8th Maths : Chapter 1 : Numbers: Exercise 1.7: Numerical Example solved problems and Text book back Numerical problems Questions with Solution

**Exercise 1.7**

**Miscellaneous
Practice Problems**

** **

**1. If 3/4 of a box of apples weighs 3 kg and 225 gm, how much does **

**Solution:**

Let the total weight of a box of apple = *x* kg.

Weight of 3/4 of a box apples = 3 kg 225 gm.

= 3.225 kg

(3/4) × *x* = 3225

*x* = { [3.225 × 4] / 3 } kg = [1.075 ×
4 ] kg = 4.3 kg

= 4 kg 300 gm.

Weight of the box of apples = 4 kg 300 gm.

** **

**2. Mangalam buys a water jug of capacity 3 (4/5) litre. If she buys another
jug which is 2 (2/3) ****times as large as the smaller jug, how
many litre can the larger one hold?**

**Solution:**

Capacity of the small water jug = 3 4/5 litres.

Capacity of the big jug = 2 2/3 times the small one.

= 2(2/3) × 3(4/5) = [8/3] × [19/5] = 152/15 =

= 10 (2/15) litres

Capacity of the large jug = 10 2/15 litres = litres.

** **

**3. Ravi multiplied 25/8 and 16/15 and he says that the simplest form of this
product is 10/3 ****and Chandru says the answer in the simplest
form is 3 1/3 Who is correct? (or) Are they both correct? Explain.**

**Solution:**

Product of 25/8 and 16/15 = [25/8] × [16/15]

= 10/3 = 3 (1/3)

Answer obtained = 400/120

= [400 ÷ 40] / [120 ÷ 40] = 10/3 = 3 (1/3)

∴ The product is 400/120 and its simplest form improper fraction
is 10/3.

And mixed fraction is 3 (1/3).

∴ Both are correct.

** **

**4. Find the length of a room whose area
is 153/10 sq.m and whose breadth is 2
11/20 m.**

**Solution:**

Length of the room × Breadth = Area of the room

Breadth of the room = 2 (11/20) m

Area of the room = 153/10 sq.m

Length × 2(11/20) = 153/10

Length = [153/10] ÷ [2 (11/20)] = 153/10 ÷ 51/20 = 153/10 × 20/51
= 6 m

Length of the room = 6 m

** **

**5. There is a large square portrait of
a leader that covers an area of 4489 cm^{2}.
If each side has a 2 cm liner, what would
be its area?**

**Solution:**

Area of the square = 4489
cm^{2}

(side)^{2} = 4489
cm^{2}

(side)^{2} = 67 ×
67

side^{2} = 67^{2}

Length of a side = 67

Length of a side with liner = 67 + 2 + 2 cm

= 71 cm

Area of the larger square = 71 × 71 cm^{2}

= 5041 cm^{2}

Area of the liner = Area of big square − Area of small square

= (5041 − 4489) cm^{2}

= 552 cm^{2}

** **

**6. A greeting card has an area 90 cm^{2}. Between what two whole numbers
is the length of its side?**

**Solution:**

Area of the greeting card = 90 cm^{2}

(side)^{2 }= 90 cm^{2}

(side)^{2 }= 2 × 5 × 3 × 3 = 2 × 5 × 3^{2}

√(side)^{2 }= √(2×5×3^{2})^{}

side^{ }= 3
√(2×5)

side^{ }= 3 √10 cm

side = 3 × 3.2 cm

side = 9.6 cm

∴ Side lies between the
whole numbers 9 and 10.

** **

**7. 225 square shaped mosaic tiles, each
of area 1 square decimetre exactly cover a square shaped verandah. How long is each
side of the square shaped verandah?**

**Solution:**

Area of one tile = 1 sq.decimeter

Area of 225 tiles = 225 sq.decimeter

225 square tiles exactly covers the square shaped verandah.

∴ Area of 225 tiles = Area of the verandah

Area of the verandah = 225 sq.decimeter

side × side = 15 × 15 sq.decimeter

side = 15 decimeters

Length of each side of verandah = 15 decimeters.

** **

**8. If ^{3}√1906624 **

**Solution:**

** **

**9. If 2^{m}**

**Solution:**

Given 2^{m}^{−1} + 2^{m}^{+1}
= 640

2^{m}^{−1} + 2^{m}^{+1}
= 128 + 512 [consecutive powers of 2]

2^{m}^{−1} + 2^{m}^{+1}
= 2^{7} + 2^{9}

*m *– 1 = 7

*m *= 7 + 1

*m *= 8

__Powers of 2__

2,4,8,16,32,64,128,256,512,…

** **

**10. Give the answer in scientific notation:**

**A human heart beats at an average of
80 beats per minute. How many times does it beat in i) an hour? ii) a day? iii)
a year? iv) 100 years?**

