 = 21)
= 5 cm : 3 cm
(iii) Converting both the quantities into the same unit
65 min : 45 min = (âµ HCF of 65 and 45 = 5)
= 13 min : 9 min
(iv) Converting both the quantities into the same unit, we get:
8 months : 12 months = (âµ HCF of 8 and 12 = 4)
= 2 months : 3 months
(v) Converting both the quantities into the same unit, we get:
2250g : 3000 g = (âµ HCF of 2250 and 3000 = 750)
= 3 g : 4 g
(vi) Converting both the quantities into the same unit, we get:
1000 m : 750 m = (âµ HCF of 1000 and 750 = 250)
= 4 m : 3 m
Answer:
Therefore, we have:
∴ A : C = 9 : 10
Page No 124:
Question 4:
Therefore, we have:
∴ A : C = 9 : 10
Answer:
∴ A : C = 2 : 5
Page No 124:
Question 5:
∴ A : C = 2 : 5
Answer:
A : B = 3 : 5
B : C = 10 : 13 =
Now, A : B : C = 3 : 5 :
∴ A : B : C = 6 : 10 : 13
Page No 124:
Question 6:
A : B = 3 : 5
B : C = 10 : 13 =
Now, A : B : C = 3 : 5 :
∴ A : B : C = 6 : 10 : 13
Answer:
We have the following:
A : B = 5 : 6
B : C = 4 : 7 =
∴ A : B : C = 5 : 6 : = 10 : 12 : 21
Page No 124:
Question 7:
We have the following:
A : B = 5 : 6
B : C = 4 : 7 =
∴ A : B : C = 5 : 6 : = 10 : 12 : 21
Answer:
Sum of the ratio terms = 7 + 8 = 15
Now, we have the following:
Kunal's share = Rs 360 = Rs 168
Mohit's share = Rs 360 = Rs 192
Page No 125:
Question 8:
Sum of the ratio terms = 7 + 8 = 15
Now, we have the following:
Kunal's share = Rs 360 = Rs 168
Mohit's share = Rs 360 = Rs 192
Answer:
Sum of the ratio terms =
Now, we have the following:
Rajan's share = Rs 880 = Rs 480
Kamal's share = Rs 880 = Rs 400
Page No 125:
Question 9:
Sum of the ratio terms =
Now, we have the following:
Rajan's share = Rs 880 = Rs 480
Kamal's share = Rs 880 = Rs 400
Answer:
Sum of the ratio terms is (1 + 3 + 4) = 8
We have the following:
A's share = Rs 5600
B's share = Rs 5600 = Rs 2100
C's share = Rs 5600 = Rs 2800
Page No 125:
Question 10:
Sum of the ratio terms is (1 + 3 + 4) = 8
We have the following:
A's share = Rs 5600
B's share = Rs 5600 = Rs 2100
C's share = Rs 5600 = Rs 2800
Answer:
Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3
Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.
Page No 125:
Question 11:
Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3
Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.
Answer:
Suppose that x is the number that must be subtracted.
Then, (17 − x) : (33 − x) = 7 : 15
Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.
Page No 125:
Question 12:
Suppose that x is the number that must be subtracted.
Then, (17 − x) : (33 − x) = 7 : 15
Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.
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:
Question 13:
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.
Answer:
Suppose that the numbers are 5x and 9x.
Then, (5x − 3) : (9x − 3) = 1 : 2
⇒
⇒ 10x − 6 = 9x − 3
⇒ x = 3
Hence, the numbers are (5 3 =) 15 and (9 3 =) 27.
Page No 125:
Question 14:
Suppose that the numbers are 5x and 9x.
Then, (5x − 3) : (9x − 3) = 1 : 2
⇒
⇒ 10x − 6 = 9x − 3
⇒ x = 3
Hence, the numbers are (5 3 =) 15 and (9 3 =) 27.
Answer:
Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
⇒ x = 15
∴ The numbers are (3 15 =) 45 and (4 15 =) 60.
Page No 125:
Question 15:
Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
⇒ x = 15
∴ The numbers are (3 15 =) 45 and (4 15 =) 60.
