## Select Exam

### eLitmus Most Repeated Question in Aptitude Section

Question 1] In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient ?
A.     0
B.     12
C.     13
D.     20

Explanation :
Let the quotient be 'x'.
Then, according to the question,
=> x/12=35
=> x=420
Therefore , the required quotient will be 420/21=20.

Question 2] A report has 20 sheets each of 55 lines and each line consists of 65 characters. If this report is to be retyped with each sheet having 65 lines and each line of 75 characters what will be the reduction percent in the pages ?
A.     22.5 %
B.     35 %
C.     25 %
D.     None of these

Explanation :
Let 20 sheets contains 55*65*20 characters which is equals with y pages which contains 65*75*y characters .
Both the sheets have equal characters.
So, 20 * 55 * 65 = y * 65 * 75
=> y = 14.6 pages means 15 pages.
So reduction in sheets = (20-15)/20 * 100% = 25%.

Question 3 ] The sum of how many terms of the series 6 + 12 + 18 + 24 + … is 1800 ?
A.     16
B.     24
C.     20
D.     18
E.     22

Explanation :
Let the number of terms be N.
Then, according to the formula,
=> 1800 = N/2[2A+(N-1)D] , where A = First-Term, D = Common-Difference.
=> 3600 = N[12 + (N-1)6]
=> 3600 =N[6 + 6N]
=> 6N^2 + 6N - 3600 = 0.
=> 6N^2 + 150N - 144n - 3600 = 0
=> 6N(N + 25)-144(N+25)
=> N = -25, 24
Hence, the number of terms is 24.

Question 4] The bankers discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time is Rs. 60. The sum due is:
A.     Rs. 360
B.     Rs. 432
C.     Rs. 540
D.     Rs. 1080

Explanation :
According to the formula,
Sum = (B.D*T.d./(B.D-T.d., where B.D =Bankers Discount, T.D = True Discount.
=> Sum = (72*60)/72-60 = 360.

Question 5] Find the probability that the sum of n numbers is divisible by n ,,assuming n is very large no.
A.     1
B.     1/2
C.     3/4
D.     None of these
E.     Can't be determined.

Explanation :
Sum of n numbers is n(n+1)/2
case 1) When n is odd take 3
Sum is 3*4/3 it is divisible by 3
we can conclude that n=odd sum of n numbers is divisible by n
case2) When n is even take 4
Sum is 4*5/2=>10 10/4 rem is 2 so not divisible
we can conclude that n=even sum of n numbers is not divisible by n
Total of total 2 cases 1 case is possible
Hence, the required probability = Possible/Total = 1/2.

Question 6] S(n) is sum of first n natural numbers, Z(n) = 2S(n)+41, lowest value of n such that Z(n) is NOT a prime.
A.     7
B.     6
C.     40
D.     Always a prime

Explanation :
As the sum of first N natural numbers is S(n)=n/2(1+n).
So,z(n)=2(s(n))+41=2*n/2(1+n)+41=n(1+n)+41
Now, put all the given options one by one
Take, n=7
z(7)=7(7+1)+41=56+41=97 a prime number
Take n=6
z(6)=6(6+1)+41=41+41=83 again a prime number
Now take n = 40
z(40)=40(40+1)+41=40*41+41=41(40+1)=41*41 = 1681 not a prime number.

Question 7 ] There are sixty markings on a clock, minutes hand was exactly on one of those markings. Nine markings away from the minutes hand was hours hand. What could be the time?
A.     5:14
B.     6:20
C.     7:48
D.     3:09

Explanation :
Given angle Difference = 9 min
we know, 1min = 6 deg
Therefore 9mins=9*6 = 54 deg
So the angle diff must be 54 deg.
Lets go through the options,
Option(A): 5hrs 14mins
Angle traced by hour hand=(5*30)+(14/2)=157
Angle traced by minute hand=(14*6)=84
Angle diff = 157-84 = 73
Similarly for option C:
Angle traced by hour hand=(7*30)+(48/2)=234
Angle traced by minute hand=(48*6)=288
Angle diff=(288-234)=54.

