# CAT 1996 QA | Previous Year CAT Paper

**Direction: Answer the questions based on the following information.**

In a locality, there are five small cities: A, B, C, D and E. The distances of these cities from each other are as follows.

AB = 2 km AC = 2km AD > 2 km AE > 3 km BC = 2 km

BD = 4 km BE = 3 km CD = 2 km CE = 3 km DE > 3 km

**1. CAT 1996 QA | Arithmetic - Time, Speed & Distance**

If a ration shop is to be set up within 2 km of each city, how many ration shops will be required?

- A.
2

- B.
3

- C.
4

- D.
5

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.

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**2. CAT 1996 QA | Arithmetic - Time, Speed & Distance**

If a ration shop is to be set up within 3 km of each city, how many ration shops will be required?

- A.
1

- B.
2

- C.
3

- D.
4

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.

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**3. CAT 1996 QA | Geometry - Quadrilaterals & Polygons**

If ABCD is a square and BCE is an equilateral triangle, what is the measure of ∠DEC?

- A.
15°

- B.
30°

- C.
20°

- D.
45°

Answer: Option A

**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°.

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**4. CAT 1996 QA | Arithmetic - Profit & Loss**

Instead of a metre scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?

- A.
20%

- B.
25%

- C.
40%

- D.
15%

Answer: Option A

**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 = $\frac{(1-0.833)}{0.833}=20\%.$

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**5. CAT 1996 QA | Geometry - Mensuration**

From a circular sheet of paper with a radius 20 cm, four circles of radius 5 cm each are cut out. What is the ratio of the uncut to the cut portion?

- A.
1 : 3

- B.
4 : 1

- C.
3 : 1

- D.
4 : 3

Answer: Option C

**Explanation** :

Area of the original paper = π(20)^{2} = 400π cm^{2}. The total cut portion area = 4(π)(5)^{2} = 100π cm^{2 }. Therefore, area of the uncut (shaded) portion = (400 – 100) = 300π cm^{2}.

Hence, the required ratio = 300π : 100π = 3 : 1.

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**6. CAT 1996 QA | Geometry - Mensuration**

A wooden box (open at the top) of thickness 0.5 cm, length 21 cm, width 11 cm and height 6 cm is painted on the inside. The expenses of painting are Rs. 70. What is the rate of painting per square centimetres?

- A.
Re 0.7

- B.
Re 0.5

- C.
Re 0.1

- D.
Re 0.2

Answer: Option C

**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).

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**Direction: Answer the questions based on the following information.**

A, S, M and D are functions of x and y, and they are defined as follows.

A(x, y) = x + y

S(x, y) = x – y

M(x, y) = xy

D(x, y) = $\frac{x}{y}$, y ≠ 0

**7. CAT 1996 QA | Algebra - Functions & Graphs**

What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for x = 2, y = 3?

- A.
60

- B.
140

- C.
25

- D.
70

Answer: Option D

**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.

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**8. CAT 1996 QA | Algebra - Functions & Graphs**

What is the value of S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]?

- A.
a

^{2}+ b^{2} - B.
ab

- C.
a

^{2}- b^{2} - D.
$\frac{a}{b}$

Answer: Option B

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

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

= $\frac{4ab}{2}=ab$

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**9. CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation**

The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.

- A.
Rs. 1.4 lakh

- B.
Rs. 2 lakh

- C.
Rs. 1 lakh

- D.
Rs. 2.1 lakh

Answer: Option C

**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 kx^{2}, 4kx^{2}, 9kx^{2} and 16kx^{2}. So the total price of the pieces = (1 + 4 + 9 + 16)kx^{2 }= 30kx^{2}. Hence, the difference in the price of the original diamond and its pieces = 100kx^{2} – 30kx^{2} = 70kx^{2} = 70000.

Hence, kx^{2} = 1000 and the original price = 100kx^{2 }= 100 × 1000 = 100000 = Rs. 1 lakh.

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**10. CAT 1996 QA | Algebra - Number Theory**

If n is any odd number greater than 1, then n(n^{2} – 1) is

- A.
divisible by 96 always

- B.
divisible by 48 always

- C.
divisible by 24 always

- D.
None of these

Answer: Option C

**Explanation** :

n(n^{2} – 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).

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**11. CAT 1996 QA | Geometry - Circles**

The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5 cm long, find the area of ΔABC.

