RRB NTPC 2021 Exam Phase-2 Memory Based Maths Questions with Answers: Check Mathematics Questions asked in RRB NTPC 2021 CBT
RRB NTPC 2021 Exam Phase-2 Memory Based Maths Questions with Answers: Get the memory based questions from Mathematics Section that came in RRB NTPC 2020-21 Phase-2 Exam held in online mode from 16^{th} Jan to 30^{th} Jan 2021.
RRB NTPC 2021 Exam Phase-2 Memory Based Maths Questions with Answers: RRB NTPC 2021 Phase-2 CBT Exam has been commenced from 16^{th} January 2021 and will be conducted till 30^{th} January 2021 for around 27 lakh Candidates across different exam centres in India. RRB NTPC 2021 Phase-3 will be conducted from 31st Jan to 12th Feb 2021 for around 28 Lakh candidates. Computer Based Test (CBT) is being conducted for the recruitment of RRB Non-Technical Popular Categories (NTPC) 35281 Vacancies (Graduate & Under-Graduate Posts) in two shifts.
In this article, we are going to share the important memory based General Intelligence & Mathematics section Questions as per the feedback received by the candidates who have appeared for RRB NTPC 2021 Phase-2 CBT Exam. Candidates are advised to definitely cover these questions for scoring high marks in the Exam. Let’s have a look at the Important Questions that are being covered in the RRB NTPC 2021 Exam:
RRB NTPC 2021 Mathematics Section Memory Based Questions with Answer: |
1. X alone can do a piece of work in 12 days and Y alone in 16 days. X and Y undertook to do it for Rs.6400. With the help of Z, they completed the work in 6 days. How much is to be paid to Z?
a) Rs. 375
b) Rs. 400
c) Rs. 600
d) Rs. 800
Answer: (b)
Explanation:
Answer: (c)
Explanation:
3. Find out the ratio of length and the area of a rectangle if the ratio of length & perimeter of rectangle is 1:6 and the area of rectangle is 800 sq.m?
a) 1:20
b) 2:10
c) 5:11
d) 2:13
Answer: (a)
Explanation: Let the length is x & perimeter be 6x
Let the breadth be B
Then 2(x + B) = 6x
x + B = 3x
B = 2x
Now
Length × breadth = area
x × 2x = 800
2x^{2} = 800
x = 20
Hence length = 20 m
Ratio of length: area
20 : 400
1 : 20
a) 5 : 3 : 2
b) 6 : 10 : 15
c) 2 : 3 : 5
d) 15 : 10 : 6
Answer: (d)
5. A started a business investing Rs. 50,000 in 2011. In 2012, he invested an additional amount of Rs. 20,000 and B joined him with an amount of Rs. 70,000. In 2013, B invested another additional amount of Rs. 20,000 and C joined them with an amount of Rs. 70,000. What will be A’s share in the profit of Rs. 2, 10,000 earned at the end of 3 years from the start of the business in 2011?
a) Rs. 1,00,000
b) Rs. 95,000
c) Rs. 1,20,000
d) None of these
Answer: (b)
Explanation: Ratio of A, B and C.
A: B: C = (50000 X 12 + 70000 X 12 + 70000 X 12) : (70000 X 12 + 90000) : (70000 X 12)
= 2280000: 1920000: 840000 =19: 16: 7
A’s share = Rs. (2, 10,000 × 19/42) = Rs. 95,000
6. P and Q can do a piece of work in 45 and 40 days respectively. They began the work together but P leaves after some days and Q finished the remaining work in 23 days. Then P has worked for:
a) 9
b) 21
c) 32
d) None of these
Answer: (a)
Explanation: Work done by P in one day =1/45 and by Q = 1/40
Work done by both in days = 8 + 9/360 = 17/360
Work done by Q = 9 23 = 207,
Remaining work is done by P & Q = 360 – 207/17 = 9 days.
7. If x - (1/x) = 3, then value of x^{3} – (1/x^{3}) is:
a) 32
b) 36
c) 40
d) 49
Answer: (b)
Explanation:
8. (0.16)^{7}/ (0.064)^{3} × (0.4)^{2}=(0.4)^{?}
a) 5
b) 6
c) 7
d) 8
e) None of the above
Answer: c)
Explanation:
9. X, Y and Z start a business. X invests one-third of the capital, Y invests One-fourth of the capital and the rest is invested by Z. The share of Z in the total profit of Rs. 480000 is.
a) Rs. 160000
b) Rs. 180000
c) Rs. 240000
d) Rs. 200000
Answer: d)
Explanation: X : Y : Z = 1/3 : 1/4 : 5/12 = 4: 3: 5
Z’s share = 480000 X 5/12 = 200000
10. In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.
a) 40%
b) 41%
c) 43%
d) 44%
Answer: d)
Explanation: Failed in mathematics, n (1) = 34
Failed in English, n (2) = 42
n (AUB) = n(1) + n (2) – n (A∩ B)
= 34 + 42- 20 = 56
Failed in either or both subjects are 56
Percentage passed = (100−56)% = 44%