758 + Interview Questions in GENERAL QA IN ALGEBRA Page 12 InterviewSolution

551.	यदि `x-1/x=3` है तो `x^(3)-1/(x^(3))` का मान क्या होगा?A. 32B. 36C. 40D. 49
Answer» Correct Answer - B `x-1/x=3` `x^(3)-1/(x^(3))-3(x-1/x)=27` `x^(3)-1/(x^(3))=27+3xx3` `=27+9=36`

Discussion

552.	यदि `x+y=7` है तो `x^(3)+y^(3)+21xy` का मान क्या होगा?A. 243B. 143C. 343D. 443
Answer» Correct Answer - C `x+y=7` (cubing both sides) `(x+y)^(3)=(7)^(3)` `x^(3)+y^(3)=3(x+y)xy=343` `x^(3)+y^(3)+21xy=343` ` x^(3)+y^(3)+21xy=343`

Discussion

553.	यदि `x+1/x=3` है तो `(x^(5)+1)/(x^(5))` का मान ज्ञात करें।A. 322B. 126C. 123D. 113
Answer» Correct Answer - C `x+1/x=3` (Squaring both sides) `x^(2)+1/(x^(2))=7` On cubing both sides `x^(3)+1/(x^(3))+3x xx 1/x(x+1/x)=27` `implies x^(3)+1/(x^(3))+3xx3=27` `x^(3)+1/(x^(3))=18` `:. (x^(3)+1/(x^(3)))(x^(2)+1/(x^(2)))=18xx7` `(x^(5)+1/(x^(5)))+(x+1/x)+126` `(x^(5)+1/(x^(5))+3=126` `=(x^(5)+1/(x^(5)))=123`

Discussion

554.	यदि `x=sqrt(3)+sqrt(2)` है तो `(x^(3)+1/(x^(3)))` का मान ज्ञात करें।A. `6sqrt(3)`B. `12sqrt(3)`C. `18sqrt(3)`D. `24sqrt(3)`
Answer» Correct Answer - C `x=sqrt(3)+sqrt(2)` `:. 1/x=sqrt(3)-sqrt(2)` `x^(3)+1/(x^(3))` `:. x^(3)=(sqrt(3)+sqrt(2))^(3)` `=(sqrt(3))^(3)+(sqrt(2))^(3)+3xxsqrt(3)xxsqrt(2)(sqrt(3)+sqrt(2))` `=3sqrt(2)+2sqrt(2)+3sqrt(6)(sqrt(3)+sqrt(2))` `=3sqrt(3)+2sqrt(2)+9sqrt(2)+6sqrt(3)` `x^(3)=9sqrt(3)+11sqrt(2)` `1/(x^(3))=9sqrt(3)-11sqrt(2)` `x^(3)+1/(x^(3))=9sqrt(3)+11sqrt(2)+9sqrt(3)-11sqrt(2)` `=18sqrt(3)` Alternate: `x=sqrt(3)+sqrt(2)` `1/x=sqrt(3)-sqrt(2)` and `x+1/x=sqrt(3)+sqrt(2)+sqrt(3)-sqrt(2)` `=2sqrt(3)` `x^(3)+1/(x^(3))=(x+1/x)^(3)-3(x+1/x)` `=(2sqrt(3))^(3)-3(2sqrt(3))` `=24sqrt(3)-6sqrt(3)=18sqrt(3)`

Discussion

555.	यदि `x` वास्तविक संख्या है तो `x+1/x!=0` और `x^(3)+1/(x^(3))=0` है तो `(x+1/x)^(4)` का मान ज्ञात करें।A. 4B. 9C. 16D. 25
Answer» Correct Answer - B `x^(3)+1/(x^(3))=0` `(x+1/x)^(3)-3x xx 1/x(x+1/x)=0` `(x+1/x)^(3)-3(x+1/x)=0` `(x+1/x)^(3)=3(x+1/x)` `(x+1/x)^(2)=3` `((x+1/x)^(2))^(2)=(3)^(2)` `(x+1/x)^(1)=9`

Discussion

556.	यदि `ab=2a-3b+ab` है तो `35+5**3` किसके बराबर होगा?A. 22B. 24C. 25D. 28
Answer» Correct Answer - A `35+53` `implies35=2xx3-3xx5+3xx5` `=6-15+15=6` `implies 53=2xx5-3xx3+3xx5` `=10-9+15=16` `:. 35+53` `implies6+16=22`

Discussion

557.	यदि `(p-q)` का `5%=(p+q)` का `30%` है तो `p:q` किसके बराबर है?A. `5:3`B. `4:1`C. `3:5`D. `1:4`
Answer» Correct Answer - B `50% (p-q)=30%(p+q)` `(p-q)/2=3/10(p+q)` `[50%=1/2]` `implies 5(p-q)=3(p+q)` `implies5p-5q=3p+3q` `implies 2p=8q` `implies1p=4q` `p:q` `implies4:1`

Discussion

558.	यदि `(a^(2)+b^(2))^(3)=(a^(3)+b^(3))^(2)` है तो `a/b+b/a=?`A. `1/3`B. `2/3`C. `-1/3`D. `-2/3`
Answer» Correct Answer - B `(a^(2)+b^(2))^(3)=(a^(3)+b^(3))^(2)` `implies a^(6)+b^(6)+3a^(2)b(a^(2)+b^(2))=a^(6)+` `b^(6)+2a^(3)b^(3)` `implies a^(6)+b^(6)+3a^(4)b^(2)+3a^(2)b^(4)=a^(6)+` `b^(6)+2a^(3)b^(3)` `implies 3a^(4)b^(2)+3a^(2)b^(4)=2a^(3)b^(3)` `implies a^(2)b^(2)(3a^(2)+3b^(2))=2a^(3)b^(3)` `3a^(2)+3b^(2)=2ab` `3(a^(2)+b^(2))=2ab` `implies(a^(2)+b^(2))/(ab)=2/3` `implies a/b+b/a=2/3`

Discussion

559.	यदि `2x+2/x=3` है तो `x^(3)+1/(x^(3))+2` का मान ज्ञात करें।A. `-9/8`B. `-25/8`C. `7/8`D. `11`
Answer» Correct Answer - C `2x+2/x=3` `implies x+1/x=3/2` Taking cube on the sides `=(x+1/x)^(3)=(3/2)^(3)` `{:\|((x+1/x=a),(x^(3)+1/(x^(3))=a^(3)-3a)):}` `x^(3)+1/(x^(3))+3(x+1/x)=27/8` `implies x^(3)+1/(x^(3))+3xx3/2=27/8` `implies x^(3)+1/(x^(3))=27/8-9/2` `=(27-36)/8=(-9)/8` `:. x^(3)+1/(x^(3))+2=(-9)/8+2` `=x^(3)+1/(x^(3))+2=(-9+16)/8=7/8`

