Samacheer Class 8 Maths - Life Mathematics

50% of the cake is distributed to the children
Hint:
One half is nothing but \(\frac { 1 }{ 2 }\)
as percentage, we need to multiply by 100
∴ \(\frac { 1 }{ 2 }\) × 100 = 50%
(ii) Aparna scored 7.5 points out of 10 in a competition.
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75
For percentage, we need to multiply by 100
We get 0.75 × 100 = 75%
(iii) The statue was made of pure silver .
The statue was made of 100% pure silver
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)
∴ to express as percentage, \(\frac{100}{100}\) × 100% = 100%(iv) 48 out of 50 students participated in sports.
96% students participated in sports.
Hint:
48 out of 50 students in fraction form is \(\frac{48}{50}\)
As a percentage, we need to multiply by 100(v) Only 2 persons out of 3 will be selected in the interview.
Only 66\(\frac{2}{3}\)% will be selected in the interview.
Hint:
2 out of 3 in fraction form is \(\frac{2}{3}\)
to express as percentage, we need to multiply by 100
\(\frac{2}{3} \times 100=\frac{200}{3}=66 \frac{2}{3} \%\)
To convert CGPA to Percentage , we need to multiply CGPA by 9.5, which will give us the percentage.

Given that 75% of number less 60% of number is 82.5
Let the number be ‘x’
∴ \(\frac{75}{100}\) x x – \(\frac{60}{100}\) x x = 82.5
∴ 0.75 x – 0.60 x = 82.5
∴ 0.15 x = 82.5
∴ x = \(\frac{82.5}{0.15}=\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.

Let the number be x. Given that when it is increased by 20% we get 80.x = 100It is important to learn to convert CGPA into percentage because both the systems are interconnected.

Method 1.
Let the number be x.
First it is increased by 25%
∴ It becomes x + \(\frac{25}{100}\) × x = \(\frac{125}{100}\)
Secondary it is decreased by 20%
\(\frac{125 x}{100}-\frac{20}{100} \times \frac{125}{100} x=\frac{125}{100} x \times \frac{80}{100}=x\)
Now we get back x, therefore there is no change.
Hence percentage change in that number is 0%
Method 2.
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we getNow this 125 is decreased by 20%, we get
125 – \(\frac{25}{100}\) × 125 = 125 – 25 = 100
∴ We get back 100 ⇒ No change
Hence percentage change in that number is 0%

Let number of boys be ‘b’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ \(\frac{b}{g}=\frac{5}{3}\) …… (A)
Failure in boys = 16% = \(\frac{16}{100}\) × b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) × g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% = \(\frac{84}{100} b\) …… (1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100} g\) ……. (2)
From A, we have \(\frac{b}{g}=\frac{5}{3}\) , adding I on both sides, we get
\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1
\(\frac{b+g}{g}=\frac{5+3}{3}=\frac{8}{3}\)
∴ g = \(\frac{3}{8}\)(b + g) ……. (3)
Similarly b = \(\frac{5}{8}\) (b + g) ……. (4)
Total pass = pass in girls + pass in boys
= (1) + (2) = \(\frac{84}{100} b+\frac{92}{100} g\)Objective Type Questions

(B) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes)
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500= \(\frac{245}{500}\) × 100 = 49%

(A) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which isNext 15% of the above is \(\frac{15}{100}\) × 2500 = 375

(D) 200
Hint:
Let the number be ‘X’
60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)
Given that when 60 is subtracted from 60%, we get 60
i.e \(\frac{60}{100}\) x – 60 = 60
∴ \(\frac{60}{100}\) x 60 + 60 = 120
∴ x = \(\frac{120 \times 100}{60}\) = 200

(D) 36
Hint:
Given that 48% of 48 = 64% of xx = 36

Case 1: Profit = Selling price (SP) – Cost price (CP)
Case 2: Loss = Cost price (CP) – Selling price (SP)
Given that profit of case 1 = loss of case 2
∴ P = 810 – CP
L = CP – 530
Since profit (P) = loss (L)
810 – CP = CP – 530
∴ 2CP = 810 + 530 = 1340 ⇒ C.P = \(\frac{1340}{2}\)
∴ CP = 670

Let cost price of one ruler be x
Given that selling price (SP) of 10 rulers.
i.e., same as cost price (CP) of 15 rulers
∴ SP of 10 rulers = 15 × x = 15x
∴ SP of 1 ruler = \(\frac{15 x}{10}\) = 1.5x
∴ Gain = SP of 1 ruler – CP of 1 ruler = 1.5x – x = 0.5x

