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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (1–4): Study the following table and answer the questions that follow:
The pie chart shows the percentage of employees in 5 departments of a company in 2014 having total employees 750.
What is the total number of employees in departments B and E together in 2014?
Correct
Explanation:
(14+26)/100 * 750 = 300Incorrect
Explanation:
(14+26)/100 * 750 = 300 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections (1–4): Study the following table and answer the questions that follow:
The pie chart shows the percentage of employees in 5 departments of a company in 2014 having total employees 750.
In 2015, if there is an increase of 2 % in the number of employees of department C, then how many employees are in department C in 2015?
Correct
Explanation:
(102/100) * (20/100) * 750Incorrect
Explanation:
(102/100) * (20/100) * 750 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (1–4): Study the following table and answer the questions that follow:
The pie chart shows the percentage of employees in 5 departments of a company in 2014 having total employees 750.
In 2014, the female employees in department E were 5/13 of total employees in that department. 4 females of E leave next year and equal males join that department, then what is the number of male employees in department E in 2015?
Correct
Explanation:
In 2014, total employees in E = (26/100) * 750 = 195
So females are (5/13) * 195 = 75, so males were = 195 – 75 = 120
In 2015, 4 males joined, so males become 120+4Incorrect
Explanation:
In 2014, total employees in E = (26/100) * 750 = 195
So females are (5/13) * 195 = 75, so males were = 195 – 75 = 120
In 2015, 4 males joined, so males become 120+4 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections (1–4): Study the following table and answer the questions that follow:
The pie chart shows the percentage of employees in 5 departments of a company in 2014 having total employees 750.
In 2014, 20% of employees in department A were post graduates which gets shifted to department D in 2015. If in department D 40% were post graduates in 2014, then what is the number of post graduate employees in department D in 2015?
Correct
Explanation:
Post grads in department A = (20/100) * (22/100) * 750 = 33
Post grads in department D = (40/100) * (18/100) * 750 = 54
So in 2015, in department D post grads becomes 54+33Incorrect
Explanation:
Post grads in department A = (20/100) * (22/100) * 750 = 33
Post grads in department D = (40/100) * (18/100) * 750 = 54
So in 2015, in department D post grads becomes 54+33 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (1–4): Study the following table and answer the questions that follow:
The pie chart shows the percentage of employees in 5 departments of a company in 2014 having total employees 750.
2 men or 4 women can complete a work in 11 days working 8 hours each day. In how many days 6 men and 10 women will complete a work twice as large working together 4 hours each day?
Correct
Explanation:
2 m = 4 w
So 1m = 2 w
6m + 10 w = 6 * (2w) + 10w = 22 w
So we have to find the number of days for 22 women
4 women do 1 work in 11 days working 8 hours, let 22 women do twice work in x days working 4 hrs each day, then
W1*D1*H1*W2 = W2*D2*H2*W1
4*11*8*2 = 22*x*4*1
Solve, x = 8 daysIncorrect
Explanation:
2 m = 4 w
So 1m = 2 w
6m + 10 w = 6 * (2w) + 10w = 22 w
So we have to find the number of days for 22 women
4 women do 1 work in 11 days working 8 hours, let 22 women do twice work in x days working 4 hrs each day, then
W1*D1*H1*W2 = W2*D2*H2*W1
4*11*8*2 = 22*x*4*1
Solve, x = 8 days 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe value of cot 10^{o} . cot 20^{o} . cot 60^{o} . cot 70^{o} . cot 80^{o} is
Correct
Explanation:
Tan (90 – ɸ) = cot ɸ, and tanɸ.cotɸ = 1
So cot 10^{o} . cot 20^{o} . cot 60^{o} . tan 20^{o} . tan 10^{o}
= cot 60^{o} = √3Incorrect
Explanation:
Tan (90 – ɸ) = cot ɸ, and tanɸ.cotɸ = 1
So cot 10^{o} . cot 20^{o} . cot 60^{o} . tan 20^{o} . tan 10^{o}
= cot 60^{o} = √3 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThere are 2 vertical posts, one of each side of a road, opposite to each other. One of the posts of 108 m high. From the top of this post, the angle of depression of the top and foot of the other post are 30 degree and 60 degree respectively. The height of eth above post in m is
Correct
Explanation:
AB = 108, let CD = x
In ∆ABC, tan 30 = AB/BC so √3 = 108/BC
So BC = 36√3
In ∆AED, tan 30 = AE/ED so 1/√3 = (108x)/36√3
Solve, x = 72 mIncorrect
Explanation:
AB = 108, let CD = x
In ∆ABC, tan 30 = AB/BC so √3 = 108/BC
So BC = 36√3
In ∆AED, tan 30 = AE/ED so 1/√3 = (108x)/36√3
Solve, x = 72 m 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeMixture A contains 60% milk and mixture B contains 80% milk. A milkman takes some mixture from mixture A and mixes it with thrice the amount of mixture from mixture B. What is the percentage of milk in the new mixture?
Correct
Explanation:
Let x from A, 3x form B
So milk from A = (60/100)*x, milk from B = (80/100)*3x
So milk in new mixture is (60x/100) + (240x/100) = 3x
Total mixture in third is x+3x = 4x
So % of milk is (3x/4x)*100Incorrect
Explanation:
Let x from A, 3x form B
So milk from A = (60/100)*x, milk from B = (80/100)*3x
So milk in new mixture is (60x/100) + (240x/100) = 3x
Total mixture in third is x+3x = 4x
So % of milk is (3x/4x)*100 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeConsider the data set {1, 2, 3, 8, 2, 7, 4, 9}. The mode in the set is
Correct
Explanation:
The mode in the set is value which appears the most often in data. More than 1 mode can be there.Incorrect
Explanation:
The mode in the set is value which appears the most often in data. More than 1 mode can be there. 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeConsider the data set {5, 2, 3, 8, 9}. The standard deviation of set is
Correct
Explanation:
Standard deviation = √[ ( (510)^2+(210)^2+(310)^2+(810)^2+(910)^2 )/5 ]
= √[ (25 + 64 + 49 + 4 + 1)/5 ] = √28.6 = 5.35Incorrect
Explanation:
Standard deviation = √[ ( (510)^2+(210)^2+(310)^2+(810)^2+(910)^2 )/5 ]
= √[ (25 + 64 + 49 + 4 + 1)/5 ] = √28.6 = 5.35
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