- A.
60 sq.cm

- B.
30 sq.cm

- C.
40 sq.cm

- D.
52 sq.cm

Answer: Option B

**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.

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**Direction: Answer the questions based on the following information.**

A watch dealer incurs an expense of Rs. 150 for producing every watch. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each watch for Rs. 100.

**12. CAT 1996 QA | Arithmetic - Profit & Loss**

If he is able to sell only 1,200 out of 1,500 watches he has made in the season, then he has made a profit of

- A.
Rs. 90,000

- B.
Rs. 75,000

- C.
Rs. 45,000

- D.
Rs. 60,000

Answer: Option B

**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).

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**13. CAT 1996 QA | Arithmetic - Profit & Loss**

If he produces 1,500 watches, what is the number of watches that he must sell during the season in order to break-even, given that he is able to sell all the watches produced?

- A.
500

- B.
700

- C.
800

- D.
1,000

Answer: Option B

**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.]

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**14. CAT 1996 QA | Algebra - Simple Equations**

Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?

- A.
10

- B.
9

- C.
12

- D.
8

Answer: Option A

**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).

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**15. CAT 1996 QA | Geometry - Triangles**

In ΔABC, ∠B is a right angle, AC = 6 cm, and D is the mid-point of AC. The length of BD is

- A.
4 cm

- B.
$\sqrt{6}$

- C.
3 cm

- D.
3.5 cm

Answer: Option C

**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.

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Direction: Answer the questions based on the following information.

A salesman enters the quantity sold and the price into the computer. Both the numbers are two-digit numbers. But, by mistake, both the numbers were entered with their digits interchanged. The total sales value remained the same, i.e. Rs. 1,148, but the inventory reduced by 54.

**16. CAT 1996 QA | Algebra - Simple Equations**

What is the actual price per piece?

- A.
Rs. 82

- B.
Rs. 41

- C.
Rs. 6

- D.
Rs. 28

Answer: Option B

**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).

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**17. CAT 1996 QA | Algebra - Simple Equations**

What is the actual quantity sold?

- A.
28

- B.
14

- C.
82

- D.
41

Answer: Option A

**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).

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**18. CAT 1996 QA | Venn Diagram**

In a locality, two-thirds of the people have cable TV, one-fifth have VCR, and one-tenth have both. What is the fraction of people having either cable-TV or VCR?

- A.
$\frac{19}{30}$

- B.
$\frac{2}{3}$

- C.
$\frac{17}{30}$

- D.
$\frac{23}{30}$

Answer: Option D

**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).

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**19. CAT 1996 QA | Miscellaneous**

Find the value of $\frac{1}{1+{\displaystyle \frac{1}{3-{\displaystyle \frac{4}{2+{\displaystyle \frac{1}{3-{\displaystyle \frac{1}{2}}}}}}}}}+\frac{3}{3-{\displaystyle \frac{4}{3+{\displaystyle \frac{1}{2-{\displaystyle \frac{1}{2}}}}}}}$

- A.
$\frac{13}{7}$

- B.
$\frac{15}{7}$

- C.
$\frac{11}{21}$

- D.
$\frac{17}{28}$

Answer: Option B

**Explanation** :

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

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

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

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

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

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

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**20. CAT 1996 QA | Algebra - Quadratic Equations**

Given the quadratic equation x^{2} – (A – 3)x – (A – 2), for what value of A will the sum of the squares of the roots be zero?

- A.
-2

- B.
3

- C.
6

- D.
None of these

Answer: Option D

**Explanation** :

If we write the given equation in the conventional form, i.e. ax^{2} + 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 = α^{2} + β^{2} = (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, α^{2} + β^{2} = (A – 3)^{2} – 2(2 – A) = A^{2} – 4A + 5 = 0.

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

Hence, option (d).

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**21. CAT 1996 QA | Geometry - Quadrilaterals & Polygons**

The figure shows the rectangle ABCD with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

- A.
${\left(\sqrt{2}-1\right)}^{2}:1$

- B.
$2{\left(\sqrt{2}-1\right)}^{2}:1$

- C.
${\left(\sqrt{2}-1\right)}^{2}:2$

- D.
None of these

Answer: Option D

**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

$\therefore \frac{PQ}{PB}=\frac{QS}{BC}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{R+r}{\sqrt{2}R}=\frac{R-r}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R+r=\sqrt{2}R-\sqrt{2}r\phantom{\rule{0ex}{0ex}}\Rightarrow r(1+\sqrt{2})=R(\sqrt{2}-1)$

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

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

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

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

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

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**22. CAT 1996 QA | Arithmetic - Percentage**

I bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more what I had paid. What per cent of the total amount paid by me was paid for the pens?

