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AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.3

Question 1.
Sudhakar borrows ₹ 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments’?
Solution:
P = 15,000
R = 9%
T = 8 years

A = ₹ 25800
∴ His monthly payment = \(\frac{25800}{8 \times 12}\)
= ₹268.75
∴ Monthly he has to pay = ₹268.75

Question 2.
A TV was bought at a price of ₹ 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.
Solution:
The C.P. of T.V = ₹ 21,000.
After 1 year its value
= 21000 – 5% of 21000
=21000 – \(\frac { 5 }{ 100 }\) × 21000
= 21000 – 1050
= ₹19,950

Question 3.
Find the amount and the compound interest on ₹ 8000 at 5% per annum, for 2 years
compounded annually.
Solution:
P = ₹8000
R = 5%
The interest is compounded every year.
Then 2 time periods wII be occurred.
∴ n = 2

∴ Amount (A) = ₹8820
C.I = A – P
= 8820 – 8000 = ₹ 820

Question 4.
Find the amount and the compound interest on ₹ 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.
Solution:
P = ₹ 6500
R = 5%
T = 1 years
∴ \(\frac{\mathrm{PTR}}{100}=\frac{6500 \times 5 \times 1}{100}\) = 325
∴ A = P + I = 6500 + 325 = 6825
∴ P = 6825
(At the begining of 2,id year A=P)
R = 6%
T = 1 year
∴ \(\frac{\mathrm{PTR}}{100}=\frac{6825 \times 6 \times 1}{100}\) = 409.5
∴ A = P + I = 6825 + 409.5
∴ Amount = ₹ 7234.50
C.I. = A – P
= ₹ 7234.50 – 6500
= ₹734.50

Question 5.
Prathibha borrows ₹47000 from a fmance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period. Find: (a) How much
amount Prathibha should repay the finance company at the end of five years. (b) her equal
monthly repayments.
Solution:
P = ₹ 47000
R = 17%
T =5 years
∴ I = \(\frac{\mathrm{PTR}}{100}=\frac{47000 \times 5 \times 17}{100}\)
= ₹ 39,950

a) Amount to be paid
A = P + I
= 47000 + 39,950
= 86950
∴ Amount to be pay = ₹ 86950

b) In monthly equal instalments she has to pay

= 149.1
= ₹ 1450 (approx)

Question 6.
The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.
Solution:
The population of Hyderabad
= 68,09,000
If every year increase in 4.7%.
Then the population of the city in 2015
= 68,09,000 ( 1 + \(\frac{4.7}{100}\) )⁴
100 J
[ ∵ P = 6809000, R = 4.7 %, n = 4(2015 -2011)]
= 68,09,000 x \(\frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100}\)
= 81,82,199

Question 7.
Find Compound interest paid when a sum of ₹ 10000 is invested for 1 year and 3 months at 8\(\frac{1}{2}\) % per annum compounded annually.
Solution:
P = ₹10,000; R = 8\(\frac { 1 }{ 2 }\) % = \(\frac { 17 }{ 2 }\)%
T = 1 year

= 50 × 17 = 850
∴ I = ₹ 850
∴ A = P + I = 10,000 + 850
A = 10,850
∴ P = 10,850; R = \(\frac { 17 }{ 2 }\)% % ; T = 3 months

= ₹ 230.50
∴ Compound Interest
= 850 + 230.50
= ₹ 1080.50

Question 8.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the
difference in amounts he would be paying after 1\(\frac{1}{2}\) years, if the interest is (i) compounded annually (ii) compounded half yearly.
Solution:
P = ₹ 80,000; R = 10%;
T = 1 year
∴ \(\frac{\mathrm{PTR}}{100}\) = \(\frac{80000 \times 10 \times 1}{100}\)
= ₹8000
∴ A = P + I = 80000 + 8000
= ₹ 88,000

Interest on 6 months :
P = 88,000 ; R = 10% ; T = 6 Months
= \(\frac { 1 }{ 2 }\) year

i) The amount to be paid after 1 year 6 months = P + I
= 88000 + 4400
A₁ = ₹ 92,400

ii) He has to pay compounded on
every 6 months in 1 \(\frac { 1 }{ 2 }\) years
∴ 3 time periods will be occurred.
∴ n = 3
R = \(\frac { 10 }{ 2 }\) = 5% P = ₹ 80,000


A₂ = ₹ 92610
∴ Difference between the amounts = A₂ – A₁ = 92610 – 92400 = ₹ 210

Question 9.
I borrowed ₹ 12000 from Prasad at 6°/o per annum simple interest for 2 years. Had
I borrowed this sum at 6% per annum compounded annually, what extra amount would
I have to pay9
Solution:
Sum borrowed from Prasad
P = ₹ 12000
T = 2 years;
R = 6%

= ₹144O
A = P + I
A₁ = P + I = 12000 + 1440
= ₹13440
12000 + 1440 , = ₹ 13440
∴ He has to pay the amount after 2 years at the rate of 6% on C.I.
P = ₹12,000; R = 6%; n = 2 years

A₂ = ₹13483.2
∴ The difference between the C.I and S.I = 13483.2 – 13440
= ₹ 43.20

Question 10.
In a laboratory the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000
Solution:
No. of bacteria in a laboratory = 5,06,000
If they are increased at the rate of 2.5% per hour then their number after 2 hours

Question 11.
Kamala borrowed ₹ 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
Solution:
Kanala borrowed from bank = ₹ 26400
Rah of interest (R) =15%
n = 2 years

After 4 rpnths the amount will be ₹ 34914
∴ P = 34914; R = 15%; T = 4 months
T = \(\frac { 4 }{ 12 }\) year
= \(\frac { 1 }{ 3 }\) year
∴ \(I=\frac{P T R}{100}=\frac{34914 \times 15 \times \frac{1}{3}}{100}\)
= ₹1745.7
∴ Kamala has to pay the amount after 2 years and 4 months to the bank = 34914 + 1745.7
= ₹36659.7

Question 12.
Bharathi borrows an amount of ₹ 12500 at 12% per annum for3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
Bharathi borrowed the sum
P = ₹12500
R = 12%
T = 3 years
S. I (I) = \(\frac { PTR }{ 100 }\)
= \(\frac{12500 \times 12 \times 3}{100}\)
= 125 × 36
= 4500
After 3 years she has to pay
(A₁)= P + I
= 12500 + 4500 .
A₁ = ₹17,000
Madhuri has to pay the amount on


A₂ = 16637.5
∴ A₁ > A₂
A₁ – A₂ = 17000 – 16637.5
= ₹ 362.5
∴ Bharathi has to pay ₹ 362.5 more than Madhuri.

Question 13.
Machinery worth ₹ 10000 depreciated by 5%. Find its value after 1 year.
Solution:
The value of machinery after 1 year on 5% depreciation

= 95 × 100
= ₹ 9500

Question 14.
Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.
Solution:
Present population of a city = 12,00,000 If its population increases at the rate of 4%, then the population after 2 years

= 120 × 104 × 104
= 12,97,920

Question 15.
Calculate compound interest on ₹ 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?
Solution:
compounded quarterly then 4 time periods will be there in 1 year.
∴ n = 4
C.I. on ₹ 1000 over a period of 1 year at
10% per annum A = P (1 + \(\frac{\mathrm{R}}{100}\) )ⁿ
P = 1000; n = 4; R = \(\frac{10}{4}=\frac{5}{2}\) %

= ₹ 1103.81
A = ₹ 1103.81
C.I. for 1 year
= 1103.81 – 1000
= ₹ 10.81

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.2

Question 1.
In the year 2012, it was estimated that there were 36.4 crore Internet users worldwide. In
the next ten years, that number will be increased by 125%. Estimate the number of Internet
users worldwide in 2022.
Solution:
Internet users in the year 2012
= 36.4 crores.
The number will be increased by next
10 years = 125%
∴ The no. of internet users in the year 2022
= 36.4 + 125% of 36.4
= 36.4 + \(\frac { 125 }{ 100 }\) × 36.4
= 36.4 + 45.5
= 81.9 crores.

