Quantitative Aptitude (Inequalities) Practice Questions (28 –
09 – 2017)
Directions
(1 – 15): In each of the following questions, read the given statement and
compare the Quantity I and Quantity II on its basis and give answer:
a) Quantity
I > Quantity II
b) Quantity
I < Quantity II
c) Quantity
I ≥ Quantity II
d) Quantity
I ≤ Quantity II
1. A 250
metres long train running at the speed of 100 kmph crosses another train
running in opposite direction at the speed of 60 kmph in 9 seconds.
Quantity I: The length of the other train
Quantity
II: The length of the first train
shrinks by 3/4th of that of the other train
2. Flipkart
listed two headphones for Rs. 476. One of the headphones was sold at a loss of
25% and the other at a gain of 29% and the company found that each headphone
was sold at the same price.
Quantity I: The cost
price of the headphone which was sold at 29% profit
Quantity
II: The selling price of the headphone
which was sold at 25% loss
3. In a
circle with centre O, PT and PS are tangents drawn to it from point P. If PT =
24 cm and OT = 10 cm
Quantity I: The length
of PO.
Quantity
II: Double the length of the hypotenuse of a right angled triangle the other
two sides of which are 8 cm and 15 cm respectively.
4. A box
contain 3 Samsung phones, 4 Vivo phones and 5 Lava phones.
Quantity I: If two
phones are drawn at random, the probability that both the phones are either
Lava or Samsung
Quantity
II: If two phones are drawn at random,
the probability that both the phones are either Vivo or Lava.
5. The sum
of the diameter and the circumference of circle A is 174 cm and the radius of
circle B is 14 cm less than the radius of circle A.
Quantity I: Twice the
radius of circle A
Quantity
II: The circumference of circle B
6. The
perimeter of a square is equal to twice the perimeter of a rectangle of length
13 cm and breadth 15 cm.
Quantity I: The
perimeter of a semicircle whose diameter is equal to the side of the square
Quantity
II: The perimeter of another semicircle
whose radius is 21 cm
7. Two
pipes A and B fill an empty tank in 40 minutes and 60 minutes respectively. If
both pipes are opened simultaneously
Quantity I:
After what certain time A should be closed so that the tank is filled in
36 minutes?
Quantity
II: If both are opened and A is closed after 10 minutes, how much further
time would it take for B to fill the bucket?
8. Mr.
Kapoor invested a certain amount in two schemes A and B offering compound
interest @ 8% pa and 10% pa respectively. If the total amount of interest in
two years was Rs. 6276 and the total amount invested was Rs. 33000,
Quantity I: The amount
invested in A
Quantity
II: The amount invested in B
9. The
ratio of the salary of Abdul to that of Fakir is 5 : 8. If the salary of Abdul
increases by 60% and that of Fakir decreases by 35% then the new ratio of their
salaries becomes 40 : 27.
Quantity I: Due to
extra leaves, the salary of Abdul gets deducted by 3/4
Quantity
II: Due to incentives, the salary of
Fakir gets increased by 12%
10. Quantity
I: The average of 15 number is 65, if
the average of the first eight number is 67 and that of the last eight number
is 63, the number that comes in the middle is:
Quantity
II: Ten less than the 7th number among
13 numbers the average of which is 51, moreover, the average of the first six
numbers of these 13 numbers is 52 and that of the last six is 46.
11. Quantity
I: Age of father, if age of Abhishek is 1/6th of
his father’s and 10 years after Abhinav’s age becomes half of Abhishek’s
father’s age.
Quantity
II: Age of father, if 5 years ago age of A’s father
was three times the age of A and 5 years hence his age will be double A’s age.
12. 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.
13. Quantity
I: Marks of new student, if the average marks of
25 students is55 and after the marks of new student also taken into
consideration, average increases by 2 marks
Quantity
II: Marks of new student, if the average marks of
20 students is 80 and after the marks of new student also taken into
consideration, average increases by 1.5 marks
14. 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
15. Quantity
I: x, if 6x^2 – 29 x – 20 = 0
Quantity
II: y, if 6y^2+ 13y – 15 = 0
Solutions:
1. A) Relative speed = (100+60) km/hr =
160 × 5/18 = 400/9 m/sec
Let
the length of the other train = x meters
(x
+ 250)/9 = 400/9 => (x + 250) = 400 => x = 150m
Quantity
I: The length of other train is 150 m.
Quantity
II: The length of the first train shrinks by 3/4th of that of the other train
Therefore,
3/4 of 150 = 112.5 m
Now,
length of the first train = 250 - 112.5 = 137.5 m
Hence,
Quantity I > Quantity II
2. B) Let the
cost price of Quantity II be x. Therefore, the cost price of Quantity I will be
= (476 – x)
QI QII
CP x (476 – x)
SP x × 75/100 (476 – x) × 129/100
Now
as both the SPs are equal, 3x/4 = (476 – x) × 129/100
or,
25x = (476 – x) × 43
or,
25x + 43x = 476 × 43
or,
68x = 476 × 43
or,
x = 301.
