Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A single discount equivalent to three simple discounts of 10%, 12%, and 15% is:<\/p>\n a)\u00a032.68%<\/p>\n b)\u00a034.17%<\/p>\n c)\u00a037%<\/p>\n d)\u00a035.36%<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The successive discounts are 10%,12% and 15%<\/p>\n Let the Initial value be 100.<\/p>\n According to question,\u00a0$100\\times\\ \\frac{\\left(100-10\\right)}{100}\\times\\ \\frac{\\left(100-12\\right)}{100}\\times\\ \\frac{\\left(100-15\\right)}{100}$<\/p>\n i.e;\u00a0$100\\times\\ \\frac{90}{100}\\times\\ \\frac{88}{100}\\times\\ \\frac{85}{100}$<\/p>\n Discounted value\u00a0= 67.32 %<\/p>\n Equivalent discount = initial value – discounted value<\/p>\n = 100 – 67.32 = 32.68 .<\/p>\n Hence option A is correct.<\/p>\n Another method :<\/p>\n Discount is also called successive decrease<\/p>\n if x and y are two successive discounts then ,<\/p>\n $x\\ +\\ y\\ -\\frac{\\left(x\\times\\ y\\right)}{100}$<\/p>\n first we take first two discount 10% and 12%<\/p>\n $\\therefore\\ 10\\ +\\ 12\\ -\\frac{10\\times\\ 12}{100}=\\ 20.8\\ \\%$<\/p>\n Now we take,<\/p>\n 20.8% and 15% ,<\/p>\n $\\therefore\\ 20.8\\ +\\ 15\\ -\\frac{20.8\\times\\ 15}{100}=\\ 32.68\\ \\%$<\/p>\n Question 2:\u00a0<\/b>A shopkeeper sold an article for \u20b9455 at a loss (in \u20b9). If he sells it for \u20b9490, then he would gain an amount four times the loss. At what price (in \u20b9) should he sell the article to gain 25%?<\/p>\n a)\u00a0575<\/p>\n b)\u00a0577.50<\/p>\n c)\u00a0570.50<\/p>\n d)\u00a0115.50<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the loss when the shopkeeper sold the article for\u00a0\u20b9455 = L<\/p>\n and the Cost price of the article = C<\/p>\n $\\Rightarrow$ C – 455 = L<\/p>\n $\\Rightarrow$ C = L + 455……(1)<\/p>\n According to the problem, when the shopkeeper\u00a0sells it for \u20b9490, then he would gain an amount four times the loss.<\/p>\n $\\Rightarrow$ 490 – C = 4L<\/p>\n $\\Rightarrow$ 490 – (L + 455) = 4L<\/p>\n $\\Rightarrow$ 35 = 5L<\/p>\n $\\Rightarrow$ L = 7<\/p>\n From (1),<\/p>\n C = L + 455 = 7 + 455 = \u20b9462<\/p>\n Cost price of the article =\u00a0\u20b9462<\/p>\n Selling price of the article when the shopkeeper sells at 25% gain = $\\frac{125}{100}\\times$C<\/p>\n =\u00a0$\\frac{125}{100}\\times$462<\/p>\n =\u00a0\u20b9577.50<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 3:\u00a0<\/b>A shopkeeper marks an article at a price such that after giving a discount of x%, he gains 20%. If the cost price and the marked price of the article are \u20b9920 and \u20b91472 respectively, then what is the value of x?<\/p>\n a)\u00a018<\/p>\n b)\u00a030<\/p>\n c)\u00a020<\/p>\n d)\u00a025<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cost price of the article =\u00a0\u20b9920<\/p>\n Gain = 20%<\/p>\n Selling price of the article = $\\frac{120}{100}\\times$920<\/p>\n =\u00a0\u20b91104<\/p>\n Marked price of the article = \u20b91472<\/p>\n Discount = x%<\/p>\n Selling price of the article = $\\frac{100-x}{100}\\times$1472<\/p>\n 1104 =\u00a0$\\frac{100-x}{100}\\times$1472<\/p>\n 69\u00a0= $\\frac{100-x}{100}\\times$92<\/p>\n 6900 = 9200 – 92x<\/p>\n 92x = 2300<\/p>\n x = 25<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 4:\u00a0<\/b>In festival season, a shopkeeper allows a discount of 10% on every item. Even after giving the discount, he makes a profit of 20%. If he does not give any discount, then