Discussion

560.	यदि `a+b+c=0` है तो `(a^(2)+b^(2)+c^(2))/(a^(2)-bc)` का मान क्या होगा?
Answer» Correct Answer - C `a+b+c=0` Assume values `a=2, b=-2, c=0` `a+b+c=2-2+0=0` (satisfy) `:. (a^(2)+b^(2)+ c^(2))/(a^(2)-bc)` `implies (4+4+0)/(4-0)implies8/4=2` Alternate `a+b+c=0` `b+c=-a` Squaring both sides `(b+c)^(2)=a^(2)` `b^(2)+c^(2)+2bc=a^(2)` `b^(2)+c^(2)=a^(2)-2bc` `:. (a^(2)+b^(2)+c^(2))/(a^(2)-bc)` `implies (a^(2)+a^(2)-2bc)/(a^(2)-bc)` `= (2a^(2)-2bc)/(a^(2)-bc)implies(2(a^(2)-bc))/(a^(2)-bc)=2`

Discussion

561.	यदि `a+1/b=1` और ` b+1/c=1` है तो `c+1/a` किसके बराबर है?
Answer» Correct Answer - C `a+1/b=1, b+1/c=1,c+1/a=?` Put values, `a=1/2 b=2 c=-1` `c+1/a=-1+1/((1/2))=-1+2=1` Alternate `implies a+1/b=1`……………….(i) `impliesa=1-1/b=(b-1)/b` `1/a=b/(b-1)impliesb+1/c=1` `1/c=1-b, c=1/(1-b)` `:. c+1/a=1/(1-b)+b/(b-1)` `1/(1-b)-b/(1-b)=(1-b)/(1-b)=1`

Discussion

562.	यदि `(sqrt(7)-2)/(sqrt(7)+2)=asqrt(7)+b` है तो `a` का मान ज्ञात करें?A. `11/3`B. `-4/3`C. `4/3`D. `(-4sqrt(7))/3`
Answer» Correct Answer - B `(sqrt(7)-2)/(sqrt(7)+2)=asqrt(7)+b` L.H.S `=(sqrt(7)-2)/(sqrts(7)+2)xx(sqrt(7)-2)/(sqrt(7)-2)` (Rationalisation) `=((sqrt(7)-2)^(2))/((sqrt(7))^(2)-(4))=(7+4-4sqrt(7))/(7-4)` `implies (11-4sqrt(7))/3` `11/3-4/3sqrt(7)=-4/3sqrt(7)+11/3` `=a sqrt(7)+11/3` `=asqrt(7)+b=` R.H.S Compare the cofficients of `sqrt(7)` and constant term `a=-4/3` `b=11/3`

Discussion

563.	`a+b+c=0` है `1/((a+b)(b+c))+1/((a+c)(b+a))+1/((c+a)(c+b))` का मान ज्ञात करें?A. 1B. 0C. `-1`D. `-2`
Answer» Correct Answer - B `a+b+c=0` `1/((a+b)(b+c))+1/((a+c)(b+a))+` `1/((c+a)(c+b))` `implies((a+c)+(b+c)+(a+b))/((a+b)(a+c)(b+c)` `implies (2(a+b+c))/((a+b)(a+c)(b+c))=0` `( :. a+b+c=0)`

Discussion

564.	यदि `a^(2)+b^(2)+2b+4a+5=0` है तो `(a-b)/(a+b)` का मान ज्ञात करें।A. `3`B. `-3`C. `1/3`D. `-1/3`
Answer» Correct Answer - C `a^(2)+b^(2)+2b+4a+5=0` `a^(2)+b^(2)+2b+4a+4+1=0` `a^(2)+4a+4+b^(2)+2b+1=0` `(a+2)^(2)+(b+1)^(2)=0` `a+2=0, a=-2` `b+1=0, b=-1` `(a-b)/(a+b)implies(-2+1)/(-2-1)` `implies (-1)/(-3)=1/3`

Discussion

565.	यदि `x+y+z=0` है तो `(x^(2))/(yz)+(y^(2))/(zx)+(z^(2))/(xy)=?`A. `(xyz)2`B. `x^(2)+y^(2)+z^(2)`C. `9`D. `3`
Answer» Correct Answer - D `(x^(2))/(yz)+(y^(2))/(zx)+(z^(2))/(xy)` `=(x^(2))/(yz)xx x/x +(y^(2)xxy)/(zx xxy)+(z^(2))/(xy)xx z/z` `= (x^(3)+y^(3)+z^(3))/(xyz)` `( :.` If `x+y+z=0` then `x^(3)+y^(3)+z^(3)implies3xyz)` `:.(3xyz)/(xyz)=3`

Discussion

566.	यदि `a^(2x+2)=1` जहां `a,1` के अतिरिक्त एक धनात्मक वास्तविक संख्या है तो `x` किसके बराबर है?A. `-2`B. `-1`C. `0`D. `1`
Answer» Correct Answer - B `a^(2x+2)=1` `a^(2x+2)=a^(0)` `2x+2=0` `x=-2/2=-1`

Discussion

567.	यदि `a+1/a+1=0(a!=0)` है तो `(a^(4)-a)` का मान क्या होगा?
Answer» Correct Answer - A `a+1/a=1=0` `a+1/a=-1` Squaring both sides `implies a^(2)+1/(a^(2))+2=1` `impliesa^(2)+1/(a^(2))=-1` `impliesa^(2)+1=(-1)/(a^(2))`……………i `impliesa+1/a=-1` (given) `:.a^(2)+1=-a` ………….ii `implies-a=(-1)/(a^(2))` From equation i and ii `a^(3)=1` `:. a^(3)-1=0` `impliesa^(4)-a=a(a^(3)-1)=axx0=0`

Discussion

568.	यदि `ax+by=1` और `bx+ay=(2ab)/(a^(2)+b^(2))` हो तो `(x^(2)+y^(2))(a^(2)+b^(2))` किसके बराबर है?A. 1B. 2C. 0.5D. 0
Answer» Correct Answer - A `ax+by=1` `bx+ay=(2ab)/(a^(2)+b^(2))` Let `a=b=1` then `x+y=1` …………i `x+y=(2xx1xx1)/(1+1)=1`………..ii Now put `x=y=1/2` in eq i and ii `1/2+1/2=1` and equation ii `1/2+1/2=1` Value of `(x^(2)+y^(2))(a^(2)+b^(2))` `=(1/2)^(2)+(1/2)^(2)(1+1)` `=(1/4+1/4)xx2=2/4xx2=1`