Question 8 ] There are 41 students in a class, number of girls is one more than number of guys. We need to form a team of four students. All four in the team cannot be from same gender. Number of girls and guys in the team should NOT be equal. How many ways can such a team be made?
A.     49400
B.     65040
C.     50540
D.     None of these

Explanation :
Let the number of boys = 20 and girls = 21.
Now the combinations are {girl,girl,girl,boy} or {boy,boy,boy,girl}
As number of them cannot be same and cannot be filled by any one group
So total number of ways:-
=> (20C3 * 21C1) + (20C1 * 21C3) = 50540.

Question 9 ] Probability that guy1 will successfully hit a shot is 2/3. probability of guy2 successfully hitting a shot is 2/3. if they shoot two shots each, what is the probability that they have equal number of hits?
A.     15/27
B.     11/27
C.     13/27
D.     None of these

Explanation :
Consider these cases:
For 0 hit each: (1 way)
(1/3*1/3)[guy1]*(1/3*1/3)[guy2]
for 1 hit each: (4 ways)
(2/3*1/3)*(2/3*1/3), (2/3*1/3)*(1/3*2/3), (1/3*2/3)*(2/3*1/3), (1/3*2/3)*(1/3*2/3)
for 2 hits each:(1 way)
(2/3*2/3)*(2/3*2/3)
Add all these cases, you will get answer as 11/27.

Question 10 ] Average weight of 20 boys and 15 girls in a class is 29 kg and 22 kg respectively.Find the average weight of all the students in the class ?
A.     25
B.     20
C.     26
D.     28

Explanation :
Average weight of 20 boys are = 20*29=580.
Average weight of 15 girls are= 15*22=330.
The total average weight of all students =total average weight /number of students
=> (580+330)/(15+20)
=> 910/35
=> 26

Question 11 ] 100 men started working to complete a work in 50 days.After working 8hours/day they had completed 1/3rd of the work in 25 days.How many men should be hired so that they will finish the work in time by working 10 hours/day.
A.     60 men
B.     50 men
C.     30 men
D.     None of these

Explanation :
Given, Working 8 hours/day, in 25 days, 100 men completes 8*25*100= 20000 men hours which is 1/3 of the work.
So remaining 2/3 work to be completed =20000*(2/3)/(1/3)=40000 men hours.
If with 'x' extra men hired ,remaining work is to be completed in 50-25=25 days, working 10 hours/day.
Then, (100+x)*25*10=40000
=> 25000 + 250x = 40000
=> x = 60.

Question 12 ] A boll drop from a height 40 m then it come to ground and again jump its half of back height then till it come to rest how it cover the distance from start point to its rest time.
A.     120
B.     130
C.     125
D.     None of these

Explanation :
When ball first re bounce then its distance 40 m and then in 2nd re bounce half of 40 and its jump upside first and then go below side So its distance becomes two times of half of 40 that is 40(1/2)*2.
Similarly in third time re bounce 40(1/2)^2*2 and so on.
It's G.P series becomes 40 + 40(1/2)*2 + 40(1/2)^2*2 + 40(1/2)^3*2 + ........
=> 40 + 2 { 40(1/2) + 40(1/2)^2 + 40(1/2)^3 +......+infinite term.}
=> 40 + 2 * 40 { 1/2 + 1/2^2 + 1/2^3+1/2^4......+infinite term.}
Here a = 1/2 and r=1/2
=> 40 + 80 ( 1/2/1-1/2) ( sum of infinite terms of G.P series is a/1-r ) [ a=1/2 and 1-r = 1-1/2=1/2 ]
=> 40+80 = 120.

Question 13 ] Area of the rectangle field is 144m2.if length is increased by 5 metres.its area increases by 40 m2.what is the length of the field?
A.     16
B.     22
C.     18
D.     20

Explanation :
If length = x, breadth = y, then xy=144 and
=> (x + 5)y = 144 + 40
=> xy + 5y = 184
Substituting xy = 144, y = 8,
So x = 144/8 = 18.