what will be his profit percent? (correct to 2 decimal places)<\/p>\n a)\u00a033<\/p>\n b)\u00a025<\/p>\n c)\u00a033.33<\/p>\n d)\u00a033.43<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cost price of the article = 100C<\/p>\n Profit = 20%<\/p>\n Selling price of the article = $\\frac{120}{100}\\times$100C = 120C<\/p>\n Discount = 10%<\/p>\n $\\frac{90}{100}\\times$Marked price = 120C<\/p>\n Marked price of the article = $\\frac{400}{3}$C<\/p>\n When no discount is provided,<\/p>\n Selling price =\u00a0$\\frac{400}{3}$C<\/p>\n Profit% = $\\frac{\\frac{400}{3}C-100C}{100C}\\times100$<\/p>\n = 33.33%<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 5:\u00a0<\/b>A trader bought 640 kg of rice. He sold a part of rice at 20% profit and the rest at 5% loss. He earned a profit of 15% in the entire transaction. What is the quantity (in kg) of rice that he sold at 5% loss?<\/p>\n a)\u00a0128<\/p>\n b)\u00a0132<\/p>\n c)\u00a0154<\/p>\n d)\u00a0256<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Using mixture and alligation method,<\/p>\n Ratio of the quantity of rice sold at 20% profit and 5% loss = 20 : 5<\/p>\n = 4 : 1<\/p>\n Quantity of rice sold at 5% loss = $\\frac{1}{4+1}\\times$640<\/p>\n = 128 kg<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 6:\u00a0<\/b>The cost price of an article is \u20b9280. A shopkeeper sells it by allowing 16% discount on its marked price and still gains 20%. What is the marked price(in \u20b9) of the article?<\/p>\n a)\u00a0400<\/p>\n b)\u00a0360<\/p>\n c)\u00a0420<\/p>\n d)\u00a0350<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The cost price of article = \u20b9280<\/p>\n Gain = 20%<\/p>\n Selling price of the article = $\\frac{120}{100}\\times$280<\/p>\n =\u00a0\u20b9336<\/p>\n Let the marked price of the article = M<\/p>\n Discount = 16%<\/p>\n Selling price of the article = $\\frac{84}{100}$M<\/p>\n $\\Rightarrow$\u00a0\u00a0336 =\u00a0$\\frac{84}{100}$M<\/p>\n $\\Rightarrow$\u00a0 M = 400<\/p>\n Marked price of the article =\u00a0\u20b9400<\/p>\n Hence, the correct answer is Option A<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 7:\u00a0<\/b>A shop keeper sold an article at four-fifth of the marked price and suffered a loss of $3 \\frac{1}{3}\\%$. Find the profit percent, if he sold the article at the marked price. (correct to nearest integer)<\/p>\n a)\u00a022<\/p>\n b)\u00a018<\/p>\n c)\u00a021<\/p>\n d)\u00a020<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cost price of the article = 100C<\/p>\n Loss =\u00a0$3 \\frac{1}{3}\\%$ =\u00a0$\\frac{10}{3}\\%$<\/p>\n Selling price of the article = 100C – $\\frac{\\frac{10}{3}}{100}\\times$100C<\/p>\n =\u00a0$\\frac{290}{3}$C<\/p>\n Shop keeper sold the article at four-fifth of the marked price.<\/p>\n $\\frac{290}{3}$C =\u00a0$\\frac{4}{5}\\times$Marked price of the article<\/p>\n Marked price of the article =\u00a0$\\frac{725}{6}$C<\/p>\n Profit percentage when article is sold at marked price =\u00a0$\\frac{\\frac{725}{6}C-100C}{100C}\\times100$<\/p>\n =\u00a0$\\frac{125C}{6\\times100C}\\times100$<\/p>\n = 20.833%<\/p>\n = 21% (approximately)<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>Trader A gives a single discount of 25% and Trader B gives two successive discounts of 20% and 5% on an identical item. If the discount given by A is \u20b9320 more than the discount given by B, then what is the marked price (in \u20b9) of the item?