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Class 8 Exam  >  Class 8 Notes  >  Mathematics (Maths) Class 8  >  RS Aggarwal Solutions: Exercise 11B - Compound Interest

RS Aggarwal Solutions: Exercise 11B - Compound Interest | Mathematics (Maths) Class 8 PDF Download

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 Page 1


B y u s i n g t h e f o r m u l a , f i n d t h e a m o u n t a n d c o m p o u n d i n t e r e s t o n :
( 1 ) R s 6 0 0 0 f o r 2 y e a r s a t 9 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Principal (P) =6000
Rate of the interest (R) = 9%
Time of the interest (N) = 2 years
Amount = P( 1+R/100)*n
Amount = 6000 (1+9/100)*2
Amount = 6000 (1.09)*2
Amount = 6000 (1.09×1.09)
Amount = 6000× 1.1881
Amount = Rupees 7128.60
Compounded Interest = Amount - Principal
C.I = 7128.6- 6000.0
C.I = 1128.6 
( 2 ) R s 1 0 0 0 0 f o r 2 y e a r s a t 1 1 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 10000
Rate =11%
Time =2 years
Compound=P(1+r100)2
=10000(1+11100)2
=10000×(111)210000
=12321
Amount=12321-10000=2321.
( 3 ) R s 3 1 2 5 0 f o r 3 y e a r s a t 8 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Page 2


B y u s i n g t h e f o r m u l a , f i n d t h e a m o u n t a n d c o m p o u n d i n t e r e s t o n :
( 1 ) R s 6 0 0 0 f o r 2 y e a r s a t 9 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Principal (P) =6000
Rate of the interest (R) = 9%
Time of the interest (N) = 2 years
Amount = P( 1+R/100)*n
Amount = 6000 (1+9/100)*2
Amount = 6000 (1.09)*2
Amount = 6000 (1.09×1.09)
Amount = 6000× 1.1881
Amount = Rupees 7128.60
Compounded Interest = Amount - Principal
C.I = 7128.6- 6000.0
C.I = 1128.6 
( 2 ) R s 1 0 0 0 0 f o r 2 y e a r s a t 1 1 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 10000
Rate =11%
Time =2 years
Compound=P(1+r100)2
=10000(1+11100)2
=10000×(111)210000
=12321
Amount=12321-10000=2321.
( 3 ) R s 3 1 2 5 0 f o r 3 y e a r s a t 8 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 31250
Rate r=8%
Time =3 years
Amount=p(1+r100)t
=31250(1+8100)3
=31250×(108100)3
=39,366Rs
compound=amount-principal
=39,366-31250
=Rs. 8116.
( 4 ) R s 1 0 2 4 0 f o r 3 y e a r s a t 1 2 ( 1 / 2 ) % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given p=Rs. 10240
r=1212% ?252%
Time=3 years
Amount=p(1+r100)t=10240(1+25/2100)3
=10240(1+18)3
=10240×(98)3
=14,580
Amount=14,580
( 5 ) R s 6 2 5 0 0 f o r 2 y e a r s 6 m o n t h s a t 1 2 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Principal (P)=Rs. 62500Rate (R)= 12% p.a.Period (n) = 2 years 6 months= 212 years
Here for first 2 years, we apply the compound interest formula and next 6 months, we apply
simple interest formula.
The amount for 2 years is
Page 3


