Math Mock Test in English for online practice of Competitive Exams. Objective MCQs with answer and solution.
Results
#1. Two boats A and B start towards each other from two places, which is 108 km apart. Speed of the boat A and B in still water are 12 km/hr and 15 km/hr respectively. If A proceeds down and B up the stream, they will meet after how many hours?
#2. A cricketer had a certain average of runs for his 64 innings. In his 65th innings, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is :
#3. The total number of prime factors in 
is
#4. The present worth of a bill due 7 months hence is Rs 1200 and if the bill were due at the end of 
years its present worth would be Rs 1016. The rate percent is :
#5. Three glasses are filled with a mixture of acid and water in equal volume. The ratios of acid and water are 2 : 3, 3 : 4 and 4 : 5 respectively. The contents of these glasses are poured in a large vessel. The ratio of acid and water in the large vessel is :
#6. The average of all the odd integers between 2 and 22 is :
#7. Water tax is increased by 20% but its consumption is decreased by 20%. Then the increase or decrease in the expenditure of the money is :
#8. Suppose that ‘x’ number of men can finish a piece of work in 30 days. If there were 6 men more, the work could be finished in 10 days less. The original number of men is :
#9. Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.
LCM of 18, 21 and 24
LCM = 2 × 3 × 3 × 7 × 4 = 504
Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504 K – 11 Where K is a positive integer
Since 23 × 21 = 483
We can write 504 K – 11
= (483 21) K – 11, = 483 K (21K – 11)
483 K is multiple of 23, since 483 is divisible by 23.
So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.
Put the value of K = 1, 2, 3, 4, 5,6, ….. and so on successively.
We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11)
= 115 which is divisible by 23.
Therefore, the required least number
= 504 × 6 – 11 = 3013
#10. If the square of the sum of two numbers is equal to 4 times of their product, then the ratio of these numbers is :
