# Proportion Chapter 1 Exercise 1B CA Foundation maths solutions

Mathematics CA foundation textbook chapter 1 Ratio and Proportion, Indices, Logarithms Exercise 1B solutions are given.

You should study the textbook lesson chapter 1.

You should observe example problems which are given in the textbook.

Ratio and Proportions, Indices, Logarithms

## Exercise 1B C.A. Foundation maths solutions Proportion Chapter 1

CA foundation maths solutions

Chapter 1

Proportion

Exercise 1B

**Problem 1**

**The fourth proportional to 4, 6, 8 is**

**Problem 2**

**The third proportional to 12, 18 is**

**Problem 3**

**The mean proportional between 25, 81 is**

**Problem 4**

**The number which has the same ratio to 26 that 6 has to 13 is**

**Problem 5**

**The fourth proportional to 2a,a****²****, c is**

**Problem 6**

**If four numbers 1/2, 1/3, 1/5, 1/x are proportional then x is**

**Problem 7**

**The mean proportional between 12****x²**** and 27y****²**** is**

**(Hint : Let k be the mean proportional)**

**Problem 8**

**If A =B/2 = C/2, then A : B : C is**

**Problem 9**

**If a/3 = b/4 = c/7, then a + b + c / c is**

**Problem 10**

**If p/q = r/s = 2.5/1.5, the value of ps : qr is**

**Problem 11**

**If x: y = z : w = 2.5 : 1.5, the value of (x + y)/(y + w) is**

**Problem 12**

**If (5x – 3y) / (5 – 3x) = 3/4, the value of x : y is**

**Problem 13**

**If A : B = 3:2 and B : C = 3 : 5, then A : B : C is**

**Problem 14**

**If x/2 = y/2 = z/7, then the value of (2x – 5 + 4z)/2y is**

**Problem 15**

**If x : y = 2 : 3, y : z = 4 : 3 then x : y : z is**

### Chapter 1 Proportion CA Foundation maths solutions Exercise 1B

C.A. foundation maths solutions

Chapter 1

Ratio and Proportions, Indices, Logarithms

Exercise 1B

**Problem 16**

**Division of ₹ 750 into 3 parts in the ratio 4 : 5 : 6 is**

**Problem 17**

**The sum of the ages of 3 persons is 150 years ago their ages were in the ratio 7 : 8 : 9. Their present ages are**

**Problem 18**

**The numbers 14, 16, 35, 42 are not in proportion. The fourth term for which they will be in proportion is**

**Problem 19**

**If x/y = z/w, implies y/x = w/z, then the process is called**

**Problem 20**

**If p/q = r/s = p – r/q – s, the process is called**

**Problem 21**

**If a/b = c/d, implies (a + b)/(a – b) = (c + d)/(c – d), the process is called**

**Problem 22**

**If u/v= w/p, then (u – v)/(u + v) = (w – p)/(w + p). The process is called**

**Problem 23**

**12, 16, star, 20 are in proportion. Then star is**

**Problem 24**

**4, star, 9, 13 1/2 are in proportion. Then star is**

**Problem 25**

**The mean proportional between 1.4 gms and 5.6 gms is**

**Problem 26**

**If a/4 = b/5 = c/9 then (a + b+ c)/c is**

**Problem 27**

**Two numbers are in the ratio 3 : 4; if 6 be added to each terms of the ratio, then the new ratio will be 4 : 5, then the numbers are**

**Problem 28**

**If a/4 = b/5 then**

**Problem 29**

**If a : b = 4 : 1 then √a/√b + √b /√a is**

**Problem 30**

**If x/(b + c – a) = y/(c + a – b) = z/(a + b – c) then ( b – c)x + ( c – a)y + ( a – b)z is**

Note : You should observe the solutions. You will be try them in your own method.

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