Maths MCQ

Let the number be $x.$
As, two$-$third of a number is greater than one$-$third of the number by $5.$
$\Rightarrow\frac{2}{3}\text{x}-\frac13\text{x}=5$
$\Rightarrow\frac{2\text{x}-\text{x}}{3}=5$
$\Rightarrow\frac{\text{x}}{3}=5$
$\Rightarrow\text{x}=5\times3$
$\therefore\text{x}=15$
So, the number is $15.$
Hence, the correct alternative is option $(c).$

Let the two consecutive odd numbers be $x$ and $x + 2.$
As, the sum of the two consecutive odd numbers is $36.$
$\Rightarrow x + (x + 2) = 36$
$\Rightarrow 2x + 2 = 36$
$\Rightarrow 2x = 36 - 2$
$\Rightarrow 2x = 34$
$\Rightarrow\text{x}=\frac{34}2$
$\Rightarrow x = 17$
$\therefore x + 2 = 17 + 2 = 19$
So, the larger number is $19.$
Hence, the correct alternative is option $(c).$

Let the two consecutive whole numbers be $x$ and $x + 1.$
As, the sum of the two cons cutive whole numbers is $43.$
$\Rightarrow x + (x + 1) = 43$
$\Rightarrow 2x + 1 = 43$
$\Rightarrow 2x = 43 - 1 ($By transposing $1$ to $\text{R.H.S.)}$
$\Rightarrow 2x = 42$
$\Rightarrow\text{x}=\frac{4}{22} ($By transposing $2$ to $\text{R.H.S.)}$
$\therefore\text{x}=21$
So, the smaller number is $21.$
Hence, the correct alternative is option $(a).$

Let the larger angle be $x.$
Then, the smaller angle $= (x - 40^\circ )$
As, the sum of the two supplementary angles is always $180^\circ .$
$\Rightarrow x + (x - 40^\circ ) = 180^\circ $
$\Rightarrow 2x - 40^\circ = 180^\circ $
$\Rightarrow 2x = 180^\circ + 40^\circ $
$\Rightarrow 2x = 220^\circ $
$\Rightarrow\text{x}=\frac{220^\circ}{2} ($By transposing $2$ to $\text{R.H.S.)}$
$\therefore\text{x}=1106^\circ$
So, the measure of the larger angle is $110^\circ .$
Hence, the correct alternative is option $(c).$

Let the smaller angle be $x.$
Then,The larger angle $= (x + 20^\circ )$
As, the sum of the two complementary angles is always $90^\circ .$
$\Rightarrow x + (x + 20^\circ ) = 90^\circ $
$\Rightarrow 2x + 20^\circ = 90^\circ $
$\Rightarrow 2x = 90^\circ - 20^\circ $
$\Rightarrow 2x = 70^\circ $
$\Rightarrow\text{x}=\frac{70^\circ}{2} ($By transposing $2$ to $\text{R.H.S.)}$
$\therefore\text{x}=35^\circ$
So, the smaller angle is $35^\circ .$
Hence, the correct alternative is option $(d).$

Let the number be $x.$
As, $23$ of the number is less than the original number by $20.$
$\Rightarrow\text{x}-\frac23\text{x}=20$
$\Rightarrow\frac{\text{x}}{1}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}}{\text{3}}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}-\text{2x}}{3}=20$
$\Rightarrow\frac{\text{x}}{3}=20$
$\Rightarrow\text{x}=20\times3 ($By transposing $3$ to $\text{R.H.S.)}$
$\therefore\text{x}=60$
So, the number is $60.$
Hence, the correct alternative is option $(d).$

Let the number be $x.$
As, the number is as much greater than $31$ as it is less than $81.$
$\Rightarrow x - 31 = 81 - x$
$\Rightarrow x + x = 81 + 31 ($By transposing $-x$ to $\text{L.H.S.}$ and $-31$ to $\text{R.H.S.)}$
$\Rightarrow 2x = 112$
$\therefore\text{x}=\frac{112}{2} ($By transposing $2$ to $\text{R.H.S.)}$
$\therefore\text{x}=56$
So, the number is $56.$
Hence, the correct alternative is option $(b).$