Discussion

569.	यदि `(3a+5b)/(3a-5b)=5` है तो `a:b` किसके बराबर है ?A. `2:1`B. `2:3`C. `1:3`D. `5:2`
Answer» Correct Answer - D `(3a+5b)/(3a-5b)=5` `implies 3a+5b=15a-25b` `implies12a=30b` `implies 2a=5b` `a:b` `5:2`

Discussion

570.	यदि `a+1/b=b+1/c=c+1/a` जहां `a!=b!=c` तो `abc` किसके बराबर है?A. `+1`B. `-1`C. `+1` & `-1`D. इनमें से कोई नहीं
Answer» Correct Answer - C `a+1/b=b+1/c=c+1/a` Put `a=1/2,b=2, c=-1` `1/2+1/2=2-1=-1+2` `1=1=1` `abc=-1/2xx2x-1` `abc=-1` again put `a=-1/2,b=-2,c=1` `a+1/b=b+1/c+c+1/a` `-1/2-1/2=-2+1=1-2` `-1=-1=-1` equation satisfied `abc=-1/2xx-2xx1=1` So abc can be `-1` and `+1`

Discussion

571.	यदि `a+b+c=2s`, तो `((s-a)^(2)+(s-b)^(2)+(s-c)^(2)+s^(2))/(a^(2)+b^(2)+c^(2))` किसके बराबर है?A. `a^(2)+b^(2)+c^(2)`B. `0`C. `1`D. `2`
Answer» Correct Answer - C `a+b+c=2s` `:.` Let `a=2` `b=1` `c=1` `s=2` `:. ((s-a)^(2)+(s-b)^(2)+(s-c)^(2)+s^(2))/(a^(2)+b^(2)+c^(2))` `implies ((2-2)^(2)+(2-1)^(2)+(2-1)^(2)+2^(2))/(2^(2)+1^(2)+1^(2))` `implies (0+1+1+4)/(4+1+1)=6/6=1`

Discussion

572.	यदि `x=a+1/a` और `y=a-1/a` है तो `x^(4)+y^(4)2x^(2)y^(2)` का मान ज्ञात करें?A. 214B. 18C. 16D. 12
Answer» Correct Answer - C `x=a+1/a` `y=a-1/a` `:. (x+y)=a+1/a+a-1/a=2a` `:. (x-y)=a+1/a-a+1/a=2/a` `:.x^(4)+y^(4)-2x^(2)y^(2)-(x^(2)-y^(2))^(2)` `implies ((x+y)(x-y))^(2)` `implies (2a xx 2/a)^(2)` `=(4)^(2)=16`

Discussion

573.	यदि `1ltxlt2` है तो `sqrt((x-1)^(2))+sqrt((x-3)^(2))` का मान क्या होगा?A. 1B. 2C. 3D. `2x-4`
Answer» Correct Answer - B `1ltxlt2` `sqrt((x-1)^(2))+sqrt((3-x)^(2))` (square root cancel with square) `:. x-1+3-x=2`

Discussion

574.	यदि `x=19` और `y=18` है तो `(x^(2)+y^(2)+xy)/(x^(3)-y^(3))` का मान ज्ञात करें?A. 1B. 37C. 324D. 361
Answer» Correct Answer - A `x=19 y=18` `(x^(2)+y^(2)+xy)/(x^(3)-y^(3))` `=(x^(2)+y^(2)+xy)/((x-y)(x^(2)+y^(2)+xy))` `=1/(x-y)=1/(19-18)=1`

Discussion

575.	यदि `a=11` और `b=9` है तो `((a^(2)+b^(2)+ab)/(a^(3)-b^(3)))` का मान ज्ञात करें?A. `1/2`B. `2`C. `1/20`D. `20`
Answer» Correct Answer - A `a=11` `b=9` `implies(a^(2)+b^(2)+ab)/(a^(3)-b^(3))` `implies (a^(3)-b^(3))=(a-b)(a^(2)+ab+b^(2))` `:. (a^(2)+b^(2)+ab)/((a=b)(a^(2)+ab+b^(2))` `=1/(a-b)=1/(11-9)=1/2`

Discussion

576.	यदि `p=101` है तो `root(3)(p(p^(2)-3p+3)-1)` का मान ज्ञात करें?A. 100B. 101C. 102D. 1000
Answer» Correct Answer - A `p=101` `=root(3)(p(p^(2)-3p+3)-1)` `=root(3)(p^(3)-3p^(2)+3p-1)` `:.[(p-1)^(3)=p^(3)-(1)^(3)-3p(p-1)]` `=root(3)((p-1)^(3))` `=p-1=101-1=100`

Discussion

577.	यदि `c+1/c=3`, हो तो `(c-3)^(7)+1/(c^(7))` का मान क्या है?A. 2B. 0C. 3D. 1
Answer» Correct Answer - B `C+1/C=3` `C-3+1/C=0` So `(C-3)^(7)+1/(C^(7))=[(C-3)+1/C]^(3)` `[(C-3)+1/C]^(4)-[(C-3)+1/C]=0`

Discussion

578.	The total quantity of water contained in two vessels was 16 litres. When 3 litres of water was poured from one vessel to the other vessel, both the vessels were found containing equal quantities of water. Then what was the difference between the quantities of water contained in the two vessels initially?1. 42. 53. 64. 7
Answer» Correct Answer - Option 3 : 6 Given: The total quantity of water = 16 litres 3 litres of water was poured from one vessel to the other vessel, both the vessels were found containing equal quantities of water Calculation Let the quantity of water in two vessels be 'x' and 'y' respectively So, x + y = 16 When 3 litres if poured from one vessel to the other we get, x - 3 = y + 3 (since 3 is removed from x and added to y, also given that they are equal) ⇒ x - y = 6 ∴ Difference between the quantities of water in litres = 6

578.

