DAV Class 5 Maths Chapter 13 Brain Teasers Solutions

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

DAV Class 5 Maths Ch 13 Brain Teasers Solutions

Question 1.
Tick (✓) the correct answer.
(а) The extra money paid by a bank for every ₹ 100 after one year is called-
(i) Interest
(ii) Rate of interest
(iii) Amount
(iv) Principal
Solution:
(ii) Rate of Interest

(b) Mr Gupta borrowed ₹ 1,75,000 from a bank to buy a car. This amount is called-
(i) Interest
(ii) Rate of interest
(iii) Loan
(iv) Time
Solution:
(iii) Loan

(c) Simple interest for ₹ 1,000 at 10% for 9 months is-
(i) ₹ 75
(ii) ₹ 900
(iii) ₹ 90
(iv) ₹ 1,075
Solution:
(i) ₹ 75
SI = \(\frac{1000 \times 10 \times 9}{100 \times 12}\)
= \(\frac{900}{12}\)
= ₹ 75

(d) If amount = ₹ 1,645.50, simple interest = ₹ 95, principal is-
(i) ₹ 1,740.50
(ii) ₹ 1,550
(iii) ₹ 1,740
(iv) ₹ 1,550.50
Solution:
(iv) ₹ 1,550.50
A = SI + P
1,645.50 = 95 + P
1,645.50 – 95 = P
P = ₹ 1,550.50

(e) If loan = ₹ 1,75,000, Rate of interest = 4%, Time = 2 years. Then amount to be repaid is equal to-
(i) ₹ 1,89,000
(ii) ₹ 1,76,400
(iii) ₹ 1,98,000
(iv) ₹ 1,80,000
Solution:
(i) ₹ 1,89,000
SI = \(\frac{1,75,000 \times 4 \times 2}{100}\) = ₹ 14,000
A = P + I
= 1,75,000 + 14,000
= ₹ 1,89,000

Question 2.
Name the three factors that determine simple interest.
Solution:

Principal
Rate of Interest
Time

Question 3.
Find the amount if the principal is ₹ 625 and the interest is ₹ 55.
Solution:
A = P + I
= ₹ 625 + ₹ 55
= ₹ 680

Question 4.
Calculate the simple interest on ₹ 1,200 at 3\(\frac{1}{2}\)% interest per annum for six months.
Solution:
P = ₹ 1,200
R = 3\(\frac{1}{2}\)% = \(\frac{7}{2}\)%
T = six months = \(\frac{6}{12}\) = \(\frac{1}{2}\) year
SI = PRT
= \(\frac{1,200 \times 7 \times 1}{100 \times 2 \times 2}\)
= ₹ 21

Question 5.
Sohan deposited ₹ 6,000 in a bank at 7\(\frac{1}{2}\)% per annum for three years. Mohan deposited ₹ 6,000 in another bank at 11% per annum for 2\(\frac{1}{2}\) years. Who will get more simple interest?
Solution:
P = ₹ 6,000
R = 7\(\frac{1}{2}\)% = \(\frac{15}{2}\)% = \(\frac{15}{200}\) per annum
T = 3 years
SI got by Sohan = 6,000 × \(\frac{15}{200}\) × 3 = ₹ 1350
P = ₹ 6,000
R = 11% = \(\frac{11}{100}\) per annum
T = \(\frac{5}{2}\) years
SI got by Mohan = \(\frac{6,000 \times 11 \times 5}{100 \times 2}\) = ₹ 1650
Mohan will get more simple interest.

Question 6.
A farmer borrowed ₹ 2,400 at 12% interest per annum. At the end of 3 years, he repaid ₹ 1,200 and a cow for the balance amount. Find the cost of the cow.
Solution:
P = ₹ 2,400
R = 12%
T = 3 years
SI = 2,400 × \(\frac{12}{100}\) × 3
= ₹ 864
A = P + SI
= ₹ 2,400 + ₹ 864
= ₹ 3,264
Cost of cow = ₹ 3,264 – ₹ 1200 = ₹ 2,064
Cost of cow is ₹ 2,064.

