RRB ALP 2018 Practice Test Papers | Arithmetic Questions (Day-18)
Dear Aspirants, Here we have given the Important RRB ALP & Technicians Exam 2018 Practice Test Papers. Candidates those who are preparing for RRB ALP 2018 can practice these Arithmetic Questions to get more confidence to Crack RRB 2018 Examination.
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RRB ALP & Technician | Aptitude Day-18
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Question 1 of 10
1. If a = 115, b = -113, c = –2, then what is the value of a3 + b3 + c3?
Explanation:
Answer: (D)
Note that here, a + b + c = 0 Therefore, a3 + b3 + c3 = 3abc = 3 (115) (-113) (-2) = 77970.
Explanation:
Answer: (D)
Note that here, a + b + c = 0 Therefore, a3 + b3 + c3 = 3abc = 3 (115) (-113) (-2) = 77970.
Question 2 of 10
2. 500 ÷ 50 × 5 =?
Explanation:
Answer: (C)
According to the BODMAS rule, ? = 500 ÷ 50 × 5 = 10 × 5 = 50
Explanation:
Answer: (C)
According to the BODMAS rule, ? = 500 ÷ 50 × 5 = 10 × 5 = 50
Question 3 of 10
3. X is twice as efficient as Y. X takes 20 days to complete a job. In how many days, can X and Y complete the job together?
Explanation:
Answer: (D)
X’s one-day work = 1/20
Y’s one-day work = 1/40 (because X is twice as efficient as Y) X and Y’s one-day work 1/20+1/ 40=3/40
Explanation:
Answer: (D)
X’s one-day work = 1/20
Y’s one-day work = 1/40 (because X is twice as efficient as Y) X and Y’s one-day work 1/20+1/ 40=3/40
Question 4 of 10
4. The difference between a number and its two-third is 24. Find the number.
Explanation:
Answer: (D)
Let the required number be x. According to the question,
x – 2x/3 = x/3 = 24
Therefore, x = 72
Explanation:
Answer: (D)
Let the required number be x. According to the question,
x – 2x/3 = x/3 = 24
Therefore, x = 72
Question 5 of 10
5. 10000 ÷ 100 ÷ 10 =?
Explanation:
Answer: (B)
If the expression includes more than one similar operation, solve from the left hand side first, and then the other. Therefore, ? = 10000 ÷ 100 ÷ 10 = 100 ÷ 10 = 10
Explanation:
Answer: (B)
If the expression includes more than one similar operation, solve from the left hand side first, and then the other. Therefore, ? = 10000 ÷ 100 ÷ 10 = 100 ÷ 10 = 10
Question 6 of 10
6. What will be the compound interest accrued on a sum of Rs. 7,200 at a rate of 5% per annum in 2y?
Explanation:
Answer: (A)
C.I. for 1st year = 5% of 7200 = ` 360
C.I. for 2nd year = 5% of 7200 + 5% of 360 = 360 + 18 = ` 378
C.I. for 2 years = 360 + 378 = ` 738
Explanation:
Answer: (A)
C.I. for 1st year = 5% of 7200 = ` 360
C.I. for 2nd year = 5% of 7200 + 5% of 360 = 360 + 18 = ` 378
C.I. for 2 years = 360 + 378 = ` 738
Question 7 of 10
7. Population of a country increases every year by 10%. If the population in January 2006 was 15.8 lakh, what was the population in January 2008?
Explanation:
Answer: (A)
Population in 2006 = 15,80,000
Rate of growth = 10%
Population in 2008 = 158000(1+10/100)2 = 1580000 × 1.1 × 1.1 = 19,11,800
Explanation:
Answer: (A)
Population in 2006 = 15,80,000
Rate of growth = 10%
Population in 2008 = 158000(1+10/100)2 = 1580000 × 1.1 × 1.1 = 19,11,800
Question 8 of 10
8. John travelled 240 km at a certain speed and then another 192 km at 8 km/h faster. If the entire trip was of 10 h, find his initial speed.
Explanation:
Answer: (C)
Let the initial speed of John be x km/h
∴ According to the question, 240/x+ 192/( x+ 8)
240(x + 8) + 192x = 10(x² + 8x)
10x² – 352x –1920 = 0
x = 40, –24/5
By ignoring the negative value, the required speed of John = 40 km/h.
