*Samacheer Kalvi Class 8 Maths Chapter 4 Life Mathematics Solutions*

*Samacheer Kalvi Class 8 Maths Chapter 4 Life Mathematics Solutions*

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students by providing Solutions for Samacheer Kalvi Class 8 Maths chapter 4 Life Mathematics. Here students can easily find all the solutions for Life Mathematics Exercise 4.1, 4.2, 4.3, 4.4 and 4.5. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 4 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.

**Life Mathematics Exercise 4.1 Solutions :**

**Life Mathematics Exercise 4.1 Solutions :**

**(1) Fill in the blanks :**

**(i) If 30% of x is 150, then x is** ___

**Solution : **

x × 30/100 = 150

= x = 150 × 10/3

= x = 500

So, then x is 500.

**(ii) 2 minutes is ___ % to an hour.**

**Solution : **

As we know, 1 hour = 60 min

∴ 2 min = 2/60 X 100

= 10/3%

**(iii) If x % of x = 25, then x =** ___

**Solution : **

x/100 X x = 25

= x2/100 = 25

= (x/100)² = (5)²

= x/100 = 5

= x = 50

Hence, x = 50

**(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is** ___

**Solution : **

Boys of the school,

= 1400 – 420

= 980

Percentage of the boys –

= 980/1400 × 100 = 70%

Hence, the percentage of boys in the school is 70%.

**(v) 0.5252 is** ___** %**

**Solution : **

Given, 0.5252

= 5252/10000 × 100

= 52.52%

**(2) Rewrite each underlined part using percentage language.**

**(i) One half of the cake is distributed to the children.**

**(ii) Aparna scored 7.5 points out of 10 in a competition.**

**(iii) The statue was made of pure silver.**

**(iv) 48 out of 50 students participated in sports.**

**(v) Only 2 persons out of 3 will be selected in the interview.**

**Solution : **

(i) One half

= 1/2 × 100

= 50%

(ii) 7.5 points out at 10

= 7.5/100 × 100

= 75%

(iii) Pure silver – 100%

(iv) 48 out of 50

= 48/50 × 100

= 96%

(v) 2 persons out of 3

= 2/3 ×100

= 200/3%

**Question no – (3) **

**48 is 32% of which number?**

**Solution : **

Let the number is x

x × 32/100 = 48

= x = 48 × 100/32

= x = 150

Hence, the number is 150.

**Question no – (4) **

**What is 25% of 30% of 400?**

**Solution : **

400 × 25/100 × 30/100

= 3000/100

= 30

**Question no – (5) **

**If a car is sold for ₹ 200000 from its original price of ₹ 300000, then find the percentage of decrease in the value of the car.**

**Solution : **

He loose = 300000 – 200000

= 100000

∴ Percent = 100000/300000 × 100

= 100/3

= 33.33 (Approximate)

= 33.33%

Hence, the percentage of decrease in the value of the car is 33.33%

**Question no – (6) **

**If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of that number.**

**Solution : **

Let the number is x

x × 75/100 – x × 60/100 = 82.5

= 75x – 60x/100 = 82.5

= 15x = 825/10 × 100

= x = 825 × 10/15

= x = 550

∴ 20 % of 550

= 550 × 20/100

= 110

**Question no – (7) **

**A number when increased by 18% gives 236. Find the number.**

**Solution : **

Let the number x

increased 18% = x + 18/100 x

∴ x + 18x/100 = 236

= 100x + 18x = 236 × 100

= 118x = 236 × 100

= x = 236 × 100/118

= x = 200

So, the number is 200.

**Question no – (8) **

**A number when decreased by 20% gives 80. Find the number.**

**Solution : **

Let the number is x

∴ x decreased by 20% = x – 20x/100

= x – 20x/100 = 80

= 100x – 20x/100 = 80

= 80x = 80 × 100

= x = 80 × 100/80

= x = 100

Hence, the number is 100.