Answer:
Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x + 6) : (3x + 6) = 9 : 4
⇒
⇒ 32x + 24 = 27x + 54
⇒ 5x = 30
⇒ x = 6
Now, present age of A = 8 6 yrs = 48 yrs
Present age of B = 3 6 yrs = 18 yrs
Page No 125:
Question 16:
Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x + 6) : (3x + 6) = 9 : 4
⇒
⇒ 32x + 24 = 27x + 54
⇒ 5x = 30
⇒ x = 6
Now, present age of A = 8 6 yrs = 48 yrs
Present age of B = 3 6 yrs = 18 yrs
Answer:
Suppose that the weight of zinc is x g.
Then, 48.6 : x = 9 : 5
⇒ x = = 27
Hence, the weight of zinc in the alloy is 27 g.
Page No 125:
Question 17:
Suppose that the weight of zinc is x g.
Then, 48.6 : x = 9 : 5
⇒ x = = 27
Hence, the weight of zinc in the alloy is 27 g.
Answer:
Suppose that the number of boys is x.
Then, x : 375 = 8 : 3
⇒ x = = 1000
Hence, the number of girls in the school is 1000.
Page No 125:
Question 18:
Suppose that the number of boys is x.
Then, x : 375 = 8 : 3
⇒ x = = 1000
Hence, the number of girls in the school is 1000.
Answer:
Suppose that the monthly income of the family is Rs x.
Then, x : 2500 = 11 : 2
⇒ x =
⇒ x = Rs 13750
Hence, the income is Rs 13,750.
∴ Expenditure = (monthly income − savings)
=Rs (13750 − 2500)
= Rs 11250
Page No 125:
Question 19:
Suppose that the monthly income of the family is Rs x.
Then, x : 2500 = 11 : 2
⇒ x =
⇒ x = Rs 13750
Hence, the income is Rs 13,750.
∴ Expenditure = (monthly income − savings)
=Rs (13750 − 2500)
= Rs 11250
Answer:
Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.
Total value of these coins = ()
However, the total value is Rs 750.
∴ 750 = 10x
⇒ x = 75
Hence, number of one rupee coins = 5 75 = 375
Number of fifty paise coins = 8 75 = 600
Number of twenty-five paise coins = 4 75 = 300
Page No 125:
Question 20:
Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.
Total value of these coins = ()
However, the total value is Rs 750.
∴ 750 = 10x
⇒ x = 75
Hence, number of one rupee coins = 5 75 = 375
Number of fifty paise coins = 8 75 = 600
Number of twenty-five paise coins = 4 75 = 300
Answer:
(4x + 5) : (3x + 11) = 13 : 17
Page No 125:
Question 21:
(4x + 5) : (3x + 11) = 13 : 17
Answer:
Now, we have (3x + 4y) : (5x + 6y)
= 25 : 39
Page No 125:
Question 22:
Now, we have (3x + 4y) : (5x + 6y)
= 25 : 39
Answer:
Now, we have:
∴ (8x − 3y) : (3x + 2y) = 3 : 8
Page No 125:
Question 23:
Now, we have:
∴ (8x − 3y) : (3x + 2y) = 3 : 8
Answer:
Suppose that the numbers are 5x and 7x.
The sum of the numbers is 720.
i.e., 5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
Hence, the numbers are (5 60 =) 300 and (7 60 =) 420.
Page No 125:
Question 24:
Suppose that the numbers are 5x and 7x.
The sum of the numbers is 720.
i.e., 5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
Hence, the numbers are (5 60 =) 300 and (7 60 =) 420.
Answer:
(i) The LCM of 6 and 9 is 18.
∴ (7 : 9) (5 : 6)
(ii) The LCM of 3 and 7 is 21.
∴ (4 : 7) (2 : 3)
(iii) The LCM of 2 and 7 is 14.
Clearly,
∴ (1 : 2) (4 : 7)
(iv) The LCM of 5 and 13 is 65.
∴ (3 : 5) (8 : 13)
Page No 125:
Question 25:
(i) The LCM of 6 and 9 is 18.
∴ (7 : 9) (5 : 6)
(ii) The LCM of 3 and 7 is 21.
∴ (4 : 7) (2 : 3)
(iii) The LCM of 2 and 7 is 14.
Clearly,
∴ (1 : 2) (4 : 7)
(iv) The LCM of 5 and 13 is 65.
∴ (3 : 5) (8 : 13)
Answer:
(i) We have
The LCM of 6, 9 and 18 is 18. Therefore, we have:
Hence, (11 : 18) (5 : 6) (8 : 9)
(ii)
The LCM of 14, 21, 7 and 3 is 42.