Question 14 ] How many 5 digit numbers can be formed by using 0,2,4,6,8 which is divisible by 8? (for a no. to be divisible by 8, last three digits of the number should be divisible by 8.
A.     30
B.     28
C.     26
D.     32

Explanation :
1st place can be filled in 4 no of ways (except 0)
2nd place can be filled in 3 no of ways
last 3 places can be filled in 18 no of ways
Hence,the required number = 4*3+18=30.

Question 15 ] 7 people average age is 30. youngest person age is 7. find average of the people when youngest was born ?
A.     18
B.     20
C.     25
D.     23

Explanation :
Given, Average age of people =30
So the total age = 210
before 7 years we have to deduct each person age by seven years 210-49 =161.
So, average age would be 161/7=23.

Question 16] How many 3 digits number are there which sum of digits is equal to 16 ?
A.     45
B.     50
C.     66
D.     60

Explanation :
Numbers end with 9=7
Numbers end with 8=8
Numbers end with 7=9
Numbers end with 6=9
Numbers end with 5=8
Numbers end with 4=7
Numbers end with 3=6
Numbers end with 2=5
Numbers end with 1=4
Numbers end with 0=3
Hence, required total = 66.

Question 17 ] The series of differences between consecutive prime numbers is represented as dp1, dp2, dp3, .... dpn .Whare dp1 is the difference between the second and the first prime number. Find the sum of series when n = 23, given that the 23rd prime number is 83 ?
A.     81
B.     82
C.     83
D.     87

Explanation :
Let 'p' denotes prime number.
Given, p23 = 83.
According to the question,
dp1= p2-p1, dp2 = p3-p2 .......
So, p2-p1 + p3-p2 + p4-p2 +.......... + p23-p22 = p23-p1 = 83 - 2 = 81.

Question 18] Einstein walks on an escalator at a rate of 5steps per second and reaches the other end in 10 sec. while coming back, walking at the same speed he reaches the starting point in 40secs. What is the number of steps on the escalator?
A.     40
B.     60
C.     120
D.     80
E.     data insufficient

Explanation :
Let the speed of escaltor be X steps per sec.
And length of escalator be Y.
Einstein's speed = 5 step/ sec
Then,
During upward movement
=> 5-X=Y/10
During downward movement
=> 5+X=Y/40
By solving equations we get, Y=80.

Question 19] How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9*8=72, 72=7*2=14, 14=1*4=4. Consider only 4 prime no.s (2,3,5,7).
A.     17
B.     18
C.     15
D.     16

Explanation :
According to the question,
2 = 12 or 21 So 1×2, 2×1, 3×4, 4×3, 2×6, 6×2, 3×7, 7×3
3 = 13 or 31 So 1×3, 3×1
5 = 15, 51 So 1×5, 5×1, 3×5, 5×3, 7×5, 5×7
7 = 17 or 71 So 1×7, 7×1
15 = 3×5 = 5×3
So total 18 numbers = 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75
Hence, the required number is 18.

Question 20] Ram draws a card randomly among card 1 to 23 and keep it back.Then Sam draws a card among those. What is the probability than Sam has drawn a card greater than Ram ?
A.     11/23
B.     11/25
C.     17/21
D.     None of these

Explanation :
Sum of probability of drawing a card greater than drawn by ram is (22/23) + (21/23) + .....(1/23) = 253/23
But probability for choosing 1 card out of 23 is 1/23
Therefore, total probability is (253/23)*(1/23)=(11/23).

Other Useful Questions [Try for extra Practice !]

1. A certain type of mixture is prepared by mixing brand A at Rs.9 a kg.
with brand B at Rs.4 a kg. If the mixture is worth Rs.7 a kg., how many
kg. of brand A are needed to make 40kgs. of the mixture?

Ans. Brand A needed is 24kgs.

