NCERT Exemplar Class 7 Maths Chapter 7 Comparing Quantities are part of NCERT Exemplar Class 7 Maths. Here we have given NCERT Exemplar Class 7 Maths Solutions Chapter 7 Comparing Quantities
NCERT Exemplar Class 7 Maths Solutions Chapter 7 Comparing Quantities
Directions: In questions 1 to 23, there are four options, out of which one is correct. Write the correct one.
Question 1.
20% of 700 m is
(a) 560 m
(b) 70 m
(c) 210 m
(d) 140 m
Solution:
(d) 20% of 700 m
(=left(frac{20}{100} times 700right) mathrm{m}=140 mathrm{m})
Question 2.
Gayatri’s income is ₹ 1,60,000 per year. She pays 15% of this as house rent and 10% of the remainder on her child’s education. The money left with her is
(a) ₹ 136000
(b) ₹ 120000
(c) ₹ 122400
(d) ₹ 14000
Solution:
(c) Gayatri’s income = ₹ 160000
House rent (=frac{15}{100} times 160000)= ₹ 24000
Remaining amount – 160000 – 24000 = ₹ 136000
Expenditure on child’s education
(=frac{10}{100} times 136000)
= ₹ 13600
∴ Money left with her
= ₹ (160000 – 24000 – 13600)
= ₹ (160000 – 37600) = ₹ 122400
Question 3.
The ratio of Fatima’s income to her savings is 4 : 1. The percentage of money saved by her
(a) 20%
(b) 25%
(c) 40%
(d) 80%
Solution:
(b) Let Fatima’s income be 4x and savings be x.
Savings percentage (=frac{x}{4 x} times 100=25 %)
Question 4.
0.07 is equal to
(a) 70%
(c) 0.7%
(b) 7%
(d) 0.07%
Solution:
(b)
(0.07=frac{7}{100}=7 %)
Question 5.
In a scout camp, 40% of the scouts were from Gujarat State and 20% of these were from Ahmedabad. The percentage of scouts in the camp from Ahmedabad is
(a) 25
(b) 32.5
(c) 8
(d) 50
Solution:
(c) Let the total scouts in camp be x.
Question 6.
What percent of ₹4500 is ₹ 9000?
(a) 200
(b) (frac{1}{2})
(c) 2
(d) 50
Solution:
(a) Let x percent of ₹ 4500 is ₹ 9000
Question 7.
5.2 is equal to
(a) 52%
(b) 5.2%
(c) 520%
(d) 0.52%
Solution:
(c) (5.2=frac{52}{10} times frac{10}{10}=frac{520}{100}=520 %)
Question 8.
The ratio 3 : 8 is equal to
(a) 3.75%
(b) 37.5%
(c) 0.375%
(d) 267%
Solution:
(b) (frac{3}{8}=0.375=frac{37.5}{100}=37.5 %)
Question 9.
225% is equal to
(a) 9 : 4
(b) 4 : 9
(c) 3 : 2
(d) 2 : 3
Solution:
(a) (225 %=frac{225}{100}=frac{9}{4}=9 : 4)
Question 10.
A bicycle is purchased for ₹ 1800 and is sold at a profit of 12%. Its selling price is
(a) ₹ 1584
(b) ₹ 2016
(c) ₹ 1788
(d) ₹ 1812
Solution:
(b) C.P. of bicycle = ₹ 1800
Profit % = 12%
S.P. = C.P. + Profit = 1800 + 216 = ₹ 2016
Question 11.
A cricket bat was purchased for ₹ 800 and was sold for ₹ 1600. Then profit earned is
(a) 100%
(b) 64%
(c) 50%
(d) 60%
Solution:
(a) C.P. of bat = ₹ 800
S.P. of bat = ₹ 1600
Profit = S.P. – C.P. = 1600 – 800 = 800
Question 12.
A farmer bought a buffalo for ₹ 44000 and a cow for ₹ 18000. He sold the buffalo at a loss of 5% but made a profit of 10% on the cow. The net result of the transaction is
(a) loss of ₹ 200
(b) profit of ₹ 400
(c) loss of ₹ 400
(d) profit of ₹ 200
Solution:
(c) C.P. of buffalo = ₹ 44000 .
Loss% = 5%
S.P. of buffalo = 44000 – 2200 = ₹ 41800
C.P. of cow = ₹ 18000
Profit % = 10% .
∴ S.P. of cow = 18000 + 1800 = ₹ 19800
Now, C.P. of both buffalo and cow
= 44000 + 18000 = ₹ 62000
S.P. of both buffalo and cow
= 41800+ 19800 = ₹ 61600
∴ Loss = 62000 – 61600 = ₹ 400
Question 13.
If Mohan’s income is 25% more than Raman’s income, then Raman’s income is less than Mohan’s income by
(a) 25%
(b) 80%
(c) 20%
(d) 75%
Solution:
(c) Let Mohan’s income be x and
Raman’s income be ₹ y.
According to question,
Thus, Raman’s income is 20% less than Mohan’s income.
Question 14.
The interest on ₹ 30000 for 3 years at the rate of 15% per annum is
(a) ₹ 4500
(b) ₹ 9000
(c) ₹ 18000
(d) ₹ 13500
Solution:
(d) Principal = ₹ 30000,
Time = 3 years, Rate = 15% p.a.
Question 15.
Amount received on ₹ 3000 for 2 years at the rate of 11% per annum is
(a) ₹ 2340
(b) ₹ 3660
(c) ₹ 4320
(d) ₹ 3330
Solution:
(b) Principal = ₹ 3000, Time = 2 years, Rate = 11% p.a.
Amount = Principal + Interest
= 3000 + 660 = ₹ 3660
Question 16.
Interest on ₹ 12000 for 1 month at the rate of 10% per annum is
(a) ₹ 1200
(b) ₹ 600
(c) ₹ 100
(d) ₹ 12100
Solution:
(c) Principal = ₹ 12000,
Time = 1 month (=frac{1}{12}) year
Rate = 10%
Question 17.
