Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RS Aggarwal Solutions: Ratio, Proportion and Unitary Method
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Page 1
Points to Remember :
Unitary Method. The method is which a unit item helps in getting the result about any number of
items is called unitary method.
Rules. (i) To get more, we multiply.
(ii) To get less, we divide.
750 )4000( 5
3750
250 )750( 3
750
×
(HCF of 4000 and 750 = 250)
=
3
16
= 16 : 3
(iv) 1.8 kg to 6 kg =
10
18
: 6
= 18 : 60 =
60
18
=
6 60
6 18
(HCF of 18, 60 = 6)
=
10
3
= 3 : 10
(v) 48 minutes to 1 hour
= 48 minutes : 60 minutes
= 48 : 60 = 4 : 5
(vi) 2.4 km to 900 m
= 2400 m : 900 m =
900
2400
=
300 900
300 2400
(HCF of 2400 and 900 = 300)
( ) EXERCISE 10 A
Q.1. Find each of the following ratios in the
simplest form :
(i) 24 to 56 (ii) 84 paise to Rs. 3
(iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Sol. (i) 24 to 56 =
56
24
=
8 56
8 24
(HCF of 24, 56 = 8)
=
7
3
= 3 : 7
(ii) 84 paise to Rs. 3 = 84 p : 300 p
=
300
84
=
12 300
12 84
(HCF of 84 and 300 = 12)
=
25
7
= 7 : 25
(iii) 4 kg to 750 g = 4000 g : 750 g
=
750
4000
=
250 750
250 4000
Page 2
Points to Remember :
Unitary Method. The method is which a unit item helps in getting the result about any number of
items is called unitary method.
Rules. (i) To get more, we multiply.
(ii) To get less, we divide.
750 )4000( 5
3750
250 )750( 3
750
×
(HCF of 4000 and 750 = 250)
=
3
16
= 16 : 3
(iv) 1.8 kg to 6 kg =
10
18
: 6
= 18 : 60 =
60
18
=
6 60
6 18
(HCF of 18, 60 = 6)
=
10
3
= 3 : 10
(v) 48 minutes to 1 hour
= 48 minutes : 60 minutes
= 48 : 60 = 4 : 5
(vi) 2.4 km to 900 m
= 2400 m : 900 m =
900
2400
=
300 900
300 2400
(HCF of 2400 and 900 = 300)
( ) EXERCISE 10 A
Q.1. Find each of the following ratios in the
simplest form :
(i) 24 to 56 (ii) 84 paise to Rs. 3
(iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Sol. (i) 24 to 56 =
56
24
=
8 56
8 24
(HCF of 24, 56 = 8)
=
7
3
= 3 : 7
(ii) 84 paise to Rs. 3 = 84 p : 300 p
=
300
84
=
12 300
12 84
(HCF of 84 and 300 = 12)
=
25
7
= 7 : 25
(iii) 4 kg to 750 g = 4000 g : 750 g
=
750
4000
=
250 750
250 4000
=
3
8
= 8 : 3
Q. 2. Express the following ratios in the
simplest form :
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
Sol. (i) 36 : 90 =
90
36
=
18 90
18 36
(HCF of 36, 90 = 18)
=
5
2
= 2 : 5
(ii) The given ratio = 324 : 144
324
144
To express it in simplest form, we divide
the numerator and denominator by the
H.C.F. of 324 and 144.
Now, H.C.F. of 324 and 144
144 ) 324
288
( 2
36 ) 144
144
( 4
= 36
324
144
324 36
144 36
9
4
Required ratio = 9 : 4.
(iii) The given ratio is 85 : 561
85
561
To express it in simplest form, we
divided the numerator and denominator
by the HCF of 85 and 561.
Now, HCF of 85 and 561 = 17
85 ) 561
510
( 6
51 ) 85
51
( 1
34 ) 51
34
( 1
17 ) 34
34
0
( 2
85
561
85 17
561 17
5
33
Required ratio = 5 : 33.
(iv) The given ratio = 480 : 384
480
384
To express it in simplest form, we divide
the numerator and denominator by the
HCF of 480 and 384.
Now, HCF of 480 and 384
= 96
384 ) 480
384
( 1
96 ) 384
384
( 4
0
480
384
480 96
384 96
5
4
Required rato = 5 : 4.
(v) The given ratio 186 403
186
403
:
To express it in the simplest form, we
divide its numerator and denominator by
the HCF of 186 and 403.
