Quantitative Aptitude Test - 1

The series follows the following pattern:

11 + 8 = 19 → 2³

19 + 9 = 28 → 3²

28 + 64 = 92 → 4³

92 + 25 = 117 → 5²

117 + 216 = 333 → 6³

∴ The missing term in the series is 117.

Given,

Loan amount taken by Ravi from Baker \(=\mathrm{P}_{1}=\) Rs. \(12500\)

Rate of interest for Ravi \(=\mathrm{R}_{1}=16 \%\)

Time taken by Ravi to repay his loan to Baker \(=\mathrm{t}_{1}=2\) years

Loan amount taken by Baker from Specter \(=\mathrm{P}_{2}=\) Rs. \(10000\)

Rate of interest for Baker \(=\mathrm{R}_{2}=12 \%\)

Time taken by Baker to repay his loan to Specter \(=\mathrm{t}_{2}=3\) years

We know, the formula of compound interest:-

\(\Rightarrow C I=\left[P\left\{\left(1+\frac{R}{100}\right)^{t}-1\right\}\right]\)

Where,

\(\mathrm{CI}=\) Compound interest

\(P=\) Principal

\(\mathrm{R}=\) Rate of interest

\(\mathrm{t}=\) Time period

Let, \(\mathrm{CI}_{1}\) and \(\mathrm{CI}_{2}\) be the compound interests paid by Ravi and Baker respectively.

Compound interest paid by Ravi to Baker:-

\(\Rightarrow \mathrm{CI}_{1}=\left[12500\left\{\left(1+\frac{16}{100}\right)^{2}-1\right\}\right] \)

\(=12500(1.3456-1)=\) Rs. 4,320\)

Compound interest paid by Baker to Specter:-

\(\Rightarrow \mathrm{CI}_{2}=\left[10000\left\{\left(1+\frac{12}{100}\right)^{3}-1\right\}\right]\)

\(=10000(1.40-1)=\) Rs. 4,049.28

Overall gain for Baker \(=\mathrm{CI}_{1}-\mathrm{CI}_{2}=4320-4049.28=\) Rs. 271

\(\therefore\) Overall gain for Baker \(=\) Rs. \(271\)

In order to find out the rate of interest, time for which the sum has been deposited is needed.

But this has provided neither in statement I or statement II.

Statement III is also not an informative statement because it states about a general case which is true for any Sum. This is applicable for Simple Interest illustrated below:-

Suppose x is sum, it will double in 25 years at the rate of 4 % per annum.

\(S t=\frac{(x \times 4 \times 25)}{100}=x\)

Therefore, it doubles in 25 years which is true for any sum.

From statement I,

This statement gives us idea about the possibility of number (average)

The sum of the middle numbers = 128, so possible prime numbers = 59, 61, 67, 71

From statement II,

The sum of the 1st and last no. = 130, which is 59, 61, 67, 71

From statement III,

The difference between 1st and last in the above statements is 12, but there are other numbers also viz. (29, 41), (31, 43) etc.

So, either I or II is sufficient to answer the question. 

Let, B can do the work in x days.

Now after completing \(\frac{1}{4}\) of the work, 

Work remaining \(=\left\{1-\frac{1}{4}\right\}=\frac{3}{4}\) is done by both A and B in 9 days.

\(\frac{9}{20}+\frac{9}{x}=\frac{3}{4}\)

\(⇒\frac{9}{x}=\frac{3}{4}-\frac{9}{20}=\frac{6}{20}\)

\(⇒x=\frac{9 \times 20}{6}=30\)

Given:

Let Rate of Interest \(={R} \%\)

From Statement I, Amount \((A)=P (1+\frac{R}{100})^{t}\)

Compound Interest (C.I.) \(=\) Amount \(-\) Principal

Given, Amount earned at principal Rs. 1000 for 3 years in C.I. is equal to Simple interest earned on principal Rs. 6655 for two years.

\(\Rightarrow\) Amount at C.I. for 3 yrs. = S.I for 2 yrs.

\(\Rightarrow 1000 \times (1+\frac{R}{100})^{3}=\frac{(6655 \times {R} \times 2)}{100}\)

Only \({R}\) is variable in above equation so we can easily find out the value of \({R}\).

\(\Rightarrow R=10 \%\)

So, Statement I is sufficient to answer.

From Statement II, Given, If total Interest earned in 37 years on on Rs. 3000 at simple intrest is Rs. 1110 .

S.I. \(=\frac{(P \times R \times T)}{100}\)

\(\Rightarrow 1110=\frac{(3000 \times {R} \times 37)}{100}\) , \(\Rightarrow {R}=10 \%\)

So, Statement II is sufficient to answer.

From Statement III, Since only Difference between compound interest in \(4^{\text {th }}\) year and \(3^{\text {rd }}\) year is given.

Ther is no mention of Principal(P).

That leaves us with to variables \({R}\) and \({P}\).

So, it is not sufficient to answer.

So, Statement I and II are sufficient to answer the question.

Equation I. 

x² + x – 156 = 0

⇒ x² + 13x – 12x – 156  = 0

⇒ x(x + 13) – 12(x + 13) = 0

⇒ (x + 13) (x – 12) = 0

⇒ x + 13 = 0 or x – 12 = 0

⇒ x = -13 or x = 12

Equation II.

y² + y – 182 = 0

⇒ y² + 14y – 13y – 182 = 0

⇒ y(y + 14) – 13(y + 14) =0

⇒ (y + 14) (y – 13) = 0

⇒ y + 14 = 0 or y – 13 = 0

⇒ y = -14 or y = 13

∴ x > y and x < y the relationship between x and y cannot be established.

According to the given equations:

I. x² - 18x + 77 = 0

⇒ x² - 11x - 7x + 77 = 0

⇒ x(x - 11) - 7(x - 11) = 0

⇒ (x - 11) (x - 7) = 0

⇒ x = 7, 11

II. y² + 3y - 70 = 0

⇒ y² + 10y - 7y - 70 = 0

⇒ y(y + 10) - 7(y + 10) = 0

⇒ (y + 10) (y - 7) = 0

⇒ y = -10, 7

∴ x ≥ y