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.

A. 12 days B. 20 days C. 18 days D. 15 days SSC CGL 2019 Tier 1 4 March 2020 Shift 1 Solution: In this question from Time and Work , Let us see the shortcut method first: Let us assume the efficiency of A be 4 (it means A can do 4 units of work in a day) Therefore, the efficiency of B will be = 125% of the efficiency of A = 5 (it means B can do 5 units of work in a day) Therefore, the efficiency of C will be = 120% of the efficiency of B = 6 (it means B can do 6 units of work in a day) Total units of work = 30 x 4 = 120 units Total work done by A, B and C in 3 days = (4 + 5 + 6) x 3 = 45 units Remaining work = 120 - 45 = 75 units The number of days required by B to complete the remaining work = 75/5 = 15 days. Hence, (D) is the correct answer. Let us see the basic method: The efficiency of A = One day work of A = 1/30 The Efficiency of B = One day work of B = 1/30 + (25/100) x (1/30) = 1/30 + 1/120 = 1/24 The Efficiency of C = One day work of C = 1/24 + (20/100) x (1/24) = 1/24 + 1/120...

A. 0 B. 1 C. 4 D. 2   SSC CGL 2019 Tier 1 3 March 2022 Shift 1 Maths Paper Question Solution: Divisibility Rule of 8: If a number is divisible by 8, then its last three digits should be divisible by 8. In the given number 1005x4, the last four digits are 5x4 Let us check by using the options: using x = 0 divide 504 by 8, and we will get 63. Here, putting x = 0 makes the given digit divisible by 8. Therefore, (A) is the correct answer.