Ten years ago, the age of Usman was one-third the age of Majeed at that time. The present age of Majeed is 12 years more than the present age of Usman. Find the present age of Usman ?
A. 16
B. 22
C. 43
D. 71
MATH MCQS
- Mathematics questions with answers cover topics such as averages, areas, percentages, solving math problems, all mathematical formulas, basic algebra, basic math, arithmetic, and geometry MCQs. Math MCQs range from basic to advanced levels, providing preparation for tests
Sum of four consecutive even numbers is 292. What would be the smallest number ?
A. 23
B. 45
C. 67
D. 70
Cost of 16 pens and 8 pencils is Rs.352 and the cost of 4 pens and 4 pencils is Rs.96. Find the cost of each pen ?
A. Rs.76
B. Rs.67
C. Rs.32
D. Rs.20
P, Q and R have Rs.6000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R ?
A. Rs.3600
B. Rs.4000
C. Rs.3000
D. Rs.2400
Eight years ago, Zahid’s age was 4/3 times that of Jabbar. Eight years hence, Zahid’s age will be 6/5 times that of Jabbar. What is the present age of Zahid ?
A. 40 years
B. 48 years
C. 30 years
D. 32 years
There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits ?
A. 24
B. 56
C. 98
D. None of them
Three consecutive odd integers are in increasing order such that the sum of the last two integers is 13 more than the first integer. Find the three integers ?
A. 7, 9, 11
B. 13, 15, 17
C. 9, 11, 13
D. 11, 13, 15
A bag contains equal number of Rs.5, Rs.2 and Re.1 coins. If the total amount in the bag is Rs.1152, find the number of coins of each kind ?
A. 144
B. 72
C. 432
D. 288
If ‘a’ denotes the unknown value and is increased by 2, the result will be 12 then the equation for this statement will be _____ ?
A. a = 12 + 2
B. a – 2 = 12
C. a + 2 = 12
D. 2a = 12
Four handbags and three school bags costs $2000 and cost of handbag is thrice as much as schoolbags. If cost of school bag is ‘a’ then the derived equation will be _____ ?
A. 3(4a) + 3(3a) = 2000
B. 3(4a) – 3(3a) = 2000
C. 3(4a) – 3a = 2000
D. 3(4a) + 3a = 2000