Quantitative Aptitude Quiz For IBPS PO Mains: 9th November 2018

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Quantitative Aptitude Quiz For IBPS PO Mains: 7th November 2018
Quantitative Aptitude Quiz For IBPS PO Mains


Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

Q1.  Manoj lend Rs. P for three years on S.I. at the rate of 15% per annum and Rajesh lend Rs. (P + 8000) for two years on C.I. at the rate of 8% per annum. Suresh borrowed sum equal to of what Manoj and Rajesh lend, for two years on C.I. at the rate of 20% per annum. If Suresh paid interest Rs. 5352 more than, what Manoj and Rajesh got total interest on their sums together. Find total sum borrowed by Suresh? 
(a) Rs. 34000
(b) Rs. 44000
(c) Rs. 32000
(d) Rs. 46000
(e) Rs. 30000

Q2. A train P, 180 meter long train passed a pole in 27/4 sec and also passed two trains Q and R in 9 sec and 39 sec respectively, where train Q running in opposite direction of train P and train R is running in same direction of train P. If length of train Q and R is 240 meter and 210 meter respectively, then in what time train Q will pass train R, if both runs in opposite direction ?
(a) 35 sec
(b) 9 7/11 sec
(c) 12 3/11 sec
(d) 15 sec
(e) 55 sec

Q3. A box contains 30 eggs out of which 6 are rotten. Two eggs are chosen at random. What is the probability that only one of the chosen eggs is rotten ?
(a) 53/145
(b) 63/145
(c)51/145
(d) 57/145
(e) 48/145

Q4. A, B and C entered into a partnership with some investment for one year. After one-year A got 2/5 profit and B and C got equal part of remaining profit. If total profit after one year is 15% instead of 10% then A got 900 Rs. more. Find the investment of B. 
(a) 12000
(b) 45000
(c) 27000
(d) 18000
(e) 13500

Q5.  A boat cover 60 km upstream and 60 km downstream in 22.5 hr with its usual speed. If boat double its speed then new upstream speed is 150% more than the usual upstream speed. Find the time taken by boat to cover 80 km in downstream with usual speed. 
(a) 12 hr
(b) 20 hr
(c) 5 hr
(d) 16 hr
(e) 10 hr

Direction (6–10): Bar graph given below shows percentage of a task done by each persons and line graph shows days taken by each person to do that part of task. 



Q6. If Ankit and Abhishek Start working together and work for x days after that they are replaced by Divyaraj and Veer who work for next (x + 2) days,  50/9% of total work are still left which is completed by Ayush with the efficiency of (x + 2) unit/day. Find in how many days Ayushcan  complete whole work alone? 
(a) 36 days
(b) 24 days
(c) 30 days
(d) 39days
(e) 45 days

Q7. Sameer and Abhishek start working alternatively starting with Sameer and work for 25 days, after that both are replaced by Ankit and Divyaraj. If Ankit and Divyaraj work alternatively starting with Divyaraj then in how many days remaining work will be completed? 
(a) 28 1/6 days
(b) 33 1/6 days
(c) 46 1/3 days
(d) 38 1/6 days
(e)  42 1/3 days

Q8.  Ankit and Veer work together for y days, Sameer work for (y – 4) days and remaining work is completed by Abhishek in (y – 10) days. If all four work for (y – 3) days together then what portion of work is left? 
(a) 1/36
(b) 1/124
(c) 1/128
(d) 1/144
(e) 1/148

Q9. If for first 15 days Ankit work with 25% less efficiency and Veer work with 33 1/3% more efficiency together and remaining work is completed by another person Satish in 57 days. Find in how many days Satish can complete the whole work alone? 
(a) 102 days
(b) 128 days
(c) 108 days
(d) 144 days
(e) 162  days

Q10.Five person Ankit and Abhishek, Sameer and Divyaraj and Veer work alternatively in such a manner that on first day Ankit and Abhishek work together, on second day Sameer and Divyaraj work together and third day Veer work alone, find in how many day whole work will be completed? 
(a) 54 3/4 days
(b) 33 1/4 days
(c) 22 1/2 days
(d) 36 1/4 days
(e) 42 3/4 days

Direction (11 – 15): What will come in the place of question(?) mark in given number series:

Q11. 55,       72,       91,        112,        135,       ?
(a) 156
(b) 160
(c) 144
(d) 164
(e) 172

Q12.  12.5,     13,    27.5,      85,       343.5,      ?
(a) 1722
(b) 1740
(c) 1720
(d) 1716
(e) 1748

Q13.  1783,     1776,      1805,    1744,       1873,      ?
(a) 1652
(b) 1668
(c) 1664
(d) 1680
(e) 1662

Q14. 12,    91,    552,     2765,       11064,      ?
(a) 33180
(b) 33210
(c) 33195
(d) 33200
(e) 33560

Q15.  5856,      488,       4880,       610,      3660,      ?
(a) 915
(b) 905
(c) 918
(d) 920
(e) 924

SOLUTIONS

S1.  Ans(c)
Sol.


