Quadratic Equations Quiz for IBPS PO / SBI PO 2017

Directions (1-5): In these questions two equations numbered I and II are given. You have to solve both the equations and mark answer.

(a) X > Y

(b) X ≤ Y

(c) X < Y

(d) X ≥ Y

(e) Relationship between X and Y cannot be established

(a) X ≥ Y

(b) X ≤ Y

(c) X < Y

(d) X > Y

(e) Relationship between X and Y cannot be established

(a) X ≥ Y

(b) X > Y

(c) X ≤ Y

(d) X < Y

(e) Relationship between X and Y cannot be established

(a) X ≥ Y

(b) X > Y

(c) X ≤ Y

(d) X < Y

(e) Relationship between X and Y cannot be established

(a) X ≥ Y

(b) X < Y

(c) X > Y

(d) X ≤ Y

(e) Relationship between X and Y cannot be established

Directions (6-10): In each of these questions, two equations I and II are given. You have to solve both the equations and give answer.

(a) x < y

(b) x > y

(c) x = y

(d) x ≥ y

(e) x ≤ y or no relationship can be established between x and y

Directions (11-15): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.

(a) If x > y

(b) If x ≥ y

(c) If x < y

(d) If x ≤ y

(e) x = y or relationship cannot be established

Solutions

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Quadratic Equations Quiz for IBPS PO / SBI PO 2017

Directions (Q. 1–4): In each of these questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the value of x and y. Give answer

a) if x > y 

b) if x >= y 

c) if x < y

d) if x <= y 

e) if x = y or relationship between x and y cannot be determined.

1. 

I. x^2 + x – 20 = 0

II. y^2 + 13y + 40 = 0

1. b

I. x^2 + 5x – 4x – 20 = 0

(x – 4) (x + 5) = 0

x = 4, – 5

II. y^2 + 8y + 5y + 40 = 0

(y + 8) (y + 5) = 0

y = –8, – 5

x >= y

2. 

I. x^2 – 11x + 30 = 0

II. y^2 – 13y + 40 = 0

2. e

I. x^2 – 6x – 5x + 30 = 0

(x – 5) (x – 6) = 0

x = 5, 6

II. y^2 – 8y – 5y + 40 = 0

(y – 8) (y – 5) = 0

y = 8, 5

No relationship between ‘x’ and ‘y’ exits.

3. 

I. x^2 + 10x + 25 = 0

II. 5y^2– √60 y + 3 = 0

3. c

I. (x + 5)^2 = 0

x = –5

II. (√5y – √3)^2 = 0

y = √3/√5

y>x

4. 

I. 10x^2 – 29x – 21 = 0

II. y^2 + 13y – 68 = 0

4. e

I. 10x^2 – 35x + 6x – 21 = 0

(5x + 3) (2x – 7) = 0

x = -3/5, 7/2

II. y^2 + 17y – 4y – 68 = 0

(y – 4) (y + 17) = 0

y = 4, –17

No relationship between x and y exists.

Directions (Q. 5–7): What value should come in the place of question mark (?) in the following number series?

5. 362, 452, 550, 656, ?

a) 770 

b) 772 

c) 670 

d) 870 

e) 790

5. a

 

6. 25, 28, 26, ?, 27, 30

a) 28 

b) 32 

c) 34 

d) 29 

e) 36

6. d

7. 9, 12, 30, 99, ?

a) 406 

b) 418 

c) 408 

d) 416 

e) 424

7. c

Directions (Q. 8–10): What value should come in the place of question mark (?) in the following questions?

8. 168.781 – 112.412 – 8.409 – 1.150 = ?

a) 44.81 

b) 46.81 

c) 40.81 

d) 47.81 

e) 46.61

8. b

? = 46.81

9. 2.01*8.96 + 128.12/(2.05*1.97) = ?

a) 50 

b) 11 

c) 21 

d) 44 

e) 23

9. a

2.01*8.96 + 128.12/(2.05*1.97) = ?

Or, 2*9 + 128/(2*2) = 18 + 32 = 50 (approx.)

10. 8.5% of 160 – 0.42% of 750 = ?

a) 11.45 

b) 12.45 

c) 13.45 

d) 9.45 

e) 10.45

10. e

? = 8.5/100 * 160 - 0.42/100 * 750

= 13.6 - 3.15 = 10.45

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IBPS PO MAINS 2016 - Quant Quiz-7 (Time and Work)

1. A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?

(a) A and B

(b) A and C

(c) B and C

(d) A and D

(e) C and D

Sol. Work done in one day by A, B and C are 1/4,1/8,1/16 and 1/32 respectively.

