puzzles quantitative aptitude test

Online aptitude test questions and answers for company 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:
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
There are 10 questions to complete.