amarchand singhvi international school 7... · now by cross multiplication method number of people...
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Page 1 of 9 19-08-20
AMARCHAND SINGHVI INTERNATIONAL SCHOOL
QUESTIONS AND ANSWERS
CLASS – VII (MATHEMATICS)
CHAPTER 8 – RATIO AND PROPORTION
Note: All these questions and answers are to be done in the notebook.
Ex 8(A)
Q.1)Express each of the following ratios in the simplest form:
(i) 28:63
Answer
Now students in this we will be dividing numerator and denominator by HCF of 28 and 63 to
obtain simplest form. HCF of 28 and 63 = 7
28:63 =
=
=
= 4 : 9
(iii) 0.4 : 0.6
Answer
0.4 : 0.6 =
=
=
=
= 2 : 3
(v)
:
Answer
:
=
=
÷
=
= 9 : 4
(vii)
:
:
Answer
In this type of sums we will take the LCM of denominators first and then we will multiply the
LCM with each fraction to obtain simplest form. LCM of 4,8 and 10 = 40
:
:
=
:
:
x 40
=
:
:
=
:
:
=
=
:
:
= 10: 5 : 4
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(ix) 125 g : 1 kg
Answer
Now students we know that ratio can be taken of the same quantity and same units so here
one term is in gram and other is in kilogram so we will be converting kilogram into gram and
then we will be dividing numerator and denominator by HCF of 125 and 1000 to obtain
simplest form. HCF of 125 and 1000 = 125
125 g : 1 kg = 125 g : 1000 g
= 125 : 1000 =
=
=
= 1 : 8
(xi) 36 minutes to 1
hours
Answer
Now students we know that ratio can be taken of the same quantity and same units so here
one term is in minutes and other is in hours so we will be converting hours into minutes and
then we will be dividing numerator and denominator by HCF of 36 and 90 to obtain simplest
form. HCF of 36 and 90 = 18
1
hours =
hours =
x 60 minutes
=
minutes =
minutes
= 3 x 30 minutes = 90 minutes
36 minutes : 1
hours = 36 minutes : 90 minutes
= 36 : 90 =
=
=
= 2 : 5
Q.2) There are 2500 students in a school out of them 1100 are girls and the rest are boys find
the ratio of:
(i) number of boys to number of girls.
Answer
Number of Students = 2500
Number of Girls = 1100
Number of Boys = Number of Students Number of Girls = 2500 1100 = 1400
So, Number of boys to Number of girls = 1400 : 1100 =
=
14 : 11
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××(iii) number of students to number of boys.
Answer
So, Number of Students to Number of boys = 2500 : 1400 =
=
25 : 14
Q.4) If the ratio of the length to the breadth of a rectangle be 5:3 and its perimeter is 144 m,
find the length of the rectangle.
Answer
Ratio of length : breadth = 5:3
So here students we come to understand that this ratio is in simplest reduced form and the original
length and breadth of rectangle is not given to us. So there would surely be a number that would
have been a common factor in length and breadth which was eliminated in ratio by cutting and then
5:3 was left. So if we assume that common factor number as Y then
The length of rectangle = 5 x Y = 5Y and breadth of rectangle = 3 x Y = 3Y
Now, Perimeter of Rectangle = 2 ( Length + Breadth )
144 = 2 ( 5Y + 3Y )
144 = 2 ( 8Y )
144 = 2 x 8Y
144 = 16Y
144 ÷ 16 = Y
9 = Y
The length of rectangle = 5 x Y = 5 x 9 = 45 m
Q.8) (i) if a:b = 5:7 and b:c = 14:15, find a:c.
Answer
=
and
a : c =
x
=
x
=
=
=
= 2:3
Q.13) Rina lost her weight in the ratio 5:3. Her original weight was 80 kg. What is her new
weight?
Answer
Ratio = 5:3 =
Inverse of Ratio =
New weight = Inverse of Ratio x Original weight
=
x 80
=
=
= 3 x 16
= 48 kg
Page 4 of 9 19-08-20
So, rina’s new weight is 8 kg.
Q.16) if 3A=5B and 4B= 6C, then A:C equals.
a) 5:2 b) 3:5 c) 2:5 d) 4:5
Answer
3A = 5B
A =
=
and
4B = 6C
B =
=
A : C =
x
=
x
=
=
=
=
= 5 : 2
So the correct answer is option a) 5:2
Note: Students please note that question number 3, 5, 6, 7, 9, 10, 11, 12, 14,15, 17 and 18 of
Exercise 8(A) are omitted from the syllabus.
