Mathematics Problem Solving Group

A. 16% B. 20% C. 24% D. 25% UPSC CDS 1 2021 Elementary Matematics Solution: It is given that CP of 100 Mangoes = SP of 80 Mangoes. Let CP and SP be Cost Price and Selling Price of 1 Mango. We can write that, 100CP = 80SP SP = 100CP/80 SP = 5CP/4 We know that Profit % = ((SP - CP)/CP) * 100 Profit % = ((5CP/4 - CP)/CP) * 100 Profit % = (1/4) * 100 Profit % = 25 % Hence, (d) is the correct answer.

A. 9 B. 10 C. 11 D. 12 JEE Mains 2021 Maths Paper Solution: It is given that the five-digit number is to be formed with digits 1, 2, 3, 4, 5, 6 The five-digit number should be divisible by 55, which means it should be divisible by 11 and 5. To be divisible by 5, the last digit of the number should be 5. Let the five-digit number be abcd5. Now, in order to be divisible by 11, |Sum of digits at odd places - Sum of digits at even places| = divisible by 11 |(a+c+5) - (b+d)| = divisible by 11 Case 1: (a+c+5) - (b+d) = 0 (a+c) = (b+d) - 5 This is possible for the values of (b, d) = (6, 4) and (a, c) = (2, 3) There will be 4 possible cases from these digits. Case 2: (a+c+5) - (b+d) = 11 (a+c) = (b+d) + 6 This is possible for the values of (b, d) = (1, 2) and (a, c) = (3, 6) There will be 4 possible cases from these digits. Case 3: (a+c+5) - (b+d) = 11 (a+c) = (b+d) + 6 This is possible for the values of (b, d) = (1, 3) and (a, c) = (6, 4) There will be 4 possible cases from these digits. Henc...

A. 48 minutes B. 50 minutes C. 54 minutes D. 60 minutes UPSC CDS 1 2021 Elementary Mathematics Paper Solution: Let us have a look at the shortcut method: Ratio of usual speed to 4/5th speed = 5 : 4 As distance is the same, ratio of usual time to time taken at 4/5th speed = 4 : 5 A difference of 1 unit of time corresponds to 12 minutes time difference, therefore, 4 units of time (usual time) will be 4 * 12 = 48 minutes. Hence, (a) is the correct answer. Let us have a look at the basic approach: Let the usual speed of man be x km/minute. Let the usual time taken by the man be t minutes. Let the distance be d km. We know that, speed = distance/time x = d/t d = x.t .......(1) If the speed of man becomes 4x/5 km/minute, then the time taken by the man will be (t - 12) minutes Using, speed = distance/time 4x/5 = d/(t - 12) d = 4x.(t - 12)/5 ....(2) Equating equation 1 and 2, we get x.t = 4x.(t - 12)/5 t = 4(t - 12)/5 5t = 4t - 48 t = 48 minutes. Hence, (a) is the correct answer. This is a qu...

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.