S2. Ans.(c)
Sol.
Lets speed of train P, Q and R be S₁, S₂ and S₃ respectively 

S3. Ans.(e)
Sol. 

S4. Ans.(e)
Sol.
A got 40% of profit
B & C got 30% each
So investment ratio of A, B and C is 4 : 3 : 3
Now,
They earn 10% profit
⇒(10x×10)/100=x
If they earn 15% profit
=(10x×15)/100=3/2 x
A got 900 Rs. more
⇒3/2 x× 40/100–(x×40)/100=900
⇒ x = 4500
Total investment = 45000
B’s investment= (45000×3)/10
=13500

S5. Ans.(e)
Sol.
Let usual speed of boat in still water = x
River speed = y
ATQ,
(x-y)250/100=(2x-y)
5x-5y=4x-2y
x = 3y
Now,
60/(x –y)+60/(x+y)=22.5
60/2y+60/4y=22.5
y = 2 km/hr
x = 6 km/hr
Required time= 80/(6+2)
=10 hr

S6. Ans.(e)
Sol.
Number of days taken by Ankit to complete the whole work alone =24/40×100 = 60 days
Number of days taken by Sameer to complete the whole work alone =4.5/12.5×100 = 36 days
Number of days taken by Divyaraj to complete the whole work alone =18/25×100 = 72 days
Number of days taken by Veer to complete the whole work alone =6/12.5×100= 48 days
Number of days taken by Abhishek to complete the whole work alone =9/10×100=90 days


S7. Ans.(b)
Sol.
If Sameer and Abhishek work alternatively
First day by Sameer = 20 unit work
Second day by Abhishek = 8 unit work
So, in two day = 28 unit work
Total work complete by Sameer & Abhishek together
= 20 × 13 + 8 × 12
= 260 + 96
= 356 unit
Remaining work = 720 – 356 = 364 unit
Remaining work complete by Ankit and Divyaraj alternatively
First day by Divyaraj = 10 unit
Second day by Ankit = 12 unit
So, in two days = 22 unit
Total work  in 32 days
=32/2×22
= 352 unit
Remaining work = 364 – 352 = 12 unit
33 days work =352+ 10 = 362 unit
Remaining 2 unit by Ankit = 2/12 = 1/6 days
Total time = (32+1+1/6)=331/6 days

S8. Ans.(d)
Sol.
Efficiency of Ankit and Veer × y + Efficiency of Sameer × (y – 4) + Efficiency of Abhishek × (y – 10) = 720
(12 + 15) y + 20 (y – 4) + 8 (y – 10) = 720
27y + 20y – 80 + 8y – 80 = 720
55y = 880
y = 16 days
ATQ—
Efficiency of (Ankit + Sameer + Veer + Abhishek) × (y – 3) = (12 + 20 + 15 + 8) × (16 – 3) = 715 unit
 Work Remaining= 720 – 715
= 5 unit
Required portion = 5/720 = 1/144

S9. Ans.(d)
Sol.
First 15 days work of Ankit and Veer
=[(12×3/4)+(15×4/3) ]×15 
= 29 × 15
= 435 unit
Remaining work = 720 – 435= 285 unit
Efficiency of Satish = 285/57 = 5 unit/day
Satish alone can complete whole work in = 720/5 = 144 days

S10. Ans.(b)
Sol.
ATQ—
First day work by Ankit & Abhishek together = (12 + 8) = 20 unit
Second day work by Sameer &Divyaraj together = (20 + 10) = 30 unit
Third day work by Veer alone = 15 unit
Total work in three days = (20 + 30 + 15) = 65 unit
 33 days work = 33/3 × 65 = 715 unit
Remaining work by Ankit & Abhishek together= (720 –715)/20 = 1/4 days
Total time = 33 1/4 days

S11. Ans(b)
Sol.
Pattern of series –
6 ×9+1=55
7 ×10+2=72
8 ×11+3=91
9 ×12+4=112
10 ×13+5=135
? = 11 ×14+6=160

S12. Ans(a)
Sol.
Pattern of series –
12.5 ×1+0.5=13
13 ×2+1.5=27.5
27.5 ×3+2.5=85
85 ×4+3.5=343.5
? = 343.5 ×5+4.5=1722

S13.Ans(e)
Sol.
Pattern of series –
1783 – (23 – 1) = 1776
1776 + (33 + 2) = 1805
1805 – (43 – 3) = 1744
1744 + (53 + 4) =1873
? = 1873 – (62 – 5) = 1662

S14. Ans(c)
Sol.
Pattern of series –
12 ×7+7=91
91 ×6+6=552
552 ×5+5=2765
2765 ×4+4=11064
? = 11064 ×3 + 3 = 33195

S15. Ans(a)
Sol.
Pattern of series –
5856 ÷12=488
488 ×10=4880
4880 ÷8=610
610 ×6=3660
? = 3660 ÷4=915

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