NCERT Solution class 7 Mathematics EXERCISE- 2.1

NCERT Solution class 7 Mathematics EXERCISE- 2.1

                             CHAPTER – 02

                         EXERCISE – 2.1

Q1. SOLVE:

(i)2 – (3/5)

SOLUTION:-

For subtraction of the two unlike fractions , first change them to the like fractions.

LCM of 1,5 = 5

Now let us change each of the given fraction into an equivalent fraction having 5 as the denominator .

= [(2/1)] x (5/5)] = (10/5)

= [(3/5) x (1/1)] = (3/5)

Now,

= (10/5) – (3/5)

= [(10- 3 ) /5]

= (7/5)

(ii) 4 + (7/8)

S0LUTON:-

For addition of two unlike fractions first change them to like the fractions.

LCM OF 1,8 = 8

Now,

let us change each of the given fraction into an equivalent fraction having 8 as the denominator

= [(4/1) x (8/8)] = (32/8)

= [ (7/8) x (1/1)] = (7/8)

Now ,

= (32 /8) + (7/8)

= [(32+7) /8]

= (39/8)

= 4.87

(iii) (3/5) + (2/7)

SOLUTION:-

For the addition of two unlike fractions first change them to the like fractions .

LCM of 5,7=35

Now  let  us change of each of the given fraction into an equivalent fraction having 35 as the denominator.

= [(3/5) x (7/7)] = (21/35)

= [ (2/7) x (5/5)] = (10/35)

Now,
= (21/35) + (10/35)

= [(21+10)/35]

= (31/35)

(iv) (9/11) – (4/15)

SOLUTION:-

For the subtraction of two  unlike fractions first change them to the like fractions.

LCM OF 11,15 = 165

Now, let us change each of the given fraction into an equivalent fraction having 165 as the denominator

= [(9/11) x (15/15)]= (135/165)

= [(4/15) x (11/11] = (44/165)

Now,

= (135/165) – (44/165)

= [(135 – 44)/165]

= (91/165)

(v) (7/10) + (2/5) + (3/2)

SOLUTION:-

For the addition of two unlike fractions, first change them to the like fractions,

LCM of 10 , 5 , 2 = 10

Now let us change each of the given fraction into an equivalent fraction having 35 as the denominator.

= [(7/10) x (1/1)] = (7/10)

= [ (2/5) x (2/2)] = (4/10)

= [(3/2) x (5/5)] = (15/10)

Now,

= (7/10) + (4/10) + (15/10)

=[(7+4+15) / 10]

= (26/10)

= (13/5)

= 2.6

(vi)   8/3 + 7/2

SOLUTION:-

For addition of two unlike fractions, first change them to the like fractions,

LCM OF 3,2 = 6

Now let us change each of the given fraction into an equivalent fraction having 6 as the denominator .

= [(8/3) x (2/2) ] = (16/6)

= [(7/2) x (3/3)] = (21/6)

Now,

= (16/6) + (21/6)

= [(16 + 21)/6]

= (37/6)

= 6.1

(vii)  17/2 – 29/8

SOLUTION:-

For the subtraction of two unlike fraction first change  them to the like fractions

LCM of 2,8= 8

Now let us change each of the given fraction into an equivalent fraction having 35 as the denominator.

= [(17/2) x (4/4)] = (68/8)

=[(29/8) x (1/1)] = ( 29/8)

Now,

= (68/8) –(29/8)

=[(68 – 29)/8]

= (39/8)

= 4.8

Q2. Arrange the following in descending order:

SOLUTION:-

(i) 2/9 , 2/3 , 8/21

SOLUTION:-

LCM of 9,3,21 = 63

Now let us change each of the given fraction into an equivalent fraction having 63 as the denominator

[(2/9) x (7/7)] = (14/63)

[(2/3) x (21/21)] = (42/63)

[(8/21) x (3/3)] = (24/63)

Clearly

(42/63) > (24/63) > (14/63)

Hence,

(2/3) > ( 8/21) > (2/9)

Hence the given fraction in descending order are (2/3) , (8/21) , (2/9)

 

(ii) 1/5 , 3/7 , 7/10

SOLUTION:-

LCM OF 5,7,10 = 70

Now let us change each of the given fraction into an equivalent fraction having 70 as the denominator

[(1/5) x (14/14)] = (14/70)

[(3/7) x(10/10)] = (30/70)

[(7/10) x (7/7)] = (49/70)

Clearly

(49/70) > ( 30/70) > ( 14/70)

Hence,

(7/10) > ( 3/7) > (1/5)

Hence the given fraction in descending order are  (7/10) (3/7) (1/5)

 

Q3. In a magic square the sum of the numbers in each column and along the diagonal is the same. is this a magic square ?

