# Exercise : 5

(b) 10.25%

(c) 10.5%

(d) None of these

**Ans.b**

Let the sum be Rs 100. Then,

S.I. for first 6 months = \(\left ( \frac{100 \times 10 \times 1}{100 \times 2} \right )\) = 5

S.I. for last 6 months = \(\left ( \frac{105 \times 10 \times 1}{100 \times 2} \right )\) = 5.25

So, amount at the end of 1 year = (100 + 5 + 5.25) = 110.25.

∴ Effective rate = (110.25 – 100) = 10.25%.

(b) 7%

(c) 7

^{1}⁄

_{8}%

(d) 10%

(e) None of these

**Ans.d**

Let the rate be R% p.a. Then,

\(\left ( \frac{5000 \times R \times 2}{100} \right ) + \left ( \frac{3000 \times R \times 4}{100} \right ) = 2200\)⇒ 100R + 120R = 2200 ⇒ R = (^{2200}⁄_{220}) = 10.

(b) 4.5%

(c) 5%

(d) 3.46%

(e) None of these

**Ans.d**

Let the original rate be R%. Then, new rate = (2R)%

∴ \(\left ( \frac{725 \times R \times 1}{100} \right ) + \left ( \frac{362.50 \times 2R \times 1}{100 \times 3} \right ) = 33.50\)

⇒ (2175 + 725)R = 33.50 × 100 × 3 = 10050

⇒ R = ^{10050}⁄_{2900} = 3.46%

(b) 0.2%

(c) 0.3%

(d) 0.4%

(e) None of these

**Ans.c**\(\left ( \frac{1500 \times R_{1} \times 3}{100} \right ) - \left ( \frac{1500 \times R_{2} \times 3}{100} \right ) = 13.50\)

⇒ 4500(R_{1} - R_{2}) = 1350 ⇒ R_{1} - R_{2} = ^{1350}⁄_{4500} = 0.3%

(b) 4 : 5

(c) 5 : 2

(d) 5 : 4

(e) None of these

**Ans.b**

Let the savings be X and Y and the rates of simple interest be 5x and 4x respectively.

Then, X × 5x × ^{1}⁄_{2} × ^{1}⁄_{100} = Y × 4x × ^{1}⁄_{2} × ^{1}⁄_{100} or ^{X}⁄_{Y} = ^{4}⁄_{5}

i.e., X : Y = 4 : 5.

(b) 2050

(c) 3000

(d) None of these

(e) 2020

**Ans.d**

Money paid in cash = 1000.

Balance payment = (20000 – 1000) = 19000.

(b) 6500

(c) 7200

(d) 7500

(e) None of these

**Ans.a**

Let the sum invested in Scheme A be x and that in Scheme B be (13900 – x).

Then, \(\left ( \frac{x \times 14 \times 2}{100} \right ) + \left [ \frac{\left ( 13900 - x \right ) \times 11 \times 2}{100} \right ] = 3508\)

⇒ 28x – 22x = 350800 – (13900 × 22) ⇒ 6x = 45000

⇒ x = 7500.

So, sum invested in Scheme B = (13900 – 7500) = 6400.

^{3}⁄

_{4}% , then the amount invested in each share was:

(b) 62,500; 37,500

(c) 72,500: 27,500

(d) 82,500; 17,500

(e) None of these

**Ans.b**

Let the sum invested at 9% be x and that invested at 11% be (100000 – x).

Then,

\(\left ( \frac{x \times 9 \times 1}{100} \right ) + \left [ \frac{\left ( 100000 - x \right ) \times 11 \times 1}{100} \right ]\)= \(\left ( 100000 \times \frac{39}{4} \times \frac{1}{100} \right )\)

⇒ \(\frac{9x + 1100000 - 11x}{100} = \frac{39000}{4} = 9750\)

⇒ 2x = (1100000 – 975000) = 125000 ⇒ x = 62500.

∴ Sum invested at 9% = 62500.

Sum invested at 11% = (100000 – 62500) = 37500.

(b) 6500

(c) 8000

(d) cannot be determined

(e) None of these

**Ans.a**

Let x, y and z be the amounts invested in schemes A, B and C respectively. Then,

\(\left ( \frac{x \times 10 \times 1}{100} \right ) + \left ( \frac{y \times 12 \times 1}{100} \right ) + \left ( \frac{z \times 15 \times 1}{100} \right ) = 3200\)⇒ 10x + 12y + 15z = 320000 ..... (i)

Now, z = 240% of y = ^{12}⁄_{5}y..... (ii)

And, z = 150% of x = ^{3}⁄_{2}x

⇒ x = ^{2}⁄_{3}z = (^{2}⁄_{3} × ^{12}⁄_{5})y = ^{8}⁄_{5}y......(iii)

From (i),(ii) and (iii), we have :

16y + 12y + 36y = 320000 ⇒ 64y = 320000 ⇒ y = 5000.

∴ Sum invested in Scheme B = 5000.

(b) 600

(c) 800

(d) 1200

(e) None of these

**Ans.d**

Let the parts be x, y and [2600 – (x + y)].Then,

\(\frac{x \times 4 \times 1}{100} = \frac{y \times 6 \times 1}{100} = \frac{\left [ 2600 - \left ( x + y \right ) \right ] \times 8 \times 1}{100}\)∴ ^{y}⁄_{x} = ^{4}⁄_{6} = ^{2}⁄_{3} or y = ^{2}⁄_{3}x.

So, \(\frac{x \times 4 \times 1}{100} = \frac{\left [ 2600 - \frac{5}{3}x \right ] \times 8}{100}\)

⇒ 4x = \(\frac{\left ( 7800 - 5x \right ) \times 8}{3} \Rightarrow 52x = \left ( 7800 \times 8 \right )\)

⇒ x = \(\left ( \frac{7800 \times 8}{52} \right ) = 1200.\)

∴ Money invested at 4% = 1200.

(b) 792

(c) 818

(d) 828

(e) None of these

**Ans.d**

Let the parts be x, y and [2379 – (x + y)].

x + (x × 2 × ^{5}⁄_{100}) = y + (y × 3 × ^{5}⁄_{100})

= z + (z × 4 × ^{5}⁄_{100})

⇒ ^{11x}⁄_{10} = ^{23y}⁄_{20} = ^{6z}⁄_{5} = k ⇒ x = ^{10k}⁄_{11}, y = ^{20k}⁄_{23}, z = ^{5k}⁄_{6}

But x + y + z = 2379.

⇒ ^{10k}⁄_{11} + ^{20k}⁄_{23} + ^{5k}⁄_{6} = 2379

⇒ 1380 k + 1320 k + 1256 k = 2379 × 11 × 23 × 6

⇒ k = \(\frac{2379 \times 11 \times 23 \times 6}{3965} = \frac{3 \times 11 \times 23 \times 6}{5}\)

∴ x = \(\left (\frac{10}{11} \times \frac{3 \times 11 \times 23 \times 6}{5} \right ) = 828\)

Hence, the first part is 828.