Question 3. Which of the following numbers are divisible by 3? (i) 106 (ii) 726 (iii) 915 (iv) 1008 Solution: (i) Sum of the digits of 106 = 1 + 0 + 6 = 7 which is not divisible by 3. Hence 106 is not divisible by 3. (ii) Sum of the digits of 726 = 7 + 2 + 6 = 15 which is divisible by 3. Hence 726 is divisible by 3. (iii) Sum of the digits of 915 = 9 + 1 + 5 = 15 which is divisible by 3. Hence 915 is divisible by 3. (iv) Sum of the digits of 1008 = 1 + 0 + 0 + 8 = 9 which is divisible by 3. Hence 1008 is divisible by 3.
Question 4. Prove that the sum of the given numbers and the numbers obtained by reversing their digits is divisible by 11. (a) 89 (b) ab (c) 69 (d) 54 Solution: (a) Given number = 89 Number obtained by reversing the order of digits = 98 Sum = 89 + 98 = 187 ÷ 11 = 17 Hence, the required number is 11.
(b) Given number = ab = 10a + b Number obtained by reversing the digits = 10b + a Sum = (10a + b) + (10b + a) = 10a + b + 10b + a = 11a + 11b = 11(a + b) ÷ 11 = a + b
(c) Given number = 69 Number obtained by reversing the digits = 96 Sum = 69 + 96 = 165 ÷ 11 = 15 Hence, the required number is 11.
(d) Given number = 54 Number obtained by reversing the digits = 45 Sum = 54 + 45 = 99 ÷ 11 = 9 Hence, the required number is 11.
Question 5. Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9. (i) 59 (ii) xy (iii) xyz (iv) 203 Solution: (i) Given number = 59 Number obtained by reversing the digits = 95 Difference = 95 – 59 = 36 ÷ 9 = 4 Hence, the required number is 9.
(ii) Given number = xy = 10x + y Number obtained by reversing the digits = 10y + x Difference = (10x + y) – (10y + x) = 10x + y – 10y – x = 9x – 9y = 9(x – y) ÷ 9 = x – y Hence, the required number is 9.
(iii) Given number = xyz = 100x + 10y + z Number obtained by reversing the digits = 100z + 10y + x Difference = (100x + 10y + z) – (100z + 10y + x) = 100x + 10y + z – 100z – 10y – x = 99x – 99z = 99(x – z) = 99(x – 2) ÷ 9 = 11 (x – z) Hence, the required number is 9.
(iv) Given number = 203 Number obtained by reversing the digits = 302 Difference = 302 – 203 = 99 ÷ 9 = 11 Hence, the required number is 9.
Question 6. If a, b, c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37. Solution: We have abc + cab + bca abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a Adding abc + cab + bca = 111a +111b + 111c = 111 (a + b+ c) = 37 × 3 (a + b + c) which is a multiple of 37. Hence proved.
Playing with Numbers Class 8 Extra Questions Short Answer Type
Question 7. Complete the magic square given below so that the sum of the numbers in each row or in each column or along each diagonal is 15. Solution: (i) A = 15 – (8 + 1) = 15 – 9 = 6 (ii) F = 15 – (8 + 5) = 15 – 13 = 2 (iii) C = 15 – (A + F) = 15 – (6 + 2) = 15 – 8 = 7 (iv) E = 15 – (1 + 5) = 15 – 6 = 9 (v) D = 15 – (E + F) = 15 – (9 + 2) = 15 – 11 = 4 (vi) B = 15 – (8 + 4) = 15 – 12 = 3 Hence the required square is
Question 8. Find the values of P and Q from the given addition problem Solution: Here, 3 + Q = 7 ⇒ Q = 7 – 3 = 4 Now taking second column, we get 4 + 7 = 11 i.e. 1 is carried over to third column ⇒ 1 + P + 2 = 9 ⇒ 3 + P = 9 P = 9 – 3 = 6 Hence the value of P = 6 and Q = 4
Question 9. Find the values of p, q and r in the following multiplication problem. Solution: 6 × 4 = 24, Here 2 is carried over second column ⇒ 6 × p + 2 – 3 × 10 = 2 [∵ 21 – 3 × 6 = 3] ⇒ 6p – 30 = 0 ⇒ p = 5 Now the multiplication problem becomes, Here 2 + r = 4 ⇒ r = 2 q × 354 = 1062 ⇒ q = 3 Hence, p = 5, q = 3, r = 2
Question 10. Observe the following patterns: 1 × 9 – 1 = 8 21 × 9 – 1 = 188 321 × 9 – 1 = 2888 4321 × 9 – 1 = 38888 Find the value of 87654321 × 9 – 1 Solution: From the pattern, we observe that there are as many eights in the result as the first digit from the right which is to be multiplied by 9 and reduced by 1. 87654321 × 9 – 1 = 788888888
Playing with Numbers Class 8 Extra Questions High Order Thinking Skills (HOTS) Type
Question 11. Complete the cross number puzzle with the given column. Solution: Hence the complete square is
Question 12. The product of two 2-digit numbers is 1431. The product of their tens digits is 10 and the product of their units digits is 21. Find the numbers. Solution: Let the required two 2-digit numbers be 10a + b and 10p + q as per the condition, we have a × p = 10 and b × q = 21 a = 2 and p = 5 or a = 5 and p = 2 Similarly b × q = 21 b = 3 and q = 7 or b = 7 and q = 3 10p + q = 57 or 10p + q = 53 and 10a + b = 23 or 10a + b = 27 Since the units digit of product 1431 is 1. Numbers are 57 and 23 or 53 and 27. Now 57 × 23 = 1311 and 53 × 27 = 1431 which is given. Hence, the required numbers are 53 and 27.
13. Which of the following is quotient obtained on dividing –18 xyz2 by –3 xz?
(a) 6 yz
(b) –6 yz
(c) 6 xy2
(d) 6 xy ► (a) 6 yz
14. Divide the given polynomial by the given monomial: (5x2− 6x) ÷ 3x
(a) (5x – 6)
(b) 1/3
(c) 1/3(5x – 6)
(d) none of these
► (c) 1/3(5x – 6)
15. Factorise 6xy – 4y + 6 – 9x.
(a) (3x – 2) (2y – 3)
(b) (3x – 2)
(c) (2y – 3)
(d) (2x – 3) (3y – 2)
► (a) (3x – 2) (2y – 3)
16. When we factorise an expression, we write it as a ________ of factors.
(a) product
(b) difference
(c) sum
(d) none of these
► (a) product
17. Factorise: 4y2 −12y + 9
(a) (7y− 5)2
(b) (5y− 3)2
(c) (2y− 5)2
(d) (2y− 3)2
► (d) (2y− 3)2
18. Amrit and Pankaj expanded (x−5)2. Amrit’s answer is x2−25 and Pankaj’s answer is x2−10x+25. Which of the following statements is correct? (a) Amrit’s answer is correct. (b) Pankaj’s answer is wrong. (c) Both got correct answer. (d) Pankaj’s answer is correct. ► (d) Pankaj’s answer is correct.
Factorisation Class 8 Extra Questions Very Short Answer Type
Question 1. Find the common factors of the following terms. (a) 25x2y, 30xy2 (b) 63m3n, 54mn4 Solution: (a) 25x2y, 30xy2 25x2y = 5 × 5 × x × x × y 30xy2 = 2 × 3 × 5 × x × y × y Common factors are 5× x × y = 5 xy
(b) 63m3n, 54mn4 63m3n = 3 × 3 × 7 × m × m × m × n 54mn4 = 2 × 3 × 3 × 3 × m × n × n × n × n Common factors are 3 × 3 × m × n = 9mn
Question 2. Factorise the following expressions. (a) 54m3n + 81m4n2 (b) 15x2y3z + 25x3y2z + 35x2y2z2 Solution: (a) 54m3n + 81m4n2 = 2 × 3 × 3 × 3 × m × m × m × n + 3 × 3 × 3 × 3 × m × m × m × m × n × n = 3 × 3 × 3 × m × m × m × n × (2 + 3 mn) = 27m3n (2 + 3mn)
Question 13. Simplify the following expressions: Solution:
Question 14. Factorise the given expressions and divide that as indicated. (a) 39n3(50n2 – 98 ) ÷ 26n2(5n – 7) (b) 44(p4 – 5p3 – 24p2) ÷ 11p(p – 8) Solution:
Question 15. If one of the factors of (5x2 + 70x – 160) is (x – 2). Find the other factor. Solution: Let the other factor be m. (x – 2) × m = 5x2 + 70x – 160
MCQ Questions for Class 8 Maths: Ch 13 Direct and Inverse Proportions
1. A garrison of 500 persons had provisions for 27 days. After 3 days a reinforcement of 300 persons arrived. For how many more days will the remaining food last now?
(a) 12 days
(b) 14 days
(c) 16 days
(d) 15 days
► (d) 15 days
2. When one quantity is increased, the other quantity is also increased. This proportion is called _______
(a) Kally proportion
(b) Direct proportion
(c) Inverse proportion
(d) None of these
► (b) Direct proportion
3. 72 books are packed in 4 cartons of the same size. How many cartons are required for 360 books?
