ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion

 

(i) 2: 3 and 4: 9 

As We know that 

Compound ratio = 2/3 × 4/9 

 = 8/27 

 = 8: 27 

 

(ii) 4: 5,5: 7 and 9: 11 

As We know that 

Compound ratio = 4/5 × 5/7 × 9/11 

=36/77= 36: 77 

 

(i) 2: 3 

As We know that 

Duplicate ratio of 2: 3 = 2²: 3²  = 4: 9 

 

(ii) √5: 7 

As We know that 

Duplicate ratio of √5: 7 = √5² ∶ 7²  = 5: 49

 

(iii) 5a: 6b 

As We know that 

Duplicate ratio of 5a: 6b = (5a)² ∶ (6b)²  = 25a² ∶ 36b²

 

 

(i) 3: 4 

As We know that 

Triplicate ratio of 

 3: 4 = 33∶ 43 = 27: 64 

 

(ii) ½: 1/3 

As We know that 

 

(i) 9: 16 

As We know that 

Sub-duplicate ratio of 9: 16 = √9: √16 = 3: 4 

 

(ii) ¼: 1/9 

As We know that 

Sub-duplicate ratio of ¼: 1/9 = √1/4: √1/9 

So we  will get = ½:¹/_³

 = 3: 2

 

(iii) 9a²: 49b²  

As We know that 

Sub-duplicate ratio of 9a² ∶ 49b²  = √9a² ∶ √49b²  = 3a: 7b

 

(i) 1: 216 

As We know that 

Sub-triplicate ratio of 1: 216 = ∛1: ∛216 

By further calculation 

 = (1³  )^(1/3)  ∶ (6³  )^(1/3)  

 = 1: 6 

 

(ii) 1/8: 1/125 

As We know that 

 

(i) 4: 7 

As We know that 

Reciprocal ratio of 4: 7 = 7: 4 

 

(ii) 3² ∶ 4^2 

As We know that 

Reciprocal ratio of 3² ∶ 4²    = 4²   ∶ 3²  = 16: 9 

 

(iii) 1/9: 2 

As We know that 

Reciprocal ratio of 1/9: 2 = 2: 1/9 = 18: 1

 

 

As It is given that 

 2: 3,17: 21,11: 14 and 5: 7 

 

 thus We can write it in fractions as 

 2/3,17/21,11/14,5/7 

 Here the LCM of 3,21,14 and 7 is 42 

 

 By converting the ratio as equivalent 

 2/3 = (2 × 14)/ (3 × 14) = 28/42 

 17/21 = (17 × 2)/ (21 × 2) = 34/ 42 

 11/14 = (11 × 3)/ (14 × 3) = 33/42 

 5/7 = (5 × 6)/ (7 × 6) = 30/42 

 

 Now writing it in ascending order 

 28/42,30/42,33/42,34/42 

 

 By further simplification 

 2/3,5/7,11/14,17/21 

 

 So we will get 

 2: 3,5: 7,11: 14 and 17: 21  

 

 

We will get 

 (x + 2y)/ (2x + y) = 11/10 or 14/13 

 (x + 2y): (2x + y) = 11: 10 or 14: 13

 

 (ii) y (3x – y): x (4x + y) = 5: 12 

 Thus It can be written as 

Taking common terms 

 5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0 

 [4 (x/y) – 3] [5 (x/y) – 4] = 0 

 

Here  

 4 (x/y) – 3 = 0 

 So we get 

 4 (x/y) = 3 

 x/y = 3/4 

 

 Similarly 

 5 (x/y) – 4 = 0 

 So we will get 

 5 (x/y) = 4 

 x/y = 4/5

 

 (a) x/y = 3/4 

 We know that 

 

 

 

 

(i) As we have given that 

Perimeter of triangle = 30 cm 

Ratio among sides = 7: 5: 3

Here the sum of ratios = 7 + 5 + 3 = 15 

 

As We know that 

Length of first side = 30 × 7/15 = 14 cm 

Length of second side = 30 × 5/15 = 10 cm 

Length of third side = 30 × 3/15 = 6 cm 

 

Therefore, the sides are 14 cm, 10 cm and 6 cm. 

