Mod
$|x+3|-|4-x|=|8+x| \text { How many solutions will this equation have? }$ A. 0 B. 1 C. 2 D. 3 A. 0 Take 4 cases (i) x < –8 –7 = –x – 8 i.e. x = –1, not satisfies assumed condition. (ii) –8 < x < –3 –7 = x+8 i.e. x = –15, not satisfies assumed condition (iii) –3 < x < 4 2x–1 = x + 8 i.e. x = 9, not satisfies assumed condition (iv) x > 4 7 = x + 8 i.e. x = 1, not satisfied assumed condition So 0 solution.
Area of Triangle in a Parallelogram
Q. In a parallelogram ABCD, points F and E are on AD and DC respectively. Point F divides AD in the ratio 2 : 1 and point E divides CD in the ratio 1 : 3. If the area of triangle DFE is 120 sq. units, then find the area (in sq units) of triangle BFE. A. 480 B. 450 C. 400 D. 500 C. 400 Area of triangle DEF = 120 \(\frac{1}{2} \times x \times 3y \times \sin\theta = 120\) ⇒ xy sinθ = 80 Now, area of parallelogram ABCD = 3x × 4y × sinθ = 960 We know, sin(180 – θ) = sinθ Area of triangle AFB = \(\frac{1}{2} \times 2x \times 4y \times \sin\left(180 - \theta \right) = 320\) Area of triangle BCE = \(\frac{1}{2} \times 3x \times y \times \sin\left(180 - \theta \right) = 120\) So, area of traingle [...]
Square and hexagon
Q. PQRS is a square of area 24 sq. cm. T and V are the midpoints of PQ and RS, respectively. X and Y are points on the line joining T and V such that angle PXQ = RYS = 120°. Find the area (in sq. cm) of the hexagon PXQRYS. A. \(4\left(2\sqrt{3}-1\right)\) B. \(2\left(6-2\sqrt{3}\right)\) C. \(12\left(2+\sqrt{3}\right)\) D. \(4\left(6-\sqrt{3}\right)\) D. \(4\left(6-\sqrt{3}\right)\)
(x+6)(x–6) is a integer
Q. Find the largest integral value of x such that \(\large \frac{x+6}{x-6}\) is a positive integer. TITA type i.e. Type In The Answer type 18 \(\frac{x+6}{x-6}\) = integer \(\Rightarrow \frac{x-6+12}{x-6}\) = integer \(\Rightarrow 1+\frac{12}{x-6}\) = integer So, (x – 6) is a factor of 12 x – 6 = 1, 2, 3, 4, 6, 12 Largest possibility of x = 12 + 6 = 18
4 digit perfect squares
Q. How many 4-digit perfect squares can be formed using the digits 1, 2, 5 and 9, without repetition? TITA type i.e. Type In The Answer type 0 The digit sum of a perfect square is always 1, 4, 7 or 9. Sum of the digits of the given numbers (1, 2, 5, 9) is 8. So, there will be no perfect square possible.
Points on a circle forming an equilateral triangle
Q. A, B, C and D are points on the circumference of a circle. ABD is an equilateral triangle and AC is the diameter of the circle. Find the ratio of the perimeter of ABCD and the length of AC. A. \(1+\frac{\sqrt[2]{3}}{2}\) B. \(\frac{1}{2}+\sqrt[2]{3}\) C. \(1+\sqrt[2]{3}\) D. \(2+\sqrt[2]{3}\) C. 1+\(\sqrt[2]{3}\)
Series
Evaluate $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$ upto n terms A. $2^n -n +1$ B. $n-2^{-n} +1$ C. $n+2^{-n} -1$ D. $2^n -n-1$ C.$n-1+2^{-n}$ Given $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$ $$ \begin{aligned} & =\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots \\ & =\mathrm{n}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots \ldots \ldots+\frac{1}{2^n}\right) \end{aligned} $$ $=\mathrm{n}-\frac{\frac{1}{2}\left(1-2^{-\mathrm{n}}\right)}{1-\frac{1}{2}} \quad\left[\right.$ Sum of $\mathrm{GP}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}$ when $\left.\mathrm{r}<1\right]$ $$ =n-1+2^{-n} $$
Permutation of 3-digit number > 100
Q. The number of three-digit numbers strictly greater than hundred, containing only digits 0, 1, 2, 3, 4 is ________. TITA type i.e. Type In The Answer type 99 Digits to be used: 0, 1, 2, 3, 4. Further we can repeat digits since nothing is mentioned. _ _ _ > 100 Case 1: 1st position: 1 2nd position: 5 possibilities 3rd position: 5 possibilities Case 2: 1st position: 2/3/4 2nd position: 5 possibilities 3rd position: 5 possibilities Total possibilities = 5 × 5 + 3 × 5 × 5 – 1(for 100) = 99
Numbers
Which of the following cannot be the sum of a 6-digit number and the number formed by reversing its digits? A.777777 B.1234442 C.424523 D.951467 D.951467 Let the numbers be abcdef and fedcba. When we add them we get 100000(a+f) + 10000(b+e) + 1000(c+d) + 100(d+c) +10(b+e) + (f+a) = 100001(a+f) + 10010(b+e) + 1100(c+d) Whatever (a+f) is, that will be the first digit and the last digit. (a+f) can be anything from 2 (=1+1) to 18 (=9+9) Moreover,Add the numbers, club them, and you'll see that the sum is divisible by 11. Only option D satisfies this
SI, CI
What rate of simple interest is equal to 40% p.a compound interest, compounded every quarter? TITA type i.e. Type In The Answer type 46.41% Let the principal be Rs. 100 Rate = 40%p.a. If the rate of interest is quarterly, then rate= 40/4 = 10% Time = 4 A= $100(1+\frac{10}{100})^4$ A=146.41 CI= 146.41-100 =46.41 Simple interest= 46.41/100 *100 = 46.41%