(a) 22
(b) 18
(c) 20
(d) None of these
► (c) 20
4. 8 g of sandal wood cost Rs 40. What will 10 g cost ?
(a) Rs 30
(b) Rs 36
(c) Rs 48
(d) Rs 50
► (d) Rs 50
5. The price of 357 mangoes is Rs.1517.25. What will be the approximate price of 49 dozens of such mangoes?
(a) Rs.3000
(b) Rs.3500
(c) Rs.4000
(d) Rs.2500
► (a) Rs.3000
6. 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used?
(a) 56 minutes
(b) 72 minutes
(c) 96 minutes
(d) 80 minutes
► (c) 96 minutes
7. A van covers 432 km with 36 litres of diesel. How much distance would it cover with 25 litres of diesel?
(a) 200 km
(b) 300 km
(c) 100 km
(d) 350 km
► (b) 300 km
8. 120 copies of a book cost Rs 600. What will 400 copies cost ?
(a) Rs 1000
(b) Rs 2000
(c) Rs 3000
(d) Rs 2400
► (b) Rs 2000
9. A boy runs 1 km in 10 minutes. How long will he take to ran 600 m ?
(a) 2 minutes
(b) 3 minutes
(c) 4 minutes
(d) 6 minutes
► (d) 6 minutes
10. The cost of 5 metres of a particular quality of cloth is Rs 210. Find the cost of 2 metres of cloth of the same type.
(a) Rs 100
(b) Rs 84
(c) Rs 90
(d) Rs 60
► (b) Rs 84
11. An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.
(a) 21 metres
(b) 30 metres
(c) 25 metres
(d) None of these
► (a) 21 metres
12. A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
(a) 15 days
(b) 20 days
(c) 25 days
(d) 30 days
► (c) 25 days
13. The scale of a map is 1:3×107 Two cities are 5 cm apart on the map. Find the actual distance between them in kilometres.
(a) 1500 km
(b) 1000 km
(c) 1100 km
(d) 2000 km ► (a) 1500 km
14. A journey by bus takes 45 minutes at 40 km/hour. How fast must a car go to undertake the same journey in 25 minutes?
(a) 36 km/h
(b) 48 km/h
(c) 72 km/h
(d) None of these
► (c) 72 km/h
15. Raju earns Rs 1440, if he works for 12 days. If he works for 30 days, he will earn
(a) Rs 2400
(b) Rs 3600
(c) Rs 4800
(d) None of these
► (b) Rs 3600
16. x varies inversely as square of y. Given that y = 3 for x = 1.find the value of x for y = 4.
(a) 3
(b) 9
(c) 1/3
(d) 9/16
► (d) 9/16
17. If the cost of 27 bags of paddy is Rs.9450, what is the cost of 36 bags of paddy?
(a) Rs.12000
(b) Rs.12600
(c) Rs.16200
(d) Rs.10620
► (b) Rs.12600
18. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work ?
(a) 24 days
(b) 28 days
(c) 34 days
(d) 35 days
► (a) 24 days
19. If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same piece of work?
(a) 4 days
(b) 5 days
(c) 3 days
(d) 2 days
► (c) 3 days
20. There is enough food to last for 40 people for 10 days. If 10 more people join them, the food will last for
(a) 10 days
(b) 12 days
(c) 8 days
(d) None of these
► (c) 8 days
21. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. How many parts of base will be used in mixture by mixing 7 part of red pigment?
(a) 70
(b) 56
(c) 63
(d) 49
► (b) 56
22. Two quantities x and y are said to be in ___________ if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant.
(a) inverse proportion
(b) mix proportion
(c) direct proportion
(d) None of these
► (c) direct proportion
important questions
Question 1:
Observe the tables given below and in each one find whether x and y are proportional: (i)
x
3
5
8
11
26
y
9
15
24
33
78
(ii)
x
2.5
4
7.5
10
14
y
10
16
30
40
42
(iii)
x
5
7
9
15
18
25
y
15
21
27
60
72
75
ANSWER:
(i) Clearly, xy=39=515=824=1133=2678=13(constant)Therefore, x and y are proportional.Clearly, xy=39=515=824=1133=2678=13(constant)Therefore, x and y are proportional.
(ii) Clearly, xy=2.510=416=7.530=1040=14, while 1442=13i.e., 2.510=416=7.530=1040is not equal to 1442.Therefore, x and y are not proportional.Clearly, xy=2.510=416=7.530=1040=14, while 1442=13i.e., 2.510=416=7.530=1040is not equal to 1442.Therefore, x and y are not proportional.
(iii) Clearly, xy=515=721=927=2575=13, while 1560=1872=14i.e., 515=721=927=2575is not equal to 1560 and1872.Therefore, x and y are not proportional.Clearly, xy=515=721=927=2575=13, while 1560=1872=14i.e., 515=721=927=2575is not equal to 1560 and1872.Therefore, x and y are not proportional.
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Question 2:
If x and y are directly proportional, find the values of x1 , x2 and y1 in the table given below:
x
3
x1
x2
10
y
72
120
192
y1
ANSWER:
Since x and y are directly propotional, we have: 372=x1120=x2192=10y1Now, 372=x1120⇒x1=120×372 = 5Since x and y are directly propotional, we have: 372=x1120=x2192=10y1Now, 372=x1120⇒x1=120×372 = 5
A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?
ANSWER:
Let the required distance be x km. Then, we have:
Quantity of diesel (in litres)
34
20
Distance (in km)
510
x
Clearly, the less the quantity of diesel consumed, the less is the distance covered. So, this is a case of direct proportion. Now, 34510=20x⇒115=20x⇒x×1=20×15=300Now, 34510=20x⇒115=20x⇒x×1=20×15=300
Therefore, the required distance is 300 km.
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Question 4:
A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?
ANSWER:
Let the charge for a journey of 124 km be ₹x.
Price(in ₹)
2550
x
Distance(in km)
150
124
More is the distance travelled, more will be the price. So, it is a case of direct proportion. ∴2550150=x124⇒x=2550×124150=2108∴2550150=x124⇒x=2550×124150=2108 Thus, the taxi charges ₹2,108 for the distance of 124 km.
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Question 5:
A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?
ANSWER:
Let the required distance be x km. Then, we have: 1 h=60 mini.e., 5 h=5×60=300 min1 h=60 mini.e., 5 h=5×60=300 min.
Distance (in km)
16
x
Time (in min)
25
300
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion. Now, 1625=x300⇒x=(16×30025)⇒x = 192Now, 1625=x300⇒x=16×30025⇒x = 192 Therefore, the required distance is 192 km.
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Question 6:
If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?
ANSWER:
Let the required number of dolls be x. Then, we have:
No of dolls
18
x
Cost of dolls (in rupees)
630
455
Clearly, the less the amount of money, the less will be the number of dolls bought. So, this is a case of direct proportion. Now, 18630=x455⇒135=x455⇒x=45535⇒x=13Now, 18630=x455⇒135=x455⇒x=45535⇒x=13
Therefore, 13 dolls can be bought for Rs 455.
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Question 7:
If 9 kg of sugar costs ₹ 238.50, how much sugar can be bought for ₹ 371?
ANSWER:
Let the quantity of sugar bought for ₹371 be x kg.
Quantity(in kg)
9
x
Price(in ₹)
238.50
371
The price increases as the quantity increases. Thus, this is a case of direct proportion. ∴9238.50=x371⇒x=9×371238.50=14∴9238.50=x371⇒x=9×371238.50=14 Thus, the quantity of sugar bought for ₹371 is 14 kg.
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Question 8:
The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?
ANSWER:
Let the length of cloth be x m. Then, we have:
Length of cloth (in metres)
15
x
Cost of cloth (in rupees)
981
1308
Clearly, more length of cloth can be bought by more amount of money. So, this is a case of direct proportion. Now, 15981=x1308⇒x=15×1308981⇒x=20Now, 15981=x1308⇒x=15×1308981⇒x=20
Therefore, 20 m of cloth can be bought for Rs 1,308.
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Question 9:
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?
ANSWER:
Let x m be the length of the model of the ship. Then, we have: 1 m = 100 cmTherefore, 15 m= 1500 cm35 m= 3500 cm1 m = 100 cmTherefore, 15 m= 1500 cm35 m= 3500 cm
Length of the mast (in cm)
Length of the ship (in cm)
Actual ship
1500
3500
Model of the ship
9
x
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more. So, this is a case of direct proportion. Now, 15009=3500x⇒x=3500×91500⇒x=21 cmNow, 15009=3500x⇒x=3500×91500⇒x=21 cm Therefore, the length of the model of the ship is 21 cm.
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Question 10:
In 8 days, the earth picks up (6.4 × 107) kg of dust from the atmosphere. How much dust will it pick up in 15 days?
