# CAT 1994 QA | Previous Year Questions

**1. CAT 1994 QA | Arithmetic - Percentage**

The number of votes not cast for the Praja Party increased by 25% in the National General Election over those not cast for it in the previous Assembly Polls, and the Praja Party lost by a majority twice as large as that by which it had won the Assembly Polls. If a total 2,60,000 people voted each time. How many voted for the Praja Party in the Assembly Elections.

- A.
1,10,000

- B.
1,50,000

- C.
1,40,000

- D.
1,20,000

Answer: Option C

**Explanation** :

Let x be the number of votes not cast for Praja Party in the previous polls.

So the number of votes not cast for the party in this assembly polls would be 1.25x.

Margin of victory in the previous polls

= (Votes cast) – (Votes not cast)

= (260000 – x) – x = (260000 – 2x).

Margin of loss in this years polls

= 1.25x – (260000 – 1.25x)

= (2.5x – 260000).

As per the given information, margin of loss this year = 2 × Margin of victory last year.

Therefore, (2.5x – 260000) = 2(260000 – 2x).

∴ x = 120000.

So the number of votes cast for the party in assembly election = 260000 – 120000 = 140000.

Hence, option (c).

Workspace:

Ghoshbabu is staying at Ghosh Housing Society, Aghosh Colony, Dighospur, Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; in his colony there is no person who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighospur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Colony and Dighospur is 33 and 121 respectively; while the persons who read Anand Bazar Patrika in Aghosh Colony and Dighospur are 32 and 117 respectively.

**2. CAT 1994 QA | Venn Diagram**

Number of persons in Dighospur who read only Ganashakti is

- A.
121

- B.
83

- C.
79

- D.
127

Answer: Option B

**Explanation** :

The data can be represented in the following Venn diagrams.

Number of persons in Dighoshpur who read only Ganashakti = 121 - 38 = 83.

Hence, option (b).

Workspace:

Ghoshbabu is staying at Ghosh Housing Society, Aghosh Colony, Dighospur, Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; in his colony there is no person who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighospur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Colony and Dighospur is 33 and 121 respectively; while the persons who read Anand Bazar Patrika in Aghosh Colony and Dighospur are 32 and 117 respectively.

**3. CAT 1994 QA | Venn Diagram**

Number of persons in Aghosh Colony who read both of these newspapers is

- A.
13

- B.
20

- C.
19

- D.
14

Answer: Option A

**Explanation** :

The data can be represented in the following Venn diagrams.

Number of persons in Aghosh Colony who read both the newspapers = 13.

Workspace:

Ghoshbabu is staying at Ghosh Housing Society, Aghosh Colony, Dighospur, Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; in his colony there is no person who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighospur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Colony and Dighospur is 33 and 121 respectively; while the persons who read Anand Bazar Patrika in Aghosh Colony and Dighospur are 32 and 117 respectively.

**4. CAT 1994 QA | Venn Diagram**

Number of persons in Aghosh Colony who read only one paper

- A.
29

- B.
19

- C.
39

- D.
20

Answer: Option C

**Explanation** :

The data can be represented in the following Venn diagrams.

Number of persons in Aghosh Colony who read only 1 newspaper = 20+19 = 39.

Workspace:

**5. CAT 1994 QA | Algebra - Logarithms**

If log_{7} log_{5} $\left(\sqrt{x+5}+\sqrt{x}\right)$ = 0, find the value of x.

- A.
1

- B.
0

- C.
2

- D.
None of these

Answer: Option D

**Explanation** :

log_{7} log_{5} $\left(\sqrt{x+5}+\sqrt{x}\right)$ = 0

We know

log_{a} b = x ⇒ a^{x} = b

∴log_{5} $\left(\sqrt{x+5}+\sqrt{x}\right)$ = 7^{0}

⇒ log_{5}$\left(\sqrt{x+5}+\sqrt{x}\right)$ = 1

⇒ $\sqrt{x+5}+\sqrt{x}$ = 5^{1}

⇒ $\sqrt{x+5}=5-\sqrt{x}$

Squaring both sides

⇒ x + 5 = 25 + x − 2 × 5$\sqrt{x}$

⇒ x + 5 = 25 + x + 10$\sqrt{x}$

⇒ 10$\sqrt{x}$ = 20

⇒ $\sqrt{x}$ = 2

⇒ ${\left(\sqrt{x}\right)}^{2}$ = (2)^{2}

⇒ x = 4

Workspace:

**6. CAT 1994 QA | Geometry - Mensuration**

A right circular cone, a right circular cylinder and a hemisphere, all have the same radius, and the heights of the cone and cylinder equal to their diameters. Then their volumes are proportional, respectively to

- A.
1 : 3 : 1

- B.
2 : 1 : 3

- C.
3 : 2 : 1

- D.
1 : 2 : 3

Answer: Option A

**Explanation** :

If the diameters and the heights of a cone and a cylinder are same, then the volume of cone is always 1/3rd the volume of the cylinder.

