## Exercise 2.7: Geometric Progression

2. Write the first three terms of the G.P. whose first term and the common ratio are given below.

(iii) *a* = 1000, *r* = 2/5

3. In a G.P. 729, 243, 81,… find *t* 7.

4. Find *x* so that *x* + 6, *x* + 12 and *x* + 15 are consecutive terms of a Geometric Progression.

5. Find the number of terms in the following G.P.

(ii)

6. In a G.P. the 9 th term is 32805 and 6 th term is 1215. Find the 12 th term.

7. Find the 10 th term of a G.P. whose 8 th term is 768 and the common ratio is 2.

8. If *a, b, c* are in A.P. then show that 3 *a* , 3 *b* , 3 *c* are in G.P.

9. In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms.

10. A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

11. Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4*th* year with respect to the offers *A* and *B*?

## Class 7 Maths Exercise 2.7

**Ex 2.7 Class 7 Maths Question 1. Find: (i) 0.4 ÷ 2 (ii) 0.35 ÷ 5 (iii) 2.48 ÷ 4 (iv) 65.4 ÷ 6 (v) 651.2 ÷ 4 (vi) 14.49 ÷ 7 (vii) 3.96 ÷ 4 (viii) 0.80 ÷ 5**

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

**(viii) 0.80 ÷ 5**

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

** 2. Find: (i) 4.8 ÷ 10 (ii) 52.5 ÷ 10 (iii) 0.7 ÷ 10 (iv) 33.1 ÷ 10 (v) 272.23 ÷ 10 (vi) 0.56 ÷ 10 (vii) 3.97 ÷10**

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

**(v) 272.23 ÷ 10**

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

** 3. Find: (i) 2.7 ÷ 100 (ii) 0.3 ÷ 100 (iii) 0.78 ÷ 100 (iv) 432.6 ÷ 100 (v) 23.6 ÷100 (vi) 98.53 ÷ 100**

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

**(iii) 0.78 ÷ 100**

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

**(iv) 432.6 ÷ 100**

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

**(vi) 98.53 ÷ 100**

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

** 4. Find: (i) 7.9 ÷ 1000 (ii) 26.3 ÷ 1000 (iii) 38.53 ÷ 1000 (iv) 128.9 ÷ 1000 (v) 0.5 ÷ 1000**

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

**(ii) 26.3 ÷ 1000**

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

**(iii) 38.53 ÷ 1000**

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

**(iv) 128.9 ÷ 1000**

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

** 5. Find: (i) 7 ÷ 3.5 (ii) 36 ÷ 0.2 (iii) 3.25 ÷ 0.5 (iv) 30.94 ÷ 0.7 (v) 0.5 ÷ 0.25 (vi) 7.75 ÷ 0.25 (vii) 76.5 ÷ 0.15 (viii) 37.8 ÷ 1.4 (ix) 2.73 ÷ 1.3**

**(iii) 3.25 ÷ 0.5**

**(iv) 30.94 ÷ 0.7**

**(vi) 7.75 ÷ 0.25**

**(vii) 76.5 ÷ 0.15**

**(viii) 37.8 ÷ 1.4**

** 6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?**

## Solved Exercise 2.7 Class 10 Maths Solution Notes (With Free PDF)

Are you finding the solution of Exercise 2.7 of 10th Class Mathematics?

If yes, you are in the right hands.

- Federal Board (FBISE)
- Punjab Board (Lahore, Faisalabad, Multan, Rawalpindi Boards)
- KPK Board (Peshawar Board)
- Sindh Board (Karachi and Hyrderboard Boars)

## 2.7: Exercises - Mathematics

This page lists exercises corresponding to the ** second edition** of our textbook, for students who wish to use that edition. The sections correspond very closely, but not exactly, to the first edition.

#### Second edition:

Elementary Linear Algebra, 2/E , by Lawrence E. Spence, Arnold J. Insel, and Stephen H. Friedberg. Prentice Hall, 2008. ISBN-10: 0131871412, ISBN-13: 9780131871410. University bookstore ($130) or AddALL ($70 and up).

### Exercises

Exercises are not collected. Below, I am listing all exercises that you certainly should be able to solve. Use your judgment in deciding which exercises to actually work. It is a good idea to work as many exercises as you can find time to solve your goal is to achieve fluency, so that you can complete exams comfortably in the time allowed.