B y u s i n g t h e f o r m u l a , f i n d t h e a m o u n t a n d c o m p o u n d i n t e r e s t o n :
( 1 ) R s 6 0 0 0 f o r 2 y e a r s a t 9 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Principal (P) =6000
Rate of the interest (R) = 9%
Time of the interest (N) = 2 years
Amount = P( 1+R/100)*n
Amount = 6000 (1+9/100)*2
Amount = 6000 (1.09)*2
Amount = 6000 (1.09×1.09)
Amount = 6000× 1.1881
Amount = Rupees 7128.60
Compounded Interest = Amount - Principal
C.I = 7128.6- 6000.0
C.I = 1128.6 
( 2 ) R s 1 0 0 0 0 f o r 2 y e a r s a t 1 1 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 10000
Rate =11%
Time =2 years
Compound=P(1+r100)2
=10000(1+11100)2
=10000×(111)210000
=12321
Amount=12321-10000=2321.
( 3 ) R s 3 1 2 5 0 f o r 3 y e a r s a t 8 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 31250
Rate r=8%
Time =3 years
Amount=p(1+r100)t
=31250(1+8100)3
=31250×(108100)3
=39,366Rs
compound=amount-principal
=39,366-31250
=Rs. 8116.
( 4 ) R s 1 0 2 4 0 f o r 3 y e a r s a t 1 2 ( 1 / 2 ) % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given p=Rs. 10240
r=1212% ?252%
Time=3 years
Amount=p(1+r100)t=10240(1+25/2100)3
=10240(1+18)3
=10240×(98)3
=14,580
Amount=14,580
( 5 ) R s 6 2 5 0 0 f o r 2 y e a r s 6 m o n t h s a t 1 2 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Principal (P)=Rs. 62500Rate (R)= 12% p.a.Period (n) = 2 years 6 months= 212 years
Here for first 2 years, we apply the compound interest formula and next 6 months, we apply
simple interest formula.
The amount for 2 years is
? Amount (A)=P (1+R100)n=Rs. 62500 (1+12100)2=Rs. 62500×(2825)2×5350=Rs.
62500×2825×2825100×784=Rs. 78400
The amount for next 6 months: 78400×(1+12×1100×2)
=78400×5350
=83104
Therefore, amount after 212 years is Rs. 83104
? C.I = A- P=Rs. 83104-Rs. 62500=Rs. 20604
( 6 ) R s 9 0 0 0 f o r 2 y e a r s 4 m o n t h s a t 1 0 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Given: Principal (P)=Rs. 9000
Rate (R)=10% per annum
Time (n)=2 years 4 months
=213 years [ ?1 year =12 months]
Amount for 2 years
? Amount (A)=P(1+R100)n
=Rs. 9000 (1+10100)2
=Rs. 9000×1110×1110
=Rs. 10890
For the next 13 years, Principal = Amount for 2 years =Rs. 10890.
Interest for 13 years =P×R×n100
=(10890×10×13100)
=Rs. 363
So, the total Amount for 213 years =Rs. 10890+363=Rs. 11253
?C.I.=A-P=Rs. 11253-Rs.9000
=Rs. 2253
Page 4


B y u s i n g t h e f o r m u l a , f i n d t h e a m o u n t a n d c o m p o u n d i n t e r e s t o n :
( 1 ) R s 6 0 0 0 f o r 2 y e a r s a t 9 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Principal (P) =6000
Rate of the interest (R) = 9%
Time of the interest (N) = 2 years
Amount = P( 1+R/100)*n
Amount = 6000 (1+9/100)*2
Amount = 6000 (1.09)*2
Amount = 6000 (1.09×1.09)
Amount = 6000× 1.1881
Amount = Rupees 7128.60
Compounded Interest = Amount - Principal
C.I = 7128.6- 6000.0
C.I = 1128.6 
( 2 ) R s 1 0 0 0 0 f o r 2 y e a r s a t 1 1 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 10000
Rate =11%
Time =2 years
Compound=P(1+r100)2
=10000(1+11100)2
=10000×(111)210000
=12321
