# QUANTITATIVE APTITUDE -- PO AND CLERICAL LEVEL/SSC RRB WITH ANSWERS

ONE 1 4

**NUMBER SYSTEMS**

1. Find the maximum value of n such that 157!
Is perfectly divisible by 10

^{n}.
1) 37
2) 38 3) 16 4) -1.15

Ans. (2) [157/5] = 31. [31/5] = 6.[6/5] =
1.31+6+1 = 38. Option (2) is correct.

2. Find the maximum value of n such that 157!
Is perfectly divisible by 12n.

1) 77
2) 76 3) 75 4) 78

Ans. (3) No. of 2’s in 157! = [157/2] + [157/4]
+ [157/8] …. + [157/128] = 78 + 39 + 19 + 9 + 4 + 2 + 1 = 152. Hence, the
number of 2

^{2}s would be [152/2] = 76. Number of 3’s in 157! = 52 + 17 + 5 + 1 = 75. The answer would be given by the lower of these values. Hence, 75 (Option c) is correct.
3. Find the maximum value of n such that 157!
Is perfectly divisible by 18n.

1) 37
2) 38 3) 39 4) 40

Ans. (1) From the above solution:

Number of 2’s in 157! = 152

Number of 3

^{2}s in 157! = [75/2] = 37.
Hence, option (1) is correct.

4. Find the maximum value of n such that 50! Is
perfectly divisible by 2520n.

1) 6 2)
8 3) 7
4) None of these

Ans. (2) 2520 = 7 x 3

^{2}x 2^{3}x 5.
The value of n would be given by the value of
the number of 7s in 50!

This value is equal to [50/7] + [50/49] = 7 + 1
= 8

Option (2) is correct

5. Find the maximum value of n such that 50! is
perfectly divisible by 12600n.

1) 7 2)
6 3) 8
4) None of these

Ans. (2) 12600 = 7 x 3

^{2}x 2^{3}x 5^{2}
The value of ‘n’ would depend on which of
number of 7s and number of 5

^{2}is lower in 50!.
Number of 7’s in 50! = 8. Note here that if we
check for 7’s we do not need to check for 3

^{2}s as there would be at least two 3’s before a 7 comes in every factorial’s value. Similarly, there would always be at least three 2’s before a 7 comes in any factorial’s value. Thus, the number of 3^{2}s and the number of 2^{3}s can never be lower than the number of 7s in any factorial’s value.
Number of 5s in 50! = 10 + 2 = 12. Hence, the
number of 52s in 50! = [12/2] = 6.

6 will be the answer as the number of 5

^{2}s is lower than the number of 7’s.
Option (2) is correct.

**AVERAGES 1 2 3**

1. The average age of 24 students and the
principal is 15 years. When the principal’s age is excluded, the average age
decreases by 1 year. What is the age of the principal?

1) 38
2) 40 3) 39 4) 37

Ans. (3) P =

2. The average weight of 3 men A, B and C is 84
kg. Another man D joins the group and the average now becomes 80 Kg. If another
man E, whose weight is 3 kg more than that of D, replaces A then the average
weight of B, C, D and E becomes 78 kg. The weight of A is

1) 70 kg
2) 72 kg 3) 79 kg 4) 78 kg

Ans. (3) D’s weight =
E’s weight = 68 + 3 = 71.

Now, we know that A + B + C + D =
and B + C + D + E =
Hence, A’s weight is 8 kg more than E’s
weight. A = 71 + 8 = 79.

3. The mean temperature of Monday to Wednesday
was
and of Tuesday to Thursday was
. If the
temperature on Thursday was 4/5 that of Monday, the temperature on Thursday was

1)
2)
3)
4)

Ans. (2) Monday + Tuesday + Wednesday =
Tuesday + Wednesday + Thursday =
Thus, Monday – Thursday = 9 and Thursday =
Monday/5
Thursday = 36 and Monday = 45.

4. Three years ago, the average age of A, B and
C was 27 years and that of B and C 5 years ago was 20 years. A’s present age is

1) 30 years
2) 35 years 3) 40 years 4) 48 years

Ans. (3) Today’s total age of A, B and C =
Today’s total age for B and C =
C’s age = 90 - 50 = 40.