You are also invited to look at the more advanced exercises that are not listed.

The following exercises are for the ** second edition** of our textbook.

## Class 7 Maths Exercise 2.4

**Ex 2.4 Class 7 Maths Question 1. Find**

**(v) 3 ÷ **

= = 7/3

**(vi) 5 ÷ **

== 25/7

**Ex 2.4 Class 7 Maths Question 2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.**

(i) Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]

Whole numbers are collection of all positive integers including 0.

Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]

Whole numbers are collection of all positive integers including 0.

**Ex 2.4 Class 7 Maths Question 3. Find :**

=

**(iv) ÷ 3**

== 13/3

**(iv)** **÷ 4**

= = 7/2

**(iv) ÷ 7**

== 31/7

**Ex 2.4 Class 7 Maths Question 4. Find :**

**(iv) ÷ (3/5)**

= = 7/3

**(v) ÷ (8/3)**

**(vi) (2/5) ÷ **

= = 3/2

**(vii) ÷ **

== 16/5

== 5/3

**(viii) ÷ **

== 11/5

== 6/5

## 2.7: Exercises - Mathematics

Exercise 1.2.7. Start with equation (4) from p. 8 of the text:

where a solution can be found for b = (2, 5, 7) and cannot be found for b = (2, 5, 6). Find two more values of b for which a solution can be found, and two more for which no solution is possible.

Answer: As noted in the text, the three columns (1, 2, 3), (1, 0, 1), and (1, 3, 4) lie in a plane, since three times (1, 2, 3) plus (1, 0, 1) equals twice (1, 3, 4). In other words, one of the three columns can be expressed as a linear combination of the other two. The value b = (2, 5, 7) is also in the same plane, being expressible as (1, 2, 3) plus (1, 3, 4).

So to find additional b for which a solution is possible, we can simply take any b in the same plane as the other vectors. One possibility is twice (2, 5, 7) or (4, 10, 14). Another is (1, 0, 1) plus (1, 3, 4) or (2, 3, 5).

We can then take those solutions and find values of b for which the equations cannot be solved, by looking for values of b not in the same plane as the previous vectors. One possibility is (4, 10, 15), another is (2, 3, 6).

UPDATE: Corrected error in first paragraph of answer thanks to freaky4you for the fix!

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Question 1.

Which of the following sequences are in G.P?

(i) 3, 9, 27, 81, ……..

(ii) 4, 44, 444, 4444, ………

(iii) 0.5, 0.05, 0.005, ……..

(iv) (frac < 1 > < 3 >),(frac < 1 > < 6 >),(frac < 1 > < 12 >)………

(v) 1, -5, 25, -125, …….

(vi) 120, 60, 30, 18, …….

(vii) 16, 4, 1, (frac < 1 > < 4 >), ……

Solution:

(i) 3, 9, 27, 81

r = Common ratio

Question 2.

Write the first three terms of the G.P. whose first term and the common ratio are given below.

(i) a = 6, r = 3

(ii) a = (sqrt < 2 >) , r = (sqrt < 2 >)

(iii) a = 1000, r = (frac < 2 > < 5 >)

Solution:

(i) a = 6, r = 3

t_{n} = ar n-1

t_{1} = ar 1-1 = ar 0 = a = 6

t_{2} = ar 2-1 = ar 1 = 6 × 3 = 18

t_{3} = ar 3-1 = ar 2 = 6 × 3 2 = 54

∴ The 3 terms are 6, 18, 54, ….

The 3 terms are 1000, 400, 160, ……………

Question 3.

In a G.P. 729, 243, 81,… find t_{7}.

Solution:

G.P = 729, 243, 81 ……

t_{7} = ?

Question 4.

Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.

Answer:

(frac

Since it is a G.P.

(frac < x+12 > < x+6 >) = (frac < x+15 > < x+12 >)

(x + 12) 2 = (x + 6) (x + 15)

x 2 + 24x + 144 = x 2 + 21x + 90

3x = -54 ⇒ x = (frac < -54 > < 3 >) = -18

Question 5.

Find the number of terms in the following G.P.

(i),4, 8, 16, …, 8192

(ii) (frac < 1 > < 3 >),(frac < 1 > < 9 >),(frac < 1 > < 27 >),……(frac < 1 > < 2187 >)

Solution:

(i) 4, 8, 16, …… 8192

Question 6.

