Practice Sold brought - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   By selling a tape-recorder Rs. for 950, I lose 5%. What per cent shall I gain by selling it for Rs.1040?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

If S.P > C.P. then there will be profit

Profit = S.P. - C.P.

Profit% = ${\text"Profit" × 100}/{\text"C.P."}$

C.P. of the tape recorder

= $100/95 × 950$ = Rs.1000

Gain = 1040 - 1000 = Rs.40

% Gain = $40/1000 × 100 = 4%$


Q-2)   A shopman bought pens at the rate of 7 for Rs.10 and sold them at a profit of 40%. How many pens would a customer get for Rs.10 ?

(a)

(b)

(c)

(d)

Explanation:

S.P. of 7 pens = ${10 × 140}/100$ = Rs.14

S.P. of 1 pen = $14/7$ = Rs.2

Clearly, 5 pens were sold for Rs.10

Using Rule 13,

Here, a = 7, x = 10, b = ?, y = 10, Gain% = 40%

Gain% = $({ay - bx}/{bx})$ × 100%

40 = $({7 × 10 - b × 10}/{b × 10}) × 100%$

4b = 70 - 10b

14b = 70 ⇒ b = $70/14$ ⇒ b = 5


Q-3)   Some articles were bought at 6 for Rs.5, and sold at 5 for Rs.6. Gain per cent is :

(a)

(b)

(c)

(d)

Explanation:

Let number of articles bought = 6 × 5 = 30

C.P. of 30 articles = Rs.$(5/6 × 30)$ = Rs.25

S.P. of 30 articles = Rs.$(6/5 × 30)$ = Rs.36

Gain % = ${36 - 25}/25 × 100$ = 44%

Using Rule 13,

Here, a = 6, x = 5, b = 5, y = 6

Gain% = $({ay - bx}/{bx})$ × 100%

= $({6 × 6 - 5 × 5}/{5 × 5})$ × 100%

= $({36 - 25}/25) × 100% = 11/25 ×100%$ = 44%


Q-4)   A person buys 100 cups at Rs.10 each. On the way 10 cups are broken. He sells the remaining cups at Rs.11 each. His loss per cent is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 2

If C.P > S.P., then there will be Loss

Loss = C.P. - S.P., Loss% = ${Loss × 100}/{C.P.}$

CP of Rs.100 cups = RS.100 × 10 = Rs.1000

10 cups are broken.

SP of 90 cups = Rs.(90 ×11) = Rs.990

Loss = Rs.(1000 - 990) = Rs.10

Loss per cent = $10/1000 × 100 = 1%$


Q-5)   By selling 12 oranges for Rs.60, a man loses 25%. The number of oranges he has to sell for Rs.100, so as to gain 25% is

(a)

(b)

(c)

(d)

Explanation:

C.P. of 12 oranges = $60 × 100/75$ = Rs.80

For a gain of 25%,

S.P. of 12 oranges = ${80 × 125}/100$ = Rs.100

Hence, 12 Orange has to sell,

[You can also check through options]


Q-6)   A man sold 20 apples for Rs.100 and gained 20%. How many apples did he buy for Rs.100?

(a)

(b)

(c)

(d)

Explanation:

If the CP of 20 apples be Rs.x,

then ${x × 120}/100 = 100$

$x = {100 × 100}/120$ = Rs.$250/3$

Rs.$250/3$ = 20 apples

Rs.100 = ${20 × 3 × 100}/250$ = 24 apples

Using Rule 13,

Here, a = ?, x = 100, b = 20, y = 100, Gain% = 20%

Gain% = $({ay - bx}/{bx})$ × 100%

20% = $({a × 100 - 20 × 100}/{20 × 100}) × 100%$

400 = 100a - 2000

2400 = 100a ⇒ a = 24


Q-7)   A man buys 12 articles for Rs.12 and sells them at the rate of Rs.1.25 per article. His gain percentage is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

If S.P > C.P. then there will be profit

Profit = S.P. - C.P.

Profit% = ${\text"Profit" × 100}/{\text"C.P."}$

C.P. = 12; S.P. = 12 × 1.25 = 15

Total Profit = 15 - 12 = 3

% gain = $3/12 × 100$ = 25%


Q-8)   If I would have purchased 11 articles for Rs.10 and sold all the articles at the rate of 10 for Rs.11, the profit per cent would have been :

(a)

(b)

(c)

(d)

Explanation:

C.P. of an article = Rs.$10/11$

S.P. of an article = Rs.$11/10$

Profit = $11/10 - 10/11$

= ${121 - 100}/110 = Rs.21/110$

Profit % = ${{21/110} × 100}/{10/11}$

= $2100/110 × 11/10$ = 21%

Using Rule 13,
If the total cost of 'a' articles having equal cost is Rs.x
and the total selling price of 'b' articles is Rs.y,
then in the transaction gain or loss per cent is given by
$({ay - bx}/{bx})$ × 100%
Where positive value signifies 'profit'
and negative value signifies 'loss'

Here, a = 11, x = 10, b = 10, y = 11

Gain% = $({ay - bx}/{bx}) × 100%$

= $({11 × 11 - 10 × 10}/{10 × 10}) × 100%$

= $({121 - 100}/100) × 100%$ = 21%


Q-9)   Oranges are bought at the rate of 10 for Rs.25 and sold at the rate of 9 for Rs.25. The profit percent is

(a)

(b)

(c)

(d)

Explanation:

Suppose the number of oranges bought

= LCM of 10 and 9 = 90

C.P. of 90 oranges = $25/10 × 90$ = Rs.225

S.P. of 90 oranges = $25/9 × 90$ = Rs.250

Profit % = $25/225 × 100 = 100/9 = 11{1}/9%$

Using Rule 13,

Here, a = 10, x = 25, b = 9, y = 25

Gain% = $({ay - bx}/{bx})$ × 100%

= $({10 × 25 - 9 × 25}/{9 × 25}) × 100%$

= $({250 - 225}/{9 × 25}) × 100% = 100/9 = 11{1}/9%$


Q-10)   A man bought pencils at the rate of 6 for Rs.4 and sold them at the rate of 4 for Rs.6. His gain% in the transaction is :

(a)

(b)

(c)

(d)

Explanation:

Let the number of pencils bought = LCM of 4, 6 = 12

CP of 6 pencils = Rs.4

CP of 12 pencils = Rs.8

S.P. of 4 pencils = Rs.6

S.P. of 12 pencils = Rs.18

Profit = Rs. (18 - 8) = Rs.10

Profit % = $10/8 × 100 = 125%$

Using Rule: 13

Here, a = 6, x = 4, b = 4, y = 6

Gain% = $({ay - bx}/{bx})$ × 100%

= $({6 × 6 - 4 × 4}/{4 × 4}) × 100%$

= $({36 - 16}/16) × 100%$

= $20/16 × 100% = 125%$