ANSWER:
Let x kg be the required amount of dust. Then, we have:
No. of days
8
15
Dust (in kg)
6.4×1076.4×107
x
Clearly, more amount of dust will be collected in more number of days. So, this is a case of direct proportion. Now, 86.4×107=15x⇒x=15×6.4×1078⇒x=12×107Now, 86.4×107=15x⇒x=15×6.4×1078⇒x=12×107
Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.
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Question 11:
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?
Clearly, more distance will be covered in more time. So, this is a case of direct proportion. Now, 5060=x72⇒x=50×7260⇒x=60Now, 5060=x72⇒x=50×7260⇒x=60 Therefore, the distance travelled by the car in 1 h 12 min is 60 km.
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Question 12:
Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?
ANSWER:
Let x km be the required distance covered by Ravi in 2 h 24 min. Then, we have: 1 h=60mini.e., 2 h 24 min=(120+24) min=144 min1 h=60mini.e., 2 h 24 min=(120+24) min=144 min
Distance covered (in km)
5
x
Time (in min)
60
144
Clearly, more distance will be covered in more time. So, this is a case of direct proportion. Now,560=x144⇒x=5×14460⇒x=12Now,560=x144⇒x=5×14460⇒x=12
Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.
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Question 13:
If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.
ANSWER:
Let x mm be the required thickness. Then, we have:
Thickness of cardboard (in mm)
65
x
No. of cardboards
12
312
Clearly, when the number of cardboard is more, the thickness will also be more. So, it is a case of direct proportion. Now, 6512=x312⇒x=65×31212⇒x=1690Now, 6512=x312⇒x=65×31212⇒x=1690
Therefore, the thickness of the pile of 312 cardboards is 1690 mm.
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Question 14:
11 men can dig 634634-metre-long trench in one day. How many men should be employed for digging 27-metre-long trench of the same type in one day?
ANSWER:
Let x be the required number of men.
Now, 634 m=274 mNow, 634 m=274 m
Then, we have:
Number of men
11
x
Length of trench (in metres)
274274
27
Clearly, the longer the trench, the greater will be the number of men required. So, it is a case of direct proportion. Now, 11274=x27⇒11×427=x27⇒x=44Now, 11274=x27⇒11×427=x27⇒x=44
Therefore, 44 men should be employed to dig a trench of length 27 m.
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Question 15:
Reenu types 540 words during half an hour. How many words would she type in 8 minutes?
ANSWER:
Let Reenu type x words in 8 minutes.
No. of words
540
x
Time taken (in min)
30
8
Clearly, less number of words will be typed in less time. So, it is a case of direct proportion. Now,54030=x8⇒x=540×830⇒x=144Now,54030=x8⇒x=540×830⇒x=144
Therefore, Reenu will type 144 words in 8 minutes.
Page No 165:
Question 1:
Observe the tables given below and in each case find whether x and y are inversely proportional: (i)
x
6
10
14
16
y
9
15
21
24
(ii)
x
5
9
15
3
45
y
18
10
6
30
2
(iii)
x
9
3
6
36
y
4
12
9
1
ANSWER:
(i) Clearly, 6×9≠ 10×15 ≠ 14×21≠ 16×24Therefore, x and y are not inversely proportional.Clearly, 6×9≠ 10×15 ≠ 14×21≠ 16×24Therefore, x and y are not inversely proportional.
(ii) Clearly, 5×18= 9×10=15×6=3×30=45×2=90=(consant)Therefore, x and y are inversely proportional.Clearly, 5×18= 9×10=15×6=3×30=45×2=90=(consant)Therefore, x and y are inversely proportional.
(iii) Clearly, 9×4=3×12=36×1=36, while 6×9=54i.e., 9×4=3×12=36×1≠6×9Therefore, x and y are not inversely proportional.Clearly, 9×4=3×12=36×1=36, while 6×9=54i.e., 9×4=3×12=36×1≠6×9Therefore, x and y are not inversely proportional.
Page No 165:
Question 2:
If x and y are inversely proportional, find the values of x1, x2, y1 and y2 in the table given below:
x
8
x1
16
x2
80
y
y1
4
5
2
y2
ANSWER:
Since x and y are inversely proportional, xy must be a constant.Since x and y are inversely proportional, xy must be a constant. Therefore, 8×y1=x1×4=16×5=x2×2=80×y2Now, 16×5=8×y1⇒808=y1∴ y1=1016×5=x1×4⇒804=x1∴ x1=2016×5=x2×2⇒802=x2∴ x2=4016×5=80×y2⇒8080=y2∴ y2=1Hence, y1=10, x1=20, x2=40 and y2=1Therefore, 8×y1=x1×4=16×5=x2×2=80×y2Now, 16×5=8×y1⇒808=y1∴ y1=1016×5=x1×4⇒804=x1∴ x1=2016×5=x2×2⇒802=x2∴ x2=4016×5=80×y2⇒8080=y2∴ y2=1Hence, y1=10, x1=20, x2=40 and y2=1
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Question 3:
If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field?
ANSWER:
Let x be the required number of days. Then, we have:
No. of days
8
x
No. of men
35
20
Clearly, less men will take more days to reap the field. So, it is a case of inverse proportion.
Therefore, 20 men can reap the same field in 14 days.
Page No 165:
Question 4:
12 men can dig a pond in 8 days. How many men can dig it in 6 days?
ANSWER:
Let x be the required number of men. Then, we have:
No. of days
8
6
No. of men
12
x
Clearly, more men will require less number of days to dig the pond. So, it is a case of inverse proportion. Now, 8 × 12 = 6 × x⇒x=8 × 126 ⇒x=16Now, 8 × 12 = 6 × x⇒x=8 × 126 ⇒x=16
Therefore, 16 men can dig the pond in 6 days.
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Question 5:
6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field?
ANSWER:
Let x be the number of days. Then, we have:
No. of days
28
x
No. of cows
6
14
Clearly, more number of cows will take less number of days to graze the field. So, it is a case of inverse proportion. Now, 28 × 6 = x × 14⇒x=28 × 614 ⇒x=12Now, 28 × 6 = x × 14⇒x=28 × 614 ⇒x=12
Therefore, 14 cows will take 12 days to graze the field.
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Question 6:
A car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 75 km/hr?
ANSWER:
Let x h be the required time taken. Then, we have:
Speed (in km/h)
60
75
Time (in h)
5
x
Clearly, the higher the speed, the lesser will be the the time taken. So, it is a case of inverse proportion. Now, 60×5=75×x⇒x=60×575⇒x=4Now, 60×5=75×x⇒x=60×575⇒x=4
Therefore, the car will reach its destination in 4 h if it travels at a speed of 75 km/h.
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Question 7:
A factory requires 42 machines to produce a given number of articles in 56 days. How many machines would be required to produce the same number of articles in 48 days?
ANSWER:
Let x be the number of machines required to produce same number of articles in 48. Then, we have:
No. of machines
42
x
No. of days
56
48
Clearly, less number of days will require more number of machines. So, it is a case of inverse proportion. Now, 42×56=x×48⇒x=42×5648⇒x=49Now, 42×56=x×48⇒x=42×5648⇒x=49
Therefore, 49 machines would be required to produce the same number of articles in 48 days.
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Question 8:
7 teps of the same size fill a tank in 1 hour 36 minutes. How long will 8 taps of the same size take to fill the tank?
ANSWER:
Let x be the required number of taps. Then, we have: 1 h = 60 min i.e., 1 h 36 min = (60+36) min = 96 min
No. of taps
7
8
Time (in min)
96
x
Clearly, more number of taps will require less time to fill the tank. So, it is a case of inverse proportion. Now, 7×96=8×x⇒x=7×968⇒x=84Now, 7×96=8×x⇒x=7×968⇒x=84
Therefore, 8 taps of the same size will take 84 min or 1 h 24 min to fill the tank.
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Question 9:
8 taps of the same size fill a tank in 27 minutes. If two taps go out of order, how long would the remaining taps take to fill the tank?
ANSWER:
Let x min be the required number of time. Then, we have:
No. of taps
8
6
Time (in min)
27
xx
Clearly, less number of taps will take more time to fill the tank . So, it is a case of inverse proportion.
A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle?
ANSWER:
Let x be the required number of days. Then, we have:
No. of days
9
x
No. of animals
28
36
Clearly, more number of animals will take less number of days to finish the food. So, it is a case of inverse proportion. Now, 9×28=x×36⇒x=9×2836⇒x=7Now, 9×28=x×36⇒x=9×2836⇒x=7
Therefore, the food will last for 7 days.
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Question 11:
A garrison of 900 men had provisions for 42 days. However, a reinforcement of 500 men arrived. For how many days will the food last now?
ANSWER:
Let x be the required number of days. Then, we have:
No. of men
900
1400
No. of days
42
x
Clearly, more men will take less number of days to finish the food. So, it is a case of inverse proportion.
In a hostel, 75 students had food provision for 24 days. If 15 students leave the hostel, for how many days would the food provision last?
