Objective Type Question
Question 1:
Mark the correct alternative in the following question:
The zero of 3x + 2 is
(a) 23 (b) 32 (c) -23 (d) -32
Answer 1:
If 3x+2=0, then3x=-2 (Transposing +2 to RHS)⇒x=-23
So, the zero of 3x + 2 is -23.
Note: A zero is that number, when put in place of the variable, makes the expression equal to zero.
Hence, the correct alternative is option (c).
Question 2:
Mark the correct alternative in the following question:
If 2x-32=5x+34, then x=(a) 34 (b) -34 (c) 43 (d) -43
Answer 2:
As, 2x-32=5x+34⇒2x-5x=32+34 (By transposing -32 to RHS and 5x to LHS)⇒-3x=64+34⇒-3x=6+34⇒x=94×(-3) (By transposing -3 to RHS)⇒x=34×(-1)⇒x=3-4∴ x=-34
Hence, the correct alternative is option (b).
Question 3:
Mark the correct alternative in the following question:
If x2-4=x3-1, then x=(a) 3 (b) 6 (c) 18 (d) 2
Answer 3:
As, x2-4=x3-1⇒x2-x3=4-1 (By transposing x3 to LHS and -4 to RHS)⇒3x6-2x6=3⇒3x-2x6=3⇒x6=3⇒x=3×6 (By transposing 6 to RHS)∴ x=18
Hence, the correct alternative is option (c).
Question 4:
Mark the correct alternative in the following question:
If x+2x-2=23, then x=(a) -10 (b) 10 (c) 43 (d) -43
Answer 4:
As, x+2x-2=23⇒3(x+2)=2(x-2) (By cross multiplication)⇒3x+6=2x-4⇒3x-2x=-6+4 (By transposing 2x to LHS and 6 to RHS)∴ x=-10
Hence, the correct alternative is option (a).
Question 5:
Mark the correct alternative in the following question:
If x6+x4=x2+34, then x=(a) 9 (b) 6 (c) -9 (d) 4
Answer 5:
As, x6+x4=x2+34⇒x6+x4-x2=34 (By transposing x2 to LHS)⇒2x12+3x12-6x12=34⇒2x+3x-6x12=34⇒-x12=34⇒-x×4=3×12 (By cross multiplication)⇒-4x=36⇒x=36-4∴ x=-9
Hence, the correct alternative is option (c).
Question 6:
Mark the correct alternative in the following question:
If 2x+53=14x+4, then x=(a) 3 (b) 4 (c) 34 (d) 43
Answer 6:
As, 2x+53=14x+4⇒2x-14x=4-53 (By transposing 53 to RHS and 14x to LHS)⇒2x1-x4=41-53⇒8x4-x4=123-53⇒8x-x4=12-53⇒7x4=73⇒7x×3=4×7 (By cross multiplication)⇒21x=28⇒x=2821∴ x=43
Hence, the correct alternative is option (d).
Question 7:
Mark the correct alternative in the following question:
If x2-x3=5, then x=(a) 8 (b) 16 (c) 24 (d) 30
Answer 7:
As, x2-x3=5⇒3x6-2x6=5⇒3x-2x6=5⇒x6=5⇒x=5×6 (By transposing 6 to RHS)∴ x=30
Hence, the correct alternative is option (d).
Question 8:
Mark the correct alternative in the following question:
If x-23=2x-13-1, then x=(a) 2 (b) 4 (c) 6 (d) 8
Answer 8:
As, x-23=2x-13-1⇒x-23-2x-13=-1 (By transposing 2x-13 to LHS)⇒(x-2)-(2x-1)3=-1⇒x-2-2x+13=-1⇒-x-13=-1⇒-x-1=-1×3 (By transposing 3 to RHS)⇒-x-1=-3⇒-x=-3+1 (By transposing -1 to RHS)⇒-x=-2∴ x=2
Hence, the correct alternative is option (a).
Question 9:
Mark the correct alternative in the following question:
The sum of two consecutive whole numbers is 43. The smaller number is
(a) 21 (b) 22 (c) 23 (d) 24
Answer 9:
Let the two consecutive whole numbers be x and x+1.As, the sum of the two consecutive whole numbers is 43.⇒x+(x+1)=43⇒2x+1=43⇒2x=43-1 (By transposing 1 to RHS)⇒2x=42⇒x=422 (By transposing 2 to RHS)∴ x=21
So, the smaller number is 21.
Hence, the correct alternative is option (a).
Question 10:
Mark the correct alternative in the following question:
The sum of two consecutive odd numbers is 36. The larger number is
(a) 17 (b) 15 (c) 19 (d) 21
Answer 10:
Let the two consecutive odd numbers be x and x+2.As, the sum of the two consecutive odd numbers is 36.⇒x+(x+2)=36⇒2x+2=36⇒2x=36-2⇒2x=34⇒x=342⇒x=17∴ x+2=17+2=19
So, the larger number is 19.
Hence, the correct alternative is option (c).
