General Aptitude Test 1

The correct answer is option 2 i.e. suffers.

Explanation      

The given sentence is a zero conditional sentence. 

• In the case of zero conditional sentences, we use the simple present tense in both the clauses.
• The only option in which the verb is in the simple present tense is option 2 i.e. 'suffers'

The correct answer is option 3 i.e. good enough.

Explanation 

• When 'enough' is used with an adjective, the adjective comes before 'enough'.
• The use of 'as' will be incorrect in this sentence.
• Hence, only option 3 is correct.

The correct answer is option 2 i.e. tree.

Explanation   

Let us find the relation between the first two words-

• A baby grows into an 'adult'.
• Now, we need to find a word which is related in a similar way to 'sapling'. A sapling is a young tree.
• Clearly, 'tree' is correct as 'a sapling grows into a tree'.

Apprise: inform or tellAppraise: evaluate, judge the quality ofContrast: compareContract (verb): shrink

The usage of 'apprises' in the first blank will render the sentence redundant as what is being informed/ told has not been mentioned. However, 'appraises' fits perfectly in the blank space. Similarly, the usage of 'contracts' in the second blank will also render the sentence meaningless as recent films cannot be shrunk with independent films. The writer may compare the former with the latter. Thus option 4 has the appropriate pair of words.

Since the sentence is comparing fruits bought today and yesterday, the comparative degree of the adjectives must be used. The comparative degree uses the word ‘than’ to separate the two entities being compared.

The comparative degree of ‘fresh’ is ‘fresher’.

Thus, option 3 is the correct answer.

Explanation:

All ratios are equal as given in the question.

Let’s assume that they are equal to k

Now,

log x = k(l + m – 2n)

log y = k (m + n – 2l)

log z = k(n + l – 2m)

Now,

log xyz = k(l + m – 2n + m + n – 2l + n + l – 2m)

log xyz = k(2l – 2l + 2n – 2n + 2m – 2m)

log xyz = k(0)

xyz = e⁰

∴ xyz = 1

Explanation:

Let the cars meet up after x hours (i.e.) distances travelled are the same because they are starting from the same point.

\(12x = \left[ {2 \times 6 + \left( {x - 1} \right)} \right] \times \frac{x}{2}\)

12 + x – 1 = 24

∴ x = 13

∴ Distance travelled = 12x = 12 × 13

Explanation:

Let \(x = \sqrt {156 + \sqrt {156 + \ldots \infty } } \) 

x² = 156 + x ⇒ x² – x – 156 = 0

x² – 13x + 12x – 156 = 0

x(x – 13) + 12(x - 13) = 0

(x + 12)(x - 13) = 0

∴ x = -12 and 13

Data:

x + 2y = 30  (1)

Find

\((\frac{2y}{5} + \frac{x}{3}) +(\frac{x}{5} + \frac{2y}{3})\)

Calculation:

Let y = 0 and x = 30

Since it satisfies equation (1)

Hence

\((\frac{2y}{5} + \frac{x}{3}) +(\frac{x}{5} + \frac{2y}{3})\)

\(= (\frac{2\times 0}{5} + \frac{30}{3}) +(\frac{30}{5} + \frac{2\times 0}{3})\)

= 0 + 10 + 6 + 0 = 16

. If x + 2y = 30, then  \((\frac{2y}{5} + \frac{x}{3}) +(\frac{x}{5} + \frac{2y}{3})\) will be equal  is 16

Assume the total number of students is 100:

• 60% of the students are boys, so:
  • Boys: 60
  • Girls: 40
• Number of girls who scored more than 40 marks:
  • 80% of girls = 80% of 40 = 32
• Number of students who scored more than 40 marks:
  • 60% of total students = 60
• Therefore, the number of boys who scored more than 40 marks:
  • 60 - 32 = 28
• Number of boys who scored 40 marks or less:
  • Total boys - Boys (scored more) = 60 - 28 = 32
• Fraction of boys who scored 40 marks or less:
  • 32/60 = 8/15

Let the total number of voters be x.

People who voted for the winner: 0.47x

People who voted for the loser: 0.47x - 308

People who cast blank votes: 60

People who did not vote: 0.1x

We can set up the following equation:

0.47x + (0.47x - 308) + 60 + 0.1x = x

• Combine like terms:
• 0.47x + 0.47x - 308 + 60 + 0.1x = x
• 1.04x - 248 = x

Rearranging gives:

• 1.04x - x = 248
• 0.04x = 248
• x = 6200

Thus, the total number of voters on the list is 6200.

The correct answer is option 4.

Explanation      

• The first event that occurred was the 'seven years war' which was a war between the British and the French.
• Following this, the Britishers were in debt.
• Due to this reason, the British demanded more from the American colonies.
• Last event was 'the Royal Proclamation'.

