Solution NCERT Exercise 2.3 part2
Fractions and Decimals 7th Math
Solution NCERT Exercise 2.3 part2
Question (3) For the fractions given below:
(a) Multiply and reduce the product to lowest form (if possible)
(b) Tell whether the fraction obtained is proper or improper
(c) If the fraction obtained is improper then convert it into a mixed fraction
(i) `2/5xx5 1/4`
Solution :
Given, `2/5xx5 1/4`
`=2/5xx21/4`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(2xx21)/(5xx4)`
`=(cancel2xx21)/(5xx2xxcancel2)`
`=21/10`
This is an improper fraction, because numerator is greater than denominator.
`=2 1/10` Answer
(ii) `6 2/5xx7/9`
Solution :
Given, `6 2/5xx7/9`
`=32/5xx7/9`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(32xx7)/(5xx9)`
`=224/45`
This is an improper fraction, because numerator is greater than denominator.
`=4 44/45` Answer
(iii) `3/2xx5 1/3`
Solution :
Given,` 3/2xx5 1/3`
`=3/cancel2xx(cancel16 8)/3`
`=cancel3/1xx8/cancel3`
`=1/1xx8/1`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(1xx8)/(1xx1)`
`= 8/1`
This is an improper fraction, because numerator is greater than denominator.
= 8 Answer
(iv) `5/6xx2 3/7`
Solution :
Given, `5/6xx2 3/7`
`=5/6xx17/7`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(5xx17)/(6xx7)`
`=85/42`
This is an improper fraction, because numerator is greater than denominator.
`=2 1/42` Answer
(v) `3 2/5xx4/7`
Solution :
Given, `3 2/5xx4/7`
`= 17/5xx4/7`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(17xx4)/(5xx7)`
`=68/35`
This is an improper fraction, because numerator is greater than denominator.
`=1 33/35` Answer
(vi) `2 3/5xx3`
Solution :
Given, `2 3/5xx3`
`=13/5xx3`
We know that denominator is always 1, if there is no denominator for a number.
`=13/5xx3/1`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(13xx3)/(5xx1)`
`=39/5`
This is an improper fraction, because numerator is greater than denominator.
`= 7 4/5` Answer
(vii) `3 4/7xx3/5`
Solution :
Given, `3 4/7xx3/5`
`=25/7xx3/5`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(cancel25 5xx3)/(7xxcancel5)`
`=15/7`
This is an improper fraction, because numerator is greater than denominator.
`=2 1/7` Answer
Question (4) Which is greater
(i) `2/7` of `3/4` or `3/5` of `5/8`
Solution :
Given, to find greater `2/7` of `3/4` or `3/5` of `5/8`
We know that, the word "of" stands for multiplication in mathematics, thus after replacing "of" with sign of multiplication (×), we get
`=cancel2/7xx3/(cancel2 xx2)` or `3/cancel5xxcancel5/8`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(1xx3)/(7xx2)` or `(3xx1)/(1xx8)`
`=3/14` or `3/8`
Now, LCM of denominators of given fractions = 56
To compare given fractions, their denominators are made equal after multiplying numerators and denominators with suitable numbers.
`=(3xx4)/(14xx4)` or `(3xx7)/(8xx7)`
`=12/56` or `21/56`
Thus, clearly, `=12/56 < 21/56`
This means `21/56` is greater.
Thus, `3/5` of `5/8` i.e. `3/8` is greater Answer
(ii) `1/2` of `6/7` or `2/3` of `3/7`
Solution :
Given, `1/2` of `6/7` or `2/3` of `3/7`
Thus, greater = ?
We know that, the word "of" stands for multiplication in mathematics, thus after replacing "of" with sign of multiplication (×), we get
`=1/cancel2xx(cancel6 3)/7` or `2/cancel3xxcancel3/7`
`=1/1xx3/7` or `2/1xx1/7`
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(1xx3)/(1xx7)` or `(2xx1)/(1xx7)`
`=3/7` or `2/7`
Clearly, `3/7` is greater.
Thus, `1/2` of `6/7` i.e. `3/7` is greater Answer
Question (5) Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is `3/4` m. Find the distance between the first and the last sapling.
Solution :
Number of plants saplings = 4
Distance between two adjacent saplings `=3/4` m
Thus, distance between first and last sapling = ?
Now, since there are 4 saplings, thus
Distance between 1st and 4th (last) sapling = distance between 1st and 2nd sapling + distance between 2nd and 3rd sapling + distance between 3rd and 4th sapling
`=3/4 m+3/4 m+3/4 m`
`=(3+3+3)/4` m
`=9/4` m
Thus, distance between 1st and last sapling `=2 1/4` m Answer
Question (6) Lipika reads a book for `1 3/4` hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution :
Given, Number of hours read by Lipika every day`=1 3/4` hours
Number of days taken by Lipika to finish the book = 6 days
Thus, total hours in all required to finish the book = ?
Total number of hours required by Lipika to finish the book = Hours every day × Number of days
`=1 3/4xx6` hours
`= 7/4xx6` hours
We know that denominator is always 1, if there is no denominator for a number.
`= 7/4xx6/1` hours
Now, to multiply two or more fractions numerator is multiplied with numerator and denominator is multiplied with denominator.
`=(7xx6)/(4xx1)` hours
`=(7xx3xxcancel2)/(cancel2xx2xx1)`
`=21/2`
`=10 1/2` hours Answer
Question (7) A car runs 16 km using 1 litre of petrol. How much distance will it cover using `2 3/4` litres of petrol.
Solution :
Given, distance cover in 1 litre of petrol = 16 km
Total petrol `=2 3/4` litre
`=11/4` litre
Thus, distance cover in given petrol = ?
Since, In 1 litre of petrol distance cover = 16 km
Thus, in `11/4` litre of petrol distance cover `=16xx11/4` km
`=cancel4xx4xx11/cancel4` km
= 11 × 4 km = 44 km
Thus, distance covered in given amount of petrol = 44 km Answer
Question (8) (a) (i) Provide the number in box `square`, such that `2/3xxsquare=10/30`
Solution :
Given, `2/3xxsquare=10/30`
`= 2/3xxsquare=(2xx5)/(3xx10)`
`= 2/3xxsquare= 2/3xx5/10`
Thus, number in the box `=5/10` Answer
(ii) The simplest form of the number obtained in `square` is _________
Solution :
The missing number in square `=5/10`
`=cancel5/(2xxcancel5)`
`=1/2` Answer
(b) (i) Provide the number in the box `square`, such that `3/5xxsquare=24/75` ?
Solution :
Given, `3/5xxsquare=24/75`
=`3/5xxsquare=(3xx8)/(5xx25)`
`=3/5xx square=3/5xx8/25`
Thus, number in the square `=8/25`
(ii) The simplest form of the number obtained in `square` is ______.
Solution :
Number obtained in `8/25`
As there is no common factor in numerator and denominator other than 1 in the obtained fraction.
Thus, the simplest form is `8/25` Answer
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