<\/p>\n a)\u00a03200<\/p>\n b)\u00a032000<\/p>\n c)\u00a030000<\/p>\n d)\u00a025000<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the marked price of the item = M<\/p>\n i) Trader A gives a single discount of 25%.<\/p>\n Discount = $\\frac{25}{100}$M =\u00a0$\\frac{1}{4}$M<\/p>\n ii) Trader B gives two successive discounts of 20% and 5%.<\/p>\n Price of the item after 20% discount = $\\frac{80}{100}\\times$M<\/p>\n Price of the item after 5% discount =\u00a0$\\frac{95}{100}\\times\\frac{80}{100}\\times$M =\u00a0$\\frac{19}{25}$M<\/p>\n Total discount given trader B = M –\u00a0$\\frac{19}{25}$M =\u00a0$\\frac{6}{25}$M<\/p>\n According to the problem,\u00a0discount given by A is \u20b9320 more than the discount given by B.<\/p>\n $\\frac{1}{4}$M =\u00a0$\\frac{6}{25}$M + 320<\/p>\n $\\frac{25M-24M}{100}=320$<\/p>\n M = \u20b932000<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>A customer wanted to purchase an item marked for \u20b910000. Shopkeeper offered two types of discounts, 25% flat discount or successive discounts of 14% and 12%. Which is the better offer for the customers and by how much?<\/p>\n a)\u00a0first offer by \u20b968<\/p>\n b)\u00a0second offer by \u20b968<\/p>\n c)\u00a0first offer by \u20b932<\/p>\n d)\u00a0second offer by \u20b9100<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 25% flat discount<\/strong><\/u><\/p>\n Selling price of the item =\u00a0$\\frac{75}{100}\\times10000$ =\u00a0\u20b97500<\/p>\n Successive discounts of 14% and 12%<\/strong><\/u><\/p>\n Price of the item after 14% discount =\u00a0$\\frac{86}{100}\\times10000$ =\u00a0\u20b98600<\/p>\n Price of the item after 12% discount =\u00a0$\\frac{88}{100}\\times8600$ =\u00a0\u20b97568<\/p>\n Difference between selling prices =\u00a07568 – 7500 =\u00a0\u20b968<\/p>\n First offer is better\u00a0by \u20b968.<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 10:\u00a0<\/b>If selling price of 75 articles is equal to cost price of 60 articles, then the approximate loss or gain percent is:<\/p>\n a)\u00a0Profit of 25%<\/p>\n b)\u00a0No profit no loss<\/p>\n c)\u00a0Loss of 30%<\/p>\n d)\u00a0Loss of 20%<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cost price of 60 articles = C<\/p>\n Cost price of 1 article =\u00a0$\\frac{C}{60}$<\/p>\n Selling price of 75 articles = C<\/p>\n Selling price of 1 article =\u00a0$\\frac{C}{75}$<\/p>\n Loss% =\u00a0$\\frac{\\frac{C}{60}-\\frac{C}{75}}{\\frac{C}{60}}\\times100$<\/p>\n =\u00a0$\\frac{\\frac{5C-4C}{300}}{\\frac{C}{60}}\\times100$<\/p>\n =\u00a0$\\frac{60}{300}\\times100$<\/p>\n = 20%<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 11:\u00a0<\/b>The marked price of an article is \u20b9180. Renu sells it after 20% discount on its marked price and still gains 25%, The cost price (in \u20b9) of the article is:<\/p>\n a)\u00a0120.80<\/p>\n b)\u00a0110.80<\/p>\n c)\u00a0115.20<\/p>\n d)\u00a0125.50<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Marked price of an article is \u20b9180.<\/p>\n Discount = 20%<\/p>\n Selling price of the article = $\\frac{80}{100}\\times$180 =\u00a0\u20b9144<\/p>\n Let the cost price of the article =\u00a0C<\/p>\n Profit = 25%<\/p>\n Selling price of the article = $\\frac{125}{100}$C<\/p>\n $\\frac{125}{100}$C =\u00a0144<\/p>\n C =\u00a0$\\frac{576}{5}$<\/p>\n C = 115.2<\/p>\n Cost price of the article =\u00a0\u20b9115.20<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 12:\u00a0<\/b>The cost price of an article is \u20b9250. A shopkeeper gains 20% by selling it at a discount of 36% on its marked price. What is the marked price (in \u20b9) of the article?