SSC CGL 2018 Practice Test Papers | Quantitative Aptitude (Day-38)
Dear Aspirants, Here we have given the Important SSC CGL Exam 2018 Practice Test Papers. Candidates those who are preparing for SSC CGL 2018 can practice these questions to get more confidence to Crack SSC CGL 2018 Examination.
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Question 1 of 10
1. Question
A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the river current in km/hr.
Correct
Answer: a)
Downstream speed= 15/(3 ¾) = 4 km/hr
Upstream speed = 5/(2 ½) = 2 km/hr
Speed of River= ½(4-2) = 1 km/hr
Incorrect
Answer: a)
Downstream speed= 15/(3 ¾) = 4 km/hr
Upstream speed = 5/(2 ½) = 2 km/hr
Speed of River= ½(4-2) = 1 km/hr
Question 2 of 10
2. Question
In a right angled isosceles triangle ABC, find the value of sinA + sinB + sinC.
Correct
Answer: b)
In right angle isosceles triangle, angle will be 45º, 45º and 90º.
sinA + sinB + sinC = sin45º + sin45º + sin90º
= 1/√2 + 1/√2 + 1
= 1+√2
Incorrect
Answer: b)
In right angle isosceles triangle, angle will be 45º, 45º and 90º.
sinA + sinB + sinC = sin45º + sin45º + sin90º
= 1/√2 + 1/√2 + 1
= 1+√2
Question 3 of 10
3. Question
The average of 50 numbers is 42. If two numbers, 64 and 46 are discarded, the average of remaining number is
Correct
Answer: b)
Average of 50 numbers = 42
Sum of 50 numbers = 50×42 = 2100
Sum of remaining numbers = 2100 – 64 – 46 = 2000
Average of remaining numbers = 2000/48 = 41.67
Incorrect
Answer: b)
Average of 50 numbers = 42
Sum of 50 numbers = 50×42 = 2100
Sum of remaining numbers = 2100 – 64 – 46 = 2000
Average of remaining numbers = 2000/48 = 41.67
Question 4 of 10
4. Question
In the adjoining diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle ∆BCF and that of the rectangle?
Correct
Answer: b)
Area of triangle = ½ × Base × Height
Area of triangle CEF =1/2 × BC × EF
=1/2 ×BC× AB/3
= AB×BC/6
The area of ABCD = AB × BC
Required ratio =AB×BC/6: AB × BC=1:6
Incorrect
Answer: b)
Area of triangle = ½ × Base × Height
Area of triangle CEF =1/2 × BC × EF
=1/2 ×BC× AB/3
= AB×BC/6
The area of ABCD = AB × BC
Required ratio =AB×BC/6: AB × BC=1:6
Question 5 of 10
5. Question
A began a business with Rs. 1500 and B joined with Rs.2000 after few months. When did B Join, if the profit at the end of the year where divided in the ratio 3 : 2?
Correct
Answer: d)
Ratio of profit = 3 : 2
So, ratio of investment = 3 :2
Total investment of A= 1500 × 12
So, total investment of B= 1500 × 12 ×2/3×1/ 2000 = 6
So, B joined after = 12 – 6 = 6 months
Incorrect
Answer: d)
Ratio of profit = 3 : 2
So, ratio of investment = 3 :2
Total investment of A= 1500 × 12
So, total investment of B= 1500 × 12 ×2/3×1/ 2000 = 6
So, B joined after = 12 – 6 = 6 months
Question 6 of 10
6. Question
A lent Rs. 16000 on simple interest for 3 years at 7 ½ % p.a. How much more should have A gained, had A given it at compound interest at the same rate and for the same time?
Correct
Answer: c)
Incorrect
Answer: c)
Question 7 of 10
7. Question
(7-10)Study the pie–chart carefully and answer the questions based on it.
Percentage of student enrolled in different game classes in a school
Total number of student = 3600
The number of student enrolled in cricket classes is what percent of that in Basket Ball classes?
Correct
Answer: b)
Required percentage = 22/21 × 100 = 104.76%
Incorrect
Answer: b)
Required percentage = 22/21 × 100 = 104.76%
Question 8 of 10
8. Question
(7-10)Study the pie–chart carefully and answer the questions based on it.
Percentage of student enrolled in different game classes in a school
Total number of student = 3600
What is the total number of student enrolled in swimming class and running class together?