Now, HCF of 186 and 403 = 31
186 ) 403
372
( 2
31 ) 186
186
( 6
0
186
403
186 31
403 31
6
13
Page 3
Points to Remember :
Unitary Method. The method is which a unit item helps in getting the result about any number of
items is called unitary method.
Rules. (i) To get more, we multiply.
(ii) To get less, we divide.
750 )4000( 5
3750
250 )750( 3
750
×
(HCF of 4000 and 750 = 250)
=
3
16
= 16 : 3
(iv) 1.8 kg to 6 kg =
10
18
: 6
= 18 : 60 =
60
18
=
6 60
6 18
(HCF of 18, 60 = 6)
=
10
3
= 3 : 10
(v) 48 minutes to 1 hour
= 48 minutes : 60 minutes
= 48 : 60 = 4 : 5
(vi) 2.4 km to 900 m
= 2400 m : 900 m =
900
2400
=
300 900
300 2400
(HCF of 2400 and 900 = 300)
( ) EXERCISE 10 A
Q.1. Find each of the following ratios in the
simplest form :
(i) 24 to 56 (ii) 84 paise to Rs. 3
(iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Sol. (i) 24 to 56 =
56
24
=
8 56
8 24
(HCF of 24, 56 = 8)
=
7
3
= 3 : 7
(ii) 84 paise to Rs. 3 = 84 p : 300 p
=
300
84
=
12 300
12 84
(HCF of 84 and 300 = 12)
=
25
7
= 7 : 25
(iii) 4 kg to 750 g = 4000 g : 750 g
=
750
4000
=
250 750
250 4000
=
3
8
= 8 : 3
Q. 2. Express the following ratios in the
simplest form :
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
Sol. (i) 36 : 90 =
90
36
=
18 90
18 36
(HCF of 36, 90 = 18)
=
5
2
= 2 : 5
(ii) The given ratio = 324 : 144
324
144
To express it in simplest form, we divide
the numerator and denominator by the
H.C.F. of 324 and 144.
Now, H.C.F. of 324 and 144
144 ) 324
288
( 2
36 ) 144
144
( 4
= 36
324
144
324 36
144 36
9
4
Required ratio = 9 : 4.
(iii) The given ratio is 85 : 561
85
561
To express it in simplest form, we
divided the numerator and denominator
by the HCF of 85 and 561.
Now, HCF of 85 and 561 = 17
85 ) 561
510
( 6
51 ) 85
51
( 1
34 ) 51
34
( 1
17 ) 34
34
0
( 2
85
561
85 17
561 17
5
33
Required ratio = 5 : 33.
(iv) The given ratio = 480 : 384
480
384
To express it in simplest form, we divide
the numerator and denominator by the
HCF of 480 and 384.
Now, HCF of 480 and 384
= 96
384 ) 480
384
( 1
96 ) 384
384
( 4
0
480
384
480 96
384 96
5
4
Required rato = 5 : 4.
(v) The given ratio 186 403
186
403
:
To express it in the simplest form, we
divide its numerator and denominator by
the HCF of 186 and 403.
Now, HCF of 186 and 403 = 31
186 ) 403
372
( 2
31 ) 186
186
( 6
0
186
403
186 31
403 31
6
13
Required ratio = 6 : 13
(vi) 777 : 1147 =
1147
777
To express it in simplest form, we divide
its numerator and denominator by the HCF
of 777 and 1147
=
37 1147
37 777
=
31
21
777 )1147( 7
777
370 )777( 2
740
37 )370( 10
370
×
(HCF of 777 and 1147 is 37)
= 21 : 31
Q. 3. Find the following ratios in simplest
form :
(i) Rs. 6.30 : Rs. 16·80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm.
(iv) 48 min : 2 hours 40 min.
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g.
Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80
Rs. 6
.
30
Rs.16
.
80
630
1680
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 630 and 1680.
Now, HCF of 630 and 1680 = 210
630 ) 1680
1260
( 2
420 ) 630
420
( 1
210 ) 420
420
( 2
0
630
1680
630 210
1680 210
3
8
Required ratio = 3 : 8
(ii) The given ratio = 3 weeks : 30 days
= (3 × 7) days : 30 days
= 21 days : 30 days
= 21 : 30
= 7 : 10
(iii) To given ratio = 3m 5 cm : 35cm
= (3 100 + 5) cm : 35 cm
= 305 cm : 35 cm
= 305 : 35
= 61 : 7.