SOLUTION:-

4/11 9/11 2/11
3/11 5/11 7/11
8/11 1/11 6/11

 

SOLUTION:-

Sum along the first row = (4/11) + (9/11) + (2/11) = (15/11)

sum along the second row = (3/11) + (5/11) + (7/11) = (15/11)

Sum along the third row = (8/11) + (1/11) + (6/11) = (15/11)

Sum along the first column = (4/11) + (3/11) + (8/11)= (15/11)

Sum along the second column = (9/11) + (8/11) + (1/11) = (15/11)

Sum along the third column = (2/11) + (7/11) + (6/11) = (15/11)

Sum along the first diagonal = (4/11) + (5/11) + ( 6/11) = (15/11)

Sum along the second diagonal  = ( 2/11) + (5/11) + (8/11) =  (15/11)

Yes the sum of the numbers in each column and along the diagonal is the same so it is magic square

Q4. A rectangular sheet of paper is 12.5 cm long and 102/3 cm wide find its perimeter

SOLUTION:-

From the question it is  given that

Length = 12.5 cm

Breadth = 32/3 cm

We know that

Perimeter of the rectangle = 2 x ( length + breadth)

= 2 x [(25/2) + (32/3)]

= 2 x {[(25 x 3 )+(32 x 2 )]/6}

= 2 x [(75+64)/6]

= 2 x [139/6]

= 139/3 cm

Hence the perimeter of the sheet of  paper is 461/3  cm.

Q5. Find the perimeter of (i) Triangle ABE (ii) the rectangle BCDE in this figure whose perimeter is greater

SOLUTION:-

From the figure

AB = 5/2 cm

AE = 18/5

BE = 11/4

ED = 7/6

(i) We know that

Perimeter of the triangle = Sum of all sides

Then,

Perimeter of triangle ABE =  AB+ BE + EA

= (5/2) + (11/4) + (18/5)

The LCM of 2,4,5,= 20

Now let us change each of the given fraction into an equivalent fraction having 20 as the denominator

= {[(5/2) x (10/10)] + [(11/4) x (5/5)] + [18/5) x (4/4)]}

= (50/20) + (55/20) + (72/20)

= ( 50+55+72)/20

= 177/20

= 817/20

(ii) Now we have to find the perimeter of the rectangle

We know that

Perimeter of the rectangle = 2 x ( length + breadth)

Then,

Perimeter of rectangle BCDE = 2 x ( BE + ED)

= 2 X [(11/4) + (7/6)]

The LCM of 4,6 = 12

Now let us change each of the given fraction into an equivalent fraction having 20 as the denominator

= 2 x {[(11/4) x (3/3) ] + [(7/6)x (2/2)]}

= 2 x [ (33/12) + (14/12)]

= 2 x (47/12)

= 47/6

Finally we have find which one is having greater perimeter

Perimeter of triangle ABE = (177/20)

Perimeter of rectangle BCDE = (47/6)

The two perimeters are in the form of unlike fractions

Changing perimeters into like fractions we have

(177/20) =(177/20) x (3/3) =531/60

(43/6) = (43/6) x (10/10) = 430/60

CLEARLY (531/60) > (430/60)

Hence,(177/20) > (43/6)

Perimeter of triangle ABE> Perimeter of rectangle ( BCDE)

Q6. Sail wants to put a picture in a frame . the picture is 75/3 cm wide to fit in the frame the picture cannot be more than 710/3 cm wide how much should the pictured be trimmed?

SOLUTION:-

From the question it is given that

picture having a width of = 38/5 cm

frame have a width of = 73/10

the picture should be trimmed by = [(38/5) – (73/10)]

= The LCM of  5,10 = 10

Now let us change each of the given fraction into an equivalent fraction having 10 as the denominator.

= [(38/5) x (2/2)] – [(73/110) x (1/1)]

= (76/10) – (73/10)

= (76 – 73)/10

= 3/10 cm

Q7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother somu. how much part of the apple did somu eat? who had the large share? by how much?

SOLUTION:-

From the question it is the given that

part of apple eaten by ritu is = (3/5)

part of apple eaten by somu is = 1- part of apple eaten by ritu

= 1 –(3/5)

The LCM OF 1,5 = 5

Now let us change each of the given fraction into an equivalent fraction having 10 as the denominator.

= [(1/1) x (5/5)] – [(3/5) x (1/1)]

= (5/5) – (3/5)

= (5-3) /5

= 2/5

part of apple eaten by somu is (2/5)

so, (3/5)> (2/5) hence ritu ate larger size of apple

now, the difference between the 32 shares = (3/5) – (2/5)

= (3-2) /5

= 1/5

thus, ritu shares is larger than share of somu by (1/5)

Q8.  Michael finished colouring a picture in (7/12) hour vaibhav finished colouring the same picture in ¾ hours . who worked longer? by what fraction was it longer?

SOLUTION:-

From the question it is the given that time taken by the Michael to colour the picture is = (3/4)

The LCM of 12, 4 = 12

Now let us change each of the given fraction into an equivalent fraction having 12 as the denominator

(7/12) = (7/12) x (1/1) = 7/12

(3/4) = (3/4) x (3/3) = 9/12

Clearly (7/12) < (9/12)

Hence (7/12) < (3/4)

Thus vaibhav worked longer time by = (3/4) – (7/12)

= (9/12) – (7/12)

= (9-7)/12

= (2/12)

= (1/6) of an hour.

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