Directions : These questions are based on the data given below.
Solution :
Pqrpqr= pqr x 1001
1001 = 7 x 11 X 13
234 X 1001 = 234234
23 X 1001 = 23023
Directions: Refer the following data to answer the questions given below.
Visual question. Notice that the difference between BA and MBA is in the denominator. Since the denominator in MBA2 is higher, it implies that this quantity must be smaller. But MBA1 could be greater than MBA2 but less than BA. Hence statement (d) is correct.
Directions: Read the following information to answer the question.
Solution:
C’s one day’s work =(1/3)−(1/6+1/8)=1/24
Therefore, A:B:C = Ratio of their one day’s work =1/6:18:1/24
=4:3:1
A’s share =Rs.600×(4/8)=300
B’s share =Rs.600×(3/8)=225
C’s share =Rs[600−(300+225)]= Rs 75
Directions: Read the following information to answer the question.
Solution:
If the amount of work done per day by Sara, Kyna, Riddhi per day are S/6, K/9, R/15 then K/9 + R/15 = S/n, R/15 + S/6 = K/n and S/6 + K/9 = R/n.
Eliminating S, K and R we get n^3 + 15n^2 = 405. How? Can you use determinants here and get this in just 1 step?
Let f(n) = n^3 + 15n^2 - 405 -> and is an increasing function for n > 0, f(4.5) < 0 and f(5) > 0.
Directions: Read the following information to answer the question.
Length of wire all 3 of them can lay in a day = 432/8 = 54 m
On an average, 54/3 = 18 m will be laid by each which as per question should be the length Bibhuti can lay in a day.
Therefore, Anuj = 18 – x ; Chandu = 18 + x
Given,
7 * (18 – x) = 5 * (18 + x)
126 – 7x = 90 + 5x
=> 36 = 12x
=> x = 3
Therefore, Anuj in a day = 18 – 3 = 15 m