Visual lens G Strategy (Concept of similarities)
In a triangle ABC, AB=AC=12 cm and D is a point on side BC such that AD= 8 cm. If AD is extended to point E such that angle ACB = angle AEB , then length in cm of AE is :
1.16
2.18
3.14
4.20
Answer & Explanation
Based on the information in the question triangle ABD ≈ triangle AEB , Thus :
AB/AE = AD/AB ,
12/AE = 8/12 = 2/3
Only option 2 satisfies the condition. Thus option 2 is the answer
Visual G Strategy | Triplets + Area of Triangles | Moderate
A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the length, in cm , of all three altitudes of the triangle ABC is….. …
Solution & Explanation
Give AB = AC = 50 cm , BC = 80 cm
The altitude from A to BC
Half of BC: 80 / 2 = 40 cm
Use Pythagoras’ theorem to find altitude h:
h² + 40² = 50²
h² + 1600 = 2500
h² = 900 , h = 30 cm
Altitude from A to BC = 30 cm
The altitudes from B and C
Area of the triangle using the altitude from A:
Area = (1/2) × 80 × 30 = 1200 cm²
Using area to find the altitude from B:
1200 = (1/2) × 50 × h
h = 1200 / 25 = 48 cm
Altitude from B to AC = 48 cm
Altitude from C to AB = 48 cm
Sum = 30 + 48 + 48 = 126
Answer 126
Square ±2 G Strategy | Easy
If (x²+ 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is:
1.44856√3
2.44853√3
3.44859√3
4.44850√3
Solution & Explanation
Correct Answer: 44853√3
Given x² + 1/x² = 25 , x > 0
Using identity (a + b)² = a² + b² + 2ab
I. x + 1/x:
(x + 1/x)² = x² + 2 + 1/x² = 25 + 2 = 27
So, x + 1/x = 3√3
II. x³ + 1/x³:
x³ + 1/x³ = (x + 1/x) * (x² + 1/x²) – (x + 1/x)
x³ + 1/x³ = (3√3) * 25 – 3√3 = 72√3
III.x⁴ + 1/x⁴:
x⁴ + 1/x⁴ = (x² + 1/x²)² – 2
x⁴ + 1/x⁴ = 25² – 2 = 625 – 2 = 623
IV.x⁷ + 1/x⁷:
x⁷ + 1/x⁷ = (x³ + 1/x³) * (x⁴ + 1/x⁴) – (x + 1/x)
x⁷ + 1/x⁷ = (72√3) * 623 – 3√3 = 44853√3
Visual lens G Strategy | Set theory + Maximum/Minimum | Moderate
In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is equal to half the number of students who chose both mathematics and physics. What is the maximum possible number of students who chose physics but not mathematics?
1.40
2.35
3.55
4.30
Solution & Explanation
Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40
Assume: x – number of students who chose both Physics and Mathematics , y – students who chose both Mathematics and Chemistry , z – student who chose both Physics and Chemistry , t – number of students who chose all three subjects
And z = y , y = x/2 (Given)
Using Set formula (the total number of students who chose at least one subject)
|P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C|
Substitute the values:
150 = 75 + 111 + 40 – x – y – z + t
Now, substitute z = y and y = x/2:
150 = 75 + 111 + 40 – x – (x/2) – (x/2) + t
150 = 226 – x – x + t
150 = 226 – 2x + t
2x – 76 = t (Equation 1)
We want to maximize 75 – x , (students who chose only Physics but not mathematics). Using options :
1.40 , 75 – x = 40 , x = 35 , t can not be negative
2.35 , 75 – x = 35 , x = 40 , t = 4
3.55 , 75 – x = 55 , x= 20 , t can not be negative
4.30 , 75 – x = 30 , x = 45 , t = 14
Since 75 – x is maximum at option 2 thus option 2 is the answer
Visual Lens G Strategy | Cyclic quadrilateral+ Area of trapezium | Moderate
ABCD is a trapezium in which AB is parallel to DC and AD is perpendicular to AB and AB = 3DC. If a circle inscribed in a trapezium touching all the sides has a radius 3cm , then the area in sq cm , of the trapezium is :
1.48
2.36√2
3.54
4.30√3
Solution & Explanation
Given:
Let the trapezium be ABCD.
AB || DC (The parallel sides are the bases).
AD perpendicular to AB (The trapezium is a right-angle trapezium). AB = 3DC. If AB = x , then DC = 3x
The circle inscribed in the trapezium has a radius r = 3cm
Height = 2r = 2 × 3 = 6cm
For cyclic quadrilateral Sum of parallel sides= Sum of non-parallel sides
AB + DC = AD + BC
3x + x = 6 + BC
BC = 4x – 6 —– I
In Triangle AEB with base BE , height CE and hypotenuse BC
BE = DC – AB = 3x – x = 2x , CE = 6cm , BC = 4x – 6
Using Pythagoras:
(2x)² + (6)² = (4x – 6)²
On solving we have x = 4cm
Area of trapezium = ½ * (AB + DC) × AD
= ½ .(4x) .6 = 48cm² , This option 1 is the answer
Breakup G Strategy | DPAC Logarithm + Equations | Hard
The sum of all possible real values of x for which
logx-3(x² – 9) = logx-3(x + 1) + 2 is :
1.√33
2.-3
3.(3+√33)/2
4.3
