Hello Aspirants,
Welcome to Online Quant Test in AffairsCloud.com. We are starting SSC CGL 2016 Course and we are creating sample questions in Quant section, type of which will be asked in SSC CGL 2016 !!!
 Click Here to View Stratus: SSC CGL 2016 Tier 1 Course
 SSC CGL 2016: Reasoning Test
 SSC CGL 2016: Quants Test
 SSC CGL 2016: English Test
 SSC CGL 2016: General Awareness Test
Help: Share Our SSC CGl 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 AptitudeIf x = [ (m+1)^{1/3} + (m1)^{1/3} ] / [ (m+1)^{1/3} – (m1)^{1/3} ], then the value of x^{3} – 3mx^{2} + 3x – m is
Correct
Explanation:
x = [ (m+1)^{1/3} + (m1)^{1/3} ] / [ (m+1)^{1/3} – (m1)^{1/3} ]
apply componendo and dividendo(x+1)/(x1) = (m+1)^{1/3}/(m1)^{1/3}
Cube both sides
[x^{3} + 3x^{2} + 3x + 1]/ [x^{3} – 3x^{2} + 3x – 1] = (m+1)/(m1)
Again apply componendo and dividendo
[2x^{2} + 6x]/[ 6x^{2}+2] = 2m/2
Solve this, we get x^{3} – 3mx^{2} + 3x – m = 0Incorrect
Explanation:
x = [ (m+1)^{1/3} + (m1)^{1/3} ] / [ (m+1)^{1/3} – (m1)^{1/3} ]
apply componendo and dividendo(x+1)/(x1) = (m+1)^{1/3}/(m1)^{1/3}
Cube both sides
[x^{3} + 3x^{2} + 3x + 1]/ [x^{3} – 3x^{2} + 3x – 1] = (m+1)/(m1)
Again apply componendo and dividendo
[2x^{2} + 6x]/[ 6x^{2}+2] = 2m/2
Solve this, we get x^{3} – 3mx^{2} + 3x – m = 0 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA person got an article after getting two successive discounts of 15% and 10% at Rs 3060. What is the marked price of this article?
Correct
Explanation:
Single equivalent dance is (15 + 10 – 15*10/100) = 23.5%
So MP is 100/(10023.5) * 3060Incorrect
Explanation:
Single equivalent dance is (15 + 10 – 15*10/100) = 23.5%
So MP is 100/(10023.5) * 3060 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeIn a rectangle ABCD, the side AB is divided into 4 equal parts by the points X, Y and Z. Find the ratio of: (area of ∆XYC)/(area of rect. ABCD)
Correct
Explanation:
Let AB = 4x and BC = y
Area of rect ABCD = 4xy
Now XY = 4x/4 = x
So area of ∆XYC = ½ * x * y = xy/2
So required ratio = (xy/2)/(4xy)Incorrect
Explanation:
Let AB = 4x and BC = y
Area of rect ABCD = 4xy
Now XY = 4x/4 = x
So area of ∆XYC = ½ * x * y = xy/2
So required ratio = (xy/2)/(4xy) 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeIf the length of a chord of a circle which makes an angle 30 degrees with the tangent drawn at one end point of the chord is 8cm, then the radius of the circle is
Correct
Explanation:
Angle ABC = 30
So angle ABO = 90 – 30 = 60
BD = 8/2 = 4
Cos OBD = BD/OB
So cos 60 = 4/OB, so OB = 8Incorrect
Explanation:
Angle ABC = 30
So angle ABO = 90 – 30 = 60
BD = 8/2 = 4
Cos OBD = BD/OB
So cos 60 = 4/OB, so OB = 8 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeIf tanɸ + cotɸ = 2, then the value of tan^{6}ɸ + cot^{12}ɸ is
Correct
Explanation:
tanɸ + cotɸ = 2
tanɸ + 1/tanɸ = 2
tan^{2}ɸ + 1 = 2tanɸ
tan^{2}ɸ – 2tanɸ + 1 = 0
(tanɸ 1) ^{2} = 0
So tanɸ = 1, also then cotɸ = 1
So tan^{6}ɸ + cot^{12}ɸ = 1+1 = 2Incorrect
Explanation:
tanɸ + cotɸ = 2
tanɸ + 1/tanɸ = 2
tan^{2}ɸ + 1 = 2tanɸ
tan^{2}ɸ – 2tanɸ + 1 = 0
(tanɸ 1) ^{2} = 0
So tanɸ = 1, also then cotɸ = 1
So tan^{6}ɸ + cot^{12}ɸ = 1+1 = 2 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA cube of side 12 feet is to be painted from outside. The cost of paint is Rs 21.50 per kg. If 32 square feet are covered by 1 kg, then what is the cost of painting the cube?