So the ratio of the volume of cone to the volume of cylinder = 1 : 3.

The only answer choice that supports this is (a).

Workspace:

**7. CAT 1994 QA | Arithmetic - Time, Speed & Distance**

Two towns A and B are 100 km apart. A school is to be built for 100 students of town B and 30 students of Town A. Expenditure on transport is Rs. 1.20 per km per student. If the total expenditure on transport by all 130 students is to be as small as possible, then the school should be built at

- A.
33 km from Town A.

- B.
33 km from Town B

- C.
Town A

- D.
Town B

Answer: Option D

**Explanation** :

Hence we find that the least expenditure will be incurred if the school is located in town B.

**HINT:**

Students please note that since there are more number of students from Town B, to minimise the total expenditure the school should be located as closer to town B as possible.

Workspace:

**8. CAT 1994 QA | Arithmetic - Time & Work**

One man can do as much work in one day as a woman can do in 2 days. A child does one third the work in a day as a woman. If an estate-owner hires 39 pairs of hands, men, women and children in the ratio 6 : 5 : 2 and pays them in all Rs. 1113 at the end of the days work. What must the daily wages of a child be, if the wages are proportional to the amount of work done?

- A.
Rs. 14

- B.
Rs. 5

- C.
Rs. 20

- D.
Rs. 7

Answer: Option D

**Explanation** :

There are 18 men, 15 women and 6 children.

Working efficiency of man : woman : child = 6 : 3 : 1.

So the ratio of the work done in a day by 18 men, 15 women and 6 children = (18 × 6) : (15 × 3) : (6 ×1) = 108 : 45 : 6.

Hence, the daily wage of Rs.1113 should be divided in this ratio. That makes it, Rs.756 for men, Rs.315 for women and Rs.42 for children.

Hence, 6 children earn Rs.42 in a day.

So the daily wage of a child should be Rs.7

Hence, option (d).

Workspace:

**9. CAT 1994 QA | Geometry - Mensuration**

A right circular cone of height h is cut by a plane parallel to the base and at a distance h/3 from the base, then the volumes of the resulting cone and the frustum are in the ratio

- A.
1 : 3

- B.
8 : 19

- C.
1 : 4

- D.
1 : 7

Answer: Option B

**Explanation** :

Let the radius and height of original cone be ‘r’ and ‘h’ respectively.

∴ The volume of the original cone (V) = $\frac{\pi {r}^{2}h}{3}$.

The height and radius of the smaller cone are $\frac{2h}{3}$ and $\frac{2r}{3}$ respectively.

So its volume = $\frac{\pi}{3}\times {\left(\frac{2r}{3}\right)}^{2}\times \frac{2h}{3}=\frac{8V}{27}$.

∴ Volume of frustum = $\left(V-\frac{8V}{27}\right)=\frac{19V}{27}.$

∴ Ratio of the volumes = 8 : 19.

Workspace:

**10. CAT 1994 QA | Algebra - Number Theory**

If a + b + c = 0, where a ≠ b ≠ c, then $\frac{{a}^{2}}{2{a}^{2}+bc}+\frac{{b}^{2}}{2{b}^{2}+ac}+\frac{{c}^{2}}{2{c}^{2}+ab}$ is equal to

- A.
zero

- B.
1

- C.
-1

- D.
abc

Answer: Option B

**Explanation** :

Assume some values of a, b & c such that sum of a, b and c is 0 where a ≠ b ≠ c, and find the value of the given expression.

Let a = 1, b = -1 and c = 0.

$\Rightarrow \frac{{a}^{2}}{2{a}^{2}+bc}+\frac{{b}^{2}}{2{b}^{2}+ac}+\frac{{c}^{2}}{2{c}^{2}+ab}=\frac{1}{2}+\frac{1}{2}$ + 0 = 1.

Workspace:

**11. CAT 1994 QA | Algebra - Progressions**

If the harmonic mean between two positive numbers is to their geometric mean as 12 : 13; then the numbers could be in the ratio

- A.
12 : 13

- B.
1/12 : 1/13

- C.
4 : 9

- D.
2 : 3

Answer: Option C

**Explanation** :

The harmonic mean of two numbers x and y is $\frac{2xy}{(x+y)}$and the geometric mean is $\sqrt{xy}$

∴ $\frac{{\displaystyle \frac{2xy}{(x+y)}}}{\sqrt{xy}}$ = $\frac{12}{13}$

⇒ $\frac{4xy}{{(x+y)}^{2}}$ = $\frac{144}{169}$

Although this can be simplified to get the answer, the best way to proceed from here would be to look out for the answer choices and figure out which pair of x & y satisfies the above equation.

