 # Mixed Quantitative Aptitude Questions (Memory based SBI PO Mains)

Below are Memory Based Questions SBI PO Mains 2017 Exam in Quantitative Aptitude Section.

1. MLKJ is a trapezoid. ABCD and PQRS are two rhombus. Diagonals of ABCD are 6 cm and 8 cm. One of the angle of PQRS is 120 degree and diagonal bisecting that angle measures 15 cm. Side of ABCD = ML, Side of PQRS = JK. FInd ON (median of trapezoid). (single question on this) A) 10 cm
B) 12 cm
C) 8 cm
D) 15 cm
E) 9 cm
Option A
Solution:
Rhombus ABCD:
Side of ABCD = (√(62 + 82)/2 = 10/2 = 5 cm
Rhombus PQRS: Angle PQO is 60. Because diagonals of rhombus bisect each other.
Also diagonals bisect each other at 90 degree. SO angle POQ is 90
Hence angle OPQ is 30 degree. Also SQ = 15, so OQ = 7.5
In triangle OPQ. With OP as base

sine 30 = perpendicular/Hypotenuse(or side of PQRS)
1/2 = (15/2)/Hypotenuse(or side of PQRS)

Solve, Side of PQRS  = 15 cm [sine 30 = 1/2]
Now  ML = 5 cm, JK = 15 cm
So Median ON = (5+15)/2 = 10 cm

Directions (2-4): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains:
15 red balls. Y yellow balls and G green balls. Probability of drawing one yellow ball is 2/9. The ratio of number of green and yellow balls is 9 : 4

Bag 2 contains:
Number of green balls is 2/3rd of G. Total number of balls in bag is 5/6th of balls in Bag A. Number of yellow balls is 3 greater than number of red balls

Bag 3 contains:
Number of red balls is 1/3rd of total number of red balls in bags A and B. Number of yellow balls is 20% more than number of yellow balls in bad B. Probability of drawing one green ball is 7/16. (3 questions asked as below)

1. One ball from each bag is drawn. Find the probability that these are yellow balls.
A) 5/16
B) 3/16
C) 7/36
D) 1/36
E) 9/16
Option D
Solution:
R = 15, Y = 4x, G = 9x. Prob of Y = 2/9
So 4x/(15+9x+4x) = 2/9
Solve, x = 3
So in bag A: R = 15, Y = 12, G = 27. Total = 54
Bag B:
G = 2/3 * 27 = 18
Total = 5/6 * 54 = 45
Now 18 + Y + Y – 3 = 45
So, yellow balls = 15, and then red = 12
So in bag B: R = 12, Y = 15, G = 18. Total = 45
Bag C:
R = 1/3 (15+12) = 9
Yellow = 120/100 * 15 = 18
Probability of green ball = 7/16
So z/(9+18+z) = 7/16 [Let z green balls]
Solve, z = 21
So in bag C: R = 9, Y = 18, G = 21. Total = 48
Now probability of yellow ball from each bag = 12/54 * 15/45 * 18/48 = 1/36
2. 5/6th of red balls, 4/9th of green balls from bag B are placed in bad D. What is the probability that 2 out of 3 balls from bag D are red?
A) 15/37
B) 15/34
C) 17/33
D) 11/34
E) 19/34
Option B
Solution:
In bag D:
Red balls = 5/6 * 12 = 10
Green balls = 4/9 * 18 = 8
So required probability = 10C2 * 8C1/18C3 = 15/34
3. P yellow balls are transferred from bag A to C. One ball is picked at random from bag C and probability of yellow ball is 2/5. Find P.
A) 3
B) 1
C) 2
D) 4
E) Other than those given in options.
Option C
Solution:
In bag C:
Probability of yellow ball = (18+P)/(48+P) = 2/5
Solve, P = 2

Directions (5-6): Distance between stations P and Q is 516 km. Train A starts from station P at 9:45 AM with speed ‘X’ km/hr. Train B starts in opposite direction from station Q half an hour later with speed ‘1.5X’ km/hr. They meet at 12:55 PM. (2 questions asked as below)

1. If both trains were moving in same direction, find the time at which they will meet.
A) 50 minutes
B) 60 minutes
C) 108 minutes
D) 40 minutes
E) 72 minutes
Option B
Solution:
Speed of A = x km/hr
So distance it traveled in 1/2 hr = x/2 km
Now remaining distance = (516 – x/2) km
Time taken to meet = 12:55 – 10:15 = 8/3 hours
So 8/3 = (516 – x/2)/(x+1.5x) [speed is added when in opposite direction]
Solve, x = 72 km/hr
So speed of A = 72 km/hr, B = 108 km/hr
Now when moving in same direction:
In half hour, A covered = 1/2 * 72 = 36 km
and relative speed = 108-72 = 36 km/hr
So time = 36/36 = 1 hour
2. What is the relative speed of B with respect to A in m/sec?
A) 50
B) 60
C) 36
D) 48
E) 72
Option A
Solution:
Relative speed =  108+72 = 180 km/hr = 180 * 5/18 = 50 m/sec

Directions (7-10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. (5 questions asked, 1 was based on mensuration, 1 on geometry, rest 3 are:)