5. Ajit Tendulkar has a certain average for 9
innings. In the tenth inning, he scores 100 runs thereby increasing his average
by 8 runs. His new average is

1) 20 2)
24 3) 28 4) 32

Ans. (3)
(average after 9 innings). Hence, new average
= 20 + 8 = 28.

**ALLIGATIONS**

1. If 5 kg of salt costing
5/kg and 3 kg of salt costing
4/kg are mixed, find
the average cost of the mixture per kilogram.

1)
4.5 2) 4.625
3)
4.75 4)
4.125

Ans. (2) Solving the following allegation
figure:

3 : 5

The answer would be 4.625/kg.

2. Two types of oils having the rates of
4/kg and
5/kg respectively are
mixed in order to produce a mixture having the rate of
4.60/kg. What should be
the amount of the second type of oil if the amount of the first type of oil in
the mixture is 40 kg?

1) 75 kg
2) 50 kg 3) 60 kg 4) 40 kg

Ans. (3) Mixing
4/kg and
5/kg to get
4.6 per kg we get that
the ratio of mixing is 2:3. If the first oil is 40 kg, the second would be 60
kg.

3. How many kilograms of sugar worth
3.60 per kg should be
mixed with 8 kg of sugar worth
4.20 per kg, such that
by selling the mixture at
4.40 per kg, there may
be a gain of 10%?

1) 6 kg
2) 3 kg 3) 2 kg 4) 4 kg

Ans. (4) Since by selling at
4.40 we want a profit
of 10%, it means that the average cost required is
4 per kg. Mixing sugar
worth
3.6/kg and
4.2/kg to get
4/kg means a mixture
ratio of 1:2. Thus, to 8 kg of the second variety we need to add 4 kg of the
first variety to get the required cost price.

4. A mixture of 125 gallons of wine and water
contains 20% water. How much water must be added to the mixture in order to
increase the percentage of water to 25% of the new mixture?

1) 10 gals
2) 8.5 gals 3) 8 gals 4) 8.33 gals

Ans. (4) In 125 gallons we have 25 gallons
water and 100 gallons wine. To increase the percentage of water to 25%, we need
to reduce the percentage of wine to 75%. This means that 100 gallons of wine =
75% of the new mixture. Thus the total mixture = 133.33 gallons. Thus, we need
to mx 133.33 – 125 = 8.33 gallons of water in order to make the water
equivalent to 25% of the mixture.

5. Ravi lends
3600 on simple interest
to Harsh for a period of 5 years. He lends a part of the amount at 4% interest
and the rest at 6% and receives
960 as the amount of
interest. How much money did he lend on 4% interest rate?

1)
2800 2)
2100 3)
2400 4)
1200

Ans. (4) Since, Ravi earns
960 in 5 years, it
means that he earns an interest of 960/5 =
192 per year. On an
investment of 3600, an annual interest of 192 represents an average interest
rate of 5.33%.

Then using the allegation figure below:

Three

We get the ratio of investments as 1:2. Hence,
he lent
3600/3 = 1200 at 4% per annum.

**PERCENTAGE 1 2**

1. Which of the following is the largest
number?

1) 20% of 200
2) 7% of 500 3) 1300% of 3 4) 700% of 9

Ans. (4) It can be clearly seen that 700% of 9
= 63 is the highest number.

2. If 25% of a number is 75, then 45% of that
number is:

1) 145 2)
125 3) 150 4) 135

Ans. (4) 0.25 N = 75
= 300. Thus, 0.45
300 = 135.

3. What is 20% of 50% of 75% of 70?

1) 5.25
2) 6.75 3) 7.25 4) 5.5

Ans. (1) 20% of 50% of 75% of 70% = 20/100
50/100
75/100
70 = 0.2
0.5
0.75
70 = 5.25.

A quicker way to think here would be: 20% of 70
= 14
50% of 14 = 7
75% of 7 = 5.25

4. If we express 41(3/17)% as a fraction, then
it is equal to

1)
2)
3)
4)

Ans. (2) 41(3/17)% = 700/17%. As a fraction,
the value = 700/(17
100) = 7/17.

5. Mr. Abhimanyu Banerjee is worried about the
balance of his monthly budget. The price of pertrol has increased by 40%. By
what percent should he reduce the consumption of petrol so that he is able to
balance his budget?