**Objective Type Question Solution : **

**Question no – (11) **

**12% of 250 litre is the same as** __** of 150 litre**

**(A) 10%**

**(B) 15%**

**(C) 20%**

**(D) 30%**

**Solution : **

Here, alternative (C) is the correct option

**Question no – (13) **

**15% of 25% of 10000 =** ___

**(A) 375**

**(B) 400**

**(C) 425**

**(D) 475**

**Solution : **

= 15/100 × 25/100 × 10000

= 375

Hence, alternative (A) is the correct option.

**Question no – (14) **

**When 60 is subtracted from 60% of a number to give 60, the number is**

**(A) 60**

**(B) 100**

**(C) 150**

**(D) 200**

**Solution : **

60/100x – 60 = 60

= 60x/100 = 120

= x = 120 × 100/60

= x = 200

Hence, alternative (D) is the correct option.

**Question no – (14) **

**If 48% of 48 = 64% of x , then x =**

**(A) 64**

**(B) 56**

**(C) 42**

**(D) 36**

**Solution : **

48/100 × 48 = 64/100 × x

= x = 48 × 48/64

= x = 36

Here, alternative (D) is the correct option.

**Life Mathematics Exercise 4.2 Solutions :**

**Life Mathematics Exercise 4.2 Solutions :**

**(1) Fill in the blanks :**

**(i) Loss or gain percentage is always calculated on the** ___

**Solution : **

Loss or gain percentage is always calculated on the “Cost price”.

**(ii) A mobile phone is sold for ₹ 8400 at a gain of 20%. The cost price of the mobile phone is** ___

**Solution : **

S.P = 8400 and Profit = 20%

∴ C.P = 100/(100+profit)% × S.P

= 100/120 × 8400

= 7000

Hence, the cost price of the mobile phone is 7000.

**(iii) An article is sold for ₹ 555 at a loss of 7 1/2 %. The cost price of the article is** ___

**Solution : **

S.P = 555 Loss = 7 1/2 % = 15/2 %

C.P = 100/(100 – loss) × S.P

= 100/(100 – 15/2) × 555

= 100 × 555/(200 -15)/2

= 100 × 555 × 2/185

= 600

So, the cost price of the article is 600.

**(iv) A mixer grinder marked at ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is** ___

**Solution : **

M.P = 4500, S.P = 4140

S.P = M.P – Discount

= 4140 = 4500 – D

= D = 4500 – 4140

= 360

Discount = Discount/M.P X 100%

= 360/4500 X 100

= 80%

So, the rate of discount is 80%.

**Question no – (2) **

**If selling an article for ₹820 causes 10% loss on the selling price, then find its cost price.**

**Solution : **

S.P = 820 Loss = 10%

C.P = 100/(100-10) × 820

= 100 × 820/90

= 8200/9

= 911.11 (Approximately)

**Question no – (3) **

**If the profit earned on selling an article for ₹ 810 is the same as loss on selling it for ₹ 530, then find the cost price of the article.**

**Solution : **

Let the profit and loss be = x%

Profit 810 and loss S.P = 530

Profit C.P = 100/(100+x) × 810

And loss; C.P = 100/(100–x) × 530

∴ 100/100+x × 810 = 100/(100-x) × 530

100-x/100+x = 530/810

5300 + 53x = 8100 – 81x

53x + 81x = 8100 – 5300

134x = 2800

= x = 2800/134

= 1400/67

= 2 60/67 %

∴ C.P = 100/100 – 1400/67 × 530

= 100/(6700 -1400/67) × 530

= 100 × 67/5300 × 530

= 670

Therefore, the cost price of the article is 670.

**Question no – (6) **

**By selling a speaker for ₹ 768, a man loses 20%. In order to gain 20%, how much should he sell the speaker?**

**Solution : **

C.P = 100/(100 – 20) × 768

= 100/80 × 768

= 960

∴ In order to given 20%

S.P = 100 + 20/100 × 960

= 120/100 × 960

= 1152

Hence, he should sell the speaker at 1152.

