CAT 1997 QA | Previous Year CAT Paper

Answer: Option C

Explanation :

If he travels at minimum speed over stretch A (i.e. 40 km/hr), the total time taken to cover this stretch $=\left(\frac{2}{40}\right)=\frac{1}{20}$ hr = 3 min.

If he then travels at the fastest speed over stretch B (i.e. 50 km/hr), the total time taken to cover this stretch = $\left(\frac{2}{50}\right)=\frac{1}{25}$ = 2.4 min.

Thus, total time taken over the first two stretches = (3 + 2.4) = 5.4 min.
In order to break the previous record he will have to cover the third stretch in (10 – 5.4) = 4.6 min. To do this he will have to cover the third stretch at $\left(\frac{2}{4.6}\right)$ = 0.434 km per minute or 26.08 km/hr. But the maximum speed over the stretch C is 20 km/hr. Hence, it is not possible for C to break the previous record.

Answer: Option B

Explanation :

The minimum speed in stretch A is 40 km/hr. If Mr Hare travels the first stretch at this speed, then the time taken by him to cover this stretch = $\left(\frac{2}{40}\right)$ = 3 min. Also he takes 3 min to cover stretch B. And he covers the entire race in (1.5 × 10) = 15 min. This means that he should have taken (15 – 3 – 3) = 9 min to cover stretch C. Hence, his speed over this stretch should be $\left(\frac{2}{9}\right)$ = 0.22 km per minute or 13.3 km/hr.

Answer: Option C

Explanation :

Let his average speed over the last stretch be x.

Hence, his average speed for first two stretches = 4x. So the total time taken to cover the three stretches $=\left(\frac{4}{4x}\right)+\left(\frac{2}{x}\right)$

His average speed over the race is 20 km/hr.

Hence, the time taken to complete the race = $\frac{6}{20}$

Hence, we have the equation $\frac{4}{4x}+\frac{2}{x}=\frac{6}{20}$ Solving this equation, we get x = 10 km/hr.

Answer: Option D

Explanation :

Although the number of students in group D is more than in any other group, we still cannot say anything about the average weight of this group as nothing is mentioned about the average weights of any of the groups or of the individual students.

Answer: Option C

Explanation :

Although one student is shifted from group A to group B, the number of students in the class and the total weight of the students remain the same. Therefore, the average weight of the class remains the same.

Answer: Option C

Explanation :

The total weight of any group will vary according to the number of students in that group. Hence, the total weight of group A and C which has (15 + 25) = 40 will be twice that of students in group B which has 20 students. However, it is clear that if all the students are of same weight, then the average weight of all groups remains same irrespective of how many students are present in each group. Hence, clearly the statement 3 is false

Answer: Option C

Explanation :

Let his marks be 100, 90, 80, 70 and 60 in the five subjects. Hence, totally he has scored 400 marks. This constitutes only 60% of the total marks. Hence, total marks = $\frac{400}{0.6}=667,$ Since the total marks in each subject is the same, hence maximum marks in each subject will be $\left(\frac{667}{5}\right)$ ≃ 133. Out of this 50% is the passing marks. In other words, to pass in a subject he needs to score 66.5 marks. We can see that only in 
one subject he scored less than this, viz. 60. Hence, he passed in 4 subjects.

Answer: Option B

Explanation :

If we consider the Chairman and the vice chairman as one set, we can see that this set can fit 8 slots in between the 8 directors. Hence, this can be done in 8! ways. Between themselves, the chairman and the
vice chairman can be arranged in 2 ways. Hence, the required answer = 2 x 8!.

Answer: Option C

Explanation :

We know that if log_a x = y , then x = a^y . So comparing this form with our equation, we can get log₇ (x² – x + 37) = 2¹ = 2 and furthermore from this we can say that
(x² – x + 37) = 7² = 49
Thus, we have the equation
x² – x – 12 = 0
The solutions of this equations are,
x = 4 or x = –3.
The value that satisfies the given answer-choices is
x = 4.