Question 2.
A owner increases the rent of his house by 5% at the end of each year. If currently its rent is ₹ 2500 per month, how much will be the rent after 2 years’?
Solution:
Present house rent = ₹ 2500 If the owner increases the rent by 5% on every year then the rent of the house after 2 years

= 2500 × \(\frac { 21 }{ 20 }\) x \(\frac { 21 }{ 20 }\)
= ₹ 2756.25

Question 3.
On Monday, the value of a company’s shares was ₹ 7.50. The price increased by 6% on Tuesday, decreased by 1.5% on Wednesday, and decreased by 2% on Thursday. Find the value of each share when trade opened on Friday.
Solution:
The value of the share when trade opened on Friday
= ₹ 7.674

Question 4.
With most of the Xerox machines. you can reduce or enlarge your original by entering a percentage for the copy. Reshma wanted to enlarge a 2 cm by 4 cm drawing. She set the Xerox machine for 150% and copied her drawing. What will be the dimensions of the
copy of the drawing be’?
Solution:
Length of the copy = 2 cm
breadth = 4 cm
If the length is increase in 150% then its
measure = 150 % of 2 cm
= \(\frac { 150 }{ 100 }\) × 2 = 1.5 × 2 = 3 cm

If the breadth is increase in 150% then
its measure = 150 % of 4 cm 150
= \(\frac { 150 }{ 100 }\) × 4 = 1.5 × 4 = 6 cm
100
∴ New length = 3 cm
breadth = 6 cm

Question 5.
The printed price of a book is ₹ 150. And discount is 15%. Find the actual amount to be paid.
Solution:
The printed price of a book = ₹ 150
Discount % = 15%
∴ Discount = 15% of 150
= \(\frac { 15 }{ 100 }\) × 150 = ₹22.5
∴ The C.P. of a book = 150 – 22.5
= ₹127.50/-

Question 6.
The marked price of an gift item is ₹ 176 and sold it for ₹ 165. Find the discount percent.
Solution:
Marked price of a gift = ₹176
S.P. = 165
Discount = 176 – 165 = ₹ 11

Question 7.
A shop keeper purchased 200 bulbs for ₹10 each. However 5 bulbs were fused and put
them into scrap. The remaining were sold at ₹12 each. Find the gain or loss percent.
Solution:
The C.P. of 200 bulbs at the rate of ₹10 for each = 200 × 10 = ₹ 2000
If 5 bulbs are fused then remaining are
= 200 – 5 = 195
∴ TheS.P. of 195 bulbs at the rate of ₹12 for each = 195 × 12 = ₹ 2340
∴ S.P. > C.P.
∴ Profit = S.P.-C.P.
= 2340 – 2000 = 340
Profit = \(\frac { Profit }{ C.P }\) × 100 = \(\frac { 340 }{ 2000 }\) × 100
Profit = 17%

Question 8.
Complete the following table with appropriate entries (Wherever possible)

Solution:

Question 9.
A table was sold for ₹2,142 at a gain of 5%. At what price should it be sold to gain 10%.
Solution:
S.P. of a table = ₹ 2142
Profit = 5%

∴ The C.P. of buyyer = ₹ 2040
Profit % = 10%

∴ S.P. = ₹2244

Question 10.
Gopi sold a watch to Ibrahim at 12% gain and Ibrahim sold it to John at a loss of 5%. If John paid ₹1,330, then find how much did Gopi sold it?
Solution:
. Let Gopi’s cost price = ₹100
Gain = 12%
∴ Gopi’s selling price to Ibrahim or Ibrahim’s cost price = ₹100 + ₹12 = ₹112
∴ Ibrahim’s loss = 5%
∴ Ibrahim’s selling price =
\(112\left(\frac{100-5}{100}\right)=\frac{112 \times 95}{100}\) = ₹106.40
For ₹100 we get = ₹106.40
For ₹1330 how much we get ?

Question 11.
Madhu and Kavitha purchased a new house for ₹3,20,000. Due to some economic
problems they sold the house for ₹2,80,000. Find (a) The loss incurred (b) the loss percentage.
Solution:
C.P. of a house = ₹ 3,20,000
S.P. of a house = ₹ 2,80,000
∴ C.P. > S.P.
a) Loss
= C.P. – S.P.
= 3,20,000 – 2,80,000 = 40,000

Question 12.
A pre-owned car show-room owner bought a second hand car for ₹ 1,50,000. He spent ₹20,000 on repairs and painting, then sold it for ₹ 2,00,000. Find whether he gets profit or loss. If so, what percent?
Solution:
After repair, the C.P of a car
= 1,50,000 + 20,000 = 1,70,000
S.P. of a ear = ₹ 2,00,000
∴ S.P. > C.P.
∴ Profit = S.P.-C.P.
= 2,00,000-1,70,000= 30,000
Profit = \(\frac { Profit }{ C.P }\) × 100
\(\frac { 30,000 }{ 1,70,000 }\) × 100 = Profit = \(\frac { 300 }{ 17 }\)
Profit% = 17.64%

Question 13.
Lalitha took a parcel from a hotel to celebrate her birthday with her friends. It was billed with ₹ 1,450 including 5% VAT. Lalitha asked for some discount, the hotel owner gave 8% discount on the bill amount. Now find the actual amount that lalitha has to pay to the hotel owner
Solution:
After allowing 5% VAT, the total bill = ₹ 1450
If 8% discount is allowed on bill, then
Discount = 8% of 1450
\(\frac { 8 }{ 100}\) × 1450 = ₹116
Discount = ₹116
∴ Lalitha has to pay the bill = 1450 -116
= ₹ 1334

Question 14.
If VAT is included in the price, find the original price of each of the following.

Solution:

Question 15.
Find the buying price of each of the following items when a sales tax of 5% is added on them.
(1) a towel of ₹50 (ii) Two bars of soap at ₹35 each.
Solution:
Given that Sales tax = 5%
(i) Cost of a towel = ₹ 50
Sales Tax = 5% of 50
= \(\frac { 5 }{ 100 }\) x 50 = \(\frac { 5 }{ 2 }\) = ₹ 2.50
∴ C.P. = Net Price + Sales
Tax = 50 + 2.50 = ₹ 52.50

(ii) The cost of two soaps at the rate of
₹ 35 each = 2 × 35 = ₹ 70
Sales Tax = 5% of 70
= \(\frac { 5 }{ 100 }\) × 70 = \(\frac { 7 }{ 2 }\) = ₹ 3.50
∴ C.P. = Net Price + Sales
Tax = 70 + 3.50 = ₹ 73.50

Question 16.
A Super-Bazar prices an item in rupees and paise so that when 4% sales tax is added, no rounding is necessary because the result is exactly in ‘n’ rupees, where ‘n’ is a positive integer. Find the smallest value of ’n’.
Solution:
Let the cost price = x say
∴ If x is increased 4% sales tax is added then
x + 4% of x = n
x + \(\frac { 4 }{ 100 }\) × x = n
\(\frac { 140x }{ 100 }\) = n
x = x = n × \(\frac { 100 }{ 104 }\) = \(\frac{25 \times \mathrm{n}}{26}\)
∴ n should be a least multiple of 26, then only the value of the article should be represented in only rupees.

[∵n = 13, 26, 39. from them 13 should betaken]
:. Required value of the article
=12.50 + \(\frac { 4 }{ 100 }\) × 12.5
12.50 + 0.5 = ₹13

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.1

Question 1.
Find the ratio of the following
(i) Smita works in office for 6 hours and Kajal works for 8 hours in her office. Find the
ratio of their working hours.
Solution:
The ratio of working hours of smita and kajal = 6:8
= (2 × 3 ) : (2 × 4) = 3 : 4

(ii) One pot contains 8 litre of milk while other contains 750 milliliter.
Solution:
8lit : 750ml
8 × 1000 : 750
= \([latex]\frac { 8000 }{ 750 }\)[/latex] = 32 : 3

(iii) speed of a cycle is 15km/h and speed of the scooter is 30km/h.
Solution:
The ratio of speeds of a cycle and a sector
= 15 : 30 = (15 × 1) : 15 × 2 = 1 : 2

Question 2.
If the compound ratio of 5:8 and 3:7 is 45:x. Find the value of x.
Solution:
The compound ratio of 5:8 and 3:7
= \(\frac{5}{8} \times \frac{3}{7}=\frac{15}{56}\)
According to the sum
15 : 56 = 45 : x
∴ x = 168

Question 3.
If the compound ratio of 7:5 and 8:x is 84:60. Find x.
Solution:
The compound ratio of 7:5 and 8:x
= \(\frac{7}{5} \times \frac{8}{x}=\frac{56}{5 x}\)
According to the sum
56 : 5x = 84 : 60

∴ x = 8

Question 4.
The compound ratio of 3:4 and the inverse ratio of 4:5 is 45:x. Find x.
Solution:
The inverse ratio of 4:5 is 45 : x
The compound ratio of 3:4 and 5 : 4
= 45 : x

⇒ x = 16 × 3 = 48
∴ x = 48

Question 5.
In a primary school there shall be 3 teachers to 60 students. If there are 400 students
enrolled in the school, how many teachers should be there in the school in the same ratio?
Solution:
No. of teachers are required for 400 students at the rate of 3 teachers to 60
students are ⇒ 60 : 3 400 : x

∴ x = 20

Question 6.
In the given figure, ABC is a triangle. Write all possible ratios by A
taking measures of sides pair wise.
8cm 10cm
(Hint: Ratio of AB : BC =8 : 6)

Solution:

In ΔABC
AB : BC = 8 : 6 = 4:3
⇒ BC : AB = 6 : 8 = 3: 4
BC : CA = 6 : 10 = 3 : 5
⇒ CA : BC = 10 : 6 = 5 : 3
CA : AB = 10:8=5:4
⇒ AB : CA = 8: 10 = 4: 5

Question 7.
If 9 out of 24 students scored below 75% marks in a test. Find the ratio of student scored below 75% marks to the student scored 75% and above marks.
Solution:
Out of 24 students who got below 75% of marks = 9
Who got 75% and above marks =24 – 9 = 15
∴ The ratio between no. of students
who got less than 75% of marks and
who got 75% and above marks
= 9 : 15 =(3 × 3):(3 × 5) = 3 : 5