Rajni and Mohini deposited ₹ 3000 and ₹ 4000 in a company at the rate of 10% per annum for 3 years and (2 frac{1}{2}) years respectively. The difference of the amounts received by them will be
(a) ₹ 100
(b) ₹ 1000
(c) ₹ 900
(d) ₹ 1100
Solution:
(d) For Rajni : Principal = ₹ 3000, Rate = 10% p.a., Time = 3 years
Amount = 3000 + 900 = ₹ 3900
For Mohini : Principal = ₹ 4000,
Rate = 10% p.a.
Time (=2 frac{1}{2} text { years }=frac{5}{2}) years
Amount = 4000 + 1000 = ₹ 5000
∴ Difference in amounts = 5000 – 3900 = ₹ 1100
Question 18.
If 90% of x is 315 km, then the value of x is
(a) 325 km
(b) 350 km
(c) 405 km
(d) 340 km
Solution:
(b) According to question,
90% of x = 315 km
⇒ x = 350 km
Question 19.
On selling an article for ₹ 329, a dealer lost 6%. The cost price of the article is
(a) ₹ 310.37
(b) ₹ 348.74
(c) ₹ 335
(d) ₹ 350
Solution:
(d) S.P. = ₹ 329, Loss % = 6%
= ₹ 350
Question 20.
(frac{25 % text { of } 50 % text { of } 100 %}{25 times 50}) is equal to 25 x 50
(a) 1.1%
(b) 0.1%
(c) 0.01%
(d) 1%
Solution:
(c)
Question 21.
The sum which will earn a simple interest of ₹ 126 in 2 years at 14% per annum is
(a) ₹ 394
(b) ₹ 395
(c)₹ 3450
(d) ₹ 540
Solution:
(c) I = ₹ 126, T = 2 years, R = 14% p.a.
Question 22.
The percent that represents the unshaded region in the figure.
(a) 75%
(b) 50%
(c) 40%
(d) 60%
Solution:
(c) Total parts = 100
Unshaded parts = 40
∴ Unshaded fraction (=frac{40}{100}=40 %)
Question 23.
The percent that represents the shaded region in the figure is
(a) 36%
(b) 64%
(c) 27%
(d) 48%
Solution:
(a) Total parts = 100
Shaded parts = 36 .
∴ Unshaded fraction (=frac{36}{100}=36 %)
Directions: In each of the questions 24 to 59, fill in the blanks to make the statements true.
Question 24.
2 : 3 = _______%
Solution:
Question 25.
(18 frac{3}{4} %=) _______ : _______
Solution:
Question 26.
30% of ₹ 360 = _______
Solution:
₹ 108
Question 27.
120% of 50 km = _________
Solution:
60 km:
Question 28.
2.5 = _______%
Solution:
Question 29.
(frac{8}{5}=) __________ %
Solution:
Question 30.
A _______ with its denominator 100 is called a per cent.
Solution:
Fraction: A fraction with its denominator 100 is called a percent.
Question 31.
15 kg is ________ % of 50 kg.
Solution:
30: Let 15 kg is x% of 50 kg
∴ (quad frac{x}{100} times 50=15)
⇒ x = 15 × 2 = 30
Question 32.
Weight of Nikhii increased from 60 kg to 66 kg. Then, the increase in weight is _______ %
Solution:
10: The increase in weight of Nikhil = 66 – 60 = 6 kg
Question 33.
In a class of 50 students, 8 % were absent on one day. The number of students present on that day was _________
Solution:
46: Total number of students = 50
Number of absent students (=frac{8}{100} times 50=4)
∴ Number of present students = 50 – 4 = 46
Question 34.
Savitri obtained 440 marks out of 500 in an examination. She secured _________% marks in the examination.
Solution:
88: Total marks = 500
Marks obtained by Savitri = 440
Question 35.
Out of a total deposit of ₹ 1500 in her bank account, Abida withdrew 40% of the deposit. Now the balance in her account is ________
Solution:
900: Total deposit in Abida’s bank account = ₹ 1500
∴ Balance in Abida’s bank account = 1500 – 600 = ₹ 900
Question 36.
_________ is 50% more than 60.
Solution:
90: Let x is 50% more than 60
Question 37.
John sells a bat for ₹ 75 and suffers a loss of ₹ 8. The cost price of the bat is _________
Solution:
₹ 83: S.P. of bat = ₹ 75
Loss = ₹ 8
∴ C.P. = 75 + 8 = ₹ 83
Question 38.
If the price of sugar is decreased by 20%, then the new price of 3 kg sugar originally costing ₹ 120 will be _______
Solution:
96: Original cost price of 3 kg sugar = ₹ 120
∴ Original cost price of 1 kg sugar
Decrease percentage = 20%
Now, new price of 1 kg sugar = 40 – 8 = ₹ 32
∴ New price of 3 kg sugar = 32 × 3 = ₹ 96
Question 39.
Mohini bought a cow for ₹ 9000 and sold it at a loss of ₹ 900. The selling price of the cow is _______
Solution:
8100: C.P. of the cow = ₹ 9000
Loss = ₹ 900
∴ S.P. of the cow = 9000 – 900 = ₹ 8100
Question 40.
Devangi buys a chair for ₹ 700 and sells it for ₹ 750. She earns a profit of _____ % _____in the transaction.
Solution:
(7 frac{1}{7}): C.P. of chair = ₹ 3700
S.P. of chair = ₹ 750
Profit = 750 – 700 = ₹ 50
Question 41.
Sonal bought a bed sheet for ₹ 400 and sold it for ₹ 440. Her________% is ________
Solution:
Profit, 10: C.P. of bed sheet = ₹ 400
S.P. of bed sheet = ₹ 440
Now, S.P. > C.P.
Thus, profit = 440 – 400 = ₹ 40
Question 42.