Let cost price of one article be C.P
Given that 2 are bought for ₹ 15
∴ 2 × CP = 15 ⇒ CP = \(\frac{15}{2}\)
Let selling price of one article be SP
Given that 3 are sold for ₹ 25
∴ 3 × SP = 25 ⇒ SP = \(\frac{25}{3}\)
∴ Gain = SP – CP = \(\frac{25}{3}-\frac{15}{2}=\frac{50-45}{6}=\frac{5}{6}\)= \(11 \frac{1}{9}\)

Selling price (SP) of speaker = ₹ 768
Loss % = 20 %
as per formula∴ CP = \(\frac{768 \times 100}{80}\) = 960
For gain of 20%, we should now calculate the selling price= 96 × 12 = ₹ 1152

Marked price of AC = ₹ 38,000
Option 1:
Selling price = ₹ 38000 & gifts worth ₹ 3000
∴ Net gain for customer = ₹ 3000 as there is no discount on AC
Option 2:
Discount of 8%, but no gift
∴ Discounted value = MP × \(\left(\frac{(100-d \%)}{100}\right)\)
38000 × \(\frac{(100-8)}{100}\) = 38000 × 0.92 = 34960
∴ Savings for customer = 38000 – 34960 = 3040
Therefore, the customer gets 3000 gift in option I where as he is able to save only ₹ 3040 in option 2. Therefore, option 2 is better.

(C) ₹ 176.40
Hint:
Cost price of article = ₹ 150
Over head expenses = 12% of cost price
= \(\frac{12}{100}\) × 150 = ₹ 18
∴ Effective cost of article = 150 + 18 = ₹ 168
Now, to gain 5%, he has to sell at

(D) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:

(A) 40%
Hint:
Let marked price be MP, after discount 1 of 20%,Comparing with formula, we get
∴ This is equivalent to a single discount of 40%

Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp. annually
∴ Amount (A) = \(\left(1+\frac{r}{100}\right)^{n}\)
= 3200 \(\left(1+\frac{25}{100}\right)^{2}\)
= 3200 × (1.025) 2 = 3362
Compound interest (CI) = Amount – Principal
= 3362 – 3200 = 162

Principal (P) = ₹ 4000
r = 10 %p.a
Compounded yearly
n = 2\(\frac { 1 }{ 2 }\) years. Since it is of the form a \(\frac{b}{c}\) years∴ CI = Amount – principal = 5082 – 4000 = 1082

Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.aLet no. of years be n
for compounding half yearly, formula isTaking cubic root on both sides,

Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years isSubstituting in the above formula, we get∴ Compound Interest(CI) = A – P = 25,875 – 15,000 = ₹ 10,875
CI = ₹ 10.875

Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2%p.a
for half yearly r = 1%
Difference between CI & SI is given by the formula

Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between CI & SI is given by the formula
CI – SI = \(p\left(1+\frac{r}{100}\right)^{n}\)
Difference between CI & SI is given as 20
∴ 20 = 8000 × \(\left(\frac{r}{100}\right)^{2}\)
∴ \(\left(\frac{r}{100}\right)^{2}=\frac{20}{8000}=\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}=\sqrt{\frac{1}{400}}=\frac{1}{20}\)
∴ r = \(\) = 5 %

Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between CI & SI is given as 1134
Principal = ? → required to find
Using formula for difference
Simple Interest SI = \(\frac{P n r}{100}\)
Compound Interest CI = p(1 + i) n – p1134 = P[(1.15) 3 – 1 – 0.45] = P(1.52 – 1.45) = P (0.07)
∴ p = \(\frac{1134 \times 100}{0.07 \times 100}=\frac{113400}{7}\)
P = ₹ 16200Objective Type Questions

(C) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6\(\left(\frac{12}{2}\right)\) conversion periods.

(B) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = \(\frac{10}{2}\) = 5 %
Compounded half yearly, so the formula is
A = P\(P\left(1+\frac{r}{100}\right)^{2 n}\)
Substituting in the above formula, we getTaking square root on both sides, we get
\(\left(\frac{21}{20}\right)^{2 n}=\left(\frac{21}{20}\right)^{2}\)
Equating power on both sides
∴ 2n = 2, n = 1

(B) ₹ 12500
Hint:
Cost of machine = 18000
Depreciation rate = 16\(\frac{2}{3}\)% = \(\frac{50}{3}\)%p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value \(\left(1-\frac{r}{100}\right)^{n}\)
Substituting in above formula, we get
Depreciated value after 2 years

(A) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10 % p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?

