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Solution: Let third number be x. :. Second number = 3x and first number = 3x/2 Now , x+3x+3x/2= 3*44 ⇒ 8x+3x/2 = 3*44 ⇒ 11x= 6*44 ⇒ x= 6*44/11= 24 :. The largest number = 3x = 3*24= 72 Then the option in (2) 72

Solution Let no of cars = x and no of motorcycles = yAccording to our conditionx + y = 20X = 20 – y (i)4x + 2y = 56 (ii)Replacing (i) in (ii) we get4(20 – y) + 2y = 5680 – 4y + 2y = 56-2y = -24Thus y = 12Now replacing y =

(1) 39 minutes (2) 46 minutes (3) 40 minutes (4) 45 minutes Solution: Part of the tank filled by pipes A and B in 1 minute = 1/36+1/45= 5+4/180 = 9/180 = 1/20 Part of the tank filled by these pipes in 7 minutes =7/20 Remaining unfilled part = 1-7/20= 20-7/20= 13/20 When all three

Solution Given: 5x²−2 , x=3We need to find the value of 5x²−2 at x=3We will put 3 in place of x in the expression 5x²−2Therefore, the value of the expression will be5x²−2=5(3)²−2=5(3)−2=45−2=43

(1) 100 (2) 92 (3) 91 (4) 50 Solution: Number of digits from 1 to 9 = 9Number of digits used in writingfrom 10 to 50= 41 x 2 =82:. Total digits =82 + 9 = 91 Then Option is (3) 91

Solution Let the total number of matches played be x.The team won around 10 matches, and the team’s winning percentage was 40%.40/100 × x = 1040x = 10 × 10040x = 1000x = 1000/40= 100/4= 25Hence, the team played 25 matches.

(1) 15 (2) 60 (3) 36 (4) 45 Solution: If the numbers be 3x and 4x, then their LCM = 12x =12x= 180⇒x=225/15=15 =First number = 3x = 45 ⇒Then the option is (4) 45

Solution x²+ (x + 1)² = 145⇒2x ²+ 2x + 1 = 145⇒2x ² + 2x – 144 = 0⇒x ² + x – 72 = 0⇒x(x + 9) – 8(x + 9) = 0⇒x = -9,8

(1) 136 square cm. (2) 120 square cm. (3) 125 square cm. (4) 100 square cm. Solution: Let the length, breadth and height of the parallelopiped be a, b and c cm respectively. :. a+ b+ c= 20 2 (ab + bc + ca) = 264 :. (a+b+c)2= a2+b2+c2+2(ab + bc+ ca) ⇒400= a2+b2+c2+264 ⇒a2+b2+c2=

Solution: Let the last number be x.According to the question,18 x 10 + 11 x 20 + x= 30 x 15⇒ 180 + 220 + x = 450⇒400 +x= 450 ⇒x=450-400=50 Then the answer is 50