SSC CGL level Solution Set 50 on Simple Interest and Compound Interest 1

50th SSC CGL level Solution Set, 1st on Simple Interest and Compound Interest problems

SSC CGL solution set 50 simple interest compound interest

This is the 50th solution set of 10 practice problem exercise for SSC CGL exam and 1st on Simple interest and Compound interest problems. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set for extracting maximum benefits from this resource.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts in the topic area
  • is adequately fast in mental math calculation
  • should try to solve each problem using the basic and rich concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use your problem solving abilities to gain an edge in competition.

Before going through the solutions you should go through our sessions on Basic and rich concepts on Simple Interest and Compound Interest for detailed explanation of the efficient Simple Interest and Compound Interest problem solving techniques and related concepts that we will use for this present set of 10 problems.

50th solution set- 10 problems for SSC CGL exam: 1st on Simple interest and Compound interest problems - time 15 mins

Problem 1.

A man borrowed Rs.16000 from two persons. He paid 6% interest per annum to one and 10% interest per annum to the other. In the first year he paid a total interest of Rs.1120. How much did he borrow at each rate?

  1. Rs.12500 : Rs.3500
  2. Rs.11000 : Rs.5000
  3. Rs.12000 : Rs.4000
  4. Rs.10000 : Rs.6000

Solution 1 : Problem analysis

As this problem involves interest on the first 1 year only, the type of interest, that is, simple or compound, does not matter. We can assume the problem as a problem on simple interest only.

Here Rs.16000 is the total sum of money borrowed. If we assume Rs. $x$ as the sum borrowed from the first person at the interest rate of 6%, then sum borrowed from the second person becomes $\text{Rs.}(16000-x)$, which he borrowed at the interest rate of 10%.

Applying the interest rates on the amounts and adding the two amounts of interest we will get the total interest which is Rs.1120. From the single linear equation in one variable it will be a short way off to the solution.

A small but important point is, the percentages need to be converted to equivalent decimals for efficient deduction of the value of $x$.

Solution 1 : Problem solving execution

Interest in 1 year for the money $\text{Rs.}x$ borrowed from the first person at the interest rate 6% is,

$I_1=0.06x$.

Interest in 1 year for the money $\text{Rs.}(16000-x)$ borrowed from the second person at the interest rate 10% is,

$I_2=0.1(16000-x)$.

Sum of these two interests is, Rs.1120. So,

$I_1+I_2=1600-0.04x=1120$,

Or, $0.04x=480$,

Or, $x=12000$, and

$16000-x=4000$.

Answer: Option c: Rs.12000 : Rs.4000

Key concepts used: Basic simple interest concepts -- Basic algebraic concepts.

Problem 2.

A certain sum of money amounts to Rs.1680 in 3 years and to Rs.1800 in 5 years. Find the sum and rate of simple interest.

  1. Rs.1200 : 4%
  2. Rs.1500 : 4%
  3. Rs.1800 : 5%
  4. Rs.1600 : 5%

Solution 2 : Problem analysis and execution

In simple interest calculations, as the interest remains fixed for each year, the difference in amounts of the two years, Rs.120 is the interest for 2 years. So per year interest is,

$Ar=\text{Rs.}60$, where $A$ is the amount and $r$ the interest rate.

Taking the first amount after 3 years,

$1680=A +3Ar=A +180$,

Or, Amount $A=\text{Rs.}1500$, and,

Interest $r=\displaystyle\frac{60}{1500}=4$%.

Answer: Option b : Rs.1500 : 4%.

Key concepts used: Basic simple interest concepts -- Change analysis technique -- Efficient simplification.

Problem 3.

A sum of money at compound interest amounts to Rs.650 at the end of the first year and Rs.676 at the end of the second year. The sum of money is,

  1. Rs.1300
  2. Rs.650
  3. Rs.1250
  4. Rs.625

Solution 3 : Problem analysis and execution

In compounding of interest, interest accrued in a year will be rate of interest applied on the amount in the beginning of the year. So in this case of our problem, interest accrued in the second year is,

$676-650=26=650r$, where $r$ is the rate of interest,

Or, $r=\displaystyle\frac{26}{650}=\frac{1}{25}=4$%.