1) 33.33
2) 28.56 3) 25 4) 14.28

Ans. (2) The following PCG will give the
answer:

40%

Price effect Consumption
effect

Hence, the percentage reduction required is
28.56% (40/140)

**PROFIT & LOSS 1 4**

1. By selling a watch for
495, a shopkeeper
incurs a loss of 10%. Find the cost price of the watch for the shopkeeper.

1)
545 2)
550 3)
555 4)
565

Ans. (2) 0.9
Price = 495
Price 550.

2. By selling a cap for
34.40, a man gains
7.5%. What will be the CP of the cap?

1)
32.80 2)
32 3)
32.40 4)
28.80

Ans. (2) The SP = 107.5% of the CP. Thus, CP =
34.4/1.075 =
32.

3. A cellular phone when sold for
4600 fetches a profit
of 15%. Find the cost price of the cellular phone.

1)
4300 2)
4150 3)
4000 4)
4500

Ans. (3) 1.15
Price = 4600
Price = 4000.

4. A machine costs
375. If it is sold at a
loss of 20%, what will be its cost price as a percentage of its selling price?

1) 80% 2)
120% 3) 110% 4) 125%

Ans. (4) A lost of 20% means a cost price of
100 corresponding to a selling price of 80. CP as a percentage of the SP would
then be 125%

5. A shopkeeper sold goods for
2400 and made a profit
of 25% in the process. Find his profit per cent if he had sold his goods for
2040.

1) 6.25%
2) 7% 3) 6.20% 4) 6.5%

Ans. (1) 2400 = 1.25
cost price
Cost price = 1920 Profit at 2040 =
120

Percentage profit = (120/2040)
100 =
6.25%

Four

**INTEREST 4 5**

1.
1200 is lent out at 5%
per annum simple interest for 3 years. Find the amount after 3 years.

1)
1380 2)
1290 3)
1470 4)
1200

Ans. (1) The annual interest would be
60. After 3 years the
total value would be 1200 + 60
3 = 1380

2. Interest obtained on a sum of
5000 for 3 years is
1500. Find the rate
percent.

1) 8% 2)
9% 3) 10% 4) 11%

Ans. (3) The interest earned per year would be
1500/3 = 500. This represents a 10% rate of interest.

3.
2100 is lent at
compound interest of 5% per annum for 2 years. Find the amount after two years.

1)
2300 2)
2315.25 3)
2310 4)
2320

Ans. (2) 2100 + 5% of 2100 = 2100 + 105 = 2205
(after 1 year). Next year it would become: 2205 + 5% of 2205 = 2205 + 110.25 = 2315.25

4.
1694 is repaid after
two years at compound interest. Which of the following is the value of the
principal and the rate?

1)
1200,20% 2)
1300,15% 3)
1400,10% 4)
1500,12%

Ans. (3) 10%
10%

5. Find the difference between the simple and
the compound interest at 5% per annum for 2 years on a principal of
2000.

1) 5 2)
105 3) 4.5 4) 5.5

Ans. (1) Simple Interest for 2 years = 100 +
100 = 200. Compound interest for 2 years: Year 1 = 5% of 2000 = 100.

Year 2:5% of 2100 = 105
Total compound interest =
205.

Difference between the Simple and Compound
interest = 205 – 200 =
5

**RATIO, PROPORTION AND VARIATION**

1. Divide
1870 into three parts
in such a way that half of the first part, one-third of the second part and
one-sixth of the third part are equal.

1) 241,343,245
2) 400,800,670 3) 470,640,1160 4) None of these

Ans. (4) Solve this question using options. ½
of the first part should equal 1/3
of the thid part. This means that the first
part should be divisible by 2, the second one by 3, and the third one by 6.
Looking at the options, none of the first 3 options has its third number
divisible by 6. Thus, option (4) is correct.

^{rd}of the second part and
2. Divide
500 among A, B, C and D
so that A and B together get thrice as much as C and D together, B gets four
times of what C gets and C gets 1.5 times as much as D. Now the value of what B
gets is

1) 300 2)
75 3) 125 4) 150

Ans. (a) (A + B) = 3 (C + D)
A + B =
375 and C + D = 125. Also, since C gets 1.5 times D we have

C = 75 and D = 50, and B = 4 C = 300.

