General Test

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General Test - 27

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Weekly Quiz Competition

Question 1
4 / -1

A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:

A
Rs. 8400

B
Rs. 11,900

C
Rs. 13,600

D
Rs. 14,700

Solution

Let C = x

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000

So, x + x + 5000 + x + 9000 = 50000

⇒ 3x = 36000

⇒ x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12

So A's Share

$=$ Rs. $35000 \times \frac{21}{50}$

$=$ Rs. 14700
Question 2
4 / -1

There was a simple interest of Rs. $4016.25$ on a principal amount at the rate of $9\%$ p.a. in $5$ years. Find the principal amount.

A
Rs. $8925$

B
Rs. $8295$

C
Rs. $9285$

D
Rs. $6335$

Solution

Given,

Simple interest in $5$ years $ =4016.25$

Rate $= 9\%$

Principal$ = P$

As we know,

$P=\frac{S . I . \times 100}{R \times T}$

So, by putting values, we get

$P=\frac{4016.25 \times 100}{9 \times 5}$

$=$ Rs. $8925$
Question 3
4 / -1

Find the simple interest on Rs. $7000$ at $\frac{50}{3} \%$ for $9$ months.

A
Rs. $1075$

B
Rs. $975$

C
Rs. $875$

D
Rs. $775$

Solution

Given,
Principle $(P)=7000$, Rate $(R)=\frac{50}{3}$, time $(T)=9$ months $=\frac{9}{12}$ year
As we know,
S.I. $=\frac{P \times R \times T}{100}$
$=\frac{7000 \times 50 \times 9}{3 \times 12 \times 100}$
$=$ Rs. $875$
Question 4
4 / -1

A person makes a profit of 20% by selling an article. If he purchased the articles at 20% less and sell it 10% more, what would be the new profit percentage?

A
65%

B
8%

C
15%

D
20%

Solution

A person makes a profit = 20%

Suppose a person buys the item for Rs 100.

New Cost Price = $100 \times \frac{100 - 20}{100} = R s . 80$

New Selling price = $\frac{120 (100 + 10)}{100} = R s . 132$

New profit percent = $(\frac{132 - 80}{80}) \times 100$ = 65%
Question 5
4 / -1

By selling an article for Rs 255, there is a loss of 15%. When the article is sold for Rs 357, the net result in the transaction is:

A
19% profit

B
17% loss

C
19% loss

D
17% profit

Solution

Cost Price $\mathrm{=\frac{100}{100-L o s s \%} \times S P}$

Cost Price of an article $=\frac{100}{100-15} \times 255 $$=\frac{100}{85} \times 255=300$ Rs

When article is sold at Rs 357 then,

Profit = SP - CP =357-300= Rs 57

Profit $ \%=\frac{57}{300} \times 100=19 \%$
Question 6
4 / -1

A number is increased by $20 \%$. To get back to the original number, the increased number is to be reduced by:

A
$20 \%$

B
$21 \%$

C
$16 \frac{2}{3} \%$

D
$14 \frac{1}{3} \%$

Solution

Let the number be 100.

Number after increased $=100 \times(\frac{120}{100})=120$

Increased number $=120-100=20$

Required percentage $=\frac{20}{120} \times 100=16 \frac{2}{3} \%$
Question 7
4 / -1

The population of a town increased from 268000 in 2007 to 375200 in 2010. What is the percentage increase in the population of the town in three years?

A
40%

B
32%

C
30%

D
28%

Solution

Percentage increase $=\frac{\text { Final-Initial }}{\text { Initial }} \times 100 \%$

Now, the population percentage increase$= \frac{375200-268000}{268000} \times 100 \%$

$= \frac{1072}{2680} \times 100 \%$$= 40 \%$
Question 8
4 / -1

A water tank has two pipes. The empty tank is filled in 12 min by the first and the full tank is emptied by the second in 20 min. The time required to fill the $\frac12$ full tank when both pipes are in action, is:

A
16 min

B
15 min

C
20 min

D
30 min

Solution

Since, first pipe takes 12 min to completely fill tank

Thus, tank filled by first pipe in 1 min $=\frac{1}{12}$

Now, second pipe takes 20 min to empty the full tank.

Thus time taken by both pipes to fill complete tank

$=\frac{1}{12}-\frac{1}{20}$

$=\frac{5-3}{60}$

$=\frac{2}{60}$

$=\frac{1}{30}$

Since, total tank will be filled in $30$ min.

Hence, time taken to fill the half tank

$=\frac{30}{2}$

$=15$ min
Question 9
4 / -1

If $x^{100}+2 x^{99}+k$ is fully divisible by $(x+1)$ then value of ${k}$ will be:

A
$7$

B
$-3$

C
$2$

D
$1$

Solution

Given,

$x+1$ is a factor of $x^{100}+2 x^{99}+k$

So, $x+1=0$

$x=-1$,

Vlaue is the zero of $x^{100}+2 x^{99}+k$

Substituting the value of $x=-1$,

then $(-1)^{100}+2(-1)^{99}+k=0$

$1-2+k=0$

$k=1$
Question 10
4 / -1

Find the value: 2$\frac{1}{4}+\frac{3}{4}=~?$

A
$\frac{5}{4}$

B
$\frac{10}{4}$

C
$\frac{1}{3}$

D
3

Solution

Let 2$\frac{1}{4}+\frac{3}{4}$ be $x$

$x =~ \frac{9}{4}+\frac{3}{4}=~ \frac{12}{4}= ~3$

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Re-Attempt Weekly Quiz Competition

General Test - 27

Result

General Test - 27

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Rank

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SHARING IS CARING

Question 1

Rs. 8400

Rs. 11,900

Rs. 13,600

Rs. 14,700

Solution

Question 2

Rs. \(8925\)

Rs. \(8295\)

Rs. \(9285\)

Rs. \(6335\)

Solution

Question 3

Rs. \(1075\)

Rs. \(975\)

Rs. \(875\)

Rs. \(775\)

Solution

Question 4

65%

8%

15%

20%

Solution

Question 5

19% profit

17% loss

19% loss

17% profit

Solution

Question 6

\(20 \%\)

\(21 \%\)

\(16 \frac{2}{3} \%\)

\(14 \frac{1}{3} \%\)

Solution

Question 7

40%

32%

30%

28%

Solution

Question 8

16 min

15 min

20 min

30 min

Solution

Question 9

\(7\)

\(-3\)

\(2\)

\(1\)

Solution

Question 10

\(\frac{5}{4}\)

\(\frac{10}{4}\)

\(\frac{1}{3}\)

3

Solution

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