As amount at the end of first year is 650, if $A$ is the starting amount,

$650=A(1+r)=\displaystyle\frac{26}{25}A$,

Or, $A=25\times{25}=625$.

Answer: Option d: Rs.625.

Key concepts used: Basic compound interest concepts -- Change analysis technique -- Efficient simplification.

Note: We have kept the interest in the form of fraction for ease of calculation. That is efficient simplification.

Problem 4.

A sum of money placed at compound interest doubles itself in 5 years. In how many years it would be 8 times of itself at the same rate of interest?

  1. 10 years
  2. 15 years
  3. 20 years
  4. 7 years

Solution 4 : Problem analysis

The interest rate here is such that a sum of money doubles itself in 5 years. If this doubled sum is invested again in the beginning of the sixth year it will again be doubled, that is, will be 4 times the original sum after another 5 years, that is, a total of 10 years. So after another 5 years, that is, after a total period of 15 years, the original sum will be 8 times of itself.

Alternately, by the principle of compounding, in a period of every 5 years any invested sum is doubled at the rate of interest as stated. So to make itself 8 times, that is, three times doubling, a total of 3 times 5, that is, 15 years will be required.

Mathematically,

$A_5=A(1+r)^5=2A$,

Or, $(1+r)^5=2$.

Assuming that after $x$ years the original amount will make itself 8 times,

$A_x=8A=A(1+r)^x$,

Or, $2^3=(1+r)^x$, substituting for 2,

Or, $(1+r)^{15}=(1+r)^x$.

So $x=15$ years.

Answer: Option b: 15 years.

Key concepts used: Basic compound interest concepts -- conceptual reasoning -- Base equalization technique.

Problem 5.

The compound interest on Rs.2000 in 2 years, if the rate of interest is 4% per annum for the first year and 3% per annum for the second year, will be,

  1. Rs.143.40
  2. Rs.141.40
  3. Rs.140.40
  4. Rs.142.40

Solution 5 : Problem analysis

As the rate of interest is different for the first and second year, we need to enumerate the actual problem state of each year using basic compound interest concepts. This process of enumeration of each year status is a rich compound interest concept, namely, Time breakdown technique.

Solution 5 : Problem solving execution

The interest for the first year will be,

$I_1=Ar=2000\times{0.04}=80$.

And the amount after 1 year,

$A_1=A +Ar=2080$.

Interest at 3% for the second year will be on this amount,

$I_2=A_1r=2080\times{0.03}=62.40$.

So total interest is,

$80+62.40=142.40$.

Answer: Option d: Rs.142.40.

Key concepts used: Basic compound interest concepts -- Rich compound interest concept -- Time breakdown technique -- Enumeration technique.

Problem 6.

An amount of money at compound interest grows up to Rs.3840 in 4 years and up to Rs.3936 in 5 years. Find the rate of interest.

  1. 2%
  2. 2.05%
  3. 3.5%
  4. 2.5%

Solution 6 : Problem analysis and execution

For the 5th year the interest is,

$A_4r=3936-3840=96$, where $A$ is the original amount, $r$ is the rate of interest, and $A_4r$ is the amount after 4 years.

Or, $3840r=96$,

Or, $r=\displaystyle\frac{1}{40}=2.5$%.

Answer: Option d : 2.5%.

Key concepts used: Basic compound interest concepts -- Change analysis technique.

Problem 7.

The time in which Rs.80000 amounts to Rs.92610 at 10% per annum compound interest, interest being compounded semi-annually, is

  1. 1.5 years
  2. 2 years
  3. 3 years
  4. 2.5 years

Solution 7 : Problem analysis and execution

As the annual interest rate is 10% and the compounding is done semi-annually, that is, every 6 months, we can assume the period unit for deductions as 6 months and the interest rate for 1 period unit as 5%.

For the first 6 month period the interest will be,

$I_1=80000\times{0.05}=4000$.

As the final amount of Rs 92610 differs from the initial amount of Rs.80000 by Rs.12610, which is very near to, $3\times{4000}$, the interest that would have accrued if the interest would have been of simple type without compounding. Even at this point examining the choice values we can choose 3 periods of 6 months or 1.5 years as the answer.

Mathematically,

Interest for the first period is Rs. 4000, and the amount at the end of 1st period, Rs.84000. Interest for the second period will be on this amount,

$I_2=84000\times{0.05}=4200$.

