Simple Interest Class 7 RS Aggarwal Exe-11B MCQs Goyal Brothers ICSE Foundation Maths Solutions. We provide step by step Solutions of lesson/ Chapter-11 Simple Interest for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.
Simple Interest Class 7 RS Aggarwal Exe-11B MCQs Goyal Brothers ICSE Maths Solutions
Board | ICSE |
Publications | Goyal brothers Prakashan |
Subject | Maths |
Class | 7th |
Chapter-11 | Simple Interest |
Writer | RS Aggrawal |
Book Name | Foundation |
Topics | Solution of Exe-11B MCQs |
Academic Session | 2024 – 2025 |
Simple Interest MCQs
Class 7 RS Aggarwal Exe-11B Goyal Brothers ICSE Foundation Maths Solutions
Choose the correct option in each of the following:
Question-1: Which of the following statements is incorrect?
(a) T = (100 × S1) / (P x R)
(b) R = (P * T) / (100 * SI)
(c) SI = (P*R*T) / 100
(d) P = (100 * SI) / (R * T)
Answer- statement (b) is incorrect because, R = (SI * 100) / (P * T).
so, option (b) is in correct.
Question-2: If the principal is ₹600, then the amount to be paid at the end of three years at 7(1/2)% pa. simple interest will be
(a) 735 (b) 750 (c) 775 (d) 785
Answer- P = 600, T = 3 years, R = 7(1/2)% = 7.5%
S.I. = P×R×T/100
(600×7.5×3)/100 = 135
A = P + SI =600 + 135 = 735.
so, option (a) is correct.
Question-3: At, what rate Rs 800 gives Rs 208 as simple interest in 2 years?
(a) 11% , (b) 12% , (c) 13%, (d) 14%
Answer- SI = 208 P = 800 T = 2
SI= (P×R×T)/100
R = (208×100)/(800×2) = 13%
so, option (c) is correct.
Question-4: In how many years will ₹900 give ₹351 as simple interest at 13% p.a.?
(a) 1(1/2) years, (b) 2 years , (c) 2(1/2) years , (d) 3 years
Answer- SI = 351 P = 900 R = 13%
SI = PRT/100
351 = (900 × 13 × T) /100
27 = 9 × 1 × T
T = 3 years so, option (d) is correct.
Question-5: At what rate per cent per annum simple interest will a sum be double of itself in 8 years?
(a) 10% , (b) 12(1/2)% , (c) 15% , (d) 17(1/2)
Answer- Let the sum of money be x
Time=8 years
Amount after 8 years becomes double=2x
Therefore,
Simple interest=Amount -principal=2x−x=x
Rate per annum =S.I×100P×T = x × 100x × 8 = 100/8 = 25 /2 = 12(1/2)
so,option (b) is correct.
Question-6: At simple interest a sum becomes 7/4 of itself int 5 years. The rate of interest is
(a) 10%p.a , (b) 12% p.a , (c) 12(1/2)%pa., (d) 15%pa.
Answer-6: Let say Sum = 4X
sum becomes 7/4 of itself = (7/4) 4X = 7X
Simple interest = 7X – 4X = 3X
Simple interest = P * R * T /100
3X = 4X * R * 5 /100
3 = R/5
R = 15%
so, option (d) is correct.
MENTAL MATHS
Class 7 RS Aggarwal Exe-11B Goyal Brothers ICSE Foundation Maths Solutions
Question-1: Fill in the blanks:
(i)The money borrowed for a certain period is called _____
Ans-(i) principal
(ii) The rate per cent per annum is the interest on_____ for 1 year.
Ans-(ii) Rs100
(iii) The simple interest on ₹800 invested at 13% per annum for 3 years is _____
Ans-(iii) P = 800 R = 13% T = 3
SI = PRT/100
SI =(800* 13* 3) /100
SI = Rs312.
(iv) If ₹2600 becomes 3900 in 5 years, then the rate of interest is _____
Ans-(iv) SI = 3900-2600 = 1300 T = 5 P = 2600
SI =PRT/100
R= (1300 * 100 ) / (2600 * 5)
R = 10%.
(v) If the principal is 3400, then the amount to be paid at the end of 5 years at 8% p.a. simple interest will be ____
Ans-(v) P = 1400 R = 8 % T = 5
SI = PRT/100
SI = (3400 * 8 * 5) / 100
SI = 1360
A =SI + P = 1360 + 3400 = Rs 4760.
Question-2: Write true (T) or false (F):
(i) The total money to be paid back to the lender is called interest.___________
Ans-(i) False
(ii) Amount = Principal + Interest ____________
Ans-(ii) True
(iii) The rate per cent per annum is the interest on Rs 1 for 1 year. ____________
Ans-(iii) False
(iv) S.I. = (100 * P) / (R * T) ____________
Ans-(iv) False
(v) If a man borrows ₹5200 at 6% p.a. simple interest, the amount he has to return at the end of 5 years is 6760. ___
Ans-(v) True
— : End of Simple Interest Class 7 RS Aggarwal Exe-11B MCQs Goyal Brothers ICSE Foundation Maths Solutions :–
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