(D) ₹ 2500
Hint:
Difference between CI and SI is given as Re I
Time period (n) = 2 yrs.
Rate of interest (r) = 2 % p.a
Formula for difference is
CI – SI = \(P \times\left(1+\frac{r}{100}\right)^{n}\)
Substituting the values in above formula, we get
1 = p × \(\left(\frac{2}{100}\right)^{2}\)
∴ p = 1 × \(\left(\frac{100}{2}\right)^{2}\) = 1 × (50) 2 = ₹ 2500

Let the required number of men be x.
Hours
Day
Men
12
18
210
14
20
x
More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = \(210 \times \frac{12}{14} \times \frac{18}{20}\)x = 162 men
162 men are required.

Let he required number of cement bags be x.
Days
Machines
Cement bags
12
36
7000
18
24
x
Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = \(7000 \times \frac{18}{12} \times \frac{24}{36}\)x = 7000 cement bags
7000 cement bags can be made

Let the required number of days be x.
Soaps
Hours
Days
9600
15
6
14400
(15 + 3) = 18
x
To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = \(6 \times \frac{14400}{9600} \times \frac{15}{18}\)x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Let the number of lorries required more = x.
Container lorries
Goods (tonnes)
Days
6
135
5
6 + x
180
4
As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = \(6 \times \frac{180}{135} \times \frac{5}{4}\)6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Time taken by A to complete the work = 12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\) —— (1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\) —— (2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}+\frac{1}{3}=\frac{1+4}{12}=\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A+ B + C)’s 1 hour work – (A + C)’s 1 hr work
\(=\frac{5}{12}-\frac{1}{6}=\frac{5-2}{12}=\frac{3}{12}=\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\) —— (1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\) —— (2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\) —— (3)
Now (1) + (2) + (3) =
[(A + B)+ (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\)
(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}+\frac{4}{60}+\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)LCM = 5 × 4 × 3 = 60
(A + B + C)’s 1 day work = \(\frac{12}{60 \times 2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s I day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(=\frac{1}{10}-\frac{1}{15}=\frac{3}{30}-\frac{2}{30}=\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work – (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
\(=\frac{1}{10}-\frac{1}{20}=\frac{6}{60}-\frac{3}{60}\)
\(=\frac{6-3}{60}=\frac{3}{60}=\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work (A + B + C)’s I day work – (A + B)’s I day work
\(=\frac{1}{10}-\frac{1}{12}=\frac{6}{60}-\frac{5}{60}=\frac{6-5}{60}=\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 118 minutes
∴ As 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A+B)’s 1 minutes work = \(\frac{1}{15}+\frac{1}{18}\)
\(\frac{12}{180}+\frac{10}{180}=\frac{22}{180}=\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chairLCM = 3 × 5 × 6 = 180∴ Time taken by (A + B) to fit a chair
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

A completes the work in 45 days.
∴ A’s 1 day work = \(\frac{1}{45}\)Remaining work = \(1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}\)
B finishes \(\frac{2}{3}\) rd work in 24 daysLet x days required

If B does the work in 3 days, A will do it in I day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{a b}{a+b}\) days
= \(\frac{24 \times 8}{24+8}\) days
= \(\frac{24 \times 8}{32}\) days = 6 daysThey together complete the work in 6 days.

Let the number of mangoes bought by fruit seller initially be x.
Given that 10% or mangoes were rotten
∴ Number of rotten mangoes = \(\frac{10}{100}\) × x
Number of good mangoes = x – no. of rotten mangoes
= \(x-\frac{10}{100} x=\frac{100 x-10 x}{100}=\frac{90}{100} x\) …….. (1)
Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%
∴ Mangoes sold = \(\frac{100}{3} \times \frac{90}{100} \times \times \frac{1}{100}=\frac{30}{100} x\) ……. (2)
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)∴ Intially he had 400 mangoes

Let the maximum marks in the exam be ‘x’
Pass percentage is given as 35%
∴ Pass mark = \(\frac{35}{100} \times x=\frac{35}{100} x\)
Student gets 31% marks = \(\frac{31}{100} \times x=\frac{31}{100} x\)
But student fails by 12 marks → meaning his mark is 12 less than pass mark.Maimum mark is 300

Let Q’s income be 100.
P’s income is 25% more than that of Q.
∴ P’s income = 100 + \(\frac{25}{100}\) × 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}}\) × 100 = \(\frac{125-100}{125}\) × 100 = \(\frac{25}{125}\) × 100 = 20%

