2 Marks Questions

Question 12 Marks

$(B)$ The $\text{H.C.F}$ of $85$ and $238$ is expressible in the form $85m -238.$ Find the value of $m.$

Answer

$(B)85=5 \times 17,238$
$=2 \times 7 \times 17$
$\operatorname{HCF}(85,238)=17$
$17=85xm-238$
$m =3$

Question 22 Marks

(B)How many positive three digit integers have the hundredths digit 8 and unit’s digit 5? Find the probability of selecting one such number out of all three digit numbers.

Answer

(B)Total number of three-digit numbers $=900$.
Numbers with hundredth digit 8 & and unit's digit 5 are 805,815,
$825, \ldots ., 895$
Number of favourable outcomes $=10$
$P ($ selecting one such number $)=\frac{10}{900}$ or $\frac{1}{90}$

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Question 32 Marks

Show that the points $A(-5,6), B(3, 0)$ and $C( 9, 8)$ are the vertices of an isosceles triangle.

Answer

$A B=\sqrt{(3+5)^2+(0-6)^2}=10$
$B C=\sqrt{(9-3)^2+(8-0)^2}=10$
$A C=\sqrt{(9+5)^2+(8-6)^2}=10 \sqrt{2}$
Since $A B=B C$, therefore $\triangle A B C$ is isosceles

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Question 42 Marks

Find the point$(s)$ on the $x-$axis which is at a distance of $\sqrt{41}$ units from the point $(8,-5)$.

Answer

Let the required point be $(x, 0)$
$\sqrt{(8-x)^2+25}=\sqrt{41}$
$\Rightarrow(8-x)^2=16$
$=>8-x= \pm 4$
$\Rightarrow x=4,12$
Two points on the $x-$axis are $(4,0) \ (12,0)$.

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Question 52 Marks

Evaluate: $\frac{2 \sin ^2 60^{\circ}-\tan ^2 30^{\circ}}{\sec ^2 45^{\circ}}$

Answer

$\frac{2\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{\sqrt{3}}\right)^2}{(\sqrt{2})^2}$
$=\frac{7}{12}$

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Question 62 Marks

Two dice are rolled together bearing numbers 4, 6, 7, 9, 11, 12. Find the probability that the product of numbers obtained is an odd number

Answer

Total number of possible outcomes $=6 \times 6=36$
For a product to be odd, both the numbers should be odd.
Favourable outcomes are $(7,7)(7,9)(7,11)(9,7)(9,9)(9,11)(11,7)(11,9)$ $(11,11)$
no. of favourable outcomes $=9$
$P ($ product is odd $)=\frac{9}{36}$ or $\frac{1}{4}$

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Question 72 Marks

Find the $\text{H.C.F}$ and $\text{L.C.M}$ of $480$ and $720$ using the Prime factorisation method.

Answer

$480=2^5 \times 3 \times 5$
$720=2^4 \times 3^2 \times 5$
$\operatorname{LCM}(480,720)=2^5 \times 3^2 \times 5=1440$
$\operatorname{HCF}(480,720)=2^4 \times 3 \times 5=240$

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Maths 2 Marks Questions

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