NCERT Exemplar Class 7 Maths Chapter 2 Fractions and Decimals are part of NCERT Exemplar Class 7 Maths. Here we have given NCERT Exemplar Class 7 Maths Solutions Chapter 2 Fractions and Decimals.
NCERT Exemplar Class 7 Maths Solutions Chapter 2 Fractions and Decimals
Directions: In questions 1 to 20, out of four options, only one is correct. Write the correct answer.
Question 1.
(frac{2}{5} times 5 frac{1}{5}) is equal to
Solution:
(b) (frac{2}{5} times 5 frac{1}{5}=frac{2}{5} times frac{26}{5}=frac{52}{25})
Question 2.
(3 frac{3}{4} div frac{3}{4}) is equal to
Solution:
(c) (3 frac{3}{4} div frac{3}{4}=frac{15}{4} div frac{3}{4}=frac{15}{4} times frac{4}{3}=5)
Question 3.
A ribbon of length (5 frac{1}{4}) m is cut into small pieces each of length (frac{3}{4}) m. Number of pieces will be:
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
Question 4.
The ascending arrangement of (frac{2}{3}, frac{6}{7}, frac{13}{21}) is:
Solution:
(b) The given numbers are (frac{2}{3}, frac{6}{7}, frac{13}{21})
LCM of denominators, i.e., 3, 7, 21 is 21.
Question 5.
Reciprocal of the fraction (frac{2}{3})is:
(a) 2
(b) 3
(c) (frac{2}{3})
(d) (frac{3}{2})
Solution:
(d) Reciprocal of the fraction (frac{2}{3} text { is } frac{3}{2})
Question 6.
The product of (frac{11}{13}) and 4 is:
Solution:
(a) (frac{11}{13} times 4=frac{11}{13} times frac{4}{1}=frac{44}{13}=3 frac{5}{13})
Question 7.
The product of 3 and (4 frac{2}{5}) is:
Solution:
(c) (3 times 4 frac{2}{5}=frac{3}{1} times frac{22}{5}=frac{66}{5}=13 frac{1}{5})
Question 8.
Pictorial representation of (3 times frac{2}{3}) is;
Solution:
(b) (3 times frac{2}{3}=frac{2}{3}+frac{2}{3}+frac{2}{3})
Hence, pictorial representation is
Question 9.
(frac{1}{5} div frac{4}{5}) is equal to:
Solution:
(d) (frac{1}{5} div frac{4}{5}=frac{1}{5} times frac{5}{4}=frac{1}{4})
Question 10.
The product of 0.03 × 0.9 is:
(a) 2.7
(b) 0.27
(c) 0.027
(d) 0.0027
Solution:
(c) (0.03 times 0.9=frac{3}{100} times frac{9}{10}=frac{27}{1000}=0.027)
Question 11.
(frac{5}{7} div 6) is equal to:
Solution:
(b) (frac{5}{7} div 6=frac{5}{7} times frac{1}{6}=frac{5}{42})
Question 12.
(5 frac{1}{6} div frac{9}{2}) is equal to:
Solution:
(d) (begin{aligned} 5 frac{1}{6} div frac{9}{2}=frac{31}{6}+frac{9}{2} &=frac{31}{6} times frac{2}{9} \ &=frac{31}{27} end{aligned})
Question 13.
Which of the following represents (frac{1}{3} text { of } frac{1}{6} ?)
Solution:
(c) (frac{1}{3} text { of } frac{1}{6}=frac{1}{3} times frac{1}{6})
Question 14.
(frac{3}{7} text { of } frac{2}{5}) is equal to
Solution:
(d) (frac{3}{7} text { of } frac{2}{5}=frac{3}{7} times frac{2}{5}=frac{6}{35})
Question 15.
One packet of biscuits requires (2 frac{1}{2}) cups of flour and (1 frac{2}{3}) cups of sugar. Estimated total quantity of both ingredients used in 10 such packets of biscuits will be
(a) less than 30 cups
(b) between 30 cups and 40 cups
(c) between 40 cups and 50 cups
(d) above 50 cups
Solution:
(c) : The requirement of flour and sugar for one packet of biscuits
∴ The requirement of both ingredients for 10 packets
i.e., between 40 cups and 50 cups
Question 16.
The product of 7 and (6 frac{3}{4}) is:
Solution:
(b) (7 times 6 frac{3}{4}=frac{7}{1} times frac{27}{4}=frac{189}{4}=47 frac{1}{4})
Question 17.
On dividing 7 by (frac{2}{5}), the result is
Solution:
(d) (7 div frac{2}{5}=7 times frac{5}{2}=frac{35}{2})
Question 18.
(2 frac{2}{3} div 5) is eQuestionual to
Solution:
(a) (2 frac{2}{3}+5=frac{8}{3}+5=frac{8}{3} times frac{1}{5}=frac{8}{15})
Question 19.
(frac{4}{5}) of 5 kg apples were used on Monday. The next day (frac{1}{3}) of what was left was used. Weight (in kg) of apples left now is
Solution:
(c): Weight of apples used on Monday
(=frac{4}{5} text { of } 5 mathrm{kg}=frac{4}{5} times 5 mathrm{kg}=4 mathrm{kg})
Weight of apples left on next day = (5 – 4) kg
= – 1 kg
Weight of apples used on next day
(=frac{1}{3} text { of } 1 mathrm{kg})
(=frac{1}{3} times 1 mathrm{kg}=frac{1}{3} mathrm{kg})
∴ Weight of apples left now (=left(1-frac{1}{3}right) mathrm{kg})
(=frac{2}{3} mathrm{kg})
Question 20.