Correct
Explanation:
Surface area of cube = 6a^{2} = 6*12^{2} = 864 sq. feet
So quantity of paint required = (864/32) = 27 kg
So cost of painting = 21.50*27Incorrect
Explanation:
Surface area of cube = 6a^{2} = 6*12^{2} = 864 sq. feet
So quantity of paint required = (864/32) = 27 kg
So cost of painting = 21.50*27 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeP varies directly with the square of Q when R is a constant and inversely with R when Q is constant. When Q = 12 and R = 9, then P = 48. Find P when Q = 20 and R = 15.
Correct
Explanation:
Let k be constant of proportionality,
If P directly varies with Q^{2}, then P = k Q^{2}
And if P inversely varies with R, then P = k/R
So in all P = k*Q^{2}/R
When Q = 12 and R = 9, then P = 48
So 48 = k*12^{2}/9
Solve, k = 3
Now when Q = 20 and R = 15,
P = 2*20^{2}/15Incorrect
Explanation:
Let k be constant of proportionality,
If P directly varies with Q^{2}, then P = k Q^{2}
And if P inversely varies with R, then P = k/R
So in all P = k*Q^{2}/R
When Q = 12 and R = 9, then P = 48
So 48 = k*12^{2}/9
Solve, k = 3
Now when Q = 20 and R = 15,
P = 2*20^{2}/15 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe students of section A and section B have average weights of 48 kg and 51 kg respectively. Find the approximate average weight of both the sections together such that the number of students in the sections A and B is in the ratio 5 : 3.
Correct
Explanation:
Students in A section = 5x, students in B section = 3x
So total weight of students of A section = 48*5x = 240x
And total weight of students of B section = 51*3x = 153x
So average weight of both sections = [(240x+153x)/(5x+3x)]Incorrect
Explanation:
Students in A section = 5x, students in B section = 3x
So total weight of students of A section = 48*5x = 240x
And total weight of students of B section = 51*3x = 153x
So average weight of both sections = [(240x+153x)/(5x+3x)] 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeA and B can complete work in 20 and 12 days respectively. A starts the work, and after 4 days B joins him. How many days they took to complete the whole work?
Correct
Explanation:
Let the work be completed in x days, so A worked for all x days, and B for (x4) days
1/20 * x + 1/12 (x – 4) = 1Incorrect
Explanation:
Let the work be completed in x days, so A worked for all x days, and B for (x4) days
1/20 * x + 1/12 (x – 4) = 1 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThere are two trains of lengths 300 m and 240 m respectively. Both start at same time. If they travel in same direction they cross each other in 36 hours and if they travel in opposite direction towards each other, they cross each other in 4 hours. What is the speed of the fastest train?
Correct
Explanation:
Let speed of 1st train = s1 km/hr, of 2nd train = s2 km/hr
Then when they travel in same direction, relative speed = s1s2
So (300+240) = (s1s2)*36. This gives s1 – s2 = 15
When they travel in opposite direction, relative speed = s1+s2
So (300+240) = (s1+s2)*4. This gives s1 + s2 = 135
Solve both equations, s1 = 75Incorrect
Explanation:
Let speed of 1st train = s1 km/hr, of 2nd train = s2 km/hr
Then when they travel in same direction, relative speed = s1s2
So (300+240) = (s1s2)*36. This gives s1 – s2 = 15
When they travel in opposite direction, relative speed = s1+s2
So (300+240) = (s1+s2)*4. This gives s1 + s2 = 135
Solve both equations, s1 = 75
_____________________________________________________________________
 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 “Hi, Can You Explain 8th Question?”