**Question no – (7) **

**Find the unknowns x, y and z.**

S. NO |
Name of the item |
Marked Price |
Selling price |
Discount |

(i) |
Book |
₹ 225 |
x |
8% |

(ii) |
LED TV |
y |
₹ 11970 |
5% |

(iii) |
Digital clock |
₹ 750 |
₹ 615 |
z |

**Solution : **

(i) M.P = 225 Discount = 8%

∴ S.P = M.P – Discount

∴ 8% = Discount/M.P × 100

Discount = 8 × 225/100

= 18

(ii) M.P = y, S.P = 11970, Discount = 5%

Discount = Discount % X M.P/100

= 5 × y/100

= y/20

M.P = S.P + Discount

= y = 11970 + y/20

= y – y/20 = 11970

= 20y – y/20 = 11970

= 19y = 11970 × 20

= y = 11970 × 20/19

= 12600

(iii) M.P = 750, S.P = 615, Discount = 7%

S.P = MP – Discount

Discount = – S.P + M.P

= – 615 + 750

= 136

Discount% = Discount/M.P × 100

= 135/750 ×100

= 18%

**Question no – (8) **

**Find the total bill amount for the data given below :**

S. No |
Name of the item |
Marked Price |
Discount |
GST |

(i) |
School bag |
₹ 500 |
5% |
12% |

(ii) |
Hair dryer |
₹ 2000 |
10% |
28% |

**Solution : **

(i) Discount % = Discount/M.P × 100

= Discount = Discount% × M.P/100

= 18 × 500/100

= 25

(ii) Discount = Discount% × M.P/100

= 10 × 2000/100

= 200

∴ S.P = M.P – Discount

= 2000 – 200

= 1800

G.S.T = 1800 × 28/100

= 504

Total bill amount

= (475 + 57 + 1800 + 504)

= 2836

**Question no – (9) **

**A branded Air-Conditioner (AC) has a marked price of ₹ 38000. there are 2 options given for the customer**

**(i) Selling Price is the same ₹38000 but with attractive gifts worth ₹3000**

**(or)**

**(ii) Discount of 8% on the marked price but no free gifts. Which offer is better?**

**Solution : **

(i) S.P = M.P = 38000 Gift = 3000

∴ Total amount = 38000 – 3000

= 35000

(ii) 8% discount

∴ Selling price = 38000 × 100-8/100

= 380 × 92

= 34960

Therefore, (ii) is better offer.

**Question no – (10) **

**If a mattress is marked for ₹7500 and is available at two successive discounts of 10% and 20%, find the amount to be paid by the customer.**

**Solution : **

M.P = 7500

In 10% discount

10% = discount/7500 × 100

= Discount = 7500 × 10/100

= 750

∴ The amount to be paid by the customer

= 7500 – 750

= 6750

In 20% Discount

20% = discount/6750 ×100

Discount

= 67500 × 20/100

= 1350

∴ The amount to be paid by the customer,

= 6750 – 1350

= 5400

Hence, the amount to be paid by the customer is 5400.

**Objective Type Question Solutions : **

**Question no – (11) **

**A fruit vendor sells fruits for ₹200 gaining ₹40. his gain percentage is**

**(A) 20%**

**(B) 22%**

**(C) 25%**

**(D) 16 2/3%**

**Solution : **

Profit % = Profit/C.P ×100

= 40/200 ×100

= 20%

Hence, alternative (A) 20% is the correct option.

**Question no – (12) **

**By selling a flower pot for 528, a woman gains 20%. at what price should she sell it to gain 25%**

**(A) ₹ 500**

**(B) ₹ 550**

**(C) ₹ 553**

**(D) ₹ 573**

**Solution : **

S.P of 20% gain = 100+20/100 × C.P

= 528 = 120/100 × C.P

= C.P 528 × 100/120

= C.P = 440

To gain 25% –

S.E = 100+25/100 × 440

= 125/100 × 440

= 550

Hence, alternative (B) ₹550 is the correct option.