2. A wizard named Nepo says “I am only three times my son’s age. My
father   is 40 years more than twice my age. Together the three of us are
a mere 1240   years old.” How old is Nepo?

Ans. 360 years old.

3. One dog tells the other that there are two dogs in front of me. The
other one also shouts that he too had two behind him. How many are they?

Ans. Three.

4. A man ate 100 bananas in five days, each day eating 6 more than the
previous day. How many bananas did he eat on the first day?

Ans. Eight.

5. If it takes five minutes to boil one egg, how long will it take to
boil four eggs?

Ans. Five minutes.

6. Three containers A, B and C have volumes a, b, and c respectively; and
container A is full of water while the other two are empty. If from
container A water is poured into container B which becomes 1/3 full, and
into container C which becomes 1/2 full, how much water is left in
container A?

7. ABCE is an isosceles trapezoid and ACDE is a rectangle. AB = 10 and EC
= 20. What is the length of AE?

Ans. AE = 10.

8. In the given figure, PA and PB are tangents to the circle at A and B
respectively and   the chord BC is parallel to tangent PA. If AC = 6 cm,
and length of the tangent AP   is 9 cm, then what is the length of the
chord BC?

Ans. BC = 4 cm.

9 Three cards are drawn at random from an ordinary pack of cards. Find
the probability that they will consist of a king, a queen and an ace.

Ans. 64/2210.

10. A number of cats got together and decided to kill between them 999919
mice. Every cat killed an equal number of mice. Each cat killed more mice
than there were cats. How many cats do you think there were ?

Ans. 991.

11. If Log2 x – 5 Log x + 6 = 0, then what would the value / values of x
be?

Ans. x = e2 or e3.

12. The square of a two digit number is divided by half the number. After
36 is added to the quotient, this sum is then divided by 2. The digits of
the resulting number are the same as those in the original number, but
they   are in reverse order. The ten’s place of the original number is
equal to twice   the difference between its digits. What is the number?

Ans. 46

13.Can you tender a one rupee note in such a manner that there shall be
total 50 coins but none of them would be 2 paise coins.?

Ans. 45 one paisa coins, 2 five paise coins, 2 ten paise coins, and 1
twenty-five paise coins.

14.A monkey starts climbing up a tree 20ft. tall. Each hour, it hops 3ft.
and slips back 2ft. How much time would it take the monkey to reach the
top?

Ans.18 hours.

15. What is the missing number in this series?   8 2 14 6 11 ? 14 6 18 12

Ans. 9

16. The minute hand of a clock overtakes the hour hand at intervals of 64
minutes of correct time. How much a day does the clock gain or lose?

Ans. 32 8/11 minutes.

17. Solve for x and y:   1/x – 1/y = 1/3, 1/x2 + 1/y2 = 5/9.

Ans. x = 3/2 or -3 and y = 3 or -3/2.

18. Daal is now being sold at Rs. 20 a kg. During last month its rate was
Rs. 16 per kg. By how much percent should a family reduce its consumption
so   as to keep the expenditure fixed?

Ans. 20 %.

19. Find the least value of 3x + 4y if x2y3 = 6.

Ans. 10.

20. Can you find out what day of the week was January 12, 1979?

Ans. Friday.

21. A garrison of 3300 men has provisions for 32 days, when given at a
rate of 850 grams per head. At the end of 7 days a reinforcement arrives
and it was found that now the provisions will last 8 days less, when
given at the rate of 825 grams per head. How, many more men can it feed?

Ans. 1700 men.

22. From 5 different green balls, four different blue balls and three
different red balls, how many combinations of balls can be chosen taking
at least   one green and one blue ball?

Ans. 3720.

23. Three pipes, A, B, & C are attached to a tank. A & B can fill it in
20   & 30 minutes respectively while C can empty it in 15 minutes. If A,
B & C   are kept open successively for 1 minute each, how soon will the
tank be filled?

Ans. 167 minutes.

24. A person walking 5/6 of his usual rate is 40 minutes late. What is
his usual time?