Question 7.
Mr Kannan had deposited ₹ 2,500 in a bank at 12% per annum. After 3\(\frac{1}{2}\) years, he withdraws this amount. Out of this money, he want to buy a dressing table costing ₹ 4000. How much more money does he need?
Solution:
P = ₹ 2,500
R = 12% = latex]\frac{12}{100}[/latex] per annum
T = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = 2,500 × \(\frac{12}{100} \times \frac{7}{2}\) = ₹ 1,050
A = P + I
= ₹ 2,500 + ₹ 1,050
= ₹ 3,550
Cost of table = ₹ 4,000
Amount he needs = ₹ 4,000 – ₹ 3,550 = ₹ 450
He needs ₹ 450 more.

Additional Questions

Question 1.
Calculate simple interest for the following:
(a) P = ₹ 4,000; R = 3% per annum; T = 2 years
Solution:
P = ₹ 4,000
R = 3% per annum = \(\frac{3}{100}\) per annum
T = 2 years
Simple Interest = PRT
= 4,000 × \(\frac{3}{100}\) × 2
= ₹ 240

(b) P = ₹ 1,200; R = 2\(\frac{1}{2}\)% per annum; T = 1\(\frac{1}{2}\) years
Solution:
P = ₹ 1,200
R = 2\(\frac{1}{2}\)% = \(\frac{5}{2}\)% per annum
T = 1\(\frac{1}{2}\) years = \(\frac{3}{2}\) years
Simple Interest = PRT
= 1200 × \(\frac{5}{200} \times \frac{3}{2}\)
= ₹ 45

(c) P = ₹ 1,040; R = 12% per annum; T = 9 months
Solution:
P = ₹ 1040
R = 12% per annum
T = \(\frac{9}{12}\) year
Simple Interest = PRT
= 1,040 × \(\frac{12}{100} \times \frac{9}{12}\)
= \(\frac{468}{5}\)
= ₹ 93.6

Question 2.
A man deposited ₹ 12,000 in a bank at an interest of 9% per annum. How much simple interest will he get at the end of 24 months?
Solution:
Principal = ₹ 12,000
R = 9% per annum = \(\frac{9}{100}\) per annum
T = 24 months = 2 years
SI = PRT
= 12000 × \(\frac{9}{100}\) × 2
= ₹ 2160
He will get a simple interest of ₹ 2160 at the end of 24 months.

Question 3.
Calculate the following:
(a) P = ₹ 1,200; SI = ?; Amount = ₹ 1,450
Solution:
A = P + SI
₹ 1,450 = ₹ 1,200 + SI
SI = ₹ 1,450 – ₹ 1,200 = ₹ 250

(b) P = ?; SI = ₹ 134; Amount = ₹ 3100
Solution:
A = P + SI
P = A – SI
= ₹ 3,100 – ₹ 134
= ₹ 2,966

Question 4.
Find the amount.
(a) P = ₹ 21,000; Interest = ₹ 350
Solution:
A = P + I
= ₹ 21,000 + ₹ 350
= ₹ 21,350

(b) P = ₹ 1,750; Interest = ₹ 250
Solution:
A = P + I
= ₹ 1,750 + ₹ 250
= ₹ 2,000

Question 5.
Maya deposited ₹ 3,500 in a bank at 14% interest per annum. She withdraws her money after 9 months. Out of this amount, she purchases a centre table worth ₹ 3,800. How much money is left with her?
Solution:
Principal = ₹ 3,500
R = 14% per annum
T = 9 months = \(\frac{9}{12}\) years
SI = PRT
= 3,500 × \(\frac{14}{100} \times \frac{9}{12}\)
= ₹ 367.5
A = P + I
= ₹ 3,500 + ₹ 367.5
= ₹ 3,867.5
Cost of centre table = ₹ 3,800
Money left with her = ₹ 3,867.5 – ₹ 3,800 = ₹ 67.5

Question 6.
Tick (✓) the correct answer.
(a) Mr Sharma borrowed ₹ 20,00,000 from the bank to buy a small truck. This amount is called
(i) Principal
(ii) Rate of Interest
(iii) Loan
(iv) Time
Solution:
(iii) Loan

(b) The extra money paid by the bank is called
(i) Principal
(ii) Rate of Interest
(iii) Loan
(iv) Interest
Solution:
(iv) Interest

(c) Extra money charged by the bank for every hundred rupees deposit is called
(i) Principal
(ii) Amount
(iii) Interest
(iv) Loan
Solution:
(iii) Interest

(d) The period of time for which money is kept in a bank is called
(i) Amount
(ii) Rate of Interest
(iii) Principal
(iv) Time period
Solution:
(iv) Time period