DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions of Simple Interest offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 13 Worksheet 3 Solutions

Question 1.
Find the amount for the following:
DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions Q1
Solution:
(a) P = ₹ 1,500, I = ₹ 150
A = P + I
= ₹ 1500 + ₹ 150
= ₹ 1650

(b) P = ₹ 750, I = ₹ 35
A = P + I
= ₹ 750 + ₹ 35
= ₹ 785

DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

(c) P = ₹ 15,000, I = ₹ 980
A = P + I
= ₹15,000 + ₹ 980
= ₹ 15,980

(d) P = ₹ 4500, I = ₹ 215
A = P + I
= ₹ 4,500 + ₹ 215
= ₹ 4,715

Question 2.
Fill in the blanks:
DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions Q2
Solution:
(a) P = ₹ 800, SI = ?, A = ₹ 905
A = P + I
905 = 800 + I
905 – 800 = I
105 = I
I = ₹ 105

(b) SI = ₹ 75.50, A = ₹ 450, P = ?
A = SI + P
450 = 75.50 + P
450 – 75.50 = P
P = ₹ 374.50

(c) P = ₹ 25,000, SI = ?, A = ₹ 31,000
A = P + SI
31,000 = 25,000 + SI
31,000 – 25,000 = I
SI = ₹ 6,000

DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

(d) SI = ₹ 515, A = ₹ 1,680, P = ?
A = P + SI
1,680 = P + 515
P = 1,680 – 515 = ₹ 1,165

Question 3.
Calculate the amount for the following:
(a) ₹ 4,000 at 9% per annum for 3 years.
Solution:
A = ?
P = ₹ 4,000
R = 9%
T = 3 years
SI = ?
SI = PRT
= 4,000 × 3 × \(\frac{9}{100}\)
= ₹ 1,080
A = P + SI
= ₹ 4,000 + ₹ 1,080
= ₹ 5,080

(b) ₹ 750 at 3\(\frac{1}{2}\)% per annum for 2 years.
Solution:
P = ₹ 750
R = 3\(\frac{1}{2}\)% = \(\frac{7}{2}\)% = \(\frac{7}{200}\)
T = 2 years
SI = ?
SI = PRT
= 750 × \(\frac{7}{200}\) × 2
= ₹ 52.5
A = P + SI
= ₹ 750 + ₹ 52.5
= ₹ 802.50

(c) ₹ 3,200 at 7\(\frac{1}{2}\)% per annum for 3\(\frac{1}{2}\) years.
Solution:
P = ₹ 3,200
R = 7\(\frac{1}{2}\)% = \(\frac{15}{2}\)% = \(\frac{15}{200}\)
T = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = ?
SI = PRT
= 3,200 × \(\frac{15}{200} \times \frac{7}{2}\)
= ₹ 840
A = SI + P
= ₹ 840 + ₹ 3,200
= ₹ 4,040

DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

(d) ₹ 900 at 5% per annum for 5\(\frac{1}{2}\) years.
Solution:
P = ₹ 900
R = 5%
T = 5\(\frac{1}{2}\) years = \(\frac{11}{2}\) years
SI = ?
SI = PRT
= 900 × \(\frac{5}{100} \times \frac{11}{2}\)
= ₹ 247.50
A = P + SI
= ₹ 900 + ₹ 247.50
= ₹ 1,147.50

Question 4.
Solve the following questions:
(a) Arun took a loan of ₹ 600 for 1\(\frac{1}{2}\) years with interest at the rate of 4% per annum. Find the interest he pays after 1\(\frac{1}{2}\) years . Also, calculate the amount he pays back after 1\(\frac{1}{2}\) years.
Solution:
Principal = ₹ 600
T = 1\(\frac{1}{2}\) years = \(\frac{3}{2}\) years
R = 4% = \(\frac{4}{100}\)
SI = P × R × T
= 600 × \(\frac{4}{100} \times \frac{3}{2}\)
= ₹ 36
A = ₹ 600 + ₹ 36 = ₹ 636
He pays interest of ₹ 36 and total amount ₹ 636 after 1\(\frac{1}{2}\) years.

