MCQ Questions for Class 6 Maths Ratio And Proportion Quiz 5

Who ran at faster pace?
Misha ran

$9$ laps in

$15$ minutes and Myra ran

$12$ laps in

$25$ minutes.

0%
Misha ran faster than Myra
0%
Myra ran faster than Misha
0%
Both ran at equal pace
0%
Cannot be determined

Explanation

Misha:

$\dfrac {9}{15} \times \dfrac {5}{5} = \dfrac {45}{75}$

Myra:

$\dfrac {12}{25} \times \dfrac {3}{3} = \dfrac {36}{75}$

$\therefore$ Misha ran faster as

$\dfrac {45}{74} > \dfrac {36}{75}$ .

The antecedent is _____, for a ratio of

$12\ m$ and

$120\ cm$

0%
$12$
0%
$120$
0%
$1,200$
0%
$10$

Explanation

The ratio of two different quantities is not defined. Therefore,

$12\ m = 1200\ cm$ .

Thus, the required ratio is

$1200 : 120$ .

$\Rightarrow 1200 : 120 = 10 : 1$ .

The antecedent of the given ratio is

$10$ .

A local television station sells time slots for programs in

$30$ -minute intervals. If the station operates

$24$ hours per day, every day of the week, what is the total number of

$30$ -minute time slots the station can sell for Tuesday and Wednesday?

0%
$96$
0%
$24$
0%
$48$
0%
$12$

Explanation

$\text{Numbers of slots in 1 hour}=2$

$\text{Each day has 24 hours.}$

$\text{Numbers of slot in 2 days = }24\times 2\times2=96$

On a house blueprint,

$2$ feet is represented by

$1$ inch. If a room on the blueprint measures

$5$ inches by

$6$ inches, what is the area of the actual room?

0%
$30$ square feet
0%
$60$ square feet
0%
$90$ square feet
0%
$120$ square feet

Explanation

$1$ inch is

$2$ feet, then

$5$ inches is

$10$ feet, while

$6$ inches is

$12$ feet.

Hence, the actual dimensions of the room is

$10$ feet by

$12$ feet.

Thus the area is

$10\times 12=120$ sq feet.

The total cost of 30 identical erasers is y dollars. At this rate, find the total cost in dollars of 70 of these erasers in terms of y.

0%
$\dfrac {3y}{7}$
0%
$\dfrac {4y}{7}$
0%
$\dfrac {7y}{4}$
0%
$\dfrac {7y}{3}$
0%
$70y$

Explanation

Given that,

The cost of

$30$ erasers is

$y$ dollars.

To find out,

The cost of

$70$ erasers.

The cost of

$30$ erasers

$=y$ dollars

Hence, the cost of one eraser

$=\dfrac{y}{30}$ dollars [unitary method]

$\therefore\$ The cost of

$70$ erasers

$=\dfrac{y}{30}\times70$

$=\dfrac{7y}{3}$ dollars.

Hence, the cost of

$70$ erasers is

$\dfrac{7y}{3}$ dollars.

Sixteen ounces is equal to one pound, and one tonne is equal to

$2,000$ pounds. How many ounces equal two tonnes?

0%
$4,000$
0%
$16,000$
0%
$32,000$
0%
$32,016$
0%
$64,000$

Explanation

Given,

$16$ ounces

$=1$ pound

$1$ tonne

$=2000$ pound

$2$ tonne will be

$=4000$ pound
As we know,

$1$ pound

$=16$ ounces

$4000$ pound will be

$=$

$16\times 4000=64000$ ounces
So,

$2$ tonne will be

$64000$ ounces.

If the consequent is

$13$ and the antecedent is

$11$ , then the ratio is _______

0%
$11 : 13$
0%
$13 : 11$
0%
$3 : 1$
0%
None of these

Explanation

First term

$:$ Second Term

$\downarrow \quad \quad \quad \quad \quad \downarrow$

Antecedent Consequent

Since antecedent

$=11$ and consequent

$=13$

$\therefore$ antecedent

$:$ consequent

$=11:13$

In the ratio between

$12 m$ and

$13\ kg$ , the antecedent is _____

0%
$12$
0%
$13$
0%
$12\ m$
0%
Not possible

Explanation

The ratio of two unlike quantities is not defined. Hence, the ratio between

$12\ m$ and

$13\ kg$ cannot be found. Thus, the antecedent is not possible.

