Aptitude Questions

1. A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs.2.40 the number of girls is

A.35
B.40
C.45
D.50

Ex:Step (i) Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x + y = 100 ————– (i)
Step (ii) A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x +2.40y = 312 ————– (ii)
Step (iii)
Solving (i) and (ii)
3.60 x + 3.60y = 360 ——- Multiply (i) by 3.60
=> 3.60 x + 2.40y = 312 ——— (ii)
1.20y = 48
y = 48/1.20 = 40
=> Number of girls = 40

2. If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days?

A.8 men
B.12 men
C.2 men
D.24 men

Ex:Let?s try solving this Problem using ratio approach.
“Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys.”
Let us therefore, find out the amount of men required, if only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men.
The man hours required to complete the new job = 4 times the man hours required to complete the old job.(As the new job is 4 times as big as the old job)
Let ‘n’ be the number of men required.
20 * 12 * 8 = n * 5 * 12 * 4 => n = 8
8 men working will be able to complete the given job.
However, the problem states that 6 women and 2 boys are working on the job.
6 women = 6/12= 5 men and 2 boys = 1 man. The equivalent of 5 + 1 = 6 men are already working.

3. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students?

A.57
B.56.8
C.58.2
D.52.2
Ex:Let the average weight of the 59 students be A.
Therefore, the total weight of the 59 of them will be 59A.
The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs.
59A + 45/ 60 = A – 0.2
=> 59A + 45 = 60A ? 12
=> 45 + 12 = 60A – 59A
=> A = 57

4. The difference between two angles of a triangle is 240. The average of the same two angles is 540.Which one of the following is the value of the greatest angle of the triangle?

A.450
B.600
C.660
D.720

Ex:Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 240.
=> a – b = 24.
Since the average of the two angles is 540, we have (a + b)/2 = 54.
Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields
({a + (a – 24)}/2) = 54
({2a ? 24}/2) = 54
2a – 24 = 54*2
2a – 24 = 108
2a = 108 + 24
2a = 132
a = 66
Also,

b = a – 24 = 66 – 24 = 42
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is 1800, a + b + c =180.
Plugging the previous results into the equation yields 66 + 42 + c = 180.
Solving for c yields c = 72
Hence, the greatest of the three angles a, b and c is c, which equals 720.

5. If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261….491492. How many 8’s will be used to write this large natural number?

A.32
B.43
C.52
D.53
Ex:From 259 to 458, there are 200 natural numbers so there will be 2*20 =40 8’s
From 459 to 492 we have 13 more 8’s and so answer is 40+13 =53

6. If each of the three nonzero numbers a , b , and c is divisible by 3, then abc must be divisible by which one of the following the numbers?

A.8
B.27
C.81
D.121
Ex:Since each one of the three numbers a, b, and c is divisible by 3, the numbers can be represented as 3p,3q, and 3r, respectively, where p, q, and r are integers.
The product of the three numbers is 3p*3q*3r =27(pqr).
Since p, q, and r are integers, pqr is an integer and therefore abc is divisible by 27.

7. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A.45%
B.45{5/11}%
C.54{6/11}%
D.55%
Ex:Number of runs made by running = 110 – (3 * 4 + 8 * 6)
= 110 – (60)
= 50.
Required percentage = ({50/110}*100)% = 45{5/11}%

8. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A.39, 30
B.41, 32
C.42, 33
D.43, 34
Ex:Let their marks be (x + 9) and x.
Then, x + 9 = (56/100)(x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.

9. The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

A.275
B.251
C.240
D.242
Ex:The order of each letter in the dictionary is ABLORU.
Now, with A in the beginning, the remaining letters can be permuted in 5! ways.
Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.
With L in the beginning, the first word will be LABORU, the second will be LABOUR.
Hence, the rank of the word LABOUR is 5!+5!+2 =242

10. A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?

A.6! * 1440
B.18! * 1440
C.18! *2! * 1440
D.None of these
EX:Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways in which all the students can be arranged = 2! * 18! * 6!
= 18! *1440

11. Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:

A.69760
B.30240
C.99748
D.42386
Number of words which have at least one letter replaced:
= Total number of words – total number of words in which no letter is repeated”
=>105 – {16P5}
=>100000 – 30240 =69760

12. In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A.0.285
B.0.40
C.100/249
D.99/250
Ex:250 numbers between 101 and 350 i.e. n(S) = 250
n(E) = 100th digits of 2 = 299-199 = 100
P(E) = {n(E)}/{n(S)} = 100/250 =0.40

13. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:

A.20
B.30
C.48
D.58
Ex:Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = 5*{3/5} : 8*{3/5} = 3 : 24/5
=> A : B : C = 2 : 3 : 24/5 = 10 : 15 : 24
=> B = (98*{15/49}) =30

14. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s salary?

A.Rs. 17,000
B.Rs. 20,000
C.Rs. 34,000
D.Rs. 38,000
Ex:Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, {2x + 4000}/{3x + 4000} = 40/57
=> 57*(2x + 4000) = 40*(3x + 4000)
=> 6x = 68,000
=> 3x = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) =Rs. 38,000

15. Sachin can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?

A.5 km
B.6 km
C.7 km
D.8 km
EX:Let total distance = D
Distance travelled at 4 kmph speed = (2/3)D
Distance travelled at 5 kmph speed = (1 – 2/3)D = (1/3)D
Total time = 1hr 24 min = (60+24)min = (84/60)hr = (21/15)hr
We know, time = distance/speed
Total time = (21/15)={(2/3)D}/4 + {(1/3)D} /5
21/15 = {2D}/12 + {D}/15
21/15={14D}/60
84 = 14D
D=6km