Class 6Fractions
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Answer the questions
(1) What f raction is represented by shaded portion of these shapes :
A)
B)
C)
D)
E)
F)
(2) Shade the images to show the f ollowing f raction addition.
4
16
+
7
16
=
and makes
(3) Add the f ollowing f ractions:
A) 14
23
26 + 4
10
27
B) 11
7
8 + 3
11
27
C) 8
26
31 + 3
16
31
D) 14
7
12 + 3
11
27
E) 10
11
15 + 2
22
23
F) 14
8
13 + 6
19
24
(4) Simplif y the f ollowing f ractions:
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A) 28
51
80 - 7
7
10
B) 16
113
143 - 5
2
13
C) 23
63
380 - 2
6
19
D) 27
127
154 - 5
2
11
(5) There are 290 students in a school, and they are collecting money to improve f acilit ies at the
school. The average contribution f rom each student was Rs. 165. If 9
50 of the amount was
spent in ref urbishing the gym, then how much money are they lef t with?
(6)Convert the 8
7
8 to improper f raction
(7)A f actory makes 9113 Watches every week. How many items are manuf actured in
7
13 year?
(8) The table below shows the number of books read by f ive children in one month.Name Number of books
Ismail ?
Balvinder 24
Sunil 9
Saina 10
Pranav 9
If Ismail read 3 books more than 1
4 of the total number of books read by the other f our
children, then how many books did Ismail read in a month?
Choose correct answer(s) from given choice
(9)Priyanka is reading a story book and this book has 343 pages. Priyanka manages to read
9
49
of the book every day. Af ter 4 days, how many pages does she have lef t to read?
a. 92 b. 91
c. 93 d. 280
(10)
1894
1000 in decimal f orm gives us
a. 1.894 b. 0.894
c. 18.94 d. 0.1894
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(11)The parking lot of the Mall has a capacity of 468 cars. On Saturday
8
18 of the parking lot was
occupied with cars. How many additional cars could park there?
a. 260 b. 252
c. 261 d. 259
Fill in the blanks
(12)
1
5 +
1
6 - 3
11
=
(13)The local community park has space allocated f or jogging, music, and sports.
14
34 of the park
has been allocated f or sports, and out of the sports area 7
14 has been given f or playing
f ootball. The ratio of the park is given f or playing f ootball is
.
(14) The parking lot of the Theater has a capacity of 2220 cars. On Thursdays the ratio of occupiedspots to empty spots in the parking lot was 12:25. The number of additional cars that could park
there is
(15) 23
25 of a century =
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Answers
(1) A)
8
12
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 12Number of parts that are shaded in the image = 8
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 8
12
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 8
12 .
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B)
3
6
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 6Number of parts that are shaded in the image = 3
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 3
6
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 3
6
.
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C)
7
9
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 9Number of parts that are shaded in the image = 7
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 7
9
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 7
9
.
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D)
4
7
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 7Number of parts that are shaded in the image = 4
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 4
7
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 4
7
.
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E)
2
6
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 6Number of parts that are shaded in the image = 2
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 2
6
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 2
6
.
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F)
8
12
Step 1
We have been asked to f ind the f raction that represents the shaded portion ofthe f ollowing shape:
Step 2
Total number of equal parts in the image = 12Number of parts that are shaded in the image = 8
Step 3
Fraction of parts of the image that are shaded = Number of shaded parts
Total number of parts
= 8
12
Step 4
Hence, the f raction that is represented by the shaded portion of the shape is 8
12 .
(2)
4
16
+
7
16
=
+ =
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(3) A) 19
179
702
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 1423
26 into an improper f raction, we will need to
multiply the denominator 26 by 14 and add the numerator 23 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
1423
26 =
14 × 26 + 23
26 =
387
26
Similarly, the mixed f raction 410
27 can be converted to an improper f raction as:
410
27 =
4 × 27 + 10
27 =
118
27
Step 2
We now need to add the f ollowing unlike f ractions: 387
26 and
118
27 .
Let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 26 and 27.The LCM of 26 and 27 is 702.
Step 4
What should we multiply with the denominator 26 to get the LCM 702?
It is 702
26 = 27.
So, the f irst f raction can be written as:
387 × 27
702 =
10449
702
Step 5
Similarly, the second f raction can be written as:
118 × 26
702 =
3068
702
Step 6
Let us now add the two like f ractions:
10449
702 +
3068
702
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= 10449 + 3068
702
= 13517
702
= 19179
702
Step 7
Thus, the sum of the f ractions 1423
26 and 4
10
27 is 19
179
702 .
