- Question -1
a) 6 b) 4 c) 5 d) 7
Just consider the sequence 3,4,5,6,7 and start subtracting and adding 1 to consecutive numbers which will get you sequence in question.
- Question -2
2. Find the missing number in 0, 4, 5, 11_
a)12 b)13 c)14 d)15
Whenever you find a sequence like 0,4,5,11 with hardly any relation between adjacent numbers, it is wiser to try adding and subtracting small numbers like 1 or 2 to all numbers. Now alternatively adding 1and subtracting 1 from each number of sequence we get 1,3,6,10. This sequence has got a pattern!
The first digit “1” is the sum of first “1” natural numbers. second digit “3” is the sum of first “2” natural numbers, and the third digit “3” is the sum of first “2” natural numbers, and the third digit”6” is the sum of first “3” natural numbers and so on.
Hence the final digit in the modified sequence would be 15 which is the sum of first 5 natural numbers. we need to subtract 1 from 15 so that it fits the sequence in question. Hence the answer is 14.
- Question -3
3.What is the Percentage increase in area when triangle is cloned(so that we have two triangles in total) and the resulting two triangles are joined on their bases to form a parallelogram?
a.100% b.150% c.300% d.120% 7
Area of a triangle with base b and height h is given by ½(b X h)
When two triangles are mounted on their bases to form s parallelogram, the area would become (b X h)
Percentage increase of area= (new area-area of the original triangle)/ (area of the original triangle) X 100%
= (b X h)-1/2(b/h) / ½(b X h) X 100%
= (1-1/2) / (1/2)X100%
1. A 2.B 3.C D. Cant be determined
Consider WA, WB and WC be the work done per day by A, B, C respectively .then
Eq2-Eq3 will give
Eq1+Eq4 will give
Sub value of WB IN eq 1,we get
Sub value of WA in eq3,we get
Since WB (work done by B per day) is greater when compared to WA and WB clearly B will be able to do the maximum work on any given day and hence he should consume least amount of time when working independently.
- Question – 5
a.121/12 hours b.144/36 hours c.144/25 hours d.121/25 hours
A can deliver 20 parcels in 3 hourse.Hence for 1 hour
He can deliver 20/3 parcels.
B can deliver 15 parcels in 4 hr.hence for 1 hr B can deliver ,20/3+15/4 parcels=80+45/12=125/12 parcels.
Hence to deliver 60 parcels they would require: 60X12/125=720/125=144/25 hr.