The picture
interprets
Solution:
((b) : frac{1}{4}+frac{1}{4}+frac{1}{4}=3 times frac{1}{4}=frac{3}{4})
Directions: In questions 21 to 44, fill in the blanks to make the statements true.
Question 21.
Rani ate (frac{2}{7}) part of a cake while her brother Ravi ate (frac{4}{5}) of the remaining. Part of the cake left is _________
Solution:
Let whole part of the cake is 1.
Part of the cake Rani ate (=frac{2}{7}) of 1 i.e., (frac{2}{7})
∴ Part of the cake left (=1-frac{2}{7}=frac{5}{7})
Now, part of the cake Ravi ate (=frac{4}{5} times frac{5}{7}=frac{4}{7})
Thus, part of the cake left (=frac{5}{7}-frac{4}{7}=frac{1}{7})
Question 22.
The reciprocal of (frac{3}{7}) is _____
Solution:
(frac{7}{3} 🙂 Reciprocal of (frac{3}{7} text { is } frac{7}{3}) is (frac{7}{3})
Question 23.
(frac{2}{3}) of 27 is
Solution:
(18 : frac{2}{3} text { of } 27=frac{2}{3} times 27=2 times 9=18)
Question 24.
(frac{4}{5}) of 45 is _____
Solution:
(36 : frac{4}{5} text { of } 45=frac{4}{5} times 45=4 times 9=36)
Question 25.
(4 times 6 frac{1}{3}) is equal to
Solution:
(frac{76}{3} text { or } 25 frac{1}{3} : 4 times 6 frac{1}{3}=4 times frac{19}{3}=frac{76}{3}=25 frac{1}{3})
Question 26.
(frac{1}{2} text { of } 4 frac{2}{7}) is ______
Solution:
(frac{15}{7} text { or } 2 frac{1}{7} : frac{1}{2} text { of } 4 frac{2}{7}=frac{1}{2} times 4 frac{2}{7}=frac{1}{2} times frac{30}{7})
(=frac{15}{7}=2 frac{1}{7})
Question 27
(frac{1}{9} text { of } frac{6}{5}) is _____
Solution:
(frac{2}{15} : frac{1}{9} text { of } frac{6}{5}=frac{1}{9} times frac{6}{5}=frac{1}{3} times frac{2}{5}=frac{2}{15})
Question 28.
The lowest form of the product (2 frac{3}{7} times frac{7}{9}) is _______
Solution:
(frac{17}{9} text { or } 1 frac{8}{9} : 2 frac{3}{7} times frac{7}{9}=frac{17}{7} times frac{7}{9}=frac{17}{9}=1 frac{8}{9})
Question 29.
(frac{4}{5} div 4) is equal to _____
Solution:
(frac{1}{5} : frac{4}{5} div 4=frac{4}{5} times frac{1}{4}=frac{1}{5})
Question 30.
(frac{2}{5} text { of } 25) is ____
Solution:
(10 : frac{2}{5} text { of } 25=frac{2}{5} times 25=2 times 5=10)
Question 31.
(frac{1}{5} div frac{5}{6}=frac{1}{5} frac{6}{5})
Solution:
(x : frac{1}{5} div frac{5}{6}=frac{1}{5} times frac{6}{5})
Question 32.
3.2 × 10 = ______
Solution:
(32 : 3.2 times 10=frac{32}{10} times 10=32)
Question 33.
25.4 × 1000 = _____
Solution:
(25400 : 25.4 times 1000=frac{254}{10} times 1000)
= 25400
Question 34.
93.5 × 100 = _____
Solution:
(9350 : 93.5 times 100=frac{935}{10} times 100=9350)
Question 35.
4.7 ÷ 10 = _____
Solution:
(0.47 : 4.7+10=frac{47}{10} div 10=frac{47}{10} times frac{1}{10})
(=frac{47}{100}=0.47)
Question 36.
4.7 ÷ 100 = ______
Solution:
(0.047 : 4.7 div 100=frac{47}{10} div 100=frac{47}{10} times frac{1}{100})
(=frac{47}{1000}=0.047)
Question 37.
4.7 ÷ 1000 = ____
Solution:
(0.0047 : 4.7+1000=frac{47}{10} div 1000)
(=frac{47}{10} times frac{1}{1000}=frac{47}{10000}=0.0047)
Question 38.
The product of two proper fractions is _____than each of the fractions that are multiplied.
Solution:
Less
Question 39.
While dividing a fraction by another fraction, we _____ the first fraction by the _______ of the other fraction.
Solution:
Multiply, reciprocal
Question 40.
8.4 ÷ = 2.1
Solution:
4: Let x be the required number.
∴ 8.4 ÷ x = 2.1
(Rightarrow 8.4 times frac{1}{x}=2.1 Rightarrow x=frac{8.4}{2.1}=frac{84}{21}=4)
Question 41.
52.7 ÷ _____ = 0.527
Solution:
100: Let x be the required number.
∴ 52.7 ÷ x = 0.527
(Rightarrow quad 52.7 times frac{1}{x}=0.527)
(Rightarrow quad x=frac{52.7}{0.527}=frac{527}{10} times frac{1000}{527}=100)
Question 42.
0.5 ___ 0.7 = 0.35
Solution:
(x : 0.5 times 0.7=frac{5}{10} times frac{7}{10}=frac{35}{100}=0.35)
Question 43.
2 _____ (frac{5}{3}=frac{10}{3})
Solution:
(x : 2 times frac{5}{3}=frac{10}{3})
Question 44.
