Quantitative Ability Test

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Quantitative Ability Test - 13

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Weekly Quiz Competition

Question 1
4 / -1

Two trains A and B, having speeds in the ratio 3 : 4, start from two different cities 500 km apart. At 9 PM, train A interchanged its speed with train B such that they both reached their destination at the same time. How much distance did train A travel from the beginning of the journey till 9 PM?

A
1500/7 km

B
1500 / 11 km

C
1000 / 7 km

D
250 km

Solution

Let the speeds of train \(A\) and train \(B\) be \(3 x\) and \(4 x\) respectively.
Let after time 't', they exchange their speeds.
It is given that after interchanging their speeds, they took equal time to reach their destination.
So, after interchange of speeds,
Distance left to be travelled by \(A=(500-3 x t) \mathrm{km}\) And the speed of \(A=4 x\)
Distance left to be travelled by \(B=(500-4 x t)\) And the speed of \(B=3 x\)
According to the question, \(\frac{500-3 x t}{4 x}=\frac{500-4 x t}{3 x}\)
On solving we get \(x t=\frac{500}{7}\) Distance travelled by \(A\) till \(9 P M=3 x t=\frac{1500}{7}\) Hence, option \(A\) is the correct answer.
Question 2
4 / -1

A worker is hired for 7 days on the condition that he has to come to work for at least 5 days (it doesn’t matter whether he works for full-day or half-day). He will be given one gold coin for full-day work, one silver coin for half-day work and no coin for being absent. The coins given on different days will have the dates imprinted on them. How many different combinations of gold and silver coins can he accumulate?

A
1248

B
1228

C
1222

D
1828

Solution

There can be three cases:
Case I : When he works for 7 days
Each day, he can have either a gold coin or a silver coin
Total number of combinations = 2⁷ = 128
Case II : When he works for 6 days
He can be absent on any one of the seven days
Also, on each of the six working days, he can have either a gold coin or a silver coin
Total number of combinations = 7 * 2⁶= 448
Case III: When he works for 5 days
He can be absent for any two days in 7C₂= 21 ways
Also, on each of the five working days, he can have either a gold coin or a silver coin
Total number of combinations = 21 * 2⁵ = 672
Total number of combinations = 128 + 448 + 672 = 1248
Hence, 1248 is the correct answer.
Question 3
4 / -1

In a class, the number of girls was 25% more than the number of boys. Some students joined the class this year in which the ratio of boys and girls was 2 : 3. After the new additions, the total number of girls is 27.27% more than the total number of boys. What is the percentage increase in the number of students after the new additions?

A
9.09%

B
11.11%

C
16.66%

D
12.25%

Solution

Let the number of boys be \(100 x\)
Then, the number of girls \(=125 x\)
Total number of students \(=225 x\)
Let the number of new boys be 2 y and the number of new girls be 3 y.
At present,
Total number of boys \(=100 x+2 y\)
Total number of girls \(=125 x+3 y\)
According to the question, \(\frac{125 x+3 y-100 x-2 y}{100 x+2 y}=\frac{27.27}{100}\)
\(=>\frac{25 x+y}{100 x+2 y}=\frac{3}{11}\)
On solving we get \(y=5 x\)
Present total number of students \(=100 x+125 x+5 y=225 x+25 x=250 x\) Required percentage \(=\frac{250 \mathrm{x}-225 \mathrm{x}}{225 x} * 100 \%=11.11 \%\)
Hence, option \(B\) is the correct answer.
Question 4
4 / -1

If ‘X’ is a prime number such that 15! + 1 < X < 15! + 15. Then find out the sum of all values that ‘X’ can assume?

A
0

B
1

C
-1

D
2

Solution

We know that number 15! is divisible by every number from 1 to 15.
A= 15! + P (Where P is an integer from 2 to 14)
Therefore A will always be divisible by P in each case.
Therefore there will not be any prime number in between 15!+1 and 15!+15.
Hence 0 is the correct answer
Question 5
4 / -1

A three-digit even number is chosen from a group of all three-digit even numbers. If the probability that the chosen number has all its digits distinct can be written in lowest form as x/y , what is the value of (x + y)(x + y)?