Page No 128:
Question 1:
(i) We have
The LCM of 6, 9 and 18 is 18. Therefore, we have:
Hence, (11 : 18) (5 : 6) (8 : 9)
(ii)
The LCM of 14, 21, 7 and 3 is 42.
Answer:
We have:
Product of the extremes = 30 60 = 1800
Product of the means = 40 45 = 1800
Product of extremes = Product of means
Hence, 30 : 40 :: 45 : 60
Page No 128:
Question 2:
We have:
Product of the extremes = 30 60 = 1800
Product of the means = 40 45 = 1800
Product of extremes = Product of means
Hence, 30 : 40 :: 45 : 60
Answer:
We have:
Product of the extremes = 36 7 = 252
Product of the means = 49 6 = 294
Product of the extremes Product of the means
Hence, 36, 49, 6 and 7 are not in proportion.
Page No 128:
Question 3:
We have:
Product of the extremes = 36 7 = 252
Product of the means = 49 6 = 294
Product of the extremes Product of the means
Hence, 36, 49, 6 and 7 are not in proportion.
Answer:
Product of the extremes = 2 27 = 54
Product of the means = 9 x = 9x
Since 2 : 9 :: x : 27, we have:
Product of the extremes = Product of the means
⇒ 54 = 9x
⇒ x = 6
Page No 128:
Question 4:
Product of the extremes = 2 27 = 54
Product of the means = 9 x = 9x
Since 2 : 9 :: x : 27, we have:
Product of the extremes = Product of the means
⇒ 54 = 9x
⇒ x = 6
Answer:
Product of the extremes = 8 35 = 280
Product of the means = 16 x = 16x
Since 8 : x :: 16 : 35, we have:
Product of the extremes = Product of the means
⇒ 280 = 16x
⇒ x = 17.5
Page No 128:
Question 5:
Product of the extremes = 8 35 = 280
Product of the means = 16 x = 16x
Since 8 : x :: 16 : 35, we have:
Product of the extremes = Product of the means
⇒ 280 = 16x
⇒ x = 17.5
Answer:
Product of the extremes = x 60 = 60x
Product of the means = 35 48 = 1680
Since x : 35 :: 48 : 60, we have:
Product of the extremes = Product of the means
⇒ 60x= 1680
⇒ x = 28
Page No 128:
Question 6:
Product of the extremes = x 60 = 60x
Product of the means = 35 48 = 1680
Since x : 35 :: 48 : 60, we have:
Product of the extremes = Product of the means
⇒ 60x= 1680
⇒ x = 28
Answer:
(i) Let the fourth proportional be x.
Then, 8 : 36 :: 6 : x
8 [Product of extremes = Product of means]
⇒ 8x = 216
⇒ x = 27
Hence, the fourth proportional is 27.
(ii) Let the fourth proportional be x.
Then, 5 : 7 :: 30 : x
⇒ [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 5x = 210
⇒ x = 42
Hence, the fourth proportional is 42.
(iii) Let the fourth proportional be x.
Then, 2.8 [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 2.8x = 49
⇒ x = 17.5
Hence, the fourth proportional is 17.5.
Page No 128:
Question 7:
(i) Let the fourth proportional be x.
Then, 8 : 36 :: 6 : x
8 [Product of extremes = Product of means]
⇒ 8x = 216
⇒ x = 27
Hence, the fourth proportional is 27.
(ii) Let the fourth proportional be x.
Then, 5 : 7 :: 30 : x
⇒ [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 5x = 210
⇒ x = 42
Hence, the fourth proportional is 42.
(iii) Let the fourth proportional be x.
Then, 2.8 [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 2.8x = 49
⇒ x = 17.5
Hence, the fourth proportional is 17.5.
Answer:
36, 54 and x are in continued proportion.
Then, 36 : 54 :: 54 : x
⇒ [Product of extremes = Product of means]
⇒ 36x = 2916
⇒ x = 81
Page No 128:
Question 8:
36, 54 and x are in continued proportion.
Then, 36 : 54 :: 54 : x
⇒ [Product of extremes = Product of means]
⇒ 36x = 2916
⇒ x = 81
Answer:
27, 36 and x are in continued proportion.
Then, 27 : 36 :: 36 : x
⇒ [Product of extremes = Product of means]
⇒ 27x = 1296
⇒ x = 48
Hence, the value of x is 48.
