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Quadratic EQuations

Directions (1-5): Given question is followed by information given in two statements named as Quantity I and Quantity II. You have study the information along with the question and compare the value derived from Quantity I and Quantity II then answer the questions.
1. Sum of 8 consecutive even numbers is S1. (SBI PO - 2018)
Quantity I: Sum of second number and eighth number in S1.
Quantity II: Sum of third number and sixth number in S1.
Ans: Quantity I > Quantity II 
Explanation:
Let the numbers be x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12 and x + 14
Quantity I: x + 2 + x + 14 = 2x + 16
Quantity II: x + 4 + x + 10 = 2x + 14
∴ Quantity I > Quantity II

 

2. The cost price of 2 items A and B is same. The shopkeeper decided to mark the price 40% more than the CP of each item. A discount of 25% was given on item A and discount of 20% was given on item B. Total profit earn on both item was Rs.34.
Quantity I: CP of the items.
Quantity II: CP of any item which was sold at 12.5% profit and profit earned on it was Rs. 25.
Ans: Quantity I = Quantity II
Explanation:
Let the Cost Prices of A & B be Rs.100 each. Marked prices of each item will be Rs.140
SP of A = 75% of 140 = 105
SP of B = 80% of 140 = 112
Total profit on A & B = 217 - 200 = Rs.17
If the profit is Rs.17,
The cost price is Rs.100 each.
If the profit is Rs.34,
The cost price of each item is Rs.200.
Quantity I: Rs.200
Quantity II: 12.5% (profit) = Rs.25
100% (CP) =  × 25 = Rs.200
∴ Quantity I = Quantity II

 

3. There are 63 cards in a box numbered from 01 to 63. Every card is numbered with only 1 number.
Quantity I: Probability of picking up a card whose digits, if interchanged, result in a number which is 36 more than the number picked up.
Quantity II: Probability of picking up a card, the number printed on which is a multiple of 8 but not that of 16.
Ans: Quantity I > Quantity II 
Explanation:
Quantity I: When interchange the digits, number is 36 more than the original number
Possibilities are (04, 15, 26, 37, 48 and 59)
Probability = 
Quantity II: Multiple of 8 but not multiple of 16
Possibilities are (8, 24, 40 and 56)
Probability = 
∴ Quantity I > Quantity II

4. A girl walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that
she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked
back to school at a speed of 4 miles per hour. All the times, she walked in the same route.
Quantity I: Her average speed over the entire journey.
Quantity II: 4 mph.
Ans: Quantity I = Quantity II or No relation can be established
Explanation:
Quantity I: Let the distance between home and school be 'x' miles and times taken to travel three times to this distance be t1, t2 and t3

⇒ 4 mph
Quantity II: 4 mph
∴ Quantity I = Quantity II

5. If a speed of boat is 500% more than the speed of a current. (SBI PO - 2018)
Quantity I: If boat can travel a distance of 63 km in 3 hrs, in downstream then 'x' is the speed of the
boat in upstream (kmph).
Quantity II: 15 km/hr.
Ans: Quantity I = Quantity II or No relation
Explanation:
Quantity I: Let the speed of current = x
Speed of boat = x + 5x
Downstream speed = x + x + 5x = 7x
∴   = 3 ⇒ x = 3
Upstream speed = 6x - x = 5x = 15 kmph
Quantity II: 15 kmph
∴ Quantity I = Quantity II

6. Given question is followed by information given in two statements named as Quantity I and Quantity II. You have to study the information along with the question and compare the value derived from
Quantity I and Quantity II then answer the question.
Quantity I- x: x2 + x - 6 = 0
Quantity II- y: y2 + 7y + 12 = 0 (SBI PO - 2018)
Ans: Quantity I  Quantity II 
Explanation:
Quantity I: x2 + x - 6 = 0
⇒ (x + 3)(x - 2) = 0
⇒ x = -3, 2
Quantity II: y2 + 7y + 12 = 0
⇒ (x + 4)(x + 3) = 0
⇒ x = -4, -3
∴ x  y

