CAT 1996 QA | Previous Year CAT Paper

Answer: Option A

Explanation :

If there is a shop at C, all A, B, C and D are within 2 km range. Another shop is needed for E.

Hence, 2 shops are required.

Answer: Option A

Explanation :

If there is a shop at C; all A, B, D and E are within 3 km range. Hence, 1 shop is required.

Answer: Option A

Video Explanation

Explanation :

Since ΔBCE is an equilateral triangle, CE = BC = BE.
And since ABCD is a square, BC = CD. Hence, CD = CE.
So in ΔCDE, we have CD = CE. Hence, ∠EDC = ∠CED.
Now ∠BCE = 60° (since equilateral triangle) and ∠BCD = 90° (since square).
Hence, ∠DCE = ∠DCB + ∠BCE = (60 + 90) = 150°.
So in ΔDCE, ∠EDC + ∠CED = 30° (since three angles of a triangle add up to 180°). Hence, we have ∠DEC = ∠EDC = 15°.

Answer: Option A

Video Explanation

Explanation :

Let the price per metre of cloth be Re 1. The shopkeeper buys 120 cm, but pays for only 100 cm. In other words, he buys 120 cm for Rs. 100. So his CP = $\left(\frac{100}{120}\right)$ = Re 0.833 per metre. Now he sells 80 cm, but charges for 100 cm. In other words, he sells 80 cm for Rs. 100. On this he offers a 20% discount on cash payment. So he charges Rs. 80 for 80 cm cloth. In other words, his SP = $\left(\frac{80}{80}\right)$ = Re 1 per metre. So his percentage profit in the overall transaction = $\frac{(1-0.833)}{0.833}=20\%.$

Answer: Option C

Video Explanation

Explanation :

Area of the original paper = π(20)² = 400π cm². The total cut portion area = 4(π)(5)² = 100π cm² . Therefore, area of the uncut (shaded) portion = (400 – 100) = 300π cm².
Hence, the required ratio = 300π : 100π = 3 : 1.

Answer: Option C

Video Explanation

Explanation :

As it can be seen from the diagram, because of the thickness of the wall, the dimensions of the inside of the box is as follows: length = (21 – 0.5 – 0.5) = 20 cm, width

= (11 – 0.5 – 0.5) = 10 cm and height = (6 – 0.5) = 5.5 cm.
Total number of faces to be painted = 4 walls + one base (as it is open from the top).
The dimensions of two of the walls = (10 × 5.5), that of the remaining two walls = (20 × 5.5) and that of the base
= (20 × 10).
So the total area to be painted = 2 × (10 × 5.5) + 2 × (20 × 5.5) + (20 × 10) = 530 cm2.
Since the total expense of painting this area is Rs. 70, the rate of painting = $\frac{70}{530}$ = 0.13 = Re 0.1 per sq.cm. (approximately).

Answer: Option D

Video Explanation

Explanation :

M(M(A(M(x, y), S(y, x)), x), A(y, x)) = M(M(A(M(2, 3),

S(3, 2)), 2), A(3, 2)) = M(M(A((2x3), (3 - 2)), 2), A(3, 2))

= M(M(A(6, 1), 2), A(3, 2)) = M(M((6 + 1), 2), (3 + 2))

= M(M(7, 2), 5) = M((7 × 2), 5) = M(14, 5) = (14 × 5) = 70.

Answer: Option B

Video Explanation

Explanation :

S[M(D(A(a,b),2),D(A(a,b),2)),M(D(S(a,b),2), D(S(a, b),2))]

= S[M(D((a + b),2),D((a + b),2)),M(D((a – b),2),

D((a – b),2))] = $S\left[M\left\{\frac{(a+b)}{2},\frac{(a+b)}{2}\right\},M\left\{\frac{(a-b)}{2},\frac{(a-b)}{2}\right\}\right]$

$=S\left[{\left\{\frac{(a+b)}{2}\right\}}^2,{\left\{\frac{(a-b)}{2}\right\}}^2={\left\{\frac{(a+b)}{2}\right\}}^2-{\left\{\frac{(a-b)}{2}\right\}}^2\right]$

= $\frac{4ab}{2}=ab$

Answer: Option C

Video Explanation

Explanation :

Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2) . . . where k is a constant.