2.001 ÷ 0.003 = ______
Solution:
(667 : 2.001 div 0.003=frac{2001}{1000} div frac{3}{1000})
(=frac{2001}{1000} times frac{1000}{3}=667)
Directions: In each of the questions 45 to 54, state whether the statement is True or False.
Question 45.
The reciprocal of a proper fraction is a proper fraction.
Solution:
False
The reciprocal of a proper fraction is an improper fraction.
Question 46.
The reciprocal of an improper fraction is an improper fraction.
Solution:
False
The reciprocal of an improper fraction is a proper fraction.
Question 47.
(begin{array}{c}{text { Product of two fractions }} \ {=frac{text { Product of their denominators }}{text { Product of their numerators }}}end{array})
Solution:
False
Question 48.
The product of two improper fractions is less than both the fractions.
Solution:
False
The product of two improper fractions is greater than both the fractions.
Question 49.
A reciprocal of a fraction is obtained by inverting it upside down.
Solution:
True
Question 50.
To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.
Solution:
True
Question 51.
To divide a decimal number by 100, we move the decimal point in the number to the left by two places.
Solution:
True
Question 52.
1 is the only number which is its own reciprocal.
Solution:
True
Question 53.
(frac{2}{3}) of 8 is same as (frac { 2 }{ 3 } div 8).
Solution:
False
Question 54.
The reciprocal of is
Solution:
False
The reciprocal of is a (frac{4}{7} text { is } frac{7}{4})
Question 55.
If 5 is added to both the numerator and the denominator of the fraction (frac{5}{9}), will the value of the fraction be changed? If so, will the value increase or decrease?
Solution:
We have, (frac{5}{9} text { and } frac{5+5}{9+5}=frac{10}{14}=frac{5}{7})
Here, (frac{5}{9} neq frac{5}{7})
Yes, the value of the fraction is changed.
Since, 9 > 7 : (therefore frac{5}{9}
Hence, the value is increased.
Question 56.
What happens to the value of a fraction if the denominator of the fraction is decreased while the numerator is kept unchanged?
Solution:
The value of a fraction will increase if the denominator of the fraction is decreased while the numerator is kept unchanged.
Question 57.
Which letter comes (frac{2}{5}) of the way among A and J ?
Solution:
Total letters from A to , are 10 and (frac{2}{5} text { of } 10=frac{2}{5} times 10=2 times 2=4)
Now, D comes at 4th position.
∴ D comes at (frac{4}{10}) i.e., (frac{2}{5}) of the way among A and J.
Question 58.
If (frac{2}{3}) of a number is 10, then what is 1.75 times of that number?
Solution:
Let the number be x.
Question 59.
In a class of 40 students of the total number of students like to eat rice only, of the total number of students like to eat chapati only and the remaining students like to eat both. What fraction of the total number of students like to eat both?
Solution:
Total number of students = 40
Number of students who like to eat rice only of (=frac{1}{5} text { of } 40=frac{1}{5} times 40=8)
Number of students who like to eat chapati only (=frac{2}{5} text { of } 40=frac{2}{5} times 40=16)
Number of students who like to eat both rice and chapati = 40 – (8 + 16) = 16
∴ Required fraction (=frac{16}{40}=frac{2}{5})
Question 60.
Renu completed (frac{2}{3}) part of her homework in 2 hours. How much part of her homework had she completed in (1 frac{1}{4}) hours?
Solution:
Let whole homework be denoted as x.
The part of home work completed by Renu is 2 hours (=frac{2}{3} x)
∴ The part of home work completed by her in 1 hour (=frac{2}{3} x+2=frac{2}{3} x times frac{1}{2}=frac{x}{3})
So, the part of home work completed by her in (1 frac{1}{4}) hours (=frac{x}{3} times 1 frac{1}{4}=frac{x}{3} times frac{5}{4}=frac{5 x}{12})
Hence, she completed (frac{5}{12}) part of her home work in (1 frac{1}{4}) hours.
Question 61.
Reemu read the (frac{1}{5})th pages of a book. If she reads further 40 pages, she would have read (frac{7}{10} mathrm{th}) pages of the book. How many pages are left to be read?
Solution:
Let total number of pages be x.
Number of pages read by Reemu (=frac{1}{5} x)
According to question,
Question 62.
Write the number in the box (square ) such that (frac{3}{7} times square=frac{15}{98})
Solution:
Let the number in the box be x.
(therefore frac{3}{7} times x=frac{15}{98} Rightarrow x=frac{15}{98} div frac{3}{7})
(Rightarrow x=frac{15}{98} times frac{7}{3}=frac{5}{14})
∴ The required number is (frac{5}{14})
Question 63.
Will the quotient (7 frac{1}{6}+3 frac{2}{3}) be a fraction greater than 1.5 or less than 1.5? Explain.
Solution:
We have, (7 frac{1}{6} div 3 frac{2}{3}=frac{43}{6} div frac{11}{3})
(=frac{43}{6} times frac{3}{11}=frac{43}{22}=1.95)
Since, 1.95 > 1.5
∴ The quotient of (7 frac{1}{6} div 3 frac{2}{3}) is a fraction greater than 1.5.
Question 64.
Describe two methods to compare (frac{13}{17}) and 0.82. Which do you think is easier and why?
Solution:
Method I: Conversion of both numbers (frac{13}{17}) and 0.82 into decimals
(frac{13}{17}=0.76 text { and } 0.82)
Method II: Conversion of both numbers (frac{13}{17}) and 0.82 into fractions
(0.82=frac{82}{100}=frac{41}{50} text { and } frac{13}{17})
∴ Method of converting numbers into decimals is easier because we can easily compare the numbers in this method but in other method we will also convert the fractions into like fractions to compare the numbers.
Question 65.
