500
Hint:
Given 30% of x is 150
i.e \(\frac{30}{100}\) × x = 150
∴ x = 500
(ii) 2 minutes is _______ % to an hour.
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60mm
x% = \(\frac{2}{60} \times 100=\frac{200}{60}=\frac{10}{3}=3 \frac{1}{3}\)
x = 3\(\frac{1}{3}\)%
(iii) If x% of x = 25, then x = _______ .
50
Hint:
Given that x% of x is 25
∴ \(\frac{x}{100}\) × x = 25
∴ x 2 = 25 × 100 = 2500
∴ x = √2500 = 50
(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is _______ .
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
∴ Number of boys in school = 1400 – 420 = 980
= \(\frac{980}{14}\) = 70
% of boys = 70%
(v) 0.5252 is _______ %.
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 × 100 = 52.52%
percent difference Calculator can be found by first finding the difference between the numbers, then finding the average of the numbers, and then dividing.
500
Hint:
Given 30% of x is 150
i.e \(\frac{30}{100}\) × x = 150
∴ x = 500
(ii) 2 minutes is _______ % to an hour.
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60mm
x% = \(\frac{2}{60} \times 100=\frac{200}{60}=\frac{10}{3}=3 \frac{1}{3}\)
x = 3\(\frac{1}{3}\)%
(iii) If x% of x = 25, then x = _______ .
50
Hint:
Given that x% of x is 25
∴ \(\frac{x}{100}\) × x = 25
∴ x 2 = 25 × 100 = 2500
∴ x = √2500 = 50
(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is _______ .
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
∴ Number of boys in school = 1400 – 420 = 980
= \(\frac{980}{14}\) = 70
% of boys = 70%
(v) 0.5252 is _______ %.
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 × 100 = 52.52%
percent difference Calculator can be found by first finding the difference between the numbers, then finding the average of the numbers, and then dividing.
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.
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.
Let the number required to be found be ‘x’
Given that 32% of x is 48
i.e., \(\frac{32}{100}\) × x = 48
∴ x = 150
Let the number required to be found be ‘x’
Given that 32% of x is 48
i.e., \(\frac{32}{100}\) × x = 48
∴ x = 150
Required to find 25% of 30% of 400
The percentage decrease calculator determines the change from one amount to a lesser amount in terms of percent decrease.
Required to find 25% of 30% of 400
The percentage decrease calculator determines the change from one amount to a lesser amount in terms of percent decrease.
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.
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 18%, we get 236.
Let the number be x. Given that when it is increased by 18%, we get 236.
Let the number be x. Given that when it is increased by 20% we get 80.
x = 100
It is important to learn to convert CGPA into percentage because both the systems are interconnected.
Let the number be x. Given that when it is increased by 20% we get 80.
x = 100
It 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 get
Now 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%
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 get
Now 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
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
- A. 10%
- B. 15%
- C. 20%
- D. 30%
(C) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) × 250 = 30 lit.
Percentage: \(\frac{30}{150}\) × 100 = 20%
(C) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) × 250 = 30 lit.
Percentage: \(\frac{30}{150}\) × 100 = 20%
- A. 48%
- B. 49%
- C. 50%
- D. 45%
(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%
(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
- B. 400
- C. 425
- D. 475
(A) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is
Next 15% of the above is \(\frac{15}{100}\) × 2500 = 375
(A) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is
Next 15% of the above is \(\frac{15}{100}\) × 2500 = 375
- A. 60
- B. 100
- C. 150
- D. 200
(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) 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
- A. 64
- B. 56
- C. 42
- D. 36
(D) 36
Hint:
Given that 48% of 48 = 64% of x
x = 36
Posted in Class 8 on January 5, 2025 January 6, 2025
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(D) 36
Hint:
Given that 48% of 48 = 64% of x
x = 36
Posted in Class 8 on January 5, 2025 January 6, 2025
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Cost Price
(ii) A mobile phone is sold for ₹ 8400 at a gain of 20%. The cost price of the mobile phone is ________ .
₹ 7000
Hint:
Let cost price of mobile be ₹ x
Given that selling price is ₹ 8400 and gain is 20%
As per formula,
(iii) An article is sold for ₹ 555 at a loss of 7\(\frac { 1 }{ 2 }\)%. The cost price of the article is ________ .
₹ 600
Hint:
Given selling price is ₹ 555 & loss 7\(\frac { 1 }{ 2 }\)%
as per formula
(iv) A mixer grinder marked at ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ________ .