Answer: Option A

Explanation :

Hint: Students please note that the percentages that are given are the basic percentages derived from basic fractions. e.g. 11.11% = $\frac{1}{9}$ and 14.28 = $\frac{1}{7}.$

Hence, you should make use of the most of this kind of knowledge. So let the CP be Re 1. Since he makes a profit of $\frac{1}{7},$ his SP = $\left(1+\frac{1}{7}\right)=\mathrm{Rs}. \frac{8}{7}$.

His marked price should be $\frac{1}{9}$ above this. So if we subtract $\frac{1}{9}$ of MP from the MP, we will get the SP.
So $\left(\mathrm{MP}-\frac{1}{9}\mathrm{MP}\right)=\mathrm{SP}=\frac{8}{7}$

Hence, MP = $\frac{9}{7}$

Therefore, percentage of mark-up on CP = (MP – CP)/CP

$=\left(\frac{9}{7}-1\right)/1=\frac{2}{7}=2\left(\frac{1}{7}\right)$ = 2 × 14.28 = 28.56%

Alternative method:
We can use the formula z = x – y – $\frac{xy}{100},$ where
z = Gain percentage
x = Percentage above CP
y = Discount percentage
∴ 14.28% = x – 11.11% – $\frac{11.11x}{100}$

or 14.28 = $\frac{100x-1111-11.11x}{100}$

or 1428 - 1111 = 88.89x

or x = 28.56% (Approximately)

Answer: Option D

Explanation :

The given expression can be written as

$\left(\frac{16n^2}{n}\right)+\left(\frac{7n}{n}\right)+\left(\frac{6}{n}\right)=16n+7+\left(\frac{6}{n}\right)$

Since n is an integer, the expression (16n + 7) will always be an integer. Hence, for the entire expression to be an integer, the part $\left(\frac{6}{n}\right)$ should also be an integer.
This can be possible only if n is a factor of 6, viz. n = 1, 2, 3, 6, –1, –2, –3 and –6.

Hence, n can have eight values.

Answer: Option D

Explanation :

Let he mix 3 kg, 4 kg and 5 kg of dry fruits at Rs. 100, Rs. 80 and at Rs. 60 per kilogram  respectively. Hence, his effective cost of the dry fruits per kilogram should be the weighted average

$=\left(\frac{3\times 100+4\times 80+5\times 60}{3+4+5}\right)=\frac{920}{12}$

In order to make a 50% profit, he will have to sell it at $\left(\frac{920}{12}\times 1.5\right)=\frac{920}{12}\times \frac{3}{2}=\frac{920}{8}$ = Rs. 115 per kg.

Since none of the answer-choices confirms this, the answer is (d).

Answer: Option B

Explanation :

20 kg fresh grapes will contain (0.9 × 20) = 18 kg water and 2 kg mass. If the dry grape has to contain 2 kg mass, it should constitute 80% of that. Hence, if 80% of dry grapes corresponds to 2 kg, its total weight will be $\left(\frac{2}{0.8}\right)$ = 2.5 kg

Answer: Option C

Explanation :

Effective speed of two trains = (80 – 40) = 40 km/hr. (Since they are moving in the same direction as inferred from the word ‘overtakes’). At this speed in 54 s, they would travel an effective distance of $\frac{(40\times 54)}{3600}$ = 0.6 km or 600 m. This effective distance should be
equal to the sum of the lengths of the two trains. So, if length of the express train is L, length of the goods train will be 2L. Hence, our equation will be L + 2L = 600 or L = 200 m. So, the time taken by this train to cross a platform 400 m long will be $\frac{(200+400)}{\left(80\times {\displaystyle \frac{5}{18}}\right)}=27$ s.

(Note that we have converted the denominator in metres per second. Hence, the factor of $\frac{5}{18}$).