ANSWER:
Let x be the required number of days. Then, we have:
No. of students
75
60
No. of days
24
x
Clearly, less number of students will take more days to finish the food. So, it is a case of inverse proportion. Now, 75×24=60×x⇒x=75×2460⇒x=30Now, 75×24=60×x⇒x=75×2460⇒x=30
Therefore, the food will now last for 30 days.
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Question 13:
A school has 9 periods a day each of 40 minutes duration. How long would each period be, if the school has 8 periods a day, assuming the number of school hours to be the same?
ANSWER:
Let x min be the duration of each period when the school has 8 periods a day.
No. of periods
9
8
Time (in min)
40
x
Clearly, if the number of periods reduces, the duration of each period will increase. So, it is a case of inverse proportion. Now, 9×40=8×x⇒x=9×408⇒x=45Now, 9×40=8×x⇒x=9×408⇒x=45
Therefore, the duration of each period will be 45 min if there were eight periods a day.
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Question 14:
If x and y vary inversely and x = 15 when y = 6, find y when x = 9.
ANSWER:
xx
15
9
yy
6
y1y1
x and y vary inversely.i.e. xy = constantNow, 15×6=9×y1⇒y1=15×69⇒y1=10x and y vary inversely.i.e. xy = constantNow, 15×6=9×y1⇒y1=15×69⇒y1=10
∴ Value of y=10y=10, when x =9
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Question 15:
If x and y vary inversely and x = 18 when y = 8, find x when y = 16.
ANSWER:
xx
18
x1x1
yy
8
16
x and y vary inversely.i.e. xy = constantNow, 18×8=x1×16⇒18×816=x1⇒9=x1x and y vary inversely.i.e. xy = constantNow, 18×8=x1×16⇒18×816=x1⇒9=x1
∴ Value of x=9x=9
Page No 166:
Question 1:
If 14 kg of pulses cost ₹ 882, what is the cost of 22 kg of pulses? (a) ₹ 1254 (b) ₹ 1298 (c) ₹ 1342 (d) ₹ 1386
ANSWER:
Let 22 kg of pulses cost ₹x.
Quantity(in kg)
14
22
Price(in ₹)
882
x
As the quantity increases, the price also increases. So, it is a case of direct proportion. ∴14882=22x⇒x=22×88214=1386∴14882=22x⇒x=22×88214=1386 Thus, the cost of 22 kg of pulses is ₹1,386.
Hence, the correct answer is option (d).
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Question 2:
If 8 oranges cost ₹ 52, how many oranges can be bought for ₹ 169? (a) 13 (b) 18 (c) 26 (d) 24
ANSWER:
Let the number of oranges that can be bought for ₹169 be x.
Quantity
8
x
Price(in ₹)
52
169
As the quantity increases the price also increases. So, this is a case of direct proportion. ∴852=x169⇒x=8×16952=26∴852=x169⇒x=8×16952=26 Thus, 26 oranges can be bought for ₹169.
Hence, the correct answer is option (c).
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Question 3:
Tick (✓) the correct answer: A machine fills 420 bottles in 3 hours. How many bottles will it fill in 5 hours? (a) 252 (b) 700 (c) 504 (d) 300
ANSWER:
(b) 700
Let x be the number of bottles filled in 5 hours.
No. of bottles
420
xx
Time (h)
3
5
More number of bottles will be filled in more time.
Tick (✓) the correct answer: A car is travelling at a uniform speed of 75 km/hr. How much distance will it cover in 20 minutes? (a) 25 km (b) 15 km (c) 30 km (d) 20 km
ANSWER:
(a) 25 km
Let x km be the required distance. Now, 1 h = 60 min
Distance (in km)
75
xx
Time (in min)
60
20
Less distance will be covered in less time. Now, 7560=x20⇒x=75×2060⇒x=25 kmNow, 7560=x20⇒x=75×2060⇒x=25 km
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Question 5:
Tick (✓) the correct answer: The weight of 12 sheets of a thick paper is 40 grams. How many sheets would weight 1 kg? (a) 480 (b) 360 (c) 300 (d) none of these
ANSWER:
(c) 300 Let x sheets weigh 1 kg. Now, 1 kg = 1000 g
Tick (✓) the correct answer: A pole 14 m high casts a shadow of 10 m. At the same time, what will be the height of a tree, the length of whose shadow is 7 metres? (a) 20 m (b) 9.8 m (c) 5 m (d) none of these
ANSWER:
(b) 9.8 m Let x m be the height of the tree.
Height of object
14
xx
Length of shadow
10
7
The more the length of the shadow, the more will be the height of the tree. Now, 1410=x7⇒x=14×710⇒x=9.8Now, 1410=x7⇒x=14×710⇒x=9.8
Therefore, a 9.8 m tall tree will cast a shadow of length 7 m.
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Question 7:
Tick (✓) the correct answer: A photograph of a bacteria enlarged 50000 times attains a length of 5 cm. The actual length of bacteria is (a) 1000 cm (b) 10−3 cm (c) 10−4 cm (d) 10−2 cm
ANSWER:
(c) 10−4 cm10-4 cm Let x cm be the actual length of the bacteria. The larger the object, the larger its image will be. Now, x1=550000=10−4 cmNow, x1=550000=10-4 cm
Hence, the actual length of the bacteria is 10−4 cm10-4 cm.
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Question 8:
Tick (✓) the correct answer: 6 pipes fill a tank in 120 minutes, then 5 pipes will fill it in (a) 100 min (b) 144 min (c) 140 min (d) 108 min
ANSWER:
(b) 144 min Let x min be the time taken by 5 pipes to fill the tank.
No. of pipes
6
5
Time (in min)
120
xx
Now, 6×120=5×x⇒x=144 Now, 6×120=5×x⇒x=144
Therefore, 5 pipes will take 144 min to fill the tank.
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Question 9:
Tick (✓) the correct answer: 3 persons can build a wall in 4 days, then 4 persons can build it in (a) 513513 days (b) 3 days (c) 413413 days (d) none of these
ANSWER:
(b) 3 days Let x be number of days taken by 4 persons to build the wall.
No. of persons
3
4
No. of days
4
xx
More number of persons will take less time to build the wall. So, it is a case of inverse proportion.
Now, 3×4=4×x⇒x=3 Now, 3×4=4×x⇒x=3
Therefore, 4 persons can build the wall in 3 days.
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Question 10:
Tick (✓) the correct answer: A car takes 2 hours to reach a destination by travelling at 60 km/hr. How long will it take while travelling at 80 km/hr? (a) 1 hr 30 min (b) 1 hr 40 min (c) 2 hrs 40 min (d) none of these
ANSWER:
(a) 1 h 30 min Let x h be the time taken by the car travelling at 80 km/hr.
Speed (km/h)
60
80
Time (in h)
2
xx
The greater the speed, the lesser will be the time taken.So, it is a case of inverse proportion.Now, 60×2= 80×x⇒x= 12080⇒x= 1.5Therefore, the car will take 1 h 30 min to reach its destination if it travels at a speed of 80 km/h.The greater the speed, the lesser will be the time taken.So, it is a case of inverse proportion.Now, 60×2= 80×x⇒x= 12080⇒x= 1.5Therefore, the car will take 1 h 30 min to reach its destination if it travels at a speed of 80 km/h.
Page No 168:
Question 1:
350 boxes can be placed in 25 cartons. How many boxes can be placed in 16 cartons?
ANSWER:
Let x be the required number of boxes.
No. of boxes
350
xx
No. of cartons
25
16
Less number of boxes will require less number of cartons. So, it is a case of direct proportion. Now, 35025=x16⇒x=350×1625⇒x=224Now, 35025=x16⇒x=350×1625⇒x=224
∴ 224 boxes can be placed in 16 cartoons.
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Question 2:
The cost of 140 tennis balls is Rs 4900. Find the cost of 2 dozen such balls.
ANSWER:
Let Rs x be the cost of 24 tennis balls.
No. of balls
140
24
Cost of balls
4900
xx
More tennis balls will cost more. Now, 1404900=24x⇒x=24×4900140⇒x=840Now, 1404900=24x⇒x=24×4900140⇒x=840
∴ The cost of 2 dozen tennis balls is Rs 840.
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Question 3:
The railway fare for 61 km is Rs 183. Find the fare for 53 km.
ANSWER:
Let Rs x be the railway fare for a journey of distance 53 km.
Distance (in km)
61
53
Railway fare (in rupees)
183
xx
The lesser the distance, the lesser will be the fare. So, it is a case of direct proportion . Now, 61183=53x⇒x=53×18361⇒x=159Now, 61183=53x⇒x=53×18361⇒x=159
The railway fare for a journey of distance 53 km is Rs 159.