Page No 128:
Question 9:
27, 36 and x are in continued proportion.
Then, 27 : 36 :: 36 : x
⇒ [Product of extremes = Product of means]
⇒ 27x = 1296
⇒ x = 48
Hence, the value of x is 48.
Answer:
(i) Suppose that x is the third proportional to 8 and 12.
Then, 8 :12 :: 12 : x
⇒ 8 (Product of extremes = Product of means )
⇒ 8x = 144
⇒ x = 18
Hence, the required third proportional is 18.
(ii) Suppose that x is the third proportional to 12 and 18.
Then, 12 : 18 :: 18 : x
⇒ (Product of extremes = Product of means )
⇒ 12x = 324
⇒ x = 27
Hence, the third proportional is 27.
(iii) Suppose that x is the third proportional to 4.5 and 6.
Then, 4.5 : 6:: 6 : x
⇒ (Product of extremes = Product of means )
⇒ 4.5x = 36
⇒ x = 8
Hence, the third proportional is 8.
Page No 128:
Question 10:
(i) Suppose that x is the third proportional to 8 and 12.
Then, 8 :12 :: 12 : x
⇒ 8 (Product of extremes = Product of means )
⇒ 8x = 144
⇒ x = 18
Hence, the required third proportional is 18.
(ii) Suppose that x is the third proportional to 12 and 18.
Then, 12 : 18 :: 18 : x
⇒ (Product of extremes = Product of means )
⇒ 12x = 324
⇒ x = 27
Hence, the third proportional is 27.
(iii) Suppose that x is the third proportional to 4.5 and 6.
Then, 4.5 : 6:: 6 : x
⇒ (Product of extremes = Product of means )
⇒ 4.5x = 36
⇒ x = 8
Hence, the third proportional is 8.
Answer:
The third proportional to 7 and x is 28.
Then, 7 : x :: x : 28
⇒ 7 28 = (Product of extremes = Product of means)
⇒ x = 14
Page No 128:
Question 11:
The third proportional to 7 and x is 28.
Then, 7 : x :: x : 28
⇒ 7 28 = (Product of extremes = Product of means)
⇒ x = 14
Answer:
(i) Suppose that x is the mean proportional.
Then, 6 : x :: x : 24
⇒ (Product of extremes = Product of means)
⇒
⇒ x = 12
Hence, the mean proportional to 6 and 24 is 12.
(ii) Suppose that x is the mean proportional.
Then, 3 : x :: x : 27
(Product of extremes =Product of means)
⇒ x = 9
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
(Product of extremes =Product of means)
⇒x = 0.6
Hence, the mean proportional to 0.4 and 0.9 is 0.6.
Page No 128:
Question 12:
(i) Suppose that x is the mean proportional.
Then, 6 : x :: x : 24
⇒ (Product of extremes = Product of means)
⇒
⇒ x = 12
Hence, the mean proportional to 6 and 24 is 12.
(ii) Suppose that x is the mean proportional.
Then, 3 : x :: x : 27
(Product of extremes =Product of means)
⇒ x = 9
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
(Product of extremes =Product of means)
⇒x = 0.6
Hence, the mean proportional to 0.4 and 0.9 is 0.6.
Answer:
Suppose that the number is x.
Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x)
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:
Question 13:
Suppose that the number is x.
Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x)
Hence, 3 must be added to each of the numbers: 5, 9, 7 and 12, to get the numbers which are in proportion.
Answer:
Suppose that x is the number that is to be subtracted.
Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x)
.
Hence, 4 must be subtracted from each of the numbers: 10, 12, 19 and 24, to get the numbers which are in proportion.
Page No 128:
Question 14:
Suppose that x is the number that is to be subtracted.
Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x)
.
Hence, 4 must be subtracted from each of the numbers: 10, 12, 19 and 24, to get the numbers which are in proportion.
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:
Question 15:
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.
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 =
∴ Height of the pole = 15 cm
Page No 128:
Question 1:
(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 =
∴ Height of the pole = 15 cm
Answer:
The correct option is (d).
Hence, a : c = 2 : 3
Page No 128:
Question 2:
The correct option is (d).
Hence, a : c = 2 : 3
Answer:
(a) 15 : 8
Page No 128:
Question 3:
(a) 15 : 8
Answer:
The correct option is (d).