Correct
Answer: d)
Student enrolled in Swimming class = 11/100 × 3600 = 396
Student enrolled in Running class = 46.8/360× 3600 = 468
Total number of students enrolled = 396 + 468 = 864
Incorrect
Answer: d)
Student enrolled in Swimming class = 11/100 × 3600 = 396
Student enrolled in Running class = 46.8/360× 3600 = 468
Total number of students enrolled = 396 + 468 = 864
Question 9 of 10
9. Question
(7-10)Study the pie–chart carefully and answer the questions based on it.
Percentage of student enrolled in different game classes in a school
Total number of student = 3600
How many students are enrolled in Hockey classes?
Correct
Answer: a)
Students enrolled in Hockey = 54/ 360×3600= 540
Incorrect
Answer: a)
Students enrolled in Hockey = 54/ 360×3600= 540
Question 10 of 10
10. Question
(7-10)Study the pie–chart carefully and answer the questions based on it.
Percentage of student enrolled in different game classes in a school
Total number of student = 3600
What is the ratio of the number of students enrolled in Tennis and Basket ball classes together to the number of students enrolled in running classes?
Correct
Answer: a)
Incorrect
Answer: a)
Click “Start Quiz” to attend these Questions and view Explanation
1) A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the river current in km/hr.
a) 1 km/hr
b) 3km/hr
c) 5 km/hr
d) 9 km/hr
2) In a right angled isosceles triangle ABC, find the value of sinA + sinB + sinC.
a) 1-√2
b) 1 +√2
c) √2 -1
d) 1+√3
3) The average of 50 numbers is 42. If two numbers, 64 and 46 are discarded, the average of remaining number is
a) 43.5
b) 41.67
c) 45
d) 40
4) In the adjoining diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle ∆BCF and that of the rectangle?
a) 1 : 4
b) 1 : 6
c) 2 : 5
d) 2 : 3
5) A began a business with Rs. 1500 and B joined with Rs.2000 after few months. When did B Join, if the profit at the end of the year where divided in the ratio 3 : 2?
a) After 7 months
b) After 8 months
c) After 4 months
d) After 6 months
6) A lent Rs. 16000 on simple interest for 3 years at 7 ½ % p.a. How much more should have A gained, had A given it at compound interest at the same rate and for the same time?
a) Rs. 229.90
b) Rs. 299.20
c) Rs. 276.75
d) Rs. 292.20
(7-10)Study the pie–chart carefully and answer the questions based on it.
Percentage of student enrolled in different game classes in a school
Total number of student = 3600
7) The number of student enrolled in cricket classes is what percent of that in Basket Ball classes?
a) 101.45%
b) 104.76%
c) 113.84%
d) 110.28%
8) What is the total number of student enrolled in swimming class and running class together?
a) 684
b) 846
c) 648
d) 864
9) How many students are enrolled in Hockey classes?
a) 540
b) 480
c) 450
d) 520
10) What is the ratio of the number of students enrolled in Tennis and Basket ball classes together to the number of students enrolled in running classes?
a) 3 : 1
b) 4 : 7
c) 7 : 5
d) 3 : 5
Answers :
1). Answer: a)
Downstream speed= 15/(3 ¾) = 4 km/hr
Upstream speed = 5/(2 ½) = 2 km/hr
Speed of River= ½(4-2) = 1 km/hr
2). Answer: b)
In right angle isosceles triangle, angle will be 45º, 45º and 90º.
sinA + sinB + sinC = sin45º + sin45º + sin90º
= 1/√2 + 1/√2 + 1
= 1+√2
3). Answer: b)
Average of 50 numbers = 42
Sum of 50 numbers = 50×42 = 2100
Sum of remaining numbers = 2100 – 64 – 46 = 2000
Average of remaining numbers = 2000/48 = 41.67
4). Answer: b)
Area of triangle = ½ × Base × Height
Area of triangle CEF =1/2 × BC × EF
=1/2 ×BC× AB/3
= AB×BC/6
The area of ABCD = AB × BC
Required ratio =AB×BC/6: AB × BC=1:6
5). Answer: d)
Ratio of profit = 3 : 2
So, ratio of investment = 3 :2
Total investment of A= 1500 × 12
So, total investment of B= 1500 × 12 ×2/3×1/ 2000 = 6
So, B joined after = 12 – 6 = 6 months
6). Answer: c)
7). Answer: b)
Required percentage = 22/21 × 100 = 104.76%
8). Answer: d)
Student enrolled in Swimming class = 11/100 × 3600 = 396
Student enrolled in Running class = 46.8/360× 3600 = 468
Total number of students enrolled = 396 + 468 = 864
9). Answer: a)
Students enrolled in Hockey = 54/ 360×3600= 540
10). Answer: a)
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