Health: The directions for a pain reliever recommend that an adult of 60 kg and over take 4 tablets every 4 hours as needed, and an adult who weighs between 40 and 50 kg take only (2 frac{1}{2}) tablets every 4 hours as needed. Each tablet weighs (frac{4}{25}) gram.
(a) If a 72 kg adult takes 4 tablets, how many grams of pain reliever is he or she receivings?
(b) How many grams of pain reliever is the recommended dose for an adult weighing 46 kg?
Solution:
(a) Each tablet weighs (frac{4}{25}) gram
∴Weight of 4 tablets (=frac{4}{25} times 4) gram
(=frac{16}{25} mathrm{gram})
(b) An adult weighing 46 kg takes only (2 frac{1}{2}) tablets.
∴ weight of (2 frac{1}{2}) tablets = (=frac{4}{25} times 2 frac{1}{2}) gram
(=frac{4}{25} times frac{5}{2} g r a m=frac{2}{5} g r a m)
Question 66.
Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals?
(a) a 18 kg adult dog
(b) a 6 kg cat
(c) a 18 kg pregnant dog
Solution:
(a) Total dosage given to a 18 kg adult dog (=left( frac { 1 }{ 2 } +frac { 1 }{ 2 } right) { tsp }=1{ tsp })
(b) Total dosage given to a 6 kg cat
(=6 times frac{1}{4} mathrm{tsp}=frac{3}{2} mathrm{tsp}=1 frac{1}{2} mathrm{tsp})
(c) Total dosage given to a 18 kg pregnant dog
(=frac{1}{2} times frac{1}{4.5} times 18 text { tsp }=9 times frac{10}{45} mathrm{tsp}=2)
Question 67.
How many (frac{1}{16}) kg boxes of chocolates can be made with (1 frac{1}{2}) kg chocolates?
Solution:
(frac{1}{16} ) kg chocolates to be filled in 1 box.
∴ 1 kg chocolates to be filled in (1 div frac{1}{16})
= 1 × 16 = 16 boxes.
Now, (1 frac{1}{2}) kg chocolates to be filled in
(16 times 1 frac{1}{2}=16 times frac{3}{2}=24) boxes.
Question 68.
Anvi is making bookmarker like the one shown in the given figure. How many bookmarker can she make from a 15 m long ribbon?
Solution:
Length of one bookmarker = (=10 frac{1}{2} mathrm{cm})
(=frac{21}{2} mathrm{cm})
∴ Number of bookmarker Anvi can make from 15 m ix., 15 × 100 cm long ribbon
(=left(frac{1}{21 / 2} times 15 times 100right)=frac{2}{21} times 1500)
(=frac{3000}{21}=142.8 approx 142)
Thus, Anvi can make 142 bookmaker from a 15 m long ribbon.
Question 69.
A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by 1.414. Find the length of the diagonal when :
(a) The length of a side of the square is 8.3 cm.
(b) The length of a side of the square is exactly 7.875 cm.
Solution:
(a) Length of a side of the square = 83 cm
∴ Length of the diagonal of the square
= 1.414 × 8.3 cm = 11.7362 cm
≈ 11.74 cm
(b) Length of a side of the square = 7.875 cm
Length of the diagonal of the square
= 1.414 × 7.875 cm
= 11.13525 cm
≈ 11.14 cm
Question 70.
The largest square that can be drawn in a circle has a side whose length is 0.707 times the diameter of the circle. By this rule, find the length of the side of such a square when the diameter of the circle is
(a) 14.35 cm
(b) 8.63 cm
Solution:
(a) Diameter of the circle = 14.35 cm ..
∴ Length of the side of the square
= 0.707 × 14.35 cm
= 10.14545 = 10.15 cm
(b) Diameter of the circle = 8.63 cm
∴ Length of the side of the square
= 0.707 × 8.63 cm
= 6.10141 cm = 6.10 cm
Question 71.
To find the distance around a circular disc, multiply the diameter of the disc by 3.14. What is the distance around the disc when :
(a) the diameter is 18.7 cm?
(b) the radius is 6.45 cm?
Solution:
(a) Diameter of the disc = 18.7 cm
∴ The distance around the disc
= 3.14 × 18.7 cm = 58.718 cm
(b) Radius of the disc = 6.45 cm
∴ Diameter of the disc = 2 × 6.45 cm = 12.9 cm
∴ The distance around the disc
= 3.14 × 12.9 cm = 40.506 cm
Question 72.
What is the cost of 27.5 m of cloth at ₹ 53.50 per metre?
Solution:
Cost of 1 metre of cloth = ₹ 53.50
∴ Cost of 275 metres of cloth = ₹ (53.50 × 27.5) = ₹ 1471.25
Question 73.
In a hurdle race, Nidhi is over hurdle B and (frac{2}{6}) of the way through the race, as shown in the given figure.
Solution:
Then, answer the following:
(a) Where will Nidhi be, when she is (frac{4}{6}) of the way through the race?
(b) Where will Nidhi be when she is (frac{5}{6}) of the way through the race?
(c) Give two fractions to tell what part of the race Nidhi has finished when she is over hurdle C.
Solution:
(a) When Nidhi is (frac{4}{6}) of the way through the race, she will be at hurdle D.
(b) When Nidhi is (frac{5}{6}) of the way through the race, she will be at hurdle E.
(c) When Nidhi is over hurdle C, she would finished the (frac{3}{6} text { or } frac{1}{2}) or middle part of the race.
Question 74.
Diameter of Earth is 12756000 m. In 1996, a new planet was discovered whose diameter is (frac{5}{86}) of the diameter of Earth. Find the diameter of this planet in km.
