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1. Nishant bought an article at 20% discount on MRP, and claims to sell it at profit of 10% of MRP. When Nikhil offered him Rs. 500 banknote, he cheated again by giving him Rs. 125 instead of Rs. 225. Find overall profit% of Nishant.

(a) 87.5% (b) 37.5% (c) 100%  (d) 62.5% (e) None of these

2. A shopkeeper marked up the price of a mobile phone by 40% of its cost price, if he increases the discount from 5% to 10%, the profit would decrease by 1400 Rs. How much profit shopkeeper would earn if he gives a discount of 20% on the marked price ? Assume that, he calculates discount only on MRP.

(a) 1800 Rs.  (b) 1200 Rs.  (c) 2800 Rs.   (d) 2200 Rs.  (e) 2400 Rs.

3.  A manufacturer of cricket balls wants to earn 25% profit on manufacturing cost after giving a discount of 23 ⅓% on MRP marked by him. But due to some reasons he lost 25% of balls & he decided to offer discount of 7913% on MRP of remaining balls. Find his overall profit% or loss%.

(a) 8% Profit (b) 12% Profit (c) 12% Loss  (d) 8% Loss (e) 5% Profit

4. Rahul bought a cycle at a discount of 16⅔% on MRP. He earned half the amount of his CP by renting it for 200 days. After that he resells it at half of MRP. In this transaction he earned Rs. 200, find MRP of cycle (in Rs).

(a) 1860 (b) 2490 (c) 2400  (d) 2280 (e) 2310

5. A man purchased two items A & B and invested Rs 50 & Rs 75 on their repairing respectively. If he earns profit of 10% on A and 12% on B, overall profit earned by him is is Rs 84. But if he earns 20% on A & 10% on B, overall profit earned by him is 14% of total price of items. Find initial total purchasing price of both items.

(a) Rs 675 (b) Rs 725 (c) Rs 750  (d) Rs 625 (e) Rs 775

6.  A shopkeeper sold two articles, if he marked up second article at 11 19% above first article’s selling price and gives a discount of 20% on that, then a loss of 20% occur on first one and 3313% profit on second article. Find total selling price of both articles, if he made total loss of Rs.75 on both articles.

(a) 4000 Rs (b) 4500 Rs (c) 5100Rs.  (d) 4800 Rs (e) 5800Rs

7. Satish buy two articles i.e. type A at Rs 500 and type B at Rs 1500. He sold type A article at x% profit and mark up type B article 2x% above the cost price and gave x% discount at the time of the sale. By this Satish earn (x -6) % profit. Find the value of ‘x’.

(a) 15% (b) 18% (c) 25%  (d) 20% (e) 30%

Directions (8-10): A article is mark up above cost price such that markup percent is double of the profit percent. If discount is 12.5%, then profit percent increased by 3313%.

8. On selling 20 such article at 12.5% discount, profit is Rs.300. Find the M.P. of each article.

(a) Rs.60  (b) Rs.160  (c) Rs.80   (d) Rs.240  (e) Rs.72

9. If shopkeeper cheat his customer by giving 20% less quantity and reducing value of discount percentage by 20% then find the new profit percent.

(a) 60% (b) 75% (c) 62.5%  (d) 80% (e) 70%

10. Initially shopkeeper have 20 articles. Out of 20, 7 articles damaged and remains unsold. Marked Price should he labeled by how much percent more than cost price so that his overall profit does not change neither his discount percentage.

(a) 156% (b) 146% (c) 136%  (d) 120% (e) 125%

Solution:-

1.  (a); Nikhil gave Rs. 500 note, & Nishant have to actually pay him Rs. 225. Hence Nishant sold it at Rs. 500 – 225 = 275 to Nikhil.  Now, Let MRP is 100x.  If he will get 10% profit, then SP will becomes 110x.  110x = 275  x = 5.5  ∴ MRP is Rs. 250  And CP for Nishant is 80100×25 = Rs. 200.  He got Rs. (500 – 125)  = Rs. 375 from Nikhil. Hence Profit % = (375–200200)×100  = 78×100  = 87.5%

2.  (e);    CPofMobileM.P.mobileS.P.ofmobile100%140%133%(1stdiscount)126%(2nddiscount)(133 – 126)% = 1400   7% = 1400    C.P.=14007×100  =20000 Rs.    S.P.after 20% discount=20000×112100  =22400   Profit = 2400 Rs.

3.  (b);

4.  (c); Let the MRP of cycle is 6�  ∴ discount is 503% of 6x = x ∴ Rahul bought this cycle for 5x He earns 2.5x by renting it and resells it at 3x.  ATQ  3x + 2.5x – 5x = 200  0.5x = 200   ⇒ x = 400 ∴ MRP = 6x = 6 × 400 = Rs. 2400

5.  (d); Let the purchasing price of items A is B be Rs x – 50, Rs y – 75 respectively.                                                                                                                               When he applied Rs 50 & Rs 75 on them, their actual CP becomes x & y ATQ 10100x+12100y=84   …(i) Also,    20100x+10100y=14100(x+y)  ⇒ 20x + 10y = 14x + 14y ⇒ 6x = 4y x= 23y   Put this value in equation (i)    10100 (2y)3+12100y=84    20y+36y300=84    ⇒y=84×30056  = Rs 450    x=450×23=Rs 300 Initial purchasing price is  = (300 + 450) – (75 + 50)  = Rs 625

6.  (c); Let marked price of second article be  10x Rs.  Selling price of first article be 9x Rs.  Selling price of second article     =10x×810=8x Rs.   C.P. of fisrt article    =9x4×5=11.25x Rs.   C.P. of second article   =8�׳4   = 6x Rs.  Total C.P. of both articles  = 11.25x + 6x  = 17.25x Rs.  Total S.P. of both articles = 9x + 8x   = 17x Rs.  ATQ— 17.25x – 17x = 75    x=750.25=300 selling price of both articles = 17 × 300  = 5100 Rs.

7.  (d);

Solutions (8-10):

8.  (c); Profit on one articles = 30  Profit on 20 articles = 30 × 20 ⇒ 600  600 → 300 1 → 0.5  M.P. → 160 → 160 × 0.5 ⇒ Rs.80

9.  (e); Let total quantity ⇒ 1000 He gives → 800 → for → 100  Cost price of → 800 → 80  Now initial discount = 30160×100 = 18.75% Reduced discount = 18.75×45 = 15% S.P. = 160×85100 = 136  Profit = (136–80)80×100 = 70%

10.  (b); Let cost price of 20 articles → 20 × 100    = 2000 Actual profit on 20 articles → 20 × 30  = 600 S.P. → 2600 S.P. of each undamaged article  ⇒ 260020–7 = 200 Initial discount = 30160×100 = 18.75%   So M.P. of each article should be    = 20081.25×100≈246    Approximately markup% = 146%