Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Ratio and Proportion
RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Q u e s t i o n : 1
Express each of the following ratios in simplest form:
i
24 : 40
ii
13.5 : 15
iii
6
2
3
: 7
1
2
iv
1
6
:
1
9
v
4: 5:
9
2
vi
2.5 : 6.5 : 8
S o l u t i o n :
i
HCF of 24 and 40 is 8.
? 24 : 40 =
24
40
=
24 ÷ 8
40 ÷ 8
=
3
5
= 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
ii
HCF of 13.5 and 15 is 1.5.
13.5
15
=
135
150
The HCF of 135 and 150 is 15. =
135 ÷ 15
150 ÷ 15
=
9
10
Hence, 13.5 : 15 in its simplest form is 9 : 10.
iii
20
3
:
15
2
= 40 : 45
The HCF of 40 and 45 is 5.
? 40 : 45 =
40
45
=
40 ÷ 5
45 ÷ 5
=
8
9
= 8 : 9
Hence, 6
2
3
: 7
1
2
in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
? 9 : 6 =
9
6
=
9 ÷ 3
6 ÷ 3
= 3 : 2
Hence,
1
6
:
1
9
in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
? 4 : 5 :
9
2
= 8 : 10 : 9
The HCF of these 3 numbers is 1.
? 8 : 10 : 9 is the simplest form.
vi
2.5 : 6.5 : 8 = 25 : 65 : 80
The HCF of 25, 65 and 80 is 5.
? 25 : 65 : 80 =
25
65
80
=
25 ÷ 5
65 ÷ 5
80 ÷ 5
=
5
13
16
= 5 : 13 : 16
Q u e s t i o n : 2
Express each of the following ratios in simplest form:
i
75 paise : 3 rupees
ii
Page 2
Q u e s t i o n : 1
Express each of the following ratios in simplest form:
i
24 : 40
ii
13.5 : 15
iii
6
2
3
: 7
1
2
iv
1
6
:
1
9
v
4: 5:
9
2
vi
2.5 : 6.5 : 8
S o l u t i o n :
i
HCF of 24 and 40 is 8.
? 24 : 40 =
24
40
=
24 ÷ 8
40 ÷ 8
=
3
5
= 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
ii
HCF of 13.5 and 15 is 1.5.
13.5
15
=
135
150
The HCF of 135 and 150 is 15. =
135 ÷ 15
150 ÷ 15
=
9
10
Hence, 13.5 : 15 in its simplest form is 9 : 10.
iii
20
3
:
15
2
= 40 : 45
The HCF of 40 and 45 is 5.
? 40 : 45 =
40
45
=
40 ÷ 5
45 ÷ 5
=
8
9
= 8 : 9
Hence, 6
2
3
: 7
1
2
in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
? 9 : 6 =
9
6
=
9 ÷ 3
6 ÷ 3
= 3 : 2
Hence,
1
6
:
1
9
in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
? 4 : 5 :
9
2
= 8 : 10 : 9
The HCF of these 3 numbers is 1.
? 8 : 10 : 9 is the simplest form.
vi
2.5 : 6.5 : 8 = 25 : 65 : 80
The HCF of 25, 65 and 80 is 5.
? 25 : 65 : 80 =
25
65
80
=
25 ÷ 5
65 ÷ 5
80 ÷ 5
=
5
13
16
= 5 : 13 : 16
Q u e s t i o n : 2
Express each of the following ratios in simplest form:
i
75 paise : 3 rupees
ii
1 m 5 cm : 63 cm
iii
1 hour 5 minutes : 45 minutes
iv
8 months : 1 year
v
2kg250g
: 3kg
vi
1 km : 750 m
S o l u t i o n :
i
Converting both the quantities into the same unit, we have:
75 paise : (3 ×
100) paise = 75 : 300
=
75
300
=
75 ÷ 75
300 ÷ 75
=
1
4
? HCFof75and300 = 75
= 1 paise : 4 paise
ii
Converting both the quantities into the same unit, we have:
105 cm : 63 cm =
105
63
=
105÷21
63÷21
=
5
3
? HCFof105and63 = 21
= 5 cm : 3 cm
iii
Converting both the quantities into the same unit
65 min : 45 min =
65
45
=
65÷5
45÷5
=
13
9
? HCFof65and45 = 5
= 13 min : 9 min
iv
Converting both the quantities into the same unit, we get:
8 months : 12 months =
8
12
=
8÷4
12÷4
=
2
3
? HCFof8and12 = 4
= 2 months : 3 months
v
Converting both the quantities into the same unit, we get:
2250g : 3000 g =
2250
3000
=
2250÷750
3000÷750
=
3
4
? HCFof2250and3000 = 750
= 3 g : 4 g
vi
Converting both the quantities into the same unit, we get:
1000 m : 750 m =
1000
750
=
1000÷250
750÷250
=
4
3
? HCFof1000and750 = 250
= 4 m : 3 m
Q u e s t i o n : 3
If A : B = 7 : 5 and B : C = 9 : 14, find A : C.
