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Equations and Inequations Questions

121. In the question, two equations numbered I and II are given. Solve both the equations and give answer
(a) x≥y
(b) x≤y
(c) x<y
(d) x>y
(e) Relationship between x and y cannot be established

I. 2x2+ 5x+3 = 0
II. y2+9y+14 = 0

Correct Ans:D
Explanation:
From I,
2x2+ 5x+3 = 0
(2x+3) (x+1) = 0
x = -3/2, -1
From II,
y2+9y+14 = 0
(y+7) (y+2) = 0
x>y
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122. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Statment: There are three positive numbers a, b and c. The average of a and b is less than the average of b and c by 1.
Quantity I: value of c.

Quantity II: value of a.

Correct Ans:Quantity I > Quantity II
Explanation:
Given that, The average of a and b is less than the average of b and c by 1.
---> (a + b)/2 = [(b + c)/2] - 1
---> [(b + c)/2] - (a + b)/2 = 1
---> (b + c - a - b) / 2 = 1
---> (c - a) / 2 = 1
---> c - a = 2

---> Quantity I: c = a + 2 ---> which implies 'c' is 2 more than 'a'
and Quantity II: a = c - 2 ---> which implies 'a' is 2 less than 'c'

Hence, Quantity I > Quantity II
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123. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Quantity I: In how many ways the word 'SCOOTER' can be arranged such that 'S' and 'R' are always at two ends?

Quantity II: A tricolor flag is to be formed having three adjacent strips of three different colors chosen from six different colors. How many different colored flags can be formed with different design in which all the three strips are always in horizontal positions?

Correct Ans:Quantity I = Quantity II or relation cannot be established
Explanation:
Quantity I:
In the word SCOOTER ---- the letter 'O' is repeated 2 times.
And S and R are considered as one element.
So, (S, C, 2O, T, E, R)
Required number of ways = (5! * 2!) / 2!
= 5!
= 5 * 4 * 3 * 2 * 1
= 120 ways

Quantity II:
First strip can be coloured in 6 ways;
second strip can be coloured in 5 ways; and
third strip can be coloured in 4 ways.

Hence, all the three strips can be coloured in 6 * 5 * 4 ways = 120 ways.

So, Quantity I = Quantity II
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124. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

A hemisphere of radius 4 cm is to be shipped in a shipping box of rectangular shape. The dimension of the box are consecutive odd numbers.
Quantity I: Minimum volume of box required to have the shipment possible.
Quantity II: 960 cm3

Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
The minimum length of any side of rectangular box should be greater than the diameter of the hemisphere.
Given Radius of hemisphere = 4 cm.
Diameter of hemisphere = 2 * Radius = 8 cm

Also given that, the dimension of the rectangular box are consecutive odd numbers.
So for odd number, minimum length= 9, next side =11, next side=13
Hence, Minimum Volume of box = length * width * height
= 9 * 11 * 13
= 1287 cm3

Given Quantity II = 960 cm3
Hence, Quantity I > Quantity II
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125. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

1 > a > 0 > b
Quantity I: Value of [(a + b)2 - a2 - b2] / [(a + b)2 - (a - b)2]

Quantity II: 1/ [2(ab3 + ab]

Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
[(a + b)2 - a2 - b2] / [(a + b)2 - (a - b)2]
---> [a2 + b2 + 2ab - a2 - b2] / [a2 + b2 + 2ab - a2 - b2 + 2ab]
---> 2ab / 4ab
---> 1 / 2

Quantity II:
1/ [2(ab3 + ab]
---> 1/ [2ab (b2 + 1)]
Given that 0 > b, so if we subsitute b = -1 and
From the statement 1 > a > 0, say a = 0.5, Then the above eqn becomes,
---> 1/ [2 * 0.5 * (-1) (-12 + 1)]
---> 1/ [-1 (1 + 1)]
---> 1/ [-1 (1 + 1)]
---> 1/ (-2)

So, Quantity I > Quantity II
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126. Compare the value of 2 quantities given in the question and give answer.

Quantity I: At simple Interest, a sum becomes 3 times in 20 year. Find the time, in which the sum will be double at the same rate of interest.
Quantity II: Simple interest for the sum of Rs. 1500 is Rs. 30 in 4 year and Rs. 60 in 8 year. Find the rate of simple interest.

