Comprehension
I and II. As your answer,
Mark (a) if I > II.
Mark (b) if I < II.
Mark (c) if I = II.
Mark (d) if nothing can be said.
Question: 1
Nineteen years from now Jackson will be 3 times as old as Joseph is now. Johnson is three years younger than Jackson.
I. Johnson’s age now.
II. Joseph’s age now.
Nineteen years from now Jackson will be 3 times as old as Joseph is now. Johnson is three years younger than Jackson.
I. Johnson’s age now.
II. Joseph’s age now.
I. Johnson’s age now.
II. Joseph’s age now.
Show Hint
When comparing two unknowns in age problems, express both in terms of one variable and directly compare.
Updated On: Aug 5, 2025
- if I > II
- if I < II
- if I = II
- if nothing can be said
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Let Jackson’s present age be $x$ years, Joseph’s present age be $y$ years, and Johnson’s present age be $z$ years.
From the problem statement:
$19$ years from now, Jackson’s age = $x + 19$.
We are told that at that time, Jackson’s age will be $3$ times Joseph’s present age: \[ x + 19 = 3y \] Also, Johnson is $3$ years younger than Jackson: \[ z = x - 3 \] We need to compare $z$ (Johnson’s age now) and $y$ (Joseph’s age now).
From $x + 19 = 3y$, we get $x = 3y - 19$.
So: \[ z = (3y - 19) - 3 = 3y - 22 \] Now, compare $z$ and $y$: \[ z - y = (3y - 22) - y = 2y - 22 \] Since ages are positive, $y>0$. For $y \geq 12$, $z>y$. For realistic human ages, Joseph’s present age will be more than $12$ years, so Johnson is older. Thus $I>II$.
From the problem statement:
$19$ years from now, Jackson’s age = $x + 19$.
We are told that at that time, Jackson’s age will be $3$ times Joseph’s present age: \[ x + 19 = 3y \] Also, Johnson is $3$ years younger than Jackson: \[ z = x - 3 \] We need to compare $z$ (Johnson’s age now) and $y$ (Joseph’s age now).
From $x + 19 = 3y$, we get $x = 3y - 19$.
So: \[ z = (3y - 19) - 3 = 3y - 22 \] Now, compare $z$ and $y$: \[ z - y = (3y - 22) - y = 2y - 22 \] Since ages are positive, $y>0$. For $y \geq 12$, $z>y$. For realistic human ages, Joseph’s present age will be more than $12$ years, so Johnson is older. Thus $I>II$.
Was this answer helpful?
0
0
Question: 2
In $\triangle ACD$, $AD = AC$ and $\angle C = 2\angle E$. The distance between parallel lines $AB$ and $CD$ is $h$. Then:
I. Area of parallelogram $ABCD$
II. Area of $\triangle ADE$
In $\triangle ACD$, $AD = AC$ and $\angle C = 2\angle E$. The distance between parallel lines $AB$ and $CD$ is $h$. Then:
I. Area of parallelogram $ABCD$
II. Area of $\triangle ADE$
I. Area of parallelogram $ABCD$
II. Area of $\triangle ADE$
Show Hint
For geometric comparison, if one shape is entirely contained inside another and they share the same height, the larger one has the greater area without requiring exact lengths.
Updated On: Aug 5, 2025
- if I > II
- if I < II
- if I = II
- if nothing can be said
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
We know $AB \parallel CD$ and the perpendicular distance between them is $h$. The area of parallelogram $ABCD$ is:
\[
\text{Area}_{ABCD} = \text{Base} \times \text{Height} = CD \times h
\]
In $\triangle ADE$, $AD = AC$ means $\triangle ACD$ is isosceles with $AD = AC$. Given $\angle C = 2\angle E$, triangle $ADE$ is smaller and lies inside parallelogram $ABCD$.
Since $ABCD$ covers more area than $ADE$ (which is just a portion of it), clearly: \[ \text{Area}_{ABCD}>\text{Area}_{ADE} \] Hence I>II.
Since $ABCD$ covers more area than $ADE$ (which is just a portion of it), clearly: \[ \text{Area}_{ABCD}>\text{Area}_{ADE} \] Hence I>II.
Was this answer helpful?
0
0
Question: 3
Last week Martin received 10 in commission for selling 100 copies of a magazine. Last week Miguel sold 100 copies of this magazine. He received his salary of 5 per week plus a commission of 2 cents for each of the first 25 copies sold, 3 cents for each of the next 25 copies sold, and 4 cents for each copy thereafter. (1 = 100 cents)
I. Martin’s commission in the last week.
II. Miguel’s total income for last week.
Last week Martin received 10 in commission for selling 100 copies of a magazine. Last week Miguel sold 100 copies of this magazine. He received his salary of 5 per week plus a commission of 2 cents for each of the first 25 copies sold, 3 cents for each of the next 25 copies sold, and 4 cents for each copy thereafter. (1 = 100 cents)
I. Martin’s commission in the last week.
II. Miguel’s total income for last week.
I. Martin’s commission in the last week.
II. Miguel’s total income for last week.
Show Hint
When comparing incomes, carefully separate commission-only amounts from total earnings, especially when one includes a base salary.