Option (c) satisfies the above expression

**Alternately**,

Please note that this sum is a classic example of how you could have gone for intelligent guess work. Since we know that the denominator of the ratio is the geometric mean, which is $\sqrt{xy}$, the two numbers should be in such a ratio that their product should be a perfect square.

The only pair from the answer choices that supports this is 4 & 9, as $\sqrt{4\times 9}$ = $\sqrt{36}$ = 6

Hence, option (c).

Workspace:

**12. CAT 1994 QA | Algebra - Quadratic Equations**

If one root of x^{2} − px + 12 = 0 is 4, while the equation x^{2} − px + q = 0 has equal roots, then the value of q is

- A.
$\frac{49}{4}$

- B.
$\frac{4}{49}$

- C.
4

- D.
$\frac{1}{4}$

Answer: Option A

**Explanation** :

If one root of x^{2} + px + 12 = 0 is 4, then 4^{2} + 4p + 12 = 0, i.e. p = –7.

x^{2} – px + q = 0 has equal roots.

If the roots are α each, 2α = - $\frac{(-7)}{1}$ = 7, i.e., α = $\frac{7}{2}$, and q = α^{2}

$\Rightarrow q=\frac{49}{4}$

Hence, option (a).

Workspace:

If md(x) = x ,

mn(x,y) = minimum of x and y and

Ma(a,b,c,...) =maximum of a,b,c…

**13. CAT 1994 QA | Algebra - Functions & Graphs**

Value of Ma[md(a),mn(md(b),a),mn(ab,md(ac))] where a = -2, b = -3, c = 4 is

- A.
2

- B.
6

- C.
8

- D.
-2

Answer: Option B

**Explanation** :

Ma[md(−2),mn(md(−3),−2),mn(6,md(−8))]

= Ma[2,mn(3,−2),mn(6,8)] = Ma[2, -2, 6] = 6.

Workspace:

If md(x) = x ,

mn(x,y) = minimum of x and y and

Ma(a,b,c,...) =maximum of a,b,c…

**14. CAT 1994 QA | Algebra - Functions & Graphs**

Given that a > b then the relation Ma[md(a), mn(a,b)] = mn[a, md(Ma(a,b))] does not hold if

- A.
a < 0, b < 0

- B.
a > 0, b > 0

- C.
a > 0, b < 0, |a| < |b|

- D.
a > 0, b < 0, |a| > |b|

Answer: Option A

**Explanation** :

For a > b, the given equation reduces to

Ma[|a|, b] = mn[a, |a|].

If b < a < 0, then |b| > |a| > 0 > a > b.

∴ Ma [|a|, b] = |a| and mn [a, |a|] = a.

Hence, option (a) is correct.

Workspace:

**15. CAT 1994 QA | Arithmetic - Time & Work**

A water tank has three taps A, B, and C. A fills four buckets in 24 minutes, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together a full tank is emptied in 2 hours. If a bucket can hold 5 litres of water, what is the capacity of the tank?

- A.
120 litres

- B.
240 litres

- C.
180 litres

- D.
60 litres

Answer: Option B

**Explanation** :

Since a bucket holds 5 litres of water, water discharged in one minute by tap A, B and C is $\frac{5}{6}$ litres, $\frac{2}{3}$ litres and $\frac{1}{2}$ litres respectively.

If A, B and C are all opened simultaneously, total discharge in one minute = $\left(\frac{5}{6}+\frac{2}{3}+\frac{1}{2}\right)$ = 2 litres.

So in 2 hours, the discharge would be 240 litres, that is the capacity of the tank.

Workspace:

**16. CAT 1994 QA | Arithmetic - Time, Speed & Distance**

Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is $\frac{3}{4}$ times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that from Chandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed from Chandigarh to Shimla?

- A.
39.2 kmph

- B.
63 kmph

- C.
42 kmph

- D.
None of these

Answer: Option C

**Explanation** :

It is clear that the ratio of the distances between (Delhi- Chandigarh) : (Chandigarh-Shimla) = 3 : 4.

The ratio of the speeds between (Delhi-Chandigarh) : (Chandigarh-Shimla) = 3 : 2.

Let the distances be 3x and 4x respectively and speeds be 3y and 2y.

So the time taken will be $\left(\frac{x}{y}\right)$ and $\left(\frac{2x}{y}\right)$ respectively.

Average speed = $\frac{(TotalDis\mathrm{tan}ce)}{(TotalTime)}=\frac{\left(7x\right)}{\left({\displaystyle \frac{x}{y}}+{\displaystyle \frac{2x}{y}}\right)}=\frac{7y}{3}=49.$

Hence, y = 21. So the average speed from Chandigarh to Shimla = 2y = 42 kmph.

Workspace:

**17. CAT 1994 QA | Algebra - Progressions**

Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?

- A.
7

- B.
64

- C.
56

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Middle term of an A.P. is average of all the terms in A.P.