Answer: Option A

Explanation :

This equation is very straightforward. If the number is 'x', then $\frac{7x}{8}-\frac{7x}{18}=770.$ On solving this equation, we get x = 1584. Hint: Students please note that if the difference in $\frac{7}{8}$ and $\frac{7}{18}$ of a number is 770, then the difference in $\frac{1}{8}$ and $\frac{1}{18}$ of the number should be 110. If we express this as an equation, we get  $\frac{x}{8}-\frac{x}{18}=110$

or 10x = 110 × 18 × 8
or x = 11 × 18 × 8
You can further proceed from here in two ways: (i) the last digit of the required answer should be (1 × 8 × 8) = 4, (ii) number should be divisible by 11. In both cases, the answer that is obtained from the given choices is 1584.

Answer: Option D

Explanation :

If we were to express 64 as product of two positive integers, we can get the following combinations: (64 × 1), (32 × 2), (16 × 4), (8 × 8).
Thus, we find that P + Q cannot be 35.

Answer: Option B

Explanation :

Total marks scored by the student in 10 papers = (80 × 10) = 800. If we exclude the papers in which he scored the highest and the lowest marks, then the total marks scored by him in remaining 8 papers = (81 × 8) = 648. Hence, his total in these two papers in which he scored the highest and the lowest marks = (800 – 648) = 152. Since his highest score is 92, his lowest score is (152 – 92) = 60.

Answer: Option A

Explanation :

We know that the sum of the roots = $-\frac{b}{a}$.

Hence, x₁ + x₂ = 2. Now we have two equations, viz. x₁ + x₂ = 2 ...(i)
and 7x₂ – 4x₁ = 47 ...(ii)

Solving these two equations, we get x₁ = –3 and x₂ = 5. Since it does not satisfy options (b) and (c), we will verify it for option (a). 

The product of the roots = (–3) × 5 = –15, $\frac{c}{a}$ in our case is c. Hence, c = –15.

Alternately,
Put values of x₁, x₂ in equation (ii). Do not match. So put c = –15 in equation (i) to get the roots of equation. After finding the roots of equation (i), check whether they satisfy equation (ii) or not. 

The roots (5, –3) satisfy the equation (ii) so answer is (a).

Hence, option (a).

Answer: Option A

Explanation :

If the radii of two circles are r₁ and r₂, then the two equations can be written πr₁² + πr₂² = 153π
or (r₁² + r₂²) = 153 and r₁ + r₂ = 15.
Now r₁² + r₂² = (r₁ + r₂)² – 2r₁r₂
Therefore, 153 = (15)² – 2r₁r₂ or r₁r₂ = 36.
If 36 is to be expressed as the product of two integers, it could be (36 × 1), (18 × 2), (12 × 3), (9 × 4), (6 × 6).
The only two factors that add up to 15 are 12 and 3.

Hence, r₁ = 12, r₂ = 3. Therefore, the ratio of larger radius to the smaller one is 12 : 3 = 4.

Answer: Option C

Explanation :

The best way to solve this is the method of simulation, e.g. let m = 10 and n = 5. Therefore m – n = 5, which is divisible by 5.
m² – n² = 100 – 25 = 75, divisible by 25. m + n = 10 + 5 = 15 is not divisible by 10.

Hence, the answer is (c).

Note that for the sum of two multiples of 5 to be divisible by 10, either both of them should be odd (i.e. ending in 5) or both of them should be even (i.e. ending in 0).