Updated On: Aug 5, 2025
- if I > II
- if I < II
- if I = II
- if nothing can be said
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Martin’s commission
Martin’s commission for selling 100 copies is given as 10. That is his commission amount for the week. Thus: \[ I = 10 \ \text{dollars} \]
Step 2: Miguel’s total income
Miguel receives a fixed salary of 5 per week.
His commission structure: - First 25 copies: 2 cents each ⇒ 25 × 0.02 = 0.50 dollars.
- Next 25 copies: 3 cents each ⇒ 25 × 0.03 = 0.75 dollars.
- Remaining 50 copies: 4 cents each ⇒ 50 × 0.04 = 2.00 dollars.
Total commission = 0.50 + 0.75 + 2.00 = 3.25 dollars.
Step 3: Total income for Miguel
\[ \text{Total} = \text{Salary} + \text{Commission} = 5.00 + 3.25 = 8.25 \ \text{dollars} \]
Step 4: Compare I and II
Martin’s commission = 10.
Miguel’s total income = 8.25.
Clearly, I > II. So the correct choice is (a).
Martin’s commission for selling 100 copies is given as 10. That is his commission amount for the week. Thus: \[ I = 10 \ \text{dollars} \]
Step 2: Miguel’s total income
Miguel receives a fixed salary of 5 per week.
His commission structure: - First 25 copies: 2 cents each ⇒ 25 × 0.02 = 0.50 dollars.
- Next 25 copies: 3 cents each ⇒ 25 × 0.03 = 0.75 dollars.
- Remaining 50 copies: 4 cents each ⇒ 50 × 0.04 = 2.00 dollars.
Total commission = 0.50 + 0.75 + 2.00 = 3.25 dollars.
Step 3: Total income for Miguel
\[ \text{Total} = \text{Salary} + \text{Commission} = 5.00 + 3.25 = 8.25 \ \text{dollars} \]
Step 4: Compare I and II
Martin’s commission = 10.
Miguel’s total income = 8.25.
Clearly, I > II. So the correct choice is (a).
Was this answer helpful?
0
0
Question: 4
$k_1, k_2, k_3$ are parallel lines. $AD = 2 \ \text{cm}$, $BE = 8 \ \text{cm}$ and $CF = 32 \ \text{cm}$.
$k_1, k_2, k_3$ are parallel lines. $AD = 2 \ \text{cm}$, $BE = 8 \ \text{cm}$ and $CF = 32 \ \text{cm}$.
Show Hint
In problems involving parallel lines and transversals, the product of certain corresponding segments often turns out equal due to proportionality theorems.
Updated On: Aug 5, 2025
- if I > II
- if I < II
- if I = II
- if nothing can be said
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
We know $k_1, k_2, k_3$ are parallel, and transversal segments will be proportional.
Given $AD = 2$ cm, $BE = 8$ cm, $CF = 32$ cm. Points A, B, C lie on one transversal; D, E, F on another.
From $AD = AB + BD$ and proportionality of segments between parallel lines: \[ \frac{AB}{BE} = \frac{DE}{EF} \] Also $BC = BE$ (same transversal segment length) and $EF = CF - CE$.
By proportionality, $(AB) \times (EF) = (BC) \times (DE)$. Thus $I = II$.
Given $AD = 2$ cm, $BE = 8$ cm, $CF = 32$ cm. Points A, B, C lie on one transversal; D, E, F on another.
From $AD = AB + BD$ and proportionality of segments between parallel lines: \[ \frac{AB}{BE} = \frac{DE}{EF} \] Also $BC = BE$ (same transversal segment length) and $EF = CF - CE$.
By proportionality, $(AB) \times (EF) = (BC) \times (DE)$. Thus $I = II$.
Was this answer helpful?
0
0
Question: 5
I. The probability of encountering $54$ Sundays in a leap year.
II. The probability of encountering $53$ Sundays in a non-leap year.
I. The probability of encountering $54$ Sundays in a leap year.
II. The probability of encountering $53$ Sundays in a non-leap year.
II. The probability of encountering $53$ Sundays in a non-leap year.
Show Hint
When dealing with probabilities of extra days in a year, always break the year into full weeks plus extra days. The arrangement of those extra days determines the outcome.
Updated On: Aug 5, 2025
- if I > II
- if I < II
- if I = II
- if nothing can be said
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Let us recall:
- A non-leap year has $365$ days = $52$ full weeks + $1$ extra day.
