puzzles quantitative aptitude test

Online aptitude test questions and answers for competitive exam

Question 1
If 5 kg apples, 6 kg oranges and 2 keg potato cost rupee 186 and 13 kg apples, 6 kg orange and 4 kg potato cost rupee 288. what would be the cost fo 1kg potato, 2kg orange and 3 kg apple
A
122
B
59
C
79
D
67
Question 1 Explanation: 
79






Question 2
A B C start a business with capitals in the ratio 1/3:1/4:1/5. A withdraws half his capitals at the end of 4 months Out of a total annual profit of Rs 847, A's share is?
A
Rs 320
B
Rs 292
C
Rs 280
D
Rs 410
Question 2 Explanation: 
Ratio of capital in the beginning 1/3:1/4:1/5 = 20:15:12
suppose the originally invest Rs 20, Rs 15, Rs 12

ratio of the investment of 12 months = (20*4+10*8):15*12:12*12
=40:45:36
Therefore A's share = Rs (847*40/121) = Rs 280
Question 3
6 boys and 8 men get a total wage of Rs 12000 in 4 days. 8 boys and 6 men get a total of Rs 9000 in 3 days. The efficiency ratio of boy and a man respectively?
A
2:5
B
2:1
C
5:4
D
1:1
Question 3 Explanation: 
By solve the equation:

6X+8Y = 3000
8X+6Y = 3000
X=Y
X:Y = 1:1
Question 4
A sector of circle of radius equal to one fourth of the diagonal of square is cut from the square taking one corner of the square as centre. If side of the square is 4 cm. then area of the remaining part is
A
14.43
B
10.34
C
12.78
D
15.68
Question 4 Explanation: 
Diagonal of the square = 42 cm.
Hence radius of the circle = 42/4 cm. = 2 cm.
Area of the sector of circle = 1/4 (* radius2)
= 1/4 (* (2)2 )
= /2
Area of the square = 4*4 = =16 cm.2
Hence area of remaining part = 16- (/2) cm.2






Question 5
There are 5 eligible Gujarati grooms of which 2 know Bengali and 5 eligible Bengali grooms of which 3 know Gujarati. There are 5 eligible Gujarati and 5 eligible Bengali brides. An eligible bride is agreeable to marry a boy of his community or a boy who knows her language. Grooms have no choice . In how many different ways 10 couples can be formed.
A
16340
B
144000
C
435680
D
37200
Question 5 Explanation: 
In 5 eligible Gujarati grooms, 2 know Bengali, remaining 3 should make their pair with Gujarati girls.
No. of methods to make such couple = 5.4.3 = 60
Similarly in 5 eligible Bengali grooms, 3 know Gujarati, remaining 2 should make their pair with Bengali girls.
No. of methods to make such couple = 5.4 = 20
No. of methods to make remaining 5 couple = 5.4.3.2.1 = 120
Hence total no. of methods = 60*20*120 = 144000

Question 6
In a cricket match, Team A scored 240 runs without losing a wicket. The score consisted of byes, wides and runs scored by two opening batsmen: Ram and Shyam. The runs scored by the two batsmen are 11 times wides. There are 7 more wides than byes. If the ratio of the runs scored by Ram and Shyam is 5:6, then the percentage of run scored by Ram out of total score of Team A is:
A
46
B
43.4
C
41.8
D
39.6
Question 6 Explanation: 
Answer: 39.6
Let Runs Scored by Ram+Shyam be, X byes be y and wides be z
As per question
x = 11z and z = 7+y

So Total runs = x+y+z = 11z+z-7+z = 240
13z-7 = 240
= > z= 19
Therefore x = 11*19 = 209
Runs scored by Ram = 209*5/11 = 95

Percent of total run scored by Ram = (95/240)*100 = 39.6%
Question 7
Out of a group of ducks, 7/2 times the square root of the number are swimming in water while two remaining are playing on the shore. The total number of ducks is
A
12
B
16
C
14
D
4
Question 7 Explanation: 
Let the number of ducks be x.
Then, 7/2 sqrt(x) + 2 = x
=> 2x - 7 sqrt(x) - 4 = 0
=> (2y+1)(y-4) = 0
=> y = -1/2 or y = 4
=> sqrt(x) = -1/2 or sqrt(x) = 4
=> x = 1/4 or x = 16
Therefore
x=16 (Number of ducks can not be a fraction)






Question 8
A vendor bought 370 chocolates for 5550 rupees. How many chocolates in 5550 rupees must he sell to gain 25%?
A
290
B
296
C
256
D
300
Question 8 Explanation: 
CP of 1 Chocolate = 5550/370 = Rs. 15
SP of 1 Chocolate for 25% profit = 15*1.25 = Rs. 18.75
So in 5550 Rupess how many chocolates should be sold = 5550/18.75 = 296

Question 9
Find the next term of the series: 6, 11, 16, 21, 26, ?
A
31
B
35
C
28
D
36
Question 9 Explanation: 
Next term of series is derived
4*(n-1) + n+1 where n=1

