Q 1-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.A - 350B - 375C - 400D - 450
```Answer - C

Explanation

"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∪  B) =  n(A) + n(B) - n(A∩  B)

=  80% of x + 85% of x -  75% of x

=(80/100x+85/100x−75/100x) =  90/100x

= 9/10x

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."
```
Q 2-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 .A - 40%B - 42%C - 43%D - 44%
```Answer - A

Explanation

"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∩B) =  20

so,  n(A∪B) =  n(A) + n(B) - n(A∩B)

= (35 + 45 - 20) = 60

∴ Percentage failed in Hindi or English  or both = 60%

Hence ,percentage passed = (100 - 60)% =  40%."
```
Q 3-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 .A - 6.50 kgB - 7 kgC - 7.25kgD - 7,50 kg
```Answer - D

Explanation

"Le original rate be Rs. x per kg.

Reduced rate = Rs.[(100−[25/4])×[1/100x]]

=Rs. 15x/16 per kg.

∴120/[15x/16]−120/x=1

⇔128/x−(120/x)= 1

⇔  x=8

So , original rate =  Rs. 8 per kg.

Reduced rate = Rs.(15/16×8)  per kg. = Rs. 7.50 per kg.
"
```
Q 4-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 ?A - 1(2/3)B - 2(1/2)C - 2(1/3)D - 2(2/3)
```Answer - D

Explanation

"Amount of salt in 30 g solution = (2/100×30)  kg =  0.6 kg

Let x kg  of pure salt  be  added .

Then , 0.6+x/30+x= 10/100

⇔  60+100x= 300+10x

⇔ 90x=240

⇔x=8/3=2(2/3)
"
```
Q 5-If A's salary is 20% less than B's salary, by how much percent is B's salary more than A'sA - 25%B - 26%C - 27%D - 28%
```Answer - A

Explanation

"Required percentage = [20/(100−20)×100]%

=   25%.
"
```
Q 6-If A earns 33(1/3)% more than B,how much percent does B earn less then A?A - 24%B - 25%C - 26%D - 27%
```Answer - B

Explanation

"Required percentage = ( {[ 100 / 3] / [100 + [100/3]} x 100 )%
=([100 / 400] x  100) %
=  25%."
```
Q 7-During one year , the population of a town increased by 5% and during the next year , the population decreased by 5% . If the total population is 9975 at the end of the second year, then what was the population size in the beginning of the first year ?A - 10000B - 10010C - 10100D - 11000
```Answer - A

Explanation

"population in the beginning of the first year

=  9975 / [(1+{5/100})(1−{5/100})]

= (9975× [20/21] × [20/19] )

= 10000
"
```
Q 8-In the new budget , the price of kerosene oil rose by 25% . By how much percent must a persn reduce his consumption so that his expenditure on it does not increase ?A - 20%B - 21%C - 22%D - 23%
```Answer - A

Explanation

"Reduction in consumption =  [R/(100+R)×100]%

=  ([25/125]×100)

=  20%.
"
```
Q 9-If the numerator of a fraction be increased by 15% and its denominator be diminished by 8% , the value of the fraction is 15/16. Find the original fraction.A - 2/3B - 3/4C - 3/5D - 2/5
```Answer - B

Explanation

"Let the original fraction be. x/y

Then
115%of x/92%of y= 15/16

⇒  115x/92y= 15/16 ⇒ x/y

= (15/16×92/115)

=  3/4.
"
```
Q 10-When the price of a product was decreased by 10% , the number sld increased by 30% . What was the effect on the total revenue ?A - 17%B - 18%C - 19%D - 20%
```Answer - A

Explanation

"Let the price of the product be Rs. 100
and let original sale be 100 pieces.

Then , total Revenue =  Rs. 100 x 100 = 10000

New revenue =  Rs.  90 x 130 =   Rs.  11700

∴ Increase in revenue =  ([1700/10000]×100))%

=  17%
"
```
Q 11-The salary of a person was reduced by 10% . By what percent should his reduced salary be raised so as to bring it at per with his original salary ?A - 11(1/3)%B - 11(2/5)%C - 11(1/8)%D - 11(1/9)%
```Answer - D

Explanation

"Let the original salary be Rs. 100. New salary = Rs. 90.

Increase on 90 = 10. Increase on 100 = ([10/90]×100)%

=  11(1/9)%.
"
```
Q 12-Paulson spends 75% of his income . His income is increased by 20% and he increased his expenditure by 10%. Find the percentage increase in his savings.A - 45%B - 47%C - 49%D - 50%
```Answer - D

Explanation

"Let original income = Rs. 100 .
Then , expenditure = Rs. 75 and saving = 25

New income = Rs 120,  New expenditure  =  Rs.([110/100]×75)

=  Rs. 165/2

New saving = Rs. (120−[165/2] )

=  Rs. 75/2

Increase in savings = Rs. ([75/2−25]

=  Rs. 25/2

∴    Increase % =  ([25/2×1/25]×100)%

=     50%
"
```
Q 13-Raman's salary was decreased by 50% and subsequently increased by 50% Bow much percent does he lose ?A - 25%B - 26%C - 27%D - 28%
```Answer - A

Explanation

"Let original salary = Rs. 100

New final salary = 150%   of (50%  of 100)
= Rs. ([150/100[×[50/100]×100)

=  Rs.75

∴Decrease  =  25%.
"
```
Q 14-A salesman's commission is 5 on all sales upto Rs. 10,000 and 4% on all sales exceeding this . He remits Rs 31,100 to his parent company after deducting his commission. Find the total sales.A - 31500B - 32000C - 32500D - 32650
```Answer - A

Explanation

"Let his total sales be Rs. x . Now, (total sales ) - (Commission)

= Rs.  31,100

∴   x - [5% of 10000 + 4% of (x- 10000)] =  31100

⇔   x−[(5/100)×10000+4/100(x−10000)]

= 31100

⇔   x−500−(x−10000/25)

= 31100

⇔  x−(x/25)=31200

⇔ 24x/25=31200

⇔ x=(31200×25/24)  =  32500.

∴Total sales = Rs. 32,500.
"
```
Q 15-10% of the inhabitants of a village having died of cholera, a panic set in, during which 25% of the remaining inhabitants left the village . The population is them reduced to 4050. Find the number of original inhabitants.A - 6000B - 6100C - 6200D - 6300
```Answer - A

Explanation

"Let the total number of original  inhabitants be x

Then , (100 - 25)% of (100 - 10)% of x =  4050

⇔([75/100]×[90/100]*x)= 4050

⇔ (27/40)x =4050

⇔  x= (4050×40/27)   =   6000.

∴Number of original inhabitants =  6000.
"
```