বিসিএস প্রিলিমিনারি সাধারণ বিজ্ঞান গতি

Example 35 :
Ranjan got 82 marks in Math , 78 marks in Physics , 65 marks in Computer , 68 marks in English , The maximum marks of each subject are 85 . How much overall percetage of marks did ranjan get ?
Answer :
Total marks get in all subject ( 82 + 78 + 65 + 68 ) = 293
Maximum marks ( 85 x 4 ) = 340
Percentage = 293 x  100 /  340 = 86.17.

Example 36 :
A student scores 25% & failed by 35 marks while another student who scores 65% get 45 marks more than minimum required marks to pass . Find the maximum marks in the exam ?
Answer :
25% – 35 = 65% g+ 45
65% – 25% = 45 + 35
40% = 80
100% = 80 x 100 / 40 = 200 marks.

Example 37 :
If N is equals to 20% of M and P is equals to 30% of N, then which one of the following equals to  40% of P?
Answer :
N = 20% of M = ( 20 / 100 ) x M = 0.2M.
P = 30% of N =  ( 30 / 100 ) x N = 0. موقع 1xbet 3N =  0.3 x 0.2M.
So here 40% of P = ( 40 / 100 ) x P = ( 0. 888 sport 4 )( 0.3 )( 0.2M )
= 0.024M.

Example 38 :
In an examination it is required to get 57% of the aggregate marks to pass . A student gets 237 marks and is declared failed by 7% marks . What are the maximum aggregate marks a student can get ?
Answer :
57% = 237 + 7%
57% – 7% = 237
50% = 237
100% = 237 x 100 / 50 = 474.

Example 39 :
A student has obtain 34% of the total marks to pass in paper. He got 113 and failed by 40 marks. The maximum marks are :
Answer :
Let the maximum number is X
Then , 34% of X = 113 + 40
34 / 100 x X = 153
X =153 x 100 / 34
X = 15300 / 34 = 450.

Example 40 :
Raju has to score 60% to pass exam . He scores 225 marks & failed  by 15% . Find the maximum marks of exam .
Answer :
60% – 15% = 225
45% = 225
100% = 225 x 100 / 45
= 500 marks .

Example 41 :
Find the following 35 is what % of 105 ?
Answer :
( 35 x 100/ 105 ) % = 100 / 3 % = 33 x 1 / 3 %

Example 42 :
1 / 3 of 1206 is what % of 134 ?
Answer :
Let 1 x 1206 / 3 = y% of 134 . Then, y x 134 / 100= 402
x = ( 402 x 100 / 134 ) = 300

এই লেকচারের পরের পেইজে যেতে নিচের …. তে ক্লিক কর।