Answer: Option B

Explanation :

$7^{3^2}$ = 7⁹ and (7³)² = 7⁶. SInce 7⁹ > 7⁶ ⇒ $7^{3^2}$ > (7³)²

Answer: Option B

Explanation :

Here, a + b + c + d + e + f + g = 200 ...(i)
80% of the people watch DD implies
c + d + f + g = 160 ...(ii)
22% of the people watch BBC implies
a + d + e + g = 44 ...(iii)
15% of the people watch CNN implies
b + e + f + g = 30 ...(iv)
(ii) + (iii) + (iv) gives
a + b + c + 2(d + e + f) + 3g = 234
Subtracting (i) from this equation,
d + e + f + 2g = 34 ...(v)

To maximize g, in equation (v), we put d = e = f = 0
∴ Maximum value of g = $\frac{34}{2}$ = 17

∴ Required percentage = $\frac{17}{200}$ × 100 = 8.5%

Hence, option (b).

Answer: Option A

Explanation :

Here, a + b + c + d + e + f + g = 200 ...(i)
80% of the people watch DD implies
c + d + f + g = 160 ...(ii)
22% of the people watch BBC implies
a + d + e + g = 44 ...(iii)
15% of the people watch CNN implies
b + e + f + g = 30 ...(iv)
(ii) + (iii) + (iv) gives
a + b + c + 2(d + e + f) + 3g = 234
Subtracting (i) from this equation,
d + e + f + 2g = 34 ...(v)

5% of people watch DD and CNN implies
f + g = 10 ...(vi)
10% of people watch DD and BBC implies
d + g = 20 ...(vii)
(v) – (vi) – (vii) gives
e = 4
∴ Required percentage = $\frac{4}{200}$ × 100 = 2%

Answer: Option D

Explanation :

Here, a + b + c + d + e + f + g = 200 ...(i)
80% of the people watch DD implies
c + d + f + g = 160 ...(ii)
22% of the people watch BBC implies
a + d + e + g = 44 ...(iii)
15% of the people watch CNN implies
b + e + f + g = 30 ...(iv)
(ii) + (iii) + (iv) gives
a + b + c + 2(d + e + f) + 3g = 234
Subtracting (i) from this equation,
d + e + f + 2g = 34 ...(v)

From equation (v), we have
(d + 4 + f) + 2g = 34
⇒ (d + f) + 2g = 30
Since we cannot find the values of d and f, the value of g cannot be ascertained.

Answer: Option B

Explanation :

The two equations can be written
2000 $\left(\frac{x}{100}\right)$ + 2000 $\left(\frac{y}{100}\right)$ = 700 and 2000 $\left(\frac{x}{100}\right)$ + 3000 $\left(\frac{y}{100}\right)=900$

The equations can be simplified to x + y = 35 and 2x + 3y = 90. Solving these two equations simultaneously, we get x = 15%.

Answer: Option D

Explanation :

Since AB is the diameter of the circle, ∠ACB would be right angle. In this triangle, we know AB = 15 and AC = 12. So we can find BC. Hence, BC = 9. Since BC = BD, AD = AC (similar triangles).
Hence, area of ΔACB = Area of Δ ABD $=\frac{1}{2}$ × AC × CB = $\frac{1}{2}$ × 12 × 9 = 54 cm²

So the area of quadrilateral ACBD = 2 × 54 = 108 sq. cm.

Answer: Option C

Explanation :

As P, Q and R are consecutive odd numbers, Q = P + 2 and R = P + 4. Now 3P = 2(P + 4) – 3. On solving this equation, we get P = 5.

Therefore, R = 5 + 4 = 9

Answer: Option D

Explanation :

This question can be done by assuming some values for x, y and z, e.g. 

Let x = 4, y = 3 and z = 1.
la(4, 3, 1) = min(7, 4) = 4;
le(4, 3, 1) = max(1, 2) = 2; 
ma(x, y, z) = $\frac{1}{2}$ (4 + 2) = 3. 

Hence, we can reject options (a) & (c).

Let x = 10, y = 3 and z = 1.
la(10, 3, 1) = min(13, 4) = 4;
le(10, 3, 1) = max(7, 2) = 7; 
ma(x, y, z) = $\frac{1}{2}$ (4 + 7) = 5.5. 

Hence, we can reject option (b) also.

