Practice Reduce increase article price - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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)
(b)
(c)
(d)
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
Q-2) 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)
(b)
(c)
(d)
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
Q-3) A reduction of 15% in the price of apples would enable a purchaser to get 2 kg more apples for Rs.240. The new price (per kg) of apples is
(a)
(b)
(c)
(d)
Let the original rate = x per kg.
New rate = 85% of x = ${85x}/100 = {17x}/20$
Original quantity for Rs.240 = $240/x$
New quantity = 240 × $20/{17x} = 4800/{17x}$
$4800/{17x} – 240/x$ = 2
${4800 – 4080}/{17x}$ = 2
$720/{17x} = 2 ⇒ x = 720/{2 × 17}$
Original rate per kg = Rs.$720/34$
Reduced rate = Rs.${17x}/20$
= Rs.$(17/20 × 720/34)$ = Rs.18
Q-4) 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)
(b)
(c)
(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
Q-5) 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)
(b)
(c)
(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
Q-6) 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)
(b)
(c)
(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
Q-7) An increase of 20% in the price of mangoes enables a person to purchase 4 mangoes less for Rs.40. The price of 15 mangoes before increase was
(a)
(b)
(c)
(d)
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
Q-8) A tradesman sold an article at a loss of 20%. If the selling price had been increased by Rs.100, there would have been a gain of 5%. The cost price of the article was :
(a)
(b)
(c)
(d)
Let the C.P. of article be x.
105% of x – 80% of x = 100
25% of x = 100
$x = {100 × 100}/25$ = Rs.400
Q-9) A reduction of 20% in the price of salt enabled a purchaser to obtain 4 kg. more for Rs.100. The reduced price of salt per kg is :
(a)
(b)
(c)
(d)
Due to fall in price, there is a saving of 20% of Rs.100 i.e., Rs.20.
With this amount the purchaser purchases 4 kg. of salt.
Reduced price of salt per kg = $20/4$ = Rs.5
Q-10) 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)
(b)
(c)
(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 = $({146.3 – 112}/112) × 100$
= ${34.3 × 100}/112 = 3430/112$
= $245/8 = 30{5}/8%$