(iv) The given ratio = 48 min : 2 hours 40 min
= 48 min : (2 60 + 40) min
= 48 min : (120 + 40) min
= 48 min : 160 min
= 48 : 160
= 3 : 10
(v) The given ratio = 1L 35 mL : 270 mL
= (1 1000 + 35) mL : 270 mL
= 1035 mL : 270 mL
= 1035 : 270
1035
270
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 1035 and 270.
270 ) 1035
810
( 3
225 ) 270
225
( 1
45 ) 225
225
( 5
0
Now, HCF and 1035 and 270 = 45
Page 4
Points to Remember :
Unitary Method. The method is which a unit item helps in getting the result about any number of
items is called unitary method.
Rules. (i) To get more, we multiply.
(ii) To get less, we divide.
750 )4000( 5
3750
250 )750( 3
750
×
(HCF of 4000 and 750 = 250)
=
3
16
= 16 : 3
(iv) 1.8 kg to 6 kg =
10
18
: 6
= 18 : 60 =
60
18
=
6 60
6 18
(HCF of 18, 60 = 6)
=
10
3
= 3 : 10
(v) 48 minutes to 1 hour
= 48 minutes : 60 minutes
= 48 : 60 = 4 : 5
(vi) 2.4 km to 900 m
= 2400 m : 900 m =
900
2400
=
300 900
300 2400
(HCF of 2400 and 900 = 300)
( ) EXERCISE 10 A
Q.1. Find each of the following ratios in the
simplest form :
(i) 24 to 56 (ii) 84 paise to Rs. 3
(iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Sol. (i) 24 to 56 =
56
24
=
8 56
8 24
(HCF of 24, 56 = 8)
=
7
3
= 3 : 7
(ii) 84 paise to Rs. 3 = 84 p : 300 p
=
300
84
=
12 300
12 84
(HCF of 84 and 300 = 12)
=
25
7
= 7 : 25
(iii) 4 kg to 750 g = 4000 g : 750 g
=
750
4000
=
250 750
250 4000
=
3
8
= 8 : 3
Q. 2. Express the following ratios in the
simplest form :
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
Sol. (i) 36 : 90 =
90
36
=
18 90
18 36
(HCF of 36, 90 = 18)
=
5
2
= 2 : 5
(ii) The given ratio = 324 : 144
324
144
To express it in simplest form, we divide
the numerator and denominator by the
H.C.F. of 324 and 144.
Now, H.C.F. of 324 and 144
144 ) 324
288
( 2
36 ) 144
144
( 4
= 36
324
144
324 36
144 36
9
4
Required ratio = 9 : 4.
(iii) The given ratio is 85 : 561
85
561
To express it in simplest form, we
divided the numerator and denominator
by the HCF of 85 and 561.
Now, HCF of 85 and 561 = 17
85 ) 561
510
( 6
51 ) 85
51
( 1
34 ) 51
34
( 1
17 ) 34
34
0
( 2
85
561
85 17
561 17
5
33
Required ratio = 5 : 33.
(iv) The given ratio = 480 : 384
480
384
To express it in simplest form, we divide
the numerator and denominator by the
HCF of 480 and 384.
Now, HCF of 480 and 384
= 96
384 ) 480
384
( 1
96 ) 384
384
( 4
0
480
384
480 96
384 96
5
4
Required rato = 5 : 4.
(v) The given ratio 186 403
186
403
:
To express it in the simplest form, we
divide its numerator and denominator by
the HCF of 186 and 403.
Now, HCF of 186 and 403 = 31
186 ) 403
372
( 2
31 ) 186
186
( 6
0
186
403
186 31
403 31
6
13
Required ratio = 6 : 13
(vi) 777 : 1147 =
1147
777
To express it in simplest form, we divide
its numerator and denominator by the HCF
of 777 and 1147
=
37 1147
37 777
=
31
21
777 )1147( 7
777
370 )777( 2
740
37 )370( 10
370
×
(HCF of 777 and 1147 is 37)
= 21 : 31
Q. 3. Find the following ratios in simplest
form :
(i) Rs. 6.30 : Rs. 16·80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm.
(iv) 48 min : 2 hours 40 min.
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g.
Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80
Rs. 6
.
30
Rs.16
.
80
630
1680
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 630 and 1680.
Now, HCF of 630 and 1680 = 210
630 ) 1680
1260
( 2
420 ) 630
420
( 1
210 ) 420
420
( 2
0
630
1680
630 210
1680 210
3
8
Required ratio = 3 : 8
(ii) The given ratio = 3 weeks : 30 days
= (3 × 7) days : 30 days
= 21 days : 30 days
= 21 : 30
= 7 : 10
(iii) To given ratio = 3m 5 cm : 35cm
= (3 100 + 5) cm : 35 cm
= 305 cm : 35 cm
= 305 : 35
= 61 : 7.