Hence, option (d).

Answer: Option B

Explanation :

ma(10, 4, le(la(10, 5, 3), 5, 3))
= ma(10,4, le(min(15,8),5,3))
= ma (10, 4, le (8, 5, 3))
= ma(10, 4, max (3,2))
= ma (10, 4, 3)

= $\frac{1}{2}$ [le(10, 4, 3) + la(10, 4, 3)]

= $\frac{1}{2}$[max(6,1) + min(14, 7)]

= $\frac{1}{2}$(6 + 7) = 6.5

Answer: Option C

Explanation :

le (15,min (10, 6),le (9,8,ma (15,10,9)))
Now ma(15,10, 9) = $\frac{1}{2}$ [le(15, 10, 9) + la(15, 10, 9)]

= $\frac{1}{2}$ [max(5, 1) + min(25,19)]

= $\frac{1}{2}$ (5 + 19) = 12

Hence, our original expression would now be
le(15,min(10,6),le(9,8,12))
= le(15, 6,max(1, − 4))
= le(15, 6,1) = max(9, 5) = 9

Answer: Option C

Explanation :

The value of ABC, three-digit number is should satisfy the condition 100A + 10B + C = A! + B! + C! ....(i)
The maximum value of three digit number is 999 and minimum is 100
We observe that 7! = 5040 > 999
So the 3-digits of the number must be
           6 as 6! = 720
and/or 5 as 5! = 120
and/or 4 as 4! = 24
and/or 3 as 3! = 6
and/or 2 as 2! = 4
and/or 1 as 1! = 1
If we consider 6 at the hundred’s place digit we see that condition (1) is not satisfied as 600 < 720 (6!) So we conclude that 6 cannot occupy any position in the number.
If we place ‘5’ at the hundred’s place then the number should lie between the range of 500 and 600. Considering the RHS of equation (1) by putting A + B = C = 5 we get the sum as 360 which is less than 500. Similarly, putting 4, 3, 2 at the hundred’s place does not satisfies the given condition (1).
Only 1 can be placed at hundred place and 5 should be one of the digit at other two position in order to make it a three digit number.

Thus, only combination we satisfies the given condition (1) is (1, 4, 5) i.e. 145 = 1! + 4! + 5! = 145.

Answer: Option B

Explanation :

The diagonal of the innermost square is 2 units. The diagonal of every successive square would increase by 2 units (since corners are one unit apart). So the diagonal of the 7th square = 14 units and that of the 8th square = 16 units. Areas of the 7th square = $\frac{1}{2}$ (14)² and that of 8th square = $\frac{1}{2}$(16)², and 128 respectively. Hence, the difference in their areas = (128 – 98) = 30 sq.units.

Answer: Option D

Explanation :

$A=\frac{2.000004}{[{(2.000004)}^2+2(2.000004\left)\right]}$

$=\frac{2.000004}{2.000004\left[\right(2.000004)+2]}$

= $\frac{1}{\left[\right(2.000004)+2]}$

$=\frac{1}{4.000004}$

= $\frac{1}{4}=0.25$ (Approximately)

$B=\frac{3.000003}{[{(3.000003)}^3+3(3.000003\left)\right]}$

$=\frac{3.000003}{3.000003\left[\right(3.000003)+3]}$

$=\frac{1}{\left[\right(3.000003)+3]}=\frac{1}{6.000003}$

$C=\frac{4.000002}{[{(4.000002)}^2+2(4.000002\left)\right]}$

$=\frac{4.000002}{4.000002\left[\right(4.000002)+2]}$

$=\frac{1}{\left[\right(4.000002)+2]}=\frac{1}{6.000002}$

Looking at the answer choices, we can see that the only (d) satisfies the relationship, viz. B is the smallest.

Answer: Option B

Explanation :

Let D₁, T₁ and D₂, T₂ denote the diameters and the thickness of the two coins respectively. If V₁ and V₂ are the values of the two coins.

