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.
You have to finish following quiz, to start this quiz:
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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