The ratio of length of two cars, M and N whose lengths are

$3\ m$ and

$4\ m$ is

$3 : 4$ . Here,

$4$ is called ________

0%
Consequent
0%
Antecedent
0%
Dividend
0%
Constant

Explanation

The ratio of lengths of car M and N is

$3 : 4$ . Here,

$4$ being the second term is called consequent.

If the antecedent is

$16$ and the consequent is

$12$ , then the ratio is _________

0%
$12 : 16$
0%
$16 : 12$
0%
$6 : 2$
0%
Cannot be determined

Explanation

$Antecedent:Consequent=16:12$

The area of the circle and the area of an regular polygon of sides of perimeter equal to that of the circle are in the ratio of

0%
$\tan { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n }$
0%
$\cot { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n }$
0%
$\cos{ \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n }$
0%
$\sin { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n }$

Explanation

Let $r$ be the radius of the circle and a be the side of polygon.

The perimeter of circle $= 2\pi r$

The perimeter of a polygon with n sides $= na$

So $2\pi r = na$

So $a = \dfrac {2\pi r}{n}$

Area of the circle, $A_1= \pi r^2$

Area of a regular polygon, $A_2=\bigg( \dfrac{1}{4} \bigg) na^2 \cot \bigg( \dfrac{\pi}{n}\bigg)$

$=\dfrac14\ n\ \bigg(\dfrac {4\pi ^2 r^2}{n^2}\bigg) \cot \bigg(\dfrac{\pi}{n}\bigg)$

$=\bigg( \dfrac{\pi ^2 r^2}{n} \bigg) \cot \bigg( \dfrac{\pi}{n} \bigg)$

$\Rightarrow \dfrac{A_1}{A_2}=\dfrac{\pi r^2}{\dfrac{\pi^2 r^2}{n} \cot \bigg( \dfrac{\pi}{n} \bigg)}$

$=\dfrac{\tan \bigg( \dfrac{\pi}{n}\bigg)}{\dfrac{\pi}{n}}$

So the ratio is $\tan \bigg( \dfrac{\pi}{n}\bigg) : \dfrac{\pi}{n}$

$14$ milliliters of a certain liquid has a mass of

$16$ grams. What is the mass (in grams) of

$28$ liters of this liquid? (

$1$ liter =

$1,000$ milliliters)

0%
$8$
0%
$32$
0%
$3,200$
0%
$8,000$
0%
$32,000$

Explanation

Given that

$14$ milliliters have

$16$ grams.

Therefore

$14$ liters will have

$16000$ grams.

So,

$28$ liters will have

$16000 \times 2 = 32000$ grams.

If a

$3$ -pound pizza is sliced in half and each half is sliced into thirds, what is the weight, in ounces, of each of the slices? (

$1$ pound =

$16$ ounces)

0%
4
0%
6
0%
8
0%
16

Explanation

Total number of slices the pizza is cut into is

$= 6$ as each half is cut into three more slices.
Now, weight of each pizza slice

$= \dfrac { \text{Total weight}}{6} =\dfrac {3 \times 16}{6} = 8$ ounches

Each floor of a high rise building is fitted with

$20$ doors. There are

$12$ floors in each building. There are

$25$ such buildings in a complex. Calculate the total number of doors fitted in the complex.

0%
$2400$
0%
$3000$
0%
$6000$
0%
$5000$

Explanation

Number of doors on

$1$ floor

$=20$
Number of floors in

$1$ building

$12$

$\therefore$ Number of doors in

$1$ building

$=20\times 12=240$
Number of buildings in

$1$ complex

$=25$

$\therefore$ Number of doors in

$25$ buildings

$=240\times 25=6000$

What is the ratio of the shaded area to the unshaded area?

0%
$5:7$
0%
$3:4$
0%
$7:10$
0%
$7:3$

Explanation

Total number of equal parts

$=10\times 2=20$
Number of shaded parts

$=14$
So, the number of unshaded parts

$=20-14=6$

$\therefore$ Required ratio

$=14:6=7:3$ .

The ratio of

$\left(\displaystyle\dfrac{1}{3}\text{ of Rs. }9.30\right)$ to

$(0.6\text{ of Rs. }1.55)$ is ___________.

