NCERT Solution class 7 Mathematics EXERCISE- 2.5
CHAPTER – 2
EXERCISE – 2.5
Q1. Which is greater?
(i) 0.5 or 0.05
SOLUTION:-
= 0.5 > 0.05
(ii) 0.7 or 0.5
SOLUTION:-
= 0.7 > 0.5
(iii) 7 or 0.7
SOLUTION:-
= 7 > 0.7
(iv) 1.37 or 1.49
SOLUTION:-
= 1.37<1.49
(v) 2.03 or 2.30
SOLUTION:-
= 2.03< 2.30
(vi) 0.8 or 0.88
SOLUTION:-
= 0.8 < 0.88
Q2.Express as rupees using decimals:
(i) 7 paise
SOLUTION:-
= 0.07
(ii) 7 rupees7 paise
SOLUTION:-
= 7.07
(iii) 77 rupees 77 paise
SOLUTION:-
= 77.77
(iv) 50 paise
SOLUTION:-
= 0.50
(v) 235 paise
SOLUTION:-
= 2.35
Q3. (i) Express 5cm in meter and kilometer
SOLUTION:-
WE KONOW,
= 1 meter = 100cm
Then,
= 1 cm = (1/100) m
= 5 cm = (5/100)
= 0.05 m
Now,
= 1 km = 1000m
Then,
= 1m = (1/1000) km
= 0.05 m = (0.05/1000)
= 0.00005 km
(ii) Express 35 mm in cm , m and km
SOLUTION:-
We know that,
= 1cm = 10mm
Then,
= 1 mm = (1/10) cm
= 35mm= (35/10) cm
= 3.5 cm
And,
= 1 meter = 100 cm
Then,
= 1 cm = (1/100) m
= 3.5 cm = (3.5 /100) m
= ( 35/1000) m
= 0.035 m
Now,
= 1 km = 1000 m
Then,
= 1m = (1/1000) km
= 0.035 m = (0.035 /1000)
= 0.000035 km
Q4. Express in kg:
(i) 200g
SOLUTION:-
We know that ,
= 1 kg = 1000g
Then,
= 1g = (1/1000) kg
= 200g = (200/1000) kg
= (2/10)
= 0.2 kg
(ii) 3470 g
SOLUTION:-
We know that,
= 1 kg = 1000g
Then,
=1 g = (1/1000) kg
= 3470 g = (3470/1000)kg
= (3470/100)
= 3.470 kg
(iii) 4 kg 8 g
SOLUTION:-
We know that ,
= 1 kg = 1000g
Then,
= 1g = (1/1000) kg
= 4 kg 8 g = 4 kg + (8/1000) kg
= 4kg + 0.008
= 4.008 kg
Q5. Write the following decimal numbers in the expanded form :
(i) 20.03
SOLUTION:-
We have,
20.03 = ( 2 x 10) + ( 0 x 1) + (0 x (1/10)) + ( 3 x (1/100))
(ii) 2.03
SOLUTION:-
We have,
2.03 = ( 2 x 1) + ( 0 x ( 1/10)) + (3 x (1/100))
(iii) 200.03
SOLUTION:-
We have,
200.03 = ( 2 x 100) + ( 0 x 10) + ( 0x1) + ( 0 x (1/10)) + ( 3 x (1/100))
(iv) 2.034
SOLUTION:-
We have,
2.034 = (2 x1) + ( 0 x (1/10)) + ( 3 x (1/100)) + ( 4 x (1/1000))
Q6. Write the place value of 2 in the following decimal numbers :
(i) 2.56
SOLUTION:-
From the question, we observe that ,
The place value of 2 in 2.56 in ones
(ii) 21.37
SOLUTION:-
From the question , we observe that
The places value of 2 in 21.37 is tens
(iii) 10.25
SOLUTION:-
From the question, we obseve that ,
The place value of 2 in 10.25 is tenths.
(iv) 9.42
SOLUTION:-
From the observes that ,
The place value of 2 in 9.42 is hundredth
(v) 63.352
SOLUTION:-
From the question , we observe that,
The places value of 2 in 63.352 is thousandth.
Q7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C . Ayub went place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
SOLUTION:-
From the question, it is given that ,
Distance travelled by dinesh = AB + BC
= 7.5 + 12.7
= 20.2 KM
Dinesh travelled 20.2 km
Distance travelled by ayub = AD + DC
= 9.3 + 11.8
= 21.1 KM
Ayub travelled 21.1 km
Clearly ayub travelled more distance by = ( 21.1 + 20.2)
= 0.9 km
Ayub travelled 0.9 km more than Dinesh .
Q8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. SARALA bought 4 kg 800 g oranges and 4 kg 150 g bananas Who bought more fruits ?
SOLUTION:-
From the question , It is given that ,
Fruits bought by shyama = 5 kg 300 g
= 5 kg + ( 300/1000) kg
= 5 kg + 0.3 kg
= 5.3 kg
Fruits bought by sarala = 4 kg 800 g + 4 kg 150 g
= (4 + (800/1000) + (4 + (150/1000))
= ( 4+ 0.8) kg +(4 + 150) kg
= 4.8 kg + 4.150 kg
= 8.950 kg
So , sarala bought more fruits.
Q9. How much less is 28 km than 42.6 km?
SOLUTION:-
Now, we have to find the difference of 42.6 km and 28 km
= 42.6 – 28.0
= 14.6
14.6 km less is 28 km than 42.6 km.