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QUANT FOR SBI PO PRELIM 10



We bring you the complete and important daily QUANT  to achieve more marks in Banking, Insurance, UPSC, SSC, CLAT, Railways and all other competitive Exams. We prepare it based on our daily current affairs.Hope you like it.

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START TEST

Direction (1-10): In the given questions, two
quantities are given, one as ‘Quantity 1’ and another
as ‘Quantity 2’. You have to determine relationship
between two quantities and choose the appropriate
option:
a. Quantity I> Quantity II
b. Quantity I < Quantity II
c. Quantity II ≥ Quantity I
d. Quantity II ≤ Quantity I
e. Quantity I = Quantity II or Relation cannot be
established

1. A can do a work in 16 days. B is 60% more efficient
than A.
Quantity I: Time taken by A and B together to do the
work.
Quantity II: Time taken by A and B to do the work
together when A works at double his original efficiency
and B works at half his original efficiency.

2. Quantity I: Find amount after 4 years, if rate of
interest is 20% and the principal amount is Rs 8000
Quantity II: Find the amount after 4 years, if Rs 10,000
becomes Rs 12,000 in 2 years at compound interest.

3. Quantity I: Days in which B can complete work alone, if A and B can complete work in 40 days, B and C
in 20 days and C and A in 30 days.
Quantity II: Days in which B can complete work alone,
if A and B can complete work in 24 days and A is 50% more efficient than B.

4. Quantity I: Find the difference between the sum of roots and product of roots of the given equation.
2x^2 – 14x + 50 = 0
Quantity II: 24

5. Quantity I: Find the curved surface area of the cone of radius 7 cm and height 24cm.
Quantity II: Find the curved surface area of the cylinder of radius 7 cm and height 12 cm.

6. Quantity I: Total surface area of cylinder who radius is 7 cm and height is 10 cm
Quantity II: Total surface area of cuboid whose dimensions are 10×12×15 cm

7. Quantity I: A and B can do a piece of work in 12 hours and 18 hours resp. resp. If both started to work on alternate days with B starting the work, find the total
time taken to complete the whole work.
Quantity II: 15 hours

8. Quantity I: x, if 6x2 – 29 x – 20 = 0
Quantity II: y, if 6y2+ 13y – 15 = 0

9. Quantity I: The time taken by the boatman to cover 48 km upstream and 60 km downtream is same. If the speed of the stream is 1.5 km/hr. find the distance
covered by the boatman in still water in 8 hours.
Quantity II: 110 km

10. It takes (X-2) men to do a work in 2Y days and it takes (X+6) men to do the same work in Y days.
Quantity I: X
Quantity II: Y

SOLUTION

1. A 
I: A=16 days ; B = 16 * 100/160=10 days 
A+B together = 16*10/(26)=80/13 days 

II: A = 16 days ; B = 10 days

A(double efficiency) = 8 days ; B(half efficiency) = 20 
days
A+B together = 80/14 
hence I > II

2. B 
I: SI = 20*4 = 80% 
100% = 8000 
180% = Rs 14400 
II.
10000 : 12000 :: 12000 : x 
x = 12000*12000/10000 = Rs 14400 
II>I

3. B 
I:
A + B = 40……..3 
B + C = 20………6 ……..(LCM = 120) 
C+ A = 30……….4 
Total = 2 (A+B+C) = 3+6+4 = 13
So A+B+C = 13/2
(A+B+C) – (B+C) = 13/2 – 4 = 5/2 
So B can complete work in 120/(5/2) = 48 days 
II: Efficiency A …….. B = 3 : 2 
So days = 2 …. 3 
LCM of 2 and 3 is 6 
A = 2………6/2 = 3 
B = 3………6/3 = 2 
Total A+B = 3+2 = 5
So 6/5 == 24 
So 1 == 20 
So 3 == 60 
So II > I 

4. B 
From I: y = 2x^2 – 14x + 50 
sum of roots = -b/a = 14/2 = 7 
Product of roots = c/a = 50/2 = 25 
Required Difference = 25 – 7 = 18 
I < II

5. A
From I: CSA = (22/7)*7*[7^2 + 24^2]^1/2 = 550 cm^2 
From II: CSA = 2*(22/7)**12 = 528 cm^2 
I > II

6. B 
I: 2πr(r+h) = 748 cm2
II: 2 (lb+bh+lh) = 900 cm2 

7. B 
From I: Total work = 36 units 
Units of work done by A and B in one hour resp. = 3 
units and 2 units 
Work done by both of them in two hours = 5 units 
Work done by both of them in 14 hours = 7*5 = 35 units 
Remaining work = 36 – 35 = 1 unit 
Time taken by B to complete the remaining work = ½ 
hours 
Time taken to complete the work = 14.5 hours 
I < II

8. C 

x = 5/6, 4
y = -3, 5/6 

9. B 
From I: Let the speed of the boat in still water be x 
km/hr. 
48/(x – 1.5) = 60/(x+1.5) 
=> x = 13.5 km/hr. 
Required distance = 13.5*8 = 108 km 
I < II

10. E 
As the time is becoming half so means number of people 
have doubled; So (X+6)=2*(X-2)=>X=10
Using M1D1=M2D2 the value of Y cannot be found. Y 
can take any value. So we cannot determine a unique 

value of Y