Comparing Quantities Test

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Comparing Quantities Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

A sum of money invested at simple interest amounts to 2480 at the end of four years and 4080 at the end of eight years. Find the principal.

A
2040

B
1480

C
1240

D
880

Solution

Simple Interest $$ SI = \dfrac {PNR}{100} $$
Also Amount $$ A = P + SI $$
So, for four years,
$$ 2480 - P = \dfrac {P \times 4 \times R}{100} $$ -- (1)

And, for eight years,
$$ 4080 - P = \dfrac {P \times 8 \times R}{100} $$ -- (2)

Subtracting equation (1) from equation (2), we get
$$ => 1600 = \dfrac {4PR}{100} $$
$$ => PR = 40,000 $$

Substituting it in equation (1), we get
$$ 2480 - P = \dfrac { 4 \times 40,000}{100} $$
$$ => P = Rs$$ $$880 $$
Question 2
1 / -0

Opposite of profit of $$\text{Rs.}$$ $$250$$ is:

A
$$+ \text{ Rs }250$$

B
Loss of $$\text{Rs }250$$

C
Discount of $$\text{Rs }250$$

D
None of the above

Solution

Opposite of profit is loss
So, the opposite of profit of $$\text{Rs }250 $$ is the loss of $$\text{Rs }250 $$.
Question 3
1 / -0

A sum of money at simple interest amounts to 800 in 2 years and to 1200 in 6 years. The sum is

A
600

B
1000

C
400

D
500

Solution

Simple Interest $$ SI = \frac {PNR}{100} $$

Given,
$$ Amount A = 800 $$
$$ => P + SI = 800 $$
$$ => P + \frac {P \times 2 \times R}{100} = 800 $$
$$ => P(100+2R) = 80,000 $$ -- (2)

And $$ Amount A = 1200 $$
$$ => P + SI = 1200 $$
$$ => P + \frac {P \times 6\times R}{100} = 1200 $$
$$ => P(100+6R) = 120000 $$ - (2)

Dividing eqn 2 by 1, we get

$$ \frac {100+6R}{100+2R} = \frac {12}{8} $$
$$ => R = \frac {100}{6} $$ %

Substituting in eqn 2, we get
$$ P = Rs 600 $$
Question 4
1 / -0

A dealer earned a profit of $$5\%$$ by selling a radio for Rs.$$714$$, What is cost price of the radio?

A
$$Rs.480$$

B
$$Rs.680$$

C
$$Rs.780$$

D
$$Rs.700$$

Solution

Let the cost price of the radio be Rs.$$x$$.
And, profit$$=5$$%
Now, selling price $$=Rs.(x+5\%\ of\ x)=Rs.\cfrac{105}{100}x$$
$$\therefore$$ $$\cfrac{105}{100}x=714$$
$$\Rightarrow x=714\times \cfrac{100}{105}=680$$
Hence, the cost price of the radio is Rs.$$680$$
Question 5
1 / -0

A sum was put at SI at a certain rate for 8 years Had it been put at 4% higher rate it would have fetched Rs. 80 more Find the sum___

A
Rs. 300

B
Rs. 200

C
Rs. 250

D
Rs. 400

Solution

$$\displaystyle \frac{P\times \left ( R+4 \right )\times 8}{100}-\frac{P\times 8\times R}{100}=80$$
$$\displaystyle \frac{2PR+8P}{25}-\frac{8PR}{100}=80$$
$$\displaystyle \frac{8PR+32P-8PR}{100}=80$$
$$\displaystyle \frac{32P}{100}=80$$
P = $$\displaystyle \frac{80\times 100}{32}$$
= Rs. 250
Question 6
1 / -0

A man borrowed $$Rs,500$$ at a rate of $$6\%$$ per annum. At the end of 3$$\displaystyle \frac{1}{2}$$ years, he paid back $$Rs. 350$$ and gave his radio for the balance amount. The price of the radio is

A
$$Rs. 255$$

B
$$Rs. 250$$

C
$$Rs. 300$$

D
$$Rs. 350$$

Solution

Given principal $$P=Rs.500$$
Rate of interest $$R=6\%$$
Time $$T=3\dfrac{1}{2}$$ years$$=\dfrac{7}{2}$$ years

Interest $$I=\dfrac{PTR}{100}$$

$$\therefore \displaystyle I = \frac{\displaystyle 500 \times \frac{7}{2} \times 6}{100} = Rs. 105$$

Amount $$= P + I$$
$$= Rs.(500 + 105)$$
$$= Rs. 605$$
So, price of radio $$=Rs.( 605 - 350) = Rs. 255$$

Hence, the price of the radio is $$Rs.255$$
Question 7
1 / -0

A dealer purchases $$15$$ articles for $$\text{Rs }25$$ and sells $$12$$ articles for $$\text{Rs }30.$$ Find the profit percentage.