0%
$1:3$
0%
$10:3$
0%
$3:10$
0%
$3:1$

Explanation

$\displaystyle\frac{1}{3}$ of Rs.

$9.30=Rs. \left(\displaystyle\frac{1}{3}\times 9.30\right)=Rs. 3.10$

$0.6$ of Rs.

$1.55=$ Rs.

$(0.6\times 1.55)=$ Rs.

$0.93$

$\therefore$ Required ratio

$=3.10:0.93=10:3$ .

In the word COMPARISON, the ratio of number of consonants to the number of vowels is ___________.

0%
$4:7$
0%
$7:4$
0%
$6:4$
0%
$4:6$

Explanation

Number of consonants in the given word

$=6$
Number of vowels in the given word

$=4$

$\therefore$ Required ratio

$=6:4$ .

Cost of a candy is

$50$ paise and cost of an ice cream is Rs.

$20$ , then the ratio of the cost of a candy to the cost of an ice-cream is __________.

0%
$40:1$
0%
$1:40$
0%
$1:20$
0%
$1:15$

Explanation

Cost of

$1$ candy

$=50$ paise
Cost of

$1$ ice-cream

$=Rs. 20=20\times 100$ paise

$=2000$ paise

$\therefore$ Required ratio

$=50:2000=1:40$ .

Out of

$30$ students in a class,

$6$ like football,

$12$ like cricket and remaining like tennis. The ratio of number of students who like tennis to the total number of students, is:

0%
$2:3$
0%
$5:1$
0%
$2:5$
0%
$5:2$

Explanation

Total number of students

$=30$
Number of students who like football

$=6$
Number of students who like cricket

$=12$

$\therefore$ Number of students who like tennis

$=30-6-12=12$

$\therefore$ Required ratio

$=12:30=2:5$ .

In a cricket coaching camp,

$1200$ children are trained. Out of which

$900$ are selected for various matches. Ratio of non-selected children to the total number of children is ___________.

0%
$300:120$
0%
$4:1$
0%
$1:4$
0%
$120:300$

Explanation

Total number of children

$=1200$
Number of selected children

$=900$

$\therefore$ Number of non-selected children

$=1200-900=300$

$\therefore$ Required ratio

$=300:1200=1:4$ .

Select the ratio that is equivalent to

$12:14$

0%
$6:2$
0%
$6:7$
0%
$3:8$
0%
$5:7$

Explanation

Given ratio is

$12:14$
It can be rewritten as

$\dfrac{12}{14}=\dfrac{6\times 2}{7\times 2}$
Cancelling

$2$ from numerator and denominator, we get

$\dfrac{12}{14}=\dfrac{6}{7}$ i.e

$6:7$
Hence, the answer is

$6:7$ .

State whether it is true or false:
The given figures represent equivalent ratios.

0%
True
0%
False

Explanation

In Fig. 1 Total number of parts is

$3$
Total number of colored part is

$1$
Therefore, the figure is divided in the ratio

$1:3$

In Fig. 2 Total number of parts is

$6$
Total number of colored part is

$2$
Therefore, the figure is divided in the ratio

$2:6$ which is same as

$1:3$

Thus, the figures 1 and 2 represent equal ratios.

Sides of two similar triangles are in the ratio

$4:9$ .Area of these triangles are in the ratio

0%
$2:3$
0%
$4:9$
0%
$81:16$
0%
$16:81$

Explanation

Given:Ratio of sides of similar triangles

$=\dfrac{4}{9}$

We know that if two triangles are similar,

ratio of areas is equal to the ratio of squares of corresponding sides.

So,

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{{\left(side\,\, of\,\, triangle\,\, 1\right)}^{2}}{{\left(side\,\, of\,\, triangle\,\, 2\right)}^{2}}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}={\left(\dfrac{4}{9}\right)}^{2}=\dfrac{16}{81}$

$\dfrac{area \,\, of\,\, triangle\,\, 1}{area \,\, of\,\, triangle\,\, 2}=\dfrac{16}{81}$

A sum of Rs. 9,000 is to be distributed among A, B and C in the ratio of 4 : 5 :What will be the difference between A's and C's shares?