3. If
, then
each fraction is equal to

1)
2) ½
3) ¼ 4) 0

Ans. (2) The given condition has a, b and c
symmetrically placed. Thus, if we use a = b = c = 2 (say) we get each fraction
as ½.

4. If 6
+ 6
= 13
, what
is the ratio of
?

1) 1:4 2)
3:2 3) 4:5 4) 1:2 (Hint: Use options to solve fast)

Ans. (2) Solve using options, Since
it is clear that a ratio of
as 3:2 fits the equation.

5. If
then the value of
is

1) ½ 2)
3)
4)

Ans. (4) 1:2 = 3 : 6 So,
From the given options, only ab/cd gives us
this value.

Five

**TIME AND WORK 1 5**

1. Raju can do 25% of a piece of work in 5
days. How many days will he take to complete the work ten times?

1) 150 days
2) 250 days 3) 200 days 4) 180 days

Ans. (3) He will complete the work in 20 days.
Hence, he will complete ten times the work in 200 days.

2. 6 men can do a piece of work in 12 days. How
many men are needed to do the work in 18 days.

1) 3 men
2) 6 men 3) 4 men 4) 2 men

Ans. (3) 6 men for 12 days means 72 mandays.
This would be equal to 4 men for 18 days.

3. A can do a piece of work in 20 days and B
can do it in 15 days. How long will they take if both work together?

1)
2)
3)
4)

Ans. (2) A’s one day work will be 5%, while B
will do 6.66% of the work in one day. Hence, their total work will be 11.66% in
a day.

In 8 days they will complete
11.66
8 = 93.33% This will leave 6.66% of the work.
This will correspond to 4/7 of the ninth day since in 6.66/11.66 both the
numerator and the denominator are divisible by 1.66.

4. In question 3 if C, who can finish the same
work in 25 days, joins them, then how long will they take to complete the work?

1)
2) 12 days
3)
4)

Ans. (1) A’s
work = 5% per day

B’s work
= 6.66% per day

C’s work
= 4% per day.

Total no. of days = 100/15.66 = 300/47 =
6(18/47)

5. Nishu and Archana can do a piece of work in
10 days and Nishu alone can do it in 12 days. In how many days can Archana do
it alone?

1) 60 days
2) 30 days 3) 50 days 4) 45 days

Ans. (1) N + A = 10% N = 8.33%
Hence A = 1.66%
60 days.

**TIME, SPEED AND DISTANCE 2 3**

1. The Sinhagad Express left Pune at noon
sharp. Two hours later, the Deccan Queen started from Pune in the same
direction. The Deccan Queen started from Pune in the same direction. The Deccan
Queen overtook the Sinhagad Express at 8 p.m. Find the average speed of the two
trains over this journey if the sum of their average speeds is 70 km/h.

1) 34.28 km/h
2) 35 km/h 3) 50 km/h 4) 12 km/h

Ans. (1) The ratio of time for the travel is
4:3 (Sinhagad to Deccan Queen). Hence, the ratio of speeds would be 3:4. Since,
the sum of their average speeds is 70 kmph, their respective speeds would be 30
and 40 kmph respectively. Use allegation to get the answer as 34.28 kmph.

2. Walking at ¾ of his normal speed, Abhishek
is 16 minutes late in reaching his office. The usual time taken by him to cover
the distance between his home and his office is

1) 48 minutes
2) 60 minutes 3) 42 minutes 4) 62 minutes

Ans. (1) When speed goes down to three fourth
(i.e. 75%) time will go up to 4/3

^{rd}(or 133.33%) of the original time. Since, the extra time required is 16 minutes, it should be equated to 1/3^{rd}of the normal time. Hence, the usual time required will be 48 minutes.
3. Ram and Bharat travel the same distance at
the rate of 6 km per hour and 10 km per hour respectively. If Ram takes 30
minutes longer than Bharat, the distance travelled by each is

1) 6 km
2)10 km 3) 7.5 km 4) 20 km

Ans.(3) Since, the ratio of speeds is 3:5, the
ratio of times would be 5:3. The difference in the times would be 2 (if looked
at in the 5:3 ratio context). Further, since Ram takes 30 minutes longer, 2
corresponds to 30. Hence, using unitary method, 5 will correspond to 75 and 3
will correspond to 45 minutes. Hence at 10 kmph, Bharat would travel 7.5km.