**Question no – (13) **

**A man buys an article for ₹ 150 and makes overhead expenses which are 12% of the cost price. at what price must he sell it to gain 5%**

**(A) ₹180**

**(B) ₹168**

**(C) ₹176.40**

**(D) ₹88.20**

**Solution : **

Total C.P = C.P + overhead expenses

= 150 + 150 × 12/100

= 150 + 18

= 168

S.P = 100+5/100 × 168

= 105/100 × 168

= 1764/10 = 176.4

Hence, alternative (C) ₹176.40 is the correct option.

**Question no – (14)**

**What is the marked price of a hat which is bought for ₹ 210 at 16% discount**

**(A) ₹243**

**(B) ₹176**

**(C) ₹230**

**(D) ₹250**

**Solution : **

Discount = discount% × M.P/100

= 16 X M/100

∴ M.P = S.P + discount

= M = 210 + 1600M/100

= 100M = 21000 + 16M

= 100M – 16M = 21000

= M = 21000/84 = 21000/21×4

= 250

Thus, the correct option is – (D)

**Question no – (15)**

**The single discount in % which is equivalent to two successive discounts of 20% and 25% is**

**(A) 40%**

**(B) 45%**

**(C) 5%**

**(D) 22.5%**

**Solution : **

Total Prize – 100

1st discount 20%

= 100 × 20/100

= 20

Price of 20%

= 100 – 20

= 80

2nd discount 25%

= 80 × 25/100

= 20%

Price of 25% = (80-20) 60%

∴ Percentage after discount

= (100 – 60)%

= 40%

Hence, alternative (A) 40% is the correct option.

**Life Mathematics Exercise 4.3 Solutions :**

**Life Mathematics Exercise 4.3 Solutions :**

**Question no – (1) **

**Fill in the blanks :**

**(i) The compound interest on ₹ 5000 at 12% p.a for 2 years, compounded annually is** ___

**Solution : **

C.I = A – P

A = P(1 + R/100)^n

= 5000 (1 + 12/100)²

= 5000 (112/100)²

= 5000 × 112/100 × 112/100

= 6272

∴ C.I = 6272 – 5000 = 1272

So, compounded annually is 1272.

**(ii) The compound interest on ₹ 8000 at 10% p.a for 1 year, compounded half yearly is** ___

**Solution : **

P = 8000, R = 10, n = 1

Compound half yearly =

= P (1 + R/200)^2n

= 8000 (1 + 10/200)²

= 8000 × (21/20)²

= 8000 × 21/20 × 21/20

= 8820

C.I = 8820 – 8000 = 820

Hence, compounded half yearly is 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 ___**

**Solution : **

Let the population 3 years ago = R

∴ A = P (1 + R/100)^n

= 26620 = P (1+ 10/100)^3

= 26620 = P (11/10)^3

= P = 26620 × (10/11)^3

= 26620 × 10/11× 10/11 × 10/11

= 20000

Thus, then the population 3 years ago was 20000.

**(iv) If the compound interest is calculated quarterly, the amount is found using the formula** ___

**Solution : **

If the compound interest is calculated quarterly, the amount is found using the formula A = P (1+ R/100)^4n.

**(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** ___

**Solution : **

C.I – S.I = 5000 × (8/100)^2

= 5000 × 8/100 × 8/100

= 32

**Question no – (3) **

**Find the compound interest on ₹3200 at 2.5 % p.a for 2 years, compounded annually.**

**Solution : **

P = 3200, R = 2.5%, n = 2

A = 3200 (1 + 25/1000)²

= 3200 (41/40)²

= 3200 × 41/40 × 41/40

= 3362

∴ C.I = 3362 – 3200 = 162

**Question no – (4) **

**Find the compound interest for 2 1/2 years on ₹ 4000 at 10 % p.a, if the interest is compounded yearly.**

**Solution : **

P = 4000, r = 10, t = 2 1/2 year.