Let the width of the rectangle be $x.$
Then,the length of the rectangle $= 3x$
As, perimeter of the rectangle $= 56m$
$\Rightarrow 2 \times ($Length $+$ Breadth$) = 56$
$\Rightarrow 2 \times (3x + x) = 56$
$\Rightarrow 2 \times 4x = 56$
$\Rightarrow 8x = 56$
$\Rightarrow\text{x}=\frac{56}8{}$
$\therefore\text{x}=7$
So, the length of the rectangle $= 3x = 3 \times 7 = 21m.$
Hence, the correct alternative is option $(c).$

As, $\frac{\text{x}-2}{3}=\frac{2\text{x}-1}{3}-1$
$\Rightarrow\frac{\text{x}-2}{3}-\frac{2\text{x}-1}{3}=-1 ($By transposing $\frac{2\text{x}-1}{3}$ to $\text{L.H.S.)}$
$\Rightarrow\frac{(\text{x}-2)-(\text{2x}-1)}{3}=-1$
$\Rightarrow\frac{\text{x}-2-2\text{x}+1}{3}=-1$
$\Rightarrow\frac{-\text{x}-1}{3}=-1$
$\Rightarrow-\text{x}-1=-1\times3 ($By transposing $3$ to $\text{R.H.S.)}$
$\Rightarrow-\text{x}-1=-3$
$\Rightarrow-\text{x}=-3-1 ($By transposing $-1$ to $\text{R.H.S.)}$
$\Rightarrow-\text{x}=-2$
$\therefore\text{x}=2$
Hence, the correct alternative is option $(a).$

As, $2\text{x}+\frac53=\frac14\text{x}+4$
$\Rightarrow2\text{x}-\frac14\text{x}=4-\frac53 ($By transposing $\frac53$ to $\text{R.H.S.}$ and $\frac14\text{x}$ to $\text{L.H.S.)}$
$\Rightarrow\frac{\text{2x}}{1}-\frac{\text{x}}{4}=\frac41-\frac53$
$\Rightarrow \frac{\text{8x}}{4}-\frac{\text{x}}{4}=\frac{12}{3}-\frac53$
$\Rightarrow \frac{8\text{x}-\text{x}}{4}=\frac{12-5}{3}$
$\Rightarrow\frac{7\text{x}}{4}=\frac{7}{3}$
$\Rightarrow\text{7x}\times3=4\times7 ($By cross multiplication$)$
$\Rightarrow21\text{x}=28$
$\Rightarrow\text{x}=\frac{28}{21}$
$\therefore\text{x}=\frac{4}{3}$
Hence, the correct alternative is option $(d).$

As, $2\text{x}-\frac32=\text{5x}+\frac34$
$\Rightarrow2\text{x}-5\text{x}=\frac{3}{4}+\frac{3}{4} ($By transpoing $-\frac32$ to $\text{R.H.S.}$ and $5x$ to $\text{L.H.S.)}$
$\Rightarrow-3\text{x}=\frac{6}{4}+\frac34$
$\Rightarrow-3\text{x}=\frac{6+3}{4}$
$\Rightarrow\text{x}=\frac{9}{4\times(-3)} ($By transposing $-3$ to $\text{R.H.S.)}$
$\Rightarrow\text{x}=\frac{3}{4\times(-1)}$
$\Rightarrow\text{x}=\frac{3}{-4}$
$\therefore\text{x}=-\frac34$
Hence, the correct alternative is option $(b).$

As, $2(2n + 5) = 3(3n - 10)$
$\Rightarrow 4n + 10 = 9n - 30$
$\Rightarrow 4n - 9n = -10 - 30 ($By transposing $10$ to $\text{R.H.S.}$ and 9n to $\text{L.H.S.})$
$\Rightarrow -5n = -40$
$\Rightarrow n = -40 - 5 ($By transposing $-5$ to $\text{R.H.S.)}$
$\therefore\text{n}=8$
Hence, the correct alternative is option $(d).$