Number of terms = 7

Average of all these 7 terms = Middle term = Fourth term = 8

Therefore, sum of all the terms = 7 × 8 = 56.

Hence, option (c).

Workspace:

**18. CAT 1994 QA | Algebra - Number Theory**

It takes the pendulum of a clock 7 seconds to strike 4 o’clock. How much time will it take to strike 11 o’clock?

- A.
18 seconds

- B.
20 seconds

- C.
19.25 seconds

- D.
23.33 seconds

Answer: Option D

**Explanation** :

If a clock has to strike 4 or 4 times, there are 3 time intervals between the 4 strikes (Since the first strike happens at the zeroth second).

So in 7 seconds the pendulum elapses 3 time intervals.

To strike 11, there has to be 10 time intervals, which will take $\frac{10\times 7}{3}$ = 23.33 seconds.

Workspace:

**19. CAT 1994 QA | Algebra - Progressions**

Along a road lie an odd number of stones placed at intervals of 10m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then the number of stones is

- A.
35

- B.
15

- C.
29

- D.
31

Answer: Option D

**Explanation** :

As the person covers 4.8 km, he covers 2.4 kms on one side and 2.4 kms on other side.

Let the number of stones be ‘n’ on either side of the road excluding the middle stone.

So total distance covered by him on one side = 20 + 40 + 60 + … + 20n

∴ 2400 = $\left(\frac{n}{2}\right)$ (2 × 20 + (n - 1)20) = 10n (n + 1)

(Here n is the number of stones)

After solving, we get n = 15

∴ Total number of stones = 15 + 15 + 1 = 31

Hence, option (d).

Workspace:

**20. CAT 1994 QA | Algebra - Number Theory**

What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?

- A.
264

- B.
259

- C.
269

- D.
None of these

Answer: Option B

**Explanation** :

Required number = LCM (8, 11, 24) – 5 = 259.

Workspace:

**21. CAT 1994 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A man buys spirit at Rs. 60 per litre, adds water to it and then sells it at Rs. 75 per litre. What is the ratio of spirit to water if his profit in the deal is 37.5%?

- A.
9 : 1

- B.
10 : 1

- C.
11 : 1

- D.
None of these

Answer: Option B

**Explanation** :

Since SP of spirit and solution water = Rs.75/l and there is a profit of 37.5%, CP of spirit and water solution

$=\frac{75}{1.375}$ = Rs. 54.54/L

This should indeed be the weighted average of the costs of spirit and water. So if we alligate, we can get the ratio of spirit : water (assuming that cost of water is 0).

i.e. 10 : 1.

Workspace:

**22. CAT 1994 QA | Geometry - Quadrilaterals & Polygons**

Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point which falls inside the rectangle, after travelling distances of 40, 50 and 60 metres. The maximum distance that the fourth could have traveled is (approximately) ….

- A.
67 metres

- B.
52 metres

- C.
22.5 metres

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Let x and y be the sides of the rectangle ABCD and z

be the length of AP.

Then CR = BP = x – z

By applying Pythagoras Theorem, we have

in Δ APO, a^{2} =OP^{2} + z^{2} ......(i)

in Δ BPO, b^{2} = OP^{2} + (x – z)^{2} ......(ii)

in Δ CRO, d^{2} = OR^{2} + (x – z)^{2} ......(iii)

in Δ DRO, c^{2} = OR^{2} + z^{2} ....... (iv)

Solving above equation, we have a^{2} + d^{2} = b^{2} + c^{2}

∴ For any point inside a rectangle as shown,

a^{2} + d^{2} = b^{2} + c^{2}

∴ Pairing up the distance so that d is to be the maximum,

we get 40^{2} + d^{2} = 50^{2} + 60^{2}

⇒ d = 67 m.

Workspace:

**23. CAT 1994 QA | Arithmetic - Time, Speed & Distance**

A and B walk from X to Y, a distance of 27 km at 5 kmph and 7 kmph respectively. B reaches Y and immediately turns back meeting A at Z. What is the distance from X to Z?

- A.
25 km

- B.
22.5 km

- C.
24 km

- D.
20 km

Answer: Option B

**Explanation** :

Let they meet at a distance x kms from X.

So the total distance travelled by A = x at the speed of 5 kmph.

Total distance travelled by B = 27 + (27 – x) = (54 – x) at the speed of 7 kmph.

Time taken by A = $\frac{x}{5}$.

Time taken by B = $\frac{(54-x)}{7}$

Since they have met at the same time, they would have travelled for the same time. Hence $\frac{x}{5}=\frac{(54-x)}{7}$ or x = 22.5 kms.

