Practice Reducing exceeding prices - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A number, on subtracting 15 from it, reduces to its 80%. What is 40% of the number?
(a)
(b)
(c)
(d)
If the number be x, then
x – 15 = ${4x}/5$
5x – 75 = 4x ⇒ x = 75
40% of 75 = ${75 × 40}/100 = 30$
Q-2) The salary of a person is reduced by 20%. To restore the previous salary, his present salary is to be increased by
(a)
(b)
(c)
(d)
Required percentage increase = $x/{100 - x} × 100$
= $(20/{100 - 20}) × 100 = 20/80 × 100 = 25%$
Q-3) In 2001, the price of a building was 80% of its original price. In 2002, the price was 60% of its original price. By what percent did the price decrease ?
(a)
(b)
(c)
(d)
Original price of building = Rs. 100 (let)
Its price in 2001 = Rs. 80
Its price in 2002 = Rs. 60
Required percentage decrease = $({80 - 60}/80) × 100$
= $200/8 = 25%$
Q-4) A reduction of 20% in the price of wheat enables Lalita to buy 5 kg more wheat for Rs.320. The original rate (in rupees per kg) of wheat was
(a)
(b)
(c)
(d)
Original price of wheat = Rs.x /kg.
New price of wheat = Rs.${4x}/5 /kg$
$320/{{4x}/5} - 320/x = 5$
$320(5/{4x} - 1/x) = 5$
$320({5 - 4}/{4x}) = 5$
$320/{4x} = 5 ⇒ x = 320/{4 × 5} = Rs.16$
Using Rule 31,
New price = ${20 × 320}/{100 × 5}$
= $1280/100$ = Rs.12.8
Let the original price be Rs.x per kg.
$x - {20x}/100 = 12.8$
80x = 12.8 × 100
$x = 1280/80 ⇒ x = 16$ per kg.
Q-5) A reduction of 10% in the price of sugar enables a housewife to buy 6.2 kg more for Rs.1116. The reduced price per kg is
(a)
(b)
(c)
(d)
Reduced price of 6.2kg of sugar
= 10% of Rs.1116 = Rs.111.6
Reduced price per kg
= Rs.$(111.6/6.2)$ = Rs.18
Using Rule 31,
New price = ${10 × 1116}/{100 × 6.2}$
= $1116/62$ = Rs.18
Q-6) A reduction in the price of apples enables a person to purchase 3 apples for Rs.1 instead of Rs.1.25. What is the % of reduction in price (approximately) ?
(a)
(b)
(c)
(d)
Using Rule 1,
If x is reduced to x0 , then, Reduce % = ${x – x_0}/x × 100$
Percentage decrease = $0.25/1.25 × 100 = 20%$
Q-7) A reduction of 20% in the price of sugar enables a purchaser to obtain 8 kg more for Rs.160. Then the price per kg before reduction was
(a)
(b)
(c)
(d)
Let the original price of sugar be Rs. x per kg.
Reduced price = Rs. ${80x}/100 = Rs.{4x}/5$ per kg.
According to the question,
$160/{{4x}/5} - 160/x$ = 8
${40 × 5}/x - 160/x = 8$
$200/x - 160/x = 8$
$40/x = 8 ⇒ 8x = 40 ⇒ x = 40/8 = 5$ per kg.
Reduced Price = ${4x}/5 = {4 × 5}/5$ = Rs. 4 per kg
Using Rule 31,
Reduced price per kg. = ${21 × 160}/{100 × 8}$ = Rs. 4
Q-8) If the price of sugar increases by 20%, one can buy 2 kg less for Rs. 50. What is the amount of sugar that could be bought before price hike?
(a)
(b)
(c)
(d)
Let original price of sugar be Rs. x per kg.
New price= Rs.$({120x}/100) =Rs.({6x}/5)$per kg.
According to the question,
$50/x - 50/{{6x}/5} = 2$
$50/x - {50 × 5}/{6x} = 2$
$50/x - 125/{3x} = 2$
${150 – 125}/{3x} = 2$
$6x = 25 ⇒ x = Rs.25/6$ kg.
Required quantity of sugar = $50/x$
= $50/{25/6} = {50 × 6}/25 = 12$kg.
Q-9) A reduction of 25% in the price of rice enables a person to buy 10 kg more rice for Rs.600. The reduced per kg price of rice is
(a)
(b)
(c)
(d)
Let original price of rice per kg = Rs.x (let)
New price of rice per kg = Rs.${3x}/4$
$600/{{3x}/4} - 600/x = 10$
$600(4/{3x} - 1/x) = 10$
$600({4 - 3}/{3x})$ = 10
$600/{3x} = 10 ⇒ x = 600/30 = Rs.20$
New price = ${3x}/4 = {3 × 20}/4$ = Rs.15/kg
Method 2: Simple Approach,
If the price of an article is reduced by a% and buyer gets c kg more for some Rs.b, the new price per kg of article = ${ab}/{100 × c}$
= ${25 × 600}/{100 × 10}$ = Rs.15
Using Rule 31,
Reduced price per kg = ${25 × 600}/{100 × 10}$ = Rs.15
Q-10) When the price of sugar decreases by 10%, a man could buy 1 kg more for Rs.270. Then the original price of sugar per kg is
(a)
(b)
(c)
(d)
Let the original price of sugar be Rs.x/kg.
New price = Rs.${9x}/10$ /kg
$270/{{9x}/10} - 270/x =1$
$300/x - 270/x = 1 ⇒ 30/x = 1$ ⇒ x = Rs.30 /kg
Using Rule 31,
New price = ${10 × 270}/{100 × 1}$ = Rs.27
Let the original price be Rs. x
$x - {10x}/100 = 27 ⇒ {90x}/100 = 27$
$x = 2700/90$ ⇒ x = 30
∴ Original price = Rs.30 per kg.