1. There are 65 cards (numbered 1, 2, 3, ….65) in a box. Two cards are picked at random.
Quantity I: Probability that both cards show 2 digit numbers which increase by 36 when its digits are reversed.
Quantity II: Probability that both cards show a number divisible by 8 but not 16.
A) Quantity I > Quantity II
B) Quantity I < Quantity II
C) Quantity I > Quantity II
D) Quantity I > Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
Option A
Solution:
15+36 = 51, 26+36 = 62, 37+36 = 73, 48+36 = 82, 59+36 = 95
So there are 5 such numbers (15,26,37,48,59)
So required prob = 5/65 * 4/64 = 20/65*64
II: Such numbers are = 8, 24, 40, 56
So required prob = 4/65 * 3/64 = 12/65*64
2. There are 5 couples who want to sit at different positions.
Quantity I: Number of ways in which men and women sit at alternate positions
Quantity II: Number of ways in which all men sit together and all women sit together.
A) Quantity I > Quantity II
B) Quantity I < Quantity II
C) Quantity I > Quantity II
D) Quantity I > Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
Option A
Solution:
I: ways for men = 5!, now 6 seats for 5 women, so for choosing 5 seats 6C5, and then arrangement of these 5 women is 5!
So total number of ways = 5! * 6C5 * 5! = 5! * 5! * 6
II: Ways for all men together = 5!, for all women together is 5!, now arrangement of men and women as group = 2!
So total number of ways = 5! * 5! * 2! = 5! * 5! * 2
So I > II
3. Two parallel lines are given. s is an acute angle. Quantity I: Angle a
Quantity II: 25 degree
A) Quantity I > Quantity II
B) Quantity I < Quantity II
C) Quantity I > Quantity II
D) Quantity I > Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
Option B
Solution: s is acute means <90 degree, so x > 90
In triangle
a + (a+40)+ x = 180
2a + 40 = 140 – x
As x > 90
So 2a is < 50
So a < 25
So I < II

We will post more questions soon. Keep Checking the Quant Section

## 46 Thoughts to “Mixed Quantitative Aptitude Questions (Memory based SBI PO Mains)”

1. ****

thanku mam

2. swati

awesome thank you AZ

3. deteminedd

thank u

4. Hitesh Rana

HAVING gone through this quant which book u strongly recommend

1. Shubhra

We will find books if get any
Exam requires strong concept building.. If concepts are clear, u can solve easily

5. Shilpa Thakur

R = 15, Y = 9x, G = 4x. Prob of Y = 2/9
So 4x/(15+9x+4x) = 2/9
Solve, x = 3???????????????? is yellow 9x nhi hoga kya??????????????// please es ko explain kru

6. Shilpa Thakur

???????????????????????????
R = 15, Y = 9x, G = 4x. Prob of Y = 2/9
So 4x/(15+9x+4x) = 2/9????????????????????????????????????
Solve, x = 3
So in bag A: R = 15, Y = 27, G = 12. Total = 54
Bag B:
G = 2/3 * 27 = 18
Total = 5/6 * 54 = 45
Now 18 + Y + Y – 3 = 45
So, yellow balls = 15, and then red = 12
So in bag B: R = 12, Y = 15, G = 18. Total = 45
Bag C:
R = 1/3 (15+12) = 9
Yellow = 120/100 * 15 = 18
Probability of green ball = 7/16
So z/(9+18+z) = 7/16 [Let z green balls]
Solve, z = 21
So in bag C: R = 9, Y = 18, G = 21. Total = 48
Now probability of yellow ball from each bag = 27/54 * 15/45 * 18/48 = 1/16

1. Shubhra

No yellow is 4x only.
See in question

1. Shubhra

Not for that right now

1. Al! Z Well

Helloo ma’am..can we olso get memory based questions of reasoning section..

1. Shubhra

Check here
http://aspirantszone.com/reasoning/

3 posts

2. Al! Z Well

Ok thnkew?

7. Diya

Rhombus PQRS:

sine 30 = (15/2)/Hypotenuse(or side of PQRS)

Solve, Side of PQRS = 15 cm [sine 30 = 1/2]
//…….
sin = perpendicular/hypotenuse………is it correct?
yaha par 15/2 kaise aaya?
PQRS ka ek angle 120 degree hai and 15cm diagonal bisect it then angle 60 degree nahi hoga??means sin 60

1. Shubhra

I will give the figure for the same.wait for some time

1. Diya

ok ma’am

2. Sawan Sharma

1. Diya

PQ = 15 cos 60 + 15 cos 60……..two times why??

1. Sawan Sharma

component needs right angle and for the right angle we need to draw a perpendicular on PQ from point S ….after that we got “PQ= 15cos60 (length from perpendicular to point Q) and x (length from point P to perpendicular) … now we need to calculate x .. so we have another right angle tringle ( we have length of perpendicular and we need to determine base “x” and we have angle SPQ =60 ) so tan60 =15sin60/x …now we can determine the vallue of x and that is 15cos60 also that is why PQ=15cos60+15cos60. hope you will get it .. 🙂

1. Diya

thank u sir

2. Sawan Sharma

dont call me sir..!!
i’m also an aspirant 🙂

2. Diya

in PQRS
cos 60 = base/hypotenus = x/15
1/2 = x/15
x = 15/2 = 7.5
after this what I can do?

3. Shubhra

Check. I have updated the solution

1. Diya

got it ma’am thank you so much 🙂

8. KUCH B........

9. Sachin Shukla

ty

10. Bijli Ka Taar...

Ty:)

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