Solution:
Diameter of Earth = 12756000 m
Diameter of a new planet (=frac{5}{86} times 12756000 mathrm{m})
= 741627.90 m
i.e., (frac{741627.90}{1000} mathrm{km}=741.6 mathrm{km})
Question 75.
What is the product of (frac{5}{129}) reciprocal?
Solution:
The reciprocal of (frac{5}{129}=frac{129}{5})
(therefore quad frac{5}{129} times frac{129}{5}=1)
Question 76.
Simplify: (frac{2 frac{1}{2}+frac{1}{5}}{2 frac{1}{2} div frac{1}{5}})
Solution:
Question 77.
Simplify:(frac{frac{1}{4}+frac{1}{5}}{1-frac{3}{8} times frac{3}{5}})
Solution:
Question 78.
Divide (frac{3}{10}) by (left(frac{1}{4} text { of } frac{3}{5}right))
Solution:
(frac{3}{10} divleft(frac{1}{4} text { of } frac{3}{5}right)=frac{3}{10} divleft(frac{1}{4} times frac{3}{5}right))
(=frac{3}{10} div frac{3}{20}=frac{3}{10} times frac{20}{3}=2)
Question 79.
(frac{1}{8}) of a number equals (frac{2}{5} div frac{1}{20}) number?
Solution:
Let the number be x.
According to question,
⇒ x = 64
Hence, the required number is 64.
Question 80.
Heena’s father paid an electric bill of ₹ 95.70 out of a 500 rupee note. How much change should he have received?
Solution:
The amount of money Heena’s father has = ₹ 500
He paid an electric bill of ₹ 385.70
∴ The amount of money he should have received = ₹ (500 – 385.70) = ₹ 114.30
Question 81.
The normal body temperature is 98.6°F. When Savitri was ill her temperature rose to 103.1°F. How many degrees above normal was that?
Solution:
The normal body temperature = 98.6°F
Savitri’s temperature, when she was ill = 103.1°F
∴ Savitri’s temperature above normal
= 103.1°F – 98.6°F
= 4.5°F
Question 82.
Meteorology: One measure of average global temperature shows how each year varies from a base measure. The table shows results for several years.
Year | 1958 | 1964 | 1965 | 1978 | 2002 |
Difference from Base |
0.10°C | -0.17°C | -0.10°C | ( left(frac{1}{50}right)^{circ} c ) | 0.54°C |
See the table and answer the following:
(a) Order the five years from coldest to warmest.
(b) In 1946, the average temperature varied by -0.03°C from the base measure. Between which two years should 1946 fall when the years are ordered from coldest to warmest?
Solution:
(a) Order of five years from coldest to warmest is 1964, 1965, 1978, 1958, 2002
(b) In 1946, the average temperature varied by -0.03°C.
Since, (-0.10^{circ} mathrm{C}
∴ 1946 should fall between 1965 and 1978.
Science Application
Question 83.
In her science class, Jyoti learned that the atomic weight of Helium is 4.0030; of Hydrogen is 1.0080; and of Oxygen is 16.0000. Find the difference between the atomic weights of:
(a) Oxygen and Hydrogen
(b) Oxygen and Helium
(c) Helium and Hydrogen
Solution:
Atomic weight of Helium = 4.0030
Atomic weight of Hydrogen = 1.0080
Atomic weight of Oxygen – 16.0000
(a) Difference between the atomic weights of Oxygen and Hydrogen
= 16.0000 – 1.0080
= 14.9920
(b) Difference between atomic weights of Oxygen and Helium = 16.0000 – 4.0030 = 11.9970
(c) Difference between atomic weights of Helium and Hydrogen = 4.0030 – 1.0080 = 2.9950
Question 84.
Measurement made in science lab must be as accurate as possible. Ravi measured the length of an iron rod and said it was 19.34 cm long; Kamal said 19.25 cm; and Tabish said 19.27 cm. The correct length was 19.33 cm. How much of error was made by each of the boys?
Solution:
The correct length of the iron rod = 19.33 cm
The length of the rod measured by Ravi = 19.34 cm
∴ Error made by Ravi = (19.34 – 19.33)cm = +0.01 cm
The length of the rod measured by Kamal = 19.25 cm
∴ Error made by Kamal = (19.25 – 19.33) cm = -0.08 cm
The length of the rod measured by Tabish = 19.27 cm
∴ Error made by Tabish = (19.27 – 19.33) cm = -0.06 cm.
Question 85.
When 0.02964 is divided by 0.004, what will be the Questionuotient?
Solution:
Question 86.
What number divided by 520 gives the same Questionuotient as 85 divided by 0.625?
Solution:
Let a number x be divided by 520.
According to question,
x ÷ 520 = 85 ÷ 0.625
(Rightarrow frac{x}{520}=frac{85}{0.625})
(Rightarrow x=frac{85 times 520}{0.625}=frac{44200}{625} times 1000)
= 70.72 × 1000 = 70720
Thus, the required number is 70720.
Question 87.
A floor is 4.5 m long and 3.6 m wide. A 6 cm square tile costs ₹ 23.25. What will be the cost to cover the floor with these tiles?
Solution:
Length of the floor = 4.5 m
= (4.5 × 100) cm = 450 cm
Width of the floor – 3.6 m – (3.6 × 100) cm = 360 cm
∴ Area of the floor – (450 × 360) cm^{2} = 162000 cm^{2}
Side of a square tile = 6 cm
∴ Area of one square tile = (6 × 6) cm^{2} = 36 cm^{2}
Number of tiles reqnuired to cover the floor
(=frac{text { Area of floor }}{text { Area of one tile }}=frac{162000}{36})
= 4500
Now, cost of one tile = ₹ 23.25
∴ Cost of 4500 tiles =₹ (23.25 ×4500)
= ₹ 104625
Question 88.
