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SSC CGL Tier1 Quantitative Aptitude Percentage Test 5

2760.Arun got 30% of the maximum marks in an examination and failed by 10 marks. However, Sujith who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
90
250
75
85
Explanation:

Let maximum marks of the examination $=x$

Marks that Arun got $=30%$ of $x$ $=\dfrac{30x}{100}$

Given that Arun failed by 10 marks.

=> pass mark = $\dfrac{30x}{100}+10~~\cdots(1)$

Marks that Sujith got $=40%$ of $x$ $=\dfrac{40x}{100}$

Given that Sujith got 15 marks more than the passing marks.

=> pass mark $=\dfrac{40x}{100}-15~~\cdots(2)$

From $(1)$ and $(2)$,

$\dfrac{30x}{100}+10=\dfrac{40x}{100}-15\dfrac{10x}{100}=25\dfrac{x}{10}=25x=10×25=250$

pass mark

$=\dfrac{30x}{100}+10$

=

$\dfrac{30×250}{100}+10$

=75+10=85

2763.In a competitive examination in state A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each state?
8200
7500
7000
8000
Explanation:

Equal number of candidates appeared in each state.

In state A, 6% candidates got selected.

In state B, 7% candidates got selected.

Given that 80 more candidates got selected in state B than A.

Therefore, 1% of candidates appeared in each state =80

=> 100% of candidates appeared in each state =80×100=8000

i.e., number of candidates appeared from each state = 8000

2764.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
60%
70%
80%
90%
Explanation:

Let total number of men = 100

Then,

20 men play football.

80 men are less than or equal to 50 years old.

Remaining 20 men are above 50 years old.

Number of football players above 50 years old

=20×$\dfrac{20}{100}$=4

Number of football players less than or equal to 50 years old

$=20-4=16$

Required percentage

$=\dfrac{16}{20}×100=80%$

2766.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received?
Rs. 76,375
Rs. 34,000
Rs. 82,150
Rs. 70,000
Explanation:

Price of the car = Rs.3,25,000

Car was insured to 85% of its price

Insured price $=325000×\dfrac{85}{100}$

Insurance company paid 90% of the insurance.

Amount paid by insurance company

=325000×$\dfrac{85}{100}$×$\dfrac{90}{100}$=325×85×9=248625

Difference between the price of the car and the amount received

=$325000-248625$

=$ Rs.76375$

2770.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
2 : 3
1 : 1
3 : 4
4 : 3
Explanation:
p>5% of A + 4% of B =$ \dfrac{2}{3} $ (6% of A + 8% of B)

$\Rightarrow \dfrac{5}{100} $ A +$ \dfrac{4}{100} $ B=$ \dfrac{2}{3} \left(\dfrac{6}{100}  A +\dfrac{8}{100}  B\right) $

$\Rightarrow \dfrac{1}{20} $ A +$ \dfrac{1}{25} $ B=$ \dfrac{1}{25} $ A +$ \dfrac{4}{75} $ B

$\Rightarrow \left(\dfrac{1}{20} -\dfrac{1}{25} \right) $ A = $ \left(\dfrac{4}{75} -\dfrac{1}{25} \right) $ B

$\Rightarrow \dfrac{1}{100} $ A =$ \dfrac{1}{75} $ B

$ \dfrac{A}{B} $=$ \dfrac{100}{75} $=$ \dfrac{4}{3} $.

$\therefore$ Required ratio = 4 : 3

2771.Q as a percentage of P is equal to P as a percentage of (P + Q). Find Q as a percentage of P.
62%
50%
75%
66%
Explanation:

Given that $\dfrac{Q}{P}=\dfrac{P}{P+Q}\cdots(1)$

Since Q can be written as a certain percentage of P, we can assume that $Q=kP$

Hence $(1)$ becomes

$\dfrac{kP}{P}=\dfrac{P}{P + kP}$

$\Rightarrow k=\dfrac{1}{1+k}\\\Rightarrow k(k+1)=1~~\cdots(2)$

Q as a percentage of P

$=\dfrac{Q}{P}×100\\=\dfrac{kP}{P}×100=100k\%~~\cdots(3)$

From here we have two approaches.


