Quantitative Aptitude PYQs

Total cost price of milk = 28 × 8.5 = Rs. 238

Gain of milk vendor = 12.5% of 238 = Rs. 29.75

∵ Selling price = Cost price + Profit

⇒ Total selling price of milk & water mixture = 238 + 29.75 = Rs. 267.75

Now, let the milk vendor added ‘x’ litres of water to the milk

∵ The milk vendor sold the mixture at the same rate,

⇒ Total selling price of milk & water mixture = (28 + x) × 8.5 = 267.76

⇒ 28 + x = 267.75/8.5

⇒ 28 + x = 31.5

⇒ x = 31.5 – 28 = 3.5 litres

∴ The milk vendor added 3.5 litres of water to the milk

As we know, the minute and hour hand of a correct clock meet after every 720/11 minutes

In the given case, the minute and hour hand are meeting after every 72 minutes

This means that the clock is running slow and losing time

Time the clock loses after every 72 minutes = 72 – 720/11 = 72/11 min.

Time the clock loses after every hour = 72/11 × 60/72 = 60/11 min.

∵ A day has 24 hours,

∴ Time the clock loses in a day = 24 × 60/11 = 1440/11 min. = \(130\frac{{10}}{{11}}\) minutes

Error: In the official question paper, the question is incorrect. We have made changes in the given options and have provided the answer accordingly.

Let the owner meet the thief ‘x’ km from his house

Speed of thief = 40 kmph

⇒ Time taken by the thief to cover ‘x’ km = x/40 hrs.

Speed of owner = 60 kmph

⇒ Time taken by the owner to cover ‘x’ km = x/60 hrs.

But, the owner started half hour after the theft,

⇒ x/40 = x/60 + 1/2

⇒ 60x = 40x + 1200

⇒ 20x = 1200

⇒ x = 1200/20 = 60 km

⇒ Time elapsed after theft = x/40 = 60/40 = 1.5 hrs.

∴ The owner met the thief 60 km from his house and 1.5 hours after the theft

Given:

X and Y together can finish a job in 6 days.

X can alone do the same job in 12 days. 

Calculation:

(X + Y), 1  day work = 1/6

⇒ X, 1  day work = 1/12

⇒ Y, 1 day work = 1/6 – 1/12 = 1/12

∴ Y will alone do the same job in 12 days

Given:

12 persons can paint 10 identical rooms in 16 days.

Calculations:

To paint 10 rooms, 12 persons take = 16 days

To paint 1 room, 12 persons will take = 16/10 = 8/5 days

To paint 1 room, 1 person will take = 12 × 8/5 = 96/5 days

To paint 1 room, 8 persons will take = 1/8 × 96/5 = 12/5 days

To paint 20 rooms, 8 persons will take = 20 × 12/5 = 48 days

∴ 8 persons will take 48 days to paint 20 rooms.

According to the logarithmic rules,

⇒ log_a xy = log_a x + log_b x

⇒ log_a xⁿ = n log_a x

⇒ log₁₀ 10 = 1

Let x = (0.2)²⁵,

⇒ log₁₀ x = log₁₀ (0.2)²⁵

⇒ log₁₀ x = 25 log₁₀ (0.2)

⇒ log₁₀ x = 25 log₁₀ (2 × 10^{– 1})

⇒ log₁₀ x = 25 [log₁₀ 2 – log₁₀ 10]

⇒ log₁₀ x = 25 [0.30103 – 1]

⇒ log₁₀ x = 25 × – 0.69897

⇒ log₁₀ x = – 17.47425

Its characteristics is 17.

∴ No. of zeros after the decimal point in the value of (0.2)²⁵ = n = 17

Calculation:

Let the ages of the father & son be ‘x’ years and ‘y’ years

respecthe tively

So, ratio of the sum & difference between their ages

⇒  (x + y)/(x – y) = 11/3

⇒ 3x + 3y = 11x – 11y

⇒ 8x = 14y

⇒ x/y = 7/4             ----(1)

⇒ Ratio of their current ages = 7 ∶ 4

Clearly, Statement 1 is incorrect.

Now, a son will attain twice his present age after ‘y’ years.

The father and son's ages after y year are (x + y) & (y + y)

Hence, the ratio of father and son's present age

⇒ Ratio of their ages after ‘y’ years = (x + y)/(y + y)

⇒ (x/2y) + (y/2y)

⇒ (1/2)[x/y + 1]

From equation (1)

⇒(1/2)[7/4 + 1] = 11/8

⇒ (x + y)/(2y) = 11/8

Therefore, Statement 2 is correct

∴ Only statement 2 is correct

Concept:

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing (0,0) else the opposite side of (0,0).

Calculation:

Given: x + y ≥ 5

⇒ y ≥ – x + 5

• The given linear inequality covers the area (shaded region) above the straight line y = – x + 5 as shown below,

Similarly, x – y ≤ 3

⇒ – y ≤ – x + 3

⇒ y ≥ x – 3

• The given linear inequality covers the area (shaded region) above the straight line y = x – 3 as shown below,

• Hence, the solution of the two linear inequalities is the common area (shaded region) covered by the two inequalities as shown below,

∴ The solution to the given linear inequalities lies in the first and second quadrants.