S o l u t i o n :
A
B
=
7
5
and
B
C
=
9
14
Therefore, we have:
A
B
×
B
C
=
7
5
×
9
14
A
C
=
9
10
? A : C = 9 : 10
Q u e s t i o n : 4
If A : B = 5 : 8 and B : C = 16 : 25, find A : C.
S o l u t i o n :
A
B
=
5
8
and
B
C
=
16
25
Now, we have:
A
B
×
B
C
=
5
8
×
16
25
?
A
C
=
2
5
? A : C = 2 : 5
Q u e s t i o n : 5
If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.
S o l u t i o n :
A : B = 3 : 5
Page 3
Q u e s t i o n : 1
Express each of the following ratios in simplest form:
i
24 : 40
ii
13.5 : 15
iii
6
2
3
: 7
1
2
iv
1
6
:
1
9
v
4: 5:
9
2
vi
2.5 : 6.5 : 8
S o l u t i o n :
i
HCF of 24 and 40 is 8.
? 24 : 40 =
24
40
=
24 ÷ 8
40 ÷ 8
=
3
5
= 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
ii
HCF of 13.5 and 15 is 1.5.
13.5
15
=
135
150
The HCF of 135 and 150 is 15. =
135 ÷ 15
150 ÷ 15
=
9
10
Hence, 13.5 : 15 in its simplest form is 9 : 10.
iii
20
3
:
15
2
= 40 : 45
The HCF of 40 and 45 is 5.
? 40 : 45 =
40
45
=
40 ÷ 5
45 ÷ 5
=
8
9
= 8 : 9
Hence, 6
2
3
: 7
1
2
in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
? 9 : 6 =
9
6
=
9 ÷ 3
6 ÷ 3
= 3 : 2
Hence,
1
6
:
1
9
in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
? 4 : 5 :
9
2
= 8 : 10 : 9
The HCF of these 3 numbers is 1.
? 8 : 10 : 9 is the simplest form.
vi
2.5 : 6.5 : 8 = 25 : 65 : 80
The HCF of 25, 65 and 80 is 5.
? 25 : 65 : 80 =
25
65
80
=
25 ÷ 5
65 ÷ 5
80 ÷ 5
=
5
13
16
= 5 : 13 : 16
Q u e s t i o n : 2
Express each of the following ratios in simplest form:
i
75 paise : 3 rupees
ii
1 m 5 cm : 63 cm
iii
1 hour 5 minutes : 45 minutes
iv
8 months : 1 year
v
2kg250g
: 3kg
vi
1 km : 750 m
S o l u t i o n :
i
Converting both the quantities into the same unit, we have:
75 paise : (3 ×
100) paise = 75 : 300
=
75
300
=
75 ÷ 75
300 ÷ 75
=
1
4
? HCFof75and300 = 75
= 1 paise : 4 paise
ii
Converting both the quantities into the same unit, we have:
105 cm : 63 cm =
105
63
=
105÷21
63÷21
=
5
3
? HCFof105and63 = 21
= 5 cm : 3 cm
iii
Converting both the quantities into the same unit
65 min : 45 min =
65
45
=
65÷5
45÷5
=
13
9
? HCFof65and45 = 5
= 13 min : 9 min
iv
Converting both the quantities into the same unit, we get:
8 months : 12 months =
8
12
=
8÷4
12÷4
=
2
3
? HCFof8and12 = 4
= 2 months : 3 months
v
Converting both the quantities into the same unit, we get:
2250g : 3000 g =
2250
3000
=
2250÷750
3000÷750
=
3
4
? HCFof2250and3000 = 750
= 3 g : 4 g
vi
Converting both the quantities into the same unit, we get:
1000 m : 750 m =
1000
750
=
1000÷250
750÷250
=
4
3
? HCFof1000and750 = 250
= 4 m : 3 m
Q u e s t i o n : 3
If A : B = 7 : 5 and B : C = 9 : 14, find A : C.