Quantity
I: The cost price of the headphone at 29% profit = 476 – 301 = 175
Quantity
II: Selling price of the headphone sold at 25% loss
=
301 × 75/100 = 225.75
Hence,
Quantity I < Quantity II
3. B)
4. B)
Quantity
I:
(3C2
+ 5C2)/12C2 = (3 + 10)/66 = 13/66
Quantity
II:
(4C2
+ 5C2)/12C2 = (6 + 10)/66 = 16/66
Hence,
Quantity I < Quantity II
5. B) Quantity I:
Let
the radius of circle A be r.
Therefore,
circumfernce of the circle A = 2πr
And
diameter = 2r
Thus,
2r + 2πr = 174
or,
r(1 + π) = 87
∴ r = 87/ (1 + 22/7) = (87 ×
7)/29 = 21 cm
∴ Quantitiy I will be 42 cm.
Now,
radius of the circle B = 21 – 14 = 7 cm
Quantity II:
Circumference
of the circle B = 2 × 22/7 × 7 = 44cm = QII
Hence,
Quantity I < Quantity II
6. B) Quantity I: Perimter of the square
= 2 × 2 (13 + 15) = 2 × 56 = 112 cm
let
the side of a square be a.
Then,
4a = 112 cm
∴ a = 28 cm
Diameter
of the circle = 28 cm
∴ Radius = 28/2 = 14 cm
∴ Perimeter of the semicircle = πr + 2r = (22/7 × 14) + 2 × 14 = 44 + 28 = 72cm
Quantity II: Perimeter
of another semicircle with a radius of 21 cm
=
22/7 × 21 + 2 × 21 = 66 + 42 = 108 cm
Hence,
Quantity I < Quantity II
7. B) Quantity I: Let the tap
A remains open for x minutes
The
efficiency equation will be x/40 + 36/60 = 1
⇒ 3x + 72 =
120 ⇒ 3x = 48
∴ x = 16 minutes.
Quantity II: Let
the tap B remains open for x extra minutes.
The
efficiency equation will be: 10/40 + (10 + x)/60 = 1
(10
+ x)/60 = 1 – ¼ = ¾
⇒ (10 + x) =
45
∴ x = 35 minutes
Clearly,
Quantity I < Quantity II
8. B) Let the
amount invested by Mr. Kapoor in scheme 'A' be x and and in scheme 'B' be
(33000 – x)
CI
Rate for two different schemes = 8% and 10%
We
can calculate the effective rate of interest
@ 8% for 2 years by applying the net% effect,
We
get
=
8 + 8 + (8 × 8)/100% = 16 + 0.64 = 16.64%
Similarly,
the effective rate of interest @ 10% for 2 years
=
10 + 10 + (10 × 10)/100% = 20 + 1 = 21%
Now,
as per the question
16.64%
of x + 21% of (33000 – x) = 6276
or,
16.64x + 21 × 33000 – 21x = 627600
or,
4.36x = 693000 – 627600
or,
4.36x = 65400
or,
x = 15000
So,
invested amount in scheme A = x = 15000
And,
invested amount in scheme B = (33000 – x) = 18000
Hence,
Quantity I < Quantity II.
9. E) Since no
absolute value related to salary is given, we can't find either of the
quantities.
Hence,
no relation can be established.
10. E) Quantity
I: Value of the middle number = (Total of first eight no. + Total
of last eight no.) – Total of 15 nos
=
(8 × 67 + 8 × 63) – (15 × 65) = (536 + 504) – 975 = 1040 – 975
= 65
Quantity II: 7th
number = Total of 13 nos. – (Total of first six no. + Total of last six
no.)
=
13 × 51 – (6 × 52 + 6 × 46)
=
663 – (312 + 276) = 663 – 588 = 75
Ten
less than 75 = 75 – 10 = 65
Hence,
Quantity I = Quantity II
11. B) I:
………………………Abhishek……………Father
………………………..…..1…………………….6
(1)
10
years after. Abhinav Father
.
1 ……………………2 (2)
Now
Abhivav is 10 years old so after 10 years he will be 20
Put
in (1)
father
= 40
Now
40-10 = 30
II:
.
A A’s father
5
years ago 1 3
5
years hence 1 2
So
3-2 = 1
1
== 10
3
== 30
So
30+5 = 35
12. 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
13. B) I: 55 +
2*25 = 105
II:
80 + 1.5*20 = 110
14. B) I: 2πr(r+h)
= 748 cm^2
II:
2 (lb+bh+lh) = 900 cm^2
15. C) x = 5/6, 4
y
= -3, 5/6
