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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeThe total distance between two destinations is ‘x’ km. If the distance travelled along the stream is thrice of the total distance and the distance travelled against the stream is double of the total distance. If the time taken to cover the distance along the stream is 40% less than the time taken to cover the distance against the stream. If Sundar covers a distance of 21 km in 1 hour 24 minutes along the stream, then find the stream speed?
Correct
Answer – 4) 4.5 kmph
Explanation:
T(down)/T(up) = 60/100
T(down) = 3x/Downstream Speed
T(up) = 2x/Upstream Speed
Downstream Speed = 21/(84/60) = 15 kmph
60/100 = [3x/15]/[2x/Upstream Speed]
Upstream Speed = 6 kmph
Stream Speed = 15 – 6 / 2 = 4.5 kmphIncorrect
Answer – 4) 4.5 kmph
Explanation:
T(down)/T(up) = 60/100
T(down) = 3x/Downstream Speed
T(up) = 2x/Upstream Speed
Downstream Speed = 21/(84/60) = 15 kmph
60/100 = [3x/15]/[2x/Upstream Speed]
Upstream Speed = 6 kmph
Stream Speed = 15 – 6 / 2 = 4.5 kmph 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA sum of money is invested in Fund A at 7.5% per annum for SI. After 4 years an amount is received and that amount is invested in Fund B at 20% per annum for CI. If the interest received at the end of 2 years is 1430, find the initial invested amount?
Correct
Answer – 2) 2500
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Fund A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500Incorrect
Answer – 2) 2500
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Fund A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDevi’s present age is three times his son’s present age and ninethirteenth of his father’s present age. The sum of their ages after five years is 115 years. What is the sum of the present ages of Devi’s son and Devi’s father?
Correct
Answer – 2) 64
Explanation:
Present age of F+S+D = 11515 = 100
S = F/3; D = 13F/9
F (1+1/3+13/9) = 100
F =36; S = 12; D =52
D+S = 52+12 = 64Incorrect
Answer – 2) 64
Explanation:
Present age of F+S+D = 11515 = 100
S = F/3; D = 13F/9
F (1+1/3+13/9) = 100
F =36; S = 12; D =52
D+S = 52+12 = 64 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeFind out the wrong number in the series given below
10, 12, 22, 50, 132, 409Correct
Answer – 5) 132
Explanation :
10 * 1 + 2 = 12
12 * 1.5 + 4 = 22
22 * 2 + 6 = 50
50 * 2.5 + 8 = 133
133 * 3 + 10 = 409Incorrect
Answer – 5) 132
Explanation :
10 * 1 + 2 = 12
12 * 1.5 + 4 = 22
22 * 2 + 6 = 50
50 * 2.5 + 8 = 133
133 * 3 + 10 = 409 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative Aptitudex² – 38x + 357 = 0
y² – 36y + 304 = 0Correct
Answer – 5) X = Y or relation cannot be established
Explanation:
x² – 38x + 357 = 0
x = 17, 21
y² – 36y + 304 = 0
y = 17, 19Incorrect
Answer – 5) X = Y or relation cannot be established
Explanation:
x² – 38x + 357 = 0
x = 17, 21
y² – 36y + 304 = 0
y = 17, 19 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative Aptitude(331.01 + 29.002)*(15.9 – 11.99)*(27.04 + 22.89) = ?
Correct
Answer – 4) 72000
Explanation :
(331 + 29)*(16 – 12)*(27 + 23) = 360 * 4 * 50 = 72000Incorrect
Answer – 4) 72000
Explanation :
(331 + 29)*(16 – 12)*(27 + 23) = 360 * 4 * 50 = 72000 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections.Q(7 – 10) Study the given table carefully and answer the following questions
Total Strength = Seminar Participants + Conference Participants + Who didn’t participate in any eventWhat is the average of number of Conference Participants participated from Colleges M, O and P.
Correct
Answer – 5) 670
Explanation:
Total strength in M = 800+35M/100 + 370; M = 1800; Conference Participants = 630
Total strength in O = 680+30O/100 + 545; O = 1750; Conference Participants = 525
Total strength in P = 850+45P/100 + 195; P = 1900; Conference Participants = 855
Average = (630+525+855)/3 = 670Incorrect
Answer – 5) 670
Explanation:
Total strength in M = 800+35M/100 + 370; M = 1800; Conference Participants = 630
Total strength in O = 680+30O/100 + 545; O = 1750; Conference Participants = 525
Total strength in P = 850+45P/100 + 195; P = 1900; Conference Participants = 855
Average = (630+525+855)/3 = 670 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections.Q(7 – 10) Study the given table carefully and answer the following questions
Total Strength = Seminar Participants + Conference Participants + Who didn’t participate in any eventIf all females and 110 males in College Q didn’t participate in the any event, then total how many males have participated in the any event?
Correct
Answer – 2) 1590
Explanation:
In College Q total strength:
x = 790+40x/100+ 410
x =2000
Conference Participants = 800
Total male Participants = 790+800 =1590Incorrect
Answer – 2) 1590
Explanation:
In College Q total strength:
x = 790+40x/100+ 410
x =2000
Conference Participants = 800
Total male Participants = 790+800 =1590 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections.Q(7 – 10) Study the given table carefully and answer the following questions
Total Strength = Seminar Participants + Conference Participants + Who didn’t participate in any eventWhat is the average of total strength in all five Colleges together?
Correct
Answer – 3) 1890
Explanation:
1800+2000+1750+1900+2000 = 9450/5 = 1890Incorrect
Answer – 3) 1890
Explanation:
1800+2000+1750+1900+2000 = 9450/5 = 1890 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections.Q(7 – 10) Study the given table carefully and answer the following questions
Total Strength = Seminar Participants + Conference Participants + Who didn’t participate in any eventApproximately total number of students who didn’t participate in any event in all Colleges together forms what percent of the total strength in all Colleges together?
Correct
Answer – 1) 21%
Explanation:
Total strength who didn’t participate in any event = 370+480+545+195+410 = 2000
Total strength in all Colleges = 9450
% = 2000/9450*100 = 21%Incorrect
Answer – 1) 21%
Explanation:
Total strength who didn’t participate in any event = 370+480+545+195+410 = 2000
Total strength in all Colleges = 9450
% = 2000/9450*100 = 21%
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