So 4*1 + 2 = 6
4*2 + 3 = 11
4*3 + 4 = 16
4*4 + 5 = 21
4*5 + 6 = 26
4*6 + 7 = 31






Question 10
The average of marks obtained by 120 candidates was 35. If the average of passed candidates was 39 and that of failed candidates was 15, the number of candidates who passed the examination, is
A
90
B
100
C
80
D
160
Question 10 Explanation: 
Suppose x passed and (120-x) failed
Then
x*39 + (120-x)*15 = 120*35
or
24x = 2400
x = 100
Question 11
A five digit number has all non-zero different digits & decreasing arranged. Average of all the digits of the numbers is 5. The difference of number and the number formed by reversing the order of digits is 84942. Sum of all such possible digits are
A
195602
B
196002
C
196052
D
195588
Question 11 Explanation: 
Let number is 10000x+1000y+100z+10a+b
Where x not= y not= z not= a not= b not= 0
On reversing digits, we get the new number
10000b + 1000a + 100z + 10y + x
Difference
9999x + 990y - 990a - 999b
= 9999(x-b) + 990(y-a) = 84942
only possible case x-b = 8
so y-a = 5
if x-b = 8
x = 9, b = 1 (Only possibility)
and y-a = 5
(y, a)
(8, 3) and (7, 2)
So two possible case
x=9 y=8 z=? a=3 b=1 ---------(1)
and x=9 y=7 z=? a=2 b=1 -----(2)
Since average of digit = 45
so z=4 in (1) and z=6 in (2)
So number are
98431
97621
Sum = 196052







Question 12
What is the next number of the following sequence 12, 132, 1452, 15972, ..........
A
16141692
B
16419162
C
16146192
D
16146152
Question 12 Explanation: 
12
1 (1+2) 2 = 1 3 2 = 132
1 (1+3) (3+2) 2 = 1 4 5 2 = 1452
1 (1+4) (4+5) (5+2) 2 = 1 5 9 7 2 = 15972
Similarly
1 (1+5) (5+9) (9+7) (7+2) 2 = 1 6 14 16 9 2 = 16141692

Question 13
A dinner party is to be fixed for a group consisting of 100 persons. In this party, 50 person do not prefer Veg, 60 prefer Non-veg and 10 do not prefer either Non-veg or Veg. The number of persons who prefer both Veg and Non-veg is
A
20
B
50
C
35
D
40
Question 14
A sum fetched a total simple interest of Rs. 453 at the rate of 6% p.a. in 5 years. What is the sum?
A
Rs 1024
B
Rs 1450
C
Rs 1510
D
Rs 1475
Question 14 Explanation: 
SI = PRT/100
P= SI*100/RT = 453*100/5*6 = 1510






Question 15
The detail of a class test are as given below Marks -- No of Student 2 --------- 4 3 --------- 7 4 --------- 10 5 --------- 15 6 --------- 8 7 --------- 5 8 --------- 1 average marks of the class will be
A
5
B
5.1
C
4.7
D
4.5
Question 15 Explanation: 
Sum of all student marks = 235
Total student = 50
So average marks = 235/50
= 4.7

Question 16
Total number of methods of selecting 6 cards out of 10 different cards is, X and total number of methods of selecting 6 cards out of 10 similar cards is y. Difference of x and y is
A
59
B
209
C
0
D
60
Question 16 Explanation: 
Total number of methods of selecting 6 cards out of 10 different cards is
10*9*8*7*6*5/6*5*4*3*2*1 = 210
And total number of methods of selecting 6 cards out of 10 similar cards is only 1
Difference is 209
Question 17
The % obtained by a student is 74. If we take % on the basis of 4 subjects (leaving the marks of fifth subject) then % is 76. How many marks, he got in the fifth subject.
A
66
B
58
C
59
D
69
Question 17 Explanation: 
74*5 = 370
76*4 = 304
so his marks in fifth subject = 370-304 = 66






Question 18
The sum of 3 even positive integers is 16. If greatest integer is 5 times the least. Then product of the 3 integers will be
A
80
B
60
C
90
D
40
Question 18 Explanation: 
since greatest number is 5 times the least one so least will definitely be 2 and greatest 10, on this basis last one number will be 4
so product = 80
Question 19
Find two numbers such that one third of greater exceeds one half on the lesser by 1, and one fifth of the greater added to one sixth of the lesser equals one half of the lesser.
A
30, 18
B
28, 12
C
32, 16
D
15, 9
Question 19 Explanation: 
Now, according to question
(1/3)x=(1/2)y+1 (let larger no, X smallest no y)
and (1/5)x+(1/6)y=(1/2)y
On solving (i) and (ii), we get x=30 and y =18






Question 20
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 48. The sum is:
A
7250
B
None of these
C
7500
D
5625
Question 20 Explanation: 
Given that difference between compound interest and simple interest is Rs.48
= > 48 = P(4/100)^2
= > 48 = P(1/25)^2
= > 48 = P(1/625)
= > P = 30000
There are 20 questions to complete.






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