S o l u t i o n :
A
B
=
7
5
and
B
C
=
9
14
Therefore, we have:
A
B
×
B
C
=
7
5
×
9
14
A
C
=
9
10
? A : C = 9 : 10
Q u e s t i o n : 4
If A : B = 5 : 8 and B : C = 16 : 25, find A : C.
S o l u t i o n :
A
B
=
5
8
and
B
C
=
16
25
Now, we have:
A
B
×
B
C
=
5
8
×
16
25
?
A
C
=
2
5
? A : C = 2 : 5
Q u e s t i o n : 5
If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.
S o l u t i o n :
A : B = 3 : 5
B : C = 10 : 13 =
10÷2
13÷2
= 5 :
13
2
Now, A : B : C = 3 : 5 :
13
2
? A : B : C = 6 : 10 : 13
Q u e s t i o n : 6
If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.
S o l u t i o n :
We have the following:
A : B = 5 : 6
B : C = 4 : 7 =
4
7
=
4×
6
4
7×
6
4
= 6 :
21
2
? A : B : C = 5 : 6 :
21
2
= 10 : 12 : 21
Q u e s t i o n : 7
Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.
S o l u t i o n :
Sum of the ratio terms = 7 + 8 = 15
Now, we have the following:
Kunal's share = Rs 360 ×
7
15
= 24 ×7
= Rs 168
Mohit's share = Rs 360 ×
8
15
= 24 ×8
= Rs 192
Q u e s t i o n : 8
Divide Rs 880 between Rajan and Kamal in the ratio
1
5
:
1
6
.
S o l u t i o n :
Sum of the ratio terms =
1
5
+
1
6
=
11
30
Now, we have the following:
Rajan's share = Rs 880 ×
1
5
11
30
= Rs 880 ×
6
11
= Rs 80 ×6
= Rs 480
Kamal's share = Rs 880 ×
1
6
11
30
= Rs 880 ×
5
11
= Rs 80 ×5
= Rs 400
Q u e s t i o n : 9
Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.
S o l u t i o n :
Sum of the ratio terms is 1 +3 +4
= 8
We have the following:
A's share = Rs 5600 ×
1
8
= Rs
5600
8
= Rs 700
B's share = Rs 5600 ×
3
8
= Rs 700 × 3
= Rs 2100
C's share = Rs 5600 ×
4
8
= Rs 700 ×4
= Rs 2800
Q u e s t i o n : 1 0
What number must be added to each term to the ratio 9 : 16 to make the ratio 2 : 3?
S o l u t i o n :
Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3
?
9+x
16+x
=
2
3
? 27 + 3x = 32 + 2x ? x = 5
Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.
Q u e s t i o n : 1 1
Page 4
Q u e s t i o n : 1
Express each of the following ratios in simplest form:
i
24 : 40
ii
13.5 : 15
iii
6
2
3
: 7
1
2
iv
1
6
:
1
9
v
4: 5:
9
2
vi
2.5 : 6.5 : 8
S o l u t i o n :
i
HCF of 24 and 40 is 8.
? 24 : 40 =
24
40
=
24 ÷ 8
40 ÷ 8
=
3
5
= 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
ii
HCF of 13.5 and 15 is 1.5.
13.5
15
=
135
150
The HCF of 135 and 150 is 15. =
135 ÷ 15
150 ÷ 15
=
9
10
Hence, 13.5 : 15 in its simplest form is 9 : 10.
iii
20
3
:
15
2
= 40 : 45
The HCF of 40 and 45 is 5.
? 40 : 45 =
40
45
=
40 ÷ 5
45 ÷ 5
=
8
9
= 8 : 9
Hence, 6
2
3
: 7
1
2
in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
? 9 : 6 =
9
6
=
9 ÷ 3
6 ÷ 3
= 3 : 2
Hence,
1
6
:
1
9
in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
? 4 : 5 :
9
2
= 8 : 10 : 9
The HCF of these 3 numbers is 1.
? 8 : 10 : 9 is the simplest form.
vi
2.5 : 6.5 : 8 = 25 : 65 : 80
The HCF of 25, 65 and 80 is 5.