**Solution:**

Heart beat per minute = 80 beats

**(i) an hour**

One hour = 60 minutes

Heart beat in an hour = 60 × 80

= 4800 = 4.8 × 10^{3}

**(ii) In a day**

One day = 24 hours = 24 × 60 minutes

∴ Heart beat in one day = 24 × 60 × 80 = 24 × 4800 = 115200

= 1.152 × 10^{5}

**(iii) a year**

One year = 365 days = 365 × 24 hours = 365 × 24 × 60 minutes

Heart beats in a year = 365 × 24 × 60 × 80

= 42048000 = 4.2048 × 10^{7}

**(iv) 100 years**

Heart beats in one year = 4.2048 × 10^{7}

Heart beats in 100 years = 4.2048 × 10^{7} × 100 =
4.2048 × 10^{7} × 10^{2}

= 4.2048 × 10^{9}

** **

** Challenging Problems**

** **

**11. In a map, if 1 inch refers to 120
km, then find the distance between two cities B and C which are 4(1/6) inches and
3(1/3) inches from the city A which lies between the cities B and C.**

**Solution:**

1 inch =120 km

Distance between A and B = 4 (1/6) inches

Distance between A and C = 3 (1/3) inches

∴ Distance between B and C = 4 (1/6) + 3 (1/3) inches

= 25/6 + 10/3 = 25/6 + 20/6 = [25 + 20] / 6 = 45/6 inches

1 inch = 120 km

∴ 45/6 inches = [45/6] × 120 km = 900 km

Distance between B and C = 900 km

** **

**12. Give an example and verify each of
the following statements**

**(i) The collection of all non-zero rational
numbers is closed under division.**

**(ii) Subtraction is not commutative for
rational numbers.**

**(iii) Division is not associative for
rational numbers.**

**(iv) Distributive property of multiplication
over subtraction is true for rational numbers. That is, a **

**(v) The mean of two rational numbers
is rational and lies between them**

**Solution:
**

**(i)**** **Let *a* = 5/6 and *b*
= −4/3 be two non zero rational numbers.

*a *÷ *b* = 5/6 ÷ −4/3 = 5/6
× 3/−4 = 5/−8 is in Q

∴ Collection of non−zero rational numbers are closed under division.

**(ii)** Let *a* = 1/2 and *b*
= −5/6 be two rational numbers.

*a* – *b* = 1/2 − (−5/6) = 1/2 + (+5/6) = 3/6 + 5/6 = [3 + 5]
/ 6 = 8/6

= 1 (2/6)
= 1 1/3

*b – a* = −5/6 – 1/2 = −5/6 – 3/6 = [−5
−3] / 6 = −8/6 = −1 (1/3)

*a − b
≠ b − a*

∴ Subtraction is not commutative for
rational numbers.

**(iii)** Let *a* = 2/5, *b*
= 6/5, *c* = 3/5 be
three rational numbers.

*a* ÷ (*b ÷ c*) = 2/5 ÷ (6/5
÷ 3/5) = 2/5 ÷ (6/5 × 5/3)

= 2/5
÷ 2/1 = 2/5 × 1/2 = 1/5 ……….(1)

(*a*
÷ *b*)* ÷ c* = (2/5 ÷ 6/5) ÷ 3/5

= (2/5
× 5/6) ÷ 3/5 = 1/3 ÷ 3/5 = 1/3 × 5/3 = 5/9 ……..(2)

From
(1) and (2) 1/5 ≠ 5/9

*a ÷* (*b ÷ c*) ≠ (*a* *÷*
*b*) *÷* *c*

∴ Division is not associative for rational numbers.

**(iv)** Let* a* = 2/9, *b*
= 3/6 , *c* = 1/3 be three rational numbers

To
prove *a* × (*b* − *c*) = *ab* − *ac*

*a* × (*b − c*) = 2/9 × (3/6
− 1/3)

= [2/9]
× ( [3 − (1 × 2)] / 6 ) = [2/9] × [(3 – 2) / 6]

= 2/9
× 1/6 = 1 / 27 ……..(1)

*ab* − *ac* = (2/9 × 3/6) –
(2/9 × 1/3) = 1/9 – 2/27

= [(1
× 3) – 2] / 27 = [3 – 2] / 27 = 1/27
….(2)

∴ From (1) and (2)

*a* × (*b* − *c*) = *ab*
− *ac*.

∴ Distributivity of multiplication over subtraction is true for
rational numbers.

**(v)** Let* a* = 2/11 and *b* = 5/6 be two rational numbers.

Mean of *a* and *b* is *c* =1/2 (*a + b*) = 1/2 (2/11 + 5/6) = 1/2 ((2×6)
+ (5×11) / 66)

= 1/2 × ([12 + 55] / 66) = 1/2 × 67/66 = 67/132 is in Q.