Nasim bought a pen for ₹ 60 and sold it for ₹ 54. His _______ % is _______
Solution:
Loss, 10 : C.P. of pen = ₹ 60
S.P. of pen = ₹54
S.P.
Therefore, Nasim has a loss of 60 – 54 = ₹ 6 .
Question 43.
Aahuti purchased a house for ₹ 50,59,700 and spent ₹ 40300 on its repairs. To make a profit of 5%, she should sell the house for ₹ ________
Solution:
5355000 : Purchasing amount of house = ₹ 5059700
Amount spent on repairing the house = ₹ 40300
∴ C.P. of house = 5059700 + 40300 = ₹ 5100000
Profit % = 5%
Thus, Aahuti should sell the house for ₹ 5355000 to make a profit of 5%.
Question 44.
If 20 lemons are bought for ₹ 10 and sold at 5 for three rupees, then ______ in the transaction is ________%
Solution:
Profit, 20: Cost price of 20 lemons = ₹ 10
∴ Cost price of 1 lemon (=frac{10}{20}) = ₹ (frac{1}{2})
Selling price of 5 lemons = ₹ 3
∴ Selling price of 1 lemon = ₹ (frac{3}{5})
Now, S.P. > C.P.
Therefore, profit in the transaction
Question 45.
Narain bought 120 oranges at ₹ 4 each. He sold 60% of the oranges at 5 each and the remaining at ₹ 3.50 each. His _______ is _______ %
Solution:
Profit, 10: Total oranges = 120
60% of oranges (=frac{60}{100} times 120=72)
Remaining oranges = 120 – 72 = 48
Now, C.P. of 120 oranges = 120 × 4 = ₹ 480
S.P. of 72 oranges -= 72 × 5 = ₹ 360
S.P. of 48 oranges = 48 × 3.50 = ₹ 168
∴ S.P. of 120 oranges = 360 + 168 = ₹ 528
Now, S.P. C.P.
Therefore, Narain gets the profit of ₹ (528 – 480) = ₹ 48
Question 46.
A fruit seller purchased 20 kg of apples at ₹ 50 per kg. Out of these, 5% of the apples were found to be rotten. If he sells the remaining apples at ₹ 60 per kg, then his ______ is ________%
Solution:
Profit, 14: Total apples – 20 kg
Rotten apples (=frac{5}{100} times 20=1 mathrm{kg})
Remaining apples = 20 – 1 = 19 kg
Now, C.P. of 20 kg apples = 20 × 50 = ₹ 1000
S.P. of 19 kg apples = 19 × 60 = ₹ 1140
S.P. > C.P.
∴ Fruit seller gets profit of ₹ (1140 – 1000) = ₹ 140
Question 47.
Interest on ₹ 3000 at 10% per annum for a period of 3 years is ________
Solution:
₹ 900 : Principal = ₹ 3000, Rate = 10% p.a., Time = 3 years
Question 48.
Amount obtained by depositing ₹ 20,000 at 8% per annum for six months is _________
Solution:
₹ 20800 : Principal = ₹ 20000, Rate = 8% p.a.
Amount = Principal + Interest = 20000 + 800 = ₹ 20800
Question 49.
Interest on ₹ 12500 at 18% per annum for a period of 2 years and 4 months is
Solution:
₹ 5250: Principal = ₹ 12500, Rate = 18% p.a.
Time = 2 years + 4 months = 2 years + (frac{4}{12}) years
Question 50.
25 ml is _______ percent of 5 litres.
Solution:
0.5: Let 25 ml is x% of 5 litres.
Question 51.
If A is increased by 20%, it equals B. If B is decreased by 50%, it equals C. Then _______% of A is equal to C.
Solution:
60: According to question,
Question 52.
Interest (=frac{P times R times T}{100}), where
T is _______
R% is _______ and P is ________
Solution:
Time period, Rate of interest, Principal:
where T = Time period
R% = Rate of interest
and P = Principal
Question 53.
The difference of interest for 2 years and 3 years on a sum of ₹ 2100 at 8% per annum is _______
Solution:
₹ 168: Principal = ₹ 2100, Rate = 8% p.a.
∴ Difference = 504 – 336 = ₹ 168.
Question 54.
To convert a fraction into a percent, we _______ it by 100.
Solution:
Multiply: To convert a fraction into a percent, we multiply it by 100.
Question 55.
To convert a decimal into a percent, we shift the decimal point two places to the _________
Solution:
Right: To convert a decimal into a percent, we shift the decimal point two places to the right
Question 56.
The ______ of interest on a sum of ₹ 2000 at the rate of 6% per annum for (1 frac{1}{2}) years and 2 years is ₹ 420.
Solution:
Sum: Principal = ₹ 2000, Rate = 6% p.a.
Sum of both interest = ₹ (180 + 240) = ₹ 420
Question 57.
When converted into percentage, the value of 6.5 is _______than 100%.
Solution:
More: Given number = 6.5
And 650% > 100%
So, after converting 6.5 into percentage, the value of 6.5 is more than 100%.
Directions: In questions 58 and 59, copy each number line.
Fill in the blanks so that each mark on the number line is labelled with a per cent, a fraction and a decimal. Write all fractions in lowest terms.
Question 58.
Solution:
Question 59.
Solution:
Directions: In questions 60 to 79, state whether the statements are True or False.
Question 60.
(frac{2}{3}=66 frac{2}{3} %)
Solution:
True
Question 61.
When an improper fraction is converted into percentage then the answer can also be less than 100.
Solution:
False
When an improper fraction is converted into percentage then the answer always greater than 100
Question 62.
8 hours is 50% of 4 days.
Solution:
False
1 day = 24 hours
∴ 4 days = 24 × 4 = 96 hours
Now, 50% of 96 hours (=frac{50}{100} times 96=48) hours
Question 63.
The interest on ₹ 350 at 5% per annum for 73 days is ₹ 35.