? 25 : 65 : 80 =
25
65
80
=
25 ÷ 5
65 ÷ 5
80 ÷ 5
=
5
13
16
= 5 : 13 : 16
Q u e s t i o n : 2
Express each of the following ratios in simplest form:
i
75 paise : 3 rupees
ii
1 m 5 cm : 63 cm
iii
1 hour 5 minutes : 45 minutes
iv
8 months : 1 year
v
2kg250g
: 3kg
vi
1 km : 750 m
S o l u t i o n :
i
Converting both the quantities into the same unit, we have:
75 paise : (3 ×
100) paise = 75 : 300
=
75
300
=
75 ÷ 75
300 ÷ 75
=
1
4
? HCFof75and300 = 75
= 1 paise : 4 paise
ii
Converting both the quantities into the same unit, we have:
105 cm : 63 cm =
105
63
=
105÷21
63÷21
=
5
3
? HCFof105and63 = 21
= 5 cm : 3 cm
iii
Converting both the quantities into the same unit
65 min : 45 min =
65
45
=
65÷5
45÷5
=
13
9
? HCFof65and45 = 5
= 13 min : 9 min
iv
Converting both the quantities into the same unit, we get:
8 months : 12 months =
8
12
=
8÷4
12÷4
=
2
3
? HCFof8and12 = 4
= 2 months : 3 months
v
Converting both the quantities into the same unit, we get:
2250g : 3000 g =
2250
3000
=
2250÷750
3000÷750
=
3
4
? HCFof2250and3000 = 750
= 3 g : 4 g
vi
Converting both the quantities into the same unit, we get:
1000 m : 750 m =
1000
750
=
1000÷250
750÷250
=
4
3
? HCFof1000and750 = 250
= 4 m : 3 m
Q u e s t i o n : 3
If A : B = 7 : 5 and B : C = 9 : 14, find A : C.
S o l u t i o n :
A
B
=
7
5
and
B
C
=
9
14
Therefore, we have:
A
B
×
B
C
=
7
5
×
9
14
A
C
=
9
10
? A : C = 9 : 10
Q u e s t i o n : 4
If A : B = 5 : 8 and B : C = 16 : 25, find A : C.
S o l u t i o n :
A
B
=
5
8
and
B
C
=
16
25
Now, we have:
A
B
×
B
C
=
5
8
×
16
25
?
A
C
=
2
5
? A : C = 2 : 5
Q u e s t i o n : 5
If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.
S o l u t i o n :
A : B = 3 : 5
B : C = 10 : 13 =
10÷2
13÷2
= 5 :
13
2
Now, A : B : C = 3 : 5 :
13
2
? A : B : C = 6 : 10 : 13
Q u e s t i o n : 6
If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.
S o l u t i o n :
We have the following:
A : B = 5 : 6
B : C = 4 : 7 =
4
7
=
4×
6
4
7×
6
4
= 6 :
21
2
? A : B : C = 5 : 6 :
21
2
= 10 : 12 : 21
Q u e s t i o n : 7
Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.
S o l u t i o n :
Sum of the ratio terms = 7 + 8 = 15
Now, we have the following:
Kunal's share = Rs 360 ×
7
15
= 24 ×7
= Rs 168
Mohit's share = Rs 360 ×
8
15
= 24 ×8
= Rs 192
Q u e s t i o n : 8
Divide Rs 880 between Rajan and Kamal in the ratio
1
5
:
1
6
.
S o l u t i o n :
Sum of the ratio terms =
1
5
+
1
6
=
11
30
Now, we have the following:
Rajan's share = Rs 880 ×
1
5
11
30
= Rs 880 ×
6
11
= Rs 80 ×6
= Rs 480
Kamal's share = Rs 880 ×
1
6
11
30
= Rs 880 ×
5
11
= Rs 80 ×5
= Rs 400
Q u e s t i o n : 9
Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.
S o l u t i o n :
Sum of the ratio terms is 1 +3 +4
= 8
We have the following:
A's share = Rs 5600 ×
1
8
= Rs
5600
8
= Rs 700
B's share = Rs 5600 ×
3
8
= Rs 700 × 3
= Rs 2100
C's share = Rs 5600 ×
4
8
= Rs 700 ×4
= Rs 2800
Q u e s t i o n : 1 0
What number must be added to each term to the ratio 9 : 16 to make the ratio 2 : 3?
S o l u t i o n :
Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3
?
9+x
16+x
=
2
3
? 27 + 3x = 32 + 2x ? x = 5
Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.