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B) 15
61
216
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 117
8 into an improper f raction, we will need to
multiply the denominator 8 by 11 and add the numerator 7 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
117
8 =
11 × 8 + 7
8 =
95
8
Similarly, the mixed f raction 311
27 can be converted to an improper f raction as:
311
27 =
3 × 27 + 11
27 =
92
27
Step 2
We now need to add the f ollowing unlike f ractions: 95
8 and
92
27 .
Let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 8 and 27.The LCM of 8 and 27 is 216.
Step 4
What should we multiply with the denominator 8 to get the LCM 216?
It is 216
8 = 27.
So, the f irst f raction can be written as:
95 × 27
216 =
2565
216
Step 5
Similarly, the second f raction can be written as:
92 × 8
216 =
736
216
Step 6
Let us now add the two like f ractions:
2565
216 +
736
216
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= 2565 + 736
216
= 3301
216
= 1561
216
Step 7
Thus, the sum of the f ractions 117
8 and 3
11
27 is 15
61
216 .
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C) 12
11
31
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 826
31 into an improper f raction, we will need to
multiply the denominator 31 by 8 and add the numerator 26 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
826
31 =
8 × 31 + 26
31 =
274
31
Similarly, the mixed f raction 316
31 can be converted to an improper f raction as:
316
31 =
3 × 31 + 16
31 =
109
31
Step 2
We now need to add the f ollowing like f ractions: 274
31 and
109
31 .
Step 3
Let us add the two like f ractions:
274
31 +
109
31
= 274 + 109
31
= 383
31
= 1211
31
Step 4
Thus, the sum of the f ractions 826
31 and 3
16
31 is 12
11
31 .
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D) 17
107
108
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 147
12 into an improper f raction, we will need to
multiply the denominator 12 by 14 and add the numerator 7 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
147
12 =
14 × 12 + 7
12 =
175
12
Similarly, the mixed f raction 311
27 can be converted to an improper f raction as:
311
27 =
3 × 27 + 11
27 =
92
27
Step 2
We now need to add the f ollowing unlike f ractions: 175
12 and
92
27 .
Let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 12 and 27.The LCM of 12 and 27 is 108.
Step 4
What should we multiply with the denominator 12 to get the LCM 108?
It is 108
12 = 9.
So, the f irst f raction can be written as:
175 × 9
108 =
1575
108
Step 5
Similarly, the second f raction can be written as:
92 × 4
108 =
368
108
Step 6
Let us now add the two like f ractions:
1575
108 +
368
108
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= 1575 + 368
108
= 1943
108
= 17107
108
Step 7
Thus, the sum of the f ractions 147
12 and 3
11
27 is 17
107
108 .
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E) 13
238
345
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 1011
15 into an improper f raction, we will need to
multiply the denominator 15 by 10 and add the numerator 11 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
1011
15 =
10 × 15 + 11
15 =
161
15
Similarly, the mixed f raction 222
23 can be converted to an improper f raction as:
222
23 =
2 × 23 + 22
23 =
68
23
Step 2
We now need to add the f ollowing unlike f ractions: 161
15 and
68
23 .
Let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 15 and 23.The LCM of 15 and 23 is 345.
Step 4
What should we multiply with the denominator 15 to get the LCM 345?
It is 345
15 = 23.
So, the f irst f raction can be written as:
161 × 23
345 =
3703
345
Step 5
Similarly, the second f raction can be written as:
68 × 15
345 =
1020
345
Step 6
Let us now add the two like f ractions:
3703
345 +
1020
345
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= 3703 + 1020
345
= 4723
345
= 13238
345
Step 7
Thus, the sum of the f ractions 1011
15 and 2
22
23 is 13
238
345 .
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F) 21
127
312
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 148
13 into an improper f raction, we will need to
multiply the denominator 13 by 14 and add the numerator 8 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
148
13 =
14 × 13 + 8
13 =
190
13
Similarly, the mixed f raction 619
24 can be converted to an improper f raction as:
619
24 =
6 × 24 + 19
24 =
163
24
Step 2
We now need to add the f ollowing unlike f ractions: 190
13 and
163
24 .
Let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 13 and 24.The LCM of 13 and 24 is 312.
Step 4
What should we multiply with the denominator 13 to get the LCM 312?
It is 312
13 = 24.
So, the f irst f raction can be written as:
190 × 24
312 =
4560
312
Step 5
Similarly, the second f raction can be written as:
163 × 13
312 =
2119
312
Step 6
Let us now add the two like f ractions:
4560
312 +
2119
312
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= 4560 + 2119
312
= 6679
312
= 21127
312
Step 7
Thus, the sum of the f ractions 148
13 and 6
19
24 is 21
127
312 .
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(4) A) 20
15
16
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 2851
80 into an improper f raction, we will need to
multiply the denominator 80 by 28 and add the numerator 51 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
2851
80 =
28 × 80 + 51
80 =
2291
80
Similarly, the mixed f raction 77
10 can be converted into an improper f raction
as:
77
10 =
7 × 10 + 7
10 =
77
10
Step 2
We now need to subtract the f ollowing unlike f ractions: 2291
80 and
77
10 .