Six

4. Two trains for Mumbai leave Delhi at 6:00
a.m. and 6:45 am and travel at 100 kmph and 136 kmph respectively. How many
kilometers from Delhi will the two trains be together?

1) 262.4 km
2) 260 km 3) 283.33 km 4) 275 km

Ans. (3) The train that leaves at 6 am would be
75 km ahead of the other train when it starts. Also, the relative speed being
36 kmph, the distance from Mumbai would be : (75/36)
136 =
283.33 km

5. Two trains, Calcutta Mail and Bombay Mail,
start at the same time from stations Kolkata and Mumbai respectively towards
each other. After passing each other, they take 12 hours and 3 hours to reach Mumbai
and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48
km/h, the speed of the Bombay Mail is

1) 24 km/h
2) 22 km/h 3) 21 km/h 4) 96 km/h

Ans. (4) If you assume that the initial stretch
of track is covered by the two trains in time t each, the following figure will
give you a clearer picture.

Calcutta Mail t 12

3 t Bombay Mail

Kolkata Meeting
Point Mumbai

From the above figure, we can deduce that, t/3
= 12/t.

Hence, t

^{2}= 36, gives us t = 6.
Hence, the distance between Kolkata to the
starting point is covered by the Calcutta Mail in 6 hours, while the same
distance is covered by the Bombay Mail in 3 hours.

Hence, the ratio of their speeds would be 1:2.
Hence, the Bombay Mail would travel at 96 kmph.

**GEOMETRY AND MENSURATION**

1. A vertical stick 20 m long casts a shadow 10
m long on the ground. At the same time, a tower casts the shadow 50 m long on
the ground. Find the height of the tower.

1)100 m
2) 120 m 3) 25 m 4) 200 m

Ans.(1) When the length of stick = 20 m, then
length of shadow = 10 m i.e. in this case length = 2
shadow with the same angle of inclination of
the sun, the length of tower that casts a shadow of 50 m
2
50 m = 100 m

i.e. height of tower = 100 m

2. In the figure,
is similar to

C

E

If we have AB = 4 cm, ED = 3 cm, CE = 4.2 and
CD = 4.8 cm, find the value of CA and CB

1) 6 cm, 6.4 cm 2) 4.8 cm, 6.4 cm 3) 5.4 cm, 6.4 cm 4) 5.6 cm, 6.4 cm

Ans. (4)

Then
Then

3. The area of similar triangles, ABC and DEF
are 144 cm
be 36 cm, the the longest side of larger
be 36cm, then the longest side of smaller
is

^{2}and 81 cm2 respectively. If the longest side of larger
1) 20 cm
2) 26 cm 3) 27 cm 4) 30 cm

Ans. (3) For similar triangles ⟹ (Ratio of sides)

^{2}= Ratio of areas
Then as per question =
{Let the longest side of
DEF =

⟹

4. Two isosceles
have equal angles and their areas are in the
ratio 16:25. Find the ratio of their corresponding heights.

Seven

1) 4/5 2)
5/4 3) 3/2 4) 5/7

Ans. (1) (Ratio of corresponding sides)

^{2}= Ratio of area of similar triangle
5. The areas of two similar
are respectively 9 cm

^{2}and 16 cm^{2}. Find the ratio of their corresponding sides.
1) 3:4 2)
4:3
3) 2:3 4) 4:5

Ans. (1) Ratio of corresponding sides =

**COORDINATE GEOMETRY**

1. Find the distance between the points (3, 4)
and (8, -6).

1)
2)
3)
4)

Ans. (2)

2. Find the distance between the points (5, 2)
and (0, 0).

1)
2)
3)
4)

Ans. (3)

3. Find the value of p if the distance between
the points (8, p) and (4, 3) is 5.

1) 6 2)
0 3) Both (1) and (2) 4)
None of these

Ans. (3)

4. Find the value of c if the distance between
the point (c, 4) and the origin is 5 units.

1) 3 2)
-3 3) Both (1) and (2) 4) None of these

Ans. (3)

5. Find the mid-point of the line segment made
by joining the points (3, 2) and (6, 4).

1)
2)
3)
4)

Ans. (1)

**PERMUTATIONS AND COMBINATIONS 1**

1. How many numbers of 3-digits can be formed
with the digits 9, 8, 7, 6, 5 (repetition of digits not allowed)?

1) 125 2)
120 3) 60 4) 150

Ans. (3) The number of numbers formed would be
given by 5
4
3 (given that the first digit can be filled in
5 ways, the second in 4 ways and the third in 3 ways – MNP rule).