In a G.P. the 9 th term is 32805 and 6 th term is 1215. Find the 12th term.

Solution:

In a G.P

t_{n} = ar n-1

t_{9} = 32805

t_{6} = 1215

t_{12} = ?

Let

t_{9} = ar 8 = 32805 ………(1)

t_{6} = ar 5 = 1215 ………. (2)

Question 7.

Find the 10th term of a G.P. whose 8 th term is 768 and the common ratio is 2.

Answer:

Here r = 2, t_{8} = 768

t_{8} = 768 (t_{n} = ar n-1 )

a. r 8-1 = 768

ar 7 = 768 …..(1)

10 th term of a G.P. = a.r 10 -1

= ar 9

= (ar 7 ) × (r 2 )

= 768 × 2 2 (from 1)

= 768 × 4 = 3072

∴ 10 th term of a G.P. = 3072

Question 9.

In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is (frac < 57 > < 2 >). Find the three terms.

Solution:

Let the three consecutive terms in a G.P are (frac < a > < r >), a, ar.

Their Product = (frac < a > < r >) × a × ar = 27

a 3 = 27 = 3 3

a = 3

Sum of the product of terms taken two at a time is (frac < 57 > < 2 >)

Question 10.

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

Solution:

Starting salary = ₹ 60,000

Increase per year = 5%

∴ At the end of 1 year the increase

= 60,0,00 × (frac < 5 > < 100 >)

₹ 3000

∴ At the end of first year his salary

= ₹ 60,000 + 3000

I year salary = ₹ 63,000

II Year increase = 63000 × (frac < 5 > < 100 >)

At the end of II year, salary

= 63000 + 3150

= ₹ 66150

III Year increase = 66150 × (frac < 5 > < 100 >)

= 3307.50

At the end of III year, salary = 66150 + 3307.50

= ₹ 69457.50

IV year increase = 69457.50 × (frac < 5 > < 100 >)

= ₹ 3472.87

Question 11.

Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹ 20,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

Offer B: ₹ 22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4th year with respect to the offers A and B?

Solution:

Offer A

Starting salary ₹ 20,000

Annual increase 6%

At the end of

III year ,salary = 22472 + 1348 = 23820

∴ IV year salary = ₹ 23820

Offer B

Starting salary = ₹ 22,000

Salary as per Option A = ₹ 23820

Salary as per Option B = ₹ 24040

∴ Option B is better.

## 2.7: Exercises - Mathematics

Show Answer to the Exercise:

*P* = 12 cm, *A* = = 6.93 cm 2 , α = *β* = γ = 60°, α' = *β*' = γ' = 120°

α = *β* = 43°30', γ = 93°

6 cm

45°

*a* = 3 dm, *b* = 4 dm, α = 36°52', *β* = 53°08'

*b* = = 11.66 cm, *m _{b}* = = 10.3 cm

**a)** *h* = 8.31 m**b)** α = 77°46'

| *MN* | = 2( 6 + ) = 16.9 cm

*a* = *b* = 16 cm, *c* = 28 cm

| ∠ *ACD* | = 27°

α = 108°, *β* = 54°, γ = 18°

*a* = 15 cm, *b* = 22 cm, *c* = 17 cm, α = 42°54', *β* = 86°38', γ = 50°29'

α = 45°, *β* = 60°, γ = 75°, *b* = = 1.73, *c* = 1.93

*h* = 325.7 m

*w* = 14.6 m

γ = 61°03', *a* = 30.28, *b* = 35.87, *c* = 33.94

*P* = 3 + = 4.73, *A* = = 0.87

*a* = 18.41 cm, *b* = 14.86 cm, *c* = 6.51 cm

*b* = 15.24, *c* = 10.66, α = 32°04', γ = 42°21'

*A* = 3.85

yes, γ = 91°18' > 90°

*a* = 20.86, *b* = 55.31, *c* = 68.06, γ = 119°32'

105.16 N

*d* _{1} = 12.54 cm, *d* _{2} = 7.14 cm

*a* = = 14.2, *b* = = 9.67, *c* = = 17.18

The triangle is right-angled.

*r* = 3

*r* = 17 cm

*a* = 13, *h* = = 9.23, *e* = 10, *f* = 24

*b* = = 3.75, *c* = = 6.25