Sunita and Rehana want to make dresses for their dolls. Sunita has (frac{3}{4}) m of cloth, and she gave (frac{1}{3}) of it to Rehana. How much did Rehana have?
Solution:
Length of cloth Sunita has (=frac{3}{4}) m
∴ Length of cloth Rehana has
(=frac{1}{3} text { of } frac{3}{4} mathrm{m}=frac{1}{3} times frac{3}{4} mathrm{m}=frac{1}{4} mathrm{m})
Question 89.
A flower garden is 22.50 m long. Sheela wants to make a border along one side using bricks that are 0.25 m long. How many bricks will be needed?
Solution:
Length of the garden = 22.50 m
Length of one brick = 0.25 m
∴ Number of bricks required to make the border (=frac{text { Length of the garden }}{text { Length of one brick }})
(=frac{22.50}{0.25}=frac{2250}{25}=90)
Question 90.
How much cloth will be used in making 6 shirts, if each required (2 frac{1}{4}[)m of cloth, allowing (frac{1}{8} m) for waste in cutting and finishing in each shirt?
Solution:
Length of cloth required for one shirt
(=2 frac{1}{4} mathrm{m}+frac{1}{8} mathrm{m})
(=left(frac{9}{4}+frac{1}{8}right) mathrm{m}=left(frac{18+1}{8}right) mathrm{m}=frac{19}{8} mathrm{m})
∴ Length of cloth required for 6 shirts
(=frac{19}{8} times 6 mathrm{m}=frac{19 times 3}{4} mathrm{m}=frac{57}{4} mathrm{m}=14 frac{1}{4} mathrm{m})
Question 91.
A picture hall has seats for 820 persons. At a recent film show, one usher guessed it was (frac{3}{4}) full, another that it was (frac{2}{3}) full. The ticket office reported 648 sales. Which usher (first or second) made the better guess?
Solution:
Total number of seats = 820
Sale of total tickets = 648
Number of sold tickets guessed by first usher
(=frac{3}{4} text { of } 820=frac{3}{4} times 820=615)
Number of sold tickets guessed by second usher
(=frac{2}{3} text { of } 820=frac{2}{3} times 820)
(=frac{1640}{3}=546.66 approx 547)
Since, 615 is more close to 648 than 547.
∴ First usher made the better guess.
Question 92.
For the celebrating children day, students of Class VII bought sweets for ₹ 740.25 and cold drink for ₹ 70. If 35 students contributed eQuestionually what amount was contributed by each student?
Solution:
Total spending amount = ₹740.25 + ₹70
= ₹ 3810.25
Total number of students who contributed money = ₹ 35
∴ Contribution of each student = ₹ (frac{810.25}{35})
= ₹ 23.15
Question 93.
The time taken by Rohan in five different races to run a distance of 500 m was 3.20 minutes, 3.37 minutes, 3.29 minutes, 3.17 minutes and 3.32 minutes. Find the average time taken by him in the races?
Solution:
Total time taken by Rohan in five races
= (3.20 +3.37 +3.29 +3.17 +3.32] minutes
= 16.35 minutes
Total number of races = 5
∴ The average time taken by Rohan
(=frac{16.35}{5}) minutes = 3.27 minutes
Question 94.
A public sewer line is being installed along (80 frac{1}{4})m of road. The supervisor says that the labourers will be able to complete 7.5 m in one day. How long will the project take to complete?
Solution:
Total length of a sewer line (=80 frac{1}{4} mathrm{m})
(=frac{321}{4} mathrm{m})
7.5 m long sewer line completed in 1 day.
(therefore frac{321}{4}) m long sewer line completed in
(frac{1}{7.5} times frac{321}{4}=frac{10}{75} times frac{321}{4}=frac{3210}{300}=10.7)
≈ 11 days
Question 95.
The weight of an object on moon is (frac{1}{6}) its weight on Earth.If an object weighs (5 frac{3}{5}) kg on Earth, how much would it weigh on the moon?
Solution:
Weight of the object on Earth (=5 frac{3}{5} mathrm{kg})
(=frac{28}{5} mathrm{kg})
∴ Weight of that object on Moon
(=frac{1}{6} text { of } frac{28}{5} k g)
(=frac{1}{6} times frac{28}{5} mathrm{kg}=frac{14}{15} mathrm{kg}=0.93 mathrm{kg})
Question 96.
In a survey, 200 students were asked what influenced them most to buy their latest CD. The results are shown in the circle graph.
(a) How many students said radio influenced them most?
(b) How many more students were influenced by radio than by a music video channel?
(c) How many said a friend or relative influenced them or they heard the CD in a shop?
Solution:
Total number of students = 200
(a) Number of students who were influenced by radio
(=frac{9}{20} text { of } 200=frac{9}{20} times 200=90)
(b) Number of students who were influenced by music video channel
(=frac{2}{25} text { of } 200=frac{2}{25} times 200=16)
Number of students who were influenced by radio = 90
∴ 90-16-74 more students were influenced by radio than by a music video channel.
(c) Numberofstudents who were influenced by friend or relative
(=frac{3}{20} text { of } 200=frac{3}{20} times 200=30)
Number of students who were influenced by hearing or seeing the CD in a shop
(=frac{1}{10} text { of } 200=frac{1}{10} times 200=20)
∴Total number of students who were influenced by both friend or relative and hearing the CD in a shop = 30 + 20 – 50
Question 97.