The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx², 4kx², 9kx² and 16kx². So the total price of the pieces = (1 + 4 + 9 + 16)kx² = 30kx². Hence, the difference in the price of the original diamond and its pieces = 100kx² – 30kx² = 70kx² = 70000.
Hence, kx² = 1000 and the original price = 100kx² = 100 × 1000 = 100000 = Rs. 1 lakh.

Answer: Option C

Video Explanation

Explanation :

n(n² – 1) = (n – 1)n(n + 1). If you observe, this is the product of three consecutive integers with middle one
being an odd integer. Since there are two consecutive even numbers, one of them will be a multiple of 4 and the other one will be multiple of 2. Hence, the product will be a multiple of 8. Also since they are three consecutive integers, one of them will definitely be a multiple of 3.
Hence, this product will always be divisible by (3 × 8)= 24.
Hint: Students, please note if a number is divisible by 96, it will also be divisible by 48 and 24. Similarly, if a number is divisible by 48, it is will always divisible by 24. Since there cannot be more than one right answers, we can safely eliminate options (a) and (b).

Answer: Option B

Video Explanation

Explanation :

The radius of the circle is 6.5 cm. Therefore, its diameter = 13 cm and AB = 13 cm. Since the diameter of a circle
subtends 90° at the circumference, ∠ACB = 90°. So ΔACB is a right-angled triangle with AC = 5, AB = 13.

Therefore, CB should be equal to 12 cm (as 5-12-13 form a Pythagorean triplet).

Hence, the area of the triangle $=\frac{1}{2}$ × AC × CB = $\frac{1}{2}$ × 5 × 12 = 30 sq.cm.

Answer: Option B

Video Explanation

Explanation :

Total expense incurred in making 1,500 watches = (1500 x 150) + 30000 = Rs. 2,55,000.

Total revenue obtained by selling 1,200 of them during the season = (1200 × 250) = Rs. 3,00,000. 

The remaining 300 of them has to be sold by him during off season.
The total revenue obtained by doing that = (300 × 100) = Rs. 30,000. 

Hence, total revenue obtained = (300000 + 30000) = Rs. 3,30,000. 

Hence, total profit = (330000 – 255000) = Rs. 75,000.

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

From the previous solution, we can see that the total expense incurred by him in manufacturing 1,500 watches = Rs.2,55,000. 

In order to break-even, he has to make a minimum revenue in order to recover his expenditure. 

He gets Rs. 250 per watch sold and Rs. 100 on every watch not sold. 

Let him sell x watches to break-even.
So our equation will be 250x + 100(1500 – x) = 255000. 

Solving this, we get x = 700 watches.

Hence, option (b).

[Note: Do not be confused by the statement "he is able to sell all the watches". This is to tell you that if required to bread even he would be able to sell all watches.]

Answer: Option A

Video Explanation

Explanation :

Since I paid Rs. 20 and because of lack of change, the clerk gave me Rs. 3 worth of stamps, it can be concluded that the total value of the stamp that I wanted to buy is Rs. 17. Since I ordered initially a minimum of 2 stamps of each denominations, if I buy exactly 2 stamps each, my total value is 2(5 + 2 + 1) = Rs. 16. The only way in which I make it Rs. 17 is buying one more stamp of Re 1. Hence, the total number of stamps that I ordered = (2 + 2 + 3) = 7. In addition, the clerk gave me 3 more.
Hence, the total number of stamps that I bought = (7 + 3) = 10 (viz. 2 five-rupee, 2 two-rupee and 6 one-rupee stamps).

Answer: Option C

Video Explanation

Explanation :

In a right-angled triangle, the length median to the hypotenuse is half the length of the hypotenuse. Hence, BD = $\frac{1}{2}$ AC = 3 cm. This relationship can be verified by knowing that the diameter of a circle subtends a right
angle at the circumference e.g. in the above figure D is the centre of the circle with AC as diameter. Hence,
∠ABC should be 90°. So BD should be the median to the hypotenuse. Thus, we can see that BD = AD = CD = Radius of this circle.
Hence, BD = $\frac{1}{2}$ diameter = $\frac{1}{2}$ AC = $\frac{1}{2}$ × hypotenuse.

