# Ratio and Proportion Questions

**FACTS AND FORMULAE FOR RATIO AND PROPORTION QUESTIONS**

**1. RATIO:** The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.

In the ratio a:b, we call a as the first term or antecedent and b, the second term or consequent.

Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio. Ex. 4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.

**2. PROPORTION : **The equality of two ratios is called proportion**. **

If a: b = c: d, we write, a: b :: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a: b :: c : d <=> (b x c) = (a x d).

**3**. **(i) Fourth Proportional :** If a : b = c: d, then d is called the fourth proportional to a, b, c.

**(ii) Third Proportional :** If a: b = b: c, then c is called the third proportional to a and b.

**(iii) Mean Proportional :** Mean proportional between a and b is $\sqrt{ab}$

**4. (i) COMPARISON OF RATIOS : **We say that (a: b) > (c: d) $\iff $ (a/b)>(c /d).

**(ii) COMPOUNDED RATIO :** The compounded ratio of the ratios (a: b), (c: d), (e : f) is (ace: bdf)

**5. (i) Duplicate ratio** of (a : b) is $\left({a}^{2}:{b}^{2}\right)$.

**(ii) Sub-duplicate ratio** of (a : b) is (√a : √b).

**(iii)Triplicate ratio** of (a : b) is $\left({a}^{3}:{b}^{3}\right)$.

**(iv) Sub-triplicate ratio** of (a : b) is $\left({a}^{\frac{1}{3}}:{b}^{\frac{1}{3}}\right)$.

**(V) $If\frac{a}{b}=\frac{c}{d},then\frac{a+b}{a-b}=\frac{c+d}{c-d}$** **(Componendo and dividendo)**

**6. VARIATION:**

(i) We say that x is directly proportional to y, if x = ky for some constant k and we write, $x\propto y$

(ii) We say that x is inversely proportional to y, if xy = k for some constant k and we write, $x\propto \frac{1}{y}$

A) 360, 160, 200 | B) 160, 360, 200 |

C) 200, 360,160 | D) 200,160,300 |

Explanation:

let ratio be x.

Hence no. of coins be 5x ,9x , 4x respectively

Now given total amount = Rs.206

=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206

we get x = 40

=> No. of 50p coins = 200

=> No. of 25p coins = 360

=> No. of 10p coins = 160

A) 10 | B) 12 |

C) 15 | D) 18 |

Explanation:

Let the quantity of alcohol and water be 4x litres and 3x litres respectively

4x/(3x+5) = 4/5

20x = 4(3x+5)

8x = 20

x = 2.5

Quantity of alcohol = (4 x 2.5) litres = 10 litres.

A) Rs. 182 | B) Rs. 190 |

C) Rs. 192 | D) Rs. 204 |

Explanation:

$Rs.\left(782*\frac{6}{23}\right)$

Given ratio =$\frac{1}{2}:\frac{2}{3}:\frac{3}{4}$= 6:8:9

1st part = $Rs.\left(782*\frac{6}{23}\right)$ = Rs. 204

A) 35 | B) 40 |

C) 45 | D) 50 |

Explanation:

Step (i): Let x be the number of boys and y be the number of girls.

Given total number of boys and girls = 100

x+y=100 -------------- (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40

The amount given to 100 boys and girls = Rs. 312

3.60x + 2.40y = 312 -------------- (ii)

Step (iii):

Solving (i) and (ii)

3.60x + 3.60y = 360 --------- Multiply (i) by 3.60

3.60x + 2.40y = 312 --------- (ii)

1.20y = 48

y = 48 / 1.20

= 40

Number of girls = 40

A) 38000 | B) 46800 |

C) 36700 | D) 50000 |

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then,

(2x+4000) / (3x+4000) = 40 / 57

⇒ 57 × (2x + 4000) = 40 × (3x+4000)

⇒ 6x = 68,000

⇒ 3x = 34,000

Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

A) 1.12 | B) 1.16 |

C) 1.20 | D) 1.30 |

A) 1:2 | B) 2:1 |

C) 11:24 | D) 36:121 |

A) 1:2:3 | B) 2:3:4 |

C) 3:4:5 | D) 4:5:6 |

Explanation:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

⇒ [(140/100) × 5x],[(150/100) × 7x] and [(175/100) × 8x]

⇒ 7x, 21x/2 and 14x.

⇒ The required ratio =7x : 21x/2 : 14x

⇒ 14x : 21x : 28x

⇒ 2 : 3 : 4