Question 8.
Find the ratio of number of vowels in the word’ MISSISSIPPI’ to the number of consonants in the simplest form.
Solution:
No. of vowels in the word MI S S SS! PPI = 4 (IIII)
No. of consonants in that word = 7 (MSSSSPP)
∴ The ratio between vowels and consonants = 4: 7

Question 9.
Rajendra and Rehana own a business. Rehana receives 25% of the profit in each month. If
Rehana received ₹ 2080 in particular month, what is the total profit in that month?
Solution:
Total Profit = x say
25% of x = 2080
⇒ \(\frac{25}{100}\) × x = 2080
⇒ \(\frac{x}{4}\) = 2080
⇒ x = 2080 × 4
∴ x = ₹ 8320

Question 10.
In triangle ABC, AB = 2.2 cm, BC = 1.5 cm and AC = 2.3 cm. In triangle XYZ, XY = 4.4cm, YZ = 3cm and XZ = 4.6cm. Find the ratio AB:XY, BC:YZ, AC:XZ. Are the lengths of corresponding sides of ΔABC and ΔXYZ are in proportion?
[Hint : Any two triangles are said to be in proportion, if their corresponding sides are in the
same ratio]
Solution:

∴ The corresponding sides of both the triangles are in proportion.
∴ ΔABC ~ ΔXYZ

Question 11.
Madhuri went to a super market. The price changes are as follows. The price of rice reduced by 5% jam and fruits reduced by 8% and oil and dal increased by 10%. Help Madhuri to find the changed prices in the given table.

Item	Original price/kg	Changed price
Rice	₹ 30
Jam	₹ 100
Apples	₹ 280
Oil	₹ 120
Dal	₹ 80

Solution:

Item	Original price/kg	Changed price
Rice	₹ 30	₹28.50
Jam	₹ 100	₹ 92
Apples	₹ 280	₹ 257.6
Oil	₹ 120	₹ 132
Dal	₹ 80	₹ 88

Question 12.
There were 2075 members enrolled in the club during last year. This year enrolment is
decreased by 4%.
(a) Find the decrease in enrolment.
(b) How many members are enrolled during this year?
Solution:
No. of persons are enrolled in the last year = 2075
Present year no. of persons are enrolled
= 4% less than the previous year.
a) Decrease in enrolment = 4% of 2075

b) No.of members are enrolled this
year = 2075 – 4% of 2075
=2075 – 83 = 1992

Question 13.
A farmer obtained a yielding of 1720 bags of cotton last year. This year she expects her crop to be 20% more. How many bags of cotton does she expect this year?
Solution:
During the last year yielding the bags of
cotton = 1720
If she expects 20% crop to be more then
=20% of 1720 .
= \(\frac{20}{100}\) × 1720
= 2 × 172
= 344 bags
Her expectation of total bags
= 1720 + 344
= 2064

Question 14.
Points P and Q are both in the line segment AB and on the same side of its midpoint. P divides AB in the ratio 2: 3, and Q divides AB in the ratio 3 :4. If PQ =2, then find the length of the line segment AB.
Solution:
Given that ‘C’ is the midpoint of line segment AB.
Here ‘P’ divides AB inthe ratio 2 : 3
‘Q’ divides AB in the ratio 3: 4

PQ =2 cm [Given]
PQ = QB – PB
= 4 – 3 = 1 part = 2cm
∴ AB = AQ + QB [with respect to Ql
AB = AP+ PB [with respect to P]
L.C.M. of 5, 7 parts = 35 parts
∴ Length of AB 35 parts
= 35 × 2[ ∵ part = 2cm]
= 70cm

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers Exercise 4.2

Question 1.
Express the following numbers in the standard form.
(i) 0.000000000947
Solution:
= \(\frac{947}{1000000000000}\) = 947 × 10^-12

(ii) 543000000000
Solution:
= 543 × 1000000000 = 543 × 10⁹

(iii) 48300000
Solution:
= 483 × 100000 = 483 × 10⁵

(iv) 0.00009298
Solution:
= \(\frac{9298}{100000000}\)
= 9298 × 10^-8

(v) 0.0000529
Solution:
= \(\frac{529}{10000000}\)
= 529 × 10^-7

Question 2.
Express the following numbers in the usual form.
(i) 4.37 × 10⁵
Solution:
= 4.37 × 100000
= 437000

(ii) 5.8 × 10⁷
Solution:
= 5.8 × 10000000

(iii) 32.5 × 10^-4
Solution:
= \(\frac{32.5}{10^{4}}=\frac{32.5}{10000}\)
= 0.00325

(iv) 3.71529 × 10⁷
Solution:
= 3.71529 × 10000000
= 37152900

(v) 3789 × 10^-5
Solution:
= \(\frac{3789}{10^{5}}=\frac{3789}{100000}\)
= 0.03789

(vi) 24.36 × 10^-3
Solution:
= \(\frac{24.36}{10^{3}}=\frac{24.36}{1000}\)
= 0.02436

Question 3.
Express the following information in the standard form
(i) Size of the bacteria is 0.0000004 m
Solution:
= [klatex]\frac{4}{10000000}[/latex] m = 4 x 10^-7 m

(ii) The size of red blood cells is 0.000007mm
Solution:
= \(\frac{7}{1000000}\) = 7 × 10^-6

(iii) The speed of light is 300000000 m/sec
Solution:
= 3 × 10,00,00,000 = 3 × 10⁸ m/sec

(iv) The distance between the moon and the earth is 384467000 m(app)
Solution:
= 384467 × 1000 m
= 384467 × 10³

(v) The charge of an electron is 0.0000000000000000016 coulombs
Solution:
= 0.0000000000000000016
= \(\frac{16}{10000000000000000000}\)
= \(\frac{16}{10^{19}}\)
= 16 × 10^-19 coulombs

(vi) Thickness of a piece of paper is 0.0016 cm
Solution:
= 0.0016 cm = \(\frac{16}{10000}\)
= \(\frac{16}{10^{4}}\)
= 16 × 10^-4 cm

(vii) The diameter of a wire on a computer chip is 0.000005 cm
Solution:
= 0.000005 cm = \(\frac{5}{1000000}\) cm
= \(\frac{5}{10^{6}}\) cm = 5 × 10^-6 cm

Question 4.
In a stack, there are 5 books, each of thickness 20 mm and 5 paper sheets each of thickness 0.016mm. What is the total thickness of the stack.
Solution:
The thickness of 5 books of a pack
= (5 books × their thickness)
= 5(papers × their thickness)
= (20 mm × 5) + (0.016 mm × 5)
= 100 mm + 0.080 mm
= (100 + 0.080) mm
= 100.08 mm
= 1.0008 × 10² mm

Question 5.
Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why? Justify your argument.
(i) x^-3 × x^-2 = x^-6
Solution:
x^-3 × x^-2 = x^-6
= x^-3+(-2) = x^-6 [∵ a^m × aⁿ = a^{m + n} ]
= x^-5 = x^-6
= -5 = -6(False)
∴ In this case do not agree with Rakesh solution.

(ii) \(\frac{X^{3}}{X^{2}}\) = x⁴
Solution:
⇒ x^3-2 = x⁴ [∵ \(\frac{a^{m}}{a^{n}}\) = a^{m – n} ]
⇒ x¹ = x⁴
[∵Here bases are equal, so exponents are also equal]
⇒ 1 = 4 (It is false)
∴ I do not agree with Rakesh solution.

(iii) (x²)³ = (x²)³ = x⁸
Solution:
(x²)³ = \(x^{2^{3}}\) = x⁸
⇒ (x²)³ = \(x^{2^{3}}\)
⇒ x^{2 × 3} = \(x^{2^{3}}\)
⇒ x⁶ = x^{2 × 2 × 2}
⇒ x⁶ = x⁸
[∵ Bases are equal, so exponents are also equal]
⇒ 6 = 8
It is false
∴ Rakesh solution is wrong.

(iv) x^-2 = √x
Solution:
⇒ x^-2 = x^1/2
⇒ -2 = 1/2 (it is false)
∴ I don’t agree with Rakesh solution.