A = p (1+ r/100)2 (1+ (1/2 × r)/100)

= 4000 (1+ 10/100)² (1+ (1/2×10)/100)

= 4000 (11/10)² (1+ 5/100)

= 4000 × 11/10 × 11/10 × 105/100

= 40 × 11 × 11 × 105/100

= 2783/5 = 55616

= 5082

C.I = (5082 – 4000) = 1082

**Question no – (6) **

**In how many years will ₹ 3375 become ₹ 4096 at 13 1/3 % p.a if the interest is compounded half-yearly?**

**Solution : **

A = 4096, P = 3375, r = 13 1/3% n

4096 = 3375 (1+ (13 1/3)/200)^2n

= 4096/3375 = (1+ (40/3)/200)^2n

= 4096/3375 = (1+ 40/600)^2n

= 4096/3375 = (16/15)^2n

2n = 3

= n = 3/2

= 1 1/2

∴ The time period = 1 1/2 year.

**Question no – (7) **

**Find the C.I on ₹ 15000 for 3 years if the rates of interest are 15%, 20 % and 25 % for the I, II and III years respectively.**

**Solution : **

According to the given question,

P = 15000, n = 3, a = 15%, b = 20%, c = 25%

A = P (1 + 6/100) (1 + b/100) (1 + c/100)

= 15000 (1 + 75/100) (1 + 20/100) (1 + 25/100)

= 15000 × 115/100 × 120/100 × 125/100

= 25878

∴ C.I = 25875 – 15000 = 10875

**Question no – (8) **

**Find the difference between C.I and S.I on ₹5000 for 1 year at 2 % p.a, if the interest is compounded half yearly.**

**Solution : **

P = 5000, n = 1, r = 2%

∴ A = 5000 × (1+ 2/200)^2.1

= 5000 × (101/100)^2

= 5000 × 101/100 × 101/100

= 5100.5

∴ C.I = 5100.5 – 5000 = 100.5

**Question no – (9) **

**Find the rate of interest if the difference between C.I and S.I on ₹8000 compounded annually for 2 years is ₹20.**

**Solution : **

C.I – S.I = P (r/100)²

20 = 5000 r/100 × r/100

= 20 = 4/5 r²

= r² = 20 × 5/4

= r = 5

Therefore, the rate of interest will be 5%

**Question no – (10) **

**Find the principal if the difference between C.I and S.I on it at 15% p.a for 3 years is ₹ 1134.**

**Solution : **

C.I – S.I = P (r/100)² (3+ r/100)

= 113y = P (15/100)² (3+ 15/100)

= 113y = P × 3/20 × 3/20 × 63/20

= P = 1134 × 20/3 × 20/3 × 20/63

= P = 16000

**Objective Type Question Solutions : **

**Question no – (11) **

**The number of conversion periods in a year, if the interest on a principal is compounded every two months is** ___

**(A) 2**

**(B) 4**

**(C) 6**

**(D) 12**

**Solution : **

Correct option – (C)

The number of conversion periods in a year, if the interest on a principal is compounded every two months is 6.

**Question no – (13) **

**The cost of a machine is 18000 and it depreciates at 16 2/3% annually. its value after 2 years will be**

**(A) ₹12000**

**(B) ₹12500**

**(C) ₹15000**

**(D) ₹16500**

**Solution : **

Correct option – (B)

P = 18000 r = 16 2/3 = 50/3 n = 2

∴ A = 18000 (1 – (50/3)/100)²

= 18000 (1- 50/300)²

= 18000 (5/6)²

= 18000 × 5/6 × 5/6

= 12500

**Question no – (13) **

**The sum which amounts to ₹2662 at 10% p.a in 3 years, compounded yearly is** ____

**(A) ₹ 2000**

**(B) ₹ 1800**

**(C) ₹ 1500**

**(D) ₹ 2500**

**Solution : **

Correct option – (A)

A = 2661, r = 10, n = 3

= 2662 = P (1+ 10/100)³

= 2662 = P (11/10)³

= 2662 = P × 11/10 × 11/10 × 11/10

= P 2662 × 10/11 × 10/11 × 10/11

= ₹ 2000

**Question no – (15) **

**The difference between compound and simple interest on a certain sum of money for 2 years at 2% p.a is ₹ 1. The sum of money is** ____

