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Arithmetic I

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Interest & Growth

Interest And Growth

MODULES

Basics & Simple Interest
Advanced Simple Interest
Basics of Compound Interest
Non-Annual Compounding
Present Value & EMI
Growth & CAGR
Common Types
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Interest and growth 1
-/10
Interest and growth 2
-/10

PRACTICE

Interest & Growth : Level 1
Interest & Growth : Level 2
Interest & Growth : Level 3
ALL MODULES

CAT 2025 Lesson : Interest & Growth - Advanced Simple Interest

bookmarked

2.1 Present Value of SI

Amount = Principal + Interest

Therefore, if Amount, say
A\bm{A}A, is provided,

A=p+pnr100A = p + \dfrac{pnr}{100} A=p+100pnr​ ⇒ p×(1+nr100) p \times \left(1 + \dfrac{nr}{100} \right)p×(1+100nr​)

⇒
p=A(1+nr100) p = \dfrac{A}{\left(1 + \dfrac{nr}{100} \right)}p=(1+100nr​)A​

Therefore, where the amount, time period and rate of interest are provided, the principal can be calculated.

Once the first equation (given above) is formed, you'd naturally move to the subsequent steps, as shown in the example below.

Example 4

What investment would result in Rs. 13,44013,44013,440 in 222 years at 6%6 \%6% per annum of simple interest?

Solution

13440=p+p×2×6100=p+3p25=28p2513440 = p + \dfrac{p \times 2 \times 6}{100} = p + \dfrac{3p}{25} = \dfrac{28p}{25}13440=p+100p×2×6​=p+253p​=2528p​

⇒
p=13440×2528=480×25=p = 13440 \times \dfrac{25}{28} = 480 \times 25 =p=13440×2825​=480×25= Rs. 12,00012,00012,000

Alternatively, you can convert the interest from percentage to decimal form.

13440=p+(p×2×0.06)=1.12p13440 = p + (p \times 2 \times 0.06) = 1.12 p13440=p+(p×2×0.06)=1.12p

⇒
p=134401.12=p = \dfrac{13440}{1.12} =p=1.1213440​= Rs. 12,00012,00012,000

Answer: Rs.
12,00012,00012,000

These questions present easy scoring opportunities as the key is forming the correct equation.

Example 5

Jasprit invested Rs. 1,6001,6001,600 for a period of 333 years and Rs. 2,0002,0002,000 for a period of 555 years. Both the deposits yielded simple interest at the same rate. If the total interest received from the two deposits was Rs. 1,1101,1101,110, what was the rate of interest?

Solution

Let the rate of interest be r%r \%r%.

1110=1600×3×r100+2000×5×r1001110 = \dfrac{1600 \times 3 \times r}{100} + \dfrac{2000 \times 5 \times r}{100}1110=1001600×3×r​+1002000×5×r​

⇒
1110=48r+100r 1110 = 48r + 100r1110=48r+100r

⇒
r=1110148=304=7.5 r = \dfrac{1110}{148} = \dfrac{30}{4} = 7.5r=1481110​=430​=7.5

Answer:
7.5%7.5 \%7.5%

2.2 Simple Interest on Multiple Deposits

In these kind of questions, a relationship between the different deposits is provided. As is the case with most of Arithmetic, the key to solving these questions is using the right variables to form the correct equations.

Example 6

Moshe lent to George and Michael sums of money in the ratio of 4:54 : 54:5. If the tenure of these loans were in the ratio of 4:34 : 34:3 and the simple interest earned was in the ratio of 8:38 : 38:3, then what was the ratio of the interest rates?

Solution

As principals, tenures and interest amounts are in ratios, we can use common variables.

Let the principals lent to George and Michael be 4p4p4p and 5p5p5p respectively.
Let the tenure of the loans be
4n4n4n and 3n3n3n respectively.
Let the simple interest earned be
8s8s8s and 3s3s3s respectively.

Let the rate of interest on sums lent to George and Michael be
xxx and yyy respectively.

4p×4n×x1005p×3n×y100=8s3s\dfrac{\dfrac{4p \times 4n \times x}{100}}{\dfrac{5p \times 3n \times y}{100}} = \dfrac{8s}{3s}1005p×3n×y​1004p×4n×x​​=3s8s​

xy=83×1516=52\dfrac{x}{y} = \dfrac{8}{3} \times \dfrac{15}{16} = \dfrac{5}{2}yx​=38​×1615​=25​

Answer:
5:25 : 25:2

Example 7

Jacinda invested a total of Rs. 14,00,00014,00,00014,00,000 in three deposits yielding a simple interest of 6%6 \%6% per annum, such that the interest earned from the first deposit in 888 years equalled the interest earned from the second deposit in 101010 years, which in turn equalled the interest earned from the third deposit in 151515 years. What was the principal amount invested in the second deposit?

Solution

Let aaa, bbb and ccc be the amounts invested in the three deposits respectively.

a+b+c=1400000⟶(1)a + b + c = 1400000 \longrightarrow (1)a+b+c=1400000⟶(1)

Let
KKK be the simple interest earned from each of them from their respective time periods.

a×8×6100=b×10×6100=c×15×6100=K\dfrac{a \times 8 \times 6}{100} = \dfrac{b \times 10 \times 6}{100} = \dfrac{c \times 15 \times 6}{100} = K100a×8×6​=100b×10×6​=100c×15×6​=K

⇒
a=100K48,b=100K60,c=100K90 a = \dfrac{100K}{48}, b = \dfrac{100K}{60}, c = \dfrac{100K}{90}a=48100K​,b=60100K​,c=90100K​

Substituting in (
111),

100K48+100K60+100K90=1400000\dfrac{100K}{48} + \dfrac{100K}{60} + \dfrac{100K}{90} = 140000048100K​+60100K​+90100K​=1400000

⇒
K6×(18+110+15)=14000 \dfrac{K}{6} \times \left(\dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{5} \right) = 140006K​×(81​+101​+51​)=14000

⇒
K×(15+12+8120)=84000 K \times \left(\dfrac{15 + 12 + 8}{120} \right) = 84000K×(12015+12+8​)=84000

K=84000×12035=288000K = 84000 \times \dfrac{120}{35} = 288000K=84000×35120​=288000

b=100×28800060=480000b = \dfrac{100 \times 288000}{60} = 480000b=60100×288000​=480000

Answer: Rs.
4,80,0004,80,0004,80,000

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