Q u e s t i o n : 1 1
What number must be subtracted from each term of ratio 17 : 33 so that the ratio becomes 7 : 15?
S o l u t i o n :
Suppose that x is the number that must be subtracted.
Then, (17 - x) : (33 - x) = 7 : 15
?
17 - x
33 - x
=
7
15
? 255 - 15x = 231 - 7x ? 8x = 255 - 231 = 24 ? x = 3
Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.
Q u e s t i o n : 1 2
Two numbers are in the ratio 7 : 11. If added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.
S o l u t i o n :
Suppose that the numbers are 7x and 11x.
Then, (7x + 7) : (11x + 7) = 2 : 3
?
7x + 7
11x + 7
=
2
3
? 21x + 21 = 22x + 14
? x = 7
Hence, the numbers are (7 ×
7 =) 49 and (11 ×
7 =) 77.
Q u e s t i o n : 1 3
Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.
S o l u t i o n :
Suppose that the numbers are 5x and 9x.
Then, (5x - 3) : (9x - 3) = 1 : 2
?
5x - 3
9x -3
=
1
2
? 10x - 6 = 9x - 3
? x = 3
Hence, the numbers are (5 ×
3 =) 15 and (9 ×
3 =) 27.
Q u e s t i o n : 1 4
Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.
S o l u t i o n :
Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
? x = 15
? The numbers are (3 ×
15 =) 45 and (4 ×
15 =) 60.
Q u e s t i o n : 1 5
The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present ages.
S o l u t i o n :
Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x + 6) : (3x + 6) = 9 : 4
?
8x+6
3x+6
=
9
4
? 32x + 24 = 27x + 54
? 5x = 30
? x = 6
Now, present age of A = 8 ×
6 yrs = 48 yrs
Present age of B = 3 ×
6 yrs = 18 yrs
Q u e s t i o n : 1 6
The ratio of copper and zinc in an alloy is 9 : 5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.
S o l u t i o n :
Suppose that the weight of zinc is x g.
Then, 48.6 : x = 9 : 5
? x =
48.6×5
9
=
243
9
= 27
Hence, the weight of zinc in the alloy is 27 g.
Q u e s t i o n : 1 7
The ratio of boys and girls in a school is 8 : 3. If the total number of girls be 375, find the number of boys in the school.
Page 5
Q u e s t i o n : 1
Express each of the following ratios in simplest form:
i
24 : 40
ii
13.5 : 15
iii
6
2
3
: 7
1
2
iv
1
6
:
1
9
v
4: 5:
9
2
vi
2.5 : 6.5 : 8
S o l u t i o n :
i
HCF of 24 and 40 is 8.
? 24 : 40 =
24
40
=
24 ÷ 8
40 ÷ 8
=
3
5
= 3 : 5
Hence, 24 : 40 in its simplest form is 3 : 5.
ii
HCF of 13.5 and 15 is 1.5.
13.5
15
=
135
150
The HCF of 135 and 150 is 15. =
135 ÷ 15
150 ÷ 15
=
9
10
Hence, 13.5 : 15 in its simplest form is 9 : 10.
iii
20
3
:
15
2
= 40 : 45
The HCF of 40 and 45 is 5.
? 40 : 45 =
40
45
=
40 ÷ 5
45 ÷ 5
=
8
9
= 8 : 9
Hence, 6
2
3
: 7
1
2
in its simplest form is 8 : 9
(iv) 9 : 6
The HCF of 9 and 6 is 3.
? 9 : 6 =
9
6
=
9 ÷ 3
6 ÷ 3
= 3 : 2
Hence,
1
6
:
1
9
in its simplest form is 3 : 2.
(v) LCM of the denominators is 2.
? 4 : 5 :
9
2
= 8 : 10 : 9
The HCF of these 3 numbers is 1.
? 8 : 10 : 9 is the simplest form.
vi
2.5 : 6.5 : 8 = 25 : 65 : 80
The HCF of 25, 65 and 80 is 5.