Correct Ans:Quantity I > Quantity II
Explanation:
Formula for Simple Interest:
SI = PNR/t
SI = Simple Interest
P = Principle Amount
N = Number of times compounded in one t
R = Rate
t =Time period
Quantity-I
Let sum = P, then for 20 year
SI = 3P- 2P = 2P
2P= P* R *T/100
R=10%
For the sum to be double
SI = 2P-p= P
P = P*10*N/100
N = 10 years
Quantity-II
SI in 8 years= 60
SI in 4 years= 30
so,
1500* R * 8/100 - 1500*R*4/100 = 60 -30
1500*R*4/100 = 30
6000R/100 = 30
R = 30/60 = 0.5%
Hence, Quantity I > Quantity II
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127. Compare the value of 2 quantities given in the question and give answer.

Quantity I: Two numbers are respectively 20% and 50% more than a 3rd number. What is the percentage of 2nd with respect to 1st?
Quantity II: From 2008 to 2009, the sales of a book decreased by 80%. If the sales in 2010 was the same as in 2008, by what percent did it increase from 2009 to 2010?

Correct Ans:Quantity II > Quantity I
Explanation:
Quantity-I:
Let us assume that the 3rd number is 100, so that 1st and 2nd numbers are 20% and 50% more. so,
First number =120
Second number= 150
Third number = 100
Required percentage of 2nd with respect to 1st = (150/120)*100
= 125%
Quantity-II
Sale in 2008= 100
Sale in 2009= 20
Sale in 2010= 100
Required percentage increase= (80/20)*100
= 400%
Hence Quantity II> Quantity I
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128. Compare the value of 2 quantities given in the question and give answer.

Two trains of lengths 50m and 65m are moving in the same direction at 18 m/s and 17m/s respectively.
Quantity I: Time taken by the faster train to cross the slower train.
Quantity II: Time taken by the slower train to cross a platform of length 156 m.

Correct Ans:Quantity I > Quantity II
Explanation:
Quantity-I
Relative speed = 18-17= 1 m/s
Required time = (50+65)/1 = 115 seconds
Quantity-II
Time taken to cross a platform of length 156 m
= (65+156)/17
= 13 seconds
Hence Quantity I > Quantity II = 13 seconds
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129. Compare the value of 2 quantities given in the question and give answer.(Compare only numerical values)

Quantity A: The least prime number greater than 24
Quantity B: The greatest prime number less than 28

Correct Ans:if quantity A > quantity B
Explanation:
For the integers greater than 24, note that 25, 26, 27, and 28 are not prime numbers, but 29 is a prime number, as are 31 and many other greater integers.
Thus, 29 is the least prime number greater than 24, and Quantity A is 29.
For the integers less than 28, note that 27, 26, 25, and 24 are not prime numbers, but 23 is a prime number, as are 19 and several other lesser integers.
Thus, 23 is the greatest prime number less than 28, and Quantity B is 23.
The correct answer is Choice A, Quantity A is greater.
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130. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 2x2 -7x +3 = 0
II. 2y2 - 7y + 6 = 0

Correct Ans:E
Explanation:
I. 2x2 -7x +3 = 0
2x2 -6x - x +3 = 0
(x-3) (2x-1) = 0
x = 3, 1/2
II. 2y2 - 7y + 6 = 0
2y2 - 4y - 3y + 6 = 0
(y-2) (2y-3) = 0
y = 2, 3/2
x and y are not related in any way. The relationship between x and y cannot be determined.
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131. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 2x2 + 11x + 14 = 0
II. 2y2 + 15y + 28 = 0

Correct Ans:C
Explanation:
I. 2x2 + 11x + 14 = 0
2x2 + 4x + 7x + 14 = 0
(x+2) (2x+7) = 0
x = -2,-7/2
II. 2y2 + 15y + 28 = 0
2y2 + 8y + 7y + 28 = 0
(y+4) (2y+7) = 0
y = -4,-7/2
x ≥ y
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132. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 9x2 - 45x + 56 = 0
II. 4y2 - 17y + 18 = 0

Correct Ans:D
Explanation:
I. 9x2 - 45x + 56 = 0
9x2 - 45x + 56 = 0
3x(3x-8) - 7 (3x-8) = 0
(3x-8) (3x-7) = 0
x = 8/3, 7/3

