**Directions (1 – 5): In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.**

**1. I. 17x**

^{2}+ 48x = 9; II. 13y^{2 }– 32y –21 = 0
A) x < y

B) x > y

C) x ≥ y

D) x ≤ y

E) x = y or the relationship cannot
be established

**2. I. 10a**

^{2}– 21a – 10 = 0; II. 40b^{2}– 19b – 14 = 0
A) a < b

B) a > b

C) a = b OR the relationship cannot
be determined

D) a ≥ b

E) a ≤ b

**3. I. 17x**

^{2}– 153x + 238 = 0; II. 2y^{2}+ 3y – 14 = 0
A) x < y

B) x > y

C) x = y OR the relationship cannot
be determined

D) x ≥ y

E) x ≤ y

**4. I. 13x**

^{2}– 156x – 364 = 0; II. y^{2}– 6y – 16 = 0
A) x < y

B) x > y

C) x = y OR the relationship cannot
be determined

D) x ≥ y

E) x ≤ y

**5. I. √(899x) + √1297 = 0; II. (255)**

^{1/4}y + (215)^{1/3}= 0
A) x > y

B) x ≥ y

C) x < y

D) x ≤ y

E) x = y Or the relationship cannot
be established.

**6. Quantity A:**A is 75% of B then what % of A is B?

**Quantity B:**Isaac scores 73 marks in Mathematics, 98 marks in Physics and 75 in Chemistry. If the maximum marks in Mathematics, Physics and Chemistry are 75,100,90 respectively. What will be the % of total marks obtained with respect to the sum of maximum marks of Mathematics and Chemistry?

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B

**7. Quantity A:**Walking at 5/4 of his usual pace, a man reaches his office 25 minutes earlier. Find his usual time.

**Quantity B:**Two horses are running in an opposite direction on a circular track of 9858 m. Speed of horses are 32 km/h and 42 km/h respectively. When will they meet for the first time?

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B

**8. Quantity A:**The mean of 45 observations was 40. It was found later that an observation 36 was wrongly taken as 63. The corrected new mean is

**Quantity B:**The average age of a husband and his wife was 26 years at the time of marriage. After five years they have a one-year child. The twice of average age of the family now is

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B

**9. Based on the given information, determine the relation between the two quantities**

**Quantity A:**In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then the 13th term is:

**Quantity B:**That value of n, for which (a

^{n+1}+ b

^{n+1})/(a

^{n}+ b

^{n}) is the arithmetic mean of a and b?

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B

**10. Based on the given information, determine the relation between the two quantities**

**A function is defined as f(x) = x**

^{2}– 3x. Then**Quantity A:**The value of f(– 1).

**Quantity B:**The value of f(4).

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B

**Solutions:**

**1. E)**I. 17x

^{2}+ 48x = 9

⇒ 17x

^{2}+ 48x – 9 = 0
⇒ 17x

^{2}+ 51x – 3x – 9 = 0
⇒ 17x(x + 3) – 3(X + 3) = 0

⇒ (x + 3)(17x – 3) = 0

Then, x = - 3 or x = +
3/17

II. 13y

^{2 }– 32y –21 = 0
⇒ 13y

^{2}– 39y + 7y – 21 = 0
⇒ 13y( y – 3) + 7(y – 3) = 0

⇒ (y – 3)(13y + 7) =
0

Then, y = + 3 or y = -
7/13

So, when x = - 3, x <
y for y = + 3 and x < y for y = - 7/13

And when x = + 3/17, x
< y for y = + 3 and x > y for y = - 7/13

∴ So, we can observe that no clear relationship cannot be
determined between x and y.