MCQ Questions for Class 8 Maths: Ch 12 Exponents and Powers
1. Multiplicative inverse of 7-2 is _________
(a) 49
(b) 5
(c) 7
(d) -14
► (a) 49
2. (am)n is equal to
(a) am+n
(b) am-n
(c) amn
(d) an-m
► (c) amn
3. Very small numbers can be expressed in standard form using __________ exponents.
(a) equal
(b) negative
(c) positive
(d) none of these
► (b) negative
4. The Base in the expression 1024 is __________.
(a) 1
(b) 10
(c) 0
(d) 24
► (b) 10
5. Fill in the blank am × an = a …….. where m and n are natural numbers:-
(a) mn
(b) m + n
(c) m – n
(d) m/n
► (b) m + n
6. The value of 1/3-2 is equal to
(a) 9
(b) 1
(c) -6
(d) 1/3
► (a) 9
7. When we have to add numbers in standard form, we convert them into numbers with the ________ exponents. (a) same (b) different (c) not equal (d) None of these ► (a) same
8. The value of 30 is ________. (a) 0 (b) 3 (c) 1 (d) None of these ► (c) 1
9. Value of (30 + 20) × 50 is (a) 1 (b) 25 (c) 2 (d) 0 ► (c) 2
10. In exponential form 149,600,000,000 m is given by :
(a) 1.496 × 1011 m
(b) 1.496 × 108 m
(c) 14.96 × 108 m
(d) 14.96 × 1011 m
► (a) 1.496 × 1011 m
11. Evaluate the exponential expression (−n)4× (−n)2, for n = 5.
(a) 25
(b) 15625
(c) 3125
(d) 625 ► (b) 15625
12. Evaluate exponential expression − 25.
(a) 15
(b) -32
(c) 16
(d) none of these ► (b) -32
13. Charge of an electron is 0.000,000,000,000,000,000,16 coulomb and in exponential form it can be written as (a) 16 × 10-18 coulomb (b) 1.6 × 10-21 coulomb (c) 1.6 × 10-19 coulomb (d) 16 × 10-21 coulomb ► (c) 1.6 × 10-19 coulomb
14. Find the value of the expression a2 for a = 10.
(a) 100
(b) 1
(c) 10
(d) None of these ► (a) 100
15. In 102 the base is
(a) 1
(b) 0
(c) 10
(d) 100 ► (c) 10
16. The multiplicative inverse of 2-3 is
(a) 2
(b) 3
(c) 3
(d) 23
► (d) 23
17. 16 is the multiplicative inverse of
(a) 2-4
(b) 28
(c) 82
(d) 24
► (a) 2-4
18. The standard form of 9030000000 is given by
(a) 9.03 × 109
(b) 90.3 × 107
(c) 903 × 106
(d) 9.03 × 10-9
► (a) 9.03 × 109
19. The value of 72 is
(a) 7
(b) 49
(c) 2
(d) 14
► (b) 49
20. The value of 10000 is
(a) 0
(b) 1000
(c) 1
(d) None of these
► (c) 1
Exponents and Powers Class 8 Extra Questions Very Short Answer Type
Question 1. Find the multiplicative inverse of: (i) 3-3 (ii) 10-10 Solution:
Question 2. Expand the following using exponents. (i) 0.0523 (ii) 32.005 Solution:
Question 3. Simplify and write in exponential form. Solution:
Question 4. Simplify the following and write in exponential form. Solution:
Question 5. Express 8-4 as a power with the base 2. Solution: We have 8 = 2 × 2 × 2 = 23 8-4 = (23)-4 = 23×(-4) = 2-12
Question 6. Simplify the following and write in exponential form. (i) (36 ÷ 38)4 × 3-4 (ii) 127 × 3-3 Solution:
Question 7. Find the value of k if (-2)k+1 × (-2)3 = (-2)7 Solution: (-2)k+1 × (-2)3 = (-2)7 ⇒ (-2)k+1+3 = (-2)7 ⇒ (-2)k+4 = (-2)7 ⇒ k + 4 = 7 ⇒ k = 3 Hence, k = 3.
Question 8. Simplify the following: Solution:
Question 9. Find the value of [(−34)−2]2 Solution:
Question 10. Write the following in standard form (i) 0 0035 (ii) 365.05 Solution:
Exponents and Powers Class 8 Extra Questions Short Answer Type
Question 11. Find the value of P if Solution:
Question 12. Solution:
Question 13. Find the value of x if Solution: ⇒ 3 + x = 18 [Equating the powers of same base] x = 18 – 3 = 15
Question 17. Find x so that (-5)x+1 × (-5)5 = (-5)7 (NCERT Exemplar) Solution: (-5)x+1 × (-5)5 = (-5)7 (-5)x+1+5 = (-5)7 {am × an = am+n} (-5)x+6 = (-5)7 On both sides, powers have the same base, so their exponents must be equal. Therefore, x + 6 = 7 x = 7 – 6 = 1 x = 1.
MCQ Questions for Class 8 Maths: Ch 11 Mensuration
1. In a quadrilateral, half of the product of the sum of the lengths of parallel sides and the parallel distance between them gives the area of
(a) rectangle
(b) parallelogram
(c) triangle
(d) trapezium
► (d) trapezium
2. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
(a) 215 cm3
(b) 172 cm3
(c) 150 cm3
(d) 168 cm3
►(d) 168 cm3
3. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm. (a) 168 cm2 (b) 168 cm3 (c) 215 cm3 (d) 150 cm3 ► (b) 168 cm3
4. The formula for finding total surface area of cuboid is
(a) 2 (lb x bh x hl)
(b) 2 (lb + bh + hl)
(c) 2h (l + b)
(d) 2 lb (bh + hl)
► (b) 2 (lb + bh + hl)
5. Which of the following is an example of two dimensions
(a) cuboid
(b) cone
(c) sphere
(d) circle
► (d) circle
6. The area of a trapezium is
(a) 1/2 (sum of parallel sides) × h
(b) 2 (sum of parallel sides) × h
(c) (sum of parallel sides) × h
(d) 1/2 (sum of parallel sides) + h
► (a) 1/2 (sum of parallel sides) × h
7. The formula for finding lateral surface area of cylinder is
(a) 2πrh
(b) πr2
(c) 2πr(r+h)
(d) 2πr
► (a) 2πrh
8. Solid figures are (a) 2 D (b) 3 D (c) 1 D (d) 4 D ► (b) 3 D
9. A rectangular paper of width 7 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.
(a) 8800 cm3 (b) 8800 cm (c) 8800 cm2 (d) none of these ► (a) 8800 cm3
10. Find the total surface area of a cube whose volume is 343 cm3. (a) 350 cm2 (b) 294 cm2 (c) 494 cm2 (d) 200 cm2 ► (b) 294 cm2
11. Two dimensional figure is a
(a) solid figure
(b) plane figure
(c) cylinder figure
(d) None of these
► (b) plane figure
12. A cylindrical tank has a capacity of 5632 m3. If the diameter of its base is 16 m, find its depth.
(a) 66m
(b) 30 m
(c) 26 m
(d) 28 m
► (d) 28 m
13. The length of parallel sides of trapezium is 14 cm and 6 cm and its height is 5 cm. Its area will be
(a) 50 cm2
(b) 100 cm2
(c) 210 cmsup>2
(d) 10 cm2
► (a) 50 cm2
14. The amount of space occupied by a three dimensional objects is called its (a) area (b) surface area (c) volume (d) lateral surface area ► (c) volume
15. Surface area of a cuboid = __________ (a) 2 h (l + b) (b) 2lbh (c) 2(lb + bh + hl) (d) None of these ► (c) 2(lb + bh + hl)
16. Find the height of cuboid whose volume is 490 cmsup>3
and base area is 35 cmsup>3. (a) 12 cm (b) 14 cm (c) 10 cm (d) 16 cm ► (b) 14 cm
17. The cost of papering the wall of a room, 12 m long, at the rate of Rs. 1.35 per square meter is Rs. 340.20. The cost of matting the floor at Re. 0.85 per square metre is Rs. 91.80. Find the height of the room. (a) 12 m (b) 8 m (c) 6 m (d) 10 m ► (c) 6 m
18. The area of four walls of the room is
(a) 2 (lb + bh + hl)
(b) 2l (h + b)
(c) 2 (lb x bh x hl)
(d) 2h (l + b)
► (d) 2h (l + b)
19. The formula for lateral surface area of cuboid is
(a) 2h (l + b)
(b) 2l (h + b)
(c) 2b (l + h)
(d) 2 (lb + bh + hl)
► (a) 2h (l + b)
20. Diagonals of rhombus are
(a) equal
(b) half of one diagonal
(c) of different length
(d) none of above
► (c) of different length
Short Answer Type Questions:
1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.
Solution:
Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides
= ½ × 15 × (12 + 20)
= 1/2 × 15 × 32
= 15 × 16
= 240 cm2
2. Calculate the height of a cuboid which has a base area of 180 cm2 and volume is 900 cm3.
Solution:
Volume of cuboid = base area × height
900 = 180 × height
So, height = 900/180 = 5 cm
3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.