? 25 : 65 : 80 =
25
65
80
=
25 ÷ 5
65 ÷ 5
80 ÷ 5
=
5
13
16
= 5 : 13 : 16
Q u e s t i o n : 2
Express each of the following ratios in simplest form:
i
75 paise : 3 rupees
ii
1 m 5 cm : 63 cm
iii
1 hour 5 minutes : 45 minutes
iv
8 months : 1 year
v
2kg250g
: 3kg
vi
1 km : 750 m
S o l u t i o n :
i
Converting both the quantities into the same unit, we have:
75 paise : (3 ×
100) paise = 75 : 300
=
75
300
=
75 ÷ 75
300 ÷ 75
=
1
4
? HCFof75and300 = 75
= 1 paise : 4 paise
ii
Converting both the quantities into the same unit, we have:
105 cm : 63 cm =
105
63
=
105÷21
63÷21
=
5
3
? HCFof105and63 = 21
= 5 cm : 3 cm
iii
Converting both the quantities into the same unit
65 min : 45 min =
65
45
=
65÷5
45÷5
=
13
9
? HCFof65and45 = 5
= 13 min : 9 min
iv
Converting both the quantities into the same unit, we get:
8 months : 12 months =
8
12
=
8÷4
12÷4
=
2
3
? HCFof8and12 = 4
= 2 months : 3 months
v
Converting both the quantities into the same unit, we get:
2250g : 3000 g =
2250
3000
=
2250÷750
3000÷750
=
3
4
? HCFof2250and3000 = 750
= 3 g : 4 g
vi
Converting both the quantities into the same unit, we get:
1000 m : 750 m =
1000
750
=
1000÷250
750÷250
=
4
3
? HCFof1000and750 = 250
= 4 m : 3 m
Q u e s t i o n : 3
If A : B = 7 : 5 and B : C = 9 : 14, find A : C.
S o l u t i o n :
A
B
=
7
5
and
B
C
=
9
14
Therefore, we have:
A
B
×
B
C
=
7
5
×
9
14
A
C
=
9
10
? A : C = 9 : 10
Q u e s t i o n : 4
If A : B = 5 : 8 and B : C = 16 : 25, find A : C.
S o l u t i o n :
A
B
=
5
8
and
B
C
=
16
25
Now, we have:
A
B
×
B
C
=
5
8
×
16
25
?
A
C
=
2
5
? A : C = 2 : 5
Q u e s t i o n : 5
If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.
S o l u t i o n :
A : B = 3 : 5
B : C = 10 : 13 =
10÷2
13÷2
= 5 :
13
2
Now, A : B : C = 3 : 5 :
13
2
? A : B : C = 6 : 10 : 13
Q u e s t i o n : 6
If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.
S o l u t i o n :
We have the following:
A : B = 5 : 6
B : C = 4 : 7 =
4
7
=
4×
6
4
7×
6
4
= 6 :
21
2
? A : B : C = 5 : 6 :
21
2
= 10 : 12 : 21
Q u e s t i o n : 7
Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.
S o l u t i o n :
Sum of the ratio terms = 7 + 8 = 15
Now, we have the following:
Kunal's share = Rs 360 ×
7
15
= 24 ×7
= Rs 168
Mohit's share = Rs 360 ×
8
15
= 24 ×8
= Rs 192
Q u e s t i o n : 8
Divide Rs 880 between Rajan and Kamal in the ratio
1
5
:
1
6
.
S o l u t i o n :
Sum of the ratio terms =
1
5
+
1
6
=
11
30
Now, we have the following:
Rajan's share = Rs 880 ×
1
5
11
30
= Rs 880 ×
6
11
= Rs 80 ×6
= Rs 480
Kamal's share = Rs 880 ×
1
6
11
30
= Rs 880 ×
5
11
= Rs 80 ×5
= Rs 400
Q u e s t i o n : 9
Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.
S o l u t i o n :
Sum of the ratio terms is 1 +3 +4
= 8
We have the following:
A's share = Rs 5600 ×
1
8
= Rs
5600
8
= Rs 700
B's share = Rs 5600 ×
3
8
= Rs 700 × 3
= Rs 2100
C's share = Rs 5600 ×
4
8
= Rs 700 ×4
= Rs 2800
Q u e s t i o n : 1 0
What number must be added to each term to the ratio 9 : 16 to make the ratio 2 : 3?
S o l u t i o n :
Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3
?
9+x
16+x
=
2
3
? 27 + 3x = 32 + 2x ? x = 5
Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.
Q u e s t i o n : 1 1
What number must be subtracted from each term of ratio 17 : 33 so that the ratio becomes 7 : 15?
S o l u t i o n :
Suppose that x is the number that must be subtracted.
Then, (17 - x) : (33 - x) = 7 : 15
?