A
398

B
371

C
389

D
351

Solution

We have to find all even three digit numbers which have distinct digits.

If the unit digit is 0, other two places can be filled in 9 * 8 ways

So, total no. of even three digit numbers with unit digit 0= 72

If the unit digit is one of 2,4 6 or 8, total number of numbers = 8 * 8 * 4 = 256

So, total number of three digit even numbers with distinct digits = 72+ 256 = 328

Total number of three digits even numbers = 450

Required probability = 328/450 = 164/225

x+y = 164 + 225 = 389

Hence, option C is the correct answer.
Question 6
4 / -1

In a rectangle ABCD where AB = 24 cm and BC = 9 cm. Points E and F are midpoints of side AB and CD. A circle is drawn in such a way that its centre coincides with rectangle’s centre. Find out the length of the diameter of the circle if both the lines DE and BF are tangents to the circle?

A
7.2 cm

B
8.4 cm

C
9.2 cm

D
6.4 cm

Solution

We can see from the symmetry that DE is parallel to line FB. The diameter of circle is essentially the

perpendicular distance between line DE and FB

Equation of line DE

\(\Rightarrow y-0=\frac{9-0}{12-0}(x-0)\)

\(\Rightarrow 3 x-4 y=0\)

Similarly the equation of line FB

\(\Rightarrow y-0=\frac{9-0}{24-12}(x-12)\)

\(\Rightarrow 3 x-4 y=36 \quad \ldots(2)\)

Distance between parallel lines DE and FB

\(d=\left|\frac{36-0}{\sqrt{3^{2}+4^{2}}}\right|\)

\(d=7.2 \mathrm{cm}\)

Therefore we can say that diameter of the circle \(=7.2 \mathrm{cm}\)let us draw the diagram

We know that \(\mathrm{E}\) and \(\mathrm{F}\) are the midpoints of \(\mathrm{AB}\) and \(\mathrm{CD}\) hence \(\mathrm{AE}=\mathrm{EB}=\mathrm{DF}=\mathrm{FC}=12 \mathrm{cm}\)
Changing the given lengths in coordinate system by assuming \(\mathrm{D}\) at the origin and \(\mathrm{B}\) at \((24,9) . \mathrm{DC}\) will become \(\mathrm{x}\)
axis and DA will become y axis. We can find out the coordinate of each point.

We can see from the symmetry that DE is parallel to line FB. The diameter of circle is essentially the
perpendicular distance between line DE and FB
Equation of line DE
\(\Rightarrow y-0=\frac{9-0}{12-0}(x-0)\)
\(\Rightarrow 3 x-4 y=0\)
Similarly the equation of line FB
\(\Rightarrow y-0=\frac{9-0}{24-12}(x-12)\)
\(\Rightarrow 3 x-4 y=36 \quad \ldots(2)\)
Distance between parallel lines DE and FB
\(d=\left|\frac{36-0}{\sqrt{3^{2}+4^{2}}}\right|\)
\(d=7.2 \mathrm{cm}\)
Therefore we can say that diameter of the circle \(=7.2 \mathrm{cm}\)
Question 7
4 / -1

A square PQRS has side of length 5 cm and points T and U lie outside the square such that TQ = US = 3cm and PT = UR = 4cm. Find the value of UT ?

A
9√2 cm

B
8√2 cm

C
7√2 cm

D
None of the above

Solution

Drawing the diagram on the basis of given information.

In triangle PTQ we can see that \(P Q^{2}=P T^{2}+T Q^{2}\)
Hence we can say that \(\triangle P T Q\) is a right-angled triangle. Same is true for \(\triangle\) SUR
Drawing 2 more similar triangles to PTQ, \(\triangle\) OWR and \(\triangle\) PVS

Here we can clearly see that TWUV is a square of side length \(=7 \mathrm{cm}\) and UT is a
diagonal of the square TWUV.
\(\Rightarrow U T=7 \sqrt{2} \mathrm{cm}\)
Hence option C is the correct answer.
Question 8
4 / -1

Find the number of integer values of \(x\) for which \(\frac{x^{2}+3 x-28}{x^{2}+6 x+10}<0 ?\)