Saree 1 :
The selling price is ₹ 2200, let cost price be CP 1 , gain is 10%
Cost price? Using the formulaSaree 2 :
The selling price is 2200, let cost price be CP 2 , loss is given as 12%. We need to find CP 2
using the formula as before,∴ Cost price of both together is CP 1 + CP 2
= 2000 + 2500 = 4500 …….. (1)
Selling price of both together is 2 × 2200 = 4400 …… (2)
Since net selling price is less than net cost price, there is a loss.Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
100 100 20 2
∴ loss % = \(\frac{100}{4500}\) × 100 = \(\frac{100}{45}=\frac{20}{9}=2 \frac{2}{9}\)%
= 2\(\frac{2}{9}\)%loss

Days (D)
Hours (H)
Men (P)
15
12
32
24
10
x
Let
P 1 = 32
P 2 = x
H 1 = 12
H 2 = 10
D 1 = 15
D 2 = 24
W 1 = 1
W 2 = 1x = 24 persons
To complete the same work 24 men needed.
To complete double the work 24 × 2 = 48 men are required.

Amutha can weave a saree in 18 days.
Anjali is twice as good as Amutha.
ie. 1f Amutha weave for 2 days, Anjali do the same work in 1 day.
If Anjali weave the saree she will take
\(\frac{18}{2}\) = 9 days.
Hence time taken by them together ab
= \(\frac{a b}{a+b}\) days
= \(\frac{18 \times 9}{18+9}=\frac{18 \times 9}{27}\) = 6 daysIn 6 clays they complete weaving the saree.

P can do a piece of work in 12 days.
∴ P’s 1 day work = \(\frac{1}{12}\)
P’s 3 day’s work = 3 × \(\frac{1}{12}=\frac{3}{12}\)
Q can do a piece of work in 15 days.
∴ Q’s 1 day work = \(\frac{1}{15}\)
Remaining work after 3 days = 1 – \(\frac{3}{12}=\frac{9}{12}\)∴ Number of days required to finish the remaining work 9Remaining work lasts for 5 days
Total work lasts for 3 + 5 = 8 days.

Let the cost price of the laptop be ‘x
Gain = 12%If the selling price was 1200 more
i.e \(\frac{112}{100} x\) + 1200, the gain is 2o%Cost price of the laptop is ₹ 15,000/-

Marked price is given as ₹ 180
Let 1 discount be d 1 % = ? (to find)
2nd discount be d 2 % = 25%
Selling price is 108 (given)
Price after 1 st discount = 180 \(\left(1-\frac{d_{1}}{100}\right)\) = P …… (1)
Price after 2 nd discount = P 1 \(\left(1-\frac{d_{2}}{100}\right)\) = 108
Substituting for P 1 from (1), we get1st discount = 20%

Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ?(required)
Applying the formula,∴ rate of compound interest is 30%

Let the number of men to be appointed more be x.To produce more pumps more men required
∴ It is direct variation.
∴ The multiplying factor is \(\frac{360}{180}\)
More days means less employees needed.
∴ It is Indirect proportion. 75
∴ The multiplying factor is \(\frac{75}{75}\)
Now 40 + x = 40 × \(\frac{360}{180} \times \frac{75}{75}\)
40 + x = 80
x = 80 – 40
x = 40
40 more man should be employed to complete the work on time as per the agreement.

\(\frac{1}{2}\) of the work is done by P in 6 days\(\frac{2}{3}\) of work done byQin4days.(P + Q) will finish the whole work in \(\frac{a b}{a+b}\) days= \(\frac{12 \times 6}{12+6}=\frac{12 \times 6}{18}\)= 4 days
(P + Q) will finish \(\frac{-3}{4}\) of the work in 4 × \(\frac{3}{4}\) = 3 days.

X can do the work in 6 days.
X’s I day work = \(\frac{1}{6}\)
X’s share for 1 day = \(\frac{1}{6}\) × 48000 = ₹ 800
X’s share for 3 days = 3 × 800 = ₹ 2400
Y can complete the work in 8 days.
Y’s 1 day work = \(\frac{1}{8}\)
Y’s I day share = \(\frac{1}{8}\) × 4800 = ₹ 600
Y’s 3 days share = ₹ 600 × 3 = ₹ 1800
(X+Y)’s 3days share = ₹ 2400 + ₹ 1800 = ₹ 4200
Remaining money is Z’s share
∴ Z’s share = ₹ 4800 – ₹ 4200 = ₹ 600

In a day, there are 24 hours .
∴ 10 hrs out of 24 hrs is \(\frac{10}{24}\)
As a percentage, we need to multiply by 100
∴ Percentage = \(\frac{10}{24}\) × 100 = 41.67%

