2Hello and welcome to exampundit. Here is a set of Quantitative Aptitude Quiz for SBI PO mains 2019. The following set is based on Quadratic Equations & Misc Questions.
Quantitative Aptitude Quiz for SBI PO Mains Day 1
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Question 1 of 10
1. Question
During a trip of 200 kilometres, Stella covers the first 80 kilometres at an average speed of x kilometres per hour and the remaining distance at an average speed of (x + 15) kilometres per hour. If her average speed for the entire trip is 250/3 kilometres per hour, what is the value of x?
Correct
Explanation
Let the total time for the entire trip be T hrs.
Average speed=200/T=250/3 =>T=12/5 hrs
(Time for 1st 80 km) + (Time for remaining 120 km) = 12/5
80/x + 120/(x+15) = 12/5
3x^2 – 205x – 1500 = 0
3x^2 – 225x +20x – 1500 = 0
3x(x – 75) + 25(x – 75) = 0
(x – 75)(3x + 25) = 0
x=75 (since value of ‘x’ can’t be negative)Incorrect
Explanation
Let the total time for the entire trip be T hrs.
Average speed=200/T=250/3 =>T=12/5 hrs
(Time for 1st 80 km) + (Time for remaining 120 km) = 12/5
80/x + 120/(x+15) = 12/5
3x^2 – 205x – 1500 = 0
3x^2 – 225x +20x – 1500 = 0
3x(x – 75) + 25(x – 75) = 0
(x – 75)(3x + 25) = 0
x=75 (since value of ‘x’ can’t be negative) 
Question 2 of 10
2. Question
While driving on a highway, H and R crossed checkpoint P at 9 am and 10 am, at individual speeds of 30 mph and 40 mph respectively. Mr.X started cycling in the same direction as H and R, from point Q at 8 am at 15 mph. If H and R met each other at Q, then at what distance from Q Ron met Mr. X?
Correct
Explanations
Let us also work on the checkpoints..
When R starts at 10am, H has travelled for an hour is therefore 30 miles ahead and that point is Q..
Rcovers 4030=10 miles in each hour so he will cover 30 miles in 30/10=3hours..
So R reaches Q at 10+3 or 1pm…
And by 1 pm, X has travelled for 1pm8am or 5hrs..In 5hrs, X has travelled 5*15=75 miles..
To cover this distance Rrequires 75/(4015)=3hrs
In these 3 hours R has travelled 3*40=120 miles
So Ans 120Incorrect
Explanations
Let us also work on the checkpoints..
When R starts at 10am, H has travelled for an hour is therefore 30 miles ahead and that point is Q..
Rcovers 4030=10 miles in each hour so he will cover 30 miles in 30/10=3hours..
So R reaches Q at 10+3 or 1pm…
And by 1 pm, X has travelled for 1pm8am or 5hrs..In 5hrs, X has travelled 5*15=75 miles..
To cover this distance Rrequires 75/(4015)=3hrs
In these 3 hours R has travelled 3*40=120 miles
So Ans 120 
Question 3 of 10
3. Question
A train X departs from station A at 11.00 am for station B, which is 180 km away. Another train Y departs from station B at 11.00 am for station A. Train X travels at an average speed of 70 kms/hr and does not stop any where until it arrives at station B. Train Y travels at an average speed of 50 kms/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B en route to station A. Ignoring the lengths the train, what is the distance, to the nearest km, from station A to the point where the trains cross each other?
Correct
Explanations
Time taken by train Y to travel a distance of 60km=60/50∗60=72minutes
Train Y was stationed at point C for 15 minutes, hence total time elapsed to reach point C =72+15=87 minutes.
During this period train X covered a distance of =70∗87/60=101.5 or approx102km
so distance left between these two train to cover =180−102−60=18km
as the trains are moving in opposite direction, relative speed 70+50=120 km/h
time taken to cover 18km =18/120
so distance traveled by train X =70∗18/120=10.5
Hence train X’s distance from A =102+10.5=112.5=112 (as rounding was done earlier)Incorrect
Explanations
Time taken by train Y to travel a distance of 60km=60/50∗60=72minutes
Train Y was stationed at point C for 15 minutes, hence total time elapsed to reach point C =72+15=87 minutes.
