Infosys Quantitative Aptitude Test pattern: Before attempting the exam please check all the question given below. Here we provided a pattern for Infosys exam for reasoning and Quantitative Aptitude and verbal ability. so this jntu fats result has provided not for all the students move to the success path and follow our updates and bookmark this page to get more job updates.
Mathematical Critical thinking and Logical Reasoning:
|Testing Areas||No. of Questions||Time (in minutes)|
Infosys Quantitative Aptitude Test pattern
In how many ways team of four can be formed from four boys and three girls such that at least one boy and one girl should be there?
34 Correct Answer
One boy and one girl can be selected as (one boy & three girls) or (two boys & two girls) or (three boys & one girl)= (4C1*3C3)+(4C2*3C2)+(4C3*3C1)=4+18+12=34
Suresh can finish a piece of work by himself in 42 days. Mahesh, who is 1/5 times more efficient as Suresh, requires X days to finish the work by working all by himself. Then what is the value of X?
35 Correct Answer 1
No of days taken by Suresh to finish the work=42.
So part of the work was done by Suresh in one day =1/42.
Since Mahesh is 1/5 times more efficient as Suresh,part of work done by Mahesh in 1 day =1/42 + (1/5)×(1/42). = (1/42) + (1/210) =6/210=1/35.
So Mahesh can finish the work in 35 days.
I forgot the last digit of a 7-digit telephone number. If 1 randomly dials the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?
1/1000 Correct Answer 1
It is given that the last three digits are randomly dialed. Then each of the digits can be selected out of 10 digits in 10 ways.
Hence required probability = (1/10)3 =1/1000
The regular mathematics faculty could not teach because of being sick. As a stop-gap arrangement, different visiting faculty taught different topics on 4 different days in a week. The scheduled time for class was 7:00 am with a maximum permissible delay of 20 minutes. The monsoon made the city bus schedules erratic and therefore the classes started at different times on different days.Mr. Singh didn’t teach on Thursday. Calculus was taught in the class that started at 7:20 am. Mr. Chatterjee took the class on Wednesday, but he didn’t teach probability. The class on Monday started at 7:00 am, but Mr. Singh didn’t teach it. Mr. Dutta didn’t teach ratio and proportion.Mr. Banerjee, who didn’t teach set theory, taught a class that started five minutes later than the class featuring the teacher who taught probability. The teacher in Friday’s class taught set theory. Wednesday’s class didn’t start at 7:10 am. No two classes started at the same time.
Probability was taught by:
Mr. Dutta on Monday Correct Answer 1
Mr. Singh on Monday
Mr. Singh on Wednesday
Mr. Dutta on Thursday
The class on Wednesday started at
7:20 am and the topic was calculus Correct Answer 1
7:00 am and topic was calculus.
7:05 am and topic was ratio and proportion
7:25 am and topic was ratio and proportion
The option which gives the correct teacher- subject combination is:
Mr. Chatterjee – set theory
Mr. Singh-set Theory Correct Answer 1
Mr. Singh – calculus
Mr. Chatterjee – ratio and proportion
Mr. Banerjee – calculus
If Banerjee has a class on Wednesday then at what time Chatarjee will have his scheduled class?
7:05 Correct Answer 1
The option which gives a possible correct class time – weekday combination is:
Wednesday – 7:20 am, Thursday – 7:15 am, Friday – 7:20 am
Wednesday 7:20 am; Thursday 7:05 am; Friday 7:10 am Correct Answer 1
Wednesday – 7:05 am, Thursday – 7:20 am, Friday – 7:10 am
Wednesday – 7:10 am, Thursday – 7:15 am, Friday – 7:05 am
Wednesday – 7:10 am, Thursday – 7:20 am, Friday – 7:05 am
If NINE + FINE = WIVES, find the value of S+I+N+E if E = 5 and V=3
18 Correct Answer 1
Given E=5, V=3 => (E+E)=S=0(1 carry, since 5+5=10 ) ……….eq1
W=1(we know there should be a carry so.. )
N+N+1 should give a remainder 5(because E= 5) and I+I= V(3) so it should have a carry. So N=7 and I=6
N+F=6, so F=8 gives 7675+8675=16350, S+I+N+E= 18
A can contains a mixture of two liquids A and B in the ratio 7: 5. When 9 liters of the mixture is drawn off and the can is filled with B, the ratio of A and B becomes 7: 9. How many liters of liquid A was contained by the can initially?
Suppose the can initially contain 7x and 5x liters of mixtures A and B respectively
A quantity of A in mixture left =(7x−7/12×9)
The quantity of B in mixture left = (5x−5/12×9)
A/B = 7/9 = ( (7x−7/12×9)/ (5x−5/12×9 + 9) ⇒x=3⇒x=3 Orginal = 7*3 So, the can contain 21 liters of A.
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