A
$$50\%$$

B
$$40\%$$

C
$$20\%$$

D
$$10\%$$

Solution

C.P. of $$15$$ articles is $$\text{Rs }25$$
$$\therefore$$ C.P. of each article $$ = \dfrac {20}{15} = \dfrac {5}{3} $$
S.P. of $$12$$ articles is $$\text{Rs }30$$
$$\therefore$$ S.P. of each article $$ = \dfrac {30}{12} = \dfrac {5}{2} $$

Now, Profit $$\% = \dfrac {S.P. - C.P.}{C.P.} \times 100 = \frac {\dfrac {5}{2}- \dfrac {5}{3}}{ \dfrac {5}{3}} \times 100 = 50 \%$$
Question 8
1 / -0

If CP of 60 articles is equal to the SP of 50 articles Then gain % is__

A
20%

B
30%

C
10%

D
60%

Solution

Let the CP of each article $$ = x $$
Then CP of $$ 50 $$ articles $$ = 50 x $$

Given, CP of $$ 60 $$ articles $$ = $$ SP of $$ 50 $$ articles.
So, SP of $$ 50 $$ articles $$ = 60 x$$

Now, Gain % $$ = \dfrac {SP - CP}{CP} \times 100 = \dfrac {60x-50x}{50x} \times 100 = 20 $$ %
Question 9
1 / -0

A shopkeeper announces a discount of $$15\%$$ on his goods. If the marked price of an article, in his shop is $$Rs. 6,000$$; how much a customer has to pay for it, if the rate of Sales Tax is $$10\%$$?

A
$$Rs. 5,610$$

B
$$Rs. 5,804$$

C
$$Rs. 5,906$$

D
None of these

Solution

Given,$$M.P=Rs.6000$$
Sales tax$$=10\%$$
S.P including sales tax$$=Rs.(6000+\dfrac{10}{100}\times 6000)$$
$$= Rs(6000+600)$$
$$=Rs.6600$$
Discount$$=15\%$$
$$\therefore$$ S.P after discount$$=Rs.(6600-15\% \ of\ 6600)$$
$$=Rs.( 6600- \dfrac{15}{100}\times 6600)$$
$$=Rs.( 6600-990)$$
$$=Rs.5610$$

Hence, the customer has to pay $$Rs.5610$$
Question 10
1 / -0

When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V.; Mrs. Geeta will save Rs. 780 in buying this T.V. Find the list price of the T.V.

A
$$Rs. 20,000$$

B
$$Rs. 24,000$$

C
$$Rs. 26,000$$

D
$$Rs. 30,000$$

Solution

Let the List price of the T.V=Rs.x
Then according to the question
$$(x+9\% of x)-(x+6\% of x)=780$$
$$\Rightarrow x+\frac{9x}{100}-x-\frac{6x}{100}=780$$
$$\Rightarrow \frac{9x}{100}-\frac{6x}{100}=780$$
$$\Rightarrow 3x=78000$$
$$\Rightarrow x=\frac{78000}{3}=Rs. 26000$$

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Comparing Quantities Test - 39

Result

Comparing Quantities Test - 39

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SHARING IS CARING

Question 1

2040

1480

1240

880

Solution

Question 2

$$+ \text{ Rs }250$$

Loss of $$\text{Rs }250$$

Discount of $$\text{Rs }250$$

None of the above

Solution

Question 3

600

1000

400

500

Solution

Question 4

$$Rs.480$$

$$Rs.680$$

$$Rs.780$$

$$Rs.700$$

Solution

Question 5

Rs. 300

Rs. 200

Rs. 250

Rs. 400

Solution

Question 6

$$Rs. 255$$

$$Rs. 250$$

$$Rs. 300$$

$$Rs. 350$$

Solution

Question 7

$$50\%$$

$$40\%$$

$$20\%$$

$$10\%$$

Solution

Question 8

20%

30%

10%

60%

Solution

Question 9

$$Rs. 5,610$$

$$Rs. 5,804$$

$$Rs. 5,906$$

None of these

Solution

Question 10

$$Rs. 20,000$$

$$Rs. 24,000$$

$$Rs. 26,000$$

$$Rs. 30,000$$

Solution

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