During this period train X covered a distance of =70∗87/60=101.5 or approx102km
so distance left between these two train to cover =180−102−60=18km
as the trains are moving in opposite direction, relative speed 70+50=120 km/h
time taken to cover 18km =18/120
so distance traveled by train X =70∗18/120=10.5
Hence train X’s distance from A =102+10.5=112.5=112 (as rounding was done earlier) 
Question 4 of 10
4. Question
Three runners start running simultaneously from the same point on a 500meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
Correct
Explanations
Let, Speed of A = 4.4
Speed of B = 4.8
Speed of C = 5
Relative speed of A and B = 4.84.4 = 0.4
at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time
So time of A and B to meet on circular track = 500/.4 = 1250 second
Relative speed of C and B = 5 – 4.8 = 0.2
at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time
So time of C and B to meet on circular track = 500/.2 = 2500 seconds
i.e. All three to meet on the track times needed = LCM of 1250 and 2500 = 2500 secondIncorrect
Explanations
Let, Speed of A = 4.4
Speed of B = 4.8
Speed of C = 5
Relative speed of A and B = 4.84.4 = 0.4
at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time
So time of A and B to meet on circular track = 500/.4 = 1250 second
Relative speed of C and B = 5 – 4.8 = 0.2
at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time
So time of C and B to meet on circular track = 500/.2 = 2500 seconds
i.e. All three to meet on the track times needed = LCM of 1250 and 2500 = 2500 second 
Question 5 of 10
5. Question
The area of a square field is 24200 sq m. How long will a lady take to cross the field diagonally at the rate of 6.6 km/hr?
Correct
Explanations
Area of a square field = 24200 sq m
Let the side of square = a
a^2 = 24200
=> a = (24200)^(1/2)
Diagonal = (2)^(1/2) * a
= 1.414 * 155
= 220
Speed of lady = 6.6 km / hour
= 6600 m / hour
= 110 m / min
Time taken by lady to cross the field diagonally = 220/110 = 2 minIncorrect
Explanations
Area of a square field = 24200 sq m
Let the side of square = a
a^2 = 24200
=> a = (24200)^(1/2)
Diagonal = (2)^(1/2) * a
= 1.414 * 155
= 220
Speed of lady = 6.6 km / hour
= 6600 m / hour
= 110 m / min
Time taken by lady to cross the field diagonally = 220/110 = 2 min 
Question 6 of 10
6. Question
Directions (Q. 6 – 10) In the following questions, two equations I and II are given. You have to solve both the equations and give answer as
I. 54×2 + 123x + 65 = 0
II. 24y2 + 38y + 15 = 0Correct
Solutions So we need to break 123 is such a fashion that the addition will make 123(a+b) and their multiply will equal to multiply of 54*65(a*b=54*65).
But first look for sign the sign of coefficient B and c is(+) so the factors will come of opposite sign i.e. ().
Now 123=54*65
Lets break 123 = this will be very easy if we first factorise 54 and 65 so let do this 54=2*3*3*3 and 65=5*13
These are possible factors so now 123=2*3*3*3*5*13
Now arrange these factors in two factors in such a way that their addition will become 123 and their multiply will become 54*65 and this will get better with practice because for this there is no short cut.
So 13*3*2=78 and 3*3*5=45
78+45=123 and 78*45=54*65
So our factors are 78 and 45 but as mentioned above factors will be of opposite sign this not it coefficient of x2 is 54 so we have to divide these by 54
X= 78/54 , 45/54
Now same for y
On same line factorise 24 and 15 and you will get = 2*2*2*3*3*5
Now arrange them in 2*2*5=20 and 3*3*2=18
Addition is 38 and multiply of 20*18 = 24*15
That is all we needed
Our factors for y = – 20/24, 18/24
Now the most important step and i.e. comparison
X Y
78/54 = 20/24
78/54 < 18/24 45/54 < 20/24 45/54 < 18/24 What we did here pick one value of x and write it twice and compare it with both value of y but first simply there value then via cross multiply you can compare them easily now again do same process for second value. And here y≥x.Incorrect
Solutions So we need to break 123 is such a fashion that the addition will make 123(a+b) and their multiply will equal to multiply of 54*65(a*b=54*65).
But first look for sign the sign of coefficient B and c is(+) so the factors will come of opposite sign i.e. ().
Now 123=54*65
Lets break 123 = this will be very easy if we first factorise 54 and 65 so let do this 54=2*3*3*3 and 65=5*13
These are possible factors so now 123=2*3*3*3*5*13
Now arrange these factors in two factors in such a way that their addition will become 123 and their multiply will become 54*65 and this will get better with practice because for this there is no short cut.
