Hello Aspirants,
Welcome to Online Quant Test in AffairsCloud.com. We are starting Railway RRB 2016 and we are creating sample questions in Quant section, type of which will be asked in Railway RRB 2016!!!
 Click Here to View Stratus: Railway RRB 2016 Course
 Railway RRB 2016: Reasoning Test
 Railway RRB 2016: Quants Test
 Railway RRB 2016: General Awareness Test
Help: Share Our Railway RRB 2016 Course page to your Friends & FB Groups
_____________________________________________________________________
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
All the Best
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Quantitative Aptitude 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA and B can complete a work in 10 days and 15 days respectively. They undertook to do a job for Rs 3000. Now with the help of C they did in 4 days. What will B get as his share?
Correct
Explanation:
1/10 + 1/15 + 1/x = 1/4
So x = 12 days, C can do in 12 days
So money will be shared in ratio 1/10 : 1/15 : 1/12 = 6 : 4 : 5
So B will get = 4/15 * 3000Incorrect
Explanation:
1/10 + 1/15 + 1/x = 1/4
So x = 12 days, C can do in 12 days
So money will be shared in ratio 1/10 : 1/15 : 1/12 = 6 : 4 : 5
So B will get = 4/15 * 3000 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeSelling price of an article is set at Rs 2200. If it was sold at 20% less, then also there would have a profit of 10%. What is the actual cost price of the article?
Correct
Explanation:
SP = (80/100)*2200 = 1760
So CP = (100/110)* 1760Incorrect
Explanation:
SP = (80/100)*2200 = 1760
So CP = (100/110)* 1760 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeThere are 1000 boys and 800 girls in a school. 70% of the boys and 90% of the girls passed in an examination. What is the approximate percentage of students who did not pass that exam?
Correct
Explanation:
Boys who dint pass = (30/100)*1000 = 300
Girls who dint pass = (10/100)*800 = 80
Total who dint pass = 300+80 = 380 out of total 1000+800 = 1800 students
So required % = (380/1800)*100Incorrect
Explanation:
Boys who dint pass = (30/100)*1000 = 300
Girls who dint pass = (10/100)*800 = 80
Total who dint pass = 300+80 = 380 out of total 1000+800 = 1800 students
So required % = (380/1800)*100 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeIf 3x + 4y = 23 and 7x + 2y = 28, then 3x + 10y equals
Correct
Explanation:
Solving equations gives, x = 3, y = 3.5
So 3x + 10y = 3*3 + 10*3.5Incorrect
Explanation:
Solving equations gives, x = 3, y = 3.5
So 3x + 10y = 3*3 + 10*3.5 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA sum of money amounts to Rs 2360 in 2 years and Rs 2900 in 5 years. The rate of interest per annum is
Correct
Explanation:
P + SI for 2 yrs = 2360
P + SI for 5 yrs = 2900
Subtract these equations, So SI for 3 yrs = 540
So SI for 1 yr = 540/3 = Rs 180
So P + SI for 2 yrs = 2360 gives
P + 2*180 = 2360,
Solve, P = 2000
Now 180 = 2000*r*1/100
Solve, r = 9Incorrect
Explanation:
P + SI for 2 yrs = 2360
P + SI for 5 yrs = 2900
Subtract these equations, So SI for 3 yrs = 540
So SI for 1 yr = 540/3 = Rs 180
So P + SI for 2 yrs = 2360 gives
P + 2*180 = 2360,
Solve, P = 2000
Now 180 = 2000*r*1/100
Solve, r = 9 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA and B invested in a business as Rs 3000 and Rs 4500 respectively. After 4 months C joins with Rs 3600. If after a year, C gets Rs 3040 as his share of profit, what is the share of A?
Correct
Explanation:
3000*12 : 4500*12 : 3600*8
10 : 15 : 8
so (8)/(10+15+8) * x= 3040
solve, x = 12540
so A’s share = 10/33 * 12540Incorrect
Explanation:
3000*12 : 4500*12 : 3600*8
10 : 15 : 8
so (8)/(10+15+8) * x= 3040
solve, x = 12540
so A’s share = 10/33 * 12540 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative Aptitude2 years hence, the sum of ages of Arun and Ridhima will be 32 years. Also 5 years ago, the ratio of their ages was 7 : 11 respectively. What will be Arun’s age 8 years hence?
Correct
Explanation:
7x and 11x
Now (5+2) = 7 yrs
So (7x+7) + (11x+7) = 32
Solve, x = 1
So Arun’s present age = 7x+5 = 12
So after 8 yrs = 12+8Incorrect
Explanation:
7x and 11x
Now (5+2) = 7 yrs
So (7x+7) + (11x+7) = 32
Solve, x = 1
So Arun’s present age = 7x+5 = 12
So after 8 yrs = 12+8 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeHow many 4 letter words starting with vowel can be formed by the letters of word UPSTREAM such that no letter repeats?
Correct
Explanation:
There are 3 vowels, so we gave 3 choices for the first letter
There are total 8 letters in UPSTREAM, and one of them has occupied first place, so 7 choices for second, 6 choices for third, and 5 choices for fourth
So 3*7*6*5 = 630Incorrect
Explanation:
There are 3 vowels, so we gave 3 choices for the first letter
There are total 8 letters in UPSTREAM, and one of them has occupied first place, so 7 choices for second, 6 choices for third, and 5 choices for fourth
So 3*7*6*5 = 630 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeWhat is the LCM of 2/5, 4/25, 6/11, 10/22?
Correct
Explanation:
LCM will be LCM of numerators/HCF of denominators
So LCM = (LCM of 2,4,6,10)/(HCF of 5,25, 11, 22)
= 60/1Incorrect
Explanation:
LCM will be LCM of numerators/HCF of denominators
So LCM = (LCM of 2,4,6,10)/(HCF of 5,25, 11, 22)
= 60/1 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA boat can travel 15 km upstream in 30 minutes. If the speed of the current is 1/3 of the speed of the boat in still water, how much distance can the boat travel downstream in 30 minutes?
Correct
Explanation:
A boat can travel 15 km upstream in 30 minutes. So Upstream speed = 15 / (30/60) = 30 km/hr
Let speed of the boat in still water = x
Speed of the current = x/3
Upstream speed = x – x/3 = 2x/3
But from above Upstream speed = 30. So 2x/3 = 30
Solve, x = 45
Speed of the current = 45/3 = 15 and speed of boat = 45
Downstream speed = 45+15 = 60
Distance travelled in 30min = 60 * (30/60)= 30 kmIncorrect
Explanation:
A boat can travel 15 km upstream in 30 minutes. So Upstream speed = 15 / (30/60) = 30 km/hr
Let speed of the boat in still water = x
Speed of the current = x/3
Upstream speed = x – x/3 = 2x/3
But from above Upstream speed = 30. So 2x/3 = 30
Solve, x = 45
Speed of the current = 45/3 = 15 and speed of boat = 45
Downstream speed = 45+15 = 60
Distance travelled in 30min = 60 * (30/60)= 30 km
_____________________________________________________________________
 Note: We are providing unique questions for you to practice well, have a try !!
 Ask your doubt in comment section, AC Mod’s ll clear your doubts in caring way.
 – E.g: @shubhra or @Yogit Kumar “Hi, Can You Explain 8th Question?”