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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , 3., , (b), , 5, , Here a =, =, , 3,, , 2,, , = 4,, , = 3, a = 1, , =5, , x :y:z =(4× 5 ( 3)× 3): (3× 1, 5 ×( 2)):(( 2) × ( 3, 1×4, or,, or,, , 6, , :4, , = 15, , ×4 :, , ×4 : ×4, , ×4, , 20, , 24, , 60, , 16, , total profit of C = 16 + 15 = 31, Profit of A = 20, , x:y:z =29:13:2, =, , :, , Profit of B =24, , = = (let), a constant, , According to the question, , ∴ x = 29 ,y =13 and z = 2, , (20 + 24) – (31) = 13 unit, 13 unit = 487.50, , 4., , (a), , 1 unit = 37.5, , Let C ‘s investment =Rs. x, , total profit = 75 unit = 75 × 37.5 = 2812.50, , B’s investment =Rs. (x-3000), A’s investment =Rs. (x-3000+6000)=Rs., (x+3000), , 6., , (a), Let the number of balls in A and B be x and y,, respectively. Then,, , Now,(A+B+C)’s investment = Rs. 72000, ⟹ x +(x-3000) +(x+3000) =72000, , =, , ⟹3x =72000 ⟹ x =24000, , 3x – 9 =7y +21, , Hence A’s investment = Rs. 27000, , 3x = 7y +30, , B’s investment = Rs. 21000, , Also,, , C’s investment = Rs. 24000, , (1), , =, , 2x – 12 = 3y +18, , Ratio of the capitals of A, B and C, , 2x = 3y +30, , =27000:21000:24000, , (2), , Solving Eq. (1) and (2) we get,, , =9:7:8, A’s share = Rs. (, , 9, , 24, , 6x = 14y +60, , × 640)= Rs. 3240, , 6x = 9y +90, y = 6 and x = 24, , 5., , (a), , Difference between x and y = 24 – 6 = 18, A, , :, , 20000 :, Profit Ratio  5, , :, , B, , :, , C, , 24000 :, , 16000, , 6, , 4, , :, , 7., , The number of residents in the two categories is, 6x and 13x, and therefore the number of residents, in the remaining category will be 9.5x., , Let total profit = 75 unit, C got extra =, , 20,  75  15, 100, , The total number of residents in the colony will be, = 6x + 13x + 9.5x = 28.5x, , Remain profit = 75 – 15 = 60 unit, Now, , A, , :, , B, , (d), , We need a multiple of 28.5 greater than 100., , :C, 6

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , The required value = 28.5 × 4 =114, 11., 8., , (c), , (c), , Let the total amount contributed by, , As per the problem:, , Reyyansh and Omita together b x and so the, , Amount spent in buying Rs.2 candies = 20 × 2, =Rs.40., , amount contributed by Akshaj will be, x/2+ 35., , Remaining candies = 80 numbers, , As per the problem,, , Remaining value =Rs.160, , x+, , If he buys 20 candies of Rs.5, then he will buy 60, candies of Rupee 1., , 35 = 179, , = 179 – 35, 3x = 288 x = Rs.96, , 9., , (d), , Amount contributed by Omita = 1/6 of Rs.96, , Let the boy purchases 24 candies of Rs.5, , = Rs.16, , The value of the candies = 24× 5 = 120, Remaining amount = Rs.80, , 12., , Let the number of candies of Rs.2 and Rupee 1 be, x and y respectively., , The number of boys and girls in Section 4 is, 27 and 18 respectively., , x + y = 76, , As per the problem, had the girls in Section, , 2x + y = 80, , 4 been 27, it would have been 9 more than the, , Solving, we get, x = 4 and y = 72., , number of boys in Section 4., , Maximum number of Rs.5 candies that can be, , Therefore, the number of boys in Section 3 is, , purchased = 24., , 10., , (c), , 27–9 =18.Also, in Section 3, girls constitute 25% of, the total strength and so the boys would be 75%, of the class strength. 18 is 75% of the total, strength and so the total strength will be 24., , (c), Given that:, a = 6b = 12c and 2b = 9d = 12e, , 13., , a = 6b = 12c and 6b = 27d = 36e, , (c), , a = 6b = 12c = 27d = 36e, , If IBM initially quoted Rs. 7x lakhs , SGI quoted 4x, lakhs, , Therefore,, , IBM's final quote = (4x - 1) lakhs, , = , = 6,, , = , = 3,, , =, , =, , bd  27d   1 , , , 18  6   18, , ,x=1, , IBM's bid winning price = Rs. 3 lakhs, So, IBM wins the bid at 4x - 1 = 3 lakhs, , =, , Note: whoever quotes the minimum price, wins, the bid ., , which is not an integer., 7