(iv) The given ratio = 48 min : 2 hours 40 min
= 48 min : (2 60 + 40) min
= 48 min : (120 + 40) min
= 48 min : 160 min
= 48 : 160
= 3 : 10
(v) The given ratio = 1L 35 mL : 270 mL
= (1 1000 + 35) mL : 270 mL
= 1035 mL : 270 mL
= 1035 : 270
1035
270
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 1035 and 270.
270 ) 1035
810
( 3
225 ) 270
225
( 1
45 ) 225
225
( 5
0
Now, HCF and 1035 and 270 = 45
1035
270
1035 45
270 45
23
6
Required ratio = 23 : 6.
(vi) The given ratio = 4 kg : 2 kg 500 g
= (4 × 1000) g : (2 1000 + 500) g
= 4000 g : 2500 g
= 4000 : 2500
= 40 : 25
= 8 : 5
4. Mr Sahai and his wife are both school
teachers and earn Rs., 16800 and Rs.
10500 per month respectively. Find the
ratio of
(i) Mr Sahai’s income to his wife’s income;
(ii) Mrs Sahai’s income to her husband’s
income;
(iii) Mr Sahai’s income to the total income of
the two.
Sol. Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
(i) Ratio in Sahai’s income and his wife
= 16800 : 10500
=
10500
16800
=
2100 10500
2100 16800
(HCF of 16800 and 105000 is 2100)
=
5
8
= 8 : 5
(ii) Ratio in Sahai’s wife income and her
husband’s income
= 10500 : Rs. 16800
= 5 : 8
(Dividing by the HCF of 10500 and
16800 = 2100)
(iii) Sahai’s income and total of their income
= 16800 : 27300 =
27300
16800
=
2100 27300
2100 16800
=
13
8
= 8 : 13
5. Rohit earns Rs. 15300 and saves Rs.
1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.
Sol. Rohit monthly earnings = Rs. 15300
and his savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,
(i) Ratio in his income and savings
= 15300 : 1224 =
1224
15300
1224 )15300( 12
1224
3060
2448
612 )1224( 2
1224
×
=
612 1224
612 15300
(Dividing by HCF of 15300 and 1224)
=
2
25
= 25 : 2
(ii) Ratio in his income and expenditure
= 15300 : 14076 =
14076
15300
=
612 14076
612 15300
(HCF = 612)
Page 5
Points to Remember :
Unitary Method. The method is which a unit item helps in getting the result about any number of
items is called unitary method.
Rules. (i) To get more, we multiply.
(ii) To get less, we divide.
750 )4000( 5
3750
250 )750( 3
750
×
(HCF of 4000 and 750 = 250)
=
3
16
= 16 : 3
(iv) 1.8 kg to 6 kg =
10
18
: 6
= 18 : 60 =
60
18
=
6 60
6 18
(HCF of 18, 60 = 6)
=
10
3
= 3 : 10
(v) 48 minutes to 1 hour
= 48 minutes : 60 minutes
= 48 : 60 = 4 : 5
(vi) 2.4 km to 900 m
= 2400 m : 900 m =
900
2400
=
300 900
300 2400
(HCF of 2400 and 900 = 300)
( ) EXERCISE 10 A
Q.1. Find each of the following ratios in the
simplest form :
(i) 24 to 56 (ii) 84 paise to Rs. 3
(iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Sol. (i) 24 to 56 =
56
24
=
8 56
8 24
(HCF of 24, 56 = 8)
=
7
3
= 3 : 7
(ii) 84 paise to Rs. 3 = 84 p : 300 p
=
300
84
=
12 300
12 84
(HCF of 84 and 300 = 12)
=
25
7
= 7 : 25
(iii) 4 kg to 750 g = 4000 g : 750 g
=
750
4000
=
250 750
250 4000
=
3
8
= 8 : 3
Q. 2. Express the following ratios in the
simplest form :
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
Sol. (i) 36 : 90 =
90
36
=
18 90
18 36
(HCF of 36, 90 = 18)
=
5
2
= 2 : 5
(ii) The given ratio = 324 : 144
324
144
To express it in simplest form, we divide
the numerator and denominator by the
H.C.F. of 324 and 144.
Now, H.C.F. of 324 and 144
144 ) 324
288
( 2
36 ) 144
144
( 4
= 36
324
144
324 36
144 36
9
4
Required ratio = 9 : 4.