In the morning, a milkman filled (5 frac{1}{2})L of milk in his can. He sold to Renu, Kamla and Renuka (frac{3}{4}) L each; to Shadma he sold (frac{7}{8})L; and to Jassi he gave (1 frac{1}{2}) L. How much milk is left in the can?
Solution:
Total quantity of milk in the can
(=5 frac{1}{2} mathrm{L}=frac{11}{2} mathrm{L})
Total quantity of milk to be sold to different persons
(=left(frac{3}{4}+frac{3}{4}+frac{3}{4}+frac{7}{8}+1 frac{1}{2}right) mathrm{L})
(=left(frac{9}{4}+frac{7}{8}+frac{3}{2}right) mathrm{L}=left(frac{18+7+12}{8}right) mathrm{L}=frac{37}{8} mathrm{L})
Now, quantity of milk left in the can
(=left(frac{11}{2}-frac{37}{8}right) mathrm{L})
(=frac{44-37}{8} mathrm{L}=frac{7}{8} mathrm{L})
Question 98.
Anuradha can do a piece of work in 6 hours. What part of the work can she do in 1 hour, in 5 hours, in 6 hours?
Solution:
Let whole work be represented by W.
The part of work done by Anuradha in 6 hours = W
∴ The part of work done by her in 1 hour (=frac{W}{6})
The part of work done by her in 5 hours (=frac{W}{6} times 5=frac{5}{6} W)
The part of work done by her in 6 hours
(=frac{W}{6} times 6=W)
Hence, Anuradha can do (frac{1}{6}) part of work in 1 hour, (frac{5}{6}) part of work in 5 hours and complete work in 6 hours.
Question 99.
What portion of a ‘saree’can Rehana paint in 1 hour if it requires 5 hours to paint the whole saree? In (4 frac{3}{5}) hours? In (3 frac{1}{2}) hours?
Solution:
Let total portion of saree be represented by 1.
The portion of saree painted by Rehana in 5 hours = 1.
∴ The portion of saree painted by her in 1 hour (=frac{1}{5})
The portion of Saree painted by her in
(4 frac{3}{5} text { hours }=frac{1}{5} times 4 frac{3}{5}=frac{1}{5} times frac{23}{5}=frac{23}{25})
The portion of saree painted by her in
(3 frac{1}{2} text { hours }=frac{1}{5} times 3 frac{1}{2}=frac{1}{5} times frac{7}{2}=frac{7}{10})
Question 100.
Rama has (6 frac{1}{4})kg of cotton wool for making pillows. If one pillow takes (1 frac{1}{4}) kg, how many pillows can she make?
Solution:
Total quantity of cotton wool (=6 frac{1}{4} mathrm{kg})kg
(=frac{25}{4} mathrm{kg})
Quantity of cotton wool required for one pillow (=1 frac{1}{4} mathrm{kg}=frac{5}{4} mathrm{kg})
∴ Required number of pillows
(=frac{25}{4} div frac{5}{4})
(=frac{25}{4} times frac{4}{5}=5)
Question 101.
It takes (2 frac{1}{3}) of cloth to make a shirt. How many shirts can Radhika make from a piece of cloth (9 frac{1}{3} m) long?
Solution:
Total length of cloth (=9 frac{1}{3} mathrm{m}=frac{28}{3} mathrm{m})
Length of the cloth reQuestionuired for one shirt
(=2 frac{1}{3} mathrm{m}=frac{7}{3} mathrm{m})
Question 102.
Ravi can walk (3 frac{1}{3}) km in one hour. How long will it take him to walk to his office which is 10 km from his home?
Solution:
Time taken by Ravi to walk (3 frac{1}{3}) km = 1 hour
∴ Time taken by him to walk 1 km
Question 103.
Raj travels 360 km on three fifths of his petrol tank. How far would he travel at the same rate with a full tank of petrol?
Solution:
Distance travelled by Raj with (frac{3}{5}) petrol tank = 360 km
Distance travelled by him with a full petrol tank (=left(360 div frac{3}{5}right) mathrm{km})
(=frac{360 times 5}{3} mathrm{km}=120 times 5 mathrm{km}=600 mathrm{km})
Question 104.
Kajol has ₹ 75. This is (frac{3}{8}) of the amount she earned. How much did she earn?
Solution:
Let money earned by Kajol be ₹ x.
According to question,
(frac{3}{8} times x=75 Rightarrow x=75 times frac{8}{3})
⇒ x = 200
∴ The amount earned by Kajol is ₹ 200.
Question 105.
It takes 17 full specific type of trees to make one tonne of paper. If there are 221 such trees in a forest, then
(i) what fraction of forest will be used to make;
(a) 5 tonnes of paper.
(b) 10 tonnes of paper.
(ii) To save (frac{7}{13}) part of the forest how much of paper we have to save.
Solution:
(i) Number of trees required to make one tonne of paper = 17
Total number of trees = 221
∴ Fraction of forest will be used to make one tonne of paper (=frac{17}{221}=frac{1}{13})
(a) Fraction of forest will be used to make 5 tonnes of paper (=frac{1}{13} times 5=frac{5}{13})
(b) Fraction of forest will be used to make 10 tonnes of paper (=frac{1}{13} times 10=frac{10}{13})
(ii) Let we have to save x tonnes of paper to save (frac{7}{13}) part of forest
Fraction of forest will be used to make x tonnes of paper (=frac{1}{13} times x=frac{x}{13})
Now, (frac{x}{13}=frac{7}{13} Rightarrow x=frac{7}{13} times 13=7)
Question 106.
Simplify and write the result in decimal form:
(left(1+frac{2}{9}right)+left(1 div 3 frac{1}{5}right)+left(1 div 2 frac{2}{3}right))
Solution:
Question 107.