0%
Rs. 600
0%
Rs. 1,000
0%
Rs. 900
0%
Rs. 1,200

Explanation

Given, total money

$= Rs.\,9,000$

and

$A: B: C = 4 :5:6$

$A$ 's share

$=\dfrac{4}{(4 + 5 + 6)} \times 9000 = 2400$

$B$ 's share

$=\dfrac{5}{(4 + 5 + 6)} \times 9000 = 3000$

$C$ 's share

$=\dfrac{6}{(4 + 5 + 6)}\times 9000=3600$

Hence, the difference between

$A$ 's share and

$C$ 's share

$=3600-2400$

$= 1200$

The length and breadth of a rectangular field are

$50\ m$ and

$15\ m$ respectively. Find the ratio of the breadth to the length of the field.

0%
$3:5$
0%
$3:10$
0%
$3:15$
0%
$10:3$

Explanation

The length of the rectangular field is $50 m$ and breadth of the field is $15 m$ .

The ratio of breadth and length is,

${\rm{breadth}}:{\rm{length}} = 15:50=\dfrac{15}{50}=\dfrac{3}{10}$

$= 3:10$

Two numbers are in the ratio

$3 : 5$ . If the sum of numbers is

$144$ , then the small number is

0%
$54$
0%
$72$
0%
$90$
0%
$48$

Explanation

$\displaystyle 144\times \frac{3}{8}=54$

$\displaystyle 144\times \frac{5}{8}=90$

the smaller number is

$54$

Find the ratio

$50$ paise to Rs.

$5$

0%
$10:1$
0%
$1:1$
0%
$1:10$
0%
None of these

Explanation

${\textbf{Step - 1: Converting rupees to paise}}$

${\text{We know that, 1 Rupee = 100 paise}}$

$\therefore {\text{ 5 Rupees = }}\dfrac{{{\text{5 }} \times {\text{ 100}}}}{{\text{1}}}{\text{ paise}}$

$\Rightarrow {\text{ 5 Rupees = 500 paise}}$ $\quad \text{.........eqn(i)}$

${\textbf{Step - 2: Substitute above result and get required ratio}}$

${\text{Ratio of 50 paise to 5 Rupees = }}$ ${\text{ }}\dfrac{{{\text{50 paise}}}}{{{\text{5 Rupees}}}}$

$={\text{ }}\dfrac{{{\text{50 paise}}}}{{{\text{500 paise}}}}$ $\quad \textbf{[From eqn(i)]}$

$= {\text{ }}\dfrac{{\text{1}}}{{{\text{10}}}}$

${\textbf{Thus, the ratio of 50 paise to 5 Rupees is 1:10. Option C is correct.}}$

Find the reduced form of the ratio of the first quantity to second quantity.

$3.8$ kg,

$1900$ gm.

0%
2:1
0%
1:2
0%
1:4
0%
1:8

Explanation

Reduced form of

$3.kg$ and

$1900gm$ in ratio

firstly kg change into gms

$1kg$ =

$1000gm$

than

$3.8kg$ =

$3.8 \times 1000$ =

$3800gm$

then, ratio

$\frac{{3800}}{{1900}} = \frac{2}{1} = 2:1$

Find the reduced form of the ratio of the first quantity to second quantity.

$7$ minutes

$20$ seconds,

$5$ minutes

$6$ seconds.

0%
111:345
0%
234:151
0%
220:153
0%
None

Explanation

Reduced form of 7 min 20 sec, 5 min 6 sec is

firstly change minute in second

$\begin{array}{l} \frac { { 7\times 60+20 } }{ { 5\times 60+6 } } \\ =\frac { { 420+20 } }{ { 300+6 } } \\ =\frac { { 440 } }{ { 306 } } \\ =\frac { { 220 } }{ { 153 } } =220:153 \end{array}$

Present age of father is

$42$ years and that of his son is

$14$ years. Find the ratio of present age of father to the present age of son.

0%
3:1
0%
1:2
0%
1:3
0%
1:1

Explanation

$Ratio=\dfrac{Present\ age\ of\ father}{Present\ age\ of\ son.}$

$=\dfrac{42}{14}$

$\boxed{Ratio=3:1}$

CBSE Questions for Class 6 Maths Ratio And Proportion Quiz 5 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Ratio And Proportion Quiz 5 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

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