1. Ramesh and Rahman can do a work in 20 and 25 days respectively. After doing collectively 10 days of work, they leave the work due to illness and Suresh completes rest of the work in 3 days. How many days Suresh alone can take to complete the whole work ?

A) 32 days B) 28 days C) 29 days D) 30 days

If $a + b + c = 9$ and $a b + b c + c a = 26$ , then find $a^{2} + b^{2} + c^{2}$ ?

A.26 B.27 C.28 D.29

Correct option is D)

Given, $a + b + c = 9$ and $a b + b c + c a = 26$ .

We know, $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$\Rightarrow (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$ .

Putting the values here, we get,
$(9)^{2} = a^{2} + b^{2} + c^{2} + 2 (26)$
$\Rightarrow 81 = a^{2} + b^{2} + c^{2} + 52$
$\Rightarrow a^{2} + b^{2} + c^{2} = 81 - 52 = 29$ .

Hence, option $D$ is correct.

A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?

A. 1 hours
B. 1.5 hours
C. 2 hours
D. 3 hours
E. 4 hours

3. A and B invest Rs. 3000 and Rs. 4000, respectively in a business. A received Rs. 10 per month out of the profit as a remuneration for running the business and the rest of the profit is divided in proportion to the investments. If in a year A totally receives Rs. 390, what does B receive?

4. A can do a certain work in 12 days. B is 60% more efficient than A. A and B together can complete the whole work in?

Given:
A can do a certain job in 12 days.
B is 60% more efficient than A.
Work = efficiency × days
Calculation:
Let efficiency of A be 100.
The efficiency of B will be 160.
Ratio of efficiency A and B = 100 : 160 = 5 : 8
Total work = efficiency ratio of A × total days taken by A to finish work
⇒ 5 × 12 = 60 unit
Time taken by A+B to finish work = total work/efficiency ratio of A+ B
⇒ 60/13= $7 \frac{1}{2}$ days
$7 \frac{1}{2days.}$

5. Find the value of 0.682² – 0.318²

6. Raman's salary was decreased by 50% and subsequently increased by 50%. How much percent does he lost?
**5*5 = 25% loss**

or

Raman original price be Rs $100$ .
Decrease by $50$ $%$ $= 100 - \frac{5 0}{1 0 0} \times 100$
$= 100 - 50 = 50$
Increase by $50$ $%$ $= 50 + \frac{5 0}{1 0 0} \times 50$
$= 50 + 25 = 75$
Loss $= 100 - 75 = 25$
Loss $%$ $= \frac{2 5}{1 0 0} \times 100$ $= 25$ $%$
Thus, he lost $25 %$ .

7. What is the mode of the following list of numbers:
2, 4, 5, 6, 5, 4, 3, 5, 3, 1 and 7

Most repeated number is 5; So, Mode is 5
8. If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?
Correct option is A)

$P$ (first letter is not vowel) $= \frac{2}{4}$
$P$ (second letter is not vowel) $= \frac{1}{3}$
Therefore, probability that none of letters would be vowels is: $\frac{2}{4} \times \frac{1}{3} = \frac{2}{1 2} = \frac{1}{6}$

Complete step by step answer:
The word HOME contains letters H, O, M and E. Out of which there are two vowels i.e., O and E and two consonants i.e., H and M.
The probability of an event is the number of favourable outcomes divided by the total number of outcomes possible i.e.,
$� (an event) = \frac{number of favourable outcomes}{total number of outcomes}$
.
To find the probability that the first letter is not a vowel, we have two possibilities that either the letter is H or M. Therefore,
Number of favourable outcomes
$= 2$

Total outcomes for letters can be H, O, M or E. Therefore,
Total number of outcomes
$= 4$

$\Rightarrow � (first letter is not a vowel) = \frac{2}{4}$

The probability that the second letter is not a vowel, considering the first event has taken place.
We have now, Number of favourable outcomes
$= 1$

Total number of outcomes
$= 3$

$\Rightarrow � (second letter is not vowel | first letter is not a vowel) = \frac{1}{3}$

Now, using the multiplication rule of probability which states that the probability t

Probability that none of the letters would be vowels
$9.$ Two persons contested an election of Parliament. The winning candidate secured 57% of the total votes polled and won by a majority of 42,000 votes. The number of total votes polled is

Let, the total number of votes poled be, 'x'

The winning candidate got (x*57/100)

= 57x/100 votes

The other candidate got =(x-57x/100)

= (100x-57x)/100

= 43x/100

∴ The winning candidate won by =(57x/100)-(43/100)

=(57x-43x)/100

= 14x/100

By condition,

14x/100 = 42000

⇒x = 42000*100/14

⇒x=3000*100

⇒x=300000

So, the total number of votes polled is 300000

10. What is the average of all numbers between 100 and 200 which are divisible by 13?

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If $a + b + c = 9$ and $a b + b c + c a = 26$ , then find $a^{2} + b^{2} + c^{2}$ ?

A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?

$11.$

$= 1 6$

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If a+b+c=9 and ab+bc+ca=26, then find a2+b2+c2?

Given, a+b+c=9 and ab+bc+ca=26.We know, (a+b+c)2=a2+b2+c2+2ab+2bc+2ca⇒(a+b+c)2=a2+b2+c2+2(ab+bc+ca).Putting the values here, we get,(9)2=a2+b2+c2+2(26)⇒81=a2+b2+c2+52⇒a2+b2+c2=81−52=29.Hence, option D is correct.

A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?

11.

=16

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If $a + b + c = 9$ and $a b + b c + c a = 26$ , then find $a^{2} + b^{2} + c^{2}$ ?

$11.$

$= 1 6$