Quantitative Aptitude Interview Preparation Guide
Sharpen your Quantitative Aptitude interview expertise with our handpicked 55 questions. These questions will test your expertise and readiness for any Quantitative Aptitude interview scenario. Ideal for candidates of all levels, this collection is a must-have for your study plan. Access the free PDF to get all 55 questions and give yourself the best chance of acing your Quantitative Aptitude interview. This resource is perfect for thorough preparation and confidence building.55 Quantitative Aptitude Questions and Answers:
1 :: 36 people {a1, a2, ..., a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, ..., {a35, a36}, {a36, a1}. Find the size of the smallest set of people such that the rest have shaken hands with at least one person in the set?
For at least one person & for exactly one person =36/3
For at least two person & for exactly two person =36/2
so
ans is 36/3= 12
if question is asked lyk no "cycle" then n-1.
i.e.36-1
For at least two person & for exactly two person =36/2
so
ans is 36/3= 12
if question is asked lyk no "cycle" then n-1.
i.e.36-1
2 :: In a group of 15, 7 have studied Latin, 8 have studied Greek, and 3 have not studied either. How many of these studied both Latin and Greek?
A. 0
B. 3
C. 4
D. 5
L=7
G=8
Total number of students(U)=15
Number of students who study neither of them (N)=3
draw a simple diagram representing the above situation.
we need to find out L INTERSECTION G.
Thus we have,
L UNION G + N = 15
L UNION G = L + G - L INTERSECTION G = 7 + 8 - L INTERSECTION G
L INTERSECTION G= 7 + 8 +N-15
= 7+8+3-15
= 3
G=8
Total number of students(U)=15
Number of students who study neither of them (N)=3
draw a simple diagram representing the above situation.
we need to find out L INTERSECTION G.
Thus we have,
L UNION G + N = 15
L UNION G = L + G - L INTERSECTION G = 7 + 8 - L INTERSECTION G
L INTERSECTION G= 7 + 8 +N-15
= 7+8+3-15
= 3
3 :: Explain series: 2,7,24,77,.
Ans is:
2*3+1=7
7*3+3=24
24*3+5=77
77*3+7=238
Therefore ans is 238
2*3+1=7
7*3+3=24
24*3+5=77
77*3+7=238
Therefore ans is 238
4 :: What is the largest prime no. stored in 8bit/6bit/9bit/7bit memory?
A,B,C are the mechanisms used separately?
for 6 bit maximum allowed number is 64 prime number is 61
for 7 bit maximum allowed number is 128 prime number is 127
for 8 bit max allowed number is 256 prime number is 253
for 9 bit max allowed number is 512 prime number is 511
for 7 bit maximum allowed number is 128 prime number is 127
for 8 bit max allowed number is 256 prime number is 253
for 9 bit max allowed number is 512 prime number is 511
5 :: A circular dart board of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point X in the circle. What is the probability that X is closer to the center of the circle than the periphery?
1/4
6 :: 1/3 of a number is 3 more than 1/6 of the same number.
What is the number?
let x be a no
now 1/3 of a no is x/3
and again
x/3=x/6+3
x/3-x/6=3.
x=18
now 1/3 of a no is x/3
and again
x/3=x/6+3
x/3-x/6=3.
x=18
7 :: A person, who decided to go to weekend trip should not exceed 8 hours driving in a day. Average speed of forward journey is 40 m/h. Due to traffic in Sundays, the return journey average speed is 30 m/h. How far he can select a picnic spot?
a) 120 miles
b) Between 120 and 140 miles
c) 160 miles
solution--- t1= time for forword journey. t2 =time 4 return journey. t=t1 + t2 =8hrs; x= destination; t1 + t2 = x / 40 + x / 30 = 8; x( 7/120 ) = 8; x= 8*120/ 7 = 137 milesie answer is(b)
8 :: How many 5 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?
Divisibility rule of 4 is last two digits are divisible by 4
so in 5 digit no last two digit should always be 12,24,32,44,52;
(any one of them may be)
so for last two digits 5 ways are possible ;
since repetion are allowed then for 3rs place form right there are also 5 ways so in 3rd place either of 1,2,3,4,5 is possible;
simillarly at 4 rth place either of 1,2,3,4,5 is possible so again 5 ways are possible
simillarly at 5th place also 5 ways are possible
so total no are: 5*5*5*5=625
so in 5 digit no last two digit should always be 12,24,32,44,52;
(any one of them may be)
so for last two digits 5 ways are possible ;
since repetion are allowed then for 3rs place form right there are also 5 ways so in 3rd place either of 1,2,3,4,5 is possible;
simillarly at 4 rth place either of 1,2,3,4,5 is possible so again 5 ways are possible
simillarly at 5th place also 5 ways are possible
so total no are: 5*5*5*5=625
9 :: If A is traveling at 72 km per hour on a highway. B is
traveling at a speed of 25 meters per second on a highway.
What is the difference in their speeds in m/sec?
ya the answer is 5sec..
converting 72kmph in m/sec we get 20m/sec
so 25-20=5sec/m
converting 72kmph in m/sec we get 20m/sec
so 25-20=5sec/m
10 :: The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
Let a, b, and c denote pencil, pen and eraser respectively.
Given that the cost of one pencil, two pens and four erasers is Rs.22
This can be written as,
a + 2b + 4c = 22 ..........(1)
Similarly, the cost of 5 pencils, 4 pens and 2 erasers is Rs. 32
This implies ....
5a + 4B + 2c = 32 ......... (2)
Adding (1) and (2), we get
6a + 6b + 6c = 54
We need to find the value of 3a + 3b + 3c.
Dividing by 2, we get
3a + 3b + 3c = 27.
Hence, the cost of 3 pencils, 3 pens, and 3 erasers is Rs.27
Given that the cost of one pencil, two pens and four erasers is Rs.22
This can be written as,
a + 2b + 4c = 22 ..........(1)
Similarly, the cost of 5 pencils, 4 pens and 2 erasers is Rs. 32
This implies ....
5a + 4B + 2c = 32 ......... (2)
Adding (1) and (2), we get
6a + 6b + 6c = 54
We need to find the value of 3a + 3b + 3c.
Dividing by 2, we get
3a + 3b + 3c = 27.
Hence, the cost of 3 pencils, 3 pens, and 3 erasers is Rs.27