Below are Memory Based Questions SBI PO Mains 2017 Exam in Quantitative Aptitude Section.Â
- 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
View Answer  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 Â = 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)
- One ball from each bag is drawn. Find the probability that these are yellow balls.
A)Â 5/16
B)Â 3/16
C)Â 7/36
D)Â 1/36
E)Â 9/16
- 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
- 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)Â 3
B) 1
C)Â 2
D)Â 4
E)Â Other than those given in options.
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)
- If both trains were moving in same direction, find the time at which they will meet.
A) 50 minutes
B) 60 minutes
C) 108Â minutes
D) 40 minutes
E) 72Â minutes
- What is the relative speed of B with respect to A in m/sec?
A) 50
B) 60
C) 36
D) 48
E) 72
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:)
- 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)Â Quantity I >Â Quantity II
B) Quantity I < Quantity II
C) Quantity I >Â Quantity II
D) Quantity I >Â Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
- There are 5 couples who want to sit at different positions.
Quantity I:Â Number of ways in which men and women sit at alternate positions
Quantity II:Â Number of ways in which all men sit together and all women sit together.
A)Â Quantity I >Â Quantity II
B) Quantity I < Quantity II
C) Quantity I >Â Quantity II
D) Quantity I >Â Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
- Two parallel lines are given. s is an acute angle.
Quantity I:Â Angle a
Quantity II:Â 25 degree
A)Â Quantity I >Â Quantity II
B) Quantity I < Quantity II
C) Quantity I >Â Quantity II
D) Quantity I >Â Quantity II
E) Quantity I = Quantity II or relationship cannot be determined
View Answer  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 SectionÂ
thanku mam
awesome thank you AZ
thank u
HAVING gone through this quant which book u strongly recommend
We will find books if get any
Exam requires strong concept building.. If concepts are clear, u can solve easily
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
???????????????????????????
First find balls in each bag. Start with bag A:
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
No yellow is 4x only.
See in question
Not for that right now
Helloo ma’am..can we olso get memory based questions of reasoning section..
Your dis post is really vry helpful..!! Thnkx a ton
Check here
http://aspirantszone.com/reasoning/
3 posts
Ok thnkew?
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
I will give the figure for the same.wait for some time
ok ma’am
https://uploads.disquscdn.com/images/bc6d33803ad018c7403359dfa8c68de75ede087aa38f116c87bba70ac2ced2ff.jpg
PQ = 15 cos 60 + 15 cos 60……..two times why??
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 .. 🙂
thank u sir
dont call me sir..!!
i’m also an aspirant 🙂
in PQRS
cos 60 = base/hypotenus = x/15
1/2 = x/15
x = 15/2 = 7.5
after this what I can do?
Check. I have updated the solution
got it ma’am thank you so much 🙂
measuration b padna hoa ab:(
ty
Ty:)
Waao…. Ye toh SBI mains ki question hai…..
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