8 %
Hint:
Marked price is ₹ 4500
Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted priòe
= 4500 – 4140 = 360
(v) The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ________ .
Cost of price shirt = ₹ 575 (CP)
GST = 5%
Cost of price shirt = ₹ 325 (CP)
GST = 5%
∴ Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945
Cost Price
(ii) A mobile phone is sold for ₹ 8400 at a gain of 20%. The cost price of the mobile phone is ________ .
₹ 7000
Hint:
Let cost price of mobile be ₹ x
Given that selling price is ₹ 8400 and gain is 20%
As per formula,
(iii) An article is sold for ₹ 555 at a loss of 7\(\frac { 1 }{ 2 }\)%. The cost price of the article is ________ .
₹ 600
Hint:
Given selling price is ₹ 555 & loss 7\(\frac { 1 }{ 2 }\)%
as per formula
(iv) A mixer grinder marked at ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ________ .
8 %
Hint:
Marked price is ₹ 4500
Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted priòe
= 4500 – 4140 = 360
(v) The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ________ .
Cost of price shirt = ₹ 575 (CP)
GST = 5%
Cost of price shirt = ₹ 325 (CP)
GST = 5%
∴ Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945
Given that selling price (SP) = ₹ 820
Loss % = 10 %
Given that selling price (SP) = ₹ 820
Loss % = 10 %
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
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 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 = \(\frac{25}{3}-\frac{15}{2}=\frac{50-45}{6}=\frac{5}{6}\)
= \(11 \frac{1}{9}\)
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 = \(\frac{25}{3}-\frac{15}{2}=\frac{50-45}{6}=\frac{5}{6}\)
= \(11 \frac{1}{9}\)
vSelling 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
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
vMarked 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.
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.
Marked price of mattress = ₹ 7500
Discount d 1 = 10%
Discount d 2 = 20%
Objective Type Questions
Marked price of mattress = ₹ 7500
Discount d 1 = 10%
Discount d 2 = 20%
Objective Type Questions
- A. 20%
- B. 22%
- C. 25%
- D. 16
(C) 25%
Hint:
Selling price ₹ 200
Gain = 40
∴ CP – Selling price – gain = 200 – 40 = 160
(C) 25%
Hint:
Selling price ₹ 200
Gain = 40
∴ CP – Selling price – gain = 200 – 40 = 160
- A. ₹ 500
- B. ₹ 550
- C. ₹ 553
- D. ₹ 573
(B) ₹ 550
Hint:
If selling price (sp) = ₹ 528
Gain % = 20 %
∴ CP = ?
(B) ₹ 550
Hint:
If selling price (sp) = ₹ 528
Gain % = 20 %
∴ CP = ?
- A. ₹ 180
- B. ₹ 168
- C. ₹ 176.40
- D. ₹ 88.20
(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
(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
- A. ₹ 243
- B. ₹ 176
- C. ₹ 230
- D. ₹ 250
(D) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:
(D) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:
- A. 40%
- B. 45%
- C. 5%
- D. 22.5%
(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%
Posted in Class 8 on September 17, 2024 September 18, 2024
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(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%
Posted in Class 8 on September 17, 2024 September 18, 2024
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₹ 1272
Hint:
Compound Interest (CI) formula is
CI = Amount – Principal
∴ 6272 – 5000 = ₹ 1272
(ii) The compound interest on ₹8000 at 10% p.a for 1 year, compounded half yearly is ________ .
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
Amount = p \(\left(1+\frac{r}{100}\right)^{2 n}\) [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = \(\frac{10}{2}\) = 5
∴ A = 8000 \(\left(1+\frac{5}{100}\right)^{2 \times 1}\) = 8000 × \(\left(\frac{105}{100}\right)^{2}\) = 8820
CI = Amount – principal = 8820 – 8000 = ₹ 820
(iii) The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was ________ .
₹ 20,000
Hint:
Rate of growth of population r = 10%
Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
The population 3 years ago was ₹ 20,000
(iv) If the compound interest is calculated quarterly, the amount is found using the formula ________ .
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
(v) The difference between the C.I and S.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is ________ .
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = \(\left(\frac{r}{100}\right)^{2}\)
Principal (P) = 5000. r = 8% p.a
∴ CI – SI = 5000\(\left(\frac{8}{100}\right)^{2}\) = 5000 × \(\left(\frac{8}{100}\right)^{2}\) × \(\left(\frac{8}{100}\right)^{2}\) = ₹ 32
CBSE Class 10 Maths formulas and equations are available chapter wise.
