A man earns Rs. 10000 per month. His salary increases by 1% in every month. What will be his salary
after 6 months. Give your solution in integers only.
: First sequence is
JAN, FEB, MAR, APR, MAY JUN
On arranging these words in increasing alphabetic sequence, we get
AJN, BEF, AMR, APR, AMY, JNU
On arranging these words in dictionary order we get
AJN, AMR, AMY, APR, BEF, JNU
Taking the words in their original words we get
JAN, MAR, MAY APR, FEB, JUN as changed sequence
On taking similar steps for finding the solution we get
SUN, MON, TUE, WED, THU, FRI, SAT
NSU, MNO, ETU, DEW, HTU, FIR, AST (increasing alphabetic sequence)
AST, DEW, ETU, FIR, HTU, MNO, NSU (dictionary order)
SAT, WED, TUE, FRI, THU, MON, SUN (answer)
This sequence contains 1 digit extra in every next number.
1 | 74 | 894 | 9688 | 88997 | 897999
=> 1 | 7 + 4 |8 + 9 + 4| 9 + 6 + 8 + 8| 8 + 8 + 9 + 9 +7|8 + 9 + 7 + 9 + 9 + 9 |
=> 1| 11| 21 | 31 | 41 | 51| ….
In ninth term sum of all digit of the number will be 81.
After that it should be 91 in tenth term which is not possible because the greatest
number of ten digit is 9999999999, the sum of its all digits = 9 × 10 = 90
After dividing an amount into two parts, Mr. Suresh invests these parts in two different business.
In first, he gains 50% profit whereas in second he has a loss of 25 %.Total profit of
Mr. Suresh in first business is 1888 rupee. What is the total investment of Mr. Suresh if he gains
23 % profit over all.
(One variable solution)
Consider that 100 Rs is invested, then two parts are x & (100 - x)
Taking 50 % profit x becomes (3/2) x
& taking 25% loss (100 – x) becomes (3/4)(100 – x)
As total profit is 23 % hence
[(3/2)x +(3/4)(100 – x)] – 100 = 23
On solving x = 64
Profit = 64 * 50/100 = 32 in every 100 rupee
Overall profit in first business = 1888
Total investment = 100 * 1888 / 32 = 5900
Alternate method
Let total investment = P & part investments are x & (P – x)
According to given condition
x/2 = 1888 implies x = 3776
Similarly according to other condition
3x/2 +(3/4)(P – x) = P 123/100
=>3/2 * 3776 +(3/4)(P – 3776) = P 123/ 100
=>3 * 1888 + (3 /4) P – 3 * 944 = 123 P/100
=>5664 – 2832 = 123 P /100 – (3/4) P
=>2832 = (123 P – 75 P)/100
=>2832 = 48 P/100
=>P = 2832 * 100/ 48 = 5900.
A man purchases three objects of consecutive prices. The number of purchased objects are
1, 2 & 3. He pays Rs 115 to shopkeeper.If all prices are taken in integers, then out of 10 choices,
the price of that object, whose 3 pieces are purchased, is
Some communication provider companies and their coded numbers are given below.
NOKIA = 1100
AIRTEL = 0021
HUTCH = 13
SPICE = 210
Similar code for VODAFONE IS
In NOKIA, there are three vowels in the word, number of consonants between these vowels are obtained by the following manner.
NOKIA
(1 letter) O (1 letter) I (0 letter) A (0 letter)
Hence code = 1100
Similarly AIRTEL
(0 letter) A (0 letter) I (2 letter) E (1 letter)
Code is 0021
Taking similar steps for VODAFONE
V O D A F O N E
(1 letter) O (1 letter) A (1 letter) O (1 letter) E (0 letter)
Code is 11110