Important Quant Miscellaneous Questions for IBPS PO/Clerk Exams
Moderate Quant Miscellaneous Questions. Important Quant Miscellaneous Questions for IBPS PO/Clerk Prelims 2018. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk exams 201718, you will come across a section on Quantitative Aptitude Section. Here we are providing you “Important Quant Miscellaneous Questions for IBPS PO/Clerk Prelims 2018“ based on the latest pattern for your daily practice.
This “Important Quant Miscellaneous Questions for IBPS PO/Clerk 2018 “ is also important for other banking exams such as SBI PO, IBPS PO, IBPS Clerk, SBI Clerk, IBPS IBPS SO, SBI SO and other competitive exams
Important Quant Miscellaneous Questions for IBPS PO/Clerk Prelims  Set – 31
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
Help: Share this quiz to your Friends & FB Groups.
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)
Average score 

Your score 

Categories
 Not categorized 0%
 Quant 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
Category: QuantA can do a piece of work in ‘x’ days while B can do the same work in (x + 4) days. If A and B together complete the work in 3.75 days, then find the value of ‘x’.
Correct
B. 6 days
Number of days taken by A to complete the work = x days
Number of days taken by B to complete the work = (x + 4) days
Number of days taken by both A and B together to complete the work = 3.75 days
So, 1/x + 1/(x+4) = 1/3.75
3.75(2x + 4) = x(x + 4)
x^{2} – 3.5x – 15 = 0
x^{2} – 6x + 2.5x – 15 = 0
(x + 2.5) (x – 6) = 0
So, x = 6 days
Incorrect
B. 6 days
Number of days taken by A to complete the work = x days
Number of days taken by B to complete the work = (x + 4) days
Number of days taken by both A and B together to complete the work = 3.75 days
So, 1/x + 1/(x+4) = 1/3.75
3.75(2x + 4) = x(x + 4)
x^{2} – 3.5x – 15 = 0
x^{2} – 6x + 2.5x – 15 = 0
(x + 2.5) (x – 6) = 0
So, x = 6 days

Question 2 of 10
2. Question
A boatman covers 36 km downstream from A to B and 48 km upstream from B to C. If the ratio of the downstream speed to the upstream speed is 3:2 and the total time taken by the boatman to cover the entire distance is 9 hours, then find the time taken by the boatman to cover the distance between A and C while going upstream.
Correct
C. 1.5 hours
Distance between A and B = 36 km
Distance between B and C = 48 km
So, distance between A and C = 48 – 36 = 12 km
Let, the downstream speed and upstream speed be 3x km/hr and 2x km/hr, respectively.
According to question,
36/3x + 48/2x = 9
54x = 216
x = 4 km/hr
Downstream speed = 12 km/hr
Upstream speed = 8 km/hr
The time taken by the boatman to cover 12 km upstream from A to C = 12/8 = 1.5 hours
Incorrect
C. 1.5 hours
Distance between A and B = 36 km
Distance between B and C = 48 km
So, distance between A and C = 48 – 36 = 12 km
Let, the downstream speed and upstream speed be 3x km/hr and 2x km/hr, respectively.
According to question,
36/3x + 48/2x = 9
54x = 216
x = 4 km/hr
Downstream speed = 12 km/hr
Upstream speed = 8 km/hr
The time taken by the boatman to cover 12 km upstream from A to C = 12/8 = 1.5 hours

Question 3 of 10
3. Question
Pipe A and pipe C can fill a cistern in 12 minutes and 15 minutes, respectively while pipe B can empty the full cistern in 18 minutes. At a time, only one pipe is opened for a minute. Pipe A is opened first then pipe B and lastly pipe C. If the same process continues until the cistern is fully filled, then find the time taken to fill the cistern.
Correct
B. 92/3 minutes
Let, the total capacity of the cistern be LCM of (12, 18 and 15) = 180 units
Number of units of water filled by pipe A in one minute = 15 units
Number of units of water emptied by pipe B in one minute = 10 units
Number of units of water filled by pipe C in one minute = 12 units
Since, pipe A opens for 1^{st} minute, pipe B opens for 2^{nd} minute and pipe C opens for 3^{rd} minutes and the process continues until the cistern is filled.
So, number of units of water filled in 3 minutes = 15 – 10 + 12 = 17 units
So, number of units of water filled in 30 minutes = 17 × 10 = 170 units
Remaining water to be filled = 180 – 170 = 10 units
These 10 units of water will be filled by pipe A in 10/15 minute = 2/3 minute.
So, total time taken to fill the cistern = 30 + 2/3= 92/3 minutes
Incorrect
B. 92/3 minutes
Let, the total capacity of the cistern be LCM of (12, 18 and 15) = 180 units
Number of units of water filled by pipe A in one minute = 15 units
Number of units of water emptied by pipe B in one minute = 10 units
Number of units of water filled by pipe C in one minute = 12 units
Since, pipe A opens for 1^{st} minute, pipe B opens for 2^{nd} minute and pipe C opens for 3^{rd} minutes and the process continues until the cistern is filled.
So, number of units of water filled in 3 minutes = 15 – 10 + 12 = 17 units
So, number of units of water filled in 30 minutes = 17 × 10 = 170 units
Remaining water to be filled = 180 – 170 = 10 units
These 10 units of water will be filled by pipe A in 10/15 minute = 2/3 minute.
So, total time taken to fill the cistern = 30 + 2/3= 92/3 minutes