$\frac{V_1}{V_2}=\left(\frac{{D_1}^2{T}_1}{{D_2}^2{T}_2}\right)={\left(\frac{D_1}{D_2}\right)}^2\left(\frac{T_1}{T_2}\right)$

Therefore, $\frac{4}{1}={\left(\frac{4}{3}\right)}^2\left(\frac{T_1}{T_2}\right)\Rightarrow \left(\frac{T_1}{T_2}\right)=\frac{9}{4}$

Answer: Option C

Explanation :

In a triangle, the line joining the mid-points of any two sides is half the length of its third side. Hence, every side of ΔPQR would be half the sides of ΔABC. Hence, area of ΔPQR would be $\frac{1}{4}$ the area of ΔABC

$=\frac{1}{4}$ × 20 = 5 sq.units

Answer: Option D

Explanation :

Let L and B denote the length and the breadth of the rectangle. So the diagonal will be $\sqrt{(L^2+B^2)}$. Hence, from the condition given, (L + B) – $\sqrt{(L^2+B^2)}=\frac{1}{2}L$

$\Rightarrow \sqrt{(L^2+B^2)}=\frac{L}{2}+B$

Squaring both sides, we get

(L²+ B²) = ${\left(\frac{L}{2}+B\right)}^2$

⇒ L² = $\frac{L^2}{4}+LB$

⇒ $\frac{3L^2}{4}=LB$

⇒ $\frac{B}{L}=\frac{3}{4}$

Shortcut:
First write the relation (L + B) – $\sqrt{(L^2+B^2)}=\frac{1}{2}L.$

or $\frac{L}{2}+B=\sqrt{(L^2+B^2)}$

Put the values of options. Option (d) satisfies. So the answer is (d).

Answer: Option A

Explanation :

25th February, 1996 was a Sunday. [This can be calculated based on the concept of Calendars]

Raja worked from 25th Feb till 2nd March i.e. for 7 days. 

Assuming each person does completes 1 unit of work every day.

⇒ Work done by Raja in 7 days = 7 × 1 = 7 units.

Now this work (7 units) has to be completed by T & J together.

On 25th Feb, i.e., Sunday both T & J would be working, hence they will complete 2 units of work.
∴ Total work done by them so far = 2 units.

On 26th Feb, i.e., Monday both T & J would be working, hence they will complete 2 units of work.
∴ Total work done by them so far = 2 + 2 = 4 units.

On 27th Feb, i.e., Tuesday only J would be working, hence they will complete 1 unit of work.
∴ Total work done by them so far = 2 + 2 + 1 = 5 units.

On 28th Feb, i.e., Wednesday both T & J would be working, hence they will complete 2 units of work.
∴ Total work done by them so far = 2 + 2 + 1 + 2 = 7 units.

∴ Both of them can finish the entire work in 4 days.

Hence, option (a).

Answer: Option C

Explanation :

Now Raja has worked for (5 days in February + 31 days in March + 2 days in April) = 38 days. Let us assume his rate to be the same as in the previous question, viz. one unit a day. Hence, he completes 38 units totally. In a week, T takes holiday on Tuesday and Thursday, while S takes holiday on Saturday and Sunday. We can see that their working pattern would be as follows.

​​​​​​​

So in a week they work together 10 units work. Thus, in three weeks, they would complete 30 units work. It

can be found out that February 25 is Sunday. So the remaining 8 units of work can be completed only on Friday, i.e. March 22.

Answer: Option B

Explanation :

Let us convert all the time to same time zone, viz. Boston. So X left Frankfurt at 6 p.m. on Friday (Frankfurt time) or 10 p.m. on Friday (Boston time). X reached Boston at 10 a.m. on Saturday (Boston time). In other words, X has taken 12 hr in all to go from Frankfurt to Boston. After 2 hr wait, X leaves at 12 noon (Boston time). Now X reaches India at 1 a.m. on Sunday (Indian time) or 11 a.m. on Saturday (Boston time). Thus, X takes 11 hr in all to go from Boston to India.

