eSaral - JEE | NEET | Class 8-10 Exam Prep App

Solve:

Question: Solve: (i) 25 + 22 + 19 + 16 + ... +x= 115 (ii) 1 + 4 + 7 + 10 + ... +x= 590. Solution: (i) 25 + 22 + 19 + 16 + ... +x= 115 Here, $a=25, d=-3, S_{n}=115$ We know: $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow 115=\frac{n}{2}[2 \times 25+(n-1) \times(-3)]$ $\Rightarrow 115 \times 2=n[50-3 n+3]$ $\Rightarrow 230=n(53-3 n)$ $\Rightarrow 230=53 n-3 n^{2}$ $\Rightarrow 3 n^{2}-53 n+230=0$ By quadratic formula: $n=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ Substituting $a=3, b=-53$ and $c=230$, ...

if

Question: If $\tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2}$, find $x$ Solution: We know that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$. We have $\tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \tan ^{-1}(\sqrt{3})=\frac{\pi}{2}-\cot ^{-1} x$ $\Rightarrow \tan ^{-1}(\sqrt{3})=\tan ^{-1} x$ $\Rightarrow x=\sqrt{3}$ $\therefore x=\sqrt{3}$...

if

Question: If $\tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2}$, find $x$ Solution: We know that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$. We have $\tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \tan ^{-1}(\sqrt{3})=\frac{\pi}{2}-\cot ^{-1} x$ $\Rightarrow \tan ^{-1}(\sqrt{3})=\tan ^{-1} x$ $\Rightarrow x=\sqrt{3}$ $\therefore x=\sqrt{3}$...

In the given figure, the perimeter of Δ ABC is

Question: In the given figure, the perimeter of Δ ABC is (a) 30 cm(b) 60 cm(c) 45 cm(d) 15 cm Solution: We know that tangents from an external point will be equal in length. Therefore, AQ = AR AQis given as 4 cm. Therefore, AR= 4 Similarly, PC = CQ PC= 5 Therefore, CQ= 5 Similarly, BP = BR BR= 6 Therefore, BP= 6 Now we can find out the perimeter of the triangle. Perimeter =AB + BC + CA From the figure we have, Perimeter = AR+ RB + BP + PC + CQ + QA Perimeter = 4 + 6 + 6 + 5 + 5 + 4 Perimeter = 3...

Evaluate:

Question: Evaluate: $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$ Solution: We know that $\sin ^{-1}(\sin x)=x$. We have $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{3 \pi}{5}\right)\right\} \quad\left[\because\left(\pi-\frac{3 \pi}{5}\right) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$ $=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)$ $=\frac{2 \pi}{5}$ $\therefore \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\frac{2 \pi}{5}$...

In the given figure, RQ is a tangent to the circle with centre O.

Question: In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR = (a) 8 cm(b) 3 cm(c) 2.5 cm(d) 5 cm Solution: It is given that, SQ= 6 cm SinceSQpasses through the centre of the circleO, it is the diameter. Therefore, the radius, OQ= 3 cm Also, given is QR= 4 We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore,OQis perpendicular toOR. We can find the length ofORby using Pythagoras theorem. We have,...

In a || gm ABCD, if ∠A = (2x + 25)° and ∠B = (3x − 5)°,

Question: In a || gmABCD, ifA= (2x+ 25) and B= (3x 5), find the value ofxand the measure of each angle of the parallelogram. Solution: ABCDis a parallelogram.i.e., A=CandB=D (Opposite angles)Also, A+B= 180o (Adjacent angles are supplementary) (2x+ 25) +(3x 5) = 180⇒ 5x+20 = 180⇒5x =160⇒x= 32oA =2⨯32 + 25 = 89oandB =3⨯32 5 = 91oHence,x= 32o, A=C=89oandB=D=91o...

In the adjoining figure, ABCD is a parallelogram in which ∠BAO = 35°, ∠DAO = 40° and ∠COD = 105°.

Question: In the adjoining figure,ABCDis a parallelogram in whichBAO= 35, DAO= 40 and COD= 105. Calculate (i) ABO, (ii) ODC, (iii) ACB, (iv) CBD. Solution: ABCD is a parallelogram. AB ∣∣ DC and BC∣∣ AD(i) In∆AOB,BAO= 35,AOB= COD= 105 (Vertically opposite angels)ABO= 180o (35o+ 105o) = 40o(ii)ODCandABOare alternate interior angles.ODC= ABO =40o(iii)ACB=CAD =40o (Alternate interior angles)(iv)CBD=ABCABD ...(i) ABC=180o BAD (Adjacent angles are supplementary) ⇒ABC=180o 75o= 105o⇒CBD=105oAB...

