In compound interest, the interest accrues the initial principal and the accumulated interest from previous periods.
Amount = P[1 + (R/100)]t
Where, P = Principal R = Rate of interest per annum t = Time in years This article contains the MCQs on 'Compound Interest' that are fetched from previous competitive exams like Bank P.O, M.B.A, M.A.T, etc.
Quantitative Aptitude Test - Compound Interest
The test consists of 13 questions on Simple Interest.
No negative marking for this test.
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The pass percentage is 70%
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Complexity Level- Moderate
Given:
The sum = 12000
The time = 2 years
The compound interest for two years = 1996.80
Formula used:
\({CI}\ =\ {P\ × \ [({1\ +\ {R\over100}})^T\ -\ 1]}\)
\({SI}\ =\ {P\ ×\ T\ ×\ R\over 100}\) Where, CI = The compound interest, P = The principle, R = The rate of the interest, SI = The simple interest, and T = The time
Calculation:
Let us assume the rate of interest be R
⇒ According to the question
⇒ \({1996.8}\ =\ {12000\ × \ [({1\ +\ {R\over100}})^2\ -\ 1]}\)
⇒ \({1996.8}\ =\ {12000\ × \ [({100\ +\ R\over100})^2\ -\ 1]}\)
⇒ \({1996.8}\ =\ {12000\over10000} × \ [({100\ +\ R})^2\ -\ 10000]\)
⇒ \({1996.8}\ =\ {1.2} × \ [({100\ +\ R})^2\ -\ 10000]\)
⇒ \({1996.8}\ =\ {1.2} × \ [({100^2\ +\ R^2\ +\ 200R})\ -\ 10000]\)
⇒ \({1996.8}\ =\ {1.2} × \ [{10000\ +\ R^2\ +\ 200R}\ -\ 10000]\)
⇒ \({1996.8}\ =\ {1.2} × \ [{\ R^2\ +\ 200R}]\)
⇒ 1996.8 = 1.2R2 + 240R
⇒ 19968 = 12R2 + 2400R
⇒ 1664 = R2 + 200R
⇒ By solving the quadratic equation
⇒ R = 8, and -208
⇒ The value of R = 8
⇒ The total amount after two years = 12000 + 1996.8 = 13996.8
⇒ The simple interest for third year on 13996.8 at 8% per annum
⇒ According to the question
⇒ \({SI}\ =\ {13996.8\ ×\ 1\ ×\ 8\over 100}\)
⇒ SI = 139.968 × 8 = 1119.744
⇒ The simple interest for the third year = 1119.744
∴ The required result will be 1119.74.
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Correct Answer: C
Solution :
\[CI=P\left[ {{\left( 1+\frac{R}{100} \right)}^{T}}-1 \right]\] \[=12000\left[ {{\left( 1+\frac{4}{100} \right)}^{2}}-1 \right]\] \[=12000\left[ {{\left( \frac{26}{25} \right)}^{2}}-1 \right]\] \[=12000\left( \frac{676}{625}-1 \right)\] \[=\frac{12000\times 51}{625}=Rs.979.2\]
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Page 2
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Correct Answer: B
Solution :
Average speed of truck \[\text{=}\frac{\text{Distance covered}}{\text{Time}\,\text{taken}}\] \[\text{=}\frac{448}{8}=56\]kmph \[\therefore \]Average speed of bicycle \[=\frac{1}{4}\times 56=14\]kmph \[\therefore \]Distance covered by the bicycle in 7 hours = Speed \[\times \]Time \[=14\times 7=98km\]
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Page 3
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Correct Answer: A
Solution :
Total marks obtained-by Bina \[=110+135+140+120+115\] \[=620\] \[\therefore \]Required percentage of marks obtained \[=\frac{620}{750}\times 100\approx 83\]
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Page 4
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Correct Answer: C
Solution :
Third number \[=5\times 40.8-2\times 46-2\times 37\] \[=204-92-74=38\]
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Page 5
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Correct Answer: E
Solution :
Required difference =(28-22)% of 3400 \[=\frac{3400\times 6}{100}=204\]
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Page 6
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Correct Answer: C
Solution :
Number of copies sold in city B and city ? =(32+22)% of 3400 \[=\frac{3400\times 54}{100}=1836\]
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Page 7
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Correct Answer: B
Solution :
Number of remaining copies =( 100 - 28 - 32 - 22)% of 3400 \[=\frac{3400\times 18}{100}=612\]
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Page 8
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Correct Answer: E
Solution :
\[?\approx 0.5\times 10\times 53\approx 265\]
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Page 9
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Correct Answer: E
Solution :
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Page 10
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Correct Answer: C
Solution :
\[?=\sqrt{4590}\approx 68\]
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Page 11
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Correct Answer: D
Solution :
\[?\approx 16\times 15\times 20\approx 4800\]
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Page 12
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Correct Answer: C
Solution :
\[?\approx \frac{4005}{20}\times 2\approx 400.5\] \[\therefore \]Required answer = 400
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Page 13
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Correct Answer: A
Solution :
The pattern of the number series is: \[17+{{\text{1}}^{2}}=17+1=18\] \[18+22=18+4=22\] \[22+{{3}^{2}}=22+9=31\] \[31+42=31+16=47\] \[47+{{5}^{2}}=47+25=72\]
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Page 14