The total quantity of water contained in two vessels was 16 litres. When 3 litres of water was poured from one vessel to the other vessel, both the vessels were found containing equal quantities of water. Then what was the difference between the quantities of water contained in the two vessels initially?1. 42. 53. 64. 7

Answer» Correct Answer - Option 3 : 6

Given:

The total quantity of water = 16 litres

3 litres of water was poured from one vessel to the other vessel, both the vessels were found containing equal quantities of water

Calculation

Let the quantity of water in two vessels be 'x' and 'y' respectively

So, x + y = 16

When 3 litres if poured from one vessel to the other we get,

x - 3 = y + 3 (since 3 is removed from x and added to y, also given that they are equal)

⇒ x - y = 6

∴ Difference between the quantities of water in litres = 6

Discussion

579.

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.I. x2 + 2x –15 = 0 II. y2 + 9y + 20 = 01. y > x2. y

Answer» Correct Answer - Option 5 : x = y or the relationship between x and y cannot be established.

Calculation:

I. x2 + 2x –15 = 0

⇒ x2 + 5x –3x –15 = 0

⇒ x × (x + 5) –3 × (x + 5) = 0

⇒ (x – 3) × (x + 5) = 0

⇒ x = 3 and x = –5

II. y2 + 9y + 20 = 0

⇒ y² + 5y + 4y + 20 = 0

⇒ y × (y + 5) + 4 × (y + 5) = 0

⇒ (y + 4) × (y + 5) = 0

⇒ y = –4 and y = –5

Value of x	Value of y	Relation
3	−4	x > y
−5	−5	x = y
3	−5	x > y
−5	−4	x < y

∴ x = y or the relationship between x and y cannot be established.

Discussion

580.

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-I. x2 = 4II. y2 - 6y + 8 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relation cannot be determined

Answer» Correct Answer - Option 4 : x ≤ y

Given:

I. x² = 4

II. y² – 6y + 8 = 0

Concept used:

Using factorization method.

Calculation:

x² = 4

⇒ x² – 4 = 0

⇒ (x – 2)(x + 2) = 0

⇒ (x – 2) = 0 or (x + 2) = 0

⇒ x = 2 or x = -2

⇒ x = -2, 2

y² – 6y + 8 = 0

⇒ y² – 4y – 2y + 8 = 0

⇒ y(y – 4) – 2(y – 4) = 0

⇒ (y – 4)(y – 2) = 0

⇒ (y – 4) = 0 or (y – 2) = 0

⇒ y = 4 or y = 2

⇒ y = 2, 4

∴ x ≤ y.

Comparison between x and y (via Tabulation):

Value of x	Value of y	Relation
2	2	x = y
2	4	x < y
-2	4	x < y
-2	2	x < y

∴ x ≤ y.

Discussion

581.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answerI. 12x2 – 46x + 14 = 0II. 14y2 – 46y + 12 = 01. x > y2. x ≤ y3. x = y or the relationship cannot be established.4. x ≥ y5. x < y

Answer» Correct Answer - Option 3 : x = y or the relationship cannot be established.

I. 12x² – 46x + 14 = 0

⇒ 12x² – 42x – 4x + 14 = 0

⇒ 6x(2x – 7) – 2(2x – 7) = 0

⇒ (2x – 7)(6x – 2) = 0

⇒ x = 7/2 or 1/3

II. 14y² – 46y + 12 = 0

⇒ 14y² – 42y - 4y + 12 = 0

⇒ 14y(y – 3) – 4(y – 3) = 0

⇒ (y – 3) × (14y – 4) = 0

⇒ y = 3 or 2/7

Comparison between x and y (via Tabulation):

Value of x	Relation	Value of y
7/2	>	3
7/2	>	2/7
1/3	<	3
1/3	>	2/7

∴ The relationship cannot be established between x and y.

Discussion

582.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answerI. 8x2 – 32x + 14 = 0II. 3y2 – 37y – 26 = 01.x > y2. x ≤ y3. x = y or the relationship cannot be established.4. x ≥ y5. x < y

Answer» Correct Answer - Option 3 : x = y or the relationship cannot be established.

I. 8x² – 32x + 14 = 0

⇒ 8x² – 28x – 4x + 14 = 0

⇒ 4x(2x – 7) – 2(2x – 7) = 0

⇒ (2x – 7)(4x – 2) = 0

⇒ x = 7/2 or 1/2

II. 3y² – 37y – 26 = 0

⇒ 3y² – 39y + 2y – 26 = 0

⇒ 3y(y – 13) + 2(y – 13) = 0

⇒ (y – 13)(3y + 2) = 0

⇒ y = 13 or -(2/3)

Comparison between x and y (via Tabulation):

Value of x	Relation	Value of y
7/2	<	13
7/2	>	-(2/3)
1/2	<	13
1/2	>	-(2/3)

∴ Relationship cannot be established between x and y.

Discussion

583.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. 2x2 – 32x + 128 = 0II. 2y2 – 32y + 96 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relationship cannot be established.

Answer» Correct Answer - Option 5 : x = y or the relationship cannot be established.

I. 2x² – 32x + 128 = 0

⇒ 2x² – 16x – 16x + 128 = 0

⇒ 2x(x – 8) – 16(x – 8) = 0

⇒ (x – 8)(2x – 16) = 0

⇒ x = 8

II. 2y² – 32y + 96 = 0

⇒ 2y² – 24y – 8y + 96 = 0

⇒ 2y(y – 12) – 8(y – 12) = 0

⇒ (y – 12)(2y – 8) = 0

⇒ y = 12 or 4

Comparison between x and y (via Tabulation):

Value of x	Relation	Value of y
8	>	4
8	<	12

∴ Relationship cannot be established between x and y.

Discussion

584.

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.I. 3x2 + 2x – 1 = 0II. 11y2 + 18y + 7 = 0 1. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

Answer» Correct Answer - Option 5 : x = y or the relationship between x and y cannot be established.

Calculation:

From I,

3x ²+ 2x – 1 = 0

⇒ 3x² + 3x – x – 1 = 0

⇒ 3x(x + 1) – 1(x + 1) = 0

⇒ (3x – 1)(x + 1) = 0

Taking,

⇒ 3x – 1 = 0 or x + 1 = 0

⇒ x = 1/3 or x = –1

From II,

11y² + 18y + 7 = 0

⇒ 11y² + 11y + 7y +7 = 0

⇒ 11y(y + 1) + 7(y + 1) = 0

⇒ (y + 1)(11y + 7) = 0

Taking,

⇒ y + 1 = 0 or 11y + 7 = 0

⇒ y = –1 or y = –7/11

Comparison between x and y (via Tabulation)

Value of x	Value of y	Relation
1/3	–1	x > y
1/3	–7/11	x > y
–1	–1	x = y
–1	–7/11	x < y

∴x = y or the relationship between x and y cannot be established.