2. How many numbers between 2000 and 3000 can
be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not
allowed?

1) 42 2)
210 3) 336 4) 440

Ans. (2) The first digit can only be 2 (1 way),
the second digit can be filled in 7 ways, the third in 6 ways and the fourth in
5 ways. A total of 1
7
6
5 = 210 ways.

3. In how many ways can a person send
invitation cards to 6 of his friends if he has four servants to distribute the
cards?

1) 6

^{4}2) 4^{6}3) 24 4) 120
Ans. (2) Each invitation card can be sent in 4
ways. Thus,
.

4. In how many ways can 5 prizes be distributed
to 8 students if each student can get any number of prizes?

1) 40 2)
5

^{8}3) 8^{5}4) 120
Ans. (3) In this case since nothing is
mentioned about whether the prizes are identical or distinct we can take the
prizes to be distinct (the most logical thought given the situation). Thus,
each prize can be given in 8 ways – thus a total of 8

^{5}ways.
5. In how many ways can 7 Indians, 5 Pakistanis
and 6 Dutch be seated in a row so that all persons of the same nationality sit
together?

1) 3! 2)
7!5!6! 3) 3!7!5!6! 4) 182

Eight

Ans. (3) We need to assume that the 7 Indians
are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of
people can be arranged amongst themselves in 3! Ways. Also, within themselves
the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5!
Was respectively. Thus, the answer is
.

**PROBABILITY**

1. In throwing a fair dice, what is the
probability of getting the number ‘3’?

1)
2)
3)
4)

Ans. (2) Out of a total of 6 occurences, 3 is
one possibility = 1/6.

2. What is the chance of throwing a number
greater than 4 with an ordinary dice whose faces are numbered from 1 to 6?

1)
2)
3)
4)

Ans. (1) 5 or 6 out of a sample space of 1, 2,
3, 4, 5 or 6 = 2/6 = 1/3

3. Find the chance of throwing at least one ace
in a simple throw with two dice.

1)
2)
3)
4)

Ans. (4) Event definition is:

(1 and 1) or (1 and 2) or (1 and 3) or (1 and
4) or (1 and 5) or (1 and 6) or (2 and 1) or (3 and 1) or (4 and 1) or (5 and
1) or (6 and 1). Total 11 out of 36 possibilities = 11/36

4. Find the chance of drawing 2 blue balls in
succession from a bag containing 5 red and 7 blue balls, if the balls are not
being replaced.

1)
2)
3)
4)

Ans. (3) Event definition: First is blue and
second is blue = 7/12
6/11 = 7/22.

5. From a pack of 52 cards, two are drawn at
random. Find the chance that one is a knave and the other a queen.

1)
2)
3)
4)

Ans. (1) Knave and queen or Queen and Knave
4/52
4/51 + 4/52
4/51 = 8/663

**SET THEORY**

1. The percentage of those who like coffee or
tea but not smoking among those who like at least one of these is

1) more than 30 2) less than 30 3) less than 25 4) None of these

Ans. (1) 30/94
more than 30%. Option (1) is correct.

2. The percentage of those who like at least
one of these is

1) 100 2)
90 3) Nil 4) 94

Ans. (4) 94%. Option (4) is correct.

3. The two items having the ratio 1:2 are

1) Tea only and tea and smoking only. 2) Coffee and smoking only and tea only.

3) Coffee and tea but not smoking and smoking
but not coffee and tea. 4) None of
these

Ans. (3) Option (3) is correct as the ratio
turns out to be 10:20 in that case.

4. The number of persons who like coffee and
smoking only and the number who like tea only bear a ratio

1) 1:2 2)
1:1 3) 5:1 4) 2:1

Ans. (2) 12:12 = 1:1
Option (2) is correct.

5. Percentage of those who like tea and smoking
but not coffee is

1) 14 2)
14.9 3) less than 14 4) more than 15

Ans. (1) 14%. Option (1) is correct.

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