Let R get x, Q gets 50% of what R gets
∴ Q gets = \(\frac{50}{100} \times x=\frac{x}{2}\)
P gets 50% of what Q gets .
∴ P gets = \(\frac{50}{100} \times \frac{x}{2}=\frac{x}{4}\)
Since 350 is divided among the three
∴ 350 = \(x+\frac{x}{2}+\frac{x}{4}\)
350 = \(\frac{4 x+2 x+x}{4}=\frac{7 x}{4}\) = 350
x = \(\frac{350 \times 4}{7}\)
Q gets = \(\frac{x}{2}=\frac{200}{2}\) = 100,
P gets = \(\frac{x}{4}=\frac{200}{4}\) = 50
∴ p = 50, Q = 100, R = 200Think (Text Book Page No. 124)
With a lot of pride, the traffic police commissioner of a city reported that the accidents had decreased by 200% in one year. He came up with this number by stating that the increase in accidents from 200 to 600 is clearly a 200% rise and now that it had gone down from 600 last year to 200 this year should be a 200% fall. Is this decrease from 600 to 200, the same 200% as reported by him? Justify.Increase from original value 200 to 600Decrease from original value 600 to 200here original value is 600
% decrease = \(\frac{600-200}{600}\) × 100 = \(\frac{400}{600}\) × 100 = 66.67 % decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the sameTry These (Text Book Page No. 126)

C.P = 5x
S.P = 7x
Profit = 7x – 5x = 2xThink (Text Book Page No. 129)
A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price MP = \(\frac{15}{100}\) × CP + CP
= \(\frac{15}{100}\) × 100 + 100 = 15 + 100 = 115
Discount % = 15 %∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25Try These (Text Book Page No. 129)

As weight increases cost also increases.
∴ Weight and cost are direct proportion.
(ii) Distance travelled by bus to the price of ticket.
As the distance increases price to travel also increases,
∴ Distance and price are direct proportion.(iii) Speed of the athelete to cover a certain distance.
As the speed increases, the time to cover the distance become less.
So speed and üme are in indirect proportion.
(iv) Number of workers employed to complete a construction in a specified time.
As the number of workers increases, the amount of work become less, so they are in indirect proportion.
(v) Area of a circle to its radius.
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Gìven x = 64, y = 0.75
and also given x andy vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

As the number of radii increases angle decreases.
Hence they are in inverse proportion
∴ xy = 4 proportional constant
3 × 120° = 360° = k = 360°Try These (Text Book Page No. 147)
The percentage difference calculator is here to help you compare two numbers.
Identify the different variations present in the following questions:

Let the required no. of articles be x
Men (P)
Days (D)
Articles (W)
24
12
48
6
6
x
(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct vanables
(iii) Days and articles are also direct variables using formula.
Let
P 1 = 24
P 2 = 6
D 1 = 12
D 2 = 6
W 1 = 48
W 1 = xx = 6 Articles

Let the required no. of workers be x
Length (work)
Hours
Workers
4 km
4 hrs
15
8 km
8 hrs
x
(i) Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ——– (1)
(ü) Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is : 4 : : 15 : x ——– (2)
Combining (1) and (2)Product of the extremes = Product of the mean
4 × 8 × x = 8 × 4 × 15
x = \(\frac{8 \times 4 \times 15}{4 \times 8}\)
x = 15 workers

Let the required hours be x.
Women
Days
Hours
25
36
12
20
30
x
As women increases hours to work decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{25}{20}\)
As days increases hours needed become less
∴ It is also an indirect variation.
∴ Multiplying factor is \(\frac{36}{30}\)
∴ x = \(12 \times \frac{25}{20} \times \frac{36}{30}\)
x = 18 hours

Let the required number of days be x.
Rice (kg)
Men
Days
420
98
45
60
42
x
If amount of rice is more it will last for more days;
∴ It is Direct Proportion
∴ Multiplying factor is \(\frac{60}{420}\)
If men increases number of days the rice lasts decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{98}{42}\)
x = \(45 \times \frac{60}{420} \times \frac{98}{42}\)x = 15 daysTry These (Text Book Page No. 150)

\(\frac{1}{3}\) of the work will be done mp days.
∴ Full work will be completed in 3p days
\(\frac{3}{4}\) th of the work will be done in = 3p x \(\frac{3}{4}\)
= \(\frac{9}{4}\)p = 2\(\frac{1}{4}\) p days.

Givenm persons complete a work in n days
(i) Then work measured in terms of Man days = mn
4 m men do the work it will be completed in \(\frac{m n}{4 m}\) days = \(\frac{m}{4}\) days.