17 - x
33 - x
=
7
15
? 255 - 15x = 231 - 7x ? 8x = 255 - 231 = 24 ? x = 3
Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.
Q u e s t i o n : 1 2
Two numbers are in the ratio 7 : 11. If added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.
S o l u t i o n :
Suppose that the numbers are 7x and 11x.
Then, (7x + 7) : (11x + 7) = 2 : 3
?
7x + 7
11x + 7
=
2
3
? 21x + 21 = 22x + 14
? x = 7
Hence, the numbers are (7 ×
7 =) 49 and (11 ×
7 =) 77.
Q u e s t i o n : 1 3
Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.
S o l u t i o n :
Suppose that the numbers are 5x and 9x.
Then, (5x - 3) : (9x - 3) = 1 : 2
?
5x - 3
9x -3
=
1
2
? 10x - 6 = 9x - 3
? x = 3
Hence, the numbers are (5 ×
3 =) 15 and (9 ×
3 =) 27.
Q u e s t i o n : 1 4
Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.
S o l u t i o n :
Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
? x = 15
? The numbers are (3 ×
15 =) 45 and (4 ×
15 =) 60.
Q u e s t i o n : 1 5
The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present ages.
S o l u t i o n :
Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x + 6) : (3x + 6) = 9 : 4
?
8x+6
3x+6
=
9
4
? 32x + 24 = 27x + 54
? 5x = 30
? x = 6
Now, present age of A = 8 ×
6 yrs = 48 yrs
Present age of B = 3 ×
6 yrs = 18 yrs
Q u e s t i o n : 1 6
The ratio of copper and zinc in an alloy is 9 : 5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.
S o l u t i o n :
Suppose that the weight of zinc is x g.
Then, 48.6 : x = 9 : 5
? x =
48.6×5
9
=
243
9
= 27
Hence, the weight of zinc in the alloy is 27 g.
Q u e s t i o n : 1 7
The ratio of boys and girls in a school is 8 : 3. If the total number of girls be 375, find the number of boys in the school.
S o l u t i o n :
Suppose that the number of boys is x.
Then, x : 375 = 8 : 3
? x =
8×375
3
= 8 ×125
= 1000
Hence, the number of girls in the school is 1000.
Q u e s t i o n : 1 8
The ratio of monthly income to the savings of a family is 11 : 2. If the savings be Rs 2500, find the income and expenditure.
S o l u t i o n :
Suppose that the monthly income of the family is Rs x.
Then, x : 2500 = 11 : 2
? x =
11×2500
2
= 11 ×1250
? x = Rs 13750
Hence, the income is Rs 13,750.
? Expenditure = monthlyincome -savings
=Rs 13750 -2500
= Rs 11250
Q u e s t i o n : 1 9
A bag contains Rs 750 in the form of rupee, 50 P and 25 P coins in the ratio 5 : 8 : 4. Find the number of coins of each type.
S o l u t i o n :
Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.
Total value of these coins = (5x ×
100
100
+ 8x ×
50
100
+ 4x ×
25
100
)
? 5x +
8x
2
+
4x
4
=
20x + 16x + 4x
4
=
40x
4
= 10x
However, the total value is Rs 750.
? 750 = 10x
? x = 75
Hence, number of one rupee coins = 5 ×
75 = 375
Number of fifty paise coins = 8 ×
75 = 600
Number of twenty-five paise coins = 4 ×
75 = 300
Q u e s t i o n : 2 0
If (4x + 5) : (3x + 11) = 13 : 17, find the value of x.
S o l u t i o n :
(4x + 5) : (3x + 11) = 13 : 17
?
4x+ 5
3x + 11
=
13
17
? 68x + 85 = 39x + 143 ? 29x = 58 ? x = 2
Q u e s t i o n : 2 1
If x : y = 3 : 4, find (3x + 4y) : (5x + 6y).
S o l u t i o n :
x
y
=
3
4
? x =
3y
4
Now, we have (3x + 4y) : (5x + 6y)
=
3x +4y
5x + 6y
=
3×
3y
4
+4y
5×
3y
4
+6y
=
9y+16y
15y +24y
=
25y
39y
=
25
39
= 25 : 39
Q u e s t i o n : 2 2
If x : y = 6 : 11, find (8x - 3y) : (3x + 2y).