Using answer choices, we note that the pair of B and C does 3/16 of work in one day; the pair of A and D does 1/4+1/32=9/32 of work in one day

Hence, A and D take 32/9 days

B and C take 16/3=32/6 days

Hence, the first pair must comprise of A and D.

2. R finishes a work in 7 days. P finishes the same job in 8 days and Q in 6 days. They take turns to finish the work. R worked on the first day, P on the second day and Q on the third and then again R and so on. Who was working on the last day when work got finished?

(a) R

(b) P

(c) Q

(d) P and Q

(e) Cannot be determined

Sol. Three day’s work = 1/7+1/8+1/6=73/168

Six day’s work = 73/84

Seventh day work = 1/7, done by R

Since 73/84+1/7=85/84 > 1, therefore, R was working on the last day.

3. Construction of a road was entrusted to a civil engineer. He has to finish the work in 124 days for which he employed 120 workers. Two-third of the work was completed in 64 days. How many workers can be reduced now without affecting the completion of the work on time?

(a) 56

(b) 64

(c) 80

(d) 24

(e) None of these

Sol. 2/3rd of the work was completed in 64 days by 120 workers.

1/3rd of the work was completed in 32 days by 120 workers.

Also 1/3rd of the work is to be completed in 60 days by (120 – x) workers, where x is the number of men reduced in order to finish the work on schedule.

So, (120 – x) × 60 = 120 ⇒ x = 56.

4. Two workers earned Rs. 225 first worked for 10 days and the second for 9 days. How much did each of them get daily if the first worker got Rs. 15 more for working 5 days than the second worker got for working 3 days?

(a) Rs. 11.70; Rs. 12.00

(b) Rs. 10.80; Rs. 13.00

(c) Rs. 11.25; Rs. 12.50

(d) Rs. 12.60; Rs. 11.00

(e) None of these

Sol. Let A got Rs. x per day and B got Rs. y per day. So, 10x + 9y = 225 and 5x = 3y + 15 ⇒ x = 10.80, y = 13.

5. Two pipes A and B can fill a tank in 20 and 30 h respectively. Both the pipes are opened to fill the tank but when the tank is 1/3 full, a leak develops in the tank. Due to this leakage one-third of the water supplied by pipes A and B goes waste. What is the total time to fill the tank if the leak if not closed ?

(a) 12 h

(b) 16 h

(c) 18 h

(d) 20 h

(e) None of these

Sol. Let us assume total work = 180 (we are not assuming it to be LCM of 20 and 30 = 60 because in that case 1/3rd of A + B will be fractional)

Time taken to fill 1/3rd of the tank = 180/ (9 + 6) = 4 h

Due to leakage, net inflow = 2/3 (9 + 6) = 10 units

Time taken to fill remaining 120 units = 12 h

So total time taken = 12 + 4 = 16 h

6. P, Q and R can do a piece of work in 16, 24 and 30 days respectively. They started the work simultaneously but P stops the work after 4 days and Q called off the work 2 days before the completion. In what time the work is finished ?

a) 100/9 days
b) 100/11 days
c) 100/7 days
d) 100/13 days
e) None of these

Sol. 4/16 + (T -2)/24 + T/30 = 1 where T is the time taken to complete the job.

       T = 100/9 days.

7. If P can do 1/3 of the work in 5 days and Q can do 1/4 of the work in 6 days, then how much money will Q get if they were paid a total of 390 rupee?

a) 120
b) 150
c) 170
d) 190
e) None of these

Sol. P can alone complete the whole  work in 15 days and Q can complete the same work alone in 24 days. So ratio of work done by them 1/15: 1/24 i.e. 8: 5
Q get = (5/13)*390 = 150.

8. A does half as much work as B in one third of the time taken by B. If together they take 20 days to finish the work then what will be the share of A if 1000 rupees is given for the whole work?

a) 400
b) 500
c) 600
d) 700
e) None of these

Sol. Let B take x days to complete the work, then A will take  = x/3 + x/3 = 2x/3 days (as half work is completed in one third of the time)
3/2x + 1/x = 1/20
X = 50. So A will complete the work in 100/3 days and B will complete the work in 50 days.
Ratio of work done by A and B – 3/100: 1/50 = 3:2
So A share = 3/5*1000 = 600.

9. A does half as much work as B does in one sixth of the time. If together they take 20 days to complete the work, then what is the time taken by A to complete the work independently.

a) 80/3 days
b) 100/3 days
c) 60/3 days
d) 140/3 days
e) None of these

Sol. Let B complete the work in X days so in one day work done by B is 1/x
as A do half work in one-sixth of the time so A will complete work in 2*x/6 = x/3 days
One day work of A and B i.e. 3/x + 1/x = 1/20. So we get x = 80
So time taken by A alone = 80/3 days.