Solution:
False
Principal = ₹ 350, Rate = 5% p.a.
Question 64.
The simple interest on a sum of ₹ P for T years at R% per annum is given by the formula: Simple Interest (=frac{T times P times R}{100})
Solution:
True
where P is the principal, R is the rate of interest and Tis the time.
Question 65.
(75 %=frac{4}{3})
Solution:
False
Question 66.
12% of 120 is 100.
Solution:
False
Question 67.
If Ankita obtains 336 marks out of 600, then percentage of marks obtained by her is 33.6.
Solution:
False
Total marks = 600
Marks obtained by Ankita = 336 .
Question 68.
0.018 is equivalent to 8%.
Solution:
False
Question 69.
50% of ₹ 50 is ₹ 25.
Solution:
True
Question 70.
250 cm is 4% of 1 km.
Solution:
False
1 km = 1000 m = 1000 × 100 cm = 100000 cm
(4 % text { of } 1 mathrm{km}=left(frac{4}{100} times 100000right) mathrm{cm}=4000 mathrm{cm})
Question 71.
Out of 600 students of a school, 126 go for a picnic. The percentage of students that did not go for the picnic is 75.
Solution:
False
Total number of students = 600
Number of students went on a picnic = 126
∴ Number of students who did not go for the picnic = 600 – 126 = 474
Percentage of students who did not go for the picnic
(=frac{474}{600} times 100=79.16 %)
Question 72.
By selling a book for ₹ 50, a shopkeeper suffers a loss of 10%. The cost price of the book is ₹ 60.
Solution:
False
Selling price of a book = ₹ 50
Loss % = ₹ 10
We know that
So, the cost price of the book is ₹ 55.5.
Question 73.
If a chair is bought for ₹ 2000 and is sold at a gain of 10%, then selling price of the chair is ₹ 2010.
Solution:
False
C.P. of a chair = ₹ 2000
Gain % = 10
We know that,
Question 74.
If a bicycle was bought for ₹ 650 and sold for ₹ 585, then the percentage of profit is 10.
Solution:
False
C.P. of a bicycle = ₹ 650
S.P. of the bicycle = ₹ 585
Therefore, there is a loss of ₹ (650 – 585) = ₹ 65
Question 75.
Sushma sold her watch for ₹ 3320 at a gain of ₹ 320. For earning a gain of 10% she should have sold the watch for ₹ 3300
Solution:
True
S.P. of a watch = ₹ 3320
Gain = ₹ 320
∴ C.P. of the watch = ₹ (3320 – 320) = ₹ 3000
Required gain % = 10
So, for earning again of 10% Sushma should have sold the watch for ₹ 3300
Question 76.
Interest on ₹ 1200 for (1 frac{1}{2}) years at the rate of 15% per annum is ₹ 180.
Solution:
False
Principal = ₹ 1200,
Time (=1 frac{1}{2} text { years }=frac{3}{2}text { years })
Rate = 15% p.a.
Question 77.
The amount received after depositing ₹ 800 for a period of 3 years at the rate of 12% per annum is ₹ 896.
Solution:
False
Principal = ₹ 800, Time = 3 years
Rate = 12% p.a.
Now, amount = Principal + Interest
= ₹ (800 + 288) = ₹ 1088
Question 78.
₹ 6400 were lent to Feroz and Rashmi at 15% per annum for (3 frac{1}{2}) and 5 years respectively. The difference in the interest paid by them is ₹ 150.
Solution:
False
For Feroz: Principal = ₹ 6400,
Rate = 15% p.a.
For Rashmi : Principal = 6400,
Rate = 15% p.a.
Time = 5 years
Required difference in interests
= ₹ (4800 – 3360) = ₹ 1440
Question 79.
A vendor purchased ₹ 120 lemons at 120 per hundred. 10% of the lemons were found rotten which he sold at ₹ 50 per hundred. If he sells the remaining lemons at ₹ 125 per hundred, then his profit will be 16%.
Solution:
False
Total number of lemons = 720
C.P. of 100 lemons = ₹ 120
Rotten lemons = 10%
Number of remaining lemons which are not rotten = 720 – 72 = 648
S.P. of 720 lemons = ₹ (36 + 810) = ₹ 846
Here, C.P. > S.P.
Therefore, there is a loss of ₹ (864 – 846) = ₹ 18
Question 80.
Find the value of x if
(i) 8% of ₹ x is ₹ 100
(ii) 32% of x kg is 400 kg
(iii) 35% of ₹ x is ₹ 280
(iv) 45% of marks x is 405.
Solution:
(i) 8% of ₹ x = ₹ 100
Question 81.
Imagine that a 10 × 10 grid has value 300 and that this value is divided evenly among the small squares. In other words, each small square is worth 3. Use a new grid for each part of this problem, and label each grid “Value: 300”.
(a) Shade 25% of the grid. What is 25% of 300? Compare the two answers.
(b) What is the value of 25 squares?
(c) Shade 17% of the grid. What is 17% of 300? Compare the two answers.
(d) What is the value of (frac{1}{10}) of the grid?
Solution:
Now, each small square is worth 3
∴ 25 small square worths 25 × 3 = 75
Both the values are same.
(b) From part (a), value of 25 small squares is 75.
Now, each small square is worth 3
∴ 17 small square worths 17 × 3 = 51
Both the values are same.
(d) Total numbers in grid = 10 × 10 = 100
∴ (frac{1}{10} text { of } 100=frac{1}{10} times 100=10)
Each small square is worth 3
∴ 10 small square worths 10 × 3 = 30
Question 82.
Express (frac{1}{6}) as a percent.
Solution:
Question 83.
Express (frac{9}{40}) as a percent.
Solution:
Question 84.
Express (frac{1}{100}) as a percent.
Solution:
(frac{1}{100}=1 %)
Question 85.
Express 80% as fraction in its lowest term.
Solution:
Question 86.