(v) 3x^-1 = \(\frac{1}{3 x}\)
Solution:
⇒ 3x^-1 = \(\frac{1}{3 x}\)
⇒ 3 × 3 = \(\frac{x}{x}\)
⇒ x⁰ = 9
⇒ 1 = 9
It is false
∴ I don’t agree with Rakesh solution.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers Exercise 4.1

Question 1.
Simplify and give reasons
(i) 4^-3
(ii) (-2) ⁷
(iii) \(\left(\frac{3}{4}\right)^{-3}\)
(iv) (-3)^-4
Solution:
(i) 4^-3 \(\frac{1}{4^{3}}=\frac{1}{64}\) [∵ a^-n = \(\frac{1}{\mathrm{a}^{\prime \prime}}\)

(ii) (-2) ⁷ = -(2) ⁷ = -128
[∵ 7 is an odd number]
[∵ (-a)ⁿ = -(aⁿ) if ‘n’ is odd]

(iii) \(\left(\frac{3}{4}\right)^{-3}\) = \(\frac{3^{-3}}{4^{-3}}=\frac{4^{3}}{3^{3}}=\left(\frac{4}{3}\right)^{3}\)

(iv) (-3)^-4 = \(\frac{1}{(-3)^{4}}\) [∵a^-n = \(\frac{1}{a^{n}}\)
= \(\frac{1}{(3)^{4}}\) [∵ 4 is even ]
= \(\frac{1}{81}\)

Question 2.
Simplify the following:
(i) \(\left(\frac{1}{2}\right)^{4} \times\left(\frac{1}{2}\right)^{5} \times\left(\frac{1}{2}\right)^{6}\)
(ii) (-2)⁷ x (-2)³ x (-2)⁴
(iii) 4⁴ x \(\left(\frac{5}{4}\right)^{4}\)
(iv) \(\left(\frac{5^{-4}}{5^{-6}}\right)\) x 5³
(v) (-3) ⁴ x 7⁴
Solution:
(i) \(\left(\frac{1}{2}\right)^{4} \times\left(\frac{1}{2}\right)^{5} \times\left(\frac{1}{2}\right)^{6}\)
\(\left(\frac{1}{2}\right)^{4+5+6}=\left(\frac{1}{2}\right)^{15}=\frac{1}{2^{15}}\)
[∵ a^m x aⁿ = a^{m + n}]

(ii) (-2)⁷ x (-2)³ x (-2)⁴
(-2)^{7 + 3 + 4} = (-2) ¹⁴ = 2 ¹⁴
[∵ (-a)ⁿ = aⁿ is even]

(iii) 4⁴ x \(\left(\frac{5}{4}\right)^{4}\)
4⁴ x \(\left(\frac{5}{4}\right)^{4}\) = 5⁴
[ ∵ \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}\) ]

(iv) \(\left(\frac{5^{-4}}{5^{-6}}\right)\) x 5³
5^-4 x (5⁶ x 5³) [ ∵ \(\frac{1}{a^{-n}}=a^{n}\)
= 5^-4 x 5⁶⁺³ [ ∵ a^m x aⁿ = (a)^m+n
= 5^-4 x 5⁹
= 5^(-4)+9 = 5⁵

(v) (-3) ⁴ x 7⁴
= 3⁴ x 7⁴[.4isevennumber]
=(3 x 7)⁴ [:amxbm=(ab)m]
= (21)⁴

Question 3.
Simplify
(i) \(2^{2} \times \frac{3^{2}}{2^{-2}} \times 3^{-1}\)
(ii) (4^-1 x 3^-1) ÷ 6^-1
Solution:
(i) \(2^{2} \times \frac{3^{2}}{2^{-2}} \times 3^{-1}\)
= 2² x 2² x 3² x 3^-1
= 2²⁺² x 3^{2 + ( – 1)}
=2⁴ x 3¹ = 16 x 3 = 48

(ii) (4^-1 x 3^-1) ÷ 6^-1

Question 4.
Simplify and give reasons
(i) (4⁰ + 5^-1) x 5² x \(\frac{1}{3}\)
Solution:

(ii) \(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{4}\right)^{-3} \times\left(\frac{1}{5}\right)^{-3}\)
Solution:
= \(\left(\frac{1}{2} \times \frac{1}{4} \times \frac{1}{5}\right)^{-3}\)
= \(\left(\frac{1}{40}\right)^{-3}\) [∵ a^m x b^m x c^m = (abc)^m
= (40)³ [ ∵]\(\frac{1}{a^{-n}}\) = aⁿ ]

(iii) (2^-1 + 3^-1 + 4^-1) x \(\frac{3}{4}\)
Solution:

(iv) \(\frac{3^{-2}}{3}\) x (3⁰ – 3^-1
Solution:

(v) 1 + 2^-1 + 3^-1 + 4⁰
Solution:

(vi) \(\left[\left(\frac{3}{2}\right)^{-2}\right]^{2}\)
Solution:

Question 5.
Simplify and give reasons
(i) \(\left[\left(3^{2}-2^{2}\right) \div \frac{1}{5}\right]^{2}\)
Solution:
\(\left.\left[(9-4) \div \frac{1}{5}\right)\right]^{2}\)
= \(\left[5 \times \frac{5}{1}\right]^{2}\) = (5²)² 5⁴ = 625 [∵ (a^m)ⁿ = a^mn]

(ii) ((5²)³ x 5⁴) ÷ 5⁶
Solution:

Question 6.
Find the value of ’n’ in each of the following:
(i) \(\left(\frac{2}{3}\right)^{3} \times\left(\frac{2}{3}\right)^{5}=\left(\frac{2}{3}\right)^{\mathrm{n}-2}\)
Solution:

Here bases are equal, so exponents are
also equal.
⇒ n – 2 = 8
⇒ n = 8 + 2 = 10
∴ n = 10

(ii) (-3)ⁿ⁺¹ x (-3)⁵ = (-3)³
Solution:
⇒(-3)ⁿ⁺¹⁺⁵ = (-3)^-4 [∵ a^m x aⁿ = a^m+n ]
⇒ (-3)ⁿ⁺⁶ = (-3)^-4
⇒ n + 6 = -4
⇒ n = -4 – 6 = -10
⇒ n = -10

(iii) 7²ⁿ⁺¹ ÷ 49 = 7³
Solution:

⇒ 7^2n+1-2 = 7³ [ ∵ \(\frac{a^{m}}{a^{n}}=a^{m-n}\) ]
⇒ 7^{2n – 1}= 7³
⇒ 2n – 1 = 3
⇒ 2n = 3 + 1 = 4
⇒ n = \(\frac{4}{2}\)
∴ n = 2

Question 7.
Find ’x’ if 2^-3 = \(\frac{1}{2^{x}}\)
Solution:
2^-3 = \(\frac{1}{2^{x}}\) = 2^-x
⇒ 2^-3 = 2^-x [ \(\frac{1}{a^{n}}\) = a^-n ]
⇒ -x = -3
∴ x = 3

Question 8.
Simplify \(\left[\left(\frac{3}{4}\right)^{-2} \div\left(\frac{4}{5}\right)^{-3}\right] \times\left(\frac{3}{5}\right)^{-2}\)
Solution:

Question 9.
If m = 3 and n = 2 find the value of
(i) 9m² – 10n³
(ii) 2m² n²
(iii) 2m³ + 3n² – 5m²n
(iv) mⁿ – n^m
Solution:
1) 9m² – 10n³
= 9(3)² – 10(2)³
= 9 x 9 – 10 x8
= 81 – 80 = 1

(ii) 2m² n²
= 2(3)² (2)²
= 2 x 9 x 4 = 72

(iii) 2m³ + 3n² – 5m²n
= 2(3)³ + 3(2)² – 5(3)²(2)
= (2 x 27) + (3 x 4) – (5 x 9 x 2)
= 54 + 12 – 90
= 66 – 90 = – 24

(iv) mⁿ – n^m
= 3² – 2³
= 3 x 3 – 2 x 2 x 2
= 9 – 8 = 1

Question 10.
Simplify and give reasons \(\left(\frac{4}{7}\right)^{-5} \times\left(\frac{7}{4}\right)^{-7}\)
Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.3

Construct the quadrilateral with the measurements given below :

Question a).
Quadrilateral GOLD: OL = 7.5 cm, GL = 6 cm, LD = 5 cm, DG = 5.5 cm and OD = 10 cm.
(Ex 3.3, Page No. 72)
Solution:
In a quadrilateral GOLD, Rough Diagram
OL = 7.5 cm, GL = 6 cm
LD = 5 cm, DG = 5.5 cm, OD = 10 cm

Construction Steps:

1. Draw a line segment \(\overline{OL}\) equal to radius 7.5 cm.
2. With the centres O, L draw arcs with radius 10 cm and 5 cm respectively. These two arcs meet at point ‘D’.
3. With the centres L, D draw arcs equal to 6 cm and 5.5 cm respectively. These two arcs meet at point ‘G’.
4. Join O, G and L, G. Also join O, D and L, D and G, D.
5. ∴ The required quadrilateral GOLD is formed.

Question b).
Quadrilateral PQRS: PQ = 4.2 cm, QR = 3 cm, PS = 2.8 cm, PR = 4.5 cm and QS = 5 cm.
Solution:


In a quadrilateral PQRS
PQ = 4.2 cm PS = 2.8 cm
QR = 3 cm PR = 4.5 cm
QS = 5 cm

Construction Steps:

1. Draw a line segment \(\overline{P Q}\) with radius 4.2 cm.
2. With the centres P, Q draw arcs equal to the radius 4.5 cm, 3 cm respectively. These two arcs meet at point R’. Join P, R and Q, R.
3. With the centres Q, P draw arcs equal to the radii 5 cm and 2.8 cm respectively. These two arcs meet at point ‘S’.
4. Join P, S and Q, S and S, R.
5. ∴ The required PQRS quadrilateral is formed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.2

Construct quadrilateral with the measurements given below:

Question (a).
Quadrilateral ABCD with AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm andAC= 7cm
Solution:
In Quadrilateral ABCD with AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and AC = 7 cm.