Question 4 of 10
4. Question
If the selling price of 10 books is equal to the cost price of 12 books, then find the profit % earned on each book.
Correct
D. 20%
Given, SP of 10 books = CP of 12 books
SP/CP = 12/10
SP/CP = 6/5
So, profit% = [(6 – 5)/5] × 100 = 20%
Incorrect
D. 20%
Given, SP of 10 books = CP of 12 books
SP/CP = 12/10
SP/CP = 6/5
So, profit% = [(6 – 5)/5] × 100 = 20%

Question 5 of 10
5. Question
The fare of a cab consists of an initial booking cost (fixed) plus the charge proportional to square root of the distance travelled. Ram pays fare of Rs. 5992 for 121 km while Shyam pays Rs. 8252 for 256 km. What is the initial booking cost of the cab?
Correct
A. Rs. 1020
Let, the initial booking cost be Rs. x and the proportionality constant be y.
According to question,
Cab fare = Initial booking cost + proportionality constant × (Number of kilometres travelled)^{1/2}
x + 11y = 5992 ———(1)
x + 16y = 8252 ———–(2)
Solving (1) and (2), we get
y = 452 and x = 1020.
So, the initial booking cost = Rs. 1020
Incorrect
A. Rs. 1020
Let, the initial booking cost be Rs. x and the proportionality constant be y.
According to question,
Cab fare = Initial booking cost + proportionality constant × (Number of kilometres travelled)^{1/2}
x + 11y = 5992 ———(1)
x + 16y = 8252 ———–(2)
Solving (1) and (2), we get
y = 452 and x = 1020.
So, the initial booking cost = Rs. 1020

Question 6 of 10
6. Question
A lender lends out different sums of money to two persons A and B. He lends to A at 8% simple interest p.a. for 3 years and to B at 6% simple interest p.a. for 4 years. If the amount returned to him by A and B is same, then find the ratio of the sum lent to A to the sum lent to B.
Correct
E. None of these
Let, the sum lent to A and to B be Rs. x and Rs. y, respectively.
Amount returned by A= x + (x × 8 × 3)/100
= x + 0.24x
= Rs. 1.24x
Amount returned by B= y + (y × 6 × 4)/100
= y + 0.24y
= Rs. 1.24y
Now, 1.24x = 1.24y
x : y = 1 : 1
Incorrect
E. None of these
Let, the sum lent to A and to B be Rs. x and Rs. y, respectively.
Amount returned by A= x + (x × 8 × 3)/100
= x + 0.24x
= Rs. 1.24x
Amount returned by B= y + (y × 6 × 4)/100
= y + 0.24y
= Rs. 1.24y
Now, 1.24x = 1.24y
x : y = 1 : 1

Question 7 of 10
7. Question
The concentration of Hcl in solution X is 75% while the concentration of H_{2}So_{4} in solution Y is 80%. If the quantity of solution Y is 30 litres and the quantity of solution X is 60% more than the quantity of solution Y, then find the difference between the quantity of Hcl in solution X and H_{2}So_{4} in solution Y.
Correct
D. 12 litres
Solution X contains 75% Hcl and 25% water.
Solution Y contains 80% H_{2}So_{4} and 20% water.
Quantity of solution Y = 30 litres
Quantity of solution X = 1.6 × 30 = 48 litres
Quantity of Hcl in solution X = 75% of 48 = 36 litres
Quantity of H_{2}So_{4} in solution Y = 80% of 30 = 24 litres
Required difference = 36 – 24 = 12 litres
Incorrect
D. 12 litres
Solution X contains 75% Hcl and 25% water.
Solution Y contains 80% H_{2}So_{4} and 20% water.
Quantity of solution Y = 30 litres
Quantity of solution X = 1.6 × 30 = 48 litres
Quantity of Hcl in solution X = 75% of 48 = 36 litres
Quantity of H_{2}So_{4} in solution Y = 80% of 30 = 24 litres
Required difference = 36 – 24 = 12 litres

