1. In an election only two candidates A and B contested 30% of the voters did not vote and 1600 votes were declared as invalid. The winner, A got 4800 votes more than his opponent thus he secured 51% votes of the total voters on the voter list. Percentage votes of the loser candidate, B out of the total voters on the voter list is:
A. 5.6%
B. 3%
C. 3%
D. 5%
Answer: Option B
Explanation:
Total voters on the voter list = x
0.51x + 0.51x – 4800 = 0.70x – 1600
1.02x – 4800 = 0.70x – 1600
x = 10000
Votes of the loser candidate = 5100 – 4800 = 300
Percentage votes of the loser candidate = [latex]\frac{300}{10000}[/latex] * 100 = 3%
2. In a school there are 2000 students. On January 2nd, all the students were present in the school except 4% of the boys and on January 3rd, all the students are present in the school except 28/3% of the girls, but in both the days number of students present in the school, were same. The number of girls in the school is?
A. 400
B. 1200
C. 800
D. 600
Answer: Option D
Explanation:
From Options;
let Number of girls = 600
Number of boys = 1400
96% of 1400 + 600 = [600 – 28/3 % of 600] + 1400 = 1944[satisfies the condition; Check the condition with other options also]
3. A school has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 1200 from the parents of the students. The people already solicited represents the parents of 60% of the students. If the School is to raise exactly the amount needed for the new building, what should be the average donation from the remaining students to be solicited?
A. Rs.800
B. Rs.900
C. Rs.850
D. Rs.600
Answer: Option D
Explanation:
People already solicited = 60% of x = 0.6x
Remaining people = 40% of x = 0.4x
Amount collected from the parents solicited= 1200 *0.6x = 720x
720x = 75%; Remaining amount = 25% = 240x
Thus, Average donations from remaining parents = [latex]\frac{240x}{0.4x}[/latex] = 600
4. The monthly income of Shyama and Kamal together is Rs.28000. The income of Shyama and Kamal is increased by 25% and 12.5% respectively. The new income of Kamal becomes 120% of the new salary of Shyama. What is the new income of Shyama?
A. Rs.12000
B. Rs.18000
C. Rs.14000
D. Rs.16000
Answer: Option D
Explanation:
The monthly income of Shyama and Kamal => S + K = 28000 —(1)
Shyama’s income = x; Kamal’s income = 28000 – x.
K = [latex]\frac{120}{100}[/latex] * S —(2)
S’s new income = (28000 – x)*[latex]\frac{112.5}{100}[/latex]
K’s new income = x * [latex]\frac{125}{100}[/latex]
(28000 – x)*[latex]\frac{112.5}{100}[/latex] = x * [latex]\frac{125}{100}[/latex]
x = 12000
New Income of Shyama = 125% of 12000 = 15000
now percent of water = ([latex]\frac{(80*4 + 60*6)}{1000}[/latex])100 = 68%
5. 500 kg of ore contained a certain amount of iron. After the first blast furnace process, 200 kg of slag containing 12.5% of iron was removed. The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore. How many kg of iron were there in the original 500 kg ore?
A. 54.2
B. 58.5
C. 46.3
D. 89.2
Answer: Option D
Explanation:
Initially ‘x’ kg of iron in 500 kg ore.
Iron in the 200 kg of removed = 200*[latex]\frac{12.5}{100}[/latex]= 25 kg.
The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore
So [latex]\frac{(x-25)}{300}[/latex] = ([latex]\frac{1205}{100}[/latex])*[latex]\frac{x}{100}[/latex]
=> x – 25 = [latex]\frac{18x}{25}[/latex]
=> 7x = 625
=> x = 89.2