Approach 1 - trial and error method

Here we use the values given in the choices to find out the answer.

Take 50% from the given choice.

If 50% is the answer,

$100k=50 \Rightarrow k=\dfrac{50}{100}=\dfrac{1}{2}$

But if we substitute $k=\dfrac{1}{2}$ in $(2),$

$k(k+1)=\dfrac{1}{2}\left(\dfrac{1}{2}+ 1\right)$ $=\dfrac{1}{2}×\dfrac{3}{2}=\dfrac{3}{4} \neq 1$

Now Take another choice, say 62%

If 62% is the answer,

$100k=62 \Rightarrow k=\dfrac{62}{100}$

If we substitute $k=\dfrac{62}{100}$ in $(2),$

$k(k+1)=\dfrac{62}{100}\left(\dfrac{62}{100}+ 1\right)$ $= \dfrac{62}{100}×\dfrac{162}{100}=\dfrac{10044}{10000}\approx 1$

Hence 62% is the answer.


Approach 2

lets solve the quadratic equation $(2)$

$k(k+1)=1\\k^2+k-1=0\\k=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\dfrac{-1 \pm \sqrt{1^2-\left[4×1×(-1)\right]}}{2×1}=\dfrac{-1 \pm \sqrt{5}}{2}\\= \dfrac{-1 \pm 2.24}{2}\\ = \dfrac{1.24}{2}\text{ or }\dfrac{-3.24}{2}\\= 0.62$

avoiding -ve value

From $(3)$, Q as a percentage of P

$=100k\%=(100×.62)\%=62\%$

2772.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much percent will he reduce the quantity of petrol purchased?
8%
7%
10%
6%
Explanation:

Assume that initial price of 1 litre petrol = Rs.100,

Benson spends Rs.100 for petrol such that he buys 1 litre of petrol.

After the increase by 25%, price of 1 litre petrol

$=100+25=125$

Since Benson spends additional 15% on petrol, amount spent by him

$=100+15=115$

Hence quantity of petrol that he can purchase

$=\dfrac{115}{125} litre$

Quantity of petrol reduced

$=\left(1-\dfrac{115}{125}\right)$=$\dfrac{10}{125}$ litre

Required reduction percent

=$\dfrac{\left(\dfrac{10}{125}\right)}{1}$×100=$\dfrac{10}{125}$×100=$\dfrac{10}{5}$×4=2×4=8%

2773.Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
39, 30
41, 32
42, 33
43, 34
Explanation:

Let their marks be $\left( x + 9\right)$ and $ x $.

Then, $ x $ + 9 =$ \dfrac{56}{100}$ $\left( x + 9 + x \right)$

$\Rightarrow$ 25$\left( x + 9\right)$ = 14$\left(2 x + 9\right)$

$\Rightarrow$ 3$ x $ = 99

$\Rightarrow x $ = 33

So, their marks are 42 and 33.

44352.Weight of A and B are in the ratio of 3:5. If the weight of A is increased by 20 percent and then the total weight becomes 132 kg
with an increase of 10 percent. B weight is increased by what percent.
2%
3%
4%
5%
E) None of these
Explanation:

Weight of A and B are 3x and 5x.

Initial weight before increase = $\dfrac{132 \times 100)}{110}$ = 120

8x = 120. X = 15

Initial weight of A and B are 45 and 75 kg respectively.

New weight of A = 54 so weight of B = 132 – 54 = 78.

So % increase = $[\dfrac{78-75}{75}] \times 100$ = 4 %
44353.40% of the students like Mathematics, 50% like English and 10% like both Mathematics and English. What % of the students like
neither English nor Mathematics?
25%
10%
20%
60%
E) 80%
Explanation:

n(M or E) = n(M) + n(E) – n(M and E)

n(M or E) = 40+50-10 = 80

so % of the students who like neither English

nor Mathematics = 100 – 80 = 20%
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