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what should iread for the preparation for B.ed entrance exam

Tanya Gupta 26th Feb, 2024

Hello

For B.Ed entrance exam prep, here are following things you can do-

1 Use guidebooks tailored to your exam.

2 Review previous years' papers.

3 Study general knowledge and current affairs.

4 Brush up on teaching aptitude and educational psychology.

5 Utilize online resources for practice tests and materials.

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Question : Directions: Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it?
c_bba_cab_ac_ab_ac

Option 1: babcc

Option 2: bcacb

Option 3: acbcb

Option 4: abcbc

Team Careers360 25th Jan, 2024

Correct Answer: acbcb


Solution : Given:
c_bba_cab_ac_ab_ac

To fill the series we have to divide the series – c_b/ba_/cab/_ac/_ab/_ac
Let's check each option:
First option: babcc; cbb/baa/cab/bac/cab/cac; (No pattern has been found.)
Second option: bcacb; cbb/bac/cab/

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Question : If $x+y+z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1,$ and $xyz=-1$, then $x^3+y^3+z^3 $ is equal to:

Option 1: –1

Option 2: 1

Option 3: –2

Option 4: 2

Team Careers360 25th Jan, 2024

Correct Answer: 1


Solution : Given:
$x+y+z=1,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, $xyz=-1$
Consider, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
⇒ $xy+yz+zx=xyz=-1$
Now, $x+y+z=1$
⇒ $(x+y+z)^2=1^2$
⇒ $x^2+y^2+z^2+2(xy+yz+zx)=1$
⇒ $x^2+y^2+z^2+2(-1)=1$
⇒ $x^2+y^2+z^2=3$
We know, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx)$
So, putting all the values, we get,
⇒ $x^3+y^3+z^3-3×(-1)=1×[3-(-1)]$
⇒ $x^3+y^3+z^3=4-3$
$\therefore x^3+y^3+z^3=1$
Hence, the correct answer is 1.

25 Views

Question : Direction: Study the bar graph given below, which shows the percent distribution of total expenditures of a company under various expenses, and answer the question.

The expenditure on interest on loans is more than the expenditure on transport by:

Option 1: 5%

Option 2: 10%

Option 3: 40%

Option 4: 30%

Team Careers360 25th Jan, 2024

Correct Answer: 40%


Solution : From the graph,
The percentage of money spent on transit = 12.5
The percentage of money spent on loan interest = 17.5
Percentage increase = $\frac{\text{( Interest on loans expense – Transport expenditure)}}{\text{(Transport expenditure)}} \times 100$
So, the required percentage $=\frac{(17.5 – 12.5)}{12.5}\times100=\frac{5}{12.5} \times 100=40$%

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