Overall, X has travelled for 25 hr (including stoppages) at an average speed of 180 miles per hour. Hence, the
distance between Frankfurt and India is (25 × 180) = 4500 miles. 

Answer: Option A

Explanation :

Let us convert all the time to same time zone, viz. Boston. So X left Frankfurt at 6 p.m. on Friday (Frankfurt time) or 10 p.m. on Friday (Boston time). X reached Boston at 10 a.m. on Saturday (Boston time). In other words, X has taken 12 hr in all to go from Frankfurt to Boston. After 2 hr wait, X leaves at 12 noon (Boston time). Now X reaches India at 1 a.m. on Sunday (Indian time) or 11 a.m. on Saturday (Boston time). Thus, X takes 11 hr in all to go from Boston to India.

On the return journey, X halts at Boston for one hour less than his previous halt there. Therefore, X takes 24 hr for his return journey.

Answer: Option D

Explanation :

We do not know the distance between Frankfurt and India and hence cannot calculate the average speed.

Hence, option (d).

Answer: Option A

Explanation :

In ΔBEC and ΔAED

∠CBE = ∠CDE (
angles in the same segment of a circle are equal)
Similarly ∠BCE = ∠EAD (
angles in the same segment of a circle are equal)
∠BEC = ∠AED (Vertical angles are equal) 

∴ By AAA similarity 
ΔCEB ~ ΔAED
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the
corresponding sides

∴ $\frac{area(∆BEC)}{area (∆AED)}={\left(\frac{BC}{DA}\right)}^2={\left(\frac{12}{24}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Answer: Option B

Explanation :

We know that length of the line joining the mid-points of two sides of a triangle is half the length of third side. Hence, the required length is half the length of side AC. Since EADF is rectangle, EF = AD = 8.
CD = (22 – 16) = 6.
So in the right-angled ΔADC, AD = 8 and CD = 6.
Therefore, AC = 10. Hence, length of the line joining the mid-points of AB and BC = $\frac{1}{2}$(10) = 5.

Answer: Option C

Explanation :

Since the policeman started 15 min late, in this time the thief would have already covered $\left(\frac{60}{4}\right)$ = 15 km. To
catch the thief, the policeman will have to make up for this distance of 15 km. Every hour the policeman is travelling (65 – 60) = 5 km more than the thief. Hence, to make up the distance of 15 km, he would take 3 hr.
Since policeman started at 12.15 p.m., he would catch the thief at 3.15 p.m.

Answer: Option B

Explanation :

Every hour the second policeman covers (65 – 60) = 5 km less than the first one. Since the first policeman catches the thief in 3 hr, in this time the second policeman will be (3 × 5) = 15 km behind.

Answer: 4

Explanation :

a³ + b³ = (a + b)(a² + b² – ab). Combining statements I and II, we get the value of (a + b) = √28 or –√28. Since we do not have the unique value of (a + b), we cannot get the unique answer a³ + b³.

Answer: 2

Explanation :

11 and 9 are coprimes of 99, and hence the number divisible by 99 must be divisible by 9 and 11. Certainly statement I alone is sufficient to answer the question. Statement II says when the digits of the number are reversed, the new number formed is divisible by 9 and 11. The best way to handle this particular case is by simulation. Let us select any number which is divisible by 9 and 11. Let us select 1386 which is divisible by 9 and 11. Hence, the original number will be 6831 which, in turn, is also divisible by 99. Hence, statement II also is independently sufficient to answer the question. Since both the statements are independently sufficient to answer the question, the answer is (b).