Explanation:
Answer: (C)
Let the initial speed of John be x km/h
∴ According to the question, 240/x+ 192/( x+ 8)
240(x + 8) + 192x = 10(x² + 8x)
10x² – 352x –1920 = 0
x = 40, –24/5
By ignoring the negative value, the required speed of John = 40 km/h.
Question 9 of 10
9. The ratio of present ages between Jyoti and Ankita is 7: 5. If after six years, their ages will be in the ratio of 4: 3, what would be the present age of Ankita?
Explanation:
Answer: (C)
Let Jyoti’s present age be 7x years and Ankita’s present age be 5x years.
Then, after 6 years, their ages will be:(7x+6) /(5x+6)=4/3
3(7x+6) = 4(5x+6)
21x+18 = 20x+24
x = 6
Therefore, Ankita’s present age = 5×6 = 30 years.
Explanation:
Answer: (C)
Let Jyoti’s present age be 7x years and Ankita’s present age be 5x years.
Then, after 6 years, their ages will be:(7x+6) /(5x+6)=4/3
3(7x+6) = 4(5x+6)
21x+18 = 20x+24
x = 6
Therefore, Ankita’s present age = 5×6 = 30 years.
Question 10 of 10
10. Fraction becomes 4 when 1 is added to the numerator and becomes 5 when 1 is subtracted from the denominator. The numerator of the given fraction is
Click “Start Quiz” to attend these Questions and view Explanation
If a = 115, b = -113, c = –2, then what is the value of a3 + b3 + c3?
69000
12995
38985
77970
500 ÷ 50 × 5 =?
2
5
50
10
X is twice as efficient as Y. X takes 20 days to complete a job. In how many days, can X and Y complete the job together?
30
25
40
40
The difference between a number and its two-third is 24. Find the number.
48
36
60
72
10000 ÷ 100 ÷ 10 =?
1
10
100
1000
What will be the compound interest accrued on a sum of Rs. 7,200 at a rate of 5% per annum in 2y?
Rs. 738
Rs. 1,738
Rs. 1,268
Rs. 648
Population of a country increases every year by 10%. If the population in January 2006 was 15.8 lakh, what was the population in January 2008?
19,11,800
18,96,000
19,11,600
18,94,000
John travelled 240 km at a certain speed and then another 192 km at 8 km/h faster. If the entire trip was of 10 h, find his initial speed.
45 km/h
60 km/h
40 km/h
75 km/h
The ratio of present ages between Jyoti and Ankita is 7: 5. If after six years, their ages will be in the ratio of 4: 3, what would be the present age of Ankita?
32
36
30
35
Fraction becomes 4 when 1 is added to the numerator and becomes 5 when 1 is subtracted from the denominator. The numerator of the given fraction is
15
11
13
17
Solution:
Answer: (d)
Note that here, a + b + c = 0 Therefore, a3 + b3 + c3 = 3abc = 3 (115) (-113) (-2) = 77970.
2.Answer: (c)
According to the BODMAS rule, ? = 500 ÷ 50 × 5 = 10 × 5 = 50
Answer: (d)
X’s one-day work = 1/20
Y’s one-day work = 1/40 (because X is twice as efficient as Y) X and Y’s one-day work 1/20+1/ 40=3/40
4.Answer: (d)
Let the required number be x. According to the question,
x – 2x/3 = x/3 = 24
Therefore, x = 72
Answer: (b)
If the expression includes more than one similar operation, solve from the left hand side first, and then the other. Therefore, ? = 10000 ÷ 100 ÷ 10 = 100 ÷ 10 = 10
Answer: (a)
C.I. for 1st year = 5% of 7200 = ` 360
C.I. for 2nd year = 5% of 7200 + 5% of 360 = 360 + 18 = ` 378
C.I. for 2 years = 360 + 378 = ` 738
Answer: (a)
Population in 2006 = 15,80,000
Rate of growth = 10%
Population in 2008 = 158000(1+10/100)2 = 1580000 × 1.1 × 1.1 = 19,11,800
Answer: (c)
Let the initial speed of John be x km/h
∴ According to the question, 240/x+ 192/( x+ 8)
240(x + 8) + 192x = 10(x² + 8x)
10x² – 352x –1920 = 0
x = 40, –24/5
By ignoring the negative value, the required speed of John = 40 km/h.
Answer: (c)
Let Jyoti’s present age be 7x years and Ankita’s present age be 5x years.
Then, after 6 years, their ages will be:(7x+6) /(5x+6)=4/3