(iii) The given ratio is 85 : 561
85
561
To express it in simplest form, we
divided the numerator and denominator
by the HCF of 85 and 561.
Now, HCF of 85 and 561 = 17
85 ) 561
510
( 6
51 ) 85
51
( 1
34 ) 51
34
( 1
17 ) 34
34
0
( 2
85
561
85 17
561 17
5
33
Required ratio = 5 : 33.
(iv) The given ratio = 480 : 384
480
384
To express it in simplest form, we divide
the numerator and denominator by the
HCF of 480 and 384.
Now, HCF of 480 and 384
= 96
384 ) 480
384
( 1
96 ) 384
384
( 4
0
480
384
480 96
384 96
5
4
Required rato = 5 : 4.
(v) The given ratio 186 403
186
403
:
To express it in the simplest form, we
divide its numerator and denominator by
the HCF of 186 and 403.
Now, HCF of 186 and 403 = 31
186 ) 403
372
( 2
31 ) 186
186
( 6
0
186
403
186 31
403 31
6
13
Required ratio = 6 : 13
(vi) 777 : 1147 =
1147
777
To express it in simplest form, we divide
its numerator and denominator by the HCF
of 777 and 1147
=
37 1147
37 777
=
31
21
777 )1147( 7
777
370 )777( 2
740
37 )370( 10
370
×
(HCF of 777 and 1147 is 37)
= 21 : 31
Q. 3. Find the following ratios in simplest
form :
(i) Rs. 6.30 : Rs. 16·80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm.
(iv) 48 min : 2 hours 40 min.
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g.
Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80
Rs. 6
.
30
Rs.16
.
80
630
1680
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 630 and 1680.
Now, HCF of 630 and 1680 = 210
630 ) 1680
1260
( 2
420 ) 630
420
( 1
210 ) 420
420
( 2
0
630
1680
630 210
1680 210
3
8
Required ratio = 3 : 8
(ii) The given ratio = 3 weeks : 30 days
= (3 × 7) days : 30 days
= 21 days : 30 days
= 21 : 30
= 7 : 10
(iii) To given ratio = 3m 5 cm : 35cm
= (3 100 + 5) cm : 35 cm
= 305 cm : 35 cm
= 305 : 35
= 61 : 7.
(iv) The given ratio = 48 min : 2 hours 40 min
= 48 min : (2 60 + 40) min
= 48 min : (120 + 40) min
= 48 min : 160 min
= 48 : 160
= 3 : 10
(v) The given ratio = 1L 35 mL : 270 mL
= (1 1000 + 35) mL : 270 mL
= 1035 mL : 270 mL
= 1035 : 270
1035
270
To express it in simplest form, we divide
its numerator and denominator by the
HCF of 1035 and 270.
270 ) 1035
810
( 3
225 ) 270
225
( 1
45 ) 225
225
( 5
0
Now, HCF and 1035 and 270 = 45
1035
270
1035 45
270 45
23
6
Required ratio = 23 : 6.
(vi) The given ratio = 4 kg : 2 kg 500 g
= (4 × 1000) g : (2 1000 + 500) g
= 4000 g : 2500 g
= 4000 : 2500
= 40 : 25
= 8 : 5
4. Mr Sahai and his wife are both school
teachers and earn Rs., 16800 and Rs.
10500 per month respectively. Find the
ratio of
(i) Mr Sahai’s income to his wife’s income;
(ii) Mrs Sahai’s income to her husband’s
income;
(iii) Mr Sahai’s income to the total income of
the two.
Sol. Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
(i) Ratio in Sahai’s income and his wife
= 16800 : 10500
=
10500
16800
=
2100 10500
2100 16800
(HCF of 16800 and 105000 is 2100)
=
5
8
= 8 : 5
(ii) Ratio in Sahai’s wife income and her
husband’s income
= 10500 : Rs. 16800
= 5 : 8
(Dividing by the HCF of 10500 and
16800 = 2100)
(iii) Sahai’s income and total of their income
= 16800 : 27300 =
27300
16800
=
2100 27300
2100 16800
=
13
8
= 8 : 13
5. Rohit earns Rs. 15300 and saves Rs.
1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.