Workspace:

Alphonso, on his death bed, keeps half his property for his wife and divide the rest equally among his three sons Ben, Carl and Dave. Some years later Ben dies leaving half his property to his widow and half to his brothers Carl and Dave together, shared equally. When Carl makes his will he keeps half his property for his widow and the rest he bequeaths to his younger brother Dave. When Dave dies some years later, he keeps half his property for his widow and the remaining for his mother. The mother now has Rs. 1,575,000.

**24. CAT 1994 QA | Algebra - Simple Equations**

What was the worth of the total property?

- A.
Rs. 30 lakh

- B.
Rs. 8 lakh

- C.
Rs. 18 lakh

- D.
Rs. 24 lakh

Answer: Option D

**Explanation** :

Let us assume that Alphonso’s total property was of Rs.x.

Since Alphonso’s wife is also the mother of Dave, the total share of this lady would be $\left(\frac{x}{2}+\frac{15x}{96}\right)=\frac{63x}{96}.$

And since, $\frac{63x}{96}$ = 1,575000

⇒ x = Rs.24 lakhs.

Workspace:

Alphonso, on his death bed, keeps half his property for his wife and divide the rest equally among his three sons Ben, Carl and Dave. Some years later Ben dies leaving half his property to his widow and half to his brothers Carl and Dave together, shared equally. When Carl makes his will he keeps half his property for his widow and the rest he bequeaths to his younger brother Dave. When Dave dies some years later, he keeps half his property for his widow and the remaining for his mother. The mother now has Rs. 1,575,000.

**25. CAT 1994 QA | Algebra - Simple Equations**

What was Carl’s original share?

- A.
Rs. 4 lakh

- B.
Rs. 12 lakh

- C.
Rs. 6 lakh

- D.
Rs. 5 lakh

Answer: Option A

**Explanation** :

Let us assume that Alphonso’s total property was of Rs.x.

Carl’s original share was $\frac{x}{6}=\frac{24}{6}$ = Rs. 4 lakhs.

Workspace:

Alphonso, on his death bed, keeps half his property for his wife and divide the rest equally among his three sons Ben, Carl and Dave. Some years later Ben dies leaving half his property to his widow and half to his brothers Carl and Dave together, shared equally. When Carl makes his will he keeps half his property for his widow and the rest he bequeaths to his younger brother Dave. When Dave dies some years later, he keeps half his property for his widow and the remaining for his mother. The mother now has Rs. 1,575,000.

**26. CAT 1994 QA | Algebra - Simple Equations**

What was the ratio of the property owned by the widows of the three sons, in the end?

- A.
7 : 9 : 13

- B.
8 : 10 : 15

- C.
5 : 7 : 9

- D.
9 : 12 : 13

Answer: Option B

**Explanation** :

Let us assume that Alphonso’s total property was of Rs.x.

The ratio’s of the property owned by the widows of the 3 sons = $\left(\frac{1}{12}:\frac{5}{48}:\frac{15}{96}\right)$ = 8 : 10 : 15.

Workspace:

**27. CAT 1994 QA | Algebra - Logarithms**

log_{6} 216$\sqrt{6}$ is

- A.
3

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

- C.
$\frac{7}{2}$

- D.
None of these

Answer: Option C

**Explanation** :

Let log_{6} 216$\sqrt{6}$ = x.

Then by rule, log_{b} a = x ⇒ b^{x} = a we have,

6^{x} = 216√6

6^{x} = 6^{3} × ${6}^{\frac{1}{2}}$

⇒ 6^{x} = ${6}^{\frac{7}{2}}$

∴ x = $\frac{7}{2}$

Workspace:

**28. CAT 1994 QA | Arithmetic - Time & Work**

There is a leak in the bottom of the tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admits 6 litres per hour and the tank is now emptied in 12 hours. What is the capacity of the tank?

- A.
28.8 litres

- B.
36 litres

- C.
144 litres

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Since the leak can empty the tank in 8 hours, the rate of leak = $\frac{1}{8}$.

And since the leak along with the tap can empty it in 12 hours, the equation can be rewritten as:

$\frac{1}{x}-\frac{1}{8}=-\frac{1}{12}$(where x is the time taken by the tap to fill the tank).

Simplifying we get, $\frac{1}{x}=\frac{1}{24}$ or x = 24.

This means that the tap can fill the tank in 24 hours.

Since the tap admits 6 litres per hour, it will admit (6 × 24) = 144 litres in 24 hours, which should be the capacity of the tank

Workspace:

**29. CAT 1994 QA | Algebra - Number Theory**

Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12, and 16 is 4.

- A.
137

- B.
1361

- C.
140

- D.
172

Answer: Option D

**Explanation** :

The LCM of 7, 12 and 16 is 336. The closest multiple of 336 to 1856 is 1680. So 1684 when divided by 7, 12 and 16 leaves a remainder of 4. This is the closest such number, less than 1856.

Hence the number to be subtracted from 1856 to get 1684, must be the least such number. So the answer is (1856 – 1684) = 172.