Answer: Option B

Video Explanation

Explanation :

Hint: Students, please note that this sum could be intelligently solved by looking at both the questions together and also the answer choices. We know that the inventory has reduced by 54 units. This means two things: (i) actual quantity sold was less than the figure that was entered the computer (i.e. after interchanging digits), so the unit’s place digit of the actual quantity sold should be less than its ten’s place digit; and (ii) the difference between the actual quantity sold and the one that was entered in the computer is 54. From question 125, we can figure out that the only answer choice that supports both these conditions is (a), as (82 – 28 = 54). So the actual quantity sold = 28. Now since the total sales is Rs.1,148, actual price per piece = $\frac{1148}{28}$ = Rs. 41. 

Hence, the answer to question 124 is (b).

Answer: Option A

Video Explanation

Explanation :

Hint: Students, please note that this sum could be intelligently solved by looking at both the questions together and also the answer choices. We know that the inventory has reduced by 54 units. This means two things: (i) actual quantity sold was less than the figure that was entered the computer (i.e. after interchanging digits), so the unit’s place digit of the actual quantity sold should be less than its ten’s place digit; and (ii) the difference between the actual quantity sold and the one that was entered in the computer is 54. From question 125, we can figure out that the only answer choice that supports both these conditions is (a), as (82 – 28 = 54). So the actual quantity sold = 28. Now since the total sales is Rs.1,148, actual price per piece = $\frac{1148}{28}$ = Rs. 41. 

Hence, the answer to question 124 is (b).

Answer: Option D

Video Explanation

Explanation :

Probability of people having cable-TV = 2/3
Probability of people having VCR = 1/5
Probability of people having both = 1/10

∴ Probability of people having cable TV or VCR = $\frac{2}{3}+\frac{1}{15}-\frac{1}{10}$ = $\frac{20+6-3}{30}$ = $\frac{23}{30}$

Hence, option (d).

Answer: Option B

Video Explanation

Explanation :

$\frac{1}{1+{\displaystyle \frac{1}{3-{\displaystyle \frac{4}{2+{\displaystyle \frac{1}{3-{\displaystyle \frac{1}{2}}}}}}}}}+\frac{3}{3-{\displaystyle \frac{4}{3+{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2}}}}}}}$

$=\frac{1}{1+{\displaystyle \frac{1}{3-{\displaystyle \frac{4}{2+{\displaystyle \frac{2}{5}}}}}}}+\frac{3}{3-{\displaystyle \frac{4}{3+{\displaystyle \frac{1}{{\displaystyle \frac{3}{2}}}}}}}$

$=\frac{1}{1+{\displaystyle \frac{1}{3-{\displaystyle \frac{4}{{\displaystyle \frac{12}{5}}}}}}}+\frac{3}{3-{\displaystyle \frac{4}{3+{\displaystyle \frac{1}{{\displaystyle \frac{3}{2}}}}}}}$

$=\frac{1}{1+{\displaystyle \frac{1}{3-{\displaystyle \frac{5}{3}}}}}+\frac{3}{3-{\displaystyle \frac{4}{3+{\displaystyle \frac{2}{3}}}}}=\frac{1}{1+{\displaystyle \frac{1}{{\displaystyle \frac{4}{3}}}}}+\frac{3}{3-{\displaystyle \frac{4}{{\displaystyle \frac{11}{3}}}}}$

$=\frac{1}{1+{\displaystyle \frac{3}{4}}}+\frac{3}{3-{\displaystyle \frac{12}{11}}}=\frac{1}{3-{\displaystyle \frac{12}{11}}}+\frac{3}{3-{\displaystyle \frac{12}{11}}}=\frac{1}{{\displaystyle \frac{7}{4}}}+\frac{3}{{\displaystyle \frac{21}{11}}}$

$=\frac{4}{7}+\frac{11}{7}=\frac{15}{7}$

Answer: Option D

Video Explanation

Explanation :

If we write the given equation in the conventional form, i.e. ax² + bx + c, a = 1, b = – (A – 3), i.e. (3 – A) and c = –(A – 2), i.e. (2 – A). 

Let the roots of this equation be α and β. So the sum of the squares of the roots = α² + β² = (α + β)² – 2αβ.

Now (α + β) = Sum of the roots = $\frac{-b}{a}$ $=\frac{(A-3)}{1}=(A-3)$ 

and αβ = Product of the roots = $\frac{c}{a}$$=\frac{(2-A)}{1}=$ (2 - A).