**2. C)**I. 10a

^{2}– 21a – 10 = 0

⇒ 10a

^{2}– 25a + 4a – 10 = 0
⇒ 5a(2a – 5) + 2(2a – 5) = 0

⇒ (2a – 5)(5a + 2) =
0

Then, a = + 5/2 or a = -
2/5

II. 40b

^{2}– 19b – 14 = 0
⇒ 40b

^{2}– 35b + 16b – 2 = 0
⇒ 5b(8b – 7) + 2(8b – 7) = 0

⇒ (8b – 7)(5b + 2) =
0

Then, b = + 7/8 or b
= - 2/5

So, when a = + 5/2, a
> b for b = + 7/8 and a > b for b = - 2/5

And when a = - 2/5 , a
< b for b = + 7/8 and a = b for b = - 2/5

∴ So, we can observe that no clear relationship cannot be
determined between a and b.

**3. D)**I. 17x

^{2}– 153x + 238 = 0

⇒ x

^{2}– 9x + 14 = 0 [Dividing both sides by 17 ]
⇒ x

^{2}– 7x – 2x + 14 = 0
⇒ x(x – 7) – 2(x – 7) = 0

⇒ (x – 7)(x – 2) = 0

Then, x = + 7 or x = + 2

II. 2y

^{2}+ 3y – 14 = 0
⇒ 2y

^{2}– 4y + 7y – 14 = 0
⇒ 2y(y – 2) + 7(y – 2) = 0

⇒ (y – 2)(2y + 7) = 0

Then, y = + 2 or y = -
7/2

So, when x = + 7, x >
y for y = + 2 and x > y for y = - 7/2

And when x = + 2, x = y
for y = + 2 and x > y for y = - 7/2

∴ So, we can clearly observe that x ≥ y.

**4. C)**I. 13x

^{2}– 156x – 364 = 0

⇒ x

^{2}– 12x – 28 = 0 [Dividing both sides by 13 ]
⇒ x

^{2}– 14x + 2x – 28 = 0
⇒ x(x – 14) + 2(x – 14) = 0

⇒ (x – 14)(x + 2) =
0

Then, x = + 14 or x = - 2

II. y

^{2}– 6y – 16 = 0
⇒ y

^{2}– 8y + 2y – 16 = 0
⇒ y(y – 8) + 2(y – 8) = 0

⇒ (y – 8)(y + 2) = 0

Then, y = + 8 or y = - 2

So, when x = + 14, x >
y for y = + 8 and x > y for y = - 2

And when x = - 2, x <
y for y = + 8 and x = y for y = - 2

∴ So, we can observe that no clear relationship cannot be
determined between x and y.

**5. A)**According to the given equations,

√(899x) + √1297 = 0

899 is not a perfect
square, but 900 is a perfect square of 30. Similarily, 1297 is not a perfect
square but 1296 is a perfect square of 36. So, we are going to consider √899 as
30 and √1297 as 36, because it is not going to affect the question.

30x + 36 = 0

30x = -36

x = -6/5

II. (255)

^{1/4}y + (215)^{1/3}= 0
255 does not have a
perfect fourth root, but 256 is a perfect fourth root of 4. Similarily, 215
does not have a perfect cube root but 216 is a cube root of 6. So, we are going
to consider (255)

^{1/4}as 4 and (215)^{1/3}as 6, because it is not going to affect the question.
4y + 6 = 0

y = -6/4

y = -3/2

It is clearly seen that y
˂ x.

Hence, (A) is correct.

**6. B)**First we will find Quantity A,

**Quantity A:**

A = 75% of B

⇒ A = 0.75B

⇒ B = A/0.75 = 1.33A = 133.33% of A

∴ B = 133.33% of A

Now,

**Quantity B:**

Sum of maximum marks of
Mathematics and Chemistry = 75 + 90 = 165

Total marks obtained by
Isaac = 73 + 98 + 75 = 246

Percentage of total marks
obtained with respect to the sum of maximum marks of Mathematics and Chemistry
= (246/165) × 100 = 149.09%

∴ Percentage of total marks obtained with respect to the sum
of maximum marks of Mathematics and Chemistry = 149.09%

∴ Quantity A < Quantity B

**7. A)**First we will find Quantity A,

**Quantity A:**

Let the original speed
and time be s km/hr and t hrs respectively.

