Answer: C) 71 Explanation: To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51). Therefore, HCF of 923, 1207 and 1349 is 71.

Answer: A) Rs. 345.15 Explanation: area of the plot = 120m x 60m = 7200 sq.marea of plot excluding gravel = 117m x 57m = 6669 sq.marea of gravel = 7200 sq.m - 6669 sq.m = 531 sq.mcost of building it = 531 sq.m x 65 ps = 34515 psIn Rs. = 34515/100 = Rs. 345.15.

Answer: B) 7/15 Explanation: Given total students in the class = 60Students who are taking Economics = 24 andStudents who are taking Calculus = 32Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4Students who are taking calculus only = 32 - 4 = 28probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

Answer: C) 4 days Explanation: One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)   One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)   Divide both sides by 2 in (I) and then multiply both sides by 5 Now we get, 15b + 20g = 1/4. Therefore, 15 boys and 20 girls can do the same work in 4 days.

Answer: A) 9944 Explanation: The largest 4 digit number is = 9999After dividing 9999 with 88, we get 55 as remainderso largest 4 digit number divisible by 88 = 9999-55 = 9944.

Answer: A) 25% Explanation: He has gain = 15 - 12 = 3,Gain% = (3/12) x 100 = (100/4) = 25.He has 25% gain.

Answer: B) 29.39 cm Explanation: Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.   As speed is same, area of cross sections should be same.    Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12  i.e πR2 = 6πR12    Here R = R and R1 = 12/2 = 6    ⇒R2 = 6×6×6    R = 14.696    => D = X = 14.696 * 2 = 29.3938 cm.