- A.
37.5%

- B.
62.5%

- C.
50%

- D.
None of these

Answer: Option B

**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%.

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**23. CAT 1996 QA | Arithmetic - Time, Speed & Distance**

In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.)

- A.
Akshay, $\frac{1}{12}$ mile

- B.
Chinmay, $\frac{1}{32}$ mile

- C.
Akshay, $\frac{1}{24}$ mile

- D.
Chinmay, $\frac{1}{16}$ mile

Answer: Option D

**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.

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**24. CAT 1996 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

- A.
2 : 3

- B.
4 : 3

- C.
3 : 2

- D.
3 : 4

Answer: Option D

**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

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**25. CAT 1996 QA | Arithmetic - Time, Speed & Distance**

A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and return at a speed 5c in the same time, then

- A.
$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$

- B.
a + b = 99

- C.
$\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$

- D.
None of these

Answer: Option C

**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 = $\frac{3x}{\left(15a\right)}=\frac{x}{\left(5a\right)}$

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

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

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**26. CAT 1996 QA | Arithmetic - Time, Speed & Distance**

A man travels from A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be

- A.
3 hr

- B.
6 hr

- C.
2 hr

- D.
4 hr

Answer: Option A

**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 3x^{2} – 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.

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**27. CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation**

Out of two-thirds of the total number of basketball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win more than threefourths of the total number of matches, if it is true that no match can end in a tie?

- A.
4

- B.
6

- C.
5

- D.
3

Answer: Option A

**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.

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**28. CAT 1996 QA | Arithmetic - Percentage**

The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?

- A.
-4%

- B.
-2%

- C.
+4%

- D.
+2%

Answer: Option C

**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%

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**29. CAT 1996 QA | Algebra - Simple Equations**

The points of intersection of three lines 2X + 3Y – 5 = 0, 5X – 7Y + 2 = 0 and 9X – 5Y – 4= 0

- A.
form a triangle

- B.
are on lines perpendicular to each other

- C.
are on lines parallel to each other

- D.
are coincident

Answer: Option D

**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.

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**30. CAT 1996 QA | Modern Math - Permutation & Combination**

A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitees?

- A.
320

- B.
160

- C.
80

- D.
200

Answer: Option B

**Explanation** :

Out of the 5 girls, 3 girls can be invited in ^{5}C_{3 }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 ^{5}C_{3} × (2)^{4} = 10 × 16 = 160.

Hence, option (b).

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**31. CAT 1996 QA | Arithmetic - Profit & Loss**

I sold two watches for Rs. 300 each, one at the loss of 10% and the other at the profit of 10%. What is the percentage of loss(–) or profit(+) that resulted from the transaction?

- A.
(+)10

- B.
(-)1

- C.
(+)1

- D.
(-)10

Answer: Option B

**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).

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**Answer the questions based on the following information.**

A series S_{1} of five positive integers is such that the third term is half the first term and the fifth term is 20 more than the first term. In series S_{2}, the n^{th} term defined as the difference between the (n + 1)^{th} term and the n^{th} term of series S_{1}, is an arithmetic progression with a common difference of 30.

**32. CAT 1996 QA | Algebra - Progressions**

First term of S_{1} is

- A.
80

- B.
90

- C.
100

- D.
120

Answer: Option C

**Explanation** :

First series: (S_{1}) = x, y, $\frac{x}{2},$ z, x + 20

Second series: (S_{2}) = a_{1}, a_{2}, a_{3}, a_{4}

Now a_{1} = y – x, a_{2} = $\frac{x}{2}$ - y, a_{3 }= z - $\frac{x}{2}$and a_{4} = x + 20 – z

a_{2} – a_{1} = 30 gives 3x – 4y = 60 ... (1)

a_{4} – a_{3} = 30 gives 3x – 4z = 20 ... (2)

and a_{4} – a_{2} = 60 gives x – 2z + 2y = 80 ... (3)

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

∴ S_{1} is 100, 60, 50, 70, 120

S_{2} is –40, –10, 20, 50.

Hence, option (c).

Workspace:

**33. CAT 1996 QA | Algebra - Progressions**

Second term of S_{2} is

- A.
50

- B.
60

- C.
70

- D.
None of these

Answer: Option D

**Explanation** :

Consider the solution to first question of this set.