**(A) ₹ 2000**

**(B) ₹ 1500**

**(C) ₹ 3000**

**(D) ₹ 2500**

**Solution : **

C.I – S.I = P (r/100)²

= 1 = P (2/100)²

= 1 = P × 1/50 × 1/50

= P = 2500

Hence, the correct option – (D)

**Life Mathematics Exercise 4.4 Solutions :**

**Life Mathematics Exercise 4.4 Solutions :**

**(1) Fill in the blanks :**

**(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job working together is** __** days.**

**Solution : **

A’s one day work = 1/24

(A + B)’s one day work = 1/6

∴ B’s one day work = 1/6 – 1/24

= 4-1/24

= 8/24

= 1/8

∴ 8 days

**(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in __ days.**

**Solution : **

P_1 = 5, w_1 = 5, D_1 = 5, and P_2 = 50, D_2 =?

P_1 × D_1/W_1 = P_2 × D_2/W_2

5 × 5/5 = 50 × D_2/50

D_2 = 5

∴ 5 days

**(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.**

**Solution : **

A’s one day work = 1/24

(A + B)’s one day work = 1/6

∴ B’s one day work

= 1/6 – 1/24

= 4-1/24

= 8/24

= 1/8

∴ 8 days.

**Question no – (2) **

**210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?**

**Solution : **

P_1 = 210, H_1 = 12, D_1 = 18, W_1 = 1

P_2 = ?, H_2 = 14, D_2 = 20, W_2 = 1

∴ P_1 × W_1XD_1/W_1 = P_2 × H_2 × D_2/W_2

= 210 × 12 × 18/1 = P_2 × 14 × 20/1

= P_2 = 210 × 12 × 18/14 × 20

= 162

**Question no – (3) **

**A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?**

**Solution : **

B_1 = 7000, D_1 = 12, M_1 = 36

B_2 =?, D_1 = 18, M_2 = 24

Formula, B_1 × D_1 × M_1 = B_2 × D_2 × M_2

= 7000 × 12 × 36 = B_2 × 18 × 24

= B_2 = 7000 × 12 × 36/18 × 24

= 7000 cement bags

**Question no – (4) **

**A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 more hours a day?**

**Solution : **

S_1 = 9600, D_1 = 6, W_1 = 15

S_2 = 14400, D_2 =?, H_2 = (15+3) = 18 Hours

Formula – 9600 × 6 × 15 = 14400 × D_2 × 18

= D_2 = 9600×6×15/14400×18

= D_2 = 10/3

= 3 1/3 day

**Question no – (5) **

**If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?**

**Solution : **

Container | Tonnes | Days |

6 | 135 | 5 |

x | 180 | 4 |

More tonnes means more container, it’s direct proportion = 180/135

Less days mean less containers, its inverse proportion = 5/4

∴ x = 6 X 150/135 X 5/4 = 10

∴ (10 – 6) = 4 more lorries.

**Question no – (6) **

**A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?**

**Solution : **

A’s one hour work = 1/12

(B + C)’s one hour work = 1/3

(A + C)’s one hour work = 1/6

∴ A + C = 1/6

= 1/12 + C = 1/6

= C = 1/6 – 1/12

= 1/12

B + C = 1/3

= 1/12 + B = 1/3

= B = 1/3 – 1/12

= 4-1/12

= 3/12 = 1/4

∴ B takes 4 hours to do the same work.

**Question no – (7) **

**A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?**

**Solution : **

(A + B) work done = 1/12

(B + C)’s work done = 1/15

(B + A)’s work done = 1/20

2 (A + B + C) work done

= 1/12 + 1/15 + 1/20

= 5 + 4 + 3/60

= 12/60

∴ A + B + C work done

= 1/5 × 2

= 1/10

∴ A’s work done

= (A + B + C) – (B + C)

= A + B + C – B – C

= 1/10 – 1/15

= 3 – 2/30

= 1/30

Therefore, A can do the work in 30 days

**Life Mathematics Exercise 4.5 Solutions :**

**Life Mathematics Exercise 4.5 Solutions :**

**Question no – (2) **

**A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.**

**Solution : **

(35 – 31) % = 4% is 12 marks

1% is 12/4 = 3 marks

31% = 3 × 31 = 93

∴ 100% = 3 × 100 = 300 marks

Hence, the maximum marks of the examination is 300.