**Alternate method:**

Subtract options from 1856 and check.

Workspace:

**30. CAT 1994 QA | Arithmetic - Profit & Loss**

A dealer offers a cash discount of 20% and still makes a profit of 20%, when he further allows 16 articles to a dozen to a particularly sticky bargainer. How much percent above the cost price were his wares listed?

- A.
100%

- B.
80%

- C.
75%

- D.
66 2/3%

Answer: Option A

**Explanation** :

16 articles sold at priced 12 articles, is equivalent to discount of 25% $\left[\left(\frac{16-12}{16}\right)\times 100\right]$

Hence, shop keepers offer two discounts of 20% and 25% respectively and still makes a profit of 20%

If c is the cost price of an article and m is the marked price, then 1.2 × c = m × $\frac{3}{4}\times \frac{4}{5}$

⇒ m = 2c.

This means that he had marked his goods 100% above his cost price.

Workspace:

**31. CAT 1994 QA | Algebra - Simple Equations**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

As your answer,

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, if nothing can be said.

Nineteen years from now Jackson will be 3 times as old as Joseph is now. Johnson is three years younger than Jackson.

I. Johnson’s age now.

II. Joseph’s age now.

Answer: 4

**Explanation** :

Since the ages of none of them is mentioned and we have two equations and three unknowns.

Hence, we cannot say anything about the ages of any of them.

Workspace:

**32. CAT 1994 QA | Geometry - Quadrilaterals & Polygons**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

As your answer,

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, if nothing can be said.

In ΔACD, AD = AC and ∠C = 2∠E. The distance between parallel lines AB and CD is h. Then

I. Area of parallelogram ABCD

II. Area of ΔADE

Answer: 3

**Explanation** :

Since ∠C = 2 ∠E , therefore ∠BCA = 60°.

Also since ABCD is a parallelogram,

AB = CD and AD = BC = AC.

Hence, ΔABC and ΔACD are equilateral triangles.

Hence, area of this triangle = $\frac{{s}^{2}}{4}\sqrt{3}.$

where s is the side of the triangle = AB = AD = DC = BC.

∴ Area of the parallelogram is twice this area = $\frac{{s}^{2}}{2}\sqrt{3}.$

Since ∠CAD = 60°, ∠DAE = 90°, so ΔEAD is a right triangle with side AD = s. Since it is a 30-60-90 triangle, hence side AE = s$\sqrt{3}.$

∴ Area of this triangle = $\frac{(s\times s\sqrt{3})}{2}=\frac{{s}^{2}}{2}\sqrt{3}.$

Hence, the required two areas are equal or I = II.

Workspace:

**33. CAT 1994 QA | Algebra - Simple Equations**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

As your answer,

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, if nothing can be said.

Last week Martin received $ 10 in commission for selling 100 copies of a magazine. Last week Miguel sold 100 copies of this magazine. He received his salary of $ 5 per week plus a commission of 2 cents for each of the first 25 copies sold, 3 cents for each of next 25 copies sold and 4 cents for each copy thereafter. ($1 = 100 cents).

I. Martin’s commission in the last week.

II. Miguel’s total income for last week.

Answer: 1

**Explanation** :

Miguel’s income = 5 + (0.02 × 25) + (0.03 × 25) + (0.04 × 50) = $8.25.

Martin’s commission = $10.

Hence obviously I > II.

Workspace:

**34. CAT 1994 QA | Geometry - Basics**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, if nothing can be said.

k_{1}, k_{2}, k_{3 }are parallel lines. AD = 2 cm, BE = 8 cm and CF = 32 cm.

I. (AB) × (EF)

II. (BC) × (DE)

Answer: 3

**Explanation** :

Since the lines are parallel, $\frac{AB}{BC}=\frac{DE}{EF}$, i.e. AB × EF

= BC × DE.

Hence I = II.

Workspace:

**35. CAT 1994 QA | Modern Math - Probability**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, if nothing can be said.

I. The probability of encountering 54 Sundays in a leap year.

II. The probability of encountering 53 Sundays in a non-leap year.

Answer: 2

**Explanation** :

53 Sundays can occur in a non-leap year, if 1st January is either a Saturday or a Sunday. But 54 Sundays can never occur.

Hence, I < II.

Workspace:

**36. CAT 1994 QA | Arithmetic - Time, Speed & Distance**

The winning relay team in a high school sports competition clocked 48 minutes for a distance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 kmph, 16 kmph, 17 kmph, and 18 kmph respectively. What is the ratio of the time taken by B to than taken by D?

- A.
5 : 16

- B.
5 : 17

- C.
9 : 8

- D.
8 : 9

Answer: Option C

**Explanation** :

Since it is a relay race, all the runners ran the same distance.

Hence, for a same distance,

ratio of times = $\frac{1}{ratioofspeeds}$

Hence, ratio of times taken by B & D = 18 : 16 = 9 : 8.

