“Must do percentage Questions Based on SSC Pre 2016 pattern “

Q1. Rs. 395 are divided among A, B and C in such a manner that B gets 25% more than A and 20% more than C. The share of A is:  
(a) Rs. 198
(b) Rs. 120
(c) Rs. 180
(d) Rs. 195
S1. Ans.(b)
Sol. Suppose A get Rs. x
Then B gets 125% of
x=125/100×x=Rs.5x/4
If B gets Rs. 120
Then C gets Rs. 100
If B gets Rs. 5x/4
Then C gets
= Rs. (100/120×5x/4) = Rs. 25x/24
∴ x+5x/4+25x/24=395
⇒  (24x + 30x + 25x) = (395 × 24)
⇒ 79x = (395 × 24)
⇒ x=((395 × 24)/79)=120
Hence, A gets Rs. 120

Q2. In an examination, 65% of the total examinees passed. If the number of failures is 420, the total number of examinees is: 
(a) 500
(b) 1200
(c) 1000
(d) 1625
S2. Ans.(b)
Sol. Percentage of failures = (100 – 65)% = 35%
Let the total number of examinees be x
Then, 35% of x = 420
⇒ 35/100×x=420
⇒ x=(420×100/35)=1200

Q3. The ratio of the number of boys and girls is 3 : 2. If 20% of the boys and 30% of the girls are scholarship holders, then the percentage of students who do not get scholarship is: 
(a) 50
(b) 72
(c) 75
(d) 76
S3. Ans.(d)
Sol. Let the number of boys 3x and number of girls be 2x
Number of scholarship holders
=(20/100×3x+30/100×2x)=6x/5
Number of those who do not hold scholarship
=(5x-6x/5)=19x/5
∴ Required percentage =(19x/5×1/5x×100)%=76%

Q4. 5% of income of A is equal to 15% of income of B and 10% of income of B is equal to 20% of income of C. If the income of C is Rs. 2000, what is the total income of A, B and C? 
(a) Rs. 14000
(b) Rs. 16000
(c) Rs. 18000
(d) Rs. 12400
S4. Ans.(c)
Sol. 10/100×B=20/100×C=B=2C
⇒ B = (2 × 2000) = 4000
5/100×A=15/100×B
⇒ A = 3B = (3 × 4000) = 12000
∴ (A + B + C) = Rs. (12000 + 4000 + 2000)
= Rs. 18000

Q5. A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top of as well as at the bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing?
(a) 65%
(b) 70%
(c) 80%
(d) 60%
S5. Ans.(b)
Sol. Total area of the paper = (30×15)cm^2
=450 cm^2
Area used =[(30-5)×(15-2.5)]  cm^2
=(25×12.5)  cm^2
=(25×25/2)  cm^2=625/2  cm^2
Percentage of area used
=(625/2×1/450×100)%=69.5%
= 70% nearly

Q6. When a number is first increased by 10% and then reduced by 10% the number: 
(a) Does not change
(b) Decrease by 1%
(c) Increase by 1%
(d) None of these
 S6. Ans.(b)
Sol. Let the given number be x
Increased number = (110% of x)
=(110/100×x)=11x/10
Finally 11x/10, reduced number
=(90% of 11x/10)=(90/100×11x/10)=99x/100
Decrease =(x-99x/100)=x/100
Decrease % =(x/100×1/x×100 )%=1%

Q7. Out of 100 students, 50 failed in English and 30 in Mathematics. If 12 students fail in both English and mathematics, then the number of students who passed in both the subjects is 
(a) 26
(b) 28
(c) 30
(d) 32
S7. Ans.(d) 
Sol. Let A = set of students who fail in English and B = set of students who fail in mathematics
Then n(A) = 50, n(B) = 30 and n(A ∩B)=12
n(A∪B)=n(A)+n(B)-n(A∩B)=(50+30-12)=68
Number of students who fail in one or both the subjects = 68
Number of those who pass in both = (100 – 68) = 32

Q8. 8% of the voters in an election did not cast their votes. In the election, there were only two candidates. The winner by obtaining 48% of the total votes defeated his contestant by 1100 votes. The total number of voters in the election were 
(a) 21000
(b) 22000
(c) 23500
(d) 27500
S8. Ans.(d)
Sol. Let the total number of voters be x
Votes cast = 92% of x =(92/100×x)=23x/25
Votes in favour of winning candidate =48/100×x=12x/25
Votes polled by defeated candidate
=(23x/25-12x/25)=11x/25
12x/25-11x/25=1100
⇒ 12x – 11x = 27500
⇒ x = 27500