Hence, α² + β² = (A – 3)² – 2(2 – A) = A² – 4A + 5 = 0.

Discriminant of this equation is less than 0, hence there no real value of A possible.

Hence, option (d).

Answer: Option D

Video Explanation

Explanation :

Let radius of the semicircle be R and radius of the circle be r.
Let P be the centre of semicircle and Q be the centre of the circle.
Draw QS parallel to BC.
Now, ΔPQS ~ ΔPBC

$$\begin{gathered} \therefore \frac{PQ}{PB}=\frac{QS}{BC} \\ \Rightarrow \frac{R+r}{\sqrt{2}R}=\frac{R-r}{R} \\ \Rightarrow R+r=\sqrt{2}R-\sqrt{2}r \\ \Rightarrow r(1+\sqrt{2})=R(\sqrt{2}-1) \end{gathered}$$

$\Rightarrow r=R\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$\Rightarrow r=R{(\sqrt{2}-1)}^2$

Required Ratio = $\frac{\pi r^2}{\pi R^2}\times 2$

$=\frac{\pi R^2{(\sqrt{2}-1)}^4\times 2}{\pi R^2}$

$=2{(\sqrt{2}-1)}^4:1$

Answer: Option B

Video Explanation

Explanation :

Let us look at the two equations. Let (5 pens + 7 pencils + 4 erasers) cost Rs. x. Hence, (6 pens + 14 pencils + 8 erasers) will cost Rs. 1.5x. Had, in the second case, Rajan decided to buy 10 pens instead of 6, the quantity of each one of them would have doubled over the first case and hence it would have cost me Rs. 2x. So (10 pens + 14 pencils + 8 erasers) = Rs. 2x. Now subtracting the second equation from the third, we get 4 pens cost Rs. 0.5x. Since 4 pens cost Re 0.5x, 5 of them will cost Re 0.625x. This is the amount that I spent on pens. Hence, fraction of the total amount paid = 0.625 = 62.5%.

Answer: Option D

Video Explanation

Explanation :

In a mile race, Akshay can be given a start of 128 m by Bhairav. This means that Bhairav can afford to start after Akshay has travelled 128 m and still complete one mile with him. In other words, Bhairav can travel one mile, i.e. 1,600 m in the same time as Akshay can travel (1600 – 128) = 1,472 m. Hence, the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time = $\frac{1600}{1472}$ = 25 : 32. Bhairav can give Chinmay a start of 4 miles. This means that in the time Bhairav runs 100 m, Chinmay only runs 96 m. So the ratio of the speeds of Bhairav and Chinmay = $\frac{100}{96}$ = 25 : 24.

Hence, we have B : A = 25 : 23 and B : C = 25 : 24. So A : B : C = 23 : 25 : 24. This means that in the time Chinmay covers 24 m, Akshay only covers 23 m. In other words, Chinmay is faster than Akshay. So if they race for $1\frac{1}{2}$ miles = 2,400 m, Chinmay will complete the race first and by this time Aksahy would only complete 2,300 m. In other words, Chinmay would beat Akshay by 100 m = $\frac{1}{16}$ mile.

Answer: Option D

Video Explanation

Explanation :

We can solve this by alligation. But while we alligate, we have to be careful that it has to be done with respect to any one of the two liquids, viz. either A or B. We can verify that in both cases, we get the same result. e.g. the proportion of A in the first vessel is $\frac{5}{6}$ and that in the second vessel is $\frac{1}{4},$ and we finally require $\frac{1}{2}$ parts of
A. Similarly, the proportion of B in the first vessel is $\frac{1}{6},$ that in the second vessel is $\frac{3}{4}$ and finally we want it to be $\frac{1}{2}.$ With respect to liquid A

With respect to liquid B

Answer: Option C

Video Explanation

Explanation :

Let the total distance be x. So the man travels a distance $\frac{3x}{5}$ at a speed 3a. Therefore, total time taken to travel this distance = $\frac{3x}{\left(15a\right)}=\frac{x}{\left(5a\right)}$

$\left[time=\frac{dis\tan ce}{speed}\right]$

He then travels a distance $\frac{2x}{5}$ at a speed 2b. Hence, time taken to travel this distance = $\frac{2x}{\left(10b\right)}=\frac{x}{\left(5b\right)}.$ So total time taken in going from A to B = $\frac{x}{\left(5a\right)}+\frac{x}{\left(5b\right)}.$ Now he travels from B to A and comes back. So total distance travelled = 2x at an average speed 5c.
Hence, time taken to return = $\frac{2x}{\left(5c\right)}$.

