In triangle PQR, points A, Band C are taken on PQ, PR and QR respectively such that QC=AC and CR = CB. If ∠QPR = 40°, then ∠ACB is equal to

Solution
AC= QC
∴ ∠QAC = ∠CQA = x
CR=CB
⇒∠CBR = ∠CRB = Y
∴ From Δ PQR,
∠x+ ∠y+ 40° = 180
∠x + ∠y = 140°....(i)
Again,
∠ACQ + ∠ACB + ∠BCR = 180°
⇒ 180°  2x + ∠ACB + 180°  2y = 180°
⇒ ∠ACB = 2 (x + y)  180°
= 2 × 140  180° = 100°
If the orthocentre and the centroid of a triangle are the same, then the triangle is

Solution
In equilateral triangle orthocentre and centroid lie at the same point.
The value of A, for which the expression x^{2}+ x^{2}– 5x + λ. will be divisible by (x – 2) is

Solution
( x  2) is a factor of polynomial P(x) = x^{3} + x^{2}5x+A.
P(2) = 0 (ie, on putting x = 2)
⇒2^{3} + 2^{2}  5 x 2 + A = 0
⇒8 + 4  10 + 1.._=0
⇒λ+2=0⇔λ=2
If x = 3 + \(2\sqrt{2}\) and xy = 1, then the value of\(\frac{x^{2} +3xy + y^{2}}{x^{2} 3xy + y^{2}}\) is

Solution
\(x =3 + 2\sqrt{2},\: xy =1\)
\(⇒y=\frac{1}{3+2\sqrt{2}}=\frac{1}{3+2\sqrt{2}}×\frac{32\sqrt{2}}{32\sqrt{2}}\)
\(=\frac{32\sqrt{2}}{98}=32\sqrt{2}\)
\(∴x + y = 3 + 2\sqrt{2} +3 2\sqrt{2} = 6\)
\(∴\frac{X^{2} + 3xy+ y^{2}}{x^{2}3xy+y^{2}}=\frac{(x + y)^{2} + xy}{(x+y)^{2} 5xy}\)
\(=\frac{36 + 1}{36  5}=\frac{37}{31}\)
A grocery dealer cheats to the extent of 10% while buying as well as selling by using false weight.
What is his increase in the profit % ?

Solution
Quicker Method:
Gain per cent\(=\left ( 10+10+\frac{10\times 10}{100} \right )\)%=21%
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Question Figure
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EXAMINATION
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