<\/p>\n a)\u00a0450<\/p>\n b)\u00a0380.50<\/p>\n c)\u00a0475<\/p>\n d)\u00a0468.75<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cost price of the article is \u20b9250.<\/p>\n Profit = 20%<\/p>\n Selling price of the article = $\\frac{120}{100}\\times$250 =\u00a0\u20b9300<\/p>\n Let the marked price of the article = M<\/p>\n Discount = 36%<\/p>\n $\\frac{64}{100}\\times$M = 300<\/p>\n M = 468.75<\/p>\n Marked price of the article =\u00a0\u20b9468.75<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>Radha bought a fridge and a washing machine together for \u20b957300. She sold the fridge at a profit of 15% and washing machine at a loss of 24% and both are sold at the same price. The cost price of washing machine(in \u20b9) is:<\/p>\n a)\u00a028650<\/p>\n b)\u00a034500<\/p>\n c)\u00a024500<\/p>\n d)\u00a022800<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cost price of fridge = x<\/p>\n Profit = 15%<\/p>\n Selling price of fridge =\u00a0$\\frac{115}{100}$x<\/p>\n Cost price of\u00a0washing machine =\u00a057300 – x<\/p>\n Loss = 24%<\/p>\n Selling price of washing machine =\u00a0$\\frac{76}{100}\\left(57300-\\text{x}\\right)$<\/p>\n According to the problem, both are sold at same price.<\/p>\n $\\frac{115}{100}$x =\u00a0$\\frac{76}{100}\\left(57300-\\text{x}\\right)$<\/p>\n 115x + 76x = 76$\\times$57300<\/p>\n 191x =\u00a076$\\times$57300<\/p>\n x =\u00a076$\\times$300<\/p>\n x =\u00a022800<\/p>\n Cost price of fridge =\u00a0\u20b922800<\/p>\n Cost price of washing machine = 57300 – x<\/p>\n =\u00a057300 – 22800<\/p>\n =\u00a0\u20b934500<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>The marked price of an article is \u20b95320. It is subject to two successive discounts, the first being 15%, and the second at a rate of 20% of the first. What is the selling price (to nearest \u20b9) of the article?<\/p>\n a)\u00a0\u20b94522<\/p>\n b)\u00a0\u20b94127<\/p>\n c)\u00a0\u20b94000<\/p>\n d)\u00a0\u20b94386<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The marked price of an article is \u20b95320.<\/p>\n First discount = 15%<\/p>\n Price of the article after 15% discount =\u00a0$\\frac{85}{100}\\times5320$ =\u00a0\u20b94522<\/p>\n Second discount = 20% of 15% =\u00a0$\\frac{20}{100}\\times$15% = 3%<\/p>\n Selling price of the article after 3% discount =\u00a0$\\frac{97}{100}\\times4522$ =\u00a0\u20b94386.34<\/p>\n =\u00a0\u20b94386 (approximately)<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 15:\u00a0<\/b>Hari suffered a loss of 8% by selling an article. If he had sold it for \u20b9300 more, he would have made a profit of 4%. Find his CP (in \u20b9).<\/p>\n a)\u00a02400<\/p>\n b)\u00a02250<\/p>\n c)\u00a02575<\/p>\n d)\u00a02500<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cost price of article = 100C<\/p>\n Hari suffered a loss of 8% by selling an article.<\/p>\n Selling price of the article =\u00a0$\\frac{92}{100}\\times$100C = 92C<\/p>\n If he had sold it for \u20b9300 more, he would have made a profit of 4%.<\/p>\n 92C + 300 =\u00a0$\\frac{104}{100}\\times$100C<\/p>\n 92C + 300 = 104C<\/p>\n 12C = 300<\/p>\n C = 25<\/p>\n The cost price of article =\u00a0\u20b92500<\/p>\n Hence, the correct answer is Option D<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/380656.png\")
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