Since the time taken in both the cases remains the same, we can write $\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}$

Therefore, $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}.$

Answer: Option A

Video Explanation

Explanation :

Total time taken by the man to travel from A to D = 16 hr and total distance travelled = 36 km. The time that he would have taken had he not rested in between will be (16 – x – 2x) = (16 – 3x). But this time should be equal to the addition of the times that he takes to travel individual segments. This is given as:

$\frac{12}{x}+\frac{12}{2x}+\frac{12}{4x}=\frac{84}{4x}=\frac{21}{x}.$ Therefore, $\frac{21}{x}=$ (16 - 3x).

So we get the equation 3x² – 16x + 21 = 0. Solving this equation, we get x = 3 or x = $\frac{7}{3}$. This should be the time for which he rested at B.

Answer: Option A

Video Explanation

Explanation :

The team has played a total of (17 + 3) = 20 matches. This constitutes $\frac{2}{3}$ of the matches. Hence, total number of matches played = 30. To win $\frac{3}{4}$ of them, a team has to win 22.5, i.e. at least win 23 of them. In other words, the team has to win a minimum of 6 matches (since it has already won 17) out of remaining 10. So it can lose a maximum of 4 of them.

Answer: Option C

Video Explanation

Explanation :

This can simply be solved by multiplying the two multiplication factors to get the effective multiplication factor. e.g. multiplication factor for 30% increase = 1.30. Multiplication factor for 20% decrease = 0.8. Hence, 1.30 × 0.8 = 1.04. This multiplication factor (i.e. 1.04) indicates that there is a 4% increase in total revenue. So the answer is +4.

Alternative method:
By using the formula x + y + $\frac{xy}{100}$

∴ x = +30%; y = – 20%

⇒ 30 + 60 + $\frac{50(-20)}{100}$

= 30 – 20 – 6 = +4%

Answer: Option D

Video Explanation

Explanation :

The three lines can be expressed as $Y=\frac{5}{3}-\frac{2X}{3},Y=\frac{5X}{7}+\frac{2}{7}$ and $\frac{9X}{5}-\frac{4}{5}.$ Therefore, the slopes of the three lines are $\frac{-2}{3},\frac{5}{7}$ and $\frac{9}{5}$ respectively. For any two lines to be perpendicular to each other, the product of their slopes = –1. We find that the product of none of the slopes is –1. For any two be parallel, their slopes should be the same. This is again not the case. And finally for the two lines to be intersecting at the same point, there should be one set of values of (X, Y) that should satisfy the equations of 3 lines. Solving the first two equations, we get X = 1 and Y = 1. If we substitute this in the third equation, we find that it also satisfies that equation. So the solution set (1, 1) satisfies all three equations, suggesting that the three lines intersect at the same point, viz. (1, 1). Hence, they are coincident.

Answer: Option B

Video Explanation

Explanation :

Out of the 5 girls, 3 girls can be invited in ⁵C₃ ways. Nothing is mentioned about the number of boys that he has to invite. He can invite one, two, three, four or even no boys. Each boy can be invited or not. He can invite them in 24 ways. Thus, the total number of ways is ⁵C₃ × (2)⁴ = 10 × 16 = 160.

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

In this case, we need not use the data that SP = Rs. 300 each. This has to be used only to figure out that the SP of both the articles is the same. Also since the profit percentage on one is equal to the loss percentage on the other, viz. 10% effectively, it will be a loss given by $\frac{{\left(10\right)}^2}{100}=$ 1%. Hence, the correct answer is (–)1.

Hence, option (b).

