Important Questions for CBSE Class 6 Maths Chapter 12 - Ratio and Proportion

1 Mark

1. In a ratio, the first term is called ____ and the second term is called ______.

Ans: Antecedent, Consequent

For example, in ratio $13:15$, 13 is Antecedent and 15 is Consequent.

2. Product of ______ = Product of extremes.
Ans: Means

In a proportion, the first and last terms are known as the extremes, while the second and third terms are known as the meAns:

3. If a, b, c, d are in proportion, then

1. ac = bd

2. ad = bc

3. ab = cd

4. None of these.

Ans: (b) ad = bc

If the ratio of the first two quantities equals the ratio of the last two quantities, the numbers a, b, c, and d are proportional.

4. A ratio has _____ units

Ans: No

Because a ratio is made up of similar quantities, the units cancel each other out, and thus there is no unit for a ratio.

  2 Mark

1. Convert 80:50 into simplification.                                             

Ans: Given ratio of 80:50

Expressing as fractions 

$\dfrac{80}{50}\ = \dfrac{8 \times 10}{5 \times 10} = \dfrac{8}{5}$

2. Find the ratio of 40 cm to 2.5 m.

Ans: let’s first convert 2.5 m into cm

2.5 m = $2.5 \times 100$ = 250 cm

Now, 

$40 cm:2.5 = 40 cm:250 cm $

$= 40:250$

$= \dfrac{40}{250}$

$= \dfrac{4}{25}$

The required ratio is $4:25$

3. The length and breadth of a field are $\mathbf{80\ \text{m}}$ and $\mathbf{30\ \text{m}}$. what is the ratio of the breadth and length of the park?       

Ans:  Given, 

Length of park = 80 m

Breadth of park = 30 m

Ratio of the breadth and length = $ \dfrac{Breadth}{Length} = \dfrac{30 m}{80 m}$

$=\dfrac{3}{8}$

Required ratio is $3:8$

4. If 30 oranges cost Rs. 120. What is the cost of 50 oranges? 

Ans: Here we will use the unitary method.

Cost of 30 oranges =  Rs.120

Now, cost of one orange = $\dfrac{120}{30} = Rs. 4 $    

Cost of 50 oranges = cost of one orange $\times $ cost of fifty oranges 

$= 4 \times  50$

$= Rs. 200$

3 Mark

1. Find the ratio of 45 minutes to an hour?

Ans: To find 45 min:1 hour

We know that one hour = 60 min

Therefore, 45 min:1 hour = 45 min:60 mnt

$= \dfrac{45}{60} $

Dividing numerator and denominator by 3

$\dfrac{45}{60} \div  \dfrac{3}{3} = \dfrac{15}{20}$

Dividing numerator and denominator by 5

$\dfrac{15}{20} \div  \dfrac{5}{5} = \dfrac{3}{4}$

2. Find the ratio 250 ml to 4 l.

Ans: To find 250 ml:4 l

First, let’s convert 4 l into ml

We know 

$1 l = 1000 ml$

$ \therefore 4 l = 1000 \times 4 ml$

$\Rightarrow 4 l = 4000 ml$

Now, 

$250 ml:4 l = 250 ml:4000 ml$

$= 250:4000$

$\Rightarrow \dfrac{250}{4000} \div \dfrac{10}{10} = \dfrac{25}{400} \div \dfrac{5}{5}$

$= \dfrac{5}{80} \div \dfrac{5}{5}$

$= \dfrac{1}{16}$

$= 1:16$

The required Ratio is 1:16.

3. Find the equivalent ratio of 75:100 

Ans:  

$=75:100$

$=\dfrac{75}{100} \div \dfrac{5}{5} = \dfrac{15}{20}$ 

$=\dfrac{15}{20} \div \dfrac{5}{5} $ 

$=\dfrac{3}{4}$

The equivalent ratios for 75:100 are 15:20 and 3:4

4. Are 20, 40, 60, 120 in proportion? 

Ans: First we will find 20:40 and $60:120$

$20:40=\dfrac{20}{40}=\dfrac{2}{4}=\dfrac{1}{2} $ 

$60:120=\dfrac{60}{120}=\dfrac{6}{12}=\dfrac{1}{2} $

The numbers in the simplest form are equal. 

Yes, 20, 40, 60, 120 are in proportion. 

5. A 15 men can reap a field in 25 days. In how many days can 20 men reap the same field?

Ans: $15$ men can reap fields in =  $25$days.. 

$1$ men can reap field in  = $(25 \times 15$ ) days

$20$ men can reap field in = $\dfrac{25\times 15}{20}$ days 

$=\dfrac{25\times 15}{20} $

$=\dfrac{5\times 3}{4} $

$=\dfrac{15}{4} days $

$=3\dfrac{3}{4} days $

Therefore, $20$ men can reap the same field in $=3\dfrac{3}{4}\text{days}$.