Ex 8 (B)
Q.1)
(i)
Answers
P
=
P
=
Now we can see that Product of extremes = Product of means
So we can say that yes
are in proportion.
(iii) 2
Answers
P
=
= 9
Page 5 of 9 19-08-20
P
=
Now we can see that Product of extremes is not equal to Product of means
So we can say that yes 2
are not in proportion.
Q.2)
(i) 0.9 : 0.6 : : X : 3
Answers
Here students it is already given that all these four numbers are in proportion.
So
Product of extremes = Product of means
0.9 x 3 = 0.6 x X
2.7 = 0.6 x X
2.7 ÷ 0.6 = X
= X
= X ( I have multiplied by 10 just to remove the decimal place)
= X
= X
4.5 = X
(iii) 1.6 : X : : 0.12 : 0.24
Answers
Here students it is already given that all these four numbers are in proportion.
So
Product of extremes = Product of means
1.6 x 0.24 = X x 0.12
0.384 = X x 0.12
0.384 ÷ 0.12 = X
= X
= X ( I have multiplied by 1000 just to remove the decimal place)
= X ( Do the cutting now by table of 12 )
Page 6 of 9 19-08-20
= X
3.2 = X
Q.3)
(i) 8, 32, 17
Answers
Let the fourth proportional be Y. Then we can say that 8, 32, 17, Y are in proportion.
That means 8:32::17:Y
Now we know that
Product of extremes = Product of means
8 x Y = 32 x 17
Y x 8 = 544
Y = 544 ÷ 8
Y = 68
So , the fourth proportional is 68.
(iii)
Answers
Let the fourth proportional be Y. Then we can say that
, Y are in proportion.
That means
: Y
Now we know that
Product of extremes = Product of means
x Y =
Y x
=
Y x
=
Y =
÷
Y =
x
Y =
Y =
So , the fourth proportional is
.
Page 7 of 9 19-08-20
Q.7) If 25 persons can dig a trench 36 m long in one day, then find the number of persons
required to dig a trench 108 m long in one day
Answers
Let the number of persons required to dig a trench 108 m long in one day be Y.
Now by cross multiplication method
Number of people Length of trench (m)
Y 108
25 36
=
=
= ÷ 36
=
So, 75 people are required to dig a trench of 108 m long in one day.
Q.8) The label on a flavoured milk bottle states that there is 1.5 g of fat in 200 mL. if you drink
500 mL of that milk, how much fat would you be getting?
Answers
Let the amount of fat in 500 mL be Y grams.
Now by cross multiplication method
Amount of fat (g) Amount of milk (mL)
Y 500
1.5 200
=
=
= ÷ 200
=
So, 3.75 grams of fat is there in 500 mL of milk.
Page 8 of 9 19-08-20
Q.9) Rajat can type 3200 words in one hour. How many words can he type in 15 minutes?
Answers
Let the number of words typed in 15 minutes be Y .
Now by cross multiplication method
Number of words Time (minutes)
Y 15
3200 60
=
=
= ÷ 60
=
So, Rajat can type 800 words in 15 minutes.
Q.12) Ayushree’s goal is to practice 20 minutes of Maths for every 60 minutes of class time.
How many minutes of Maths practice should she do if the class is 90 minutes long ?
a) 24 b) 32 c)25 d) 30
Answers
Let the number of minutes of Maths practice in 90 minutes be Y .
Now by cross multiplication method
Maths practice time (minutes) Class time (minutes)
Y 90
20 60
=
=
= ÷ 60
=
So, Ayushree should do 30 minutes of Maths practice in 90 minutes class.
So, the correct answer is Option d) 30
Q.13) A building casts a shadow that is 432m long. At the same time a person who is 2m long
casts a shadow 6m long . How high is the building?
a) 216 m b) c) 108 m d) m
Page 9 of 9 19-08-20
Answers
Let the Height of building be Y .
Now by cross multiplication method
Height (m) Length of Shadow (m)
Y 432
2 6
=
=
= ÷ 6
=
So, height of building is 144 m.
So, the correct answer is Option b) 144 m
Note: Students please note that question number 2(Part V), 4, 5, 6, 10, 11, 14 and 15 of
Exercise 8(B) are omitted from the syllabus.
Ex 8 (C)
Note: Students please note that Exercise 8(C) is omitted from the syllabus.