Quantitative Aptitude Questions with Answers – Part 1
This blog explains about Quantitative Aptitude Questions with Answers – Part 1 and is given below :
1.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
A.120 metres B.180 metres C.324 metres D.150 metres
Answer :Option D
Explanation :
1 km = 1000 m and 1 hr = 3600s
Speed = 60 × 1000 / 3600
= 50/3 m/s .
Now distance = speed × time
= 50 / 3 × 9
= 50 × 3
= 150 m
Therefore the length of the train is 150 metres. ( 0.15 km )
2.A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
A.3 B.4 C.5 D.6
Answer: Option C
Explanation:
C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1
= Rs. 6/5 = Rs. 1.20
For Rs. 1.2, toffees sold = 6
For Re. 1, toffees sold = 6 x 5/6
= 5
3.The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
A.17 kg B.20 kg C.26 kg D.31 kg
Answer :Option D
Explanation :
Weight of B = (A + B)’s weight + (B + C)’s weight – (A + B + C)’s weight
= 40 * 2 + 43 * 2 – 45 * 3
= 80 + 86 – 135
= 31 kg.
Hence option [D] is correct answer.
4.A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :
A.4 days B.6 days C.8 days D.12 days
Answer : Option C
Explanation :
Let A and C complete the work in x days
(A+B)’s one day’s work= 1/8
(B+C)’s one day’s work= 1/12
(C+A)’s one day’s work= 1/x
(A+B+C)’s one day’s work = 1/6
We have C’s one day work = (A+B+C)-(A+B)
= 1/6 – 1/8
= 8-6/48
C’s one day work = 1/24
A’s one day work = (A+B+C)-(B+C)
= 1/6 – 1/12
= 2-1/12
A’s one day work = 1/12
(A+C)’s one day work = 1/24 + 1/12
= 1+2/24
= 3/24
= 1/8
x = 8 days
5.Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
A.2 : 5 B.3 : 5 C.4 : 5 D.6 : 7
Answer: Option C) 4:5
Explanation:
Let the third number be x.
Then, first number = 120% of x =120x/100 = 6x/5
Second number =150% of x = 150x/100 = 3x/2
Ratio of first two numbers = 6x/5 : 3x/2
= 12x : 15x
= 4 : 5
6.The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
A.4 years B.8 years C.10 years D .None of these
Answer: Option A
Explanation :
Since, the children were born at the interval of three years ,let us assume their age as x, x+3,x+6,x+9,x+12.
given- the sum of five children is 50
therefore,
x+x+3+x+6+x+9+x+12 = 50
5x+30 = 50
5x = 50-30
5x=20
x=4
Hence, age of the youngest child is 4 years.
7.Look at this series: 8, 6, 9, 23, 87 , … What number should come next?
A.128 B.226 C.324 D.429
Answer : Option D
Explanation :
(8×1) – 2=6.
(6×2) – 3=9
(9×3) – 4=23
(23×4) – 5=87
The next term = ( 87×5 ) – 6
= 429
8.66 games were played in a tournament where each player played against each of the rest. The number of players is
A) 33 B) 12 C) 13 D) 11
Answer : Option B
Explanation :
Let the number of players = n+1
Number of games = n(n+1)/2 = 66
= n(n+1) = 132
= 11 * 12 = 132
Thus the number of players = 12
9. A father is 30 years older than his son however he will be only thrice as old as the son after 5 years .what is father’s present age?
(A) 40 yrs (B) 30 yrs(C)50 yrs(D)None of these
Answer : Option A
Explanation :
Let the age of the son = x
Then the age of his father = x+30
Given,father will be only thrice as old as the son after 5 years
x+30+5 = 3(x+5)
x+35 = 3x+15
35-15 = 3x-x
20 = 2x
2x = 20
x = 20/2 = 10
Therefore the age of his father = x+30
= 10+30
= 45 years
10. Arun, Kamal and Vinay invested Rs. 8000, Rs. 4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months, there was a gain of Rs. 4005, then what will be the share of Kamal?
A.Rs. 890 B.Rs. 1335 C.Rs. 1602 D.Rs. 1780
Answer: Option A
Explanation:
Arun : Kamal : Vinay = (8,000 x 6) : (4,000 x 8) : (8,000 x 8)
= 48 : 32 : 64
= 3 : 2 : 4.
Kamal’s share = Rs.4005 x 2/9
= Rs. 890.