Some more

Basic types of quadratic equations:
1) ax2 + bx + c = 0
2) ax2 − bx + c = 0
3) ax2 + bx − c = 0
4) ax2 − bx − c = 0
 In the first type of quadratic equation we can get two roots are negative.
​​​​​​​ In the second type of quadratic equation we can get two roots are positive.
​​​​​​​ In the third type of quadratic equation we can get one positive root and one negative root.
​​​​​​​In the fourth type of quadratic equation we can get one positive root and one negative root.
If we compare first two types of equations second equation is greater when compare to first equation the reason is second equation can get positive roots and first equation can get negative roots
If we compare third and fourth type of quadratic equations third type and fourth type we can get one positive and one negative root.
So, the relationship can’t be established.
e.g.:
1. I) ax​​​​​​​2 + bx + c = 0
    II) ay2 − by + c = 0
Sol: x < y
2. I) ax​​​​​​​2 + bx − c = 0
   II) ax​​​​​​​2 − bx − c = 0
Sol: No relation can be established.

Model Questions

1. I. x2 + 13x + 36 = 0
II. 3y2 − 28y + 64 = 0
a) If x < y    b) If x > y    c) If x > y    d) If x < y
e) If x = y or no relation can be established.
Sol: I.  x2 + 9x + 4x + 36 = 0
(x + 9) (x + 4) = 0
x = −9, −4
II. 3y2 − 16y − 12y + 64 = 0
(3y − 16) (y − 4) = 0

Clearly, x < y                            Ans: a


2. I. x2 + 3x − 40 = 0
II. 2y2 + 7y − 60 = 0
a) If x < y   b) If x > y    c) If x > y    d) If x < y
e) If x = y or no relation can be established.
Sol: I. x + 8x − 5x − 40 = 0
(x + 8) (x − 5) = 0
x = −8, 5
II. 2y2 + 15y − 8y − 60 = 0
(2y + 15) (y − 4) = 0

Clearly, no relation can be established         Ans: e


3. I. 2x2 − 21x + 45 = 0
II. 5y2 + 42y + 88 = 0
a) If x < y    b) If x > y    c)  If x > y    d) If x < y
e) If x = y or no relation can be established.
Sol: I. 2x2 − 15x − 6x + 45 = 0
(x − 3) (2x − 15) = 0

Clearly, x > y                                 Ans: b


4. I. 2x- x − 3 = 0
II. y2 − 11y + 30 = 0
a) If x < y    b) If x > y    c) If x > y    d) If x < y
e) If x = y or no relation can be established.
Sol: I. 2x2 − 3x + 2x − 3 = 0
(x + 1) (2x − 3) = 0
x = −1,

II. y2 − 6y − 5x + 30 = 0
(y − 5) (y − 6) = 0
y = 5, 6
Clearly, x < y                                   Ans: a


5. I. x2 − 2x − 24 = 0
II. y2 + 6y − 40 = 0
a) If x < y    b) If x > y    c) If x > y    d) If x < y
e) If x = y or no relation can be established.
Sol: I. x2 − 6x + 4x − 24 = 0
(x − 6) (x + 4) = 0
x = −4, 6
II. y2 + 10y − 4y − 40 = 0
(y + 10) (y − 4) = 0 ⇒ y = −10, 4
Clearly, no relation can be established            Ans: e


6. I. x2 − 5x − 14 = 0
II. y2 − 3y − 88 = 0
a) If x > y    b) If x > y   c) If x < y    d) If x < y
e) If x = y or no relation can be established between x and y.
Sol: I. x2 − 7x + 2x − 14 = 0
(x + 2) (x − 7) = 0
x = −2, 7
II. y2 − 11y + 8y − 88 = 0
(y + 8) (y − 11) = 0 ⇒ y = −8, 11
Clearly, no relation can be established             Ans: e


7. I. 7x2 + 13x + 6 = 0
II. 5y2 + 11y + 6 = 0
a) If x > y    b) If x > y     c) If x < y d) If x < y
e) If x = y or no relation can be established between x and y.
Sol: I. 7x2 + 7x + 6x + 6 = 0
(7x + 6) (x + 1) = 0