Some pictures (a) to (f) are given below. Tell which of them show:
Solution:
- (2 times frac{1}{4}) represents the addition of 2 figures, each representing 1 shaded part out of 4 equal parts. Hence (2 times frac{1}{4}) is represented by (d).
- (2 times frac{3}{7}) represents the addition of 2 figures, each representing 3 shaded parts out of 7 equal parts. Hence (2 times frac{3}{7}) is represented by (f).
- (2 times frac{1}{3}) represents the addition of 2 figures, each representing 1 shaded part out of 3 equal parts. Hence (2 times frac{1}{3}) is represented by (c).
- (4 times frac{1}{4}) represents the addition of 4figures, each representing 1 shaded part out of 4 equal parts. Hence (4 times frac{1}{4}) is represented by (b).
- (3 times frac{2}{9}) represents the addition of 3 figures, each representing 2 shaded parts out of 9 equal parts. Hence (3 times frac{2}{9}) is represented by (a).
- (3 times frac{1}{4}) represents the addition of 3 figures, each representing 1 shaded part out of 3 equal parts. Hence (3 times frac{1}{4}) is represented by (e)
Question 108.
Evaluate : (0.3) × (0.3) – (0.2) (0.2)
Solution:
(0.3) × (0.3) – (0.2) ×(0.2)
(=frac{3}{10} times frac{3}{10}-left(frac{2}{10} times frac{2}{10}right)=frac{9}{100}-frac{4}{100})
(=frac{5}{100}=0.05)
Question 109.
Evaluate: (frac{0.6}{0.3}+frac{0.16}{0.4})
Solution:
(frac{0.6}{0.3}+frac{0.16}{0.4}=left(frac{6}{10} times frac{10}{3}right)+left(frac{16}{100} times frac{10}{4}right))
(=2+frac{4}{10}=frac{20+4}{10}=frac{24}{10}=2.4)
Question 110.
Find the value of:
(frac{(0.2 times 0.14)+(0.5 times 0.91)}{(0.1 times 0.2)})
Solution:
Question 111.
A square and an equilateral triangle have a side in common. If side of triangle is (frac{4}{3}) cm long, find the perimeter of figure formed (see figure).
Solution:
Question 112.
Rita has bought a carpet of size (4 m times 6 frac{2}{3} m) But her room size is (3 frac{1}{3} m times 5 frac{1}{3} m). What fraction of area should be cut off to fit wall to wall carpet into the room?
Solution:
Question 113.
Family photograph has length (14 frac{2}{5}) cm and breadth (10 frac{2}{5}) cm.It has border of uniform width (2 frac{3}{5}) cm. Find the area of framed photograph.
Solution:
Question 114.
Cost of a burger is ₹ (20 frac{3}{4}) and of Macpuff is ₹ (15 frac{1}{2}) Find the cost of 4 burgers and 14 macpuffs.
Solution:
Question 115.
A hill, (101 frac{1}{3})m in height, has (frac{1}{4})th of its height under water. What is the height of the hill visible above the water?
Solution:
Total height of the hill
Question 116.
Sports: Reaction time measures how Questionuickly a runner reacts to the starter pistol. In the 100 m dash at the 2004 Olympic Games, Lauryn Williams had a reaction time of 0.214 second. Her total race time, including reaction time, was 11.03 seconds. How long did it take her to run the actual distance?
Solution:
Total race time of Lauryn Williams = 11.03 seconds
Her reaction time = 0.214 second
∴ Time taken by her to run the actual distance
=(11.03 – 0.214) seconds = 10.816 seconds
Question 117.
State whether the answer is greater than 1 or less than 1. Put a’mark in appropriate box.
Solution:
Question 118.
There are four containers that are arranged in the ascending order of their heights. If the height of the smallest container given in the figure is expressed as (frac{7}{25} x=10.5) cm.. Find the height of the largest container.
Solution:
Height of the smallest container = 10.5 cm
We have given that
∴ Height of the largest container is 37.5 cm.
Directions: In questions 119 to 122, replace “?” with the appropriate fraction.
Question 119.
Solution:
Here, the pattern is
Question 120.
Solution:
Here, the pattern is
Question 121.
Solution:
Here, the pattern is 0.05, 0.05 × 10 – 0.5, 0.5 × 10 – 5,5 × 10 = 50
∴ Next number will be 50 × 10 = 500
Question 122.
Solution:
Here, the pattern is
Directions: What is the error in each of Question 123 and 124?
Question 123.
A student compared (-frac{1}{4}) and -0.3. He changed (-frac{1}{4}) to the decimal -0.25 and wrote, “Since 0.3 is greater than 0.25,-0.3 is greater than -0.25”. What was the student’s error?
Solution:
Since 0.3 is greater than 0.25.
∴ -0.3 is less than -0.25
∴ The error is -0.30 >-0.25
Question 124.
A student multiplied two mixed fractions in the following manner: (2 frac{4}{7} times 3 frac{1}{4}=6 frac{1}{7}). What error the student has done?
Solution:
(2 frac{4}{7} times 3 frac{1}{4}=frac{18}{7} times frac{13}{4}=frac{9 times 13}{7 times 2}=frac{117}{14}=8 frac{5}{14}).
but given that
(2 frac{4}{7} times 3 frac{1}{4}=6 frac{1}{7}).
∴ Error is that mixed fractions are not converted into improper fractions.
Question 125.
In the pattern (frac{1}{3}+frac{1}{4}+frac{1}{5}+ldots) which fraction makes the sum greater than 1 (first time)? Explain.
Solution:
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