 

(ii) as We know that 

Sum of all the angles of a triangle = 180° 

Here the ratio among angles = 2: 3: 4 

Sum of ratios = 2 + 3 + 4 = 9

As we have given that 

Ratio of three numbers = 1/2: 1/3: 1/4 

 = (6: 4: 3)/ 12 

 = 6: 4: 3 

 

Lets assumed first number = 6x 

Second number = 4x

Third number = 3x 

So based on the condition 

 (6x)²  + (4x)²  + (3x)²  = 244 

 36x²  + 16x²  + 9x²   = 244 

So we will get 

 61x²  = 244

  x²  = 244/61 = 4 = 2²  

 x = 2 

 

Here 

First number = 6x = 6 × 2 = 12 

Second number = 4x = 4 × 2 = 8 

Third number = 3x = 3 × 2 = 6

(i) As we have given that 

 Ratio between A,B and C = 7: 5: 4 

 Consider A share = 7x 

 B share = 5x 

 C share = 4x 

 So the total sum = 7x + 5x + 4x = 16x 

 

 Based on the condition 

 5x – 4x = 500 

 x = 500 

 So the total sum = 16x = 16 × 500 = Rs 8000

 

 

(ii) 6 months investment of A = Rs 50000 

1 month investment of A = 50000 × 6 = Rs 300000 

 

4 months investment of B = Rs 60000 

1 month investment of B = 60000 × 4 = Rs 240000 

 

5 months investment of C = Rs 80000 

1 month investment of C = 80000 × 5 = Rs 400000 

 

Here the ratio between their investments = 300000: 240000: 400000 = 30: 24: 40 

Sum of ratio = 30 = 24 + 40 = 94 

Total earnings = Rs 18800 

 

So we will get 

 A share = 30/94 × 18800 = Rs 6000 

 B share = 24/94 × 18800 = Rs 4800 

 C share = 40/94 = 18800 = Rs 8000

(i) As we have given that 

 Mixture of milk to water = 45 litres 

 Ratio of milk to water = 13: 2 

 Sum of ratio = 13 + 2 = 15 

 Here the quantity of milk = (45 × 13)/ 15 = 39 litres 

 Quantity of water = 45 × 2/15 = 6 litres 

 

Lets assumed  x litre of water to be added,then water = (6 + x) litres 

 Here the new ratio = 3: 1 

 39: (6 + x) = 3: 1 

 We can write it as

 39/ (6 + x) = 3/1 

 By cross multiplication 

 39 = 18 + 3x 

 3x = 39 – 18 = 21 

 x = 21/3 

= 7 litres 

Hence, 7 litres of water is to be added to the mixture.

 

(ii) As we have given that 

Ratio between boys and girls = 5: 3 

Number of pupils = 560 

So the sum of ratios = 5 + 3 = 8 

 

As We know that 

 Number of boys = 5/8 × 560 = 350 

 Number of girls = 3/8 × 560 = 210 

 Number of new boys admitted = 10 

 So the total number of boys = 350 + 10 = 360

 

Lets assumed x as the number of girls admitted 

 Total number of girls = 210 + x 

 Based on the condition 

 360: 210 + x = 3: 2 

 We can write it as 

 360/ 210 + x = 3/2 

 By cross multiplication 

 630 + 3x = 720 

 3x = 720 – 630 = 90 

 So we will get 

 x = 90/3 = 30 

Hence, 30 new girls are to be admitted.

 

 

(i) Lets assumed the monthly pocket money of Ravi and Sanjeev as 5x and 7x 

Their expenditure is 3y and 5y respectively. 

 5x – 3y = 80 …… (1) 

 7x – 5y = 80 …… (2) 

 Now multiply equation (1) by 7 and (2) by 5 

 Subtracting both the equations 

 35x – 21y = 560 

 35x – 25y = 400 

 So we  will get 

 4y = 160 

 y = 40 

 

 In equation (1)

  5x = 80 + 3 × 40 

 = 200 

 x = 40

 Here the monthly pocket money of Ravi = 5  × 40 = 200

 

(ii) Lets assumed x as the number of students in class 

Ratio of boys and girls = 4: 3 

Number of boys = 4x/7 

Number of girls = 3x/7 

 

Based on the problem 

 (4x/7 + 20): (3x/7 – 12) = 2: 1 

 We can write it as 

 (4x + 140)/ 7: (3x – 84)/ 7 = 2: 1 

 So we get 

 (4x + 140)/ 7 × 7/ (3x – 84) = 2/1 

 (4x + 140)/ (3x – 84) = 2/1 

 6x – 168 = 4x + 140 

 6x – 4x = 140 + 168 

 2x = 308 

 x = 308/2 

 = 154 

Therefore, 154 students were there in the class.