Clearly, x < y                            Ans: d


8. I. x2 + 16x + 60 = 0
II. y2 − 14y + 45 = 0
a) If x > y    b) If x > y    c) If x < y     d) If x <​​​​​​​ y
e) If x = y or no relation can be established between x and y.
Sol: I.  x2 + 10x + 6x + 60 = 0
(x + 6) (x + 10) = 0
x = −6, − 10
II. y2 − 9y − 5y + 45 = 0
(y − 5) (y − 9) = 0 ⇒ y = 5, 9
Clearly, x < y                               Ans: c


9. I. 7x2 + x − 8 = 0
II. 7y2 − 23y + 16 = 0
a) If x = y or no relation can be established
b) If x > y   c) If x < y   d) If x > y   e) If x < y
Sol: I. 7x2 + 8x − 7x − 8 = 0
(7x + 8) (x − 1) = 0


11. I. x2 − 23x + 132 = 0
II. y2 − 25y + 156 = 0
a) If x > y    b) If x > y
c) If x = y or no relation can be established
d) If y >​​​​​​​ x                   e) If y > x
Sol: I. x2 − 12x − 11x + 132 = 0
(x − 12) (x − 11) = 0
x = 11, 12
II. y2 − 12y − 13y + 156 = 0
(y – 12) (y – 13) = 0
y = 12, 13
Clearly, y >​​​​​​​ x                                       Ans: d


12. I. 15x2 − 8x + 1 = 0
II. 9y2 + 24y + 7 = 0
a) If x > y            b) If x > y
c) If x = y or no relation can be established
d) If y > x                 e) If y > x
Sol: I. 15x2 − 5x − 3x + 1 = 0
(5x − 1) (3x − 1) = 0


13. I. 5x2 − 4x − 105 = 0
II. 5y2 + 46y + 105 = 0
a) If x > y      b) If x > y
c) If x = y or no relation can be established
d) If y > x          e) If y > x
Sol: I. 5x2 − 25x + 21x − 105 = 0
(5x + 21) (x − 5) = 0


14. I. 4x2 − 20x + 25 = 0
II. y2 − 400 = 0
a) If x > y
b) If x = y or no relation can be established between x and y.
c) If x > y          d) If x < y               e) If x < y
Sol: I. 4x2 − 10x − 10x + 25 = 0
(2x − 5) (2x − 5) = 0
​​​​​​​

 

Posted Date : 10-02-2022

గమనిక : ప్రతిభ.ఈనాడు.నెట్లో వచ్చే ప్రకటనలు అనేక దేశాల నుండి, వ్యాపారస్తులు లేదా వ్యక్తుల నుండి వివిధ పద్ధతులలో సేకరించబడతాయి. ఆయా ప్రకటనకర్తల ఉత్పత్తులు లేదా సేవల గురించి ఈనాడు యాజమాన్యానికీ, ఉద్యోగస్తులకూ ఎటువంటి అవగాహనా ఉండదు. కొన్ని ప్రకటనలు పాఠకుల అభిరుచిననుసరించి కృత్రిమ మేధస్సు సాంకేతికతతో పంపబడతాయి. ఏ ప్రకటనని అయినా పాఠకులు తగినంత జాగ్రత్త వహించి, ఉత్పత్తులు లేదా సేవల గురించి తగిన విచారణ చేసి, తగిన జాగ్రత్తలు తీసుకొని కొనుగోలు చేయాలి. ఉత్పత్తులు / సేవలపై ఈనాడు యాజమాన్యానికి ఎటువంటి నియంత్రణ ఉండదు. కనుక ఉత్పత్తులు లేదా సేవల నాణ్యత లేదా లోపాల విషయంలో ఈనాడు యాజమాన్యం ఎటువంటి బాధ్యత వహించదు. ఈ విషయంలో ఎటువంటి ఉత్తర ప్రత్యుత్తరాలకీ తావు లేదు. ఫిర్యాదులు తీసుకోబడవు.

 

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