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{AB}}\) with radius 4.5 cms.
2. With the centres A, B draw arcs equal to 7 cm and 5.5 cm respectively. The intersection of these two arcs keep as ‘C’.
3. Join A, C and B, C.
4. With centres C, A draw arcs equal to 4 cm, 6 cm respectively. These intersecting point is keep as ’D’.
   Join D, C and A, D.
5. ∴ The required quadrilateral ABCD is formed.

Question (b).
Quadrilateral PQRS with PQ = 3.5 cm, QR = 4 cm, RS = 5 cm, PS = 4.5 cm and QS= 6.5 cm
Solution:
In a quadrilateral PQRS,
PQ = 3.5 cm, QR = 4 cm, RS = 5 cm,
PS – 4.5 cm, QS = 6.5 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{PQ}}\) with radius 3.5 cm.
2. With the centres P, Q draw arcs equal to 4.5 cm and 6.5 cm respectively.
3. These two arcs meet at point ‘S’.
4. With the centres S, Q draw arcs with radius 5 cm, 4 cm respectively. These two arcs intersected at point ‘R’.
5. Join P, S; Q, S; S, R and Q, R.
6. ∴ The required quadrilateral PQRS is formed.

Question (c).
Parallelogram ABCD with AB = 6cm, CD = 4.5 cm and BD = 7.5 cm
Solution:
In a parallelogram ABCD; AB = 6 cm, BC = 4.5 cm, BD = 7.5 cm
AB = CD (;cm
BC = AD = 4.5 cm
BD = 7.5 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{AB}}\) with radius 6 cms.
2. With the centres A, B draw arcs with radius 4.5 cm, 7.5 cm respectively. These two arcs meet at point ‘D’.
3. With the centres D, B draw arcs with radius 6 cm, 4.5 cm respectively. These two arcs meet at point ‘C’.
4. Join A, D and B, C and D, C and B, D.
5. ∴ The required parallelogram ABCD is formed.

Question (d).
Rhombus NICE with NI = 4 cm and IE = 5.6 cm
Solution:
In a rhombus NI = IC = CE = NE = 4 cm, IE = 5.6 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{NI}}\) with radius 4 cm.
2. With the centres N, I draw two arcs with radius 4 cm, 5.6 cm respectively. These two arcs meet at point E’.
3. With the centres E, I draw arcs with radius 4 cm. These two arcs meet at point ‘C’.
4. Join N, E and I, E. Also join E, C and I, C.
5. .’. The required rhombus NICE is formed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.1

Construct the quadrilaterals with the measurements given below:

Question (a).
Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.
Solution:
In Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.

Construction Steps:

1. Construct a line segment \(\overline{\mathrm{AB}}\) with radius 5.5 cm
2. With the centre A draw a ray and an arc which are equL1 to 45° and 5 cm.
3. These intersecting point is keep as ‘D’.
4. With centres D, B draw two arcs equal to radius 4 cm, 3.5 cm respectively.
5. The intersecting point of these two arcs is keep as ‘C’.
6. Join DC and BC. A F
7. ∴ The required quadrilateral ABCD is formed.

Question (b).
Quadrilateral BEST with BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.
Solution:
In Quadrilateral BEST
BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{BE}}\) with radius 2.9 cm.
2. With the centre B, draw a ray of 75° and draw 2.9
   an arc with radius 3.4 cm, keep the intersecting point of these two as T.
3. With the centres T, E draw arcs with radius 2.7 cm, 3.2 cm respectively. These intersection point is keep as S’.
4. Join T, S and E,S.
5. ∴ The required quadrilateral BEST is formed.

Question (c).
Parallelogram PQRS with PQ = 4.5 cm, QR =3 cm and ∠PQR = 60°.
Solution:

In a parallelogram PQRS
PQ = 4.5 cm, QR = 3 cm, ZPQR = 60°.
=RS4.5cmzPS=3crn
[: Opposite sides of a I)aralielograrn are equal]

Construction Steps:

1. Draw a line segment ¡i with radius 4.5 cm.
2. With the centre Q draw a ray and an arc equal to 60° and 3 cm.
3. The intersecting point of these two keep as R’.
4. With the centres R, P draw arcs with 4.5 cm, 3 cm respectively. Keep ‘S’ as the intersecting point of these two arcs.
5. Join P, S and R, S.
6. ∴ The required parallelogram PQRS is formed.

Question (d).
Rhombus MATH with AT =4 cm, ∠MAT =120°.
Solution:

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{MA}}\) with radius 4 cm.
2. With the centre A draw a ray and an arc equal to 120°, 4 cm. These two intersecting point be keep as T.
3. With the centres M, T draw arcs equal to 4 cms.
   These two arcs intersected at the point ‘H’.
4. Join M, H and T, H.
5. ∴ The required rhombus MATH is formed.

Question (e).
Rectangle FLAT with FL =5 cm, LA= 3 cm.
Solution:


In a rectangle FLAT
FL=AT=5cm, LA = TF = 3cm, ∠F = ∠L = ∠A = ∠T = 90°

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{FL}}\) with radius 5 cm.
2. With the centre F draw a ray and an arc equal to 900, 3 cm.
   These to meet at point T.
3. With the centres T, L draw arcs equal to 5 cm, 3 cm respectively.
4. These two arcs meet at the point ‘A’.
5. Join T, A and L, A.
6. ∴ The required rectangle FLAT is formed.

Question (f).
Square LUDO with LU = 4.5 cm.
Solution:


In a square LUDO
LU = UD = DO = OL = 4.5 cm
∠L = ∠U = ∠D = ∠O = 90°

Construction Steps: 45

1. Draw a line segment \(\overline{\mathrm{LU}}\) with radius 4.5 cm.
2. With the centre ‘L’, draw a ray of 90° and an arc with radius 4.5 cm. These two meet at the point ‘O’.
3. Now with the centre U’, draw another ray of 90° and an arc with radius 4.5 cm. These two meet at the point “D”.
4. Join O, D.
5. ∴ The required square LUDO is formed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.5

Question 1.
Solve the following equations.
i) \(\frac{n}{5}-\frac{5}{7}=\frac{2}{3}\)
Solution:

ii) \(\frac{x}{3}-\frac{x}{4}=14\)
⇒ \(\frac{4 x-3 x}{12}\) = 14
⇒ \(\frac{x}{12}\) = 14
⇒ x = 12 × 14 = 168
∴ x = 168

iii) \(\frac{z}{2}+\frac{z}{3}-\frac{z}{6}=8\)

iv) \(\frac{2 p}{3}-\frac{p}{5}=11 \frac{2}{3}\)

v) \(9 \frac{1}{4}=y-1 \frac{1}{3}\)

vi) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)

vii) \(\frac{x}{2}-\frac{1}{4}=\frac{x}{3}+\frac{1}{2}\)

viii) \(\frac{2 x-3}{3 x+2}=\frac{-2}{3}\)
⇒ 3(2x – 3) = – 2(3x + 2)
⇒ 6x – 9 = -6x – 4
⇒ 6x + 6x = -4 + 9
⇒ 12x = 5
∴ x = \(\frac{5}{12}\)

ix) \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
Solution:
⇒ \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
⇒ 2(8p – 5) = – (7p + 1)
⇒ 16p – 10 = – 7p – 1
⇒ 16p + 7p = – 1 + 10
⇒ 23p = 9
∴ x = \(\frac{9}{23}\)

x) \(\frac{7 y+2}{5}=\frac{6 y-5}{11}\)
⇒ 11 (7y + 2) = 5 (6y-5)
⇒ 77y + 22 = 30y – 25
⇒ 77y – 30y = – 25 – 22
⇒ 47y = – 47
∴ y = \(\frac{-47}{47}\)
∴ y = -1

xi) \(\frac{x+5}{6}-\frac{x+1}{9}=\frac{x+3}{4}\)

⇒ 4(x + 13) = 18 (x + 3)
⇒ 4x + 52 = 18x + 54
⇒ 4x – 18x = 54-52
⇒ – 14x = 2
⇒ x = \(\frac{2}{-1}\) = \(\frac{-1}{7}\)
∴ x = \(\frac{-1}{7}\)

xii) \(\frac{3 t+1}{16}-\frac{2 t-3}{7}=\frac{t+3}{8}+\frac{3 t-1}{14}\)
Solution:

⇒ -11t + 55 = 2(19t + 17) = 38t + 34
⇒ -11t – 38t = 34 – 55
⇒ -49t = – 21
⇒ \(\frac{-21}{-49}\) = \(\frac{3}{7}\)
∴ t = \(\frac{3}{7}\)

Question 2.
What number is that of which the third part exceeds the fifth part by 4?
Solution:
Let the number be ‘x’ say.
\(\frac{1}{3}\) rd of a number = \(\frac{1}{3}\) x x = \(\frac{x}{3}\)
\(\frac{3}{7}\) th of a number = \(\frac{1}{5}\) x x = \(\frac{x}{5}\)
According to the sum

∴ The required number is 30.

