**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.**