Bef ore we subtract, let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 80 and 10.The LCM of 80 and 10 is 80.
Step 4
What should we multiply with the denominator 80 to get the LCM 80?
It is 80
80 = 1.
So, the f irst f raction can be written as:
2291 × 1
80 × 1 =
2291
80
Step 5
Similarly, the second f raction can be written as:
77 × 8
10 × 8 =
616
80
Step 6
Let us now subtract the two like f ractions:
2291
80 -
616
80
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= 2291 - 616
80
= 1675
80
= 1675 ÷ 5
80 ÷ 5 ...[ Dividing both numerator and denominator by their HCF to bring ]
= 335
16
= 2015
16
Step 7
Thus, the simplif ied value of the given f ractions is 2015
16 .
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B) 11
7
11
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 16113
143 into an improper f raction, we will need to
multiply the denominator 143 by 16 and add the numerator 113 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
16113
143 =
16 × 143 + 113
143 =
2401
143
Similarly, the mixed f raction 52
13 can be converted into an improper f raction
as:
52
13 =
5 × 13 + 2
13 =
67
13
Step 2
We now need to subtract the f ollowing unlike f ractions: 2401
143 and
67
13 .
Bef ore we subtract, let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 143 and 13.The LCM of 143 and 13 is 143.
Step 4
What should we multiply with the denominator 143 to get the LCM 143?
It is 143
143 = 1.
So, the f irst f raction can be written as:
2401 × 1
143 × 1 =
2401
143
Step 5
Similarly, the second f raction can be written as:
67 × 11
13 × 11 =
737
143
Step 6
Let us now subtract the two like f ractions:
2401
143 -
737
143
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= 2401 - 737
143
= 1664
143
= 1664 ÷ 13
143 ÷ 13 ...[ Dividing both numerator and denominator by their HCF to bring
]
= 128
11
= 117
11
Step 7
Thus, the simplif ied value of the given f ractions is 117
11 .
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C) 20
17
20
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 2363
380 into an improper f raction, we will need to
multiply the denominator 380 by 23 and add the numerator 63 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
2363
380 =
23 × 380 + 63
380 =
8803
380
Similarly, the mixed f raction 26
19 can be converted into an improper f raction
as:
26
19 =
2 × 19 + 6
19 =
44
19
Step 2
We now need to subtract the f ollowing unlike f ractions: 8803
380 and
44
19 .
Bef ore we subtract, let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 380 and 19.The LCM of 380 and 19 is 380.
Step 4
What should we multiply with the denominator 380 to get the LCM 380?
It is 380
380 = 1.
So, the f irst f raction can be written as:
8803 × 1
380 × 1 =
8803
380
Step 5
Similarly, the second f raction can be written as:
44 × 20
19 × 20 =
880
380
Step 6
Let us now subtract the two like f ractions:
8803
380 -
880
380
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= 8803 - 880
380
= 7923
380
= 7923 ÷ 19
380 ÷ 19 ...[ Dividing both numerator and denominator by their HCF to bring
]
= 417
20
= 2017
20
Step 7
Thus, the simplif ied value of the given f ractions is 2017
20 .
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D) 22
9
14
Step 1
Let us f irst convert the given mixed f ractions into improper f ractions. In order to
convert the mixed f raction 27127
154 into an improper f raction, we will need to
multiply the denominator 154 by 27 and add the numerator 127 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:
27127
154 =
27 × 154 + 127
154 =
4285
154
Similarly, the mixed f raction 52
11 can be converted into an improper f raction
as:
52
11 =
5 × 11 + 2
11 =
57
11
Step 2
We now need to subtract the f ollowing unlike f ractions: 4285
154 and
57
11 .
Bef ore we subtract, let us f irst convert them to like f ractions.
Step 3
Let us f ind the LCM of the denominators 154 and 11.The LCM of 154 and 11 is 154.
Step 4
What should we multiply with the denominator 154 to get the LCM 154?
It is 154
154 = 1.
So, the f irst f raction can be written as:
4285 × 1
154 × 1 =
4285
154
Step 5
Similarly, the second f raction can be written as:
57 × 14
11 × 14 =
798
154
Step 6
Let us now subtract the two like f ractions:
4285
154 -
798
154
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= 4285 - 798
154
= 3487
154
= 3487 ÷ 11
154 ÷ 11 ...[ Dividing both numerator and denominator by their HCF to bring
]
= 317
14
= 229
14
Step 7
Thus, the simplif ied value of the given f ractions is 229
14 .