∵ Distance = Speed × Time

⇒ The distance travelled by the man = s × t = st km

At 5/4 of his usual
speed, the time taken by the man = t – (25/60) = [t – (5/12)]
hrs

(∵ 1 hour = 60 min)

⇒ The distance traveled = (5s/4) × [t – (5/12)] = 1.25st – (25s/48)

But the distance traveled
in both the cases will be the same.

⇒ st = 1.25st – (25s/48)

⇒ 0.25st = 25s/48

⇒ t = 100/48 = 2.08 hrs

∴ His usual time = 2.08 hrs

Now,

**Quantity B:**

Let both horses meet
after T minutes.

32000 m is covered by
first horse in 60
min. (∵ 1 km = 1000 m and 1 hr = 60 min)

Distance covered by the
first horse in T min = (32000 × T)/60

Likewise, Distance
covered by the second horse in T min = (42000 × T)/60

Given that the two horses
are running in an opposite direction on a circular track of 9858 m.

⇒32000×T/60+42000×T/60=9858

⇒ T ≈ 7.993 minutes

∴ The two horses will meet for the first time in 7.993
minutes.

∴ Quantity A > Quantity B

**8. B)**First we will find Quantity A,

**Quantity A:**

Sum of 45 observations =
40 × 45 = 1800

Correct sum of the 45
observations = 1800 – 63 + 36 = 1773

Corrected mean = 1773/45
= 39.4

∴ Corrected mean = 39.4

Now,

**Quantity B:**

Sum of the ages of husband
and wife at time of marriage = 26 × 2 = 52

Sum of the ages of the
family after 5 years when they have a one-year old child = 52 + 5 + 5 + 1 = 63

Average age of the family
after 5 years when they have a one-year old child = 63/3 = 21

∴ Twice of average age of the family now = 21 × 2 = 42

∴ Quantity A < Quantity B

**9. E)**First we will find Quantity A,

**Quantity A:**

We know that for an
A.P.(Arithmetic progression),

The n

^{th}term, T_{n}= a + (n – 1)d
Where,

a = First term

d = Common difference

According to the
question,

5T

_{5}= 8T_{8}
⇒ 5[a + (5 – 1)d] = 8[a +
(8 – 1)d]

⇒ 5a + 20d = 8a + 56d

⇒ 3a + 36d = 0

⇒ 3(a + 12d) = 0

⇒ a + (13 – 1)d = 0

⇒ T

_{13}= 0
∴ T

_{13}= 0
Now,

**Quantity B:**

∵ Arithmetic mean of a and b is (a + b)/2.

(a

^{n+1}+ b^{n+1})/(a^{n}+ b^{n}) = (a + b)/2
⇒ 2(a

^{n+1}+ b^{n+1}) = (a^{n}+ b^{n})(a + b)
⇒ 2a

^{n+1}+ 2b^{n+1}= a^{n+1}+ a^{n}b + b^{n}a + b^{n+1}
⇒ a

^{n+1}+ b^{n+1}= a^{n}b + b^{n}a
⇒ a

^{n}(a – b) = b^{n}(a – b)
⇒ a

^{n}= b^{n}
⇒ (a/b)

^{n}= 1
⇒ (a/b)

^{n}= (a/b)^{0}(∵ m^{0}= 1)
∵ Bases are equal, equate the powers.

⇒ n = 0

∴ Quantity A = Quantity B.

**10. E)**Given,

f(x) = x

^{2}– 3x
First we will find
Quantity A,

**Quantity A:**

f(– 1) = (– 1)

^{2}– 3(–1) = 1 + 3 = 4**Quantity B:**

f(4) = 4

^{2}– 3(4) = 16 – 12 = 4
∴ f(– 1) = 4 = f(4)

∴ Quantity A = Quantity B.