• The number affected by famine always increases with the population. Therefore, if the population increased in 2005, then the number of those affected by famine also increased.
• If there was an increase in 2005, there must have been more people affected in that year than the previous year of 2004.
• The available information does not allow us to draw any of the other inferences.
• The number of those affected by famine could increase without a corresponding percentage of the population increase.
• Neither can we draw inferences about any particular year between 2000 and 2010.

Note:

Option 2 cannot be correct because considering the case where in 2000 the population is 100, and people affected are 50 in 2001 the population grew to say 200, and people affected is 100.

So in both cases,the percentage of people affected is 50% but the number of people affected grew.

Calculation:

Effective plan area of the room = {(15 m × 12 m) – (3 m × 2 m)}

Total volume of the room neglecting the area of ditch = {(15 m × 12 m) – (3 m × 2 m)} × 8 m = 1392 m³

Total volume of one box of chocolate = 1 m × 1 m × 0.5 m = 0.5 m³

Consider three different one-digit natural numbers as x, y and z

Now, different two-digit number possible with these one digit numbers are:

1)10x + y

2)10y + x

3) 10x + z

4) 10z + x

5) 10y + z

6) 10z + y

Sum of all these two-digit numbers = (10x + y) + (10y + x) + (10x + z) + (10z + x) + (10y + z) + (10z + y)

= 10x + y + 10y + x + 10x + z + 10z + x + 10y+ z + 10z + y

= 20x + 20 y + 20 z + 2x + 2y + 2z

= 20 (x + y + z) + 2 (x + y + z)

= 22(x + y + z)

Now divide this with sum of one-digit numbers, that is, x, y and z

\({\rm{Result}} = \frac{{22\left( {{\rm{x}} + {\rm{y}} + {\rm{x}}} \right)}}{{{\rm{x}} + {\rm{y}} + {\rm{z}}}} = 22\)

Initial working hours = (6 × 3.5) + (5 × 3) = 36 hours

Let hourly wages are of Rs. 1 and the production in one hour is 1 unit.

Total money they are getting initially = Rs. 36

Total initial production = 36 units

New Policy:

Now for new policy salary increment in weekly wages is of 8%

Total money they are getting = 36 × 0.08 = Rs. 38.88

Total working hours = 7.5 × 5 = Rs. 37.5

As per old policy management has to pay = 37.5 × 1 = Rs. 37.5

But they are paying 38.88

As per old policy, the production in 37.5 hours = 37.5 units

But they are paying Rs.38.88. So, the expected production = 38.88 units

Assume Deepak’s salary =10000
original hike(50%) amount = 5000 ; Revised salary =15000

Wrongly typed(80%) hike amount = 8000
Diff = 3000; For three months = 9000

Fourth Month Salary = 15000-9000=6000
15000*x/100 = 6000 => x=40%

Differentiation gives the rate of change of a function.

f₁(n) = n¹⁰⁰, f₂(n) = (1.2)ⁿ

By differentiating with respect to n, 100 times, we will get a constant number, 100!, for n¹⁰⁰ but for 1.2ⁿ we will get n (n – 1)(n – 2) … (n – 99) 1.2^{(n – 100)} which is ever increasing with n.

So, f₂(n) is greater than f₁(n).

f₂(n) = (1.2)ⁿ

f₃(n) = 2^n/2 = (√2)ⁿ = (1.414)ⁿ

f₄(n) = 3^n/3 = \({\left( {\sqrt[3]{3}} \right)^n}\) = (1.44)ⁿ

Let the price of two articles are 3X and 4X.
After increment the ratio will be:
110% of 3x/(4X+4) = 3/4
x=10
Thus the CP of second article = 4X = 4*10 = Rs. 40.

A population of 2500 persons requires a minimum area of 3000 m² for primary schools.

Minimum area required for one person = 3000/2500 = 1.2 m²

From the above table, it is clear that Sector 3 is having maximum shortage of school area per person.

Assuming a total of 100 students in the school:

• The ratio of boys to girls is 3:2, which means:
  • Boys: 60
  • Girls: 40
• 20% of boys receive scholarships:
  • 60 boys × 20% = 12 boys
• 25% of girls receive scholarships:
  • 40 girls × 25% = 10 girls
• Total students who receive scholarships:
  • 12 boys + 10 girls = 22 students
• Percentage of students who do not receive scholarships:
  • 100% - (22 students / 100 students × 100%) = 78%

Total amount = x
Savings(%)
[100 – (23 + 33 + 19 + 16 )]% = 9 %
9% of x = 504
=> x = 504 * 100/9 = 5600
Amount spend on food and insurance policy together = 56% of 5600 = Rs.3136