Discussion

585.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. x2 – 3x – 10 = 0II. y2 – 18y + 45 = 0 1. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

Answer» Correct Answer - Option 5 : x = y or the relationship between x and y cannot be established.

Calculation:

From I,

x ²– 3x – 10 = 0

⇒ x² – 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5)(x + 2) = 0

Taking,

⇒ x – 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

From II,

y² – 18y + 45 = 0

⇒ y² – 15y – 3y + 45 = 0

⇒ y(y – 15) – 3(y – 15) = 0

⇒ (y – 15) (y – 3) = 0

Taking,

⇒ y – 15 = 0 or y – 3 = 0

⇒ y = 15 or y = 3

Comparison between x and y (via Tabulation)

Value of x	Value of y	Relation
5	15	x < y
5	3	x > y
-2	15	x < y
-2	3	x < y

∴ x = y or the relationship between x and y cannot be established.

Discussion

586.	A man buys 2 apples and 3 kiwi fruits for Rs. 37. If he buys 4 apples and 5 kiwi fruits for Rs. 67, then what will be the total cost of 1 apple and 1 kiwi fruit?1. Rs. 152. Rs. 283. Rs. 184. Rs. 20
Answer» Correct Answer - Option 1 : Rs. 15 Given A man buys 2 apples and 3 kiwi = Rs.37 He buys 4 apples and 5 Kiwi for Rs.67 Concept used We solve these type of question by equation method Calculation Let the price of 1 apple be Rs.x Let the price of 1 Kiwi be Rs.y According to question ∴ Price of 2 apples and 3 kiwi ⇒ 2x + 3y = Rs.37 ----(1) ⇒ Price of 4 Apples and 5 Kiwi ⇒ 4x + 5y = Rs.67 ----(2) ⇒ Multiply equation (1) by 2 and then subtract it from equation (2), we get ⇒ y = Rs.7 ⇒ put the value of y in equation (1), we get ⇒ 2x = 37- 21 ⇒ x = Rs.8 ∴ price of 1 apple and 1 kiwi ⇒ 8 + 7 ⇒ Rs.15

586.

A man buys 2 apples and 3 kiwi fruits for Rs. 37. If he buys 4 apples and 5 kiwi fruits for Rs. 67, then what will be the total cost of 1 apple and 1 kiwi fruit?1. Rs. 152. Rs. 283. Rs. 184. Rs. 20

Answer» Correct Answer - Option 1 : Rs. 15

Given

A man buys 2 apples and 3 kiwi = Rs.37

He buys 4 apples and 5 Kiwi for Rs.67

Concept used

We solve these type of question by equation method

Calculation

Let the price of 1 apple be Rs.x

Let the price of 1 Kiwi be Rs.y

According to question

∴ Price of 2 apples and 3 kiwi

⇒ 2x + 3y = Rs.37 ----(1)

⇒ Price of 4 Apples and 5 Kiwi

⇒ 4x + 5y = Rs.67 ----(2)

⇒ Multiply equation (1) by 2 and then subtract it from equation (2), we get

⇒ y = Rs.7

⇒ put the value of y in equation (1), we get

⇒ 2x = 37- 21

⇒ x = Rs.8

∴ price of 1 apple and 1 kiwi

⇒ 8 + 7

⇒ Rs.15

Discussion

587.

In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.I. 7x2 + 15x + 8 = 0II. y2 + y/2 – 1/2 = 01. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

Answer» Correct Answer - Option 4 : x ≤ y

I. 7x² + 15x + 8 = 0

⇒ 7x² + 7x + 8x + 8 = 0

⇒ 7x(x + 1) + 8(x + 1) = 0

⇒ (7x + 8)(x + 1) = 0

⇒ x = -8/7, -1

II. y² + y/2 – 1/2 = 0

⇒ y² + y – 1/2y – 1/2 = 0

⇒ y(y + 1) – 1/2(y + 1) = 0

⇒ (y – 1/2)(y + 1) = 0

⇒ y = 1/2, -1

Value of x	Value of y	Relation
-8/7	1/2	X < y
-8/7	-1	X < y
-1	1/2	X < y
-1	-1	X = y

∴ x ≤ y

Discussion

588.	Umesh goes on a tour of three cities of Malaysia. In every city, he spends Rs. 40 more than 1/4th of the money he has with him. At the end of the tour, Umesh has only Rs. 16400 left with him. Find the amount with him before he started the tour of Malaysia.1. Rs. 386722. Rs. 443513. Rs. 354364. Rs. 389815. Rs. 38781
Answer» Correct Answer - Option 4 : Rs. 38981 Calculation: Let the amount with Umesh, before he started the trip to Malaysia be x. In the first city he spends = x/4 + 40 He would be left with = x – (x/4 + 40) ⇒ 3x/4 – 40 In the second city he spends = ¼(3x/4 - 40) + 40 ⇒ 3x/16 + 30 He would be left with = (3x/4 - 40) – (3x/16 + 20) ⇒ 9x/16 – 60 In the third city he spends = ¼(9x/16 - 60) + 40 ⇒ 9x/64 + 15 He would be left with = (9x/16 - 60) – (9x/64 + 15) ⇒ 27x/64 – 45 ∴ total amount left with him = 16400 ⇒ 27x/64 – 45 = 16400 ⇒ 27x/64 = 16445 ⇒ x = (16445 × 64)/27 ⇒ x = 38980.74 ≈ Rs. 38,981 The amount with Umesh before he started the tour of Malaysia is Rs. 38,981.

588.

Umesh goes on a tour of three cities of Malaysia. In every city, he spends Rs. 40 more than 1/4th of the money he has with him. At the end of the tour, Umesh has only Rs. 16400 left with him. Find the amount with him before he started the tour of Malaysia.1. Rs. 386722. Rs. 443513. Rs. 354364. Rs. 389815. Rs. 38781

Answer» Correct Answer - Option 4 : Rs. 38981

Calculation:

Let the amount with Umesh, before he started the trip to Malaysia be x.