Express (33 frac{1}{3} %) as a ratio in the lowest term.
Solution:
Question 87.
Express (16 frac{2}{3} %), as a ratio in the lowest form.
Solution:
Question 88.
Express 150% as a ratio in the lowest form.
Solution:
Question 89.
Sachin and Sanjana are calculating 23% of 800.
Now calculate 52% of 700 using both the ways described above. Which way do you find easier?
Solution:
I : 52% of 700 = (1% of 700) × 52
= (0.01 × 700) × 52
= 7 × 52 = 364
We find that the way II is easier.
Question 90.
Write 0.089 as a percent.
Solution:
Question 91.
Write 1.56 as a percent.
Solution:
Question 92.
What is 15% of 20?
Solution:
Question 93.
What is 800% of 800?
Solution:
Question 94.
What is 100% of 500?
Solution:
Question 95.
What percent of 1 hour is 30 minutes?
Solution:
Let x% of 1 hour is 30 minutes.
Now, 1 hour = 60 minutes
Question 96.
What percent of 1 day is 1 minute?
Solution:
Let x% of 1 day is 1 minute,
Now, 1 day – 24 hours = (24 × 60) minutes
Question 97.
What percent of 1 km is 1000 metres?
Solution:
Let % of 1 km is 1000 metres.
Now, 1 km = 1000 metres
Question 98.
Find out 8% of 25 kg.
Solution:
Question 99.
What percent of ₹ 80 is ₹ 100?
Solution:
Let x% of ₹ 80 is ₹ 100
Question 100.
45% of the population of a town are men and 40% are women. What is the percentage of children?
Solution:
Total percentage = 100
Percentage of men = 45
Percentage of women = 40
∴ Percentage of children = 100 – (45 + 40) = 100 – 85 = 15
Question 101.
The strength of a school is 2000. If 40% of the students are girls then how many boys are there in the school?
Solution:
Total number of students = 2000
∴ Number of boys = 2000 – 800 = 1200
Question 102.
Chalk contains 10% calcium, 3% carbon and 12% oxygen. Find the amount of carbon and calcium (in grams) in (2 frac{1}{2}) kg of chalk.
Solution:
Question 103.
800 kg of mortar consists of 55% sand, 33% cement and rest lime. What is the mass of lime in mortar?
Solution:
Total mass of mortar = 800 kg
Mass of sand in mortar = 55% of 800 kg
Mass of cement in mortar = 33% of 800 kg
∴ Mass of lime in mortar = [800 – (440 + 264)] kg = 96 kg
Question 104.
In a furniture shop, 24 tables were bought at the rate of ₹ 450 per table. The shopkeeper sold 16 of them at the rate of ₹ 600 per table and the remaining at the rate of ₹ 400 per table. Find her gain or loss percent.
Solution:
C.P. of 1 table = ₹ 450
∴ C.P. of 24 tables = ₹ (24 × 450) = ₹ 10800
S.P. of 16 tables = ₹ {(16 × 600) = ₹ 9600
S.P. of remaining tables i.e., 8 tables
= ₹ (8 × 400) = 3200
S.P. of 24 tables = ₹ (9600 + 3200) = ₹ 12800
Here, S.P. > C.P.
Therefore, there is a gain of ₹ (12800 – 10800) = ₹ 2000
Question 105.
Medha deposited 20% of her money in a bank. After spending 20% of the remainder, she has ₹ 4800 left with her. How much did she originally have?
Solution:
Let the amount of money Medha had ₹ x.
Amount of money deposited in bank = 20% of x
Question 106.
The cost of a flower vase got increased by 12%. If the current cost is ₹ 896, what was its original cost?
Solution:
Let the original cost of a flower vase = ₹ x
Cost of the flower vase after increased by
Question 107.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years?
Solution:
Total amount borrowed by Radhika = ₹ 12000
I: Principal = ₹ 4000, Rate = 18% p.a.,
Time = 3 years
II: Principal = ₹ (12000 – 4000) = ₹ 8000,
Rate = 15%, Time = 3 years
Total interest = ₹ (2160 + 3600) = ₹ 5760
Question 108.
A man travelled 60 km by car and 240 km by train. Find what percent of total journey did he travel by car and what percent by train?
Solution:
Distance travelled by car = 60 km
Distance travelled by train = 240 km
Total distance travelled = (60+240) km = 300 km
Percentage of journey travelled by car
(=frac{60}{300} times 100=20 %)
Percentage of journey travelled by train
(=frac{240}{300} times 100=80 %)
Question 109.
By selling a chair for ₹ 1440, a shopkeeper loses 10%. At what price did he buy it?
Solution:
S.P. of a chair = ₹ 1440
Loss% = 10
So, the shopkeeper bought the chair at ₹ 1600.
Question 110.
Dhruvika invested money for a period from May 2006 to April 2008 at rate of 12% per annum. If interest received by her is ₹ 1620, find the money invested.
Solution:
Let the principal be ₹ P.
Rate = 12% p.a., Interest = ₹ 1620
Time = 2 years
So, amount of money invested by Dhruvika is ₹ 6750.
Question 111.
A person wanted to sell a scooter at a loss of 25%. But at the last moment he changed his mind and sold the scooter at a loss of 20%. If the difference in the two SP’s is ₹ 4000, then find the CP of the scooter.
Solution:
Let the C.P. of a scooter = ₹ x
S.P. of the scooter for a loss of 25%
(=left(frac{100-25}{100}right) times x=frac{3}{4} x)
S.P. of the scooter for a loss of 20%
(=left(frac{100-20}{100}right) times x=left(frac{80}{100}right) x=frac{4}{5} x)
Difference in both S.P.’s of the scooter
(=frac{4}{5} x-frac{3}{4} x=frac{1}{20} x)
According to question,
(frac{1}{20} x=4000)
⇒ x = 4000 × 20 = 80000
Question 112.
The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the ratio of the number of literate women to the total population.
