# Quantitative Aptitude Miscellaneous Questions for IBPS Clerk : Set-6 Quantitative Aptitude Miscellaneous Questions for IBPS Clerk. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Quantitative Aptitude Section. Here we are providing you “Quantitative Aptitude Miscellaneous Questions” for IBPS Clerk based on the latest pattern for your daily practice.

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## Quantitative Aptitude Miscellaneous Questions for IBPS Clerk : Set-6

1.12 men and 16 women together can complete a work in 26 days while 10 women and 18 children together can complete the work in 20 days. Also, 8 men and 16 children can complete the work in 25 days. 1 man, 1 woman and 1 child start the work and from 2nd day onwards every day 1 new man, 1 new woman and 1 new child join the work. On which day the work shall be completed?
a. 20th day
b. 21st day
c. 22nd day
d. None of the above
e. Cannot be determined

B. 21st day

LCM of 26, 20 and 25 = 1300
Let the total work be 1300 unit
Let ‘m’ be the units of work done by 1 man in 1 day
‘w’ be the units of work done by 1 woman in 1 day
and ‘c’ be the units of work done by 1 child in 1 day
26(12m + 16w) = 1300 ………(i)
20(10w + 18c) = 1300 ………(ii)
25(8m + 16c) = 1300 ……….(iii)
On solving equation (i), (ii) and (iii);
We get, m = 1.5 units, w = 2 units and c = 2.5 units
Work done on day 1 = 1.5+2+2.5 = 6*1 units
Work done on day 2 = (6+6*1) = 6*2 units

Work done on day 3 = 6*3 units
Let the work is completed on nth day
6*1 + 6*2 + …………. + 6n ≥ 1300
6n(n+1)/2 ≥ 1300
3n(n+1) ≥ 1300
The smallest integer value of n satisfying the above equation will be the required number of days
For n = 20
3n(n+1) = 1260
For n = 21
3n(n+1) = 1386 ≥ 1300
Therefore, the work shall be finished on 21st day.

2.The ages of A, B, C, D and E are 35, 33, 30, 27 and 25 years respectively. A, B, C, D and E bought furniture carrying MRP in the ratio of 2:3:4:5:6 respectively. A bought furniture carrying Rs. 1500 MRP. At the time of billing each of them received a surprise discount equal to their age but not more than 30%. Had they known the discount scheme earlier and exchanged their purchase among themselves to maximize combined discount, how much combined saving could have been done apart from the discount given?
a. Rs. 105
b. Rs. 135
c. Rs. 165
d. Rs. 195
e. Rs. 215

D. Rs. 195

MRP of A’s purchase = 1500 (Discount: 30%)
Discount amount = 1500*30% = Rs. 450
MRP of B’s purchase = 2250 (Discount: 30%)
Discount amount = 2250*30% = Rs. 675
MRP of C’s purchase = 3000 (Discount: 30%)
Discount amount = 3000*30% = Rs. 900
MRP of D’s purchase = 3750 (Discount: 27%)
Discount amount = 3750*27% = Rs. 1012.5
MRP of E’s purchase = 4500 (Discount: 25%)
Discount amount = 4500*25% = Rs. 1125
Total discount = 450+675+900+1012.5+1125 = 4162.5

Maximum combined discount is possible when C,D and E receive 30% discount, A receives 25% discount and B receives 27% discount.
Maximum combined discount = 4500*30% + 3750*30% + 3000*30% + 2250*27% + 1500*25% = 4357.5
Therefore, the apart from the given discount, Rs. (4357.5-4162.5) = Rs. 195 could have been saved.

3.A class comprises of 6 boys and some girls. If there are 281 ways of forming a group of 5 students such that group constitutes at most 2 girls, what is the probability of forming a group of 5 students with at least 2 girls?
A. 81/154
B. 73/154
C. 127/154
D. 27/154
E. None of the above

C. 127/154

Let the number of girls be n
A/Q
6C5 + 6C4 * nC1 +6C3 * nC2 = 281
6 + 15n + 10n(n-1) = 281
=> 10n2+ 5n – 275 = 0
=> 2n2+ n – 55 = 0
=> 2n2+ 11n – 10n – 55 = 0
=> n(2n+11) – 5(2n+11) = 0
=> (n-5) (2n+11)= 0
Here n is a positive integer so, n = 5
Probability of forming a group of 5 students with at most 1 girl =(6C5 + 6C4 * 5C1 )/11C5 = 81/462= 27/154
Required probability = 1 – 27/154 = 127/154

4.A and B are standing on points P and Q respectively which are 200 km apart. A starts moving to and fro between P and Q at a speed of 20 km/h while B starts moving to and fro between Q and P at a speed of 30 km/h. What will be the difference between the total distance travelled by A and B when they meet for the 10th time?
a. 380 km
b. 570 km
c. 650 km
d. 760 km
e. 820 km

5.The ratio of Ekta’s and Reema’s income last year was 10:3. The ratio of Ekta’s this year income and last year income is 6:5 and the ratio of Reema’s this year income and last year income is 2:3. If the sum of Ekta’s and Reema’s present incomes is Rs. 5,124, what was Reema’s income last year?
a. Rs. 1464
b. Rs. 1830
c. Rs. 1372
d. Rs. 1654
e. Rs. 1098