Second term of S_{2} = -10.

Hence, option (d).

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**34. CAT 1996 QA | Algebra - Progressions**

What is the difference between second and fourth terms of S_{1}?

- A.
10

- B.
20

- C.
30

- D.
60

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Difference between second and fourth term of S_{1} = 70 - 60 = 10

Hence, option (a).

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**35. CAT 1996 QA | Algebra - Progressions**

What is the average value of the terms of series S_{1}?

- A.
60

- B.
70

- C.
80

- D.
Average is not an integer

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Average of S_{1} = (100 + 60 + 50 + 70 + 120)/5 = 80.

Hence, option (c).

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**36. CAT 1996 QA | Algebra - Progressions**

What is the sum of series S_{2}?

- A.
10

- B.
20

- C.
30

- D.
40

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Sum of S_{2} = (-40) + (-10) + 20 + 50 = 20.

Hence, option (b).

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**37. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

1. if the question cannot be answered even with the help of both the statements taken together.

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**A tractor travelled a distance 5 m. What is the radius of the rear wheel?**

I. The front wheel rotates ‘N’ times more than the rear wheel over this distance.

II. The circumference of the rear wheel is ‘t’ times that of the front wheel.

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.

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**38. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

1. if the question cannot be answered even with the help of both the statements taken together.

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**What is the ratio of the two liquids A and B in the mixture finally, if these two liquids kept in three vessels are mixed together? (The containers are of equal volume.)**

I. The ratio of liquid A to liquid B in the first and second vessel is 3 : 5, 2 : 3 respectively.

II. The ratio of liquid A to liquid B in vessel 3 is 4 : 3.

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.

Workspace:

**39. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

1. if the question cannot be answered even with the help of both the statements taken together.

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**If a, b and c are integers, is (a – b + c) > (a + b – c)?**

I. b is negative.

II. c is positive.

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).

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**40. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**If α and β are the roots of the equation (ax ^{2} + bx + c = 0), then what is the value of (α^{2} + β^{2})?**

I. α + ß = -$\left(\frac{b}{a}\right)$

II. 2αß = $\left(\frac{c}{a}\right)$

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.

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**41. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**What is the cost price of the article?**

I. After selling the article, a loss of 25% on cost price is incurred.

II. The selling price is three-fourths of the cost price.

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.

Workspace:

**42. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**What is the selling price of the article?**

I. The profit on sales is 20%.

II. The profit on each unit is 25% and the cost price is Rs. 250.

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.

Workspace:

**43. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**How many different triangles can be formed?**

I. There are 16 coplanar, straight lines.

II. No two lines are parallel.

Answer: 1

**Explanation** :

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

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**44. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**What is the total worth of Lakhiram's assets?**

I. A compound interest at 10% on his assets, followed by a tax of 4% on the interest, fetches him Rs. 1,500 this year.

II. The interest is compounded once every four months.

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.

Workspace:

**45. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**How old is Sachin in 1997?**

I. Sachin is 11 years younger than Anil whose age will be a prime number in 1998.

II. Anil's age was a prime number in 1996.

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.

Workspace:

**46. CAT 1996 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Mark the answer as

2. if the question can be answered by any one of the two statements.

3. if each statement alone is sufficient to answer the question, but not the other one (e.g. statement I alone is required to answer the question, but not statement II and vice versa).

4. if both statements I and II together are needed to answer the question.

**What is the number of type-2 widgets produced, if the total number of widgets produced is 20,000?**

I. If the production of type-1 widgets increases by 10% and that of type-2 decreases by 6%, the total production remains the same.

II. The ratio in which type-1 and type-2 widgets are produced is 2 : 1.

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.

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**47. CAT 1996 QA | Algebra - Number Theory**

If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

- A.
7 and 8

- B.
8 and 0

- C.
5 and 8

- D.
None of these

Answer: Option B

**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.

Workspace:

**48. CAT 1996 QA | Geometry - Quadrilaterals & Polygons**

Which of the following values of x do not satisfy the inequality (x^{2} – 3x + 2 > 0) at all?

- A.
1 ≤ x ≤ 2

- B.
–1 ≥ x ≥ –2

- C.
0 ≤ x ≤ 2

- D.
0 ≥ x ≥ –2

Answer: Option A

**Explanation** :

If we simplify the expression x^{2} – 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.

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