In an examination, 80% of the students passed in English, 85% in Mathematics and 75 % in both English and
Mathematics. If 40 students failed in both the subjects , find the total number of students.
Answer : Option C
Let the total number of students be `x`
Let A and B represent the sets of students who passed in English and Mathematics respectively.
Then, number of students passed in one or both the subjects
= n (A`uu` B) = n(A) + n(B) - n(A`nn` B) = 80% of `x` + 85% of `x` - 75% of `x`
=`(80/100 x + 85/100 x - 75/100 x)` = `90/100 x = 9/10 x`
Students who failed in both the subjects = `(x - (9x)/(10)) = x/10`
So, `x/10`= 40 of x = 400. Hence total number of students = 400.
In an examination 35% of total students failed in Hindi , 45% failed English and 20% in both .Find the percentage of
those who passed in both the subjects .
Answer : Option A
Then A and B be the sets of students who failed in Hindi and English respectively
Then , n(A) = 35, n(B) = 45, n`(A nn B)` = 20
so, n`(A uu B)` = n(A) + n(B) - n`(A nn B)` = (35 + 45 - 20) = 60
`:.` percentage failed in Hindi or English or both = 60%
Hence ,percentage passed = (100 - 60)% = 40%.
Due to a reduction of `6 1/4`% in the price of sugar , a man is able to buy 1 kg more for Rs. 120. Find the original and reduced rate of sugar .
Answer : Option D
Le original rate be Rs. `x` per kg.
Reduced rate = Rs.`[(100 - 25/4) xx 1/100 x] = Rs. (15x)/(16)` per kg.
`:.` ` (120)/((15x)/(16)) - 120/x = 1 hArr 128/x - 120/x = 1 hArr x = 8`
So , original rate = Rs. 8 per kg.
Reduced rate = Rs.`(15/16 xx 8)` per kg. = Rs. 7.50 per kg.
How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to a 10% solution ?
Answer : Option D
Amount of salt in 30 g solution = `(2/100 xx 30)`kg = 0.6 kg
Let `x` kg of pure salt be added .
Then , `(0.6 + x)/(30 + x) = 10/100 hArr 60 + 100x = 300 + 10x hArr 90x = 240 hArr x = 8/3 = 2 2/3`