Question 8 of 10
8. Question
A metallic cylinder of radius 12 cm and height 20 cm is melted to form cones of height 10 cm. If 24 such cones can be formed with the melted metallic cylinder, then find the slant height of the cones so formed?
Correct
B. 2√34 cm
Let, the radius of each cone formed by melting metallic cylinder be ‘r’ cm.
Number of cones formed = 24
(π × 12^{2} × 20) ÷ (1/3)π × r^{2} × 10 = 24
r = 6 cm
Now, in a cone:
r^{2} + h^{2} = l^{2}
=>l^{2} = 6^{2} + 10^{2}
l = (36 + 100)^{0.5} = 2√34 cm
Incorrect
B. 2√34 cm
Let, the radius of each cone formed by melting metallic cylinder be ‘r’ cm.
Number of cones formed = 24
(π × 12^{2} × 20) ÷ (1/3)π × r^{2} × 10 = 24
r = 6 cm
Now, in a cone:
r^{2} + h^{2} = l^{2}
=>l^{2} = 6^{2} + 10^{2}
l = (36 + 100)^{0.5} = 2√34 cm

Question 9 of 10
9. Question
The cost price of pulse1 is Rs. 45/kg and the cost price of pulse2 is Rs. 35/kg. If both the pulses are mixed to form a mixture of 60 kg and the average cost price of the mixture formed is Rs. 39, then find the quantity of pulse1 mixed to form the mixture.
Correct
B. 24 kg
Let, the quantity of pulse1 mixed be x kg.
Then, the quantity of pulse2 mixed = (60 – x) kg
According to question,
45x + 35(60 – x) = 39 × 60
45x + 2100 – 35x = 2340
10x = 240
x = 24 kg
So, the quantity of pulse1 mixed = 24 kg
Incorrect
B. 24 kg
Let, the quantity of pulse1 mixed be x kg.
Then, the quantity of pulse2 mixed = (60 – x) kg
According to question,
45x + 35(60 – x) = 39 × 60
45x + 2100 – 35x = 2340
10x = 240
x = 24 kg
So, the quantity of pulse1 mixed = 24 kg

Question 10 of 10
10. Question
A coin is tossed by a man. If head appears, he will throw a card from a well shuffled deck of 52 cards and if tail appears, he throws a dice. Find the probability of getting a black card.
Correct
C. 1/4
In a pack of 52 cards, there are 26 black cards and 26 red cards.
Probability of getting a head = probability of getting a tail = 1/2
Probability of getting a black card = 26/52
Required probability = (1/2) × (26/52) = 1/4
Incorrect
C. 1/4
In a pack of 52 cards, there are 26 black cards and 26 red cards.
Probability of getting a head = probability of getting a tail = 1/2
Probability of getting a black card = 26/52
Required probability = (1/2) × (26/52) = 1/4
Note:
 Click view Questions button to view Explanation
 Ask your doubt in comment section we’ ll clear your doubts in caring way.
 If you find any mistake, please tell us in the comment section and also mail us support@letsstudytogether.co
Crack IBPS PO Prelims 2018  Target 30 days Practice Sets of All Sections & Free PDF’s
High Level Quant & Data Interpretation PDF for Bank Exams
Books Name  Download Link 
Data Interpretation PDF for SBI PO Prelims 2018 – Based on SBI PO/Clerk Exams 2018  Click Here 
Quantitative Aptitude EBook for Banking, SSC and Insurance Exams  Click Here 
300 High Level Data Interpretation (DI) Questions & Answers PDF – Download Here  Click Here 
हाई लेवल डाटा इंटरप्रिटेशन प्रैक्टिस वर्कबुक (नवीनतम पैटर्न पर आधारित )  Click Here 
100 High Level Data Interpretation Questions & Answers PDF (English) Download Here  Click Here 
300 High Level Data Interpretation (DI) Questions & Answers PDF – Download Here
SBI PO/Clerk  RRB PO/Clerk  Railway Exams  SSC CGL 2018
S. No.  Exams  Direct Links 
1.  SBI PO/Clerk 2018  Click Here 
2.  IBPS RRB PO/Clerk 2018  Click Here 
3.  Railway RRB ALP & Group D 2018  Click Here 
4.  SSC CGL 2018  Click Here 