In the first city he spends = x/4 + 40

He would be left with = x – (x/4 + 40)

⇒ 3x/4 – 40

In the second city he spends = ¼(3x/4 - 40) + 40

⇒ 3x/16 + 30

He would be left with = (3x/4 - 40) – (3x/16 + 20)

⇒ 9x/16 – 60

In the third city he spends = ¼(9x/16 - 60) + 40

⇒ 9x/64 + 15

He would be left with = (9x/16 - 60) – (9x/64 + 15)

⇒ 27x/64 – 45

∴ total amount left with him = 16400

⇒ 27x/64 – 45 = 16400

⇒ 27x/64 = 16445

⇒ x = (16445 × 64)/27

⇒ x = 38980.74 ≈ Rs. 38,981

The amount with Umesh before he started the tour of Malaysia is Rs. 38,981.

Discussion

589.

Ambani bought a total of 30 Aircraft, Lamborghini, and Bentley for Rs.136 million. He bought one Lamborghini more than Bentley. The cost of each Aircraft was Rs. 8 million each, Lamborghini was Rs. 3 million each and Bentley was Rs. 4 million each. How many Aircraft did he buy?1. 72. 63. 84. 95. 5

Answer» Correct Answer - Option 1 : 7

Calculation:

Let the number of Aircraft, Lamborghini and Bentley bought by Ambani be x, y, and z respectively.

According to the question:

x + y + z = 30 ----(i)

8x + 3y + 4z = 136 ----(ii)

From (ii) – (i) × 4, we get

4x – y = 16 ----(iii)

From (iii)

When y = 0 then x = 4

∴ z = (30 – 4 - 0)

⇒ z = 26

The other solutions are obtained by adding 1 repeatedly to x.

When x = 5 then y = 4

∴ z = (30 – 5 - 4)

⇒ z = 21

When x = 6 then y = 8

∴ z = (30 – 6 - 8)

⇒ z = 16

When x = 7 then y = 12

∴ z = (30 – 7 - 12)

⇒ z = 11

When x = 8 then y = 16

∴ z = (30 – 8 - 16)

⇒ z = 6

When x = 9 then y = 20

∴ z = (30 – 9 - 20)

⇒ z = 1

When x = 10 then y = 34

∴ z = (30 – 10 - 34)

⇒ z = -14

This is not possible.

Calculate results in a table.

X	4	5	6	7	8	9
Y	0	4	8	12	16	20
Z	26	21	16	11	6	1

Since he bought at least one of each thing and one more Lamborghini than Bentley.

So, 4^th combination is the right combination.

No. of Aircrafts = 7 ; No. of Lamborghinis = 12 ; No. of Bentleys = 11

Ambani bought 7 Aircraft.

Discussion

590.	The value of \(\frac{1}{{(7\; + \;4\sqrt 3 })^2} + \frac{1}{{(7\; - \;4\sqrt 3 )^2}}\) is1. 1942. 1543. 1844. 149
Answer» Correct Answer - Option 1 : 194 Given: \(\frac{1}{{(7\; + \;4√ 3 })^2} + \frac{1}{{(7\; - \;4√ 3 )^2}}\) Concept used: Componendo and dividendo Calculation: \(\frac{1}{{(7\; + \;4√ 3 })^2} + \frac{1}{{(7\; - \;4√ 3 )^2}}\) ⇒ \(\frac{(7\; - \;4√3)^2}{{(7\; + \;4√ 3 })^2(7\; - \;4√3)^2} + \frac{(7\; + \;4√3)^2}{{(7\; - \;4√ 3 )^2(7 \;+\;4√3)^2}}\) ⇒ \(\frac{(7\; - \;4√3)^2}{{1}} + \frac{(7\; + \;4√3)^2}{1}\) ⇒ 7² + (4√3)² – 2 × 7 × 4√3 + 72 + (4√3)2 + 2 × 7 × 4√3 ⇒ 49 + 48 – 56√3 + 49 + 48 + 56√3 ⇒ 49 + 48 + 49 + 48 ⇒ 194 ∴ Required value is 194

590.

The value of \(\frac{1}{{(7\; + \;4\sqrt 3 })^2} + \frac{1}{{(7\; - \;4\sqrt 3 )^2}}\) is1. 1942. 1543. 1844. 149

Answer» Correct Answer - Option 1 : 194

Given:

\(\frac{1}{{(7\; + \;4√ 3 })^2} + \frac{1}{{(7\; - \;4√ 3 )^2}}\)

Concept used:

Componendo and dividendo

Calculation:

\(\frac{1}{{(7\; + \;4√ 3 })^2} + \frac{1}{{(7\; - \;4√ 3 )^2}}\)

⇒ \(\frac{(7\; - \;4√3)^2}{{(7\; + \;4√ 3 })^2(7\; - \;4√3)^2} + \frac{(7\; + \;4√3)^2}{{(7\; - \;4√ 3 )^2(7 \;+\;4√3)^2}}\)

⇒ \(\frac{(7\; - \;4√3)^2}{{1}} + \frac{(7\; + \;4√3)^2}{1}\)

⇒ 7² + (4√3)² – 2 × 7 × 4√3 + 72 + (4√3)2 + 2 × 7 × 4√3

⇒ 49 + 48 – 56√3 + 49 + 48 + 56√3

⇒ 49 + 48 + 49 + 48

⇒ 194

∴ Required value is 194

Discussion

591.	If a2 + b2 = 2a + 4b – 5, then what is the value of (a + b)?1. 52. 43. 24. 3
Answer» Correct Answer - Option 4 : 3 Given: a2 + b2 = 2a + 4b – 5 Formula used: (a – b)² = a² + b² – 2ab Calculation: a2 + b2 = 2a + 4b – 5 ⇒ a2 + b2 – 2a – 4b + 5 = 0 ⇒ a2 + b2 – 2a – 4b + 1 + 4 = 0 ⇒ (a2 – 2a + 1) + (b2 – 4b + 4) = 0 ⇒ (a – 1)² + (b – 2)² = 0 The above expression is possible only when, (a – 1) = 0, (b – 2) = 0 ⇒ a = 1, b = 2 a + b = 1 + 2 = 3 ∴ The value of a + b is 3

591.

If a2 + b2 = 2a + 4b – 5, then what is the value of (a + b)?1. 52. 43. 24. 3

Answer» Correct Answer - Option 4 : 3

Given:

a2 + b2 = 2a + 4b – 5

Formula used:

(a – b)² = a² + b² – 2ab

Calculation:

a2 + b2 = 2a + 4b – 5

⇒ a2 + b2 – 2a – 4b + 5 = 0

⇒ a2 + b2 – 2a – 4b + 1 + 4 = 0

⇒ (a2 – 2a + 1) + (b2 – 4b + 4) = 0

⇒ (a – 1)² + (b – 2)² = 0

The above expression is possible only when,

(a – 1) = 0, (b – 2) = 0

⇒ a = 1, b = 2

a + b = 1 + 2 = 3

∴ The value of a + b is 3

Discussion

592.