Question 3.
The difference between two positive integers is 36. The quotient when one integer is
divided by other is 4. Find the integers.
(Hint: If one number is ‘X’, then the other number is ‘x – 36’)
Solution:
Let the two positive numbers be x, (x – 36) say.
If one number is divided by second tten the quotient is 4.
∴ \(\frac{x}{x-36}=4\)
⇒ x = 4(x – 36) = 4x – 144
⇒ 4x – x = 144
3x = 144
x = 48
∴ x – 36 = 48 – 36 = 12
∴ The required two positive intgers are 48, 12.

Question 4.
The numerator of a fraction is 4 less than the denominator. If 1 is added to both its
numerator and denominator, it becomes 1/2 . Find the fraction.
Solution:
Let the denominator of a fractin be x.
The numerator of a fraction is 4 less than the denominator.
∴ The numerator = x – 4
∴ Fraction \(\frac{x-4}{x}\)
If ‘1’ is added to both, its numerator and denominator, it becomes \(\frac{1}{2}\)
∴ \(\frac{1+x-4}{1+x}=\frac{1}{2}\)
2 + 2x – 8 = 1 + x
2x – x = 1 + 6 = 7
x = 7
∴ The denominator = 7
The numerator = 7 – 4 = 3
∴ Fraction = \(\frac{3}{7}\)

Question 5.
Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively,
the sum of their quotients will be 10.
(Hint: Let the consecutive numbers = x, x+ 1, x+ 2, then \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\))
Solution:
Let the three consecutive numbers be assume that x, (x + 1), (x + 2) respectively.
Given that x, (x + 1), (x + 2) are divided by 10, 17, 26 respectively, the sum of the quotients is 10. Then
⇒ \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\)
⇒ \(\frac{x \times 221+130(x+1)+85(x+2)}{2210}=10\)
⇒ 221x + 130x + 85x + 130 + 170 = 22,100
⇒ 436x + 300 = 22,100
⇒ 436x = 22,100 – 300
⇒ 436x = 21,800
⇒ \(\frac{21800}{436}\)
∴ x = 50
∴ The required three consecutive num-bers are x = 50
x + 1 =50+ 1 = 51
x + 2 = 50 + 2 = 52

Question 6.
In class of 40 pupils the number of girls is three-fifths of the number of boys. Find the
number of boys in the class.
Solution:
Let the number of boys = x say.
Total number of students = 40
Number of girls = \(\frac{3}{5}\) × x = \(\frac{3x}{5}\)
According to the sum 3x

∴ x = 25
∴ Number of boys in the class room = 25

Question 7.
After 15 years , Mary’s age will be four times of her present age. Find her present age.
Solution:
Let the present age of Mary = x years say.
After 15 years Mary’s age = (x + 15) years
According to the sum
(x + 15) = 4 x x
⇒ x + 15 = 4x
⇒ 4x – x =15
⇒ 3x = 15
⇒ x = 5
∴ The present age of Mary = 5 years.

Question 8.
Aravind has a kiddy bank. It is full of one-rupee and fifty paise coins. It contains 3 times
as many fifty paise coins as one rupee coins. The total amount of the money in the bank is
₹ 35. How many coins of each kind are there in the bank?
Solution:
Number of 1 rupee coins = x say.
Number of 50 – paise coins = 3 x x = 3x
The value of total coins = \(\frac{3x}{2}\) + x
[∵50 paisa coins of 3x = ₹\(\frac{3x}{2}\)
According to the sum
⇒ \(\frac{3x}{2}\) + x = 35
⇒ \(\frac{3 x+2 x}{2}\) = 35
⇒ 5x = 2 × 35
⇒ x = 2 × \(\frac{35}{5}\)
∴ x = 14
∴ Number of 1 rupee coins = 14
Number of 50 paisa coins = 3 × x = 3 × 14 = 42

Question 9.
A and B together can finish a piece of work in 12 days. If ‘A’ alone can finish the same work in 20days , in how many days B alone can finish it?
Solution:
A, B can do a piece of work in 12 days.
(A + B)’s 1 day work = \(\frac{1}{12}\) th part.
A can complete the same work in 20 days.
Then his one day work = \(\frac{1}{20}\)
B’s one day work = (A+B)’s 1 day work – A’s 1 day work
\(=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}\)
∴ Number of days to take B to com¬plete the whole work = 30.

Question 10.
If a train runs at 40 kmph it reaches its destination late by 11 minutes . But if it runs at 50 kmph it is late by 5 minutes only. Find the distance to be covered by the train.
Solution:
Let the distance to be reached = x km say. Time taken to travel ‘x’ km with speed x
40 km/hr = \(\frac{x}{40}\) hr.
Time taken to travel ‘x’ km with speed 50 km/hr = \(\frac{x}{50}\) hr.
According to the sum the difference between the times

∴ The required distance to be trav¬elled by a train = 20 kms‘.

Question 11.
One fourth of a herd of deer has gone to the forest. One third of the total number is
grazing in a field and remaining 15 are drinking water on the bank of a river. Find the total
number of deer.
Solution:
Number of deer = x say.
Number of deer has gone to the forest
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of deer grazing in the field
= \(\frac{1}{3}\) × x =\(\frac{x}{3}\)
Number of remaining deer =15
According to the sum

∴ x = 36
∴ The total number of deer = 36

Question 12.
By selling a radio for ₹903, a shop keeper gains 5%. Find the cost price of the radio.
Solution:
The selling price of a radio = ₹ 903
Profit % = 5%
C.P = ?
C.P = \(\frac{\mathrm{S.P} \times 100}{(100+\mathrm{g})}\)
= \(\frac{903 \times 100}{(100+5)}\)
= \(\frac{903\times 100}{105}\)
C.P. = 8.6 × 100 = 860
∴ The cost price of the radio = ₹ 860

Question 13.
Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has 7 sweets left. How many did he have to start with?
Solution:
Number of sweets with Sekhar = x say.
Number of sweets given to Renu
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of sweets given to Raji = 5
Till he has 7 sweets left.
x – ( \(\frac{x}{4}\) + 5) = 7
⇒ x – \(\frac{x}{4}\) – 5 = 7
⇒ x – \(\frac{x}{4}\) = 7 + 5 = 12
⇒ \(\frac{4 x-x}{4}\) = 12
⇒ \(\frac{3x}{4}\) = 12
⇒ x = 12 × \(\frac{4}{3}\) = 16
∴ x = 4 × 4 = 16
∴ Number of sweets with Sekhar at the beginning = 16

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.4

Question 1.
Find the value of ’x’ so that l || m

Solution:
Given l|| m. Then 3x – 10° = 2x + 15°
[Vertically opposite angles and corresponding angles are equal.]
⇒ 3x – 10 = 2x + 15
⇒ 3x – 2x = 15 + 10
∴ x = 25°

Question 2.
Eight times of a number reduced by 10 is equal to the sum of six times the number and 4. Find the number.
Solution:
Let the number be ‘x’ say.
8 times of a number = 8 × x = 8x
¡f10 is reduced from 8x then 8x – O
6 times of a number = 6 × x = 6x
If 4 is added to 6x then 6x + 4
According to the sum,
8x – 10 = 6x + 4
⇒ 8x – 6x = 4 + 10
⇒ 2x = 14
⇒ x = 7
∴ The required number = 7

Question 3.
A number consists of two digits whose sum is 9. If 27 is subtracted from the number its digits are reversed. Find the number.
Solution:
Let a digit of two digit number be x.
The sum of two digits = 9
∴ Another digit = 9 – x
The number = 10 (9 – x) + x
= 90 – 10x + x
= 90 – 9x
If 27 is subtraçted from the number its digits are reversed.
∴ (90 – 9x) – 27 = 10x + (9 – x)
63 – 9x = 9x + 9
9x + 9x = 63 – 9
18x = 54
∴ x = \(\frac { 54 }{ 18 }\) = 3
∴ Units digit = 3
Tens digit = 9 – 3 = 6
∴ The number = 63

Question 4.
A number is divided into two parts such that one part is 10 more than the other. If the two parts are in the ratio 5:3, find the number and the two parts.
Solution:
If a number is divided into two parts in he ratio of 5 : 3, let the parts be 5x, 3x say.
According to the sum,
5x = 3x + 10
⇒ 5x – 3x = 10
⇒ 2x = 10
∴ x = \(\frac { 10 }{ 2 }\)
∴ x = 5
∴ The required number be
x + 3x = 8x
= 8 × 5 = 40
And the parts of number are
5 = 5 × 5 = 25
3 = 3 × 5 = 15

Question 5.
When I triple a certain number and add 2, I get the same answer as I do when I subtract the number from 50. Find the number.
Solution:
Let the number be x’ say.
3 times of a number = 3 × x = 3x
If 2 is added to 3x then 3x + 2
If ‘xis subtracted from 50 then it becomes 50 – x.
According to the sum,
3x + 2 = 50 – x
3x + x = 50 – 2
4x = 48 .
x = 12
∴ The required number 12

Question 6.
Mary is twice older than her sister. In 5 years time, she will be 2 years older than her sister. Find how old are they both now.
Solution:
Let the age of Marys sister = x say.
Mary’s age = 2 × x = 2x
After 5 years her sister’s age
= (x + 5) years
After 5 years Mary’s age
= (2x + 5) years
According to the sum,
2x + 5 = (x + 5) + 2
= 2x – x = 5 + 2 – 5
∴ The age of Mary’s sister = x = 2 years
Mary’s age = 2x = 2 x 2 = 4 years

Question 7.
In 5 years time, Reshma will be three times old as she was 9 years ago. How old is she now?
Solution:
Reshma’s present age = ‘x’ years say.
After 5 years Reshmats age
= (x + 5) years
Before 9 years Reshma’s age
=(x – 9) years
According to the sum
= x+ 5 = 3(x – 9) = 3x – 27
x – 3x = -27-5
-2x = -32
x = \(\frac{-32}{-2}\) = 16
∴ x = 16
∴ Reshma’s present age = 16 years.