The amount at the end of the second period will be, $84000+4200=88200$.

Interest for the 3rd period will be,

$I_3=88200\times{0.05}=4410$, and the total amount, $88200+4410=92610$.

Answer: Option a: 1.5 years.

Key concepts used: Basic compound interest concepts -- Compounding period concept -- Estimation technique -- Free resource use principle -- Time breakdown technique -- Enumeration technique.

Problem 8.

A man borrows Rs.2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly installments. How much will each installment be?

  1. Rs.1352
  2. Rs.1283
  3. Rs.1377
  4. Rs.1275

Solution 8 : Problem analysis and execution

Let us visualize how two equal installments will wipe out the debt at the same time ensure the interest accrual.

At the end of the first year the total amount due will be,

$A_1=2550\left(1+\displaystyle\frac{1}{25}\right)=102\times{26}$, dividing 2550 by 25 was instantaneous and so we did it but left the factor of 26 for later use.

Assuming $x$ is returned back as the first instalment, the amount on which the second year interest accrual would take place will be,

$A_{1n}=(102\times{26})-x$.

The amount at the end of second year will then be,

$A_2=\left(102\times{26}-x\right)\left(1+\displaystyle\frac{1}{25}\right)=x$, as this is the second instalment that has to complete the transaction.

So,

$\left(102\times{26}-x\right)\left(\displaystyle\frac{26}{25}\right)=x$,

Or, $\displaystyle\frac{51}{25}x=26\times{102}\times{\frac{26}{25}}$,

Or, $51x=26\times{102}\times{26}$,

Or, $x=2\times{676}=1352$

Answer: Option a: Rs.1352.

Key concepts used: Installment concept -- Basic compound interest concept-- Delayed evaluation -- Efficient simplification.

Note: As decimal calculation usually gives a bit of trouble, we have kept the interest in fraction form, and did the larger multiplication as late as possible. Though the steps shown are many, actual deduction is inherently simple and quick.

Problem 9.

The difference between the compound interest and the simple interest at the same rate of interest, for the amount Rs.5000 in 2 years, is Rs.32. The rate of interest is,

  1. 5%
  2. 10%
  3. 12%
  4. 8%

Solution 9 : Problem analysis and execution:

By the given condition,

$I_{compound}-I_{simple}=32$,

Or, $(Ar^2+2Ar)-2Ar=32$, where $A$ is the original amount and $r$ is the same rate of interest over 2 years,

Or, $Ar^2=32$,

Or, $5000\times{r^2}=32$,

Or, $r^2=\displaystyle\frac{32}{5000}=\frac{64}{10000}=\left(\frac{8}{100}\right)^2$,

Or, $r=8$%.

Answer: Option d: 8%.

Key concepts used: Basic compound interest concept -- Basic simple interest concept -- Basic algebraic concepts.

Problem 10.

If the compound interest on a sum for 2 years at 12.5% per annum is Rs.510, the simple interest for the same sum at the same rate for the same period of time is,

  1. Rs.480
  2. Rs.450
  3. Rs.400
  4. Rs.460

Solution 10 : Problem analysis and execution:

If $A$ is the amount invested for 2 years and $r$ be the interest, the compound interest is,

$I_{compound}=A(1+r^2)-A=Ar(r+2)$,

Or, $Ar\left(\displaystyle\frac{1}{8}+2\right)=510$, interest 12.5% is equivalent to the fraction $\frac{1}{8}$,

Or, $Ar=\displaystyle\frac{510\times{8}}{17}=240$.

So the simple interest for the same period on the same amount at the same interest rate is,

$I_{simple}=2Ar=480$.

Answer: Option a: Rs.480.

Key concepts used: Basic compound interest concept -- Basic simple interest concept -- Efficient simplification.

Note: To ease calculation we have kept the interest as fraction equivalent to 12.5%. Also knowing beforehand that simple interest per year is $Ar$, we have evaluated $Ar$ itself, not the components, avoiding evaluation of important component $A$ and the final multiplication with $r$. This is efficient simplification. You need to evaluate compound variables consisting of more than one variable, if that is directly used in the final solution. This approach speeds up the solution process considerably.


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Basic and rich concepts on Simple Interest and Compound Interest

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