Hence, A : C = 15 : 8
Page No 128:
Question 4:
The correct option is (d).
Hence, A : C = 15 : 8
Answer:
The correct option is (b).
Hence, A : B = 4 : 3
Page No 128:
Question 5:
The correct option is (b).
Hence, A : B = 4 : 3
Answer:
(a) 1 : 3 : 6
Page No 129:
Question 6:
(a) 1 : 3 : 6
Answer:
(b) 30 : 42 : 77
Page No 129:
Question 7:
(b) 30 : 42 : 77
Answer:
(c) 6 : 4 : 3
Page No 129:
Question 8:
(c) 6 : 4 : 3
Answer:
(a) 3 : 4 : 5
= 3 : 4 : 5
Page No 129:
Question 9:
(a) 3 : 4 : 5
= 3 : 4 : 5
Answer:
(b) 15 : 10 : 6
Page No 129:
Question 10:
(b) 15 : 10 : 6
Answer:
Hence, (7x + 3y) : (7x − 3y) = 11 : 3
The correct option is (c).
Page No 129:
Question 11:
Hence, (7x + 3y) : (7x − 3y) = 11 : 3
The correct option is (c).
Answer:
(c) 5 : 2
∴ a : b = 5 : 2
Page No 129:
Question 12:
(c) 5 : 2
∴ a : b = 5 : 2
Answer:
(c) 9
Page No 129:
Question 13:
(c) 9
Answer:
(b) 7
Suppose that x is the number that is to be added.
Then, (3 + x) : (5 + x) = 5 : 6
Page No 129:
Question 14:
(b) 7
Suppose that x is the number that is to be added.
Then, (3 + x) : (5 + x) = 5 : 6
Answer:
(d) 40
Suppose that the numbers are x and y.
Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7
Hence, sum of numbers = 15 + 25 = 40
Page No 129:
Question 15:
(d) 40
Suppose that the numbers are x and y.
Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7
Hence, sum of numbers = 15 + 25 = 40
Answer:
(a) 3
Suppose that x is the number that is to be subtracted.
Then, (15 − x) : (19 − x) = 3 : 4
Page No 129:
Question 16:
(a) 3
Suppose that x is the number that is to be subtracted.
Then, (15 − x) : (19 − x) = 3 : 4
Answer:
(a) Rs 180
A's share =
Page No 129:
Question 17:
(a) Rs 180
A's share =
Answer:
(d) 416
Let x be the number of boys.
Then, 8 : 5 = x : 160
Page No 129:
Question 18:
(d) 416
Let x be the number of boys.
Then, 8 : 5 = x : 160
Answer:
(a) (2 :3)
LCM of 3 and 7 = 21
Page No 129:
Question 19:
(a) (2 :3)
LCM of 3 and 7 = 21
Answer:
(c) 16
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
Page No 129:
Question 20:
(c) 16
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
Answer:
(b) 12
Suppose that the mean proportional is x.
Then, 9 : x :: x : 16
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Question 21:
(b) 12
Suppose that the mean proportional is x.
Then, 9 : x :: x : 16
Answer:
(a) 18 years
Suppose that the present ages of A and B are 3x yrs and 8x yrs, respectively.
After six years, the age of A will be (3x+6) yrs and that of B will be (8x+6) yrs.
Then, (3x +6) : (8x + 6) = 4 : 9
Page No 131:
Question 1:
(a) 18 years
Suppose that the present ages of A and B are 3x yrs and 8x yrs, respectively.
After six years, the age of A will be (3x+6) yrs and that of B will be (8x+6) yrs.
Then, (3x +6) : (8x + 6) = 4 : 9
Answer:
The given fractions are .
LCM of 5 and 9 = 5 9 = 45
Page No 131:
Question 2:
The given fractions are .
LCM of 5 and 9 = 5 9 = 45
Answer:
The sum of ratio terms is 10.
Then, we have:
A's share = Rs
Page No 131:
Question 3:
The sum of ratio terms is 10.
Then, we have:
A's share = Rs
Answer:
Product of the extremes = 25 6 = 150
Product of the means = 36 5 = 180
The product of the extremes is not equal to that of the means.
Hence, 25, 36, 5 and 6 are not in proportion.
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Question 4:
Product of the extremes = 25 6 = 150
Product of the means = 36 5 = 180
The product of the extremes is not equal to that of the means.
Hence, 25, 36, 5 and 6 are not in proportion.