So 13*3*2=78 and 3*3*5=45
78+45=123 and 78*45=54*65
So our factors are 78 and 45 but as mentioned above factors will be of opposite sign this not it coefficient of x2 is 54 so we have to divide these by 54
X= 78/54 , 45/54
Now same for y
On same line factorise 24 and 15 and you will get = 2*2*2*3*3*5
Now arrange them in 2*2*5=20 and 3*3*2=18
Addition is 38 and multiply of 20*18 = 24*15
That is all we needed
Our factors for y = – 20/24, 18/24
Now the most important step and i.e. comparison
X Y
78/54 = 20/24
78/54 < 18/24 45/54 < 20/24 45/54 < 18/24 What we did here pick one value of x and write it twice and compare it with both value of y but first simply there value then via cross multiply you can compare them easily now again do same process for second value. And here y≥x. 
Question 7 of 10
7. Question
I. 2×2 (10+√14)x + 5√14 = 0
II. 2y2 +(12√14)y – 6V14 =0Correct
Solutions – now this question looks tough but very easy because question has answer itself
2×2 (10+√14)x + 5√14 = 0
Multiply coefficient of x2 and constant
2*5√14=10√14
And now we have to break 10+√14 in such a fashion that their addition will become number itself and multiply will become 10V14(the step by step process explained in solution of question 1.
Now look for sign we have – and + so the root will come in + and +( try to remember this with every question)
So our factors for x= 10/2 , V14/2
2y2 +(12√14)y – 6√14 =0
Now first sign we have + and – factors will be in + and –( greater value will take – sign.
Y= 12/2, √14/2( here greater is 12 so 12 took – sign)
X y
10/2=5 > 12/2= 6
5 > √14/2
√14/2 > 6
√14/2 = √14/2
So X ≥ YIncorrect
Solutions – now this question looks tough but very easy because question has answer itself
2×2 (10+√14)x + 5√14 = 0
Multiply coefficient of x2 and constant
2*5√14=10√14
And now we have to break 10+√14 in such a fashion that their addition will become number itself and multiply will become 10V14(the step by step process explained in solution of question 1.
Now look for sign we have – and + so the root will come in + and +( try to remember this with every question)
So our factors for x= 10/2 , V14/2
2y2 +(12√14)y – 6√14 =0
Now first sign we have + and – factors will be in + and –( greater value will take – sign.
Y= 12/2, √14/2( here greater is 12 so 12 took – sign)
X y
10/2=5 > 12/2= 6
5 > √14/2
√14/2 > 6
√14/2 = √14/2
So X ≥ Y 
Question 8 of 10
8. Question
I. 5/Va + 6/Va = Va
II. b2 = 115/2/VbCorrect
Solutions :
5/Va + 6/Va = Va
Solve by multiplying Va on both the sides, LHS Va will get cancelled. On RHS Va*Va= a
Hence, 5+6=a
A= 11
b2 = 115/2/Vb
after solving we get
b2 + b1/2 = 115/2
b5/2 = 11 5/2
b = 11
hence A=BIncorrect
Solutions :
5/Va + 6/Va = Va
Solve by multiplying Va on both the sides, LHS Va will get cancelled. On RHS Va*Va= a
Hence, 5+6=a
A= 11
b2 = 115/2/Vb
after solving we get
b2 + b1/2 = 115/2
b5/2 = 11 5/2
b = 11
hence A=B 
Question 9 of 10
9. Question
I. 63×2 85x +28 =0
II. 49y2 + 35y 36 = 0Correct
Solutions :
refer solution of 1st question it is on same line give it try and solve.
X = 7/9, 4/7
Y= 4/7 , 9/7
X≥yIncorrect
Solutions :
refer solution of 1st question it is on same line give it try and solve.
X = 7/9, 4/7
Y= 4/7 , 9/7
X≥y 
Question 10 of 10
10. Question
If the roots of equation 2×2 – 5x + c = 0 are in the ratio of 4:6, then find the value of c
Correct
Solutions :
As ratio is 2:3 so let roots of equation are 2R and 3R
As we know in quadratic equation(ax2 + bx + c=0)
B = sum of roots so 2R + 3R = 5R
But in equation value of b is 5 so putting value of R = ½
Satisfy it. So value of R = ½
Now C = product of roots ( 2R*3R= 6R2)
Value of c is given C( but as coefficient of x2 is 2 so we have to divide it by 2
6R2 = C/2 (R=1/2)
6*1/4 = C/2
C = 3
So value of C is 3Incorrect
Solutions :
As ratio is 2:3 so let roots of equation are 2R and 3R
As we know in quadratic equation(ax2 + bx + c=0)
B = sum of roots so 2R + 3R = 5R
But in equation value of b is 5 so putting value of R = ½
Satisfy it. So value of R = ½
Now C = product of roots ( 2R*3R= 6R2)
Value of c is given C( but as coefficient of x2 is 2 so we have to divide it by 2
6R2 = C/2 (R=1/2)
6*1/4 = C/2
C = 3
So value of C is 3
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