A
10

B
-9

C
9

D
-10

Solution

We have to find the range of \(x\)
\(\Rightarrow \frac{x^{2}+3 x-28}{x^{2}+6 x+10}<0\)
\(\Rightarrow \frac{x^{2}+7 x-4 x-28}{x^{2}+6 x+10}<0\)
\(\Rightarrow \frac{(x+7)(x-4)}{x^{2}+6 x+10}<0\)
We can see that discriminant of the quadratic equation present in denominator
\(\Rightarrow D=b^{2}-4 a c=6^{2}-4 \times 1 \times 10=-4\)
We can see that \(D \leq 0\) and \(a>0\) hence the value of this quadratic equation will be positive.
Hence the inequality will hold true when the numerator is negative. Therefore, \(x\) can take from any value from
(-7,4)
Therefore possible integer values of \(x=\{-6,-5,-4,-3,-2,-1,0,1,2,3\}\)
We can say that \(x\) can take any of these 10 integer values.
Question 9
4 / -1

A trader mixes two varieties of rice-one at Rs.30 per kg and another at Rs.40 per kg, in the ratio 2 : 3. He uses a false weight which displays 1 Kg while weighing 800 g. What price should he quote to the customers while selling the mixture to have an overall gain of 30%?

A
Rs. 37.44

B
Rs. 39.41

C
Rs. 33.54

D
Rs. 41.70

Solution

Let us assume that he mixes 2 kg of rice at Rs.30per kg and 3 kg of rice at Rs.40 per kg.
Total CP of 5 kg rice = Rs.(2 * 30 + 3 * 40) = Rs.180
CP per kg = Rs.180/5 = Rs.36
CP of 800 g of mixture = Rs.28.8
As he uses a false weight, he will sell only 800 g saying that it is one kg
Required SP = Rs.28.8 + 30% of Rs.28.8 = Rs.37.44
Hence, option A is the correct answer.
Question 10
4 / -1

Bronze is an alloy which contains copper, zinc and tin in the ratio 20 : 1 : 4. Brass is an alloy of copper and zinc. A mixture of brass and bronze in a certain ratio contains 74% of copper, 16% of zinc and 10% of tin. What is the ratio of copper and zinc in brass?

A
5 : 3

B
13 : 11

C
16 : 9

D
cannot be determined

Solution

Let the total volume of the mixture of brass and bronze be 100 units.
Volume of copper = 74 units, the volume of zinc = 16 units and volume of tin = 10 units
The ratio of tin in bronze = 4 / 25
Let x units of bronze was mixed.
Then, 4x / 25 = 10
On solving, we get x = 62.5 units.
Thus, 62.5 units of bronze was mixed.
Volume of brass = (100 - 62.5) units = 37.5 units
The ratio of copper, zinc and tin in bronze is 20 : 1: 4
Volume of copper from bronze in the mixture = 20 / 25*62.5 = 50 units
Volume of zinc from bronze in the mixture =1 / 25*62.5 = 2.5 units
The remaining volume of copper and zinc in the mixture must be from brass
Remaining volume of copper = 74 - 50 = 24 units
Remaining volume of zinc = 16 - 2.5 = 13.5
Required ratio = 24 : 13.5 = 16 : 9
Hence, option C is the correct answer.

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Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 13

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Quantitative Ability Test - 13

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SHARING IS CARING

Question 1

1500/7 km

1500 / 11 km

1000 / 7 km

250 km

Solution

Question 2

1248

1228

1222

1828

Solution

Question 3

9.09%

11.11%

16.66%

12.25%

Solution

Question 4

0

1

-1

2

Solution

Question 5

398

371

389

351

Solution

Question 6

7.2 cm

8.4 cm

9.2 cm

6.4 cm

Solution

Question 7

9√2 cm

8√2 cm

7√2 cm

None of the above

Solution

Question 8

10

-9

9

-10

Solution

Question 9

Rs. 37.44

Rs. 39.41

Rs. 33.54

Rs. 41.70

Solution

Question 10

5 : 3

13 : 11

16 : 9

cannot be determined

Solution

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