Question 8.
A town’s population increased by 1200 people, and then this new population decreased 11%. The town now had 32 less people than it did before the 1200 increase. Find the original population.
Solution:
Let th population of a town after the increase of 1200 is x say.
11% of present population

The present population of town
= 11,200 – 1200 = 10,000

Question 9.
A man on his way to dinner shortly after 6.00 p.m. observes that the hands of his watch form an angle of 110°. Returning before 7.00 p.m. he notices that again the hands of his watch form an angle of 1100. Find the number of minutes that he has been away.
Solution:
Let the number be ‘x ray.
\(\frac { 1 }{ 3 }\) rd of a number = \(\frac { 1 }{ 3 }\) x x = \(\frac { x }{ 3 }\)
\(\frac { 1 }{ 5 }\) th of a number = \(\frac { 1 }{ 5 }\) x x = \(\frac { x }{ 5 }\)
According to the sum

∴ x = 30
∴ The required number is 30.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.3

Question
Solve the following equations:
1. 7x – 5 = 2x
2. 5x – 12 = 2x – 6
3. 7p- 3 = 3p + 8
4. 8m + 9 = 7m + 8
5. 7z + 13 = 2z + 4
6. 9y + 5 = 15y – 1
7. 3x + 4 = 5(x – 2)
8. 3(t – 3) = 5(2t – 1)
9. 5(p – 3) = 3(p – 2)
10. 5(z + 3) = 4(2z + 1)
11. 15(x – 1) + 4(x + 3) = 2 (7 + x)
12. 3 (5z – 7) +2 (9z – 11) = 4 (8z – 7) – 111
13. 8(x – 3) – (6 – 2x) = 2(x + 2) – 5 (5 – x)
14. 3(n – 4)+2(4n – 5) = 5(n + 2) + 16
Solution:
1. 7x – 5 = 2x
⇒ 7x – 2x = 5
⇒ 5x = 5
⇒ x = \(\frac { 5 }{ 5 }\) = 1
∴ x = 1

2. 5x – 12 = 2x – 6
⇒ 5x – 2x = – 6 + 12
⇒ 3x = 6
⇒ x = \(\frac { 6 }{ 3 }\) = 2
∴ x = 2

3. 7p – 3 = 3p + 8
⇒ 7p – 3p = 8 + 3
⇒ 4p = 11
⇒ p = \(\frac { 11 }{ 4 }\)

4. 8m + 9 = 7m + 8
⇒ 8m – 7m = 8 – 9
∴ m = – 1

5. 7z + 13 = 2z + 4
⇒ 7z – 2z = 4 – 13
⇒ 5z = – 9
∴ z = \(\frac { -9 }{ 5 }\)

6. 9y + 5 = 15y – 1
⇒ 9y – 15y = – 1 – 5
⇒ -6y = -6
⇒ y = \(\frac { -6 }{ -6 }\)
∴ y = 1

7. 3x + 4 = 5(x – 2)
⇒ 3x + 4 =5x – 10
⇒ 3x – 5x= – 10 – 4
⇒ – 2x = – 14
:. x = 71

8. 3(t – 3) = 5(2t – 1)
⇒ 3t – 9 = 10t – 5
⇒ 3t – 10t = – 5+ 9
⇒ – 7t = 4
∴ t = \(\frac { -4 }{ 7 }\)

9. 5 (p – 3) = 3 (p – 2)
⇒ 5p – 15 = 3p – 6
⇒ 5p – 3p = -6 + 15
⇒ 2p = 9
∴ p = \(\frac { 9 }{ 2 }\)

10. 5(z + 3) = 4(2z + 1)
⇒ 5z + 15 = 8z + 4
⇒ 5z – 8z = 4 – 15
⇒ – 3z = – 11
⇒ z = \(\frac { -11 }{ 3 }\)
∴ z = \(\frac { -11 }{ 3 }\)

11. 15(x – 1) + 4(x + 3) = 2(7 + x)
⇒ 15x – 15 + 4x + 12= 14 + 2x
⇒ 19x – 3 = 14 + 2x
⇒ 19x – 2x = 14 + 3
⇒ 17x = 17 ,
x = \(\frac { 17 }{ 17 }\) = 1
∴ x = 1

12. 3(5z – 7)+2(9z – 11) = 4(8z – 7) – 111
⇒ 15z – 21 + 18z – 22 = 32z – 28 – 111
⇒ 33z – 43 = 32z – 139
⇒ 33z – 32z = – 139 + 43
∴ z = – 96

13. 8(x – 3) – (6 – 2x)=2(x+2)-.5(5 – x)
⇒ 8x – 24 – 6 + 2x = 2x + 4 – 25 +5x
⇒ 8x – 30 = 5x – 21
⇒ 8x – 5x= – 21 +30
⇒ 3x = 9
∴ x = 3

14. 3(n – 4) + 2(4n – 5) = 5(n + 2) + 16
⇒ 3n – 12 + 8n – 10 = 5n + 10 + 16
⇒ 11n – 22 = 5n + 26
⇒ 11n – 5n = 26 + 22
⇒ 6n =48
⇒ n = \(\frac { 48 }{ 6 }\) = 8
∴ n = 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.2

Question 1.
Find ‘x’ in the following figures?


Solution:
1) In a triangle the exterior angle is equal to the sum of its opposite interior angles.
∴ ∠ACD = ∠B + ∠A
⇒ 123°= x + 56°
⇒ x = 123°- 56° = 67°
∴ x = 67°

ii) Sum of three angles of a triangle = 180°
∠P + ∠Q +∠R = 180°
⇒ 45° + 3x + 16°+ 68° = 180°
⇒ 3x + 129° = 180°
3x = 180 – 129 = 51
∴ x = \(\frac{51}{3}\)
∴ x = 17°

iii) ∠A + ∠B + ∠C = 180°
⇒ 25° + x + 30° = 180°
x + 55°= 180°
x = 180 – 55 = 125°
∴ x = 125°

iv) In ΔXYZ, \(\overline{\mathrm{XY}}=\overline{\mathrm{XZ}}\) then ∠Y = ∠Z
∴ 2x + 7° = 45°
⇒ 2x = 45 – 7
⇒ 2x = 38
⇒ x = \(\frac{38}{2}\)
∴ x = 19°

v) From ΔBOA
\(\overline{\mathrm{AB}}=\overline{\mathrm{OA}} \) ⇒ ∠B = ∠O = 3x + 10° ………(1)
From ΔCOD
\(\overline{\mathrm{OC}}=\overline{\mathrm{CD}} \) ⇒ ∠O = ∠D …………………(2)
[∵ The angles which are opposite to the equal sides are equal].
from (1) & (2)
∠BOA = ∠COD
[∵ Vertically opposite angles are equal.]
But ∠COD = 90 – x
(∵ 2x + ∠O + ∠D = 180
⇒ 2x + ∠O + ∠O = 180 (∵∠O = ∠D)
⇒ 2∠O = 180 – 2x
∠O = \(\frac{180-2 x}{2}\) = 90 – x]
∴ From ∠BOA = ∠COD
⇒ 3x + 10 = 90 – x
⇒ 3x + x = 90 – 10
⇒ 4x = 80
∴ x = 20°

Question 2.
The difference between two numbers is 8. if 2 is added to the bigger number the result will be three times the smaller number. Find the numbers.
Solution:
Let the bigger number be x.
The difference between two numbers 8
∴ Smaller number = x – 8
If 2 is added to the bigger number the result will be three times the smaller
number.
So x + 2 = 3(x – 8)
x + 2 = 3x – 24
x – 3x = -24 – 2
– 2x = -26
∴ x = \(\frac{26}{2}\) = 13
∴ Bigger number = 13
Smaller number = 13 – 8 = 5

Question 3.
What are those two numbers whose sum is 58 and difference is 28’?
Solution:
Let the bigger number be ‘x’.
The sum of two numbers = 58
∴ Smaller number = 58 – x
The difference of two numbers = 28
∴ x – (58 – x) = 28
x – 58 + x = 28
2x = 28 + 58 = 86
∴ x = \(\frac{86}{2}\) = 43
∴ Bigger number or one number = 43
Smaller number or second number = 58 – 43 = 15

Question 4.
The sum of two consecutive odd numbers is 56. Find the numbers.
Solution:
Let the two consecutive odd numbers
be 2x 1, 2x + 3 say.
Sum of the odd numbers
=(2x + 1) + (2x + 3) = 56
= 4x + 4 = 56
⇒ 4x = 56 – 4 = 52
x = \(\frac{52}{4}\) = 13
∴ x = 13
∴ 2x + 1 = 2 × 13 + 1
= 26 + 1 = 27
2x + 3 = 2 × 13 + 3
= 26 + 3 = 29
∴ The required two consecutive odd numbers be 27, 29.