Solution:
Perimeter of square formula = 4 × side of the square
Hence, P (square) = 4 × 60 = 240 cm
Perimeter of rectangle formula = 2 × (Length + Breadth)
Hence, P (rectangle) = 2 (80 + Breadth)
= 160 + 2 × Breadth
According to the given question, 160 + 2 × Breadth = 240 cm 2 × Breadth = 240 – 160 Breadth = 80/2 The breadth of the rectangle = 40 cm
Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm2
4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.
Solution:
Given, length of lawnmower = 1m = 100cm
Its circumference = π × D = 22/7 × 84 = 264 cm
Length of field will be = 264 × 750 = 198000 cm
Here, the width of field = length of the lawnmower i.e. 100 cm
So, area of field = 198000 × 100 = 19,800,000 cm²
Or, 1980 m²
5. The area of a rhombus is 16 cm2 and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.
Solution:
Area of rhombus = ½ × d1 × d2
⇒ 16 = ½ × 4 × d2
So, d2 = 32/4 = 8 cm
Long Answer Type Questions:
6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.
Solution:
The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm
We know,
Area of circle = πr²
So,
Area of the entire circle = π(4)² = 16π cm2
And,
Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm2
Thus, the remaining area = 16π – 9π = 7π cm2
7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.
Solution:
Given,
Length of the box, l = 2 m,
Breadth of box, b = 1 m
Height of box, h = 1.5 m
We know that the surface area of a cuboid = 2(lb + lh + bh)
But here the bottom part is not to be painted.
So,
Surface area of box to be painted = lb + 2(bh + hl)
= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)
= 2 + 2 (1.5 + 3.0)
= 2 + 9.0
= 11
Hence, the required surface area of the cuboidal box = 11 m2
8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.
Solution:
From the question statement draw the diagram.
Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.
Now, CM will be the distance between the two parallel sides or the height of the trapezium.
We know,
Area of trapezium = ½ × sum of parallel sides × height.
So, height has to be found.
In the diagram, draw CL || AD
Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm
As AD = CB,
CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.
Here, BL = AB – AL = (40 – 20) = 20 cm. So,
LM = MB = ½ BL = ½ × 20 = 10 cm
Now, in ΔCLM,
CL2 = CM2 + LM2 (Pythagoras Theorem)
262 = CM2 + 102
CM2 = 262 – 102
Using algebraic identities, we get; 262 – 102 = (26 – 10) (26 + 10)
MCQ Questions for Class 8 Maths: Ch 10 Visualising Solid Shapes
1. Which is the three-dimensional figure formed by rotating a triangle?
(a) Cone
(b) Quadrilateral
(c) Prism
(d) Square
► (a) Cone
2. How many congruent edges does a cube have?
(a) 12
(b) 6
(c) 8
(d) none of these
► (a) 12
3. A square pyramid always has ___.
(a) Four lateral faces, which are parallel to each other.
(b) Four lateral faces, which are congruent equilateral triangles and a rectangular base.
(c) Two bases which are congruent and parallel.
(d) Four lateral faces, which are congruent isosceles triangles and a square base.
► (d) Four lateral faces, which are congruent isosceles triangles and a square base.
4. How many circular bases does a cylinder have?
(a) 1
(b) 2
(c) 3
(d) 4
► (b) 2
5. How many vertices does a triangular pyramid have ?
(a) 1
(b) 2
(c) 3
(d) 4
► (d) 4
6. The lateral faces of a pyramid are____.
(a) triangles
(b) pentagons
(c) rectangles
(d) None of these
► (a) triangles
7. How many triangles would you find in a net that folds into a square pyramid?
(a) 3
(b) 4
(c) 5
(d) 6
► (b) 4
8. Identify the statement that is false for a prism.
(a) A right prism has rectangular lateral faces.
(b) A right prism has two bases.
(c) A right prism has triangular lateral faces.
(d) A right prism has identical parallel faces.
9. What is the difference between a rectangle and a cube?
(a) A rectangle is 3-dimensional and a cube is 2-dimensional.
(b) A rectangle and a cube are both 2-dimensional.
(c) A rectangle is 2-dimensional and a cube is 3-dimensional.
(d) A rectangle and a cube are both 3-dimensional.
► (c) A rectangle is 2-dimensional and a cube is 3-dimensional.
10. Which of the following statements is true?
(a) The lateral faces of a square prism are triangles.
(b) The lateral faces of a triangular prism can be squares or rectangles.
(c) The lateral faces of a square pyramid can be squares.
(d) The lateral faces of a triangular pyramid can be squares or rectangles.
► (b) The lateral faces of a triangular prism can be squares or rectangles.
11. How many vertices does a cone have?
(a) 2
(b) 1
(c) 3
(d) 4
► (b) 1
12. For a polyhedron, if ‘F’ stands for number of faces, V stands for number of vertices and E stands for number of edges, then which of the following relationships is named as Euler’s formula ?
(a) F + V = E + 2
(b) F + E = V + 2
(c) V + E = F + 2
(d) F+ V = E – 2
► (a) F + V = E + 2
13. The top-view of a cube looks like a:
(a) Circle
(b) Square
(c) Rectangle
(d) Triangle
► (b) Square
14. The top-view of a cuboid looks like a:
(a) Circle
(b) Square
(c) Rectangle
(d) Triangle
► (c) Rectangle
15. How many circular bases does a cylinder have?
(a) 2
(b) 3
(c) 4
(d) 1
► (a) 2
16. The faces of a triangular pyramid consist of
(a) 1 square and 3 triangles
(b) 2 triangles and 3 rectangles
(c) 4 triangles
(d) 2 rectangles and 3 triangles
► (c) 4 triangles
17. A three dimensional shape is _________ object.
(a) solid
(b) 2d
(c) plane
(d) None of these
► (a) solid
18. What do you call solid figures with line segments as their edges?
(a) Polygons
(b) Squares
(c) Cylinders
(d) Polyhedrons
► (d) Polyhedrons
19. Which of the following can be other name of a cylinder?
8. The expression x + y + z is in (a) one variable (b) no variable (c) three variables (d) two variables ► (c) three variables
9. Which of the following is a monomial ? (a) 4x² (b) a + 6 (c) a + 6 + c (d) a + b + c + d
► (a) 4x²
10. How many terms are there in the expression 5 – 3xy ?
(a) 1
(b) 2
(c) 3
(d) 5
► (b) 2
11. How many terms are there in the expression 5xy + 9yz + 3zx + 5x – 4y ?
(a) 1
(b) 3
(c) 4
(d) 5
► (d) 5
12. If x = 3 is solution of x² + kx + 15, value of k is
(a) k = -8, x = 3
(b) k = 8, x = 5
(c) k = 6, x = 5
(d) None of above
► (a) k = -8, x = 3
13. The coefficient in the term -5x is
(a) 5
(b) -5
(c) 1
(d) 2
► (b) -5
14. n (4 + m) = 4n + ___
(a) 4m
(b) 4n
(c) 4mn
(d) nm
► (d) nm
15. The like terms of the following are
(a) x, 3x
(b) x, 2y
(c) 2y, 6xy
(d) 3x, 2y
► (a) x, 3x
16. Value of expression ‘a(a2+a +1)+5’ for ‘ a’ = 0 is
(a) a+5
(b) 1
(c) 6
(d) 5 ► (d) 5
17. Which of the following is like term as 7xy? (a) 9 (b) 9x (c) 9y (d) 9xy ► (d) 9xy
18. The area of triangle is’ xy’ where’ x’ is length and ‘y’ is breadth. If the length of rectangle is increased by 5 units and breadth is decreased by 3 units, the new area of rectangle will be (a) (x-y)(x+3) (b) (xy+15) (c) (x+5)(y-3) (d) (xy+5-3) ► (c) (x+5)(y-3)
19. The expression 7xy has the factors (a) 7,x,y (b) x,y (c) 7,x (d) 7,y ► (a) 7,x,y
20. Which of the following is a binomial? (a) 4x + y + 2 (b) 2x + 7 (c) 3x + 4y – 6 (d) 3x ► (b) 2x + 7
21. When numbers/literals are added or subtracted, they are called _________. (a) identities (b) expressions (c) variables (d) terms ► (d) terms
22. The volume of rectangular box whose length, breadth and height is 2p,4q.8r respectively is (a) 14pqr (b) 2p+4q+8r (c) 64pqr (d) 64 ► (c) 64pqr
Algebraic Expressions and Identities Class 8 Extra important Questions Short Answer Type
Question 18. Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2 (NCERT Exemplar) Solution: LHS = (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q) [using a2 -b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q] = (22pq) (8q) = 176 pq2 = RHS. Hence Verified.