Sol. Rohit monthly earnings = Rs. 15300
and his savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,
(i) Ratio in his income and savings
= 15300 : 1224 =
1224
15300
1224 )15300( 12
1224
3060
2448
612 )1224( 2
1224
×
=
612 1224
612 15300
(Dividing by HCF of 15300 and 1224)
=
2
25
= 25 : 2
(ii) Ratio in his income and expenditure
= 15300 : 14076 =
14076
15300
=
612 14076
612 15300
(HCF = 612)
=
23
25
= 25 : 23
(iii) Ratio in his expenditure and savings
= 14076 : 1224 =
1224
14076
=
612 1224
612 14076
=
2
23
= 23 : 2
Q. 6. The ratio of the number of male and
female workers in a textile mill is 5 : 3.
If there are 115 male workers, what is
the number of female workers in the
mill ?
Sol. Let the number of male and female
workers in the mill be 5x and 3x
respectively. Then,
5x = 115
5
5
115
5
x
(Dividing both sides by 5)
x = 23
Number of female workers in the mill
= 3x
= 3 23 = 69.
Q. 7. The boys and the girls in a school are in
the ratio 9 : 5. If the total strength of the
school is 448, find the number of girls.
Sol. Let the number of boys and girls in the
school be 9x and 5x respectively.
According to the question,
9x + 5x = 448
14x = 448
14
14
448
14
x
(Dividing both sides by 14)
x = 32.
Number of girls = 5x = 5 32 = 160.
Q.8. Divide Rs. 1575 between Kamal and
Madhu in the ratio 7 : 2.
Sol. Total amount = Rs. 1575
Ratio in Kamal and Madhu’s share
= 7 : 2
Sum of ratios = 7 + 2 = 9
Kamal’s share = Rs.
9
7 1575
= Rs. 175 × 7 = Rs. 1225
Madhu’s share = Rs.
9
2 1575
= Rs. 175 × 2 = Rs. 350
Q.9. Divide Rs. 3450 among A, B and C in
the ratio 3 : 5 : 7.
Sol. Total amount = Rs. 3450
Ratio in A, B and C shares = 3 : 5 : 7
Sum of share = 3 + 5 + 7 = 15
A’s share = Rs.
15
3 3450
= Rs. 230 × 3 = Rs. 690
B’s share = Rs.
15
5 3450
= Rs. 230 × 5 = Rs. 1150
C’s share = Rs.
15
7 3450
= Rs. 230 × 7 = Rs. 1610
Q. 10. Two numbers are in the ratio : 11 : 12.
If their sum is 460, find the numbers.
Sol. Let the numbers be 11x and 12x. Then,
11x + 12x = 460 23x = 460
23
23
460
23
x
(Dividing both sides by 23)
x = 20.
Required numbers are (11 × 20) and
12 × 20, that is , 220 and 240.
Q.11. A 35-cm line segment is divided into
two parts in the ratio 4 : 3. Find the
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FAQs on RS Aggarwal Solutions: Ratio, Proportion and Unitary Method - Mathematics (Maths) Class 6
| 1. What is the importance of learning ratio, proportion, and unitary method in Class 6? |  |
Ans. Learning ratio, proportion, and unitary method in Class 6 is important as it lays the foundation for understanding more complex mathematical concepts in higher grades. These topics help students develop problem-solving skills, logical reasoning, and critical thinking abilities.
| 2. How can I improve my understanding of ratio and proportion in Class 6? | |
Ans. To improve your understanding of ratio and proportion in Class 6, you can practice solving a variety of problems related to these topics. Use visual aids, such as diagrams or objects, to understand the concept visually. Additionally, seeking help from your teacher or referring to reliable study materials can also enhance your understanding.
| 3. Are there any real-life applications of ratio, proportion, and unitary method that I can relate to? | |
Ans. Yes, there are several real-life applications of ratio, proportion, and unitary method. For example, when cooking, you can use ratios and proportions to adjust the quantity of ingredients based on the number of servings. In shopping, you can compare prices using ratios to find the best deal. Unitary method helps in solving problems related to time, speed, distance, and more.
| 4. How can I use the unitary method to solve problems effectively? | |
Ans. To use the unitary method effectively, you need to understand the relationship between the given quantities and the desired quantity. Break down the problem into smaller steps, identify the unitary value, and then apply it to find the desired quantity. Practice solving various types of unitary method problems to improve your skills.
| 5. Are there any online resources or practice tests available for ratio, proportion, and unitary method in Class 6? | |
Ans. Yes, there are several online resources and practice tests available for ratio, proportion, and unitary method in Class 6. You can find interactive tutorials, video lessons, and practice quizzes on educational websites. Additionally, you can also download e-books or use mobile applications specifically designed for practicing these concepts.
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