Question 5.
The sum of three consecutive multiples of 7 is 777. Find these multiples.
(Hint: Three consecutive multiples of 7 are ‘x’, ‘x+ 7’, ‘x+ 14’)
Solution:
Let the three consecutive multiples of 7 be x, x + 7, x + 14 say.
According to the sum,
The sum of three consecutive multiples of 7 is 777.
⇒ x + (x + 7) + (x + 14)= 777
⇒ 3x + 21 = 777
⇒ 3x = 777 – 21 = 756
x = \(\frac { 756 }{ 3 }\) = 252
x+ 7 = 252 + 7 = 259
x + 14 252 + 14 = 266
∴ The required three consecutive multiples of 7 are 252, 259, 266

Question 6.
A man walks 10 km, then travels a certain distance by train and then by bus as far as twice by the train. 1f the whole journey is of 70km, how far did he travel by train?
Solution:
The distance travelled by walk = 10 km
Let the distance travelled by train = x km say.
The distance travelled by bus
= 2 × x = 2x km
∴ 10 + x + 2x = 70
⇒ 3x = 70 – 10
⇒ 3x = 60
⇒ x = \(\frac { 60 }{ 3 }\) = 20
⇒ x = 20
∴ The distance travelled by train = 20 km.

Question 7.
Vinay bought a pizza and cut it into three pieces. When the weighed the first piece he found that it was 7g lighter than the second piece and 4g.heavier than the third piece. If the whole pizza weighed 300g. How much did each of the three pieces weigh?

(Hint: weight of normal piece be ‘x’ then weight of largest piece is ‘x+ 7’, weight of the smallest piece is ‘x-4’)
Solution:
If pizza is cut into three pieces.
Let the weight of first piece he ‘x’ gm say.
Weight of the second piece = (x + 7) gm
Weight of the third piece = (x – 4) gm
According to the sum
∴ x + (x + 7) + (x – 4) = 300
⇒ 3x + 3 = 300
⇒ 3x = 300 – 3 = 297
⇒ x = \(\frac { 297 }{ 3 }\) = 99
∴ x= 99
x + 7= 99 + 7 =106
x – 4 = 99 – 4 = 95
∴ The required 3 pieces of pizza weighs 95 gm. 99 gm, 106 gm.

Question 8.
The distance around a rectangular field is 400 meters.The length of the field is 26 meters more than the breadth. Calculate the length and breadth of the field’?
Solution:
Let the breadth of a rectangular field = x m
Length = (x + 26) m.
Perimeter of a rectangular field
= 2(l + b) = 400
l + b = 200
x + 26 + x = 200
2x = 200 – 26 = 174
x = \(\frac { 174 }{ 2 }\)
∴ x = 87
∴ The length = x +26
= 87 + 26
= 113 m
Breadth = x = 87 m.

Question 9.
The length of a rectangular field is 8 meters less than twice its breadth. If the perimeter of the rectangular field is 56 meters, find its length and breadth’?
Solution:
Let the breadth of a rectangular field = xm.
Length = 2 × x – 8 = (2x – 8) m.
Perimeter of a field = 56 m.
∴ 2(l + b) = 56
l + b = 28
2x – 8 + x = 28
3x = 28 + 8 = 36
x = \(\frac { 36 }{ 3 }\)
∴ x= 12
∴ Breadth = 12 m
Length = 2x – 8
= 2 × 12 – 8
= 24 – 8 = 16m.

Question 10.
Two equal sides of a triangle are each 5 meters less than twice the third side. if the perimeter of the triangle is 55 meters, find the length of its sides’?
Solution:
A triangle in which the length of the third side = x m. say.
The length of remaining two equal sides = 2 × x – 5 = (2x – 5) m.
Perimeter of a triangle = 55 m.
∴ (2x – 5) + (2x -5) + x = 55
⇒ 5x – 10 = 55
⇒ 5x = 65
⇒ x = \(\frac { 65 }{ 5 }\)
∴ x = 13m
2x – 5 = 2 × 13 – 5 = 26 – 5 = 21m.
∴ The lengths of three sides of a triangle are 13, 21, 21. (in m.)

Question 11.
Two complementary angles differ by 12°, find the angles’?
Solution:
Let one angle in two complementary angles be x.
Sum of the two complementary angles = 90°
∴ Second angle = 90° – x
Here two complementary angles differ by 12°.
∴ x – (90°- x) = 12°
x – 90° + x = 12°
2x = 12° + 90° = 102°
∴ \(\frac{102^{\circ}}{2}\) = 51°
one angle = 51°
Second angle 90° – 51° = 39°

Question 12.
The ages of Rahul and Laxmi arc in the ratio 5:7. Four years later, the sum of their ages will
be 56 years. What are their present ages’?
Solution:
The ratio of ages of Rahul and Lakshmi = 5:7
Let their ages be 5x, 7x say.
After 4 years Rahuls agt. = 5x + 4
After 4 years Lakshmis age = 7x + 4
According to the sum,
After 4 years the sum of their ages = 56
⇒ (5x + 4) + (7x + 4) = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 = 48
⇒ x= \(\frac { 48 }{ 12 }\) = 4
∴ x = 4
∴ Rahuls present age
= 5x = 5 × 4 = 20 years
∴ Lakshmi’s present age
= 7x = 7 × 4 = 28 years

Question 13.
There are 180 multiple choice questions in a test. A candidate gets 4 marks for every correct answer, and for every un-attempted or wrongly answered questions one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the
test how many questions did he answer correctly ?
Solution:
Number of questions attempted for correct answers = x say
umber of questions attempted for wrong answers = 180 – x
4 marks are awarded for every correct answer.
Then numl)er of marks ohtained for correct answers 4 × x = 4x
1 mark is deducted for every wrong answer.
∴ Number of marks deducted for wrong answers
=(180 – x) × 1 = 180 – x
According to the sum.
4x – (180 – x) = 450
⇒ 4x – 180 + x = 450
⇒ 5x = 450 + 180
⇒ 5x = 630
x = \(\frac { 630 }{ 5 }\)
∴ x = 126
∴ Number of questions attempted for correct answers = 126

Question 14.
A sum of ₹ 500 is in the form of denominations of ₹ 5 and ₹ 10. If the total number of notes is 90 find the number of notes of each denomination.
(Hint: let the number of 5 rupee notes be ‘x’, then number of 10 rupee notes = 90 – x)
Solution:
Number of ₹ 5 notes = x say.
Number of ₹ 10 notes = 90 – x
5x + 10(90 – x) = 500
5x + 900 – 10x = 500
– 5x = 500 – 900 = – 400
x = \(\frac { -400 }{ -5 }\)
∴ x = 80
∴ Number of ₹ 5 notes = 80
Number of ₹ 10 notes
= 90 – x = 90 – 80 = 10

Question 15.
A person spent ₹ 564 in buying geese and ducks,if each goose cost ₹ 7 and each duck ₹ 3 and if the total number of birds bought was 108, how many of each type did he buy?

Solution:
Let the number of pens be x.
The total number of things = 108
∴ The number of pencils = 108 – x
Thecostofpens of x = ₹7 × x = ₹ 7x
The cost of pencils of (108- x)
= ₹3(108 – x) = ₹ (324 – 3x)
Total amount to t)uy eflS and Pencils = ₹564
∴ 7x + (324 – 3x) 5M
7x + 324 – 3x = 564
4x = 564 – 324 = 240
∴ x = \(\frac { 240 }{ 4 }\) = 60
The number of pens = 60
The number of pencils = 108 – 60 = 48

Question 16.
The perimeter ofa school volleyball court is 177 ft and the length is twice the width. What are the dimensions of the volleyball court’?

Solution:
Breadth of a volleyball court = x feet say.
∴ Its length = 2 × x = 2 x feet.
The perimeter of a court = 177 feet.
⇒ 2(l + b) = 177
⇒ 2(2x + x) = 177
⇒ 2 x 3x = 177
⇒ 6x = 177
⇒ x =\(\frac { 177 }{ 6 }\)
∴ x = 29.5
∴ The breadth of a volleyball court = 29.5 ft
The length of a volleyball court = 2x = 2 × 29.5 = 59ft

Question 17.
The sum of the page numbers on the facing pages of a book is 373. What are the page numbers?
(Hint :Let the page numbers of open pages are x and (x + 1)

Solution:
Number of first page of a opened book x
Number ol second page = x + 1
∴ The surï of the numbers of two pages = 373
⇒ x + (x + 1) = 373
⇒ 2x + 1 = 373
2x = 372
⇒ x = 186
∴x + 1 = 186 + 1 = 187
∴ Numbers of two c(nlsecutive pages = 186, 187