If (a2 – b2) = 30, a – b = 6 then, what is the value of a + b?1. 72. 93. 54. 12

Answer» Correct Answer - Option 3 : 5

Given:

(a2 – b2) = 30

a – b = 6

Formula required:

(a + b)(a – b) = a² – b²

Calculations:

(a + b) × 6 = 30

⇒ (a + b) = 30/6

⇒ (a + b) = 5

∴ The value of (a + b) is 5

Discussion

593.

If x + y + z = 5, x2 + y2 + z2 = 21 and y2 = zx, then the value of y is:1. 1/52. 2/53. 1/24. 1/4

Answer» Correct Answer - Option 2 : 2/5

Given:

x + y + z = 5

x² + y² + z² = 21

y² = zx

Formula used:

(x + y + z)² = x2 + y2 + z² + 2(xy + yz + xz)

Calculations:

(x + y + z)² = 5²

⇒ x2 + y2 + z2 + 2(xy + yz + xz) = 25

⇒ 21 + 2(xy + yz + xz) = 25

⇒ 2(xy + yz + xz) = 25 - 21

⇒ xy + yz + xz = 4/2

⇒ xy + yz + xz = 2

⇒ xy + yz+ y² = 2

⇒ y(x + y + z) = 2

⇒ y(5) = 2

⇒ y = 2/5

∴ The required value of y = 2/5

Discussion

594.

The cost of 4 pencil and 5 pen is Rs 47 and cost of 2 pencil and 7 pen is Rs. 55, then what is the cost of the one pencil?1. Rs. 52. Rs. 33. Rs. 74. Rs. 8

Answer» Correct Answer - Option 2 : Rs. 3

Given:

The cost price of 4 pencils and 5 pens = 47

The cost price of 2 pencils and 7 pens = 55

Calculation:

Let the cost price of pencil be 'x' and pen be 'y'

4x + 5y = 47 ------(1)

2x + 7y = 55 -------(2)

Now, 2 × eq (2) – eq (1)

2(2x + 7y) – (4x + 5y) = 2 × 55 – 47

⇒ 4x + 14y – 4x – 5y = 110 – 47

⇒ 9y = 63

⇒ y = 7

Substituting the value of y in eq (1) then,

⇒ 4x + 5 × 7 = 47

⇒ 4x = 47 – 35

⇒ 4x = 12

⇒ x = 3

∴ The cost price of one pencil is Rs. 3

Discussion

595.

If a3 + b3 = 62 and a + b = 2, then the value of ab is:1. -62. 93. 64. -9

Answer» Correct Answer - Option 4 : -9

Given:

a + b = 2

a3 + b3 = 62

Formula used:

(a + b)³ = a3 + b3 + 3ab(a + b)

Calculation:

a + b = 2

Cubing the given equation

(a + b)³ = 2³

⇒ a³ + b³ + 3ab(a + b) = 8 ----[(a + b)3 = a3 + b3 + 3ab(a + b)]

⇒ 62 + 3ab × 2 = 8 ----(a3 + b3 = 62, a + b = 2)

⇒ 6ab = 8 – 62

⇒ 6ab = -54

⇒ ab = -54/6 = -9

∴ The value of ab is -9.

Discussion

596.

18x2 – 12x – 2y + 3xy = 1. (6x + y)(3x – 2)2. (6x – y)(3x – 2)3. (3x – y)(6x + 2)4. (3x + y)(6x – 2)

Answer» Correct Answer - Option 1 : (6x + y)(3x – 2)

Calculation:

18x2 – 12x – 2y + 3xy

⇒ 6x(3x – 2) + y( 3x – 2)

⇒ (6x + y) (3x – 2)

∴ The required answer is (6x + y) (3x – 2)

Discussion

597.

If \(x=\frac{{\sqrt 3 }}{2}\), then the value of \(\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }}\) is equal to:1. √22. √33. 34. 2

Answer» Correct Answer - Option 2 : √3

Calculation:

\({√(1 + x) + √(1 – x)}\over{√(1 + x) – (√1 – x)}\) × \({√(1 +x) + √(1 – x)}\over {√(1 + x) + √(1 – x)}\)

⇒ \((1 + x) + (1 – x) + 2√(1 – x^2) \over(1 + x) – (1 – x)\)

⇒ \(2 + 2√(1 – (√3/2)^2)\over2x\)

⇒ \(2 + 2 × (1/2)\over 2 × (√3/2) \)

⇒ (3/√3) × (√3/√3)

⇒ √3

Discussion

598.

If 9x + 7y = 130, 2x – y = 11, then what will be the value of (x + y)?1. 152. 143. 214. 16

Answer» Correct Answer - Option 4 : 16

Given:

9x + 7y = 130

2x – y = 11

Calculation:

Our given equation are

9x + 7y = 130 ----(1)

2x – y = 11 ----(2)

Multiply eq (2) with 7 and equation (1) with 1

9x + 7y = 130 ----(3)

7(2x – y) = 11 × 7 = 77 ----(4)

Add equation (4) and equation (3)

9x + 7y + 14x – 7y = 130 + 77

⇒ 23x = 207

⇒ x = 207/23 = 9

Now, from eq (2)

2x – y = 11

⇒ 2 × 9 – y = 11

⇒ y = 18 – 11 = 7

Now, x + y = 9 + 7 = 16

∴ The required answer is 16

Discussion

599.

x - 1/x = 5 Find the value of x4 + 1/x41. 7312. 7243. 7274. 745

Answer» Correct Answer - Option 3 : 727

Given:

x - 1/x = 5 ----(1)

Formula Used:

x² + 1/x² = (x - 1/x)² + 2

x² + 1/x² = (x + 1/x)² - 2

Calculation:

x2 + 1/x2 = (x - 1/x)2 + 2

⇒ x² + 1/x² = 5² + 2 [Using (1)]

⇒ x² + 1/x² = 27 ----(2)

x4 + 1/x⁴

⇒ (x²)² + (1/x²)²

⇒ (x² + 1/x²)² - 2 [Using (2)]

⇒ (27)² - 2

⇒ 729 - 2

⇒ 727

Discussion

600.