- A leap year has $366$ days = $52$ full weeks + $2$ extra days.
Case 1: Leap year — Probability of 54 Sundays
In a leap year, there are $52$ Sundays guaranteed (because there are $52$ weeks), plus there are $2$ extra days.
These extra days could be any of the combinations: \[ (\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday}) \] For there to be $54$ Sundays, both extra days must contain exactly $2$ Sundays — but that’s impossible because the two extra days are consecutive. The only way to get $54$ Sundays is if the first extra day is a Sunday (then the second is Monday).
Out of $7$ possible starting days, exactly $1$ gives $54$ Sundays.
Thus: \[ P(\text{54 Sundays in leap year}) = \frac{1}{7} \]
Case 2: Non-leap year — Probability of 53 Sundays
In a non-leap year, there are $52$ Sundays guaranteed, plus $1$ extra day.
If that extra day is a Sunday, then total Sundays = $53$.
Out of $7$ possibilities for the extra day, $1$ is Sunday.
Thus: \[ P(\text{53 Sundays in non-leap year}) = \frac{1}{7} \]
Comparison: We have: \[ I = \frac{1}{7}, \quad II = \frac{1}{7} \] Clearly, $I = II$.
Therefore, the answer is (c) if I = II.
- A leap year has $366$ days = $52$ full weeks + $2$ extra days.
Case 1: Leap year — Probability of 54 Sundays
In a leap year, there are $52$ Sundays guaranteed (because there are $52$ weeks), plus there are $2$ extra days.
These extra days could be any of the combinations: \[ (\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday}) \] For there to be $54$ Sundays, both extra days must contain exactly $2$ Sundays — but that’s impossible because the two extra days are consecutive. The only way to get $54$ Sundays is if the first extra day is a Sunday (then the second is Monday).
Out of $7$ possible starting days, exactly $1$ gives $54$ Sundays.
Thus: \[ P(\text{54 Sundays in leap year}) = \frac{1}{7} \]
Case 2: Non-leap year — Probability of 53 Sundays
In a non-leap year, there are $52$ Sundays guaranteed, plus $1$ extra day.
If that extra day is a Sunday, then total Sundays = $53$.
Out of $7$ possibilities for the extra day, $1$ is Sunday.
Thus: \[ P(\text{53 Sundays in non-leap year}) = \frac{1}{7} \]
Comparison: We have: \[ I = \frac{1}{7}, \quad II = \frac{1}{7} \] Clearly, $I = II$.
Therefore, the answer is (c) if I = II.
Was this answer helpful?
0
0
Top Questions on Quantitative Aptitude
The table shows the percentage distribution of students in four departments:
Department Percentage of Students Physics 25% Chemistry 30% Mathematics 20% Biology 25% If there are 400 students in total, and 40% of Physics students are female, how many female students are in the Physics department?
- CAT - 2016
- General Aptitude
- Quantitative Aptitude
- In a class, 60% of students are boys, and 40% are girls. If 30% of boys and 20% of girls passed an exam, what percentage of the class passed?
- CAT - 2016
- General Aptitude
- Quantitative Aptitude
- A rectangle has an area of 60 cm² and a perimeter of 32 cm. What is the length if it is greater than the width?
- CAT - 2016
- General Aptitude
- Quantitative Aptitude
- A sum of Rs. 10,000 is invested at 5% per annum compound interest for 2 years. What is the total amount?
- CAT - 2016
- General Aptitude
- Quantitative Aptitude
- A bag contains 4 red, 3 white, and 2 blue balls. Two balls are drawn. What is the probability both are red?
- CAT - 2016
- General Aptitude
- Quantitative Aptitude
View More Questions
Questions Asked in CAT exam
- \( 10^{68} \) is divided by 13, the remainder is is
- CAT - 2024
- Divisibility and Remainder
- ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of \(\triangle ADE\) is
- CAT - 2024
- Mensuration
- Two places A and B are 45 kms apart and connected by a straight road. Anil goes from A to B while Sunil goes from B to A. Starting at the same time, they cross each other in exactly 1 hour 30 minutes. If Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A, the speed of Anil, in km per hour, is
- CAT - 2024
- Time, Speed and Distance
- The selling price of a product is fixed to ensure 40% profit. If the product had cost 40% less and had been sold for 5 rupees less, then the resulting profit would have been 50%. The original selling price, in rupees, of the product is
- CAT - 2024
- Profit and Loss
- Suppose $x_1, x_2, x_3, \dots, x_{100}$ are in arithmetic progression such that $x_5 = -4$ and $2x_6 + 2x_9 = x_{11} + x_{13}$. Then, $x_{100}$ equals ?
- CAT - 2024
- Arithmetic Progression
View More Questions