Answer: Option C

Video Explanation

Explanation :

First series: (S₁) = x, y, $\frac{x}{2},$ z, x + 20

Second series: (S₂) = a₁, a₂, a₃, a₄

Now a₁ = y – x, a₂ = $\frac{x}{2}$ - y, a₃ = z - $\frac{x}{\mathrm{2 }}$and a₄ = x + 20 – z

a₂ – a₁ = 30 gives 3x – 4y = 60 ... (1)

a₄ – a₃ = 30 gives 3x – 4z = 20 ... (2)

and a₄ – a₂ = 60 gives x – 2z + 2y = 80 ... (3)

Solving these equations we get the values of x = 100, y = 60, z = 70

∴ S₁ is 100, 60, 50, 70, 120

S₂ is –40, –10, 20, 50.

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

Consider the solution to first question of this set.

Second term of S₂ = -10.

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

Consider the solution to first question of this set.

Difference between second and fourth term of S₁ = 70 - 60 = 10

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

Consider the solution to first question of this set.

Average of S₁ = (100 + 60 + 50 + 70 + 120)/5 = 80.

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Consider the solution to first question of this set.

Sum of S₂ = (-40) + (-10) + 20 + 50 = 20.

Hence, option (b).

Answer: 1

Explanation :

None of the statements is useful in finding the radius of the rear wheel. In the question, distance travelled is given. But the number of rotations taken by it is not given.

Answer: 1

Explanation :

Given that containers are in equal volume, that does not mean that quantities in each container are in equal volumes. Since we do not know the quantity of the liquid, we cannot find the ratio of the final mixture.

Answer: 4

Explanation :

This question can be answered by using the two statements.
Given (a – b + c) > (a + b – c).
It is nothing but is (–b + c) > (b – c).
Since b is negative and c is positive,
⇒ c > b
Using both statements
c > 0
b < 0
c > b
So always (a – b + c) > (a + b – c).

Answer: 4

Explanation :

Using statement II

2αß = $\frac{c}{a}$ = αß

⇒ α = 0 or β = 0 or α and β = 0
Hence, cannot be answered.

Answer: 1

Explanation :

Both the statements are telling the same, that selling price is 75% of cost price.
So we cannot determine the actual cost of the article.

Answer: 3

Explanation :

By using statement II we can determine the selling price of the article.
Selling price = 1.25 × 250 = 312.5
But by using statement I we cannot determine the selling price.

Answer: 1

Explanation :

The question cannot be answered until and unless number of concurrent lines are known.

Answer: 4

Explanation :

Both the statements are needed to answer the questions. Since in statement I all the dates are given except the time to compound the interest. That date is given in the second statement.

Answer: 1

Explanation :

We cannot answer the question using both the statements.
Given that Anil’s ages are prime numbers in 1998 and 1996. It is of difference 2. There are so many prime numbers with difference 2. They are (17, 19), (41, 43) . . . so on.

So we cannot find out exact age of Sachin.

Answer: 2

Explanation :

Consider the statement I:
Let number of type-1 widgets = x.
Number of type-2 widgets = y.
From the given question, x + y = 20000.
From statement I, 1.1 x + 0.94y = 20000.
So we can get x and y.
From statement II, number of type-2 widgets produced

$=\frac{1}{3}$ × 20000 = 6667.

The question can be answered by using either of the statements alone.

Answer: Option B

Video Explanation

Explanation :

For the number to be divisible by 9, the sum of the digits should be a multiple of 9.

We find that the sum of all the digits (excluding A and B) = (7 + 7 + 4 + 9 + 5 + 8 + 9 + 6) = 55. The next higher multiple of 9 is 63 or 72.

Hence, the sum of A and B should either be 8 or 17. We find that (a) and (c) cannot be the answer.

For a number to be divisible by 8, the number formed by its last three digits should be divisible by 8. The last three digits are 96B. The multiples of 8 beginning with 96 are 960 and 968. Hence, B can either be 0 or 8. Both of which satisfy our requirement of the number being divisible by 9 as well.

Therefore, A and B could either be 0 and 8 or 8 and 0 respectively.

Answer: Option A

Video Explanation

Explanation :

If we simplify the expression x² – 3x + 2 > 0, we get (x – 1)(x – 2) > 0. For this product to be greater than zero, either both the factors should be greater than zero or both of them should be less than zero. Therefore, (x – 1) > 0 and (x – 2) > 0 or (x – 1) < 0 and (x – 2) < 0.

Hence, x > 1 and x > 2 or x < 1 and x < 2. If we were to club the ranges, we would get either x > 2 or x < 1. So for any value of x equal to or between 1 and 2, the above equation does not follow.