Question

pratah borrowed a sum of money from Arun at simple interest at the rate of 12% per annum for first three years 16% the next five years and 20% per annum for the period beyond 8 years. if at the end of 11 years the total invest is 6080 more than the sum the sum borrowed was​

Answer 1

Answer:

Correct option is

Correct option isB

Correct option isBRs.8000.

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 +

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 +

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8)

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8)

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x +

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x +

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080⇒176x=100x+608000⇒76x=608000⇒x=

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080⇒176x=100x+608000⇒76x=608000⇒x= 76

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080⇒176x=100x+608000⇒76x=608000⇒x= 76608000

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080⇒176x=100x+608000⇒76x=608000⇒x= 76608000

Correct option isBRs.8000.Let the sum borrowed be Rs.x. Then,100x×12×3 + 100x×16×5 + 100x×20×(11−8) =x+6080⇒ 10036x + 10080x + 10060x =x+6080⇒ 100176x =x+6080⇒176x=100x+608000⇒76x=608000⇒x= 76608000 =Rs.8000.

Step-by-step explanation:

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Answer 2

Answer:

Pratap borrowed a sum of money from Arun at simple interest, at the rate of 12% p.a. for the first three years, 16% p.a. for the next five years and ...