Lets assumed the number of passes = 4x 

Number of failures = x 

Total number of students appeared = 4x + x = 5x

 

In case 2 

 Number of students appeared = 5x – 30 

 Number of passes = 4x – 20 

 So the number of failures = (5x – 30) – (4x – 20) 

 By further calculation 

 = 5x – 30 – 4x + 20 

 = x – 10 

 

Based on the condition 

c(4x – 20)/ (x – 10) = 5/1 

 By cross multiplication 

 5x – 50 = 4x – 20 

 5x – 4x = - 20 + 50 

 x = 30 

 

Hence No. of students appeared = 5x = 5 × 30 = 150

 (i) 10: 35 = x: 42 

 Thus We can write it as 

 35 × x = 10 × 42 

 So we will get 

 x = (10 × 42)/ 35 

 x = 2 × 6 x = 12 

 

 (ii) 3: x = 24: 2 

 Thus We can write it as 

 x × 24 = 3 × 2 

 So we will get 

 x = (3 × 2)/ 24 

 x = ¼ 

 

 iii) 2.5: 1.5 = x: 3 

 Thus We can write it as 

 1.5 × x = 2.5 × 3 

 So we will get 

 x = (2.5 × 3)/ 1.5

  x = 5.0  

 

 (iv) x: 50∶: 3: 2 

 Thus We can write it as 

 x × 2 = 50 × 3 

 So we will get 

 x = (50 × 3)/ 2 

 x = 75 

 (i) 3,12,15 

 Lets assumed x as the fourth proportional to 3,12 and 15 

 3: 12∶: 15: x 

 Thus We can write it as 

 3 × x = 12 × 15 

 So we will get 

 x =   (12 × 15)/ 3 

 x = 60 

 

 (ii) 1/3,1/4,1/5 

 Lets as sumed  x as the fourth proportional to 1/3,1/4 and 1/5 

 1/3: 1/4:: 1/5: x 

 thus We can write it as 

 1/3 × x = 1/4 × 1/5 

 So we will get 

 x = 1/4 × 1/5 × 3/1 

 x = 3/20 

 

 (iii) 1.5,2.5,4.5 

 Lets assumed  x as the fourth proportional to 1,5,2.5 and 4.5 

 1.5: 2.5∶: 4.5: x 

 Thus We can write it as 

 1.5 × x = 2.5 × 4.5 

 So we will get 

 x = ( 2.5 × 4.5)/ 1.5 

 x = 7.5 

 

 (iv) 9.6 kg,7.2 kg,28.8 kg 

 Lets asumed  x as the fourth proportional to 9.6,7.2 and 28.8 

 9.6: 7.2∶: 28.8: x 

 Thus We can write it as 

 9.6 × x = 7.2 × 28.8 

  So we willget 

 x = (7.2 × 28.8)/ 9.6 

 x = 21.6  

 

(i) Lets assumed  x as the third proportional to 5, 10 

 5: 10∶: 10: x 

 Thus It can be written as 

 5 × x = 10 × 10 

 x = (10 × 10)/ 5 = 20 

 

Hence, the third proportional to 5, 10 is 20. 

 

(ii) Lets assumed  x as the third proportional to 0.24,0.6 

  0.24: 0.6∶: 0.6: x 

  Thus It can be written as 

 0.24 × x = 0.6 × 0.6

 x = (0.6 × 0.6)/ 0.24 = 1.5 

 

Hence, the third proportional to 0.24, 0.6 is 1.5. 

