Question

. A man invested 6,000 in three parts at simple interest—first part at 3% p.a. for 4 years
second part at 4% p.a. for 5 years and the third part at 5% p.a. for 3 years. If the interest received on each part is equal, find that part which was invested at 5% p.a.​

Answer 1

Step-by-step explanation:

Let one part is x and other part is (6000−x)

Now,

S.I on first part =S.I. on second part

x×

100

12

×

12

9

=(6000−x)×

100

10

×

2

3

∴x=3750

and (6000−x)

⇒6000−3750=2250

Therefore two parts of 6000 are 3750 and 2250.

Answer 2

Step-by-step explanation:

Let the two parts be x ans 6000−x

Simple interest = 100

principle×rate×time

⇒ 100x(6)(2)

= 100/(6000−x)(8)(3)

⇒x=12000−2x

⇒3x=12000

⇒x=4000

Two parts are Rs.4000, 2000

Simple interest =100

Principal× Rate ×Time

=>100×(6)(2)

=>100/(6000-x)(8)(3)

=>X=12000-2x

=>3x=12000

=>X=4000

Two part are ₹ 4000,2000