In the given figure, if AD, AE and BC are tangents to the circle at D,

Question: In the given figure, if AD, AE and BC are tangents to the circle at D, E and F respectively, Then, (a) AD = AB + BC + CA(b) 2AD = AB + BC + CA(c) 3AD = AB + BC + CA(d) 4AD = AB + BC + CA Solution: In the given problem, the Right Hand Side of all the options is same, that is, AB + BC + CA So, we shall find outAB + BC + CAand check which of the options has the Left Hand Side value which we will arrive at. By looking at the figure, we can write, AB + BC + CA = AB + BF + FC + CD We know th...

Show that

Question: Show that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$. Solution: We have LHS $=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\sin a$, we get $=\sin ^{-1}\left(2 \sin a \sqrt{1-\sin ^{2} a}\right)$ $=\sin ^{-1}(2 \sin a \cos a)=\sin ^{-1}(\sin 2 a)=2 a=2 \sin ^{-1} x \quad(\because x=\sin a)$...

Show that

Question: Show that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$. Solution: We have LHS $=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\sin a$, we get $=\sin ^{-1}\left(2 \sin a \sqrt{1-\sin ^{2} a}\right)$ $=\sin ^{-1}(2 \sin a \cos a)=\sin ^{-1}(\sin 2 a)=2 a=2 \sin ^{-1} x \quad(\because x=\sin a)$...

AB and CD are two common tangents to circles which touch each other at C.

Question: AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to(a) 4 cm(b) 6 cm(c) 8 cm(d) 12 cm Solution: Let us first put the given data in the form of a diagram. We know that tangents drawn from an external point will be equal. Therefore, AD = CD Since CD is given as 4 cm, AD= 4 cm Similarly, BD = CD Therefore, BD= 4 cm AB = AD + BD AB= 4 + 4 AB= 8 cm Therefore, the answer to this question is option (c)...

Find the sum of all those integers between 100 and 800

Question: Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7. Solution: The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are: 103, 119...791 Here, we have: a= 103 d= 16 $a_{n}=791$ We know: $a_{n}=a+(n-1) d$ $\Rightarrow 791=103+(n-1) \times 16$ $\Rightarrow 688=16 n-16$ $\Rightarrow 704=16 n$ $\Rightarrow 44=n$ Also, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{44}=\frac{44}{2}...

The length of the tangent drawn from a point 8 cm away form the centre of a circle of radius 6 cm is

Question: The length of the tangent drawn from a point 8 cm away form the centre of a circle of radius 6 cm is (a) $\sqrt{7} \mathrm{~cm}$ (b) $2 \sqrt{7}$ (c) $10 \mathrm{~cm}$ (d) $5 \mathrm{~cm}$ Solution: Let us first put the given data in the form of a diagram. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,OPis perpendicular toQP. We can now use Pythagoras theorem to find the length ofQP. $Q P^{2}=O Q^{2}-O P^{2}$ $Q P^{2}...

Find the sum of all those integers between 100 and 800

Question: Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7. Solution: The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are: 103, 119...791 Here, we have: a= 103 d= 16 $a_{n}=791$ We know: $a_{n}=a+(n-1) d$ $\Rightarrow 791=103+(n-1) \times 16$ $\Rightarrow 688=16 n-16$ $\Rightarrow 704=16 n$ $\Rightarrow 44=n$ Also, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{44}=\frac{44}{2}...

If four sides of a quadrilateral ABCD are tangential to a circle, then

Question: If four sides of a quadrilateralABCDare tangential to a circle, then(a) AC + AD = BD + CD(b) AB + CD = BC + AD(c) AB + CD = AC + BC(d) AC + AD = BC + DB Solution: The figure of the Quadrilateral is drawn below. We know that length of the tangents drawn from an external point will be equal. Therefore, AP = PQ DP = DS CR = CS BR = BQ Let us add all the above four equations. We get AP + DP + CR + BR = PQ + DS + CS + BQ By looking at the figure, we can rewrite the above equation as, AD + B...

Find the sum of all those integers between 100 and 800

Question: Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7. Solution: The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are: 103, 119...791 Here, we have: a= 103 d= 16...

