model 10 reduced/increased price of an article Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on profit & loss topic of quantitative aptitude

Questions : If the cost of pins reduces by Rs.4 per dozen, 12 more pins can be purchased for Rs.48. The cost of pins per dozen after reduction is:

(a) Rs.16

(b) Rs.20

(c) Rs.8

(d) Rs.12

The correct answers to the above question in:

Answer: (d)

Let the original price = x per dozen

New price = (x – 4) per dozen

Original number of pins = $48/x$ dozens

New number of pins = $48/{x – 4}$ dozens

According to the question,

$48/{x -4} – 48/x = 1$

⇒ $48({x – x + 4}/{x(x - 4)})$= 1

⇒ x (x – 4) = 48 × 4

⇒ $x^2 – 4x – 192 = 0$

⇒ $x^2 – 16x + 12x – 192$ = 0

⇒ x (x – 16) + 12 (x – 16) = 0

⇒ (x – 16) (x + 12) = 0

⇒ x = 16, because the price of pins can not be negative.

x ≠–12

New price = 16 – 4 = Rs.12 per dozen

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Read more reduce increase article price Based Quantitative Aptitude Questions and Answers

Question : 1

A manufacturer fixes his selling price at 33% over the cost of production. If cost of production goes up by 12% and manufacturer raises his selling price by 10%, his percentage profit is

a) 36$5/9%$

b) 35%

c) 28$3/8$%

d) 30$5/8%$

Answer: (d)

Cost of production of article = Rs.100 (let)

S.P. = Rs.133

New cost of production = Rs.112

S.P. = ${133 × 110}/100$ = Rs.146.30

Profit per cent = $({146.3 – 112}/112) × 100$

= ${34.3 × 100}/112 = 3430/112$

= $245/8 = 30{5}/8%$

Question : 2

If the price of eraser is reduced by 25% a person can buy 2 more erasers for a rupee. How many erasers are available for a rupee ?

a) 4

b) 2

c) 8

d) 6

Answer: (c)

Cost of 2 erasers = 25% of 1

= $25/100$ × 1= Rs.$1/4$

Cost of one eraser = Rs.$1/8$

8 erasers will be available for Rs.1

Question : 3

If a man reduces the selling price of a fan from Rs.1,250 to Rs.1,000, his loss increases by 20%. The cost price of the fan is

a) Rs.2,500

b) Rs.2,350

c) Rs.2,400

d) Rs.2,450

Answer: (a)

Let the cost price of fan be Rs. x,

According to the question,

10% of x = 1250 – 1000

${x × 10}/100$ = 250

$x = {250 × 100}/10$ = Rs.2500

Note : Here, increase in loss should be 10%.

Question : 4

A trader bought 10 kg of apples for Rs.405 out of which 1 kg of apples were found to be rotten. If he wishes to make a profit of 10%, at what rate should he sell the remaining apples per kg?

a) Rs.50

b) Rs.51

c) Rs.45

d) Rs.49.50

Answer: (d)

Selling price = 405 × 110% = Rs.445.50

Remaining apples = 10 – 1= 9 kg

Therefore, the remaining apples (per kg) cost

=$445.50/9$ = Rs.49.50

Question : 5

The reduction of 12 in the selling price of an article will change 5% gain into 2$1/2%$ loss. The cost price of the article is

a) Rs.80

b) Rs.100

c) Rs.140

d) Rs.160

Answer: (d)

If the C.P. of article be Rs.x, then

$x × (105 – {195}/2)$% = 12

$x × 15/200 = 12 ⇒ x = {12 × 200}/15$ = Rs.160

Question : 6

A reduction of 20% in the price of rice enables a buyer to buy 5 kg more for rupees 1200. The reduced price per kg of rice will be:

a) 48

b) 60

c) 36

d) 45

Answer: (a)

Original price of rice = Rs.x per kg.

New price = ${80x}/100$ = Rs.${4x}/5$ per kg

According to the question,

$1200/{{4x}/5} – 1200/x$ = 5

${1200 × 5}/{4x} – 1200/x$ = 5

$1500/x – 1200/x$ = 5

$300/x$ = 5 ⇒ 5x = 300

$x = 300/5$ = Rs.60 per kg

New price of rice

= Rs.$({4 × 60}/5)$ per kg = Rs.48 per kg

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