Amount=12321-10000=2321.
( 3 ) R s 3 1 2 5 0 f o r 3 y e a r s a t 8 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 31250
Rate r=8%
Time =3 years
Amount=p(1+r100)t
=31250(1+8100)3
=31250×(108100)3
=39,366Rs
compound=amount-principal
=39,366-31250
=Rs. 8116.
( 4 ) R s 1 0 2 4 0 f o r 3 y e a r s a t 1 2 ( 1 / 2 ) % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given p=Rs. 10240
r=1212% ?252%
Time=3 years
Amount=p(1+r100)t=10240(1+25/2100)3
=10240(1+18)3
=10240×(98)3
=14,580
Amount=14,580
( 5 ) R s 6 2 5 0 0 f o r 2 y e a r s 6 m o n t h s a t 1 2 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Principal (P)=Rs. 62500Rate (R)= 12% p.a.Period (n) = 2 years 6 months= 212 years
Here for first 2 years, we apply the compound interest formula and next 6 months, we apply
simple interest formula.
The amount for 2 years is
? Amount (A)=P (1+R100)n=Rs. 62500 (1+12100)2=Rs. 62500×(2825)2×5350=Rs.
62500×2825×2825100×784=Rs. 78400
The amount for next 6 months: 78400×(1+12×1100×2)
=78400×5350
=83104
Therefore, amount after 212 years is Rs. 83104
? C.I = A- P=Rs. 83104-Rs. 62500=Rs. 20604
( 6 ) R s 9 0 0 0 f o r 2 y e a r s 4 m o n t h s a t 1 0 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Given: Principal (P)=Rs. 9000
Rate (R)=10% per annum
Time (n)=2 years 4 months
=213 years [ ?1 year =12 months]
Amount for 2 years
? Amount (A)=P(1+R100)n
=Rs. 9000 (1+10100)2
=Rs. 9000×1110×1110
=Rs. 10890
For the next 13 years, Principal = Amount for 2 years =Rs. 10890.
Interest for 13 years =P×R×n100
=(10890×10×13100)
=Rs. 363
So, the total Amount for 213 years =Rs. 10890+363=Rs. 11253
?C.I.=A-P=Rs. 11253-Rs.9000
=Rs. 2253
( 7 ) F i n d t h e a m o u n t o f R s 8 0 0 0 f o r 2 y e a r s c o m p o u n d e d a n n u a l l y a n d t h e r a t e s b e i n g 9 %
p e r a n n u m d u r i n g t h e f i r s t y e a r a n d 1 0 % p e r a n n u m d u r i n g t h e s e c o n d y e a r .
Solution
Principal (P) = Rs. 8000 Period
(n) = 2 years Rate (R1)=9% for the first year R2=10% the second year
? Amount (A)= P (1+R1100)1(1+R2100)1=8000 (1+9100) (1+10100)
=Rs 8000×109100×110100=Rs. 9592
( 8 ) A n a n d o b t a i n e d a l o a n o f R s 1 2 5 0 0 0 f r o m t h e A l l a h a b a d B a n k f o r b u y i n g c o m p u t e r s .
T h e b a n k c h a r g e s c o m p o u n d i n t e r e s t a t 8 % p e r a n n u m , c o m p o u n d e d a n n u a l l y . W h a t
a m o u n t w i l l h e h a v e t o p a y a f t e r 3 y e a r s t o c l e a r t h e d e b t ?
Solution
Principal (p) = 1,25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years
? Amount (A) = P(1+r100)n
=Rs. 125000×(1+8100)3
=Rs. 125000×(2725)3
=Rs. 125000×2725×2725×2725
=Rs. 157464 Ans.
( 9 ) T h r e e y e a r s a g o , B e e r u p u r c h a s e d a b u f f a l o f r o m S u r j e e t f o r R s 1 1 0 0 0 . W h a t p a y m e n t
w i l l d i s c h a r g e h i s d e b t n o w , t h e r a t e o f i n t e r e s t b e i n g 1 0 % p e r a n n u m , c o m p o u n d e d
a n n u a l l y ?
Solution
Price of a buffalo (P) = Rs. 11000
Rate of interest (R)= 10% p.a.
Page 5