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , 17., 14., , (a), , Mileage when petrol is adulterated, , 6a2 = 2(lb + bh + lh), , = of (16) = 12 km/litre, , l : b : h = 1 : 2 : 4. Therefore, b = 2l ; h = 4l, , Hence the cost increases to a factor ( ) of the, , Hence, 6a2 = 2[2l2 + 8l2 + 4l2], , original cost of maintenance. Hence, it increases, , 6a2 =, , by × 100 = 33.33%, , 28l2, , a=(, , l∴a3 = (, , l3, 18., , Volume ratio of cube and cuboid, = a3 : lbh = (, 15., , (c), , (b), Let x be the number of children employed., , :8, , Then, number of men and women are 8x and 5x, respectively. Let y be the common factor for wags, then wages of men, women and children are Rs. 5y, , 2y and 3y repectively. Total wages = (40 + 10 +, 3) xy = 53xy = 318, , (a), If the 3 kinds of birds are taken to be 3x , 7x and, 5x respectively, then 7x - 3x = 63y (where y is any positive, integer)., , So xy = 6, , ⇒x=, , So, they will be getting 40 × 6, 10 × 6 and 3 × 6, respectively., , Minimum value of y for which x is a natural, number is 4., , Alternate method:, , ⇒ x = 63, , Sum of all three wages have to be equal to Rs. 318., Only (b) displays this property., , Hence, number of birds = 15x = 945., 19., 16., , (b), , Ratio of price =, , Let the smaller number be x, , =, , ∴Larger number = x, ∴ x2 +, , (d), , =, , (, , ∴ Ratio of depreciation = 1 -, , = 502, , Hence, loss =, , Put the value of x form the choices and see which, one satisfies it., Alternate method:, , 20., , Lets ratio of 100 nos. be 3x., , × 28000, , (c), , Given 6x2 + 5x + 10 = 256 (i.e. ( )), , a c e,   =k, b d f, , 6x2 + 5x = 246, , a = bk, c = dk, e = fk, , . ....(i), , 1, , Now check from options (b) 2x = 12 , x =, ,  a n p  c n q  en r  n, =,  n, n, n , b pd qf r , , 6 Put x = 6 in equation (i) , it satisfies ., , 8, , =, 14,000

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, 1, ,  bn k n p  dn k n q  f n k n r  n, , , bn p  dn q  f n r, , , , , , , b pd qk r,  kn n, , b p  dn q  f n r, , ,  , kn, , 1, n, , n, , n, , n, ,  , , 23., , (a), Let the present age of son and father x years and, y years respectively., , 1, n, , , , , Now,, According to question, , =k, , Condition – I, , option C, , [y+(y-x)] + [x+(y-x)] =, , ck c,  = k (given), dk d, , 136, 3y – x = 136 . ................ (i), Condition – II, , 21., , (c), , [y-(y-x)]+[x-(y-x)] = 40, , Let the marks be 6x,7x,8x ,9x and 10x,, respectively, making a total of 40x., , 3x-y = 40, y = 3x – 40 ................... (ii), , This represents 60% of the total marks:, ×, , = 40 ⟹, , Put the value in eq. (i) we get, , =, , 3(3x-40)-x = 136, 200x/3 is the total for five papers, and so the total, marks per paper = 40x/3., Now 50% of the marks per paper= (, , x = 32, and y = 96 – 40 = 56 years, , ) × 0.5 =, , The age of father and son = 56 and 32 years., , = 6.66, There are four papers in which the student has got, more than 6.66x., , 22., , 24., , (c), , 55% of X’s income = Rs.11000, , let Ratio of1consecutive, no.P= 3 : 5 :, Rs. :, 50old, P prime, :, 25, Coins → 3, :, 5, :, 7, 7, 7, 5, Value → 3, :, :, 4, 2, Value → 12, :, 10, :, 7, = 29, ×2, ×2, ×2, ×2, Rs. 24, Rs. 20, Rs. 14 58, , Therefore, X’s income =Rs.20000, , Now of coins → 24, 40, 56, , The income of X and Y is in the ratio 5:4, , Now reverse the number of coins . than, , (d), It is given that X spends 45% of his income and so, must be saving 55% of the income., , If X’s income is Rs.20000, the income of Y will be, Rs.16000, , Number of coins, , = Value, , 56, , = 20, , 50 Paise → 40, , = 20, , 25 Paise → 24, , =6, , 1 Rs. →, , Also, the amount of savings of Y = 25% of, Rs.11000 = Rs.2750, Expenditure of Y =Rs.16000-Rs.2750 =Rs.13250, , Now total value of coins = 56 + 20 + 6 = Rs. 82, , Expenditure of Y as a percentage of the income =, (13250/16000)×100 = 82.82%, 9