₹ 1272
Hint:
Compound Interest (CI) formula is
CI = Amount – Principal
∴ 6272 – 5000 = ₹ 1272
(ii) The compound interest on ₹8000 at 10% p.a for 1 year, compounded half yearly is ________ .
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
Amount = p \(\left(1+\frac{r}{100}\right)^{2 n}\) [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = \(\frac{10}{2}\) = 5
∴ A = 8000 \(\left(1+\frac{5}{100}\right)^{2 \times 1}\) = 8000 × \(\left(\frac{105}{100}\right)^{2}\) = 8820
CI = Amount – principal = 8820 – 8000 = ₹ 820
(iii) The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was ________ .
₹ 20,000
Hint:
Rate of growth of population r = 10%
Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
The population 3 years ago was ₹ 20,000
(iv) If the compound interest is calculated quarterly, the amount is found using the formula ________ .
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
(v) The difference between the C.I and S.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is ________ .
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = \(\left(\frac{r}{100}\right)^{2}\)
Principal (P) = 5000. r = 8% p.a
∴ CI – SI = 5000\(\left(\frac{8}{100}\right)^{2}\) = 5000 × \(\left(\frac{8}{100}\right)^{2}\) × \(\left(\frac{8}{100}\right)^{2}\) = ₹ 32
CBSE Class 10 Maths formulas and equations are available chapter wise.
True
Hint:
Depreciation formula is \(P\left(1-\frac{r}{100}\right)^{n}\)
(ii) If the present population ola city is P and it increases at the rate of r% p.a, then the population n years ago would be \(P\left(1-\frac{r}{100}\right)^{n}\)
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
∴ Present popuLation (P) = x × \(\left(1+\frac{r}{100}\right)^{n}\)
∴ x = \(\frac{P}{\left(1+\frac{r}{100}\right)^{n}}\)
(iii) The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450.
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
(iv) The time taken for ₹ 1000 to become ₹ 1331 at 20% p.a, compounded annually is 3 years.
False
Hint:
Pnncipal money = 1000
rate of interest = 20%
Amount = 1331, applying in formula we get
(v) The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522.
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula,
Amount(A) = P × \(\left(1+\frac{r}{100}\right)^{4 n}\)
Since quarterly we have to divide ‘r’ by 4
∴ Interest A – P = 18522 – 16000 = 2522 (True)
True
Hint:
Depreciation formula is \(P\left(1-\frac{r}{100}\right)^{n}\)
(ii) If the present population ola city is P and it increases at the rate of r% p.a, then the population n years ago would be \(P\left(1-\frac{r}{100}\right)^{n}\)
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
∴ Present popuLation (P) = x × \(\left(1+\frac{r}{100}\right)^{n}\)
∴ x = \(\frac{P}{\left(1+\frac{r}{100}\right)^{n}}\)
(iii) The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450.
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
(iv) The time taken for ₹ 1000 to become ₹ 1331 at 20% p.a, compounded annually is 3 years.
False
Hint:
Pnncipal money = 1000
rate of interest = 20%
Amount = 1331, applying in formula we get
(v) The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522.
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula,
Amount(A) = P × \(\left(1+\frac{r}{100}\right)^{4 n}\)
Since quarterly we have to divide ‘r’ by 4
∴ Interest A – P = 18522 – 16000 = 2522 (True)
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) = ₹ 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 (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
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.a
Let no. of years be n
for compounding half yearly, formula is
Taking cubic root on both sides,
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.a
Let no. of years be n
for compounding half yearly, formula is
Taking 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 is
Substituting in the above formula, we get
∴ Compound Interest(CI) = A – P = 25,875 – 15,000 = ₹ 10,875
CI = ₹ 10.875
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 is
Substituting 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) = ₹ 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 %
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 – p
1134 = 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 = ₹ 16200
Objective Type Questions
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 – p
1134 = 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 = ₹ 16200
Objective Type Questions
- A. 2
- B. 4
- C. 6
- D. 12
(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.
(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.
- A. 6 months
- B. 1 year
- C. 1\(\frac{1}{2}\) years
- D. 2 years
(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 get
Taking 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) 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 get
Taking 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
- A. ₹ 2000
- B. ₹ 12500
- C. ₹ 15000
- D. ₹ 16500
(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
(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
- B. ₹ 1800
- C. ₹ 1500
- D. ₹ 2500
(A) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10 % p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?