 

(iii) Lets assumed  x as the third proportional to Rs. 3 and Rs. 12 

 3: 12∶: 12: x 

 thus It can be written as 

 3 × x = 12 × 12 

 x = (12 × 12)/ 3 = 48 

 

Hence, the third proportional to Rs. 3 and Rs. 12 is Rs. 48 

 

(iv) Lets assumed x as the third proportional to 5 ¼ and 7 

 5 ¼: 7∶: 7: x 

 thus It can be written as 

 21/4 × x = 7 × 7 

 x = (7 × 7 × 4)/ 21 = 28/3 = 9 1/3  

 

Hence, the third proportional to 5 ¼ and 7 is 9 1/3.

 

(i) Lets assumed  x as the mean proportion of 5 and 80 

 5: x∶: x: 80 

 thus It can be written as 

 x²  = 5 × 80 = 400 

 x = √400 = 20 

 

Therefore, mean proportion of 5 and 80 is 20. 

 

(ii) Lets assumed x as the mean proportion of 1/12 and 1/75 

 1/12: x∶: x: 1/75 

 thus It can be written as 

 x²  = 1/12 × 1/75 = 1/900 

 x = √1/900 = 1/30

 

Therefore, mean proportion of 1/12 and 1/75 is 1/30. 

 

(iii) Lets assumed  x as the mean proportion of 8.1 and 2.5 

 8.1: x∶: x: 2.5 

Thus It can be written as 

 x²  = 8.1 × 2.5 = 20.25 

 x = √20.25 = 4.5 

Therefore, mean proportion of 8.1 and 2.5 is 4.5.

 

(iv) lets assumed  x as the mean proportion of (a – b) and (a³  – a²  b),a ˃ b 

 (a – b): x∶: (a³  – a² b) 

 thus It can be written as 

 x²  = (a – b) (a³   – a²  b) 

 So we will get 

 x²  = (a – b) a²  (a – b)

 x²   = a²  (a – b)²  

 Here 

 x = a (a – b) 

Therefore, mean proportion of (a – b) and (a³ – a² b),a ˃ b is a (a – b).

 

As we have given that 

a, 12, 16 and b are in continued proportion 

 a/12 = 12/16 = 16/b 

 

as We know that 

 a/12 = 12/16 

 By cross multiplication 

 16a = 144 

 a = 144/16 

 = 9 

 

Similarly 

 12/16 = 16/b 

 By cross multiplication 

 12b = 16 × 16 = 256 

 b = 256/12 

 = 64/3 

 = 21 1/3 

 Therefore,a = 9 and b = 64/3 or 21 1/3.

 Lets assumed  x to be added to 5,11,19 and 37 to make them in proportion

 5 + x: 11 + x∶: 19 + x: 37 + x 

 thus It can be written as 

 (5 + x) (37 + x) = (11 + x) (19 + x) 

 By further calculation 

 185 + 5x + 37x + x²  = 209 + 11x + 19x + x²  

 185 + 42x + x²   = 209 + 30x + x²   

 So we will get 

 42x – 30x +  x²   – x²   = 209 – 185 

 12x = 24 

 x = 2 

Hence, the least number to be added is 2.

 

 

Lets assumed  x be subtracted from each term 

 23 – x,30 – x,57 – x and 78 – x are proportional 

 thus It can be written as 

 23 – x: 30 – x∶: 57 – x: 78 – x 

 (23 – x)/ (30 – x) = (57 – x)/ (78 – x) 

 By cross multiplication 

 (23 – x) (78 – x) = (30 – x) (57 – x) 

 By further calculation 

 1794 – 23x – 78x + x²   = 1710 – 30x – 57x + x² 

 x²   – 101x + 1794 – x²   + 87x – 1710 = 0 

 So we will get 

 - 14x + 84 = 0 

 14x = 84 

 x = 84/14 

 = 6 

Therefore, 6 is the number to be subtracted from each of the numbers.

As we have given that 

 2x – 1,5x – 6,6x + 2 and 15x – 9 are in proportion 

 Thus We can write it as 

 (2x – 1) (15x – 9) = (5x – 6) (6x + 2) 

 By further calculation 

 30x²   – 18x – 15x + 9 = 30x²   + 10x – 36x – 12 

 30x²  – 33x + 9 = 30x²   – 26x – 12 

 30x²   – 33x – 30x²   + 26x = - 12 – 9 

 So we will get 

 -7x = - 21 

 x = -21/-7 

 = 3 

Therefore, the value of x is 3.