Solution:
Total population of a village = 8000
Number of literate people = 80% of 8000
Number of literate women = 40% of 6400
Question 113.
In an entertainment programme, 250 tickets of ₹ 400 and 500 tickets of ₹ 100 were sold. If the entertainment tax is 40% on ticket of ₹ 400 and 20% on ticket of ₹ 100, find how much entertainment tax was collected from the programme.
Solution:
Cost of 250 tickets of ₹ 400 per ticket
= ₹ (250 × 400) = ₹ 100000
Cost of 500 tickets of ₹ 100 per ticket
= ₹ (500 × 100) = ₹ 50000
Tax on tickets of ₹ 400
Tax on tickets of 100
So, total tax collected from the programme
= ₹ (40000 + 10000) = ₹ 50000
Question 114.
Bhavya earns ₹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings.
Solution:
Monthly income of Bhavya = ₹ 50000
Spending amount = 80% of ₹ 50000
New income – 20% of 50000 + 50000
= ₹ 60000
New spending amount = 40000 + 20% of 40000
= 40000 + 8000 = ₹ 48000
∴ New savings of Bhavya
= ₹ (60000 – 48000) = ₹ 12000
Question 115.
In an examination, there are three papers each of 100 marks. A candidate obtained 53 marks in the first and 75 marks in the second paper. How many marks must the candidate obtain in the third paper to get an overall of 70 percent marks?
Solution:
Total marks = 100 + 100 + 100 = 300
Marks obtained by the candidate in first paper = 53
Marks obtained by the candidate in second paper = 75
Let the marks obtained by the candidate in third paper = x
Total marks obtained by the candidate = 53 + 75 + x – 128 + x
According to question,
(frac{(128+x)}{300} times 100=70)
⇒ 128 + x = 3x 70
⇒ x = 210 – 128 = 82
So, the candidate must obtain 82 marks in the third paper.
Question 116.
Health Application A doctor reports blood pressure in millimetres of mercury (mm Hg) as a ratio of systolic blood pressure to diastolic blood pressure (such as 140 over 80). Systolic pressure is measured when the heart beats, and diastolic pressure is measured when it rests. Refer to the table of blood pressure ranges for adults.
Blood Pressure Ranges | |||
Normal | Prehypertension | Hypertension (Very High) | |
Systolic | Under 120 mm Hg | 120-139 mm Hg | 140 mm Hg and above |
Diastolic | Under 80 mm Hg | 80-89 mm Hg | 90 mm Hg and above |
Manohar is a healthy 37 years old man whose blood pressure is in the normal category.
(a) Calculate an approximate ratio of systolic to diastolic blood pressures in the normal range.
(b) If Manohar’s systolic blood pressure is 102 mm Hg, use the ratio from part (a) to predict his diastolic blood pressure.
(c) Calculate ratio of average systolic to average diastolic blood pressure in the prehypertension category.
Solution:
(a) Systolic blood pressure in the normal range = 120 mm Hg
Diastolic blood pressure in the normal range = 80 mm Hg
(b) Systolic blood pressure of Manohar = 102 mm Hg
⇒ Diastolic blood pressure
(c) Average systolic blood pressure in the prehypertension category
Average diastolic blood pressure in the prehypertension category
Question 117.
(a) Science Application: The king cobra can reach a length of 558 cm. This is only about 60 per cent of the length of the largest reticulated python. Find the length of the largest reticulated python.
(b) Physical Science Application: Unequal
masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down. Unequal masses will balance when the following proportion is true:
Two children can be balanced on a Seesaw when (frac{text { mass } 1}{text { length } 2}=frac{text { mass } 2}{text { length } 1}) The child on the left and child on the right are balanced. What is the mass of the child on the right?
(c) Life Science Application: A DNA model was built using the scale 2 cm : 0.0000001 mm. If the model of the DNA chain is 17 cm long, what is the length of the actual chain?
Solution:
(a) Let the length of the largest reticulated python – x cm
According to question,
60% of x = 558
⇒ x = 930
(b) We have,
(c) Length of actual chain 0.0000001 mm is measured as 2 cm in the model.
∴ Length of actual chain when length of chain is 17 cm in the model
(=frac{0.0000001}{2} times 17=0.00000085 mathrm{mm})
Question 118.
Language Application
Given below are few Mathematical terms.
Find
(a) The ratio of consonants to vowels in each of the terms.
(b) The percentage of consonants in each of the terms.
Solution:
(a) (i) Number of consonants in Hypotenuse’ = 6
Number of vowels in ‘Hypotenuse’ = 4
(ii) Number of consonants in ‘Congruence’ = 6
Number of vowels in ‘Congruence’ = 4
(iii) Number of consonants in ‘Perpendicular’ = 8
Number of vowels in ‘Perpendicular = 5
(iv) Number of consonants in ‘Transversal’ = 8
Number of vowels in ‘Transversal’ = 3
(v) Number of consonants in ‘Correspondence’ = 9
Number of vowels in ‘Correspondence’ = 5
(b)(i) Percentage of consonants in ‘Hypotenuse’
(ii) Percentage of ‘consonants in Congruence’
(iii) Percentage of consonants in ‘Perpendicular’
(iv) Percentage of consonants in ‘Transversal’
(v) Percentage of consonants in ‘Correspondence’
Question 119.
What’s the Error? An analysis showed that 0.06 per cent of the T-shirts made by one company were defective. A student says this is 6 out of every 100. What is the student’s error?
Solution:
Therefore 6 out of every 10000 are defective T-shifts.
Question 120.
What’s the Error? A student said that the ratios (frac{3}{4} text { and } frac{9}{16}) were proportional. What error did the student make?
Solution:
(frac{3}{4}=frac{3 times 4}{4 times 4}=frac{12}{16} neq frac{9}{16})
Therefore (frac{3}{4} text { and } frac{9}{16}) are not proportional.
Question 121.