II. 4y<sup>2</sup> — 17y + 18 = 0
4y<sup>2</sup> — 8y - 9y + 18 = 0
(y-2) (4y-9) = 0
y =2, 9/4
x>y

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133. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I. 4x2 + 16x + 15 = 0
II. 2y2 +3y +1 = 0

Correct Ans:A
Explanation:
I. 4x2 + 16x + 15 = 0
4x2 + 10x + 6x + 15 = 0
2x(2x+5) + 3(2x+5) = 0
(2x+5) + (2x+3) = 0
x = -5/2, -3/2
II. 2y2 + 3y + 1 = 0
2y2 + 2y + y + 1 = 0
(y+1) (2y+1) = 0
y=-1, -1/2
y>x
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134. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.
(a) if x < y
(b) if x ≤ y
(c) if x ≥ y
(d) if x > y
(e) relationship between x and y cannot be determined

I.9x2 - 36x + 35 = 0
II. 2y2 - 15y- 17 = 0

Correct Ans:E
Explanation:
I. 9x2 - 36x + 35 = 0
9x2 - 21x - 15x + 35 = 0
3x (3x-7) - 5 (3x-7) = 0
(3x-7) (3x-5) = 0
x = 5/3 , 7/3
II. 2y2- 15y- 17 = 0
2y2 -17y + 2y - 17 = 0
(y+1) (2y-17) = 0
y = -1, 17/2
x and y are not related in any way. The relationship between x and y cannot be determined.
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135. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Quantity I: How many numbers greater than 1000 but less than 5000 can be made from digits 0, 1, 3, 5, 6?

Quantity II: There are 5 blue flags, 4 red flags and 3 green flags, in Debu’s wardrobe. He has to select 4 flags from this set. In how many ways can he select these four flags such that there is at least one blue flag and exactly one green flag in them (Do not consider that the flags are in pairs)?

Correct Ans:Quantity I > Quantity II
Explanation:
Quantity I:
When not written about repetition, take that repetition to be allowed.
Numbers between 1000 to 5000 means 4 digit numbers. So 4 boxes.
Now 0, 5 and 6 cannot be placed in Box 1 (Thousandth place). Because, 0 will make it 3-digit number and 5 & 6 will make number greater than 5000.
So there are 2 choices (ie., numbers 1 and 3) for Box 1 (Thousandth place), and all 5 choices (ie., numbers 0, 1, 3, 5, 6) for Box 2, 3, and 4 (Hundredths, Tenths and units places).
So total numbers = 2C1 * 5C1 * 5C1 * 5C1 ---> (since 4 digit number)
= 2*5*5*5
= 250

Quantity II:
No. of Blue flags = 5;
No. of Red flags = 4;
No. of Green flags = 3

In the selected four flags ---> there is at least one blue flag and exactly one green flag
So, the possible selections are:
Way 1: 1 blue flag out of 5 * 1 green flag out of 3 * 2 Red flags out of 4
----> 5C1 * 3C1 * 4C2
----> 5 * 3 * 6
----> 90

Way 2: 2 blue flag out of 5 * 1 green flag out of 3 * 1 Red flags out of 4
----> 5C2 * 3C1 * 4C1
----> 10 * 3 * 4
----> 120

Way 3: 3 blue flag out of 5 * 1 green flag out of 3
----> 5C3 * 3C1
----> 10 * 3
----> 30

So, Total No. of ways = 90 + 120 + 30 = 240

here, Quantity I > Quantity II.
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136. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Ram invested Rs. P in scheme A and Rs. 2P in scheme B, for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% per annum. Respective ratio between the interest earned from scheme A and that earned from scheme B was 8 : 21.
Quantity I: Rate of interest offered by scheme A.

Quantity II: Rate of interest offered by scheme C (simple interest per annum), when Rs. 1600 is invested for 3 years earns an interest of Rs. 384.

Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
Let the rate of interest offered by scheme A ie., simple interest = R% per annum.
Given Time, (n) = 2 years.