**Question no – (3) **

**Sultana bought the following things from a general store. calculate the total bill amount paid by her**

**(i) Medicines costing ₹800 with GST at 5%**

**(ii) Cosmetics costing ₹650 with GST at 12%**

**(iii) Cereals costing ₹900 with GST at 0%**

**(iv) Sunglass costing ₹1750 with GST at 18%**

**(v) Air Conditioner costing ₹28500 with GST at 28%**

**Solution : **

(i) Medicines are bought by

= (800 + 40)

= 840

(ii) 650 × 12/100

= 78

Cosmetic are bought by

= (650+78)

= 728

(iii) Cereals costing 900.

(iv) 1750 × 18/100 = 315

∴ Sunglass costing

= (1750 + 315)

= 2065.

(v) 28500 × 28/100

= 7980

∴ Conditioner costing

= (28500 + 7980)

= 36480

∴ Total amount,

= (840 + 728 + 900 + 2065 + 96480)

= Rs. 41013.

**Question no – (4) **

**P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?**

**Solution : **

Left, Q’s income = 100

P’s income = 125

∴ Difference = 25 Rs

If P’s income is Rs 100; then Q’s income loss than P,

= (25/125 × 100)

= 20 Rs

∴ Q’s income is less than P’s income 20%.

**Question no – (6) **

**If 32 men working 12 hours a day can do a work in 15 days, then how many men working 10 hours a day can do double that work in 24 days?**

**Solution : **

P_{1} = 32, H_{1} = 12, D_{1} = 15, W_{1} = 1

P_{2} = ?, H_{2} = 10, D_{2} = 24, W_{2} = 2

∴ P_{1} × H_{1} × D_{1}/W_{1} = P_{2} × H_{2} × D_{2}/W_{2}

= 32 × 12 × 15/1 = P_{2 }× 10 × 24/2

= P_{2} = 32 × 12 × 15/5 × 24

= 48

**Challenging Problem Solutions : **

**Question no – (9) **

**If the numerator of a fraction is increased by 50% and the denominator is decreased by 20%, then it becomes 3/5. find the original fraction**

**Solution : **

Let the fraction be a/b

(a × 50/100) / (b × 20/100) = 3/5

= a/2 × 5/b = 3/5

= a/b = 3/5 × 2/5

= a/b = 6/25

**Question no – (10) **

**Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20% . Find the cost price of the laptop.**

**Solution : **

Let, the cost price is x

12% gain = x ×12/100 = 12x/100

20% gain = x × 20/100 = 20x/100

∴ (12x/100 + 1200) = 20x/100

= 12x + 120000/100 = 20x/100

= 20x = 12x + 120000

= 20x – 12x = 120000

= 8x = 120000

= x = 120000/8

= 15000

Therefore, the cost price is Rs 15000.

**Question no – (11) **

**A shopkeeper gives two successive discounts on an article whose marked price is ₹ 180 and selling price is ₹ 108. Find the first discount percentage if the second discount is 25%.**

**Solution : **

M.P = 180, S.P = 108,

D_{2} = 25%, D_{1 }= ?

S.P = M.P – Discount

= 108 = 180- Discount

∴ Discount,

= 180 – 108

= Rs 72

Therefore, the first discount will be Rs 72.

**Question no – (13) **

**A small–scale company undertakes an agreement to make 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?**

**Solution : **

W_{1} = 540, D_{1} = 150, P_{1} = 40

W_{2} = 180, D_{2} = (150 + 75) = 225 day, P_{2} = ?

P_{1 }× D_{1}/W_{1} = P_{2 }× D_{2}/W_{2}

= 40×150/540 = P_{2 }× 225/180

= P_{2} = 40 × 150 × 180/540 × 225

= 80/9

**Next Chapter Solution : **

👉 *Geometry *