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

Answer: (b)

Let the original price of 1 mango be x.

New rate = 120% of x = ${6x}/5$

Number of mangoes bought in Rs.40 = $40/x$

New quantity = ${40 × 5}/{6x} = 100/{3x}$

$40/x – 100/{3x} = 4$

${120 – 100}/{3x} = 4 ⇒ 20/{3x} = 4$

3x = 5 ⇒ $x$ = Rs.$5/3$

Price of 15 mangoes before increase = $5/3 × 15$ = Rs.25

Answer: (c)

Required profit percent = $(x + y + {xy}/100)%$

by Here, x = 25%, y = - $25/2%$

= $(25 – 25/2 – {25 × 25}/200)% = (25/2 – 25/8)%$

= $({100 – 25}/8)% = 75/8 % = 9{3}/8%$

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

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 = $({146.3 – 112}/112) × 100$

= ${34.3 × 100}/112 = 3430/112$

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

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