 

 

As we have given that 

x + 5 is the mean proportion between x + 2 and x + 9 

thus We can write it as 

 (x + 5)²  = (x + 2) (x + 9) 

 By further calculation 

x²   + 10x + 25 = x²   + 11x + 18 

x²    + 10x – x²   – 11x = 18 – 25 

 So we will get 

  - x = - 7

 x = 7 

Hence, the value of x is 7.

 

Lets assumed  x be added to each number 

 16 + x ,26 + x and 40 + x are in continued proportion 

Thus It can be written as 

 (16 + x)/ (26 + x) = (26 + x)/ (40 + x) 

 By cross multiplication 

 (16 + x) (40 + x) = (26 + x) (26 + x) 

 On further calculation 

 640 + 16x + 40x + x²   = 676 + 26x + 26x + x² 

 640 + 56x + x²   = 676 + 52x +x²   

 56x + x²   – 52x – x²   = 676 – 640 

 So we will get 

   4x = 36 

 x = 36/4 

 = 9 

Hence, 9 is the number to be added to each of the numbers. 

 

 

Lets assumed  a and b as the two numbers 

As we have given that 28 is the mean proportional 

 a: 28∶: 28: b 

 We get 

 ab = 28²  = 784 

 Here a = 784/b …… (1) 

 

As We know that 224 is the third proportional 

 a: b∶: b: 224 

So we will get 

 b²  = 224a …..(2) 

 

Now by substituting the value of a in equation (2) 

 b²  = 224 × 784/b 

 So we will get 

 b³  = 224 × 784 

 b³= 175616 = 56³  

 b = 56 

 

By substituting the value of b in equation (1) 

 a = 784/56 = 14

 

Therefore, 14 and 56 are the two numbers.

 

 

As we have given that 

b is the mean proportional between a and c 

thus We can write it as 

 b²  = a × c 

 b² = ac …..(1) 

As We know that 

 a,c,a²  + b²  and  b²  + c²  are in proportion 

 thus It can be written as 

 a/c = (a²  + b²   )/ (b²   + c²  ) 

 By cross multiplication 

 a (b²  + c²  ) = c (a²  + b²  ) 

 Using equation (1) 

 a (ac + c²  ) = c (a²  + ac) 

 So we will  get 

 ac (a + c) = a² c + ac²  

 Here ac (a + c) = ac (a + c) which is true.

Therefore, it is proved.

 

 

As we have given that 

y is mean proportional between x and z 

thus We can write it as 

 y²  = xz …… (1) 

 

Lets assumed 

 LHS = xyz (x + y + z)³  

 thus It can be written as 

 = xz.y (x + y + z)³  

 Using equation (1) 

 = y²  y (x + y + z)³  

 = y³   (x + y + z)³  

 So we will get 

 = [y (x + y + z)]³  

By further calculation 

 = (xy + y²  + yz)³  

 Using equation (1) 

 = (xy + yz + zx)³  

= RHS Hence, it is proved.

 

As we have given that

 a + c = mb and 1/b + 1/d = m/c 

 a + c = mb 

 Dividing the equation by b 

 a/b + c/d = m …….(1) 

 

 1/b + 1/d = m/c 

 Multiplying the equation by c 

 c/b + c/d = m …… (2) 

 

 Using equation (1) and (2)

  a/b + c/b = c/b + c/d 

 So we get 

 a/b = c/d 

 

Therefore, it is proved that a, b, c and d are in proportion.

RHS = 27 (a + b) (c + d) (e + f) 

 thus It can be written as 

 = 27 (bk + b) (dk + d) (fk + f) 

 Taking out the common terms 

 = 27 b (k + 1) d (k + 1) f (k + 1) 

 So we will get 

 = 27 bdf (k + 1)³  

 

Therefore, LHS = RHS.

 

As we have given that 

 a,b,c,d are in proportion 

 Lets assumed a/b = c/d = k 

 a = b,c = dk 

 

 (i) LHS = (5a + 7b) (2c – 3d) 

 Substituting the values 

 = (5bk + 7b) (2dk – 3d) 

 Taking out the common terms 

 = k (5b + 7b) k (2d – 3d) 

 So we will get 

 = k²  (12b) (-d)

 = - 12 bd k² 

 

 RHS = (5c + 7d) (2a – 3b) 

 Substituting the values 

 = (5dk + 7d) (2kb – 3b) 

 Taking out the common terms 

  = k (5d + 7d) k (2b – 3b)  

 So we will get 

 = k²  (12d) (-b) 

 = - 12 bd k²  

Therefore, LHS = RHS. 