(5) Rs. 39237
Step 1
The number of students in the school = 290
Step 2
The average contribution f rom each student = Rs. 165Theref ore, the total contribution f rom all the students = 290 × 165 = Rs. 47850
Step 3
Now, the amount spent in ref urbishing the gym = 47850 × 9
50 = Rs. 8613
Step 4
The amount spent in ref urbishing the gym = Rs. 8613Theref ore, the money lef t with them = 47850 - 8613 = Rs. 39237.
(6) 71
8
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(7) 255164
Step 1
The number of weeks in a year = 52 weeks
Step 2
The number of weeks in 7
13 of a year =
7
13 × number of weeks in a year
= 7
13 × 52
= 28 weeks
Step 3
The number of Watches made by the f actory in one week = 9113
Step 4
The number of Watches manuf actured in 28 weeks ( 7
13 of a year) = 28 × 9113 = 255164.
(8) 16
Step 1
Total number of books read by Balvinder, Sunil, Saina and Pranav,= 24 + 9 + 10 + 9= 52
Step 2
It is given that Ismail reads 3 books more than 1
4 of 52. Theref ore number of books read
by Ismail,
= 1
4 × 52 + 3
= 13 + 3= 16 books
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(9) b. 91
Step 1
In one day she reads ( 9
49 × 343 = 63) pages
Step 2
Theref ore in 4 days, she would read (4 × 63 = 252) pages
Step 3
Theref ore remaining number of pages to read = (Total pages - Read pages)⇒ = 343 - 252⇒ = 91 pages
(10) a. 1.894
Step 1
Let us understand the def init ion of decimal f raction and decimals:Decimal fraction: is a f raction whose denominator is a power of ten such as 10, 100, 1000etc.A decimal number, or just decimal, ref ers to any number written in decimal notation. In anumber written in decimal notation, all digits to the lef t of the decimal point have f acevalues more than one and those on the right of the decimal point have f ace values lessthan one.
Step 2
We should remember f ollowing f acts about decimal numbers:Fact 1. When a decimal number is multiplied by ten, the result is obtained by moving thedecimal point to the right by one digit.Fact 2. When a decimal number is divided by ten, the result is obtained by moving thedecimal point to the lef t by one digit.Fact 3. When a number does not have a decimal point, we can assume that there is ahidden decimal point to the right of the unit digit (right-most digit).
Step 3
Now, let us try to convert the decimal f raction 1894/1000 to a decimal number.
Step 4
Here, the number 1894 is to be divided by 1000. In other words we can say that 1894 is tobe divided by 10 three times.
Step 5
In number 1894, the decimal point can be assumed to be hidden to the right of the unit digit.When it is divided by 10 three times, the decimal point will move three digits to the lef t.Hence the decimal number we get will be:1.894
Step 6
Theref ore, the answer is 1.894.
ID : in-6-Fractions [30]
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(11) a. 260
Step 1
Total number of parking spaces = 468
Step 2
Number of occupied parking spaces = 8
18 × 468
⇒ = 208
Step 3
Number of vacant parking spaces = Total - Occupied⇒ = 468 - 208⇒ = 260
ID : in-6-Fractions [31]
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(12) 118 315
Step 1
The given expression 1
5 +
1
6 - 3
11
can be simplif ied as:
1
5 +
1
6 × 11 - 3
11
= 1
5 +
1
66 - 3
11
= 1
5 +
1
63
11
= 1
5 +
11
63
= 1 × 63 + 11 × 5
315
= 63 + 55
315
= 118
315
Step 2
Thus, the value of 1
5 +
1
6 - 3
11
is 118
315 .
ID : in-6-Fractions [32]
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(13) 7 34
Step 1
: Let the area of the f ull park be 1.
Step 2
The area allocated f or sports = 14
34 of the total park area =
14
34 × 1 =
14
34
Step 3
The area allocated f or playing f ootball out of the sports area = 7
14 of the sports area
= 7
14 ×
14
34
= 7
34
Step 4
Thus the ratio of the park given f or playing f ootball is 7
34 .
(14) 1500
Step 1
It is given that ratio of occupied spots to empty spots is 12:25, which means if there are 37( = 12 + 25) spots, 12 are occupied and 25 are empty
Step 2
But is given that total number of parking spots are 2220. Theref ore total number of spotsare 60 times ( = 2220/37) of 37
Step 3
Theref ore available spots will also be 60 times of 25. Theref ore available parking spots,= 25 × 60 = 1500
ID : in-6-Fractions [33]
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(15) 92
Step 1
The numeric term f or century is 100.
Step 2
We need to f ind the number that is 23
25 of 100. That is, we need to f ind the value of
23
25
× 100.
Step 3
The value of 23
25 × 100 = 23 × 4 ... (Dividing 100 by 25)
= 92
Step 4
Hence, 23
25 of century is 92.
ID : in-6-Fractions [34]
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