  4 Mark

1. Fill up the blanks $\mathbf{\dfrac{2}{3}=\dfrac{14}{}=\dfrac{6}{}}$ 

Ans: Let ,                                     

$\dfrac{2}{3}=\dfrac{14}{x} $

$2x=14\times 3 $

$x=\dfrac{14\times 3}{2}=7\times 3=21$

Let’s substitute 

$\dfrac{2}{3}=\dfrac{14}{}$

Similarly, 

$\dfrac{2}{3}=\dfrac{6}{} $

$2x=3\times 6 $

$x=\dfrac{3\times 6}{2}=3\times 3=9 $ 

$\dfrac{2}{3}=\dfrac{6}{} $

Hence, $\dfrac{2}{3}=\dfrac{14}{21}=\dfrac{6}{9}$

2. Two numbers are in the ratio 3:5 and their sum is 192. Find the numbers.

Ans: Let common ratio is $x$

Therefore, numbers are 3x and 5x

According to question, 

3x + 5x = 192

$\therefore 8x=192$

$x=\dfrac{192}{8}=24$

Required numbers are, 

$3x=3\times 24=72$ 

$5x=5\times 24=120 $

3. Compare the ratios \[\mathbf{\left( 1\text{ }:\text{ }2 \right)}\] and \[\mathbf{\left( 4\text{ }:\text{ }5 \right)}\]

Ans:  Given, $1:2= \dfrac{1}{2}$

$4:5= \dfrac{4}{5}$

LCM of $5$ and $2$ is $10$ 

Therefore, 

$\dfrac{1}{2} \times \dfrac{5}{5} = \dfrac{5}{10}$ 

$\dfrac{4}{5} \times \dfrac{2}{2} = \dfrac{8}{10}$

Here, $\dfrac{5}{10} < \dfrac{8}{10}$

$\dfrac{5}{10} = \dfrac{1}{2} \dfrac{8}{10} = \dfrac{4}{5}$

Therefore, $\dfrac{1}{2}<\dfrac{4}{5} $

$1:2<4:5$

4. If $\mathbf{x:63::36:81}$

Ans:  We can use formula 

Product of means = Product of extremes

$\Rightarrow 63 \times 36 = 81 \times x $

$\Rightarrow 81 \times x = 63 \times 36 $

$\Rightarrow x = \dfrac{63 \times 36}{81} $

$\Rightarrow \dfrac{7 \times 36}{9} $

$\Rightarrow 7 \times 4 $

$ \Rightarrow x=28 $

5 Mark

1. Divide \[\mathbf{\text{Rs}\text{. 2000/-}}\] between Asha and Kiran in the ratio 4:6.

Ans: Given amount = \[\text{Rs}\text{. 2000/-}\]

Given ratio = $4:6$

Let common ratio is $x$

Then, Asha’s share = $4x$

Similarly, Kiran’s share = $6x$

According to Question, 

$4x+6x=2000 $

$\Rightarrow 10x=2000 $

$\Rightarrow x = \dfrac{2000}{10} = 200$

Now, 

Asha’s share = $4\times 200 = \text{Rs}\text{.} 800$

Kiran’s share = $6\times 200 = \text{Rs}\text{.} 1200$

2. Divide $\mathbf{\text{Rs}\text{.} \text{500}}$ among $\mathbf{\text{A,B,C}}$ in the ratio \[\mathbf{1\text{ }:\text{ }2\text{ }:\text{ }3}\]

Ans:  Given amount = $\text{Rs}\text{.} \text{500}$

Given ratio = \[1\text{ }:\text{ }2\text{ }:\text{ }3\]

Let common ratio = $x$

A’s Share = $x$

B’s Share = $2x$

C’s Share = $3x$

According to question 

$x+2x+3x=500 $

$ \Rightarrow  6x = 500 $

$ \Rightarrow  x = \dfrac{500}{6} $

$ \Rightarrow  x = 83.33 $

Now, 

A’s Share = $\text{Rs}\text{.} 83.33$

B’s  share = $\text{Rs}\text{.} (2 \times  83.33) = \text{Rs}. 166.66$ 

C’s Share = $\text{Rs}\text{.} 3 \times  83.33 =\text{Rs}\text{.} 250 $

3. If $\mathbf{4:x::x:36}$. Find the value of $\mathbf{x}$.

Ans:  We will use formula 

Product of means = Product of extremes

$x \times x=4\times 36 $

${{x}^{2}}=144 $

$ x=\sqrt{144} $

$ x=12$ 

4. If 10, 15, x are in proportion. Find the value of x.

Ans: Given, 10, 15, x are in proportion.

\[\Rightarrow \text{10:15::15:x}\]

We will use formula 

Product of means = Product of extremes

$ 15\times 15=10 \times x $

$10x=15\times  15 $

$ \Rightarrow x=\dfrac{15 \times 15}{10} $

$  =\dfrac{15\times 3}{2} $

$ =\dfrac{45}{2} $

$ \Rightarrow x=22.5$ 

5. Find the ratio of the price of coffee powder to that of milk powder when coffee powder cost  24/- per 100\,g and milk powder cost 180/- per kg.

Ans:  Given, cost of $100 g$ coffee powder =$\text{Rs}\text{.} 24 $

Cost of $1\,g$ coffee powder =$\text{Rs}\text{.} \dfrac{24}{100} $

Cost of $1000 g$ coffee powder =$ \text{Rs}\text{.} \left( \dfrac{24}{100} \times 1000 \right)= \text{Rs}\text{.} 240 $

Cost of $1 \text{kg}$of coffee powder = $\text{Rs}\text{.} 240$
Cost of $1\ \text{kg}$ of milk powder = $\text{Rs}\text{.} 180 $

Cost of $1 \text{kg}$ of coffee:Cost of $1 \text{kg}$ of milk powder

$=\text{Rs}\text{.} 240 :\text{Rs}.180$

$=\dfrac{240}{180}$

$=\dfrac{24}{18}=\dfrac{8}{6}$

$=\dfrac{4}{3}=4:3$

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