4GMAT Diagnostic test Q3 - Problem Solving - Word Problems : Algebraic Factorization
4GMAT Diagnostic Test Q8 - Problem Solving : Simple and Compound Interest
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GMAT QUANTITATIVE REASONING SIMPLE & COMPOUND INTEREST PROBLEM SOLVING Diagnostic Test
Transcript of 4GMAT Diagnostic Test Q8 - Problem Solving : Simple and Compound Interest
GMAT QUANTITATIVE REASONING
SIMPLE & COMPOUND
INTEREST
PROBLEM SOLVING
Diagnostic Test
Question
Robin invested $1000 in a 12% simple interest savings deposit for 3years. He also invested an equal amount in a 10% compound interestsavings deposit for 3 years. At the end of 3 years, how much moreinterest did he get from the simple interest deposit?
A. $31
B. $60
C. $39
D. $29
E. $390
Part 1
Simple & Compound Interest Formulae
Formulae recap
Simple Interest
Formulae recap
Simple Interest
Pnr
100Simple Interest, SI =
Formulae recap
Simple Interest
Pnr
100Simple Interest, SI =
P – Principal; n – number of years;
r – rate of interest % p.a.
Formulae recap
Simple Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Formulae recap
Simple Interest Compound Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Formulae recap
Simple Interest Compound Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Amount A = P 1+r
100
n
Formulae recap
Simple Interest Compound Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Amount A = P 1+r
100
n
P – Principal; n – number of years;
r – rate of interest % p.a.
Formulae recap
Simple Interest Compound Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Amount A = P 1+r
100
n
P – Principal; n – number of years;
r – rate of interest % p.a.
Compound Interest CI = A - P
Formulae recap
Simple Interest Compound Interest
Pnr
100Simple Interest, SI =
Amount accrued A = P + SI
P – Principal; n – number of years;
r – rate of interest % p.a.
Amount A = P 1+r
100
n
P – Principal; n – number of years;
r – rate of interest % p.a.
Compound Interest CI = A - P
Note
In simple interest, the formula given computes the simple interest.In compound interest, the formula given computes the amount accrued.
Part 2
Compute simple and compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
$1000 in a 12% simple interest savings deposit for 3 years
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
$1000 in a 12% simple interest savings deposit for 3 yearsPnr
100Simple Interest, SI =
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
$1000 in a 12% simple interest savings deposit for 3 yearsPnr
100Simple Interest, SI =
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
Simple Interest, SI =
$1000 in a 12% simple interest savings deposit for 3 yearsPnr
100Simple Interest, SI =
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
Simple Interest, SI = 1000×3×12
100
$1000 in a 12% simple interest savings deposit for 3 yearsPnr
100Simple Interest, SI =
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
Simple Interest, SI = 1000×3×12
100= $360
$1000 in a 12% simple interest savings deposit for 3 yearsPnr
100Simple Interest, SI =
$360
Simple Interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 1: Compute simple interest
Simple Interest, SI = 1000×3×12
100= $360
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
$1000 in a 10% compound interest savings deposit for 3 years
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A =
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
= 1000110
100
3
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
= 1000110
100
3
= $1331
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
= 1000110
100
3
= $1331
Compound Interest CI = A - P
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
= 1000110
100
3
= $1331
Compound Interest CI = A - P = 1331 - 1000 = $331
$1000 in a 10% compound interest savings deposit for 3 years
Amount A = P 1+r
100
n
$331
Compound Interest
Step 2: Compute compound interest
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Amount A = 1000 1+10
100
3
= 1000100+10
100
3
= 1000110
100
3
= $1331
Compound Interest CI = A - P = 1331 - 1000 = $331
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 3: Compute the difference
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 3: Compute the difference
01 Simple interest = $360
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 3: Compute the difference
01 Simple interest = $360 02 Compound interest = $331
Difference = 360 – 331 = $29
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 3: Compute the difference
01 Simple interest = $360 02 Compound interest = $331
Correct Answer choice D.
Difference = 360 – 331 = $29
At the end of 3 years, how much more interest did he
get from the simple interest deposit?
Step 3: Compute the difference
01 Simple interest = $360 02 Compound interest = $331
Alternative Method to compute CI
Do not use formula
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1
Principal for 1st year =
$1000
Rate of interest is10%
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Principal for 3nd year =
$1210
Rate of interest is10%
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Principal for 3nd year =
$1210
Rate of interest is10%
Interest for year 3 =
10% of 1210 = $121
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Principal for 3nd year =
$1210
Rate of interest is10%
Interest for year 3 =
10% of 1210 = $121
Amount at the end of year 2
= 1210 + 121 = $1331
Computing iteratively$1000 @ 10% p.a. compound interest for 3 years
I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3
Principal for 1st year =
$1000
Rate of interest is10%
Interest for year 1 =
10% of 1000 = $100
Amount at the end of year 1
= 1000 + 100 = $1100
Principal for year 2 =
Amount end of year 1
Principal for 2nd year =
$1100
Rate of interest is10%
Interest for year 2 =
10% of 1100 = $110
Amount at the end of year 2
= 1100 + 110 = $1210
Principal for year 3 =
Amount end of year 2
Principal for 3nd year =
$1210
Rate of interest is10%
Interest for year 3 =
10% of 1210 = $121
Amount at the end of year 2
= 1210 + 121 = $1331
CI = 1331 – 1000 = $331
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