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. "