Also 2/11 = [2 × 12] / [11 × 12] = 24/132

5/6 = [5 × 22] / [6 × 22] = 110 / 132

∴ 24/132 < 67/132 < 110/132

∴ The mean lies between the given rational numbers 2/11 and 5/6.

** **

**13. If 1/4 of a ragi adai weighs 120 grams, what will be the weight of 2/3 of the same
ragi adai?**

**Solution:**

Let the weight of 1 ragi adai = *x* grams

given 1/4 of *x* = 120 gm

(1/4) × *x* = 120

*x* = 120 × 4

*x* = 480 gm

∴ 2/3 of the adai = 2/3 × 480
gm = 2 × 160 gm = 320 gm

2/3 of the weight of adai = 320 gm

** **

**14. If p **

**Solution:**

**Given**: *p* + 2*q* = 18 ….(1)

*pq* = 40 ….(2)

2/*p* + 1/*q* = [(2 × *q*) + (l × *p*)] / *pq*
= [2*q* + *p* ] / *pq* = 18/40 [∵ from (1) and (2)]

2/*p* + 1/*q* = 9/20

** **

**15. Find x if 5(x/5) **

**Solution:**

[ 5 *x*/5 ] × [3 3/4] = 21

[(25 + *x*)
/ 5] × [15/4] = 21

[ 25+ *x *]* */
5 = 21 ÷ [15/4]

[25+ *x *] /
5 = 21 × [4/15]

[25+ *x *] /
5 = 28/5

25 + *x *= [28 × 5] / 5

25 + *x *= 28

*x *= 28 – 25

*x *= 3

** **

**16. By how much does 1/(10/11) exceed
(1/10) / 11 ?**

**Solution:**

The difference = { 1 / [10/11] } – { [1/10] / 11 } = (1/1 × 11/10)
– (1/10 × 1/11) = 11/10 – 1/110 = [121 – 1] / 110 = 120/110 = 12/11

{ 1 / [10/11] } exceed {
[1/10] / 11 } by 12/11

** **

**17. A group of 1536 cadets wanted to
have a parade forming a square design. Is it possible? If it is not possible, how
many more cadets would be required?**

**Solution:**

Number of cadets to form square design

1536 = __2 × 2__ × __2 × 2__ × __2 × 2__ × __2 × 2__
× 2 × 3

The numbers 2 and 3 are unpaired

∴ It is impossible to have the parade forming square design with
1536 cadets.

39 × 39 = 1521

Also 40 × 40 = 1600

∴ We have to add (1600 − 1536) = 64 to make 1536 a perfect
square.

∴ 64 more cadets would be
required to form the square design.

** **

**18. Evaluate: √286225 and use it to compute
√2862.25 ****+**** √28.6225**

**Solution:**

√286225 = 535

√2862.25 = √[286225 / 100] = √286225 / √100 = 535/10 = 53.5

√28.6225 = √[286225 / 1000] = √286225/√10000 = 535/100 = 5.35

∴ √2862.25 + √28.6225 = 53.5 + 5.35 = 58.85

** **

**19. Simplify: (3.769 × 10 ^{5}) + (4.21 × 10^{5})**

**Solution:**

(3.769 × 10^{5}) + (4.21 × 10^{5})** **=
3,76,900 + 4,21,000

= 7,97,900 = 7.979 × 10^{5}

** **

**20. Order the following from the least
to the greatest: 16 ^{25} , 8^{100} , 3^{500} , 4^{400}
, 2^{600}**

**Solution:**

16^{25} = (2^{4})^{25} = 2^{100}

8^{100} = (2^{3})^{100}
= 2^{300}

4^{400} = (2^{2})^{400} = 2^{800}

2^{600} = 2^{600}

Comparing the powers we have. 2^{100} < 2^{300}
< 2^{600} < 2^{800}

∴ The required order : 16^{25} , 8^{100}, 2^{600},
3^{500}, 4^{400}

**Exercise
1.7**

**Miscellaneous Practice
Problems**

**1. 4 kg 300 gm **

**2. 10 2/5 litre **

**3. both are correct **

**4. 6 m **

**5. 552 cm^{2}**

**6. 9 cm and 10 cm **

**7. 15 decimetre **

**8. 625 **

**9. 8 **

**10. (i) 4.8 × 10 ^{3
}(ii) 1.152 × 10^{5 }(iii) 4.2048 × 10^{7 }(iv) 4.2048 × 10^{9}**

**Challenging Problems **

**11. 900 km **

**13. 320 gm **

**14. 9/20 **

**15. x = 3 **

**16. 12/11 **

**17. No, 64**

**18. 58.85**

**19. 7.979 × 10 ^{5}
**

**20. 8 ^{100}, 2^{600},
3^{500}, 4^{400}, 16^{25} **

^{}

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8th Maths : Chapter 1 : Numbers : Exercise 1.7 | Questions with Answers, Solution | Numbers | Chapter 1 | 8th Maths

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