Mathematics Problem Solving Group

A. 300 B. 280 C. 310 D. 290 SSC CGL 2022 13 August 2021 Shift 1 Solution: Let the articles be A1 and A2 sold at 20% loss and 16% profit respectively. Let the Cost Price (CP) of the A1 for the trader be = x Then the Cost Price (CP) of the A2 for the trader will be = 490 - x It is given that the Selling Price (SP) of A1 and A2 are the same. Let the Selling Price be = y It is given that, On sale of A1, Loss % = 20% On sale of A2, Profit % = 16% We know that, Loss % = ((CP - SP)/CP) * 100 .......(1) Profit % = ((SP - CP)/CP) * 100 ........(2) Putting values in eq. 1 for A1 and in eq. 2 for A2 20 = ((x - y)/x) * 100 .......(3) 16 = ((y - (490 - x))/(490 - x)) * 100 ....(4) From eq. 3, we get 20/100 = 1 - y/x y/x = 1 - 20/100 y/x = 80/100 y = 4x/5 ....(6) Substitute eq. (5) in eq. (4) 16 = ((4x/5 - (490 - x))/(490 - x)) * 100 16/100 = (4x/5)/(490 - x) - 1 4x/(5(490 - x)) = 16/100 + 1 4x/(5(490 - x)) = 116/100 4x/(5(490 - x)) = 29/25 100x = 29 * 5 * (490 - x) 100x = 71050 - 145x 245x...

A. 60 B. 120 C. 300 D. 240 SSC CGL 2020 13 August 2021 Shift 1 Solution: Given that: Let a, d 1 and d 2 be the side, diagonal 1 and diagonal 2 of the rhombus respectively. side of rhombus (a) = 13 cm diagonal of rhombus (d 1 ) = 24 cm We know that Area of Rhombus = (product of diagonals)/ 2 = (d 1 x d 2 )/2 ……..(1) a 2 = (d 1 /2) 2 + (d 2 /2) 2 ……….(2) Putting all the values in eq. (1) to find d 2 13 2 = (24/2) 2 + (d 2 /2) 2 169 = (12) 2 + (d 2 /2) 2 169 – 144 = (d 2 /2) 2 25 x 4 = (d 2 ) 2 (d 2 ) 2 = 100 d 2 = 10 …..(3) Putting all the values in eq. (1), we get Area of Rhombus = (24 x 10)/2 Area of Rhombus = 120 cm 2 Hence, (B) is the correct answer. Study about Co-ordinate Geometry Formulas and Tricks.