Quantitative Aptitude Test - 13

Required value \(=(800-96+420)=284\)

∴ Total male engineers from both localities are 284 less than the total population of Gomti Nagar.

Required ratio = 96 : 300 = 8 : 25

∴ The ratio of male engineers from Gomti Nagar to female engineers from Indira Nagar is 8 : 25.

Required percentage = \(\frac{(1200 - 144 + 300)}{1200}\) × 100 = 63% less.

∴  Total number of female engineers from both localities is 63% less than total population of Indira Nagar.

Statement I:

According to problem,

⇒ X = 1.4Y ......(i)

We do not get the difference between X and Y from this.

Statement II:

According to problem,

\(\Rightarrow Y=Z-Z \times \frac{25}{100}\)

\(\Rightarrow Y=0.75 Z\)

\(\Rightarrow Z=\frac{Y}{0.75}\)

\(\Rightarrow Z=\frac{4 Y}{3}\) .....(ii)

We do not get the difference between X and Y from this.

Statement III:

According to problem,

\(\Rightarrow X+Z=82\) ....(iii)

We do not get the difference between \(X\) and \(Y\) from this.

Combining (i), (ii) and (iii) we get,

\(\Rightarrow 1.4 Y+\frac{4 Y}{3}=82\)

\(\Rightarrow Y=30\)

\(\therefore X=1.4 \times 30\)

\(\Rightarrow X=42\)

\(\therefore\) Difference between \(X\) and \(Y=42-30=12\)

\(\therefore\) All I, II and III are required.

Number of students from chemistry department from college A =75

Number of students from chemistry department from college B = 95

Number of students from chemistry department from college C = 70

Number of students from chemistry department from college D = 70

Total number of students = (75 + 95 + 70 + 70)=310

Number of students from physics department from college \(C\)\( =90\)

Number of students from chemistry department from college \(C\)\( =70\)

Number of students from math department from college \(C\)\( =85\)

Number of students from physics department from college \(D\)\( =60\)

Number of students from chemistry department from college \(D\)\( =70\)

Number of students from math department from college \(D\)\( =95\)

Total number of students from college \(C\)\(=(90+70+85)\)\( =245\)

Total number of students from college \(D\)\(=(60+70+95)\)\( =225\)

Now, the total number of students from college \(C\) is more than that of college \(D\).

Difference between them\(=(245-225)\)\( =20\)

So, Required percentage\(=\) \(\left(\frac{20}{225} \times 100\right) \%\)\(=\left(\frac{80}{9}\right)\)

\(=8.88 \% \approx\) 9%

\(\therefore\) The students from college \(C\) are 9% more than that of college \(D\).

Number of students from chemistry department from college B = 95

Number of students from chemistry department from college C = 70

Number of students from physics department from college A = 50

Number of students from physics department from college D = 60

Total students from the chemistry department from both college B and college C = (95 + 70)= 165

Total students from the physics department from both colleges A and college D = (50 + 60)= 110

So, the difference between the students from two departments = (165 - 110)= 55

Number of students from physics department from college \(A\) = \(50\)

Number of students from physics department from college \(B\) = \(80\)

Number of students from physics department from college \(C\) = \(90\)

Number of students from physics department from college \(D\) = \(60\)

Average= \(\frac{\text { Total sum of all numbers }}{\text { Total number of the item in the set }}\)

Average of all students from the physics department of all four colleges= \(\frac{50+80+90+60}{4}\)= \(\frac{280}{4}\)= 70

Students from the physics department from college A= 50

Students from the math department from college A= 65

Students from chemistry department from college C=70

Students from the math department from college C= 85

Total number of students from physics and math department from college A= (50 + 65)= 115

Total number of students from chemistry and math department from college C = (70 + 85)= 155

Now, the ratio between them = 115 : 155= 23 : 31

Given:

The ratio of the investment \(=3: 6: 8\)

Time period \(=2\) years

Statement (I):

The average amount of profit earned by Tom, Jerry, and BOB = Rs.17000

Total profit of Tom, Jerry and BOB \(=17000 \times 3=\) Rs. 51000

Ratio of the investment \(=3: 6: 8\)

So, Profit shared by Jerry \(=\frac{51,000}{17} \times 6=18000\)

\(\therefore\) Profit earned by Jerry \(=\) Rs. 18000

Statement (II):

The average of investment by all three \(=\) Rs. 34000

Total investment \(=34000 \times 3=\) Rs. 102000

Profit earned at the end of 2 years is \(\frac{1}{2}\) of total investment.

Total profit \(=102000 \times \frac{1}{2}=\) Rs. 51000

Ratio of the investment \(=3: 6: 8\)

So, Profit shared by Jerry \(=\frac{51000}{17} \times 6=18000\)

\(\therefore\) Profit earned by Jerry \(=\) Rs. 18000

Thus, either statement I alone or statement II alone is sufficient to answer the question.