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Fractions, Decimals and Rational Numbers > Exercise 2.1

Exercise 2.1

Write four more numbers in the following pattern: 13 , 26 , 39 , 412....

Solution:

Instruction

We have, 26 = 1x23x2 , 39 = 1x33x3
412 = 1x43x4
1x13x1 = /3, 1x23x2 = /6.
Thus, we observe a pattern in these numbers.

Exercise - 1

Compute and express the result as a mixed fraction:

(i)

2+34

=4×+34

=+34

=4

=

(ii)

79+13

=+3×9

=+9

=9

=

(iii)

147

=77

=

(iv)

223+12

=8×+23+12

=3+12

=+3×2

=6

=

(v)

5816

=3×4×24

=24

=

(vi)

223+312

=2×+23+2×+12

=3+2

=+6

=6

=

2. Arrange the following in ascending order.

(i) 58 , 56 , 12

Ascending Order

12
56
58

(ii) 25 , 13 , 310

Ascending Order

25
310
13

3. Check if the following is a magic square where the sum of numbers in each row, column, and diagonal is the same:

Sol:

Let's check each row, column, and diagonal:

Rows:

R1: 613+1313+213=++13=13

R2: 313+713+1113=++13=13

R3: 1213+113+813=++13=13

Columns:

C1: 613+313+1213=++13=13

C2: 1313+713+113=++13=13

C3: 213+1113+813=++13=13

Diagonals:

D1: 613+713+813=++13=13

D2: 1213+713+213=++13=13

Therefore, the sum of numbers in each row, column, and diagonal is equal to 13.

Hence, this is a magic square.

**{.m-red}4. Find the perimeter of a rectangular sheet of paper with length 5 2/3 cm and width 3 1/5 cm.__

Sol:

Let's first convert mixed numbers to improper fractions:

Length = 523 cm

=5×+3

=3 cm

Width = 315 cm

=3×+5

=5 cm

Perimeter = 2(length + width)

=2173+165

=2×+×3×5

=2+

=2×15

=15

=

Therefore, the perimeter of the rectangular sheet is 17 11/15 cm.

5. The recipe requires 3 1/7 cups of flour. Radha has 1 3/8 cups of flour. How many more cups of flour does she need?

Sol:

Let's first convert mixed numbers to improper fractions:

Required flour = 317 cups =3×+7 {.reveal(when="blank-1")}=7 cups

Radha has = 138 cups {.reveal(when="blank-2")}=1×+8 {.reveal(when="blank-4")}=8 cups

More flour needed = Required flour - Available flour

=227118

=××7×8

=

=56

= cups

Therefore, Radha needs 1 43/56 more cups of flour.

6. Abdul is preparing for his final exam. He has completed 5/12 part of his course content. Find out how much course content is left?

Sol:

Let's first convert to fraction:

Completed part = 512

Remaining part = 1 - Completed part

=1512

=12

=12

Therefore, 7/12 part of the course content is left.

7. Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in the figure. Which figure has greater perimeter and by how much?

Sol:

Let's first convert mixed numbers to improper fractions:

(i) For ΔABE:

First side = 225 cm

=2×+5

=5 cm

Second side = 213 cm

=2×+3

=3 cm

Third side = 412 cm

=4×+2

=2 cm

Perimeter of ΔABE = Sum of all sides

=125+73+92

=++

=30

= cm

(ii) For rectangle BCDE:

Length = 2 cm

Width = 125 cm

=1×+5

=5 cm

Perimeter of rectangle = 2(length + width)

=22+75

=2

=5

= cm

Difference in perimeter = Perimeter of ΔABE - Perimeter of rectangle BCDE

=11630345

=××30×5

=

=150

=15 cm

Therefore, the rectangle BCDE has a greater perimeter by 22/15 cm.