Answer: 2

Explanation :

The diagram will be as shown. M is Mali and P is Pali. Consider statement I. When the person covers $\frac{1}{3}$ distance, he is 3 km east and 1 km north of Mali. Based on this statement alone, we can easily find out where
he will be on line MP by using Pythagoras’ theorem. Once we find this distance, we can easily get distance MP by multiplying it by 3. Similarly, based on statement II alone also, we can find distance MP. Hence, both the statements are independently sufficient to answer the question. Hence, the answer is (b).

Answer: 4

Explanation :

Students! be careful. Generally, as we see two unknowns (i.e. x and y in this case) and two equations, we tempt to mark the answer as c, i.e. combining two statements, we can easily find the values of x and y. But have a look at the equations 3x + 2y = 45 and 10.5x + 7y = 157.5. Multiplying 1st equation by 3.5, we get 2nd equation. Hence, these are not really two different equations. Hence, data is insufficient to answer the question. In general, remember the following rule. If we have two equations Ax + By = k₁ and Cx + Dy = k₂, and A × D = B × C, then the equations cannot be solved.

Answer: 4

Explanation :

P says he can see one black and one white hat. So either Q is wearing white and R is wearing black, or Q is wearing black and R is wearing white. Q also makes same statement. Still we cannot say the colour of the hat which P is wearing.

Answer: 3

Explanation :

Let the speed of the motorcycle be x km/hr. Therefore, speed of the car will be (x + 10) km/hr.
From statement II, we can form the following equation.

$\left(\frac{100}{(x+10)}\right)=\left(\left(\frac{100}{x}\right)+2\right)$

After solving this equation, we can get the speed of the car. Hence, this question can be answered by combining both the statements.

Answer: 2

Explanation :

Let V₁ be the original volume and r₁ and h₁ be the radius and height of the cone respectively.

V₁ = $\left(\frac{1}{3}\right)$ × π × (r₁)² × h₁. Consider statement I. If the cone is cut parallel to base and dividing the height in
the ratio 1 : 2, then r₂ = $\left(\frac{1}{2}\right)\times r_1$ and h₂ = $\left(\frac{1}{2}\right)$ × h₁, where r₂ and h₂ are the radius and height of the new
cone respectively. If V₂ is the volume of new cone, then V2 = $\left(\frac{1}{3}\right)$ × π(r₂)² × h₂

$=\left(\frac{1}{3}\right)\times \pi \times {\left(\frac{1}{2}\times r_1\right)}^2\times \left(\left(\frac{1}{2}\right)h_1\right)=\left(\frac{1}{8}\right)\times V_1$

Hence, statement I alone is sufficient to answer the question (as we get the ratio as 1 : 8).

Similarly, based on statement II alone, we can find the ratio (which will be 1 : 27).

Answer: 3

Explanation :

If we solve the two given equations, we get the point of intersection as (3, 2). Let A = (3, 2). The lines of our interest (let it be L₁ and L₂) also pass through A. One of the lines passes through (0, 4). Let L₁ passes through (0, 4), but it also passes through (3, 2). Hence, we can find the slope of L₂ (which is equal to – $\frac{2}{3}$). Hence, slope of L₂ will be $\frac{3}{2}$ since L₁ and L₂ are perpendicular.
Hence, equations of L₁ and L₂ can be obtained by using slope point form. (Students! we need not really find out the equations.) After getting both the equations, we can find the area bounded by L₁ and L₂ and coordinate axes.

Answer: 4

Explanation :

Let A and B be the CP of the chair and the table respectively. So 1.15A + 1.2B = SP.
Hence, profit = 0.15A + 0.2B. Now consider statement II, CP = 1.1A + 1.2B. As per new CP, now profit will be
SP – CP = (1.15A + 1.2B) – (1.1A + 1.2B) = 0.05A + 0xB = 0.05A. Combining both statements, we get the
equation as
0.05A = 0.15A + 0.2B – 20. Still we cannot find the answer.

Answer: 4

Explanation :

None of the statements specifies the direction in which Tez and Gati are moving, which is very significant.