Write the value

Question: Write the value of $\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)$. Solution: $\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right) \neq \frac{5 \pi}{4}$ as $\frac{5 \pi}{4}$ does not lie between 0 and $\pi$. We have $\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{3 \pi}{4}\right)\right\}=\cos ^{-1}\left\{\cos \left(\frac{3 \pi}{4}\right)\right\}=\frac{3 \pi}{4}$...

PQ is a tangent drawn from a point P to a circle with centre O and

Question: PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that POR = 120, then OPQ is(a) 60(b) 45(c) 30(d) 90 Solution: Let us first put the given data in the form of a diagram. Given data is as follows: QORis the diameter. $\angle P O R=120^{\circ}$ We have to find $\angle O P Q$. Since QOR is the diameter of the circle, it is a straight line. Therefore, $\angle Q O R=180^{\circ}$ That is, $\angle P O R+\angle P O Q=180^{\circ}$ But $\angl...

Find the sum of the series:

Question: Find the sum of the series: 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3nterms. Solution: The given sequence i.e., 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 +..... to 3n terms. can be rewritten as 3 + 6 + 9 + .... to nterms + 5 + 9 + 13 + .... tonterms + 7 + 12 + 17 + .... tonterms Clearly, all these sequence forms an A.P. havingnterms with first terms 3, 5, 7 and common difference 3, 4, 5 Hence, required sum $=\frac{n}{2}[2 \times 3+(n-1) 3]+\frac{n}{2}[2 \times 5+(n-1) 4]+\frac{n}{2}[2...

Find the sum of all integers between 100 and 550,

Question: Find the sum of all integers between 100 and 550, which are divisible by 9. Solution: The integers between 100 and 550 that are divisible by 9 are: 108, 117...549 Here, we have: $a=108$ $d=9$ $a_{n}=549$ $\Rightarrow 108+(n-1)(9)=549$ $\Rightarrow 9 n-9=441$ $\Rightarrow 9 n=450$ $\Rightarrow n=50$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{50}=\frac{50}{2}[2 \times 108+(50-1) \times 9]$ $\Rightarrow S_{50}=25(657)=16425$...

ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm.

Question: ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in Δ ABC. The radius of the circle is(a) 1 cm(b) 2 cm(c) 3 cm(d) 4 cm Solution: Let us first put the given data in the form of a diagram. Let us first find out AC using Pythagoras theorem. $A C^{2}=A B^{2}+B C^{2}$ $A C^{2}=8^{2}+6^{2}$ $A C^{2}=64+36$ $A C^{2}=100$ $A C=10$ Also, we know that tangents drawn from an external point will be equal in length. Therefore w...

In the adjoining figure, ABCD is a parallelogram in which ∠A = 60°.

Question: In the adjoining figure,ABCDis a parallelogram in whichA= 60. If the bisectors of Aand BmeetDCatP, prove that (i) APB= 90, (ii)AD=DPandPB=PC= BC, (iii)DC= 2AD. Solution: ABCDis a parallelogram.A=CandB=D (Opposite angles)AndA+ B= 180o (Adjacent angles are supplementary)B= 180o A⇒180o 60o= 120o (∵A= 60o)A=C =60oandB=D =120o (i) $\ln \triangle A P B, \angle P A B=\frac{60^{\circ}}{2}=30^{\circ}$ and $\angle P B A=\frac{120^{\circ}}{2}=60^{\circ}$ $\therefore \angle A P B=180^{\c...

Find the sum of all even integers between 101 and 999.

Question: Find the sum of all even integers between 101 and 999. Solution: The even integers between 101 and 999 are: 102, 104...998 Here, we have: $a=102$ $d=2$ $a_{n}=998$ $\Rightarrow 102+(n-1) 2=998$ $\Rightarrow 2 n-2=896$ $\Rightarrow 2 n=898$ $\Rightarrow n=449$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{449}=\frac{449}{2}[2 \times 102+(449-1) \times 2]$ $\Rightarrow S_{449}=\frac{449}{2}[1100]$ $\Rightarrow S_{449}=246950$...

Find the sum of all even integers between 101 and 999.

Question: Find the sum of all even integers between 101 and 999. Solution: The even integers between 101 and 999 are: 102, 104...998 Here, we have: $a=102$ $d=2$ $a_{n}=998$ $\Rightarrow 102+(n-1) 2=998$ $\Rightarrow 2 n-2=896$ $\Rightarrow 2 n=898$ $\Rightarrow n=449$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{449}=\frac{449}{2}[2 \times 102+(449-1) \times 2]$ $\Rightarrow S_{449}=\frac{449}{2}[1100]$ $\Rightarrow S_{449}=246950$...

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