(A) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10 % p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?
- A. ₹ 2000
- B. ₹ 1500
- C. ₹ 3000
- D. ₹ 2500
(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
Posted in Class 8 on January 3, 2025 January 4, 2025
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(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
Posted in Class 8 on January 3, 2025 January 4, 2025
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Copyright © 2026 Samacheer Kalvi
2 days
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in _________ days.
5
(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in ________ days.
8
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in ___________days.
25
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for ₹ 200000. The amount that A will get is .
₹ 1,20,000
2 days
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in _________ days.
5
(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in ________ days.
8
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in ___________days.
25
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for ₹ 200000. The amount that A will get is .
₹ 1,20,000
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 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 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 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.
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.
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.
(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 chair
LCM = 3 × 5 × 6 = 180
∴ Time taken by (A + B) to fit a chair
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours
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 chair
LCM = 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 days
Let x days required
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 days
Let 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 days
They together complete the work in 6 days.
Posted in Class 8 on September 18, 2024 September 19, 2024
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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 days
They together complete the work in 6 days.
Posted in Class 8 on September 18, 2024 September 19, 2024
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vLet 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 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 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%
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%
vSaree 1 :
The selling price is ₹ 2200, let cost price be CP 1 , gain is 10%
Cost price? Using the formula
Saree 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
Saree 1 :
The selling price is ₹ 2200, let cost price be CP 1 , gain is 10%
Cost price? Using the formula
Saree 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 = 1
x = 24 persons
To complete the same work 24 men needed.
To complete double the work 24 × 2 = 48 men are required.
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 = 1
x = 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 days
In 6 clays they complete weaving the saree.
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 days
In 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 9
Remaining work lasts for 5 days
Total work lasts for 3 + 5 = 8 days.
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 9
Remaining work lasts for 5 days
Total work lasts for 3 + 5 = 8 days.
Original fraction = \(\frac{x}{y}\)
numerator increased by 50%
∴ Numerator = \(\frac{150}{100} x\)
Denominator decreased by 20%
∴ Denominator = \(\frac{80}{100} y\)
Hence
Original fraction = \(\frac{x}{y}\)
numerator increased by 50%
∴ Numerator = \(\frac{150}{100} x\)
Denominator decreased by 20%
∴ Denominator = \(\frac{80}{100} y\)
Hence
vLet 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/-
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 get
1st discount = 20%
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 get
1st 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 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.
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.
\(\frac{1}{2}\) 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
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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
Posted in Class 8 on September 18, 2024 September 19, 2024
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Try These (Text Book Page No. 124)
Try These (Text Book Page No. 124)
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%
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 = 200
Think (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 600
Decrease from original value 600 to 200
here 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 same
Try These (Text Book Page No. 126)
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 = 200
Think (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 600
Decrease from original value 600 to 200
here 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 same
Try These (Text Book Page No. 126)
Loss
Percentage difference calculator tool makes the calculation faster, and it displays the difference in the percentage in a fraction of seconds.
Loss
Percentage difference calculator tool makes the calculation faster, and it displays the difference in the percentage in a fraction of seconds.
Loss
Percentage of Loss
Loss
Percentage of Loss
C.P = 5x
S.P = 7x
Profit = 7x – 5x = 2x
Think (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 = – 2.25
Try These (Text Book Page No. 129)
C.P = 5x
S.P = 7x
Profit = 7x – 5x = 2x
Think (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 = – 2.25
Try These (Text Book Page No. 129)
\(\frac{\mathrm{PNR}}{100}\)
\(\frac{\mathrm{PNR}}{100}\)
I = 3600 – 2000 = 1600
Try These (Text Book Page No. 141)
I = 3600 – 2000 = 1600
Try These (Text Book Page No. 141)
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.
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.
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}=\frac{84}{x}\)
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}=\frac{84}{x}\)
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16
Try These (Text Book Page No. 145)
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16
Try These (Text Book Page No. 145)
I If x andy vary directly then \(\frac{x}{y}\) = k.
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1
I If x andy vary directly then \(\frac{x}{y}\) = k.
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1
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
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:
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 = x
x = 6 Articles
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 = x
x = 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 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 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 days
Try These (Text Book Page No. 150)
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 days
Try 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.
\(\frac{1}{3}\) 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.
Posted in Class 8 on January 5, 2025 January 6, 2025
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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.
Posted in Class 8 on January 5, 2025 January 6, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
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Copyright © 2026 Samacheer Kalvi