Workspace:

If f (x) = 2x + 3 and g(x) = $\frac{x-3}{2},$ then

**37. CAT 1994 QA | Algebra - Functions & Graphs**

fog(x) is equal to

- A.
1

- B.
gof(x)

- C.
$\frac{15x+9}{16x-5}$

- D.
$\frac{1}{x}$

Answer: Option B

**Explanation** :

Let g(x) = $\frac{x-3}{2}$ be y. So, fog(x) = f(y) = 2y + 3.

Substituting y = $\frac{x-3}{2},$ we get

fog(x) = (x – 3) + 3 = x = $\frac{\left[\right(2x+3)-3]}{2}$ = gof(x)

Workspace:

If f (x) = 2x + 3 and g(x) = $\frac{x-3}{2},$ then

**38. CAT 1994 QA | Algebra - Functions & Graphs**

For what value of x; f (x) = g(x −3)?

- A.
-3

- B.
$\frac{1}{4}$

- C.
-4

- D.
None of these

Answer: Option C

**Explanation** :

If 2x + 3 = $\frac{\left[\right(x-3)-3]}{2},$ then x = -4.

Workspace:

If f (x) = 2x + 3 and g(x) = $\frac{x-3}{2},$ then

**39. CAT 1994 QA | Algebra - Functions & Graphs**

What is the value of (gofofogogof) (x) (fogofog)(x)?

- A.
x

- B.
x

^{2} - C.
$\frac{5x+3}{4x-1}$

- D.
$\frac{(x+3)(5x+3)}{(4x-5)(4x-1)}$

Answer: Option B

**Explanation** :

From Question 87, fog(x) = gof (x) = x, you will realise that if you were to form a chain of these functions for even number of times, you would still end up getting x.

For eg. fogofog(x) = fog(x) = x. Since both the brackets have the functions repeated for even number of times, each of their value will be x and their product will be x^{2}.

Workspace:

If f (x) = 2x + 3 and g(x) = $\frac{x-3}{2},$ then

**40. CAT 1994 QA | Algebra - Functions & Graphs**

What is the value of fo(fog)o(gof)(x)?

- A.
x

- B.
x

^{2} - C.
2x + 3

- D.
$\frac{x+3}{4x-5}$

Answer: Option C

**Explanation** :

From question 87, gof(x) = fog(x) = x.

fo(fog)o(gof)(x) = fo(fog)(x) = f(x) = 2x + 3.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**41. CAT 1994 QA | Data Sufficiency**

Is the distance from the office to home less than the distance from the cinema hall to home?

I. The time taken to travel from home to office is as much as the time taken from home to the cinema hall, both distance being covered without stopping.

II. The road from the cinema hall to home is bad and speed reduces, as compared to that on the road from home to the office.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option C

**Explanation** :

Statement I tells us that the time taken to cover both distances is the same, but it does not tell us anything about the speeds at which these are covered. This information is given by the second statement, which says the speed from cinema hall to home is less than that between home to the office.

Hence by using both the statements we can say that the distance between cinema hall to home is less than that between home to the office.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**42. CAT 1994 QA | Data Sufficiency**

A and B work at digging a ditch alternately for a day each. If A can dig a ditch in ‘a’ days and B can dig that ditch in ‘b’ days, will work get done faster if A begins the work?

I. n is a positive integer such that n$\left(\frac{1}{a}+\frac{1}{b}\right)$ = 1

II. b > a

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option A

**Explanation** :

If the total work is one unit, work done by A and B in one day will be $\frac{1}{a}$ unit and $\frac{1}{b}$ unit respectively.

**Using statement I:** $\frac{n}{a}+\frac{n}{b}$ = 1,

Since ‘n’ is an integer, if both A and B work for n days, work will be completed no matter who starts a work.

**Using statement II, **nothing can be concluded as total amount of work is not known.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**43. CAT 1994 QA | Data Sufficiency**

If twenty sweets are distributed among some boys and girls such that each girl gets two sweets and each boy gets three sweets, what is the number of boys and girls?

I. The number of girls is not more than five.

II. If each girl gets 3 sweets and each boy gets 2 sweets, the number of sweets required for the children will still be the same.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option B

**Explanation** :

2g + 3b = 20.

Since b & g should be integers the values that satisfy this equation are (g = 10 & b =0), (g = 7 and b = 2), (g

= 4 & b = 4), and (g = 1 and b = 6).

From the statement I, we can shortlist the last two possibilities i.e. g =4 or g = 1, but cannot get a unique answer.

The statement II suggests that the number of girls and boys have to be equal. Hence we get a unique answer viz. g = 4 & b = 4. Only statement II is required to answer the question.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**44. CAT 1994 QA | Data Sufficiency**

If the selling price were to be increased by 10%, the sales would reduce by 10%. In what ratio would profits change?

I. The cost price remains constant.

II. The cost price increased 10%.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option B

**Explanation** :

P = (SP – CP) x Sales. From the data given in the question we can figure out that P1 = (1.1SP – CP) x

0.9Sales. Hence $\frac{P}{P1}=\frac{1.11(SP-CP)}{(1.1SP-CP)}$ To find this ratio we need to eliminate the variables CP & SP. This can only be done if in the denominator, CP is replaced by 1.1CP. In other words, if the CP increases by 10%, as in that case our ratio will be $\frac{1.11}{1.1}$ = 1.01. Hence only Statement II is required to answer the question.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**45. CAT 1994 QA | Data Sufficiency**

What is the average weight of the 3 new team members who are recently included into the team?

I. The average weight of the team increases by 20 kg.

II. The 3 new men substitute earlier members whose weights are 64 kg, 75 kg and 66 kg.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option D

**Explanation** :

As neither average weight of the original members is not mentioned nor the number of members in original team, question cannot be answered.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**46. CAT 1994 QA | Data Sufficiency**

Is segment PQ greater than segment RS?

I. PB > RE,BQ = ES.

II. B is a point on PQ, E is a point on RS.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option C

**Explanation** :

None of the statement alone is sufficient to answer the question.

Using both statements together:

PQ = PB + BQ and RS = RE + ES

If BQ = ES and PB > RE, PQ > PS.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**47. CAT 1994 QA | Data Sufficiency**

Three boys had a few coffee Bite toffees with them. The number of toffees with the second were four more than those with the first and the number of toffees with the third were four more than those with the second. How many toffees were there in all?

I. The number of toffees with each of them is a multiple of 2.

II. The first boy ate up four toffees from what he had and the second boy ate up six toffees from

what had and the third boy gave them two toffees each from what he had and the number of toffees remaining with each of them formed a geometric progression.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option B

**Explanation** :

Let the number of toffees with the first, second and third boy be x, (x+4) and (x+8) respectively.

Hence, total number of toffees = (3x+12)

The statement I merely suggests that (3x+12) is a multiple of 2, which means that x is a multiple of 2.

Nothing concrete can be concluded on the basis of this statement.

The statement II suggests that (x – 4 + 2 ), (x + 4 –6 + 2) and (x + 8 – 4) are in GP or (x-2), x and (x+4) is in GP.

∴ x^{2} = (x + 4)(x − 2)

⇒ x = 4

⇒ (3x + 12) = 24

Question can be answered using statement II alone.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**48. CAT 1994 QA | Data Sufficiency**

Little Beau Peep lost her sheep. She couldn’t remember how many were there. She knew she would have 400 more next year, than the number of sheep she had last year. How many sheep were there?

I. The number of sheep last year was 20% more than the year before that and this simple rate of increase continues to be the same for the next 10 years.

II. The increase is compounded annually.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option C

**Explanation** :

The statement I suggests that the number of sheep had increased by 20% last year over the previous year. But it does not suggest whether the rate of increase is annual or not.

For eg. 20% increase in a year can also be obtained by 9.5% increase ever 6 months. i.e. 1.095 x 1.095 = 1.20.

The statement II however suggests that the increase is compounded annually.

Hence, now we can find the answer.

If the number of sheep last year was x, then x + 400 = x(1.2)^{2}

Hence, x = 909.

Thus we require both statements to answer the question.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**49. CAT 1994 QA | Data Sufficiency**

What will be the total cost of creating a 1- foot border of tiles along the inside edges of a room?

I. The room is 48 feet in length and 50 fet in breadth.

II. Every tile costs Rs. 10.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option D

**Explanation** :

From the statement I, we can find out the area that needs to be bordered. And from the statement II, we can find out the cost of each tile. But to find the total cost, we require the total number of tiles and to find this we require the dimension of each tile. Since this is not known, we cannot answer the question using either statements.

Workspace:

**Each of these items has a question followed by two statements. As the answer,**

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

**50. CAT 1994 QA | Data Sufficiency**

Ten boys go to a neighbouring orchard. Each boy steals a few mangoes. What is the total number of mangoes they steal?

I. The first boy steals 4 mangoes and the fourth boy steals 16 mangoes and the eight boy 32 mangoes and the tenth boy steals 40 mangoes.

II. The first boy stole the minimum number of mangoes and the tenth boy stole the maximum number of mangoes.

- A.
a

- B.
b

- C.
c

- D.
d

Answer: Option D

**Explanation** :

From the statement I, we can only find the number of mangoes stolen by 4 of the 10 boys.

The statement II suggests that the number of mangoes stolen by each of the remaining six boys is more than 4 and less than 40. Although from the two statements that are given it is tempting to assume that the number of mangoes stolen by the boys must be in AP, since it is not mentioned explicitly we cannot answer the question.

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**