Solved Maths Questions of Marketing Organizer (Main Exam)

This post contains detailed solution for Maths questions that were asked in the Kerala PSCMarketing Organizer (Main Exam), which was conducted on November 05, 2024.

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Squaring on both the sides,

Substituting the value of ( a² +1a²),

Solution∴The mean of a² and1a² is 2M² - 1.

2. Actual weight of a rice bag is 50 kg. In a hurry, it was weighted as 50.5 kg. The errorpercentage is

Weight of the rice bag = 50kg. (Given)

Error weight of the rice bag = 50.5 kg. (Given)

∴  the error percentage is1%.

3.A printer numbers the pages of a book starting with 1 and uses 3189 digits in all.How many pages does the book have?

In here, we have to find the no. of single digit, double digit, triple digit and so on... pages.

From 1 to 9, there are 9 pages, each using 1 digit.∴ Total digits used =  9 × 1 = 9.Double-digit pagesFrom 10 to 99, there are 90 pages, each using 2 digit.Total digits used =  90 × 2 = 180.

Triple-digit pagesFrom 100 to 999, there are 90 pages, each using 3 digit.Total digits used =  900 × 3 = 2700

∴ Total digits used = 2889.

Solution∴  Total no. of pages  = 1074 pages.

5.  If ax= b, by= c and cz= a, then the value of xyz isA) 0B) 1C) 2           D) None of theseExplanation:Given,  ax= b   ----------- Iby= c ------------ IIcz= a ----------- IIIax= b can be written as,b = axSubstitute value of b in II,    (ax)y= caxy= c     (∵ (pm)n= pmn) ---------- IVBy equation III,                         a = czSubstitute value of a in IV,(cz)xy= ccxyz=c  (∵ (pm)n= pmn)xyz = 1Solution∴  The value of xyz  = 1.6.  If the selling price is doubled, the profit triples. Then the profit percent isA) 300%B)200%         C) 150%D) 100%Explanation:Let P = Profit, Cost price, CP = x and Selling Price, SP = yBy the formula , Profit % = [(SP - CP)/CP]×100P%   = [(y-x)/x]×100Given,P = y-x            --------------------------- I3P = 2y-x--------------------------- II(∵ it is given that if selling price is doubled, profit triples)Substitute I in II,                            3P = 2y-x3(y-x) = 2y-x3y-3x = 2y-xy = 2x             ---------------------------- IIISubstitute III, in the formula,    P%   = [(y-x)/x]×100= [(2x-x)x]×100= 1×100= 100Solution∴  The profit percent is 100.7.  A metal sphere of diameter 12 cm is melted to make a cone with base same asdiameter. Then the height of the cone isA) 12 cmB)18 cmC) 24 cm           D) 30 cmExplanation:Let r1 be the radius of the sphere and r2 be the radius of base of the cone.Given, the diameter of the sphere, d =12 cm.∴    r1 = d/2r1 = 12/2 = 6cm∴   r2 = 6cm (∵ it is given thatbase of the cone has the same diameter as the sphere)Step 1: Calculate the volume of the sphereVolume of the sphere, V1 =  (4/3)πr₁³Substituting the value of r1,V1 =  (4/3)π × 6³=  904.78 cm³Step 2:Calculating the volume of the cone,Volume of the sphere, V2 =  (1/3)πr₂²h    ------------- ISince it is given that the sphere is melted to form the cone, their volumes are equal,∴   V1 = V2Substituting the value of r2 & V1 in equation I,904.78 =(1/3)π×6²×h904.78=  36π/3×hh ~ 24cmSolution∴  Theheight of the cone is 24 cm.8.  A train runs at a speed of 90 kilometre per hour and crosses a bridge of length150 meter in 15 seconds. Then the length of the train in meter isA) 175B)200C) 225D) 250Explanation:Given the speed of the train  = 90km/hr= 90×  5/18(∵ Converting km/h to m/s, multiply by 5/18)= 25 m/sLet the length of the train =  'x' m.Total distance covered = Length of the train + Length of the bridge= x + 150By using the formula,Distance = Speed × TimeSubstituting the values,x + 150 = 25×  15x  = 375 -150x = 225Solution∴  Thelength of the train is 225m.9.  The unit’s digit in the product of first 50 odd natural numbers isA) 0B)1C) 3          D) 5Explanation:50 Odd Natural Numbers, N ={1,3,5,7,9,11,13, 15,17, 19,.......99}The product of 50 Odd Natural Numbers, N = (1×  3×5×  7×  9×  11×  13×15×  .........99)Method IAny number multiplied by 5 gives either 0 or 5 in units place, and 5 multiplied by even gives '0' and 5 multiplied byodd gives '5'.N = (1×  3×5×  7×  9×  11×  13×15×  .........99)= 5×(1×  3×  7×  9×  11×  13×  .........99)=5× Odd no.= 5 in the unit digit.Method IINotice a pattern in  Odd Natural Numbers, N ={1,3,5,7,9,11,13, 15,17, 19,.......99}ie,  in the unit's digits it is { 1, 3, 5, 7, 9, 1, 3, 5, 7, 9....99}This pattern repeats every 5 numbers. So, for the first50odd numbers, we have10complete cycles of this pattern.∴  for each pattern,   1   ×  3  ×  5   ×  7  ×  9 = 945∴  unit's digit of 945 is 5.Solution∴ The unit's digit in the product of the first 50 odd natural numbers is 5.Thanks for reading!!!