Answer:
x : 18 :: 18 : 108
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Question 5:
x : 18 :: 18 : 108
Answer:
Suppose that the numbers are 5x and 7x.
Then, 5x + 7x = 84
⇒ 12x = 84
⇒ x = 7
Hence, the numbers are (5 7 =) 35 and (7 7 =) 49.
Page No 131:
Question 6:
Suppose that the numbers are 5x and 7x.
Then, 5x + 7x = 84
⇒ 12x = 84
⇒ x = 7
Hence, the numbers are (5 7 =) 35 and (7 7 =) 49.
Answer:
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7
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Question 7:
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7
Answer:
Distance covered in 60 min = 54 km
Distance covered in 1 min =
∴ Distance covered in 40 min =
Page No 131:
Question 8:
Distance covered in 60 min = 54 km
Distance covered in 1 min =
∴ Distance covered in 40 min =
Answer:
Suppose that the third proportional to 8 and 12 is x.
Then, 8 : 12 :: 12 : x
⇒ 8x = 144 (Product of extremes = Product of means)
⇒ x = 18
Hence, the third proportional is 18 .
Page No 131:
Question 9:
Suppose that the third proportional to 8 and 12 is x.
Then, 8 : 12 :: 12 : x
⇒ 8x = 144 (Product of extremes = Product of means)
⇒ x = 18
Hence, the third proportional is 18 .
Answer:
40 men can finish the work in 60 days.
1 man can finish the work in 60 40 days. [Less men, more days]
75 men can finish the work in
Hence, 75 men will finish the same work in 32 days.
Page No 131:
Question 10:
40 men can finish the work in 60 days.
1 man can finish the work in 60 40 days. [Less men, more days]
75 men can finish the work in
Hence, 75 men will finish the same work in 32 days.
Answer:
(d) 6 : 4 : 3
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Question 11:
(d) 6 : 4 : 3
Answer:
(a) 2 : 3 : 4
Page No 131:
Question 12:
(a) 2 : 3 : 4
Answer:
(c) 11 : 3
Page No 131:
Question 13:
(c) 11 : 3
Answer:
(a) 3
Suppose that the number to be subtracted is x.
Then, (15 − x) : (19 − x) = 3 : 4
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Question 14:
(a) 3
Suppose that the number to be subtracted is x.
Then, (15 − x) : (19 − x) = 3 : 4
Answer:
(b) 360
Sum of the ratio terms = 4 + 3 = 7
∴ B's share = = Rs 360
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Question 15:
(b) 360
Sum of the ratio terms = 4 + 3 = 7
∴ B's share = = Rs 360
Answer:
(c) 40 years
Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively.
After 5 years, the ages of A and B will be (5x+5) yrs and (2x+5) yrs, respectively.
Then, (5x + 5) : (2x + 5) = 15 : 7
⇒
Cross multiplying, we get:
35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8
Hence, the present age of A is 5 8 = 40 yrs.
Page No 131:
Question 16:
(c) 40 years
Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively.
After 5 years, the ages of A and B will be (5x+5) yrs and (2x+5) yrs, respectively.
Then, (5x + 5) : (2x + 5) = 15 : 7
⇒
Cross multiplying, we get:
35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8
Hence, the present age of A is 5 8 = 40 yrs.
Answer:
(b) 896
Suppose that the number of boys in the school is x.
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576
Hence, total strength of the school = 576 + 320 = 896
Page No 131:
Question 17:
(b) 896
Suppose that the number of boys in the school is x.
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576
Hence, total strength of the school = 576 + 320 = 896
Answer:
(i) 15 : 8
∴ C : A=15 : 8
(ii) 5 : 4
(iii) 1 : 3 : 6
(iv) 30 : 42 : 77
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Question 18:
(i) 15 : 8
∴ C : A=15 : 8
(ii) 5 : 4
(iii) 1 : 3 : 6
(iv) 30 : 42 : 77
Answer:
(i) F
Suppose that the mean proportional is x.
Then, 0.4 : x :: x : 0.9
(ii) F
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
⇒ 9x = 144 (Product of extremes = Product of means)
⇒ x = 16
(iii) T
8 : x :: 48 : 18
⇒ 144 = 48x (Product of extremes = Product of means)
⇒ x = 3
(iv) T
⇒ 12a = 30b
⇒ a : b = 5 : 2
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