Question 11:
Mark the correct alternative in the following question:
Twice a number when increased by 7 gives 25. The number is
(a) 7 (b) 9 (c) 10 (d) 8
Answer 11:
Let the number be x.As, twice the number when increased by 7 gives 25.⇒2x+7=25⇒2x=25-7 (By transposing 7 to RHS)⇒2x=18⇒x=182 (By transposing 2 to RHS)∴ x=9
So, the number is 9.
Hence, the correct alternative is option (b).
Question 12:
Mark the correct alternative in the following question:
The length of a rectangle is three times its width and its perimeter 56 m. The length is
(a) 7 m (b) 14 m (c) 21 m (d) 28 m
Answer 12:
Let the width of the rectangle be x. Then,the length of the rectangle=3xAs, perimeter of the rectangle=56 m⇒2×(Length+Breadth)=56⇒2×(3x+x)=56⇒2×4x=56⇒8x=56⇒x=568∴ x=7So, the length of the rectangle=3x=3×7=21 m
Hence, the correct alternative is option (c).
Question 13:
Mark the correct alternative in the following question:
Two-third of a number is greater than one-third of the number by 5. The number is
(a) 10 (b) 5 (c) 15 (d) 12
Answer 13:
Let the number be x.As, two-third of a number is greater than one-third of the number by 5.⇒23x-13x=5⇒2x-x3=5⇒x3=5⇒x=5×3∴ x=15
So, the number is 15.
Hence, the correct alternative is option (c).
Question 14:
Mark the correct alternative in the following question:
If the sum of a number and its two-fifth is 70. The number is
(a) 70 (b) 50 (c) 60 (d) 90
Answer 14:
Let the number be x.As, the sum of a number and its two-fifth is 70.⇒x+25x=70⇒x1+2x5=70⇒5x5+2x5=70⇒5x+2x5=70⇒7x5=70⇒7x=70×5 (By transposing 5 to RHS)⇒7x=350⇒x=3507 (By transposing 7 to RHS)∴ x=50
So, the number is 50.
Hence, the correct alternative is option (b).
Question 15:
Mark the correct alternative in the following question:
23 of a number is less than the original number by 20. The number is
(a) 30 (b) 40 (c) 50 (d) 60
Answer 15:
Let the number be x.As, 23 of the number is less than the original number by 20.⇒x-23x=20⇒x1-2x3=20⇒3x3-2x3=20⇒3x-2x3=20⇒x3=20⇒x=20×3 (By transposing 3 to RHS)∴ x=60
So, the number is 60.
Hence, the correct alternative is option (d).
Question 16:
Mark the correct alternative in the following question:
A number is as much greater than 31 as it is less than 81. The number is
(a) 46 (b) 56 (c) 66 (d) 76
Answer 16:
Let the number be x.As, the number is as much greater than 31 as it is less than 81.⇒x-31=81-x⇒x+x=81+31 (By transposing -x to LHS and -31 to RHS)⇒2x=112⇒x=1122 (By transposing 2 to RHS)∴ x=56
So, the number is 56.
Hence, the correct alternative is option (b).
Question 17:
Mark the correct alternative in the following question:
Two complementary angles differ by 20°. The smaller angle is
(a) 55° (b) 25° (c) 65° (d) 35°
Answer 17:
Let the smaller angle be x. Then,The larger angle=(x+20°)As, the sum of the two complementary angles is always 90°.⇒x+(x+20°)=90°⇒2x+20°=90°⇒2x=90°-20°⇒2x=70°⇒x=70°2 (By transposing 2 to RHS)∴ x=35°
So, the smaller angle is 35°.
Hence, the correct alternative is option (d).
Question 18:
Mark the correct alternative in the following question:
Two supplementary angles differ by 40°. The measure of the larger angle is
(a) 70° (b) 80° (c) 110° (d) 100°
Answer 18:
Let the larger angle be x. Then,The smaller angle=(x-40°)As, the sum of the two supplementary angles is always 180°.⇒x+(x-40°)=180°⇒2x-40°=180°⇒2x=180°+40°⇒2x=220°⇒x=220°2 (By transposing 2 to RHS)∴ x=110°
So, the measure of the larger angle is 110°.
Hence, the correct alternative is option (c).
Question 19:
Mark the correct alternative in the following question:
The sum of three consecutive odd numbers is 81. The middle number is
(a) 25 (b) 27 (c) 31 (d) 29
Answer 19:
Let the three consecutive odd numbers be x, x+2 and x+4.As, the sum of the three consecutive numbers is 81.⇒x+(x+2)+(x+4)=81⇒3x+6=81⇒3x=81-6 (By transposing 6 to RHS)⇒3x=75⇒x=753 (By transposing 3 to RHS)⇒x=25∴ x+2=25+2=27
So, the middle number is 27.
Hence, the correct alternative is option (b).
Question 20:
If 2(2n + 5) = 3(3n - 10), then n =
(a) 5 (b) 3 (c) 7 (d) 8
Answer 20:
As, 2(2n+5)=3(3n-10)⇒4n+10=9n-30⇒4n-9n=-10-30 (By transposing 10 to RHS and 9n to LHS)⇒-5n=-40⇒n=-40-5 (By transposing -5 to RHS)∴ n=8
Hence, the correct alternative is option (d).
No comments:
Post a Comment