(b) Amit deposited ₹ 4,800 in his account. The bank pays an interest of 6% per annum. What amount will Amit get back after three years?
Solution:
Principal = ₹ 4,800
R = 6% per annum
T = 3 years
SI = P × R × T
= 4,800 × \(\frac{6}{100}\) × 3
= ₹ 864
A = P + I
= ₹4,800 + ₹ 864
= ₹ 5,664
Amit will get back ₹ 5,664 after three years.

DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

(c) A man borrowed ₹ 700 from his friend. He promised to return the amount after six months at an interest of 8% per annum. How much money will he pay back after six months?
Solution:
Principal = ₹ 700
R = 8%
T = 6 months = \(\frac{1}{2}\) year
SI = P × R × T
= 700 × \(\frac{8}{100} \times \frac{1}{2}\)
= ₹ 28
A = SI + P
= ₹ 700 + ₹ 28
= ₹ 728
He will pay back ₹ 728 after 6 months.

(d) Surender deposited ₹ 2,500 in a bank at 12% interest per annum. He withdraws his money after 2\(\frac{1}{2}\) years. Out of this amount, he buys an almirah for ₹ 3,000. How much money is left with him?
Solution:
P = ₹ 2,500
R = 12%
T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
SI = P × R × T
= 2,500 × \(\frac{12}{100} \times \frac{5}{2}\)
= ₹ 750
A = P + I
= ₹ 2,500 + ₹ 750
= ₹ 3,250
Cost of almirah = ₹ 3,000
Amount left = A – cost of almirah
= ₹ 3,250 – ₹ 3,000
= ₹ 250
₹ 250 is left with him.

DAV Class 5 Maths Chapter 13 Value Based Questions

Nakul’s father Mr. Gupta got transferred from Delhi to Mumbai. Mr. Gupta had deposited ₹ 55000 in the Punjab National Bank two years before at 8% per annum. He closed the account and with the interest money he purchased gifts for his friends and family members. Nakul also got two pairs of jeans, two shirts, and a video game as a gift. Nakul with the permission of his parents gifted one jean and one shirt to Rahul, the son of the milkman. Rahul was very happy and thanked Nakul for the gift.

Question 1.
How much money did Mr Gupta get from the bank?
Solution:
Principal = ₹ 55,000
T = 2 years
R = 8% = \(\frac{8}{100}\) per annum
SI = PRT
= 55,000 × \(\frac{8}{100}\) × 2
= ₹ 8,800
A = P + I
= ₹ 55000 + ₹ 8800
= ₹ 63,800
Mr. Gupta got ₹ 63,800 from the bank.

Question 2.
How much money was spent by Mr Gupta on gifts?
Solution:
Interest = ₹ 8,800
With interest money, he purchased a gift. So he purchased a shirt and jeans for ₹ 8,800.

DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions

Question 3.
How do you feel when you gift/donate anything to the needy?
Solution:
When we gift/donate anything to the needy we feel happy and self-satisfied.

DAV Class 5 Maths Chapter 13 Worksheet 3 Notes

Concept of Amount
The amount we get back after a certain period of time is the sum of Principal and Interest.

  • Amount = Principal + Interest
  • Principal = Amount – Interest
  • Interest = Amount – Principal

Example 1.
Sarita deposited ₹ 3,000 in a bank at 8% interest per annum. What amount will she get back after 3\(\frac{1}{2}\) years?
Solution:
Principal = ₹ 3,000
R = 8% = \(\frac{8}{100}\)
Time = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = P × R × T
= 3,000 × \(\frac{8}{100} \times \frac{7}{2}\)
= ₹ 840
Amount received at the end of 3\(\frac{1}{2}\) years = P + Interest
= ₹ 3,000 + ₹ 840
= ₹ 3,840