__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