Question 19. Find the value of \(\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }\), using a suitable identity. (NCERT Exemplar) Solution:
Question 20. Find the value of x, if 10000x = (9982)2 – (18)2 (NCERT Exemplar) Solution: RHS = (9982)2 – (18)2 = (9982 + 18)(9982 – 18) [Since a2 -b2 = (a + b) (a – b)] = (10000) × (9964) LHS = (10000) × x Comparing L.H.S. and RHS, we get 10000x = 10000 × 9964 x = 9964
MCQ Questions for Class 8 Maths: Ch 8 Comparing Quantities
1. Find the ratio of speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(a) It is 1:2
(b) It is 1:3
(c) It is 2:1
(d) It is 3:1
► (a) It is 1:2
2. The ratio of 10m to 10 km is:
(a) 1/10
(b) 1/100
(c) 1/1000
(d) 1000
► (c) 1/1000
3. The price of a house was Rs 34,00,000 last year. It has increased by 20% this year. What is the price now?
(a) Rs 40,40,000
(b) Rs 40,80,000
(c) Rs 30,40,000
(d) Rs 30,80,000
► (b) Rs 40,80,000
4. Calculate compound interest on Rs 10,800 for 3 years at 12.5% per annum compounded annually.
(a) 13,377.34
(b) 4577.34
(c) 14,377.34
(d) None of these
► (b) 4577.34
5. An item marked at Rs 840 is sold for Rs 714. What is the discount %?
(a) 20%
(b) 10%
(c) 15%
(d) none of these
► (c) 15%
6. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?
(a) 20%
(b) 15%
(c) 10%
(d) 5%
► (c) 10%
7. A picnic is being planned in a school. Girls are 60% of the total number of students and are 300 in number. Find the ratio of the number of girls to the number of boys in the class.
(a) It is 3:2
(b) It is 3:1
(c) It is 2:3
(d) It is 2:1
► (a) It is 3:2
8. The price of a scooter was Rs 34,000 last year. It has increased by 20% this year. What is the price now?
(a) Rs 30,800
(b) Rs 30,400
(c) Rs 40,800
(d) Rs 40,400
► (c) Rs 40,800
9. A shopkeeper purchased 500 pieces for Rs 20 each. However 50 pieces were spoiled in the way and had to be thrown away. The remaining were sold at Rs 25 each. Find the gain or loss %.
(a) 18%
(b) 15%
(c) 12.5%
(d) none of these
► (c) 12.5%
10. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find the population after 3 years.
(a) 29484
(b) 28696
(c) 24576
(d) 30184
► (a) 29484
11. Find the ratio of speed of a cycle 20 km per hour to the speed of scooter 30 km per hour.
(a) It is 3:1
(b) It is 2:1
(c) It is 2:3
(d) It is 1:3
► (c) It is 2:3
12. The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
(a) 25153
(b) 24153
(c) 23153
(d) None of these
► (c) 23153
13. Sohan bought a washing machine for Rs 40,000, then spent Rs 5,000 on its repairs and sold it for Rs 50,000. Find his loss or gain per cent.
(a) Loss 10%
(b) Loss 20%
(c) Profit 11%
(d) none of these
► (c) Profit 11%
14. ________ means comparing two quantities.
(a) Ratio
(b) Proportion
(c) Percent
(d) None of these
► (a) Ratio
15. Find selling price (SP) if a profit of 5% is made on a cycle of Rs 700 with Rs 50 as overhead charges.
(a) Rs 600
(b) Rs 787.50
(c) Rs 780
(d) None of these
► (b) Rs 787.50
16. Find the ratio of speed of a car 50 km per hour to the speed of scooter 40 km per hour.
(a) It is 4:5
(b) It is 4:1
(c) It is 5:4
(d) It is 1:5
► (c) It is 5:4
17. The sale price of a shirt is Rs.176. If a discount of 20% is allowed on its marked price, what is the marked price of the shirt?
(a) Rs.160
(b) Rs.180
(c) Rs. 200
(d) Rs. 220
► (d) Rs. 220
18. The difference in S.I. and C.I. on a certain sum of money in 2 years at 15% p.a. is Rs.144. Find the sum.
(a) Rs. 6000
(b) Rs. 6200
(c) Rs. 6300
(d) Rs. 6400
► (d) Rs. 6400
19. Rohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500 on its repairs and sold it for Rs 3,300. Find his loss or gain per cent.
(a) Loss 15% 2a
(b) Loss 10%
(c) Profit 10%
(d) None of these
► (c) Profit 10%
20. The value of an article which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs.9680, what is the price at which it was purchased?
(a) Rs.10000
(b) Rs.12500
(c) Rs.14575
(d) Rs.16250
► (b) Rs.12500
21. Find the ratio of 5 m to 10 km.
(a) It is 1:3
(b) It is 3000:1
(c) It is 2000:1
(d) It is 1:2000
► (d) It is 1:2000
22. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?
(a) 20%
(b) 15%
(c) 10%
(d) 5%
► (c) 10%
23. A shopkeeper purchased 300 bulbs for Rs 10 each. However 10 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %.
(a) 15%
(b) 13%
(c) 16%
(d) none of these
► (c) 16%
24. In what time will Rs.1000 amount to Rs.1331 at 10% p.a. compounded annually?
(a) 4 years
(b) 3 years
(c) 2 years
(d) 1 year
► (b) 3 years
Comparing Quantities Class 8 Extra Questions Very Short Answer Type
Question 1. Express the following in decimal form: (a) 12% (b) 25% Solution: (a) 12% = 12100 = 0.12 (b) 25% = 25100 = 0.25
Question 2. Evaluate the following: (a) 20% of 400 (b) 1212% of 625 Solution:
Question 3. If 20% of x is 25, then find x. Solution: 20% of x = 25 Hence x = 125
Question 4. Express the following as a fraction (a) 35% (b) 64% Solution:
Question 5. Express the following into per cent (а) 135 (b) 2 : 5 Solution:
Question 6. There are 24% of boys in a school. If the number of girls is 456, find the total number of students in the school. Solution: Let the total number of students be 100. Number of boys = 24% of 100 = 24100 × 100 = 24 Number of girls = 100 – 24 = 76 ⇒ If number of girls is 76, then total number of students = 100 ⇒ If Number of girls is 1, then total number of students = 10076 If Number of girls is 456, then total number of students = 100×45676 = 600 Hence, the total number of students in the school = 600
Question 7. The cost of 15 articles is equal to the selling price of 12 articles. Find the profit per cent. Solution: Let CP of 15 articles be ₹ 100 CP of 1 article = ₹ 10015 SP of 12 articles = ₹ 100 SP fo 1 article = ₹ 10012 SP > CP Hence, profit = 25%
Question 8. An article is marked at ₹ 940. If it is sold for ₹ 799, then find the discount per cent. Solution: MP = ₹ 940 SP = ₹ 799 Discount = MP – SP = 940 – 799 = ₹ 141 Hence, discount = 15%
Question 9. A watch was bought for ₹ 2,700 including 8% VAT. Find its price before the VAT was added. Solution: Cost of watch including VAT = ₹ 2,700 Let the initial cost of the watch be ₹ 100 VAT = 8% of ₹ 100 = ₹ 8 Cost of watch including VAT = ₹ 100 + ₹ 8 = ₹ 108 If cost including VAT is ₹ 108, then its initial cost = ₹ 100 If cost including VAT is ₹ 1, then its initial cost = ₹ 100108 If cost including VAT is ₹ 2,700, then its initial cost = ₹ 100108 × 2700 = ₹ 2500 Hence, the required cost = ₹ 2,500
Question 10. Find the amount if ₹ 2,000 is invested for 2 years at 4% p.a. compounded annually. Solution:
Comparing Quantities Class 8 Extra Questions Short Answer Tpye
Question 11. A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent. (NCERT Exemplar) Solution: Let the number be 100 20% increase = 20100 × 100 = 20 Increased value = 100 + 20 = 120 Now it is decreased by 20% Decreased value = 120 – 120100 × 20 = 120 – 24 = 96 Net decrease = 100 – 96 = 4 Decrease per cent = 4100 × 100 = 4% Hence, the net decrease per cent = 4%
Question 12. Two candidates Raman and Rajan contested an election. Raman gets 46% of the valid votes and is#defeated by 1600 votes. Find the total number of valid votes cast in the election. Solution: Let the total number of valid votes be 100 Number of votes got by Raman = 46% of 100 = 46100 × 100 = 46 Number of votes got by Rajan = 100 – 46 = 54 Difference between the votes = 54 – 46 = 8 8% of Valid votes = 1,600 ⇒ 8100 × Valid votes = 1,600 ⇒ Valid votes = 1600×1008 = 20,000 Hence, the total number of valid votes = 20,000
Question 13. A man whose income is ₹ 57,600 a year spends ₹ 43,200 a year. What percentage of his income does he save? Solution: Annual income of a man = ₹ 57,600 Amount spent by him in the year = ₹ 43,200 Net amount saved by him = ₹ 57,600 – ₹ 43,200 = ₹ 14,400 Percentage of his annual saving Saving = SavingIncome × 100 = 1440057600 × 100 = 25% Hence, the saving percentage = 25%
Question 14. A CD player was purchased for ₹ 3,200 and ₹ 560 were spent on its repairs. It was then sold at a gain of 1212 %. How much did the seller receive? Solution: Cost price of the CD player = ₹ 3,200 Amount spent on its repairing = ₹ 560 Net cost price = ₹ 3,200 + ₹ 560 = ₹ 3,760 Hence, the required amount = ₹ 4,230
Question 15. A car is marked at ₹ 3,00,000. The dealer allows successive discounts of 6%, 4% and 212 % on it. What is the net selling price of it? Solution: Marked price of the car = ₹ 3,00,000 Net selling price after the successive discounts Hence, the net selling price = ₹ 2,63,952
Question 16. Ramesh bought a shirt for ₹ 336, including 12% ST and a tie for ₹ 110 including 10% ST. Find the list price (without sales tax) of the shirt and the tie together. Solution: List price of the shirt = 110112 × 336 = ₹ 300 List price of the tie = 100110 × 110 = ₹ 100 List price of both together = ₹ 300 + ₹ 100 = ₹ 400
Question 17. Find the amount of ₹ 6,250 at 8% pa compounded annually for 2 years. Also, find the compound interest. Solution:
Question 18. Find the compound interest on ₹ 31,250 at 12% pa for 1212 years. Solution:
Question 19. Vishakha offers a discount of 20% on all the items at her shop and still makes a profit of 12%. What is the cost price of an article marked at ₹ 280? (NCERT Exemplar) Solution: Marked Price = ₹ 280 Discount = 20% of ₹ 280 = 12 × 280 = ₹ 56 So selling price = ₹ (280 – 56) = ₹ 224 Let the cost price be ₹ 100 Profit = 12% of ₹ 100 = ₹ 12 So selling price = ₹ (100 + 12) = ₹ 112 If the selling price is ₹ 112, cost price = ₹ 100 If the selling price is ₹ 224, cost price = ₹ (100112 × 224) = ₹ 200
Question 20. Find the compound interest on ₹ 48,000 for one year at 8% per annum when compounded half yearly. (NCERT Exemplar) Solution: Principal (P) = ₹ 48,000 Rate (R) = 8% p.a. Time (n) = 1 year Interest is compounded half yearly Therefore Compound Interest = A – P = ₹ (519,16.80 – 48,000) = ₹ 3,916.80.