What’s the Error? A clothing store charges ₹ 1024 for 4 T-shirts. A student says that the unit price is ₹ 25.6 per T-shirt. What is the error? What is the correct unit price?
Solution:
Cost price for 4 T-shirts = ₹ 1024 .
Therefore correct unit price is ₹ 256.
Question 122.
A tea merchant blends two varieties of tea in the ratio of 5 : 4. The cost of first variety is ₹ 200 per kg and that of second variety is ₹ 300 per kg. If he sells the blended tea at the rate of ₹ 275 per kg, find out the percentage of her profit or loss.
Solution:
Let quantity of first variety and second variety of tea be 5x kg and 4x kg respectively.
Cost of first variety of tea = ₹ (200 × 5x) = ₹ 1000 x
Cost of second variety of tea = ₹ (300 × 4x) = ₹ 1200x
Total cost price of tea = ₹ (1000x + 1200x) = ₹ 2200x
Selling price of tea = ₹ (9x × 275) = ₹ 2475x
Here, S.P. > C.P.
Therefore, there is a profit of = ₹ (2475x – 2200x) = ₹ 275x
Question 123.
A piece of cloth 5 m long shrinks 10 percent on washing. How long will the cloth be after washing?
Solution:
Length of cloth = 5m
Shrinking percentage after washing = 10%
∴ Length of cloth, after washing
= 5 – 10% of 5
(=5-frac{10}{100} times 5=5-0.5=4.5 mathrm{m})
Question 124.
Nancy obtained 426 marks out of 600 and the marks obtained by Rohit are 560 out of 800. Whose performance is better?
Solution:
Percentage of marks obtained by Nancy
(=frac{426}{600} times 100=71 %)
Percentage of marks obtained by Rohit
(=frac{560}{800} times 100=70 %)
So, Nancy’s performance is better.
Question 125.
A memorial trust donates ₹ 5,00,000 to a school, the interest on which is to be used for awarding 3 scholarships to students obtaining first three positions in the school examination every year. If the donation earns an interest of 12 percent per annum and the values of the second and third scholarships are ₹ 20,000 and ₹ 15,000 respectively, find out the value of the first scholarship.
Solution:
Total donation amount = ₹ 500000
Interest on donation amount
Value of the second and third scholarships = ₹ (20000 + 15000) = ₹ 35000
Now, value of the first scholarship = ₹ (60000 – 35000) = ₹ 25000
Question 126.
Ambika got 99 per cent marks in Mathe matics, 76 per cent marks in Hindi, 61 percent in English, 84 per cent in Science, and 95% in Social Science. If each subject carries 100 marks, then find the percentage of marks obtained by Ambika in the aggregate of all the subjects.
Solution:
Total maximum marks = 5 × 100 = 500
Total marks obtained by Ambika in all subjects = 99 + 76 + 61 + 84 + 95 = 415
Percentage of marks obtained by Ambika
(=frac{415}{500} times 100=83 %)
Question 127.
What sum of money lent out at 16 percent per annum simple interest would produce ₹ 9600 as interest in 2 years?
Solution:
S.I. = ₹ 9600, Time = 2 years,
Rate = 16% p.a.
We know that,
Question 128.
Harish bought a gas-chullah for ₹ 900 and later sold it to Archana at a profit of 5 percent. Archana used it for a period of two years and later sold it to Babita at a loss of 20 percent. For how much did Babita get it?
Solution:
For Harish: C.P. = ₹ 900, Profit % = 5
Now, S.P. for Harish would be the C.P. for Archana.
For Archana : C.P. = ₹ 945, Loss % = 20
Again, S.P. for Archana would be the C.P. for Babita.
So, Babita got the gas-chullah for ₹ 756.
Question 129.
Match each of the entries in Column with the appropriate entries in Column II
Solution:
(i) → (E)
(ii) → (H)
(iii) → (O)
(iv) → (J)
(v) → (G)
(vi) → (L)
(vii) → (B)
(viii) → (A)
(ix) → (F)
(x) → (K)
(xi) → (D)
(xii) → (I)
Question 130.
In a debate competition, the judges decide that 20 percent of the total marks would be given for accent and presentation. 60 percent of the rest are reserved for the subject matter and the rest are for rebuttal. If this means & marks for rebuttal, then find the total marks.
Solution:
Let the total marks be x,
Marks for accent and presentation -20% of x
(=frac{20}{100} times x=frac{x}{5})
Question 131.
Divide ₹ 10000 in two parts so that the simple interest on the first part for 4 years at 12 percent per annum may be equal to the simple interest on the second part for 4.5 years at 16 percent per annum.
Solution:
Total amount which has to be divide = ₹ 10000
Let first part be = ₹ x
Then, second part = ₹ (10000 – x)
Interest for second part
So, the first part = ₹ 6000
And second apart = ₹ (10000 – 6000) = ₹ 4000
Question 132.
₹ 9000 becomes ₹ 18000 at simple interest in 8 years. Find the rate percent per annum.
Solution:
Principal = ₹ 9000, Amount = ₹ 18000,
Time = 8 years
∴ Interest = ₹ (18000 – 9000) = ₹ 9000
Question 133.
In how many years will the simple interest on a certain sum be 4.05 times the principal at 13.5 percent per annum?
Solution:
Let the principal be ₹ P.
Then, simple interest = ₹ 4.05 P
Rate = 13.5%
Question 134.
The simple interest on a certain sum for 8 years at 12 percent per annum is ₹ 3120 more than the simple interest on the same sum for 5 years at 14 percent per annum. Find the sum.
Solution:
Let the sum be ₹ P.
Interest for 8 years at 12% per annum
Question 135.
The simple interest on a certain sum for 2.5 years at 12 percent per annum is ₹ 300 less than the simple interest on the same sum for 4.5 years at 8 percent per annum. Find the sum.
Solution:
Let the sum be ₹ P.