(7x + 12)2 – 336x is equal to 1. (7x + 11)22. (7x + 9)23. (7x – 9)24. (7x – 12)2

Answer» Correct Answer - Option 4 : (7x – 12)²

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

(7x + 12)2 – 336x

⇒ 49x² + 144 + 168x – 336x

⇒ 49x² – 168x + 144

⇒ 49x² – 84x – 84x + 144

⇒ 7x (7x – 12) – 12(7x –12)

⇒ (7x – 12)(7x –12)

⇒ (7x – 12)²

∴ The required answer is (7x – 12)2

Discussion

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यदि `a+1/a+1=0(a!=0)` है तो `(a^(4)-a)` का मान क्या होगा?

यदि `ax+by=1` और `bx+ay=(2ab)/(a^(2)+b^(2))` हो तो `(x^(2)+y^(2))(a^(2)+b^(2))` किसके बराबर है?A. 1B. 2C. 0.5D. 0

यदि `(3a+5b)/(3a-5b)=5` है तो `a:b` किसके बराबर है ?A. `2:1`B. `2:3`C. `1:3`D. `5:2`

यदि `a+1/b=b+1/c=c+1/a` जहां `a!=b!=c` तो `abc` किसके बराबर है?A. `+1`B. `-1`C. `+1` & `-1`D. इनमें से कोई नहीं

यदि `a+b+c=2s`, तो `((s-a)^(2)+(s-b)^(2)+(s-c)^(2)+s^(2))/(a^(2)+b^(2)+c^(2))` किसके बराबर है?A. `a^(2)+b^(2)+c^(2)`B. `0`C. `1`D. `2`

यदि `x=a+1/a` और `y=a-1/a` है तो `x^(4)+y^(4)2x^(2)y^(2)` का मान ज्ञात करें?A. 214B. 18C. 16D. 12

यदि `1ltxlt2` है तो `sqrt((x-1)^(2))+sqrt((x-3)^(2))` का मान क्या होगा?A. 1B. 2C. 3D. `2x-4`

यदि `x=19` और `y=18` है तो `(x^(2)+y^(2)+xy)/(x^(3)-y^(3))` का मान ज्ञात करें?A. 1B. 37C. 324D. 361

यदि `a=11` और `b=9` है तो `((a^(2)+b^(2)+ab)/(a^(3)-b^(3)))` का मान ज्ञात करें?A. `1/2`B. `2`C. `1/20`D. `20`

यदि `p=101` है तो `root(3)(p(p^(2)-3p+3)-1)` का मान ज्ञात करें?A. 100B. 101C. 102D. 1000

यदि `c+1/c=3`, हो तो `(c-3)^(7)+1/(c^(7))` का मान क्या है?A. 2B. 0C. 3D. 1

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.I. x2 + 2x –15 = 0 II. y2 + 9y + 20 = 01. y > x2. y

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer-I. x2 = 4II. y2 - 6y + 8 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relation cannot be determined

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answerI. 12x2 – 46x + 14 = 0II. 14y2 – 46y + 12 = 01. x > y2. x ≤ y3. x = y or the relationship cannot be established.4. x ≥ y5. x < y

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answerI. 8x2 – 32x + 14 = 0II. 3y2 – 37y – 26 = 01.x &gt; y2. x ≤ y3. x = y or the relationship cannot be established.4. x ≥ y5. x < y

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. 2x2 – 32x + 128 = 0II. 2y2 – 32y + 96 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relationship cannot be established.

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.I. 3x2 + 2x – 1 = 0II. 11y2 + 18y + 7 = 0 1. x &gt; y2. x &lt; y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. x2 – 3x – 10 = 0II. y2 – 18y + 45 = 0 1. x &gt; y2. x &lt; y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

A man buys 2 apples and 3 kiwi fruits for Rs. 37. If he buys 4 apples and 5 kiwi fruits for Rs. 67, then what will be the total cost of 1 apple and 1 kiwi fruit?1. Rs. 152. Rs. 283. Rs. 184. Rs. 20

In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.I. 7x2 + 15x + 8 = 0II. y2 + y/2 – 1/2 = 01. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

Ambani bought a total of 30 Aircraft, Lamborghini, and Bentley for Rs.136 million. He bought one Lamborghini more than Bentley. The cost of each Aircraft was Rs. 8 million each, Lamborghini was Rs. 3 million each and Bentley was Rs. 4 million each. How many Aircraft did he buy?1. 72. 63. 84. 95. 5

The value of \(\frac{1}{{(7\; + \;4\sqrt 3 })^2} + \frac{1}{{(7\; - \;4\sqrt 3 )^2}}\) is1. 1942. 1543. 1844. 149

If a2 + b2 = 2a + 4b – 5, then what is the value of (a + b)?1. 52. 43. 24. 3

If (a2 – b2) = 30, a – b = 6 then, what is the value of a + b?1. 72. 93. 54. 12

If x + y + z = 5, x2 + y2 + z2 = 21 and y2 = zx, then the value of y is:1. 1/52. 2/53. 1/24. 1/4

The cost of 4 pencil and 5 pen is Rs 47 and cost of 2 pencil and 7 pen is Rs. 55, then what is the cost of the one pencil?1. Rs. 52. Rs. 33. Rs. 74. Rs. 8

If a3 + b3 = 62 and a + b = 2, then the value of ab is:1. -62. 93. 64. -9

18x2 – 12x – 2y + 3xy = 1. (6x + y)(3x – 2)2. (6x – y)(3x – 2)3. (3x – y)(6x + 2)4. (3x + y)(6x – 2)

If \(x=\frac{{\sqrt 3 }}{2}\), then the value of \(\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }}\) is equal to:1. √22. √33. 34. 2

If 9x + 7y = 130, 2x – y = 11, then what will be the value of (x + y)?1. 152. 143. 214. 16

x - 1/x = 5 Find the value of x4 + 1/x41. 7312. 7243. 7274. 745

(7x + 12)2 – 336x is equal to 1. (7x + 11)22. (7x + 9)23. (7x – 9)24. (7x – 12)2

यदि `ab=2a-3b+ab` है तो `35+5**3` किसके बराबर होगा?A. 22B. 24C. 25D. 28

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answerI. 8x2 – 32x + 14 = 0II. 3y2 – 37y – 26 = 01.x > y2. x ≤ y3. x = y or the relationship cannot be established.4. x ≥ y5. x < y

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.I. 3x2 + 2x – 1 = 0II. 11y2 + 18y + 7 = 0 1. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.

In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. x2 – 3x – 10 = 0II. y2 – 18y + 45 = 0 1. x > y2. x < y3. x ≥ y4. x ≤ y5. x = y or the relationship between x and y cannot be established.