10. A and B can do a piece of work in 20 and 25 days respectively. They began to work together but A leaves after some days and B completed the remaining work in 12 days. Number of days after which A left the job-

a) 5.7/9 days
b) 6.7/9 days
c) 7.7/9 days
d) 11.7/9 days
e) None of these

Sol. (1/20 + 1/25)*T + 12/25 = 1
We will get T = 52/9 i.e. 5.7/9 days.

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IBPS PO MAINS 2016 - Quant Quiz-4 (P&C and Probability)

1. In how many ways "PROBATIONARY" can be written such that the vowels are always together ?

A. 2 * 7!*5!
B. 8!
C. 2 * 8!*5!
D. 8! / 2!
E. 7!

2. What is the probability that the number selected from the set {1,2,3,.......50} is a multiple of 3 or 6 ?

A. 12/25
B. 8/25
C. 4/25
D. 16/25
E. None of these

3. A bag contains 4 black and 6 white balls. 2 balls are drawn at randomn. Find the probability that atleast one of them is black ?

A. 1/3
B. 2/3
C. 3/4
D. 3/5
E. None of these

4. A bag contains 20 tickets numbered from 1-20. 2 tickets are drawn at randomn. Find the probability that both numbers are prime ?

A. 16/95
B. 15/95
C. 14/95
D. 17/95
E. None of these

5. A bag contains 3 red, 6 white and 7 black balls. Two balls are drawn at random. What is the probability that both are black ?

A. 9/40
B. 7/40
C. 11/40
D. 13/40
E. None of these

6. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, the number of points in which they intersect each other is ?

A. 200
B. 180
C. 190
D. 220
E. 230

7. A boy has 3 library tickets and 8 books of his interest are there in the library. Out of these 8, he does not want to borrow Chemistry Part II, unless Chemistry Part I is also borrowed. The number of ways in which he can choose the three books to be borrowed is ?

A. 42
B. 40
C. 41
D. 45
E. 44

8. In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship were ?

A. 20
B. 22
C. 16
D. 18
E. None of these

9. How many motor vehicle registration number plates can be formed with the digits 1, 2, 3, 4, 5 (no digits being repeated) if it is given that registration number can have 1 to 5 digits ?

A. 320
B. 325
C. 350
D. 315
E. None of these

10. If a team of four persons is to be selected from 8 males and 8 females, then in how many ways can the selections be made to include at least one male ?

A. 1750
B. 1500
C. 1450
D. 1600
E. None of these


                                                ANSWERS

1. Vowels in the word PROBATIONARY are O, A, I, O, A
     Now keeping them together and treating them as one word.
     No.  of ways of writting that will be = 8!
     Now the vowel group its self can be written in = (5!/2!*2!) 

Thus, total ways of writing PROBATIONARY is = (8!*5!)/(2!*2!) = 2*7!*5!

2. Multiples of 3 are {3,6,.......,48} = 16 numbers
    Multiples of 6 are {6,12,18,...48} = 8 numbers

Now, as multiples of 6 are already included in multiples of 3.
Thus, no. of favorable cases are = 16

Required Probability = 16/50 = 8/25

3. Total ways of selecting 2 balls from 10 = 10C2 = 45
     No. of ways of selecting 2 white balls from 6 white balls = 6C2 = 15

No. of favorable cases are = total ways - no. of ways of selecting no black ball or in                                                                           other words 2 white balls

                                              = 45 - 15 = 30

Required Probability = 30/45 = 2/3

4. Total prime no.s from 1-20 = 8
    Total ways of selecting 2 tickets from 20 = 20C2 = 190
     No. of ways of selecting two prime no.s from 8 = 8C2 = 28

Required Probability = 28/190 = 14/95.


 5. The reqd. probability is 7C2 / 16C2 = 7/40.

 6.

    

    
7. Number of Ways =  No Chemistry or Chemistry I alone or Chemistry I and
                                     Chemistry II
   
    = 6C3 + 6C2 + 6C1 = 20 + 15 + 6 = 41.

8. 
    
    
     

9. If number is 1 digit number, Number of ways= 5. 

    If number is 2 digit number, Number of ways=5×4=20

    If 3 digit number, Number of ways= 5×4×3= 60

    If 4 digit number, Number of ways=5×4×3×2=120

    If 5 digit number, Number of ways=5×4×3×2×1=120.

    By adding these we get the sum as 325.

10. If we subtract the cases in which all are females, we will be left with the cases where atleast one male is selected.

     Therefore, the total ways in which 4 members can be selected from 16 = 16C4 
      and total ways in which 4 females can be selected = 8C4.
      So required ways to select at least one male is 16C4 - 8C4 = 1750.

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