E. Rs. 1098

Let Ekta and Reema’s current income be 6x and 2y respectively.
Hence, their income last year was 5x and 3y respectively.
Since the ratio of their income last year was 10 : 3;
5x : 3y = 10 : 3
=> x = 2y

The sum of their present incomes is Rs.5,124.
=> 6x + 2y = 5124
=> 7x = 5124
=> x = 732 and y = 366
=> Reema’s income last year = 3y = Rs. 1,098

6.Two trains while moving in opposite direction on parallel tracks, take 10 seconds to cross each other. However, if they travel in the same direction, the longer train, which is faster than shorter train, crosses the shorter train in 30 seconds. If the length of the longer train is decreased by 75%, it takes 12 seconds less to cross the shorter train traveling in the same direction. Find the time taken by the longer train to cross a tunnel twice its length, if the difference between the lengths of the trains is 50 m?
a. 20 seconds
b. 22 seconds
c. 24 seconds
d. 32 seconds
e. 36 seconds

C. 24 seconds

Let the length of shorter and longer trains be ‘x’ and (x+50) respectively
And their speed be v and u m/s, respectively
In first case: If trains travel in opposite direction
(x+(x+50))/10 = u+v => (2x+50)/10 = u+v ……………..(i)
In second case: If trains travel in same direction
(2x+50)/30 = u-v ……………………(ii)
In third case: If trains travel in same direction with decreased length
[x+{(x+50)/4}]/18 = u-v …………………(iii)
On solving (i) and (ii)
u = (2x+50)*2/10*3 = 2(2x+50)/30 …………………. (iv)
v = (2x+50)/30 ……………………..(v)
on solving (iii),(iv) and (v)

x = 350 m, u = 50 m/s and v = 25 m/s
Therefore, the length of longer train = 350+50 = 400 m
Length of tunnel = 2*400 = 800 m
Total distance = (400+800) = 1200 m
Time taken to cross tunnel = 1200/50 = 24 seconds

7.Average age of a family of four people, 5 years ago, was 40 years. 5 years from now, the ratio of oldest member’s age to the youngest member’s age will be 18:7 and the ratio of present ages of other two member is 7:3. If the youngest member is 40 years younger than second oldest member, what is the present age of oldest member?
a. 57 years
b. 62 years
c. 67 years
d. 72 years
e. 77 years

C. 67 years

Let the current age be ‘a’ of oldest member, ‘b’ of the second oldest member, ‘d’ of youngest member and ‘c’ of the fourth member of the family.
a-5 + b-5 + c-5 + d-5 = 40*4
a + b + c + d = 180 ……….(i)
(a+5)/(d+5) = 18/7
=> 7a-18d = 55 ………(ii)
d+40 = b………(iii)
b/c = 7/3………(iv)
Solving the above four equations we get,
a = 67 years

8.A watch merchant bought 140 watches at the rate of Rs. 900 per dozen watches. He marked the watches at Rs. 120 per piece. He sold 75% of the watches at 10% discount on marked price and the remaining watches were sold at 20% profit on cost price. Had he not given any discount, by what percentage his profit would have increased?
a. 12.4%
b. 13.3%
c. 14.6%
d. 15.9%
e. 16.4%

D. 15.9%

Unit cost price = 900/12 = Rs. 75
Total cost price = 140*75 = Rs. 10500
Total selling price = 140*0.75*120*0.9 + 140*0.25*75*1.2 = 11340 + 3150 = 14490
Profit = 14490 – 10500 = Rs. 3990
If there will be no discount, total selling price = 120*140 = 16800
Therefore, required percentage = 100*(16800-14490)/14490 = 15.9%

9.’Bhoomi’ is a rectangular shaped construction site. The ratio of the square of the perimeter of ‘Bhoomi’ and the sum of the squares of the diagonals of ‘Bhoomi’ is 98:25. What is the ratio of the sum of the adjacent sides of ‘Bhoomi’ and difference between adjacent sides of ‘Bhoomi’?
a. 2
b. 3
c. 5
d. 7
e. 9

D. 7

Let the length and the breadth of the rectangle be l and b respectively
Then, (2(l + b))2/2(l2+ b2) = 98/25
(l2+ b2+2lb)/(l2+b2) = 49/25
50*l*b = 24(l2+b2)
12l2-25*l*b + 12b2= 0
(l/b) = (3/4) or (4/3)
But l cannot be less than b
=> (l/b) = (4/3)
Required ratio = (l + b)/(l – b) = ((4b/3 + b))/((4b/3 – b)) = 7

10.What is the remainder when 1! + 2! + 3! + 4! + … + 77! is divided by 15?
a. 3
b. 5
c. 7
d. 9
e. 11

A. 3

n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1
Thus, every term from 15! to 77! will have a ‘15’ in it.
Thus, every term from 15! to 77! is divisible by 15.
Since 15 = 5 × 3, any x! that has a term (5 × 3) in it will also be divisible by 15.
5! = 5 x 4 × 3 × 2 × 1
Since every subsequent term from 5! to 14! will have a 5! in it, each term from 5! to 14! is divisible by 15.
Therefore, required remainder = remainder of [ (1! + 2! + 3! + 4!)/15 ] = remainder of [ (1+2+6+24)/15 ] = 3

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