MCQ Questions for Class 8 Maths: Ch 7 Cubes and Cube Roots
1. Ones place digit in the cube of 5832 is ______. (a) 5 (b) 7 (c) 2 (d) 8 ► (d) 8
2. What is the cube of double of ‘a’? (a) 16a3 (b) 2a (c) 8a3 (d) 4a2 ► (c) 8a3
3. If (2744)1/3 = 2p+2, then the value of P is (a) 3 (b) 6 (c) 2 (d) 8 ► (b) 6
4. Each prime factor appears _________ times in its cube? (a) 2 (b) 3 (c) 1 (d) 4 ► (b) 3
5. The value of 53 is __________. (a) 125 (b) 15 (c) 10 (d) 75 ► (a) 125
6. A natural number is said to be a perfect cube, if it is the cube of some _________. (a) natural number (b) square number (c) cube number (d) cuboid number ► (a) natural number
7. The square of a natural number subtracts from its cube comes 100. The number is __________. (a) 2 (b) 3 (c) 5 (d) 1 ► (c) 5
8. The expansion of a3 is ___________. (a) 3 × a (b) a + a + a (c) 3 × 3 × 3 (d) a × a × a ► (d) a × a × a
9. If volume of cube is 4913cm³ then length of side of cube is (a) 16 cm (b) 17 cm (c) 18 cm (d) 19 cm ► (b) 17 cm
10. What will be the unit digit of the cube of a number ending with 6 ? (a) 4 (b) 6
(c) 2 (d) 8 ► (b) 6
11. The numbers 1, 8, 27… are ______. (a) negative numbers (b) cube numbers (c) square numbers (d) none of these ► (b) cube numbers
12. The cube of 23 is ___________ (a) 2304 (b) 23 (c) 12167 (d) 529 ► (c) 12167
13.The smallest natural number by which 704 must be divided to obtain a perfect cube is (a) 22 (b) 12 (c) 11 (d) 13 ► (c) 11
15. The symbol for cube root is __________. (a) √3 (b) ∛ (c) 2√3 (d) 3√3 ► (b) ∛
16. The cube of an odd number is always __________. (a) odd number (b) even number (c) prime number (d) none of these ► (a) odd number
17. What will be the unit digit of the cube root of a number ends with 8? (a) 2 (b) 8 (c) 4 (d) 6 ► (a) 2
18. The cube root of the 216 x (−32) x 54 is ________ (a) -36 (b) -72 (c) -48 (d) none of these ► (b) -72
19. Ones digit of cube of a number depends on the ________ of the number. (a) tens digit (b) ones digit (c) hundred digit (d) none of these ► (b) ones digit
20. If (504 + p) is a perfect cube number, whose cube root is p, then p = ______. (a) 6 (b) 4 (c) 2 (d) 8 ► (d) 8
21. How many digits will be there in the cube root of 46656 ? (a) 2 (b) 1 (c) 3 (d) 4 ► (a) 2
22. Find the cube root of 0.001331. (a) 0.111 (b) 0.101 (c) 0.11 (d) none of these ► (c) 0.11
IMPORTANT QUESTIONS
Question 1:
Which of the following numbers are notperfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
ANSWER:
(i) The prime factorisation of 216 is as follows.
2
216
2
108
2
54
3
27
3
9
3
3
1
216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.
(ii)The prime factorisation of 128 is as follows.
2
128
2
64
2
32
2
16
2
8
2
4
2
2
1
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.
(iii) The prime factorisation of 1000 is as follows.
2
1000
2
500
2
250
5
125
5
25
5
5
1
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.
(iv)The prime factorisation of 100 is as follows.
2
100
2
50
5
25
5
5
1
100 = 2 × 2 × 5 × 5
Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.
(v)The prime factorisation of 46656 is as follows.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.
Page No 114:
Question 4:
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
ANSWER:
Here, some cuboids of size 5 × 2 × 5 are given.
When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.
LCM of 5, 2, and 5 = 10
Let us try to make a cube of 10 cm side.
For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height.
Total cuboids required according to this arrangement = 2 × 5 × 2 = 20
With the help of 20 cuboids of such measures, a cube is formed as follows.
Alternatively
Volume of the cube of sides 5 cm, 2 cm, 5 cm
= 5 cm × 2 cm × 5 cm = (5 × 5 × 2) cm3
Here, two 5s and one 2 are left which are not in a triplet.
If we multiply this expression by 2 × 2 × 5 = 20, then it will become a perfect cube.
Thus, (5 × 5 × 2 × 2 × 2 × 5) = (5 × 5 × 5 × 2 × 2 × 2) = 1000 is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.
)
NCERT Solution for Class 8 math – cubes and cube roots 114 , Question 4
Page No 116:
Question 1:
Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
ANSWER:
(i) Prime factorisation of
∴
(ii) Prime factorisation of
∴
(iii) Prime factorisation of
∴
(iv) Prime factorisation of
∴
(v) Prime factorisation of
∴
(vi) Prime factorisation of
∴
(vii) Prime factorisation of
∴
(viii) Prime factorisation of
∴
(ix) Prime factorisation of
∴
(x)Prime factorisation of
∴
NCERT Solution for Class 8 math – cubes and cube roots 116 , Question 1
Page No 116:
Question 2:
State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeroes.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
ANSWER:
For finding the cube of any number, the number is first multiplied with itself and this product is again multiplied with this number.
(i) False. When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will be again an odd number.
For example, the cube of 3 (i.e., an odd number) is 27, which is again an odd number.
(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.
Foe example, the cube of 10 is 1000 and there are 3 zeroes at the end of it.
The cube of 100 is 1000000 and there are 6 zeroes at the end of it.
(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.
For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.
(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8.
The cube of 12 is 1728 and the cube of 22 is 10648.
(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.
(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.
(vii)True, as the cube of 1 and 2 are 1 and 8 respectively.
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Question 3:
You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768
ANSWER:
Firstly, we will make groups of three digits starting from the rightmost digit of the number as.
There are 2 groups, 1 and 331, in it.
Considering 331,
The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.
Taking the other group i.e., 1,
The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.
Hence,
The cube root of 4913 has to be calculated.
We will make groups of three digits starting from the rightmost digit of 4913, as. The groups are 4 and 913.
Considering the group 913,
The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.
Taking the other group i.e., 4,
We know that, 13 = 1 and 23 = 8
Also, 1 < 4 < 8
Therefore, 1 will be taken at the tens place of the required cube root.
Thus,
The cube root of 12167 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 12167, as. The groups are 12 and 167.
Considering the group 167,
167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.
Taking the other group i.e., 12,
We know that, 23 = 8 and 33 = 27
Also, 8 < 12 < 27
2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.
Thus,
The cube root of 32768 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 32768, as .
Considering the group 768,
768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.
Taking the other group i.e., 32,
We know that, 33 = 27 and 43 = 64
Also, 27 < 32 < 64
3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.
.
. The groups are 4 and 913.
. The groups are 12 and 167.
.