Given, x² – y² = 0

⇒ x² = y²      ----(1)

Also, (x – a)² + y² = 1

Substituting from (1), we get,

⇒ (x – a)² + x² = 1

⇒ x² – 2ax + a² + x² = 1

⇒ 2x² – 2ax + (a² – 1) = 0      ----(2)

As we know, if a quadratic equation of form ax² + bx + c = 0 have only one positive solution, its Discriminant = b² – 4ac = 0

Hence, for the eq. (2),

⇒ Discriminant = (– 2a)² – 4(2)(a² – 1) = 0

⇒ 4a² – 8a² + 8 = 0

⇒ 4a² = 8

⇒ a² = 8/4 = 2

⇒ a = ±√2

∵ There is only one positive solution,

∴ a = √2

Given, α, β and γ are the zeros/roots of the polynomial f(x) = ax³ + bx² + cx + d,

As we know, a cubic polynomial can be written in terms of its roots as,

⇒ a(x – α)(x – β)(x – γ) = ax³ – a(α + β + γ)x² + a(αβ + βγ + γα)x – aαβγ

Comparing with the given equation, we get,

⇒ (α + β + γ) = – b/a

⇒ (αβ + βγ + γα) = c/a

⇒ αβγ = – d/a

Now, using the algebraic identity,

⇒ (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)

Substituting values, we get,

⇒ (– b/a)² = (α² + β² + γ²) + 2(c/a)

⇒ b²/a² = (α² + β² + γ²) + 2c/a

⇒ (α² + β² + γ²) = b²/a² – 2c/a

⇒ (α² + β² + γ²) = (b² – 2ac)/a²

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4%

∴ 4 % percentage of students passed in all three subjects.

Alternate Method 

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

∴ 4 % percentage of students passed in all three subjects.

Calculation:

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Percent of student pass in exactly one subject

⇒  7 + x + 2 + x + 2 + x

⇒ 11 + 3x

⇒ 11 + 3 ×  4 = 23 %

∴ 23 % students passed in exactly one subject.

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4%

Hence, a number of students passed at least two subject

⇒ \(\frac{25-4 + 30-4 + 28 - 4 + 4}{100}\times 100\)

⇒ \(\frac{83 - 2\times 4}{100}\times 360\) = 270

∴ 270 students passed in at least two subjects.

Alternate Method 

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Hence, a number of students passed at least two subject

⇒ \(\frac{25-x + 30-x + 28 - x + x}{100}\times 100\)

⇒ \(\frac{83 - 2x}{100}\times 360\)

⇒ \(\frac{83 - 2\times 4}{100}\times 360\) = 270

∴ 270 students passed in at least two subjects.

Formula used:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

Calculation:

Let A, B & C denotes Biology, Physics & Chemistry respectively.

Consider a total 100 % students.

We know that, 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +

n(A ∩ B ∩ C)

⇒ 100 - 2 = 55 + 62 + 60 - 25 - 30 - 28 + n (A ∩ B ∩ C) 

⇒ n (A ∩ B ∩ C)  = 98 - 94 

⇒  n (A ∩ B ∩ C) = 4% = passed in all three subject.

Hence,

Percentage of students passed in Physice & Biology 

⇒  (25 – 4)% = 21%

Percentage of students passed in Physics and Chemistry = 30%

Therefore, required ratio = 30 ∶ 21 = 10 ∶ 7

∴ 10 : 7 is the required ratio.

Alternate Method  

Total passed student = 98%

⇒ 7 + x + 2 + x + 2 + x + 30 - x + 25 - x+ 28 - x + x = 98%

⇒ 94 + x = 98

⇒ x = 4 %

Therefore, required ratio = 30 ∶ (25 - 4)  = 10 ∶ 7

∴ 10 : 7 is the required ratio.

On a nominal scale, the measured data is classified using labels and they do not have any numerical value.

On an ordinal scale, the measured data is classified using ranks and orders and the difference between the values is not important.

On an interval scale, both order and the exact difference between the measured values are considered.

On a ratio scale, an absolute zero is also present along with the order and difference between measured data.

∵ The ratings of a hotel are classified using ranks in terms of stars, which is a measure of its quality

∴ Ratings of a hotel are measured on an ordinal scale

Short trick∶

Using allegation method

Required ratio = 3 ∶ 2

Detailed solution∶

Let the number of students in section A be ‘x’ and that in section B be ‘y’

Total marks of section A = 65x

Total marks of section B = 70y

But, the total marks of both sections combined = 67(x + y)

⇒ 65x + 70y = 67x + 67y

⇒ 2x = 3y

⇒ x/y = 3/2

∴ Required ratio = 3 ∶ 2

As we know, the median is the magnitude of the middle term when all the terms are arranged in order of their magnitude (either ascending or descending)

Given, the median of 19 observations = 30

When two more observations are taken, observation of value 8 is added to one side of the median as 8 < 30, while observation with value 32 is added to the other side of the median as 30 < 32

Hence, the median will remain the same

∴ The median of 21 observations = 30

A frequency polygon is a type of graphical representation of data and is obtained by joining the mid-points of the histogram

∴ As the number of observations and the number of classes of data increase, the shape of the frequency polygon becomes increasingly smooth.

Pooled mean is simply known as the weighted mean

Hence, when the two means are pooled,

Pooled mean = x̅ = (n₁x̅₁ + n₂x̅₂)/(n₁ + n₂)

⇒ x̅ = n₁x̅₁/(n₁ + n₂) + n₂x̅₂/(n₁ + n₂)

∵ n₁ > n₂

The first term will be nearly equal to x̅₁ and the second term will be less than x̅₂

∴ x̅₁ < x̅ < x̅₂

Median is the magnitude of the middle term when all the terms in the series are arranged in order of their magnitudes (either ascending or descending). Hence, the median is a positional average, where the middlemost value of the series represents the whole series.

Due to this property, the median is not dependent on the magnitude of all the terms, i.e., if a term smaller than the median is replaced by another term that is also smaller than the median, then the median remains unchanged. So, the median is not influenced by the extreme values of the distribution. Hence, the median is the best-suited measure in case of an open-ended frequency distribution.

∴ Both Statement I and Statement II are true and Statement II is the correct explanation of Statement I