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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: 2Ï€r(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