(iii)(a⁴  + c⁴   ): (b⁴  + d⁴  )= a²  c² ∶ b²  d²  

 thus We can write it as

Solution: 

As we have given that 

a, b, c, d are in continued proportion 

Here we will get 

 a/b = b/c = c/d = k 

 c = dk,b = ck = dk .k = dk²  

 a = bk = dk²   .k = dk³   

As we have given that 

 (ma + nb): b∶: (mc + nd): d 

 thus We can write it as 

 (ma + nb)/ b = (mc + nd)/ d 

 By cross multiplication 

 mad + nbd = mbc + nbd 

 Here mad = mbc 

 ad = bc 

 By further calculation 

 a/b = c/d 

Therefore, it is proved that a, b, c, d are in proportion.

 As we have given that

 (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d) 

Thus, We can write it as

By cross multiplication 

 50x – 75 = 12x + 1 

 50x – 12x = 1 + 75 

 So we will get 

 38x = 76 

 x = 76/38 

 = 2

As we have given that 

 (7p + 3q): (3p – 2q) = 43: 2 

 thus We can write it as 

 (7p + 3q)/ (3p – 2q) = 43/2 

 By cross multiplication 

 129p – 86q = 14p + 6q 

 129p – 14p = 6q + 86q 

 So we will get 

 115p = 92q 

 By division 

 As we have given that

5x + 12x + 13x = 360 cm 

By further calculation 

30x = 360 

We will get 

x = 360/30 

= 12 

 

Here the length of the longest side = 13x 

Substituting the value of x 

 = 13 × 12 

 = 156 cm

  Lets assumed 5x and 6x as the savings of Lokesh and his sister.

 Lokesh should save Rs y more 

 Based on the problem 

 (5x + y)/ (6x + 30) = 5/6 

 By cross multiplication 

 30x + 6y = 30x + 150 

 By further calculation 

 30x + 6y – 30x = 150 

 So we will get 

 6y = 150 

 y = 150/6 

 = 25  

Therefore, Lokesh should save Rs 25 more than his sister.

 

Lets assumed  x be added to each number 

 So the numbers will be 

 15 + x,17 + x,34 + x and 38 + x 

 Based on the condition 

 (15 + x)/ (17 + x) = (34 + x)/ (38 + x) 

 By cross multiplication 

 (15 + x) (38 + x) = (34 + x) (17 + x) 

 By further calculation 

 570 + 53x +  x²   = 578 + 51x + x²  

 So we will get 

  x²    + 53x –  x²   – 51x = 578 – 570 

 2x = 8 

 x = 4 

Hence, 4 must be added to each of the numbers.  

As we have given that 

 (a + 2b + c),(a – c) and (a – 2b + c) 

 are in continued proportion 

 thus We can write it as 

 (a + 2b + c)/ (a – c) = (a – c)/ (a – 2b + c) 

 By cross multiplication 

 (a + 2b + c)(a – 2b + c)= (a – c)²  

 On further calculation 

 a²  – 2ab + ac + 2av – 4b²+ 2bc + ac – 2bc + c²  = a²  – 2ac + c²  

 So we will get 

 a²  – 2ab + ac + 2ab – 4b²  + 2bc + ac – 2bc + c²  – a²  + 2ac – c²  = 0 

 4ac – 4b²  = 0 

 Dividing by 4 

 ac – b²  = 0 

 b²  = ac 

Therefore, it is proved that b is the mean proportional between a and c.

As we have given that 

2,6,p,54 and q are in continued proportion 

Thus We can write it as 

 2/6 = 6/p = p/54 = 54/q

 

(i) as We know that 

 2/6 = 6/p 

 By cross multiplication 

 2p = 36 

 p = 18 

 

(ii) as We know that 

 p/54 = 54/q 

 By cross multiplication 

 pq = 54 × 54 

 Substituting the value of p 

 q = (54 × 54)/ 18 = 162 

Therefore, the values of p and q are 18 and 162