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Correct Answer: D
Solution :
The pattern of the number series is: \[13+1=14\] \[14+3=17\] \[17+5=22\] \[22+7=29\] \[29+9=38\]
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Page 15
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Correct Answer: A
Solution :
The pattern of the number series is: \[6+2=8\] \[8-4=4\] \[4+8=12\] \[12-16=-4\] \[4+32=28\]
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Page 16
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Correct Answer: B
Solution :
The pattern of the number series Is: \[222-24=198\] \[198-20=178\] \[178-16=162\] \[162-12=150\] \[150-8=142\]
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Page 17
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Correct Answer: C
Solution :
The pattern of the number series is: \[53\times 6=318\] \[318\times 5=1590\] \[1590\times 4=6360\] \[6360\times 3=19080\] \[19080\times 2=38160\]
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Page 18
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Correct Answer: A
Solution :
Some bottles are papers.(i) All stars are bottles\[\to \]Universal Affirmative (A-type). (ii) Some bottles are papers\[\to \]Particular Affirmative (I-type). (iii) No paper is calendar\[\to \]Universal Negative (E-type). (iv) Some papers are not calendars \[\to \]Particular Negative (O- type). No paper is a calendar. \[I+E\Rightarrow \]O-type of Conclusion "Some bottles are not calendars** Venn Diagrams All stars are bottles.Some bottles are papers.No paper is calendar.We have derived the Conclusion "Some bottles are not calendars". It may be represented by Venn- Diagrams as:Therefore, Conclusion I follows.
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Page 19
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Correct Answer: A
Solution :
Some bottles are papers.(i) All stars are bottles\[\to \]Universal Affirmative (A-type). (ii) Some bottles are papers\[\to \]Particular Affirmative (I-type). (iii) No paper is calendar\[\to \]Universal Negative (E-type). (iv) Some papers are not calendars \[\to \]Particular Negative (O- type). No paper is a calendar. \[I+E\Rightarrow \]O-type of Conclusion "Some bottles are not calendars** Venn Diagrams All stars are bottles.Some bottles are papers.No paper is calendar.Combine figures andThus. Conclusion I follows.
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Page 20
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Correct Answer: B
Solution :
Some bottles are papers.(i) All stars are bottles\[\to \]Universal Affirmative (A-type). (ii) Some bottles are papers\[\to \]Particular Affirmative (I-type). (iii) No paper is calendar\[\to \]Universal Negative (E-type). (iv) Some papers are not calendars \[\to \]Particular Negative (O- type). No paper is a calendar. \[I+E\Rightarrow \]O-type of Conclusion "Some bottles are not calendars** Venn Diagrams All stars are bottles.Some bottles are papers.No paper is calendar.Conclusion II is Converse of the first Premise.
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Page 21
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Correct Answer: E
Solution :
Some pencils are blankets.(i) All stars are bottles\[\to \]Universal Affirmative (A-type). (ii) Some bottles are papers\[\to \]Particular Affirmative (I-type). (iii) No paper is calendar\[\to \]Universal Negative (E-type). (iv) Some papers are not calendars \[\to \]Particular Negative (O- type). All blankets are erasers. \[I+A\Rightarrow \]I-type of Conclusion "Some pencils are erasers.?? We have derived the Conclusion : "Some pencils are erasers.?? It may be illustrated by Venn Diagrams as :Conclusion I is derived Conclusion. Conclusion II also follows.
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Page 22
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Correct Answer: B
Solution :
Some pencils are blankets.(i) All stars are bottles\[\to \]Universal Affirmative (A-type). (ii) Some bottles are papers\[\to \]Particular Affirmative (I-type). (iii) No paper is calendar\[\to \]Universal Negative (E-type). (iv) Some papers are not calendars \[\to \]Particular Negative (O- type). All blankets are erasers. \[I+A\Rightarrow \]I-type of Conclusion "Some pencils are erasers.?? First Premise may be illustrated as :Conclusion II follows.
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Page 23
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Correct Answer: E
Solution :
From statements I and I in
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Correct Answer: E
Solution :
From statements I and II \[\begin{matrix} R & > & P & > & Q \\ \end{matrix}\] \[\underset{S,T}{\longleftrightarrow}\]
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Page 25
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Correct Answer: E
Solution :
From statements I and II Suman scored 10 to 20 marks
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Correct Answer: B
Solution :
From statements I and II Rahul went for a vacation II January. February. March April. May or June. From statement II Rahul went for a vacation II April.
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