SSC CHSL MATHS QUIZ

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Question 1:

In $\triangle A B C, \angle C=60^{\circ}, \angle A=75^{\circ}, A D \perp B C$, where $D$ is a point on $B C, B E \perp A C$ at $E$, where $E$ is a point on $A C . A D$ and $B E$ intersect each other at $\mathrm{H}$. What is the measure of $\angle \mathrm{CHD}$ ?

Question 2:

In $\triangle A B C$, the sides $\mathrm{AB}$ and $\mathrm{AC}$ are produced to $\mathrm{P}$ and $\mathrm{Q}$, respectively. The bisectors of $\angle \mathrm{PBC}$ and $\angle \mathrm{QCB}$ intersect at a point $\mathrm{O}$. If $\angle \mathrm{A}=56^{\circ}$, then $\angle \mathrm{BOC}$ equals:

Question 3:

In a quadrilateral $A B C D, \angle C=102^{\circ}$ and $\angle D=58^{\circ}$. The bisectors of $\angle A$ and $\angle B$ meet at $\mathrm{P}$. What is the measure of $\angle A \mathrm{~PB}$ ?

Question 4:

A certain sum is distributed among A, B, C and D in the ratio 3 : 4 : 8 :6 respectively. If the share of C is ₹ 2,820, more than that of B, then what is the sum of shares of A and D?

Question 5:

In $\triangle \mathrm{ABC}, \angle \mathrm{BAC}=90^{\circ}$ and $\mathrm{AD} \perp \mathrm{BC} .$ If $\mathrm{BD}=$ $4 \mathrm{~cm}$ and $\mathrm{CD}=9 \mathrm{~cm}$, then the length of the $\mathrm{AD}(\mathrm{in} \mathrm{cm})$ is?

Question 6:

The third proportional of 12 and 18 is:

Question 7:

A policeman goes from his home to police station with his bike at a speed of $50 \mathrm{~km} / \mathrm{h}$, he is late by 30 minute. If he goes at $60 \mathrm{~km} / \mathrm{h}$, he is late by only 5 minute. Find the distance between his house and police station. (in $\mathrm{km}$ )

Question 8:

The perimeters of two similar triangles $\Delta \mathrm{ABC}$ and $\Delta \mathrm{PQR}$ are $36 \mathrm{~cm}$ and $24 \mathrm{~cm}$ respectively. If $\mathrm{PQ}=10 \mathrm{~cm}$, then $\mathrm{AB}$ is:

Question 9:

If $\sin \theta+\cos \theta=\frac{\sqrt{5}}{2}$ , then  Find the value of  $\sin ^{6} \theta+\cos ^{6} \theta+$ $5 \sin ^{2} \theta \cdot \cos ^{2} \theta$ .

Question 10:

A, B, C, D are four points on a circle. AC and BD intersect at a point $\mathrm{E}$ such that $\angle \mathrm{BEC}=$ $120^{\circ}$ and $\angle \mathrm{ECD}=40^{\circ} \cdot \angle \mathrm{BAC}$ is: