Question

1. Divides10,000 into two parts so as the s.i. on first part for 2 years 6% p.a. is equal to the simple of interest on the second part for 3 years at 4% p.a.
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Answer 1

Step-by-step explanation:

Given :-

S.I. on the first part for 2 years 6% p.a. is equal to the S.I. on the second part for 3 years at 4% p.a. of the number 10,000

To find :-

Divide 10000 into two parts according to the given rules ?

Solution :-

Given number = 10,000

Let the first part of 1000 be X

The second part of 10000 = (10000-X)

S.I. on the first part for 2 years at 6% p.a.

We know that

S.I. = PTR/100

We have, P = Rs. X , T = 2 years and R = 6%

On Substituting these values in the above formula

=> S.I. = (X×2×6)/100

=> S.I. = 12X/100

=> S.I. = Rs. 3X/25 ----------------------(1)

S I. on the second part for 3 years at 4% p.a.

We know that

S.I. = PTR/100

We have, P = Rs. (10000-X) , T = 3 years and

R = 4 %

On Substituting these values in the above formula

=> S.I. = (10000-X)×3×4/100

=> S.I. = 12(10000-X)/100

=> S.I. = Rs. 3(10000-X) /25 ----------------------(2)

According to the given problem

S.I. on the first part for 2 years 6% p.a. is equal to the S.I. on the second part for 3 years at 4% p.a. of the number 10,000.

=> (1) = (2)

=> 3X/25 = 3(10000-X) /25

=> X = 10000-X

=> X+X = 10000

=> 2X = 10000

=> X = 10000/2

=> X = 5000

First part = 5000

Second part = 10000-5000 = 5000

Answer :-

The two parts of the given number 10000 are 5000 and 5000

Check :-

S.I. on the first part for 2 years at 6% p.a.

We know that

S.I. = PTR/100

We have, P = Rs. 5000 , T = 2 years and R = 6%

On Substituting these values in the above formula

=> S.I. = (5000×2×6)/100

=> S.I. = 60000/100

=> S.I. = Rs. 600 ----------------------(1)

S I. on the second part for 3 years at 4% p.a.

We know that

S.I. = PTR/100

We have, P = Rs.5000 , T = 3 years and

R = 4 %

On Substituting these values in the above formula

=> S.I. = 5000×3×4/100

=> S.I. = 60000/100

=> S.I. = Rs. 600 ----------------------(2)

(1) = (2)

Both are equal.

Verified the given relations in the given problem.

Used formulae:-

→ S.I. = PTR/100

• P = Principal
• T = Time
• R = Rate of Interest
• S.I.= Simple Interest

Answer 2

Answer:

Given :-

• The S.I on first part for 2 years at 6% per annum is equal to the S.I on the second part for 3 years at 4% per annum.

To Find :-

• What are the two parts of 10000.

Formula Used :-

$\bigstar$ Simple Interest or S.I Formula :

$\longrightarrow \sf\boxed{\bold{\pink{S.I =\: \dfrac{P \times r \times t}{100}}}}$

where,

• S.I = Simple Interest
• P = Principal
• r = Rate of Interest
• t = Time Period

Solution :-

Let,

$\mapsto \bf 1^{st}\: Part =\: x$

$\mapsto \bf 2^{nd}\: Part =\: 10000 - x$

$\sf\bold{\underline{\purple{\clubsuit\: In\: the\: first\: case\: :-}}}$

$\leadsto$ S.I on the first part for 2 years at 6% per annum.

Given :

• Principal = Rs x
• Rate of Interest = 6% per annum
• Time Period = 2 years

According to the question by using the formula we get,

$\implies \sf S.I =\: \dfrac{x \times 6 \times 2}{100}$

$\implies \sf S.I =\: \dfrac{x \times 12}{100}$

$\implies \sf S.I =\: \dfrac{\cancel{12}x}{\cancel{100}}$

$\implies \sf\bold{\green{S.I =\: \dfrac{3x}{25}\: ------\: (Equation\: No\: 1)}}$

$\sf\bold{\underline{\purple{\clubsuit\: In\: the\: second\: case\: :-}}}$

$\leadsto$ S.I on the second part for 3 years at 4% per annum.

Given :

• Principal = Rs (10000 - x)
• Rate of Interest = 4% per annum
• Time Period = 3 years

According to the question by using the formula we get,

$\implies \sf S.I =\: \dfrac{(10000 - x) \times 4 \times 3}{100}$

$\implies \sf S.I =\: \dfrac{(10000 - x) \times \cancel{12}}{\cancel{100}}$

$\implies \sf S.I =\: \dfrac{(10000 - x) \times 3}{25}$

$\implies \sf\bold{\green{S.I =\: \dfrac{3(10000 - x)}{25}\: ------\: (Equation\: No\: 2)}}$

Now, according to the question :

$\dashrightarrow \bf \dfrac{3x}{25} =\: \dfrac{3(10000 - x)}{25}$

$\dashrightarrow \sf \dfrac{\cancel{3}x}{\cancel{25}} =\: \dfrac{\cancel{3}(10000 - x)}{\cancel{25}}$

$\dashrightarrow \sf \dfrac{x}{1} =\: \dfrac{10000 - x}{1}$

By doing cross multiplication we get,

$\dashrightarrow \sf 1(x) =\: 1(10000 - x)$

$\dashrightarrow \sf x =\: (10000 - x)$

$\dashrightarrow \sf x =\: 10000 - x$

$\dashrightarrow \sf x + x =\: 10000$

$\dashrightarrow \sf 2x =\: 10000$

$\dashrightarrow \sf x =\: \dfrac{\cancel{10000}}{\cancel{2}}$

$\dashrightarrow \sf x =\: \dfrac{5000}{1}$

$\dashrightarrow \sf\bold{\blue{x =\: 5000}}$

Hence, the required two parts are :

✯ 1ˢᵗ Part :

$\leadsto \sf 1^{st}\: Part =\: x$

$\leadsto \sf\bold{\red{1^{st}\: Part =\: 5000}}$

✯ 2ⁿᵈ Part :

$\leadsto \sf 2^{nd}\: Part =\: (10000 - x)$

$\leadsto \sf 2^{nd}\: Part =\: (10000 - 5000)$

$\leadsto \sf\bold{\red{2^{nd}\: Part =\: 5000}}$

$\therefore$ The two parts of 10000 are 5000 and 5000 .