B y u s i n g t h e f o r m u l a , f i n d t h e a m o u n t a n d c o m p o u n d i n t e r e s t o n :
( 1 ) R s 6 0 0 0 f o r 2 y e a r s a t 9 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Principal (P) =6000
Rate of the interest (R) = 9%
Time of the interest (N) = 2 years
Amount = P( 1+R/100)*n
Amount = 6000 (1+9/100)*2
Amount = 6000 (1.09)*2
Amount = 6000 (1.09×1.09)
Amount = 6000× 1.1881
Amount = Rupees 7128.60
Compounded Interest = Amount - Principal
C.I = 7128.6- 6000.0
C.I = 1128.6 
( 2 ) R s 1 0 0 0 0 f o r 2 y e a r s a t 1 1 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 10000
Rate =11%
Time =2 years
Compound=P(1+r100)2
=10000(1+11100)2
=10000×(111)210000
=12321
Amount=12321-10000=2321.
( 3 ) R s 3 1 2 5 0 f o r 3 y e a r s a t 8 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given,
Principal=Rs. 31250
Rate r=8%
Time =3 years
Amount=p(1+r100)t
=31250(1+8100)3
=31250×(108100)3
=39,366Rs
compound=amount-principal
=39,366-31250
=Rs. 8116.
( 4 ) R s 1 0 2 4 0 f o r 3 y e a r s a t 1 2 ( 1 / 2 ) % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Given p=Rs. 10240
r=1212% ?252%
Time=3 years
Amount=p(1+r100)t=10240(1+25/2100)3
=10240(1+18)3
=10240×(98)3
=14,580
Amount=14,580
( 5 ) R s 6 2 5 0 0 f o r 2 y e a r s 6 m o n t h s a t 1 2 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Principal (P)=Rs. 62500Rate (R)= 12% p.a.Period (n) = 2 years 6 months= 212 years
Here for first 2 years, we apply the compound interest formula and next 6 months, we apply
simple interest formula.
The amount for 2 years is
? Amount (A)=P (1+R100)n=Rs. 62500 (1+12100)2=Rs. 62500×(2825)2×5350=Rs.
62500×2825×2825100×784=Rs. 78400
The amount for next 6 months: 78400×(1+12×1100×2)
=78400×5350
=83104
Therefore, amount after 212 years is Rs. 83104
? C.I = A- P=Rs. 83104-Rs. 62500=Rs. 20604
( 6 ) R s 9 0 0 0 f o r 2 y e a r s 4 m o n t h s a t 1 0 % p e r a n n u m c o m p o u n d e d a n n u a l l y .
Solution
Given: Principal (P)=Rs. 9000
Rate (R)=10% per annum
Time (n)=2 years 4 months
=213 years [ ?1 year =12 months]
Amount for 2 years
? Amount (A)=P(1+R100)n
=Rs. 9000 (1+10100)2
=Rs. 9000×1110×1110
=Rs. 10890
For the next 13 years, Principal = Amount for 2 years =Rs. 10890.
Interest for 13 years =P×R×n100
=(10890×10×13100)
=Rs. 363
So, the total Amount for 213 years =Rs. 10890+363=Rs. 11253
?C.I.=A-P=Rs. 11253-Rs.9000
=Rs. 2253
( 7 ) F i n d t h e a m o u n t o f R s 8 0 0 0 f o r 2 y e a r s c o m p o u n d e d a n n u a l l y a n d t h e r a t e s b e i n g 9 %
p e r a n n u m d u r i n g t h e f i r s t y e a r a n d 1 0 % p e r a n n u m d u r i n g t h e s e c o n d y e a r .
Solution
Principal (P) = Rs. 8000 Period
(n) = 2 years Rate (R1)=9% for the first year R2=10% the second year
? Amount (A)= P (1+R1100)1(1+R2100)1=8000 (1+9100) (1+10100)
=Rs 8000×109100×110100=Rs. 9592
( 8 ) A n a n d o b t a i n e d a l o a n o f R s 1 2 5 0 0 0 f r o m t h e A l l a h a b a d B a n k f o r b u y i n g c o m p u t e r s .
T h e b a n k c h a r g e s c o m p o u n d i n t e r e s t a t 8 % p e r a n n u m , c o m p o u n d e d a n n u a l l y . W h a t
a m o u n t w i l l h e h a v e t o p a y a f t e r 3 y e a r s t o c l e a r t h e d e b t ?
Solution
Principal (p) = 1,25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years
? Amount (A) = P(1+r100)n
=Rs. 125000×(1+8100)3
=Rs. 125000×(2725)3
=Rs. 125000×2725×2725×2725
=Rs. 157464 Ans.
( 9 ) T h r e e y e a r s a g o , B e e r u p u r c h a s e d a b u f f a l o f r o m S u r j e e t f o r R s 1 1 0 0 0 . W h a t p a y m e n t
w i l l d i s c h a r g e h i s d e b t n o w , t h e r a t e o f i n t e r e s t b e i n g 1 0 % p e r a n n u m , c o m p o u n d e d
a n n u a l l y ?
Solution
Price of a buffalo (P) = Rs. 11000
Rate of interest (R)= 10% p.a.
Period (n) = 3 years
? Price of buffalo at present
(A)=(1+R100)n=Rs. 11000(1+10100)3=Rs. 11000×(1110)3
=Rs. 11000×1110×1110×1110
=Rs. 14641
( 1 0 ) S h u b h a l a x m i t o o k a l o a n o f R s 1 8 0 0 0 f r o m S u r y a F i n a n c e t o p u r c h a s e a T V s e t . I f t h e
c o m p a n y c h a r g e s i n t e r e s t a t 1 2 % p e r a n n u m d u r i n g t h e f i r s t y e a r a n d 1 2 ( 1 / 2 ) % p e r
a n n u m d u r i n g t h e s e c o n d y e a r , h o w m u c h w i l l s h e h a v e t o p a y a f t e r 2 y e a r s ?
S o l u t i o n :
Amount of loan taken (P) = Rs. 18000
Rate (R1)=12% p.a. during first yearR1=1212% = 252% p.a. during second year
Period (n) = 2 years
? Total amount (A)= P (1+R1100)1 (1+R2100)1= Rs. 18000 (1+12100) (1+252×100)= Rs.
18000×2825×98=Rs. 22680
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Document Description: RS Aggarwal Solutions: Exercise 11B - Compound Interest for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation. The notes and questions for RS Aggarwal Solutions: Exercise 11B - Compound Interest have been prepared according to the Class 8 exam syllabus. Information about RS Aggarwal Solutions: Exercise 11B - Compound Interest covers topics like and RS Aggarwal Solutions: Exercise 11B - Compound Interest Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Exercise 11B - Compound Interest.
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The "RS Aggarwal Solutions: Exercise 11B - Compound Interest Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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Students of Class 8 can study RS Aggarwal Solutions: Exercise 11B - Compound Interest alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Exercise 11B - Compound Interest, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RS Aggarwal Solutions: Exercise 11B - Compound Interest is prepared as per the latest Class 8 syllabus.
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