S o l u t i o n :
x
y
=
6
11
? x =
6y
11
Now, we have:
8x -3y
3x + 2y
=
8×
6y
11
-3y
3×
6y
11
+2y
=
48y-33y
18y + 22y
=
15y
40y
=
3
8
? (8x - 3y) : (3x + 2y) = 3 : 8
Read More
|
76 videos|345 docs|39 tests
|
FAQs on RS Aggarwal Solutions: Ratio and Proportion - Mathematics (Maths) Class 7
| 1. What is the definition of ratio and proportion? |  |
Ans. Ratio is the quantitative relationship between two or more quantities, indicating how many times one quantity is contained in another. Proportion is an equation that states that two ratios are equal.
| 2. How do I solve problems involving ratio and proportion? | |
Ans. To solve problems involving ratio and proportion, you can use the following steps:
1. Identify the given ratios or proportions.
2. Set up an equation using the given ratios or proportions.
3. Solve the equation to find the unknown value.
| 3. What are the different types of ratios? | |
Ans. There are four types of ratios:
1. Simple Ratio: It compares two quantities of the same kind.
2. Compound Ratio: It compares two or more simple ratios.
3. Duplicate Ratio: It compares the square of two quantities.
4. Sub-duplicate Ratio: It compares the square root of two quantities.
| 4. Can you give an example of solving a problem using ratio and proportion? | |
Ans. Sure! Let's solve the following problem:
If the ratio of boys to girls in a class is 3:5 and there are 24 boys, how many girls are there?
Solution:
Step 1: Given ratio of boys to girls = 3:5
Step 2: Let the number of girls be x.
According to the given ratio, we can set up the proportion:
3/5 = 24/x
Step 3: Cross-multiply and solve for x:
3x = 5 * 24
3x = 120
x = 40
Therefore, there are 40 girls in the class.
| 5. How can ratio and proportion be applied in real-life situations? | |
Ans. Ratio and proportion are used in various real-life situations, such as:
1. Cooking: Recipes often require ingredients to be used in a certain ratio or proportion.
2. Finance: Calculating interest rates, investments, and loan repayments often involve ratios and proportions.
3. Maps and Scales: Maps and scale models use ratios to represent real-life distances.
4. Mixing Solutions: Chemical solutions are often mixed in specific ratios and proportions for desired outcomes.
5. Sports and Fitness: Athletes and fitness enthusiasts use ratios and proportions to track progress and set goals.
Related Exams
About this Document
|
4.61/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Ratio and Proportion for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RS Aggarwal Solutions: Ratio and Proportion have been prepared according to the Class 7 exam syllabus. Information about RS Aggarwal Solutions: Ratio and Proportion covers topics
like and RS Aggarwal Solutions: Ratio and Proportion Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Ratio and Proportion.
Introduction of RS Aggarwal Solutions: Ratio and Proportion in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RS Aggarwal Solutions: Ratio and Proportion in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Ratio and Proportion
In this doc you can find the meaning of RS Aggarwal Solutions: Ratio and Proportion defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Ratio and Proportion theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Ratio and Proportion tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Exam
,Sample Paper
,Free
,MCQs
,Summary
,study material
,RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7
,Viva Questions
,ppt
,RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7
,RS Aggarwal Solutions: Ratio and Proportion | Mathematics (Maths) Class 7
,mock tests for examination
,Semester Notes
,Previous Year Questions with Solutions
,past year papers
,practice quizzes
,shortcuts and tricks
,Objective type Questions
,Extra Questions
,video lectures
,Important questions
;
Additional Information about RS Aggarwal Solutions: Ratio and Proportion for Class 7 Preparation
RS Aggarwal Solutions: Ratio and Proportion Free PDF Download
The RS Aggarwal Solutions: Ratio and Proportion is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal Solutions: Ratio and Proportion now and kickstart your journey towards success in the Class 7 exam.
Importance of RS Aggarwal Solutions: Ratio and Proportion
The importance of RS Aggarwal Solutions: Ratio and Proportion cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal Solutions: Ratio and Proportion Notes
RS Aggarwal Solutions: Ratio and Proportion Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Ratio and Proportion.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal Solutions: Ratio and Proportion Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Ratio and Proportion Class 7 Questions
The "RS Aggarwal Solutions: Ratio and Proportion Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal Solutions: Ratio and Proportion on the App
Students of Class 7 can study RS Aggarwal Solutions: Ratio and Proportion alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Ratio and Proportion,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal Solutions: Ratio and Proportion is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup