Q.197.
1. What is the percentage increase in the gross traffic receipts in 1995-96 as compared to 1993-94?
A. 33.9%
B. 41.5%
C. 20.7%
D. 17%

2. If profit = gross traffic receipts - total expenditure, then in 1996-97, what percentage of gross traffic receipts is the profit made?
A. 5.9%
B. 6.4%
C. 7.2%
D. 8%

3. In which year was the profit as a percentage of gross traffic receipts the highest?
A. 1997-98
B. 1996-97
C. 1995-96
D. 1994-95

4.In order to make a profit of 10%. What should have been gross traffic receipts (in Rs. crores) in 1994-95, total expenditure remaining the same?
A. 5,667
B. 5,876
C. 6,444
D. 7,667

5. By what amount (in Rs. crores) has the expenditure increased over the period 1993-94 to 1997-98?
A. 4,100
B. 3,900
C. 3,850
D. 3,700

Answer.
1. B
Percentage increase in gross traffic receipts from 1993-94 to 1995-96 is:
$\text{% increase}=\frac{7500-5300}{5300} \times 100 = 41.5\%$

2. A
$\text{Profit}=8500-8000=500$
$\text{% Profit}=\frac{500}{8500} \times 100 = 5.9\% $

3. C
In 1993-94 profit % $= \frac{5300-5100}{5300} \times 100 = 3.77\%$
In 1994-95 profit % $= \frac{6400-5800}{6400} \times 100 = 9.37\%$
In 1995-96 profit % $= \frac{7500-5900}{7500} \times 100 = 21.3\%$
In 1996-97 profit % $= \frac{8500-8000}{8500} \times 100 = 5.9\%$
In 1997-98 profit % $= \frac{9400-8800}{9400} \times 100 = 6.38\%$

4. C
Let gross traffic receipts $=x$
Then
$\frac{x-5800}{x} \times 100=10=6444$

5. D
Increase in expenditure from 1993-94 to 1997-98 $= 8800-5100=3700$

Q.196. Complete the series of figures, by selecting correct answer figure from the given responses

Answer. 3

Q.195. A survey of film watching habits of people living in 5 cities, I, II, III, IV and V is summarized below. The column (1) gives percentage of film watchers in each city who see only one film a week. The column (2) gives the total number of film-watchers who see more than one film per week. Read the table and answer the following questions.

City	(1)	(2)
I	50	3200
II	30	3500
III	65	7000
IV	25	5400
V	80	2900

1. How many film-watchers in city II see only one film a week?
A. 900
B. 1500
C. 1600
D. 3200

2. Which city has the highest number of film-watchers who see only one film a week?
A. I
B. II
C. III
D. IV

3. The city with the lowest numbers of film-watchers is?
A. I
B. II
C. IV
D. V

4. The city with the highest numbers of film-watchers is?
A. I
B. V
C. IV
D. III

5. The total number of film-watchers in the five cities who see only one film a week is:
A. 31100
B. 29500
C. 28600
D. 28300

Answer.
1.
In city II 30% of film-watchers see only one film a week.
So 70% watch two films a week.
But number of people who see two films a week $=3500$
So if total number of film-watchers in city II $=x$
Then
$\frac{70}{100} \times x = 3500$
$x=5000$
So number of film-watchers who see only one film a week will be $=5000-3500=1500$
Answer. B. 1500

2.
Number of film-watchers who see only one film a week in all 5 cities are as under:
$I \rightarrow \frac{3200}{50} \times 100-3200=3200$
$II \rightarrow \frac{3500}{70} \times 100-3500=1500$
$III \rightarrow \frac{7000}{35} \times 100-7000=13000$
$IV \rightarrow \frac{5400}{75} \times 100-5400=1800$
$V \rightarrow \frac{2900}{20} \times 100-2900=11600$
Answer. C. III

3.
Number of film watchers in all cities are as under:
$I \rightarrow \frac{3200}{50} \times 100=6400$
$II \rightarrow \frac{3500}{70} \times 100=5000$
$III \rightarrow \frac{7000}{35} \times 100=20000$
$IV \rightarrow \frac{5400}{75} \times 100=7200$
$V \rightarrow \frac{2900}{20} \times 100=14500$
Answer. B. II

4.
Number of film watchers in all cities are as under:
$I \rightarrow \frac{3200}{50} \times 100=6400$
$II \rightarrow \frac{3500}{70} \times 100=5000$
$III \rightarrow \frac{7000}{35} \times 100=20000$
$IV \rightarrow \frac{5400}{75} \times 100=7200$
$V \rightarrow \frac{2900}{20} \times 100=14500$
Answer. D. III

5.
Number of film-watchers who see only one film a week in all 5 cities are as under:
$I \rightarrow \frac{3200}{50} \times 100-3200=3200$
$II \rightarrow \frac{3500}{70} \times 100-3500=1500$
$III \rightarrow \frac{7000}{35} \times 100-7000=13000$
$IV \rightarrow \frac{5400}{75} \times 100-5400=1800$
$V \rightarrow \frac{2900}{20} \times 100-2900=11600$
Total $=3200+1500+13000+1800+11600=31100$
Answer. A. 31100

Q.194. Assume that
$\sqrt{13}=3.605 \text{ (Approx.)}$
$\sqrt{130}=11.40 \text{ (Approx.)}$
Find the value of
$\sqrt{1.3}+\sqrt{1300}+\sqrt{0.013}$
A. 36.164
B. 37.304
C. 36.304
D. 37.164

Answer. B. 37.304

$\rightarrow \sqrt{1.3}+\sqrt{1300}+\sqrt{0.013}$
$\rightarrow \frac{\sqrt{1.3} \times 10}{10}+10\sqrt{13}+\frac{100 \times \sqrt{0.013}}{100}$
$\rightarrow \frac{\sqrt{130}}{10}+10\sqrt{13}+\frac{\sqrt{130}}{100}$
$\rightarrow \frac{11.40}{10}+10(3.605)+\frac{11.40}{100}$
$\rightarrow 1.140+36.05+0.1140$
$\rightarrow 37.304$

Q.193. Simplify : $1 + \frac{1}{1+\frac{2}{2+\frac{3}{1+\frac{4}{5}}}}$
A. $1\frac{11}{17}$
B. $1\frac{5}{7}$
C. $1\frac{6}{17}$
D. $1\frac{12}{17}$

Answer. A. $1\frac{11}{17}$

$\rightarrow 1 + \frac{1}{1+\frac{2}{2+\frac{3}{1+\frac{4}{5}}}}$
$\rightarrow 1 + \frac{1}{1+\frac{2}{2+\frac{5}{3}}}$
$\rightarrow 1 + \frac{1}{1+\frac{6}{11}}$
$\rightarrow 1 + \frac{11}{17}$
$\rightarrow 1\frac{11}{17}$

Q.192. In covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is:
A. $1$ hour
B. $1\frac{1}{2}$ hours
C. $2$ hours
D. $2\frac{1}{2}$ hours

Answer. C. $2$ hours

Let the time take by A $=T \text{ hrs}$
Then time taken by B $=T-\text{30 min}=(T-0.5) \text{ hrs}$
Now let distance $=d$
Speed of A $=3x$
Speed of B $=4x$
So
$Distance=Speed \times Time$
$Distance=3x \times T = 4x \times (T-0.5)$
$3T=4T-2$
$T=2$ hours

Q.191. The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?
A. 10 a.m.
B. 10:30 a.m.
C. 11 a.m.
D. 11:30 a.m.

Answer. C. 11 a.m.

Suppose trains meet in t time after 8 a.m.
Then distance covered by first train in time t $=60t$
As second train started 1 hour later the distance covered by second train in time t $=75(t-1)$
Total distance $=60t+75t-75=330$
$t=3 \text{ hrs}$
So trains will meet at $=8+3=11 \text{ a.m.}$