Q9. Because of scarcity of rainfall, the price of a land decreases by 12% and its production also decreases by 4%. What is the total effect on revenue? 
(a) Loss of 16%
(b) Gain of 15%
(c) Loss of 15.52%
(d) Gain of 15.48%
S9. Ans.(c) 

Sol. Net effect =[-12-4+(-12)(-4)/100]%
= (–16 + 0.48)% = –15.25%   
Q10. From 1980-1990, the population of a country was increased by 20%. 
From 1990 – 2000, the population of the country was increased by 20%
From 2000-2010, the population of the country was increased by 20% 
Then the overall increased population (in percentage) of the country from 1980-2010 was 
(a) 60%
(b) 62.8%
(c) 72.2%
(d) 72.8%
S10. Ans.(d)
Sol. Let the population in the year 1980 = x
Then, the population in the year 1990 = 120% of x
Then, the population in the year 2000
= 120% of 120% of x
And the population in year 2010
= 120% of 120% of 120% of x
=6/5×6/5×6/5×x=216/125 x
Hence, required percentage
=(216/125  – x)/x×100%
=91/125×100%=72.8%

Q11. A dishonest shopkeeper pretends to sell his goods at cost price but uses false weights and gains 11 1/9%. For a weight of 1kg he uses:  
(a) a weight of 875 gm
(b) a weight  of 800 gm
(c) a weight of 950 gm
(d) None of these
S11. Ans.(d)
Sol. Let the error be x gms.
Then, x/((100 – x) )×100=100/9
⇒ x/(1000 – x)=1/9
⇒ 9x = 1000 – x
⇒ 10x = 1000
⇒ x = 100
Weight used = (1000 – 100) gm = 900 gm.

Q12. An article passing through two hands is sold at a profit of 38% at the original cost price. If the first dealer makes a profit of 20%, then the profit percent made by the second is: 
(a) 5%
(b) 10%
(c) 12%
(d) 15%
S12. Ans.(d)
Sol. Let the C.P. be Rs. 100
Then, S.P = Rs. 138
Let the profit made by 2nd dealer be x%
Then, (100 + x)% of 120% of Rs. 100 = Rs. 138
⇒ ((100 + x))/100×120/100×100=138
⇒ 6(100 + x) = 690
⇒ 6x = 90 ⇒ x = 15
∴ Required% = 15%

Q13. The marked price of a shirt and trousers are in the ratio 1 : 2. The shopkeeper gives 40% discount on the shirt. If the total discount on both is 30%, the discount offered on the trousers is:  
(a) 15%
(b) 20%
(c) 25%
(d) 30%
S13. Ans.(c)
Sol. Let the M.P; of shirt be Rs. x and that of trousers be Rs. 2x
Let y% be the discount on trousers.
Then, 60/100×x+(100 – y))/50×2x=70/100×(x+2x)
⇒ 3/5+((100 – y))/50=21/10
⇒ ((100 – y))/50=(21/10-3/5)=15/10=3/2
⇒ (100-y)=(3/2×50)=75
⇒ y = 25

Q14. A shopkeeper sold an air-conditioner for Rs. 25935 with a discount of 9% and earned a profit of 3.74%. What should have been the percentage of profit earned if no discount were offered?  
(a) 12.3%
(b) 15.6%
(c) 16%
(d) None of these
S14. Ans.(d)
Sol.
 91/100×x=25935
⇒ x=(25935×100/91)=28500
⇒ M.P. = Rs. 28500
S.P. = Rs. 25935, Gain = 3.74%
∴ C.P. = Rs. (100/(103.74)×25935) = Rs. 25000
New S.P. = Rs. 28500, C.P. = Rs. 25000

∴ New Gain% =(3500/25000×100)%=14%

Q15. A fan is listed at Rs. 1500 and a discount of 20% is offered on the list price. What additional discount must be offered to the customer to bring the net price to Rs. 1104?  
(a) 8%
(b) 10%
(c) 12%
(d) 15%
S15. Ans.(a)
Sol. Let additional discount be x%
Then, 80% of (100 – x)% of Rs. 1500 = Rs. 1104
⇒ 80/100×((100-x))/100×1500
⇒ 12x = (1200 – 1104) = 96
⇒ x = 8
∴ Additional discount = 8%

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