Interest for 2.5 years at 12% per annum
(=frac{mathrm{P} times 12 times 2.5}{100}=frac{3}{10} mathrm{P})
Interest for 4.5 years at 8% per annum
(=mathrm{P} times frac{8}{100} times 4.5=frac{9}{25} mathrm{P})
According to question,
Question 136.
Designing a Healthy Diet When you design your healthy diet, you want to make sure that you meet the dietary requirements to help you grow into a healthy adult.
As you plan your menu, follow the following guidelines
1. Calculate your ideal weight as per your height from the table given below this question.
2. An active child should eat around 55.11 calories for each kilogram desired weight.
3. 55 percent of calories should come from carbohydrates. There are 4 calories in each gram of carbohydrates.
4. 15 percent of your calories should come from proteins. There are 4 calories in each gram of proteins.
5. 30 percent of your calories may come from fats. There are 9 calories in each gram of fat.
Ideal Height and Weight Proportion | |||||
Men | Women | ||||
Height | Weight | Height | Weight | ||
Feet | cm | Kilograms | Feet | cm | Kilograms |
5′ | 152 | 48 | _{4}-7″ | 140 | 34 |
5’1″ | 155 | 51 | 4’8″ | 142 | 36 |
5’2″ | 157 | 54 | 4-9- | 145 | 39 |
5’3″ | 160 | 56 | 4’10” | 147 | 41 |
5’4″ | 163 | 59 | 4’11” | 150 | 43 |
5’5″ | 165 | 62 | 5′ | 152 | 45 |
5’6″ | 168 | 65 | 5’1″ | 155 | 48 |
5’7″ | 170 | 67 | 5’2″ | 157 | 50 |
5’8″ | 173 | 70 | 5’3″ | 160 | 52 |
5’9″ | 175 | 73 | 5-4- | 163 | 55 |
5’10” | 178 | 75 | 5’5″ | 165 | 57 |
5’11” | 180 | 78 | 5’6″ | 168 | 59 |
6′ | 183 | 81 | 5’7″ | 170 | 61 |
6’1″ | 185 | 84 | 5’8″ | 173 | 64 |
6’2″ | 188 | 86 | 5’9″ | 175 | 66 |
6’3″ | 191 | 89 | 5’10” | 178 | 68 |
6’4″ | 193 | 92 | 5’11” | 180 | 70 |
Solution:
Let the height of a girl student of 7th class be 5 feet.
(1) Ideal weight – 45 kg
(2) Quantity of calories needed = 45 × 55.11 = 2479.95 calories
(3) Calories that should come from carbohydrates
(4) Calories that should come from proteins
(5) Calories that may come from fat.
Question 137.
150 students are studying English, Maths or both. 62 percent of students study English and 68 percent are studying Maths. How many students are studying both?
Solution:
Total number of students = 150
Number of students studying English
Number of students studying Maths
∴ Number of students studying both subjects = (93 + 102) – 150 = 45
Question 138.
Earth Science: The table lists the world’s 10 largest deserts.
Largest Deserts in the World | |
Desert | Area (km^{2}) |
Sahara (Africa) | 8,800,000 |
Gobi (Asia) | 1,300,000 |
Australian Desert (Australia) | 1,250,000 |
Arabian Desert (Asia) | 850,000 |
Kalahari Desert (Africa) | 580,000 |
Chihuahuan Desert (North America) | 370,000 |
Takla Makan Desert (Asia) | 320,000 |
Kara Kum (Asia) | 310,000 |
Namib Desert (Africa) | 310,000 |
Thar Desert (Asia) | 260,000 |
(a) What are the mean, median and mode of the areas listed?
(b) How many times the size of the Gobi Desert is the Namib Desert?
(c) What percentage of the deserts listed are in Asia?
(d) What percentage of the total area of the deserts listed is in Asia?
Solution:
(a) Total area of 10 Deserts = 8800000 + 1300000 + 1250000 + 850000 + 580000 + 370000 + 320000 + 310000 + 310000 + 260000 = 14350000 km^{2}
Since, n = 10 which is even
Since, area = 310000 km^{2} occurs two times.
∴ Mode = 310000 km^{2}
(b) Area of Gobi Desert = 1300000 km^{2}
Area of Namib Desert = 310000 km^{2}
Now, 1300000 = x × 310000
(Rightarrow quad x=frac{1300000}{310000}=4.19)
So, size of the Gobi Desert is 4.19 times the Namib Desert
(c) Total number of Deserts = 10
Deserts listed in Asia = 5
(d) Total area of Deserts listed in Asia = 1300000 + 850000 + 320000+ 310000 + 260000 = 3040000 km^{2}
Total area of all Deserts = 14350000 km^{2}
Percentage of Deserts in Asia
(=frac{3040000}{14350000} times 100=21.1 %)
Question 139.
Geography Application: Earth’s total land area is about 148428950 km^{2}. The land area of Asia is about 30 per cent of this total. What is the approximate land area of Asia to the nearest square km?
Solution:
Earth’s total land area = 148128950 km^{2}
Land area of Asia = 30% of 148428950 km^{2}
(=frac{30}{100} times 148428950=44528685 mathrm{km}^{2})
Question 140.
The pieces of Tangrams have been rearranged to make the given shape.
By observing the given shape, answer the following questions:
What percentage of total has been coloured?
(i) Red (R) = _______
(ii) Blue (B) = _______
(iii) Green (G) = _______
Check that the sum of all the percentages calculated above should be 100.
If we rearrange the same pieces to form some other shape, will the percentatge of colours change?
Solution:
(i) Total traction for Red (R)
∴ Percentage of red colour = 37.5%
(ii) Total fraction for Blue (B)
∴ Percentage of blue colour = 50%
(iii) Total fraction for Green (G)
∴ Percentage of green colour = 12.5%
Now, sum of all percentage
= 37.5% + 50% +12.5% = 100%
If we rearrange the same pieces to form some other shape, then percentage of colours will not change.
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