Then, Simple interest, S.I = (p x n x r)/100
= (P * 2 * R)/100
= 2PR/100

Now, Compound Interest, C.I = Amount - Principal
Amount = P x [1+ (r/100)]n
C.I = P{[1+ (r/100)]n - 1}
So, Compound Interest received by Scheme B, C.I = 2P{[1+ (10/100)]2 - 1}
= 2P{[1+ (1/10)]2 - 1}
= 2P{[(10+ 1)/10]2 - 1}
= 2P{[11/10]2 - 1}
= 2P{121/100 - 1}
= 2P{121 - 100}/100
= 2P{21/100}
= 42P/100

Given that, S.I from scheme A : C.I from scheme B = 8 : 21
----> (2PR/100) : (42P/100) = 8 : 21
----> R : 21 = 8 : 21
----> R = 8
Hence, Quantity I: Rate of interest offered by scheme A, R = 8% per annum.

Now, Quantity II:
Simple interest --> Principal, P = Rs. 1600
n = 3 years
S.I = Rs. 384
Then, Simple interest, S.I = (p x n x r)/100
---> 384 = (1600 * 3 * r)/100
---> 384 = 48r
---> r = 8
Hence, Rate of interest offered by scheme C = 8% per annum.

Here Quantity I = Quantity II.
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137. In the following question, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
I. 3x2 "“ 25x + 52 = 0
II. 3y2 "“ 17y + 20= 0

Correct Ans:If x≥y
Explanation:
I. 3x2 â€“ 25x + 52 = 0
3x2 - 12x - 13x + 52 = 0
3x(x - 4) - 13(x - 4) = 0
(x - 4)(3x - 13) = 0
x = 4, 13/3 = 4, 4.3

II. 3y2 â€“ 17y + 20= 0
3y2 - 12y - 5y + 20= 0
3y(y - 4) - 5(y - 4) = 0
(y - 4)(3y - 5) = 0
y = 4, 5/3 = 4, 1.7
Hence, x≥y
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138. If x = 332, y = 333, z = 335 then the value of x3 + y3 + z3 "“ 3xyz = ?

Correct Ans:7000
Explanation:
(x3 + y3 + z3 â€“ 3xyz) = (x + y + z) (x2 + y2 + z2 â€“ xy â€“ yz â€“ zx)
Also we know that,
(x3 + y3 + z3 â€“ 3xyz) = (x + y + z) * 1/2[(x â€“ y)2 + (y â€“ z)2 + (z â€“ x)2]
On putting the values, we get
= 1000 Ã— 1/2[(332 â€“ 333)2 + (333 â€“ 335)2 + (335 â€“ 332)2
= 1000 Ã— 1/2[1 + 4 + 9]
= 7000
Hence, option C is correct.
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139. Quantity 1: Perimeter of a Square having its side is equal to hypotenuse of a right-angled triangle having its sides 8mtrs and 6mtrs respectively.
Quantity 2: Perimeter of a Rhombus having its diagonals d1 and d2 are 12mtr and 16mtr respectively.

Correct Ans:Quantity 1 = Quantity 2 or Relationship can’t be established.
Explanation:
Quantity 1:
Hypotenuse of a triangle = √(a²+b²)
= √(8²+6²)
= √(64+36)
= √100
= 10 mtrs
And Perimeter of the square = 4a = 4 × 10 = 40mtrs.

Quantity 2:
Diagonals of rhombus intersect at 90°angle.
The diagonals and the sides of the rhombus form right triangle.
One side of the right triangle is 16/2= 8mtrs
Other side of the right triangle is 12/2= 6mtrs
Then the third side of the triangle = √(a²+b²)
= √(8²+6²)
= 10mtrs
And Perimeter of the rhombus = 4a = 4 × 10 = 40mtrs.
So Quantity 1 is equal to Quantity 2.
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140. Study the following question carefully and choose the right answer.
If x = âˆš3 + âˆš4 + âˆš5 than x4 "“ 8x3 + 8x2 + 32x = ?

Correct Ans:44
Explanation:
x = âˆš3 + âˆš4 + âˆš5
x â€“ 2 = âˆš3 + âˆš5
Squaring on both sides we get,
x2 â€“ 4x + 4 = 8 + 2âˆš15
x2 â€“ 4x â€“ 4 = 2âˆš15
Squaring on both sides we get,
(a - b - c)Â² = aÂ² + bÂ² + cÂ² - 2ab + 2bc - 2ac
x4 + 16x2 + 16 â€“ 8x3 + 32x â€“ 8x2 = 60
x4 â€“ 8x3 + 8x2 + 32x = 44
Hence, option C is correct.
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