# Exercise : 1

(b) 10,000

(c) 12,000

(d) 9,000

(e) None of these

**Ans.a**

Let the Principal = P

Then \(\frac{P \times 8 \times 4}{100} + \frac{P \times 10 \times 6}{100} + \frac{P \times 12 \times 5}{100}\)

= 12160

⇒ 152P = 12160 × 100

or \(\frac{12160 \times 100}{152} = 8000\)

(b) 8032.50

(c) 4462.50

(d) 8900

(e) None of these

**Ans.a**

Let the sums be P.

Now, 45% of P = 4016.25

or, P = 8925

(b) 1004

(c) 1028

(d) Data inadequate

(e) None of these

**Ans.c**

Rate of interest = \(\frac{956 - 800}{3 \times 800} \times 100 = 6.50 \%\)

\Amount = 800 + \(\frac{800 \times 9.5 \times 3}{100}\)

= 800 + 228 = 1028

(b) 8 p.c.p.a.

(c) 9 p.c.p.a.

(d) Data inadequate

(e) None of these

**Ans.b**

Time = \(\frac{900 \times 100}{3000 \times 6}\) = 5 years

Rate = \(\frac{1600 \times 100}{5 \times 4000}\) = 8%

(b) 14,000

(c) 18,000

(d) 12,000

(e) None of these

**Ans.d**

We have, SI = \(\frac{p \times r \times t}{100}\)

∴ 11400 = \(\frac{p \times 6 \times 2}{100} + \frac{p \times 9 \times 3}{100} + \frac{p \times 14 \times 4}{100}\)

or,12p + 27p + 56p = 11400 × 100

or, 95p = 11400 × 100

∴ p = 12000

(b) 350

(c) 245

(d) Cannot be determined

(e) None of these

**Ans.d**

Let p and r be the principal amount and rate of interest respectively.

Then, \(\frac{p \times r \times 7}{100} = 1750\)

or, pr = 25000

Now, SI = \(\frac{p \times \left(r + 2 \right) \times 7}{100} = 1750\)

We have to find the value of

\(\frac{p \times \left(r + 2 \right) \times 7}{100} - \frac{p \times r \times 7}{100}\) = M - 1750M = SI when the rate of interest is 2% more. When we solve this equation, we find that we have two variables and one equation. Therefore, can’t be determined the correct answer.

(b) 42

(c) 62

(d) 20

(e) None of these

**Ans.c**

For 3 years:

\(Diff. = \frac{Sum \times \left(rate \right)^{2} \left(300 + rate \right)}{\left(100 \right)^{3}}\)= \(\frac{2000 \times 10 \times 10 \times 310}{100 \times 100 \times 100}\) = 62

(b) 5000

(c) 6500

(d) Cannot be determined

(e) None of these

**Ans.b**

Ratio of Nikhilesh’s investments in different schemes

= \(100 : \frac{150 \times 100}{240} : 150 = 8 : 5 : 12\)

Now, according to the question,

\(\frac{8k \times 10}{100} + \frac{5k \times 12}{100} + \frac{12k \times 15}{100}\) = 3200or, 80k + 60k + 180k = 3200 × 100

or, 320k = 3200 × 100

or, k = 1000

∴ amount invested in scheme B will be = 1000 × 5 = 5000

(b) 18000

(c) 15000

(d) Data inadequate

(e) None of these

**Ans.e**

Let the amount deposited at the rate of 15% per annum be x.

15% of x + 18% of (25000 – x) = 4050

or, 15% of x + 18% of 25000 – 18% of x = 4050

or, 3% of x = 4500 – 4050 = 450 ⇒ x = 15000

\ Amount deposited at 18% = (25000 – 15000 ) = 10000

(b) 24000

(c) 25000

(d) Data inadequate

(e) None of these

**Ans.e**\(\frac{30p}{100} + 450 = p + \frac{15}{100} - p\)

⇒ p = 20,000.