CTET June 2011 Paper II Solved Question

**Sol:**

Let length of the rectangle = X and breadth = Y,

According to question,

$\frac{X}{2\left(X+Y\right)}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}=>3X=2X+2Y\phantom{\rule{0ex}{0ex}}=X=2Y\phantom{\rule{0ex}{0ex}}=\frac{X}{Y}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}=X:Y=2:1$

**Sol:**

$\left(5*2\right)={5}^{2}+{2}^{2}\phantom{\rule{0ex}{0ex}}=25+4\phantom{\rule{0ex}{0ex}}=29\phantom{\rule{0ex}{0ex}}Therefore,\left(5*2\right).25=({5}^{2}+{2}^{2}{)}^{2}-{25}^{2}\phantom{\rule{0ex}{0ex}}=(25+4{)}^{2}-{25}^{2}\phantom{\rule{0ex}{0ex}}={29}^{2}-{25}^{2}\phantom{\rule{0ex}{0ex}}=841-625\phantom{\rule{0ex}{0ex}}=216\phantom{\rule{0ex}{0ex}}$

- c
^{2}-a^{2}= b^{2} - c
^{2}-a^{2}< b^{2} - c
^{2}-a^{2}> b - c
^{2}+b^{2}= a^{2}

**Sol:**

Let a,b and c are the three natural numbers in ascending order, then
$a<b<c\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}<{b}^{2}<{c}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}<{c}^{2}-{a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {c}^{2}-{a}^{2}>{b}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {c}^{2}-{a}^{2}>b$

**Sol:**

Let the price of an item = x

price of actual items = price of 4 items = 4x

price offered to customer for the 4 items =3x

discount = 4x-43=x

% discount = $\frac{x\times 100}{4x}=25\%$

**Sol:**

√2 + √3 + √2 - √3 = √2+√2=2√2

**sol:**

Let the radius of the iron rod = r, length = h

hence its volue = πr^{2}h

A/Q,

radius after recasting = $\frac{r}{4}$

let the new length = H

Again from question, ${\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi}(\frac{\mathrm{r}}{4}{)}^{2}\mathrm{H}$

=> 16h=H

Hence new lenght = 16 times of its original length

**Sol:**

$Given,\phantom{\rule{0ex}{0ex}}x=\frac{547.527}{0.0082}\phantom{\rule{0ex}{0ex}}Dividingbothsideby10,\phantom{\rule{0ex}{0ex}}\frac{x}{10}=\frac{547.527}{0.082}\phantom{\rule{0ex}{0ex}}\frac{x}{10}=\frac{547527}{82}\phantom{\rule{0ex}{0ex}}Hence\frac{547527}{82}=\frac{x}{10}$

**Sol:**

$68600=2\times 2\times 2\times 5\times 5\times 7\times 7\times 7\phantom{\rule{0ex}{0ex}}=14\sqrt[3]{5\times 5}$

Hence **5** is the required number for the given number to be perfect cube.

**Sol:**

Given QA=4, CQ=2, CA=10

Let CP=X

∴ PQ=(X-2)

PB=(10-X)

According to question,

AP=PB

From ∆ AQP

AP ^{2} = QP ^{2}+QA ^{2}

(10-X)
^{2}=(X-2)^{2}+4^{2} (10-X+X-2)(10-X-X+2)=16

8(12-2X)=16

12-2X=2

6-X=1

∴ X= **5**

- Circle
- Spindle
- Cylinder
- trapezium

**Sol:**

**Spindle**

**Sol:**

**Sol:**

**Sol:**

**Sol:**

**Sol:**

**Sol:**

- Natural Number
- Even Number
- Odd Number
- Prime Number

**Sol:**

- 349/625
- 359/625
- 357/625
- 347/625

**Sol:**

- 2:1
- 1:1
- 2:√3
- √3:2

**Sol:**

**Sol:**

4/16 - 1/8 = 3/8

6/7 - 2/9 = 4/2

The above represents the work of a student. If this error pattern continues, the student's answer to 5/11 - 2/7 will be?

**Sol:**

- Rotation
- Reflections
- Nets
- Decimals

**Sol:**

4

6

7

- add exponents
- add exponents and multiply
- multiply numbers with same base
- multiply numbers with different bases

**Sol:**

- The two equations can be added or subtracted to solve them
- That equations can be solved by method of substitutions
- The method of solving equations graph
- That both the equations in 3 can be altered by multiplying with suitable numbers

**Sol:**