Quantitative Ability Test

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Quantitative Ability Test - 15

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Weekly Quiz Competition

Question 1
4 / -1

Let \(A\) and \(B\) are two sets with two elements each, such that all the four elements are distinct.
\(A \times B\) is defined as the set of all ordered pairs \((a, b)\) where, \(a \in A\) and \(b \in B\) that is \(A \times B=\{(a, b) \mid a \in A \text { and } b \in B\}\)
How many subsets are there of the set \(A \times B\) with at least two elements?

A
11

B
12

C
-11

D
10
Question 2
4 / -1

For what value of \(a,\) will the equation \(\left(a^{2}-9\right) x^{2}-\left(a^{2}+2 a-3\right) x+a^{2}+a-6=0\) have more than two solutions?

A
-3

B
2

C
1

D
0

Solution

The given equation will have more than two solutions when it is an identity. \(\Rightarrow\left(a^{2}-9\right)=\left(a^{2}+2 a-3\right)=\left(a^{2}+a-6\right)=0\)
\(\left(a^{2}-9\right)=0 \Rightarrow a=3\) or -3
\(\left(a^{2}+2 a-3\right)=0 \Rightarrow a=-3\) or 1
\(\left(a^{2}+a-6\right)=0 \Rightarrow a=-3\) or 2
since, for \(a=-3\) satisfies all the equations, for \(a=-3,\) the given equation will
have more than two solutions or infinite solutions.
Hence, option A is the correct answer.
Question 3
4 / -1

How many positive integral values of \(x x\) are there for which \(_{1} x^{3}-3 x^{2}-10 x+24<0 ?\)

A
1

B
0

C
2

D
-1

Solution

By hit and trial, we get to know that \(x=2\) is a root of the given equation.
So, the given equation can be factorized into \((x-2)(x-4)(x+3)<0\)
\(x \in(-\infty,-3) U(2,4)\)
So, there is only one positive integral value of \(x\) satisfying the given equation.
Hence, 1 is the correct answer.
Question 4
4 / -1

What is the value of \(x x\) for which
\(\log _{10}\left(2^{x}-x+41\right)=x\left(1-\log _{10} 5\right) ?\)

A
41

B
43

C
44

D
48

Solution

\(\log _{10}\left(2^{x}-x+41\right)=x\left(1-\log _{10} 5\right)\)
\(\Rightarrow>\log _{10}\left(2^{x}-x+41\right)=x\left(\log _{10} 10-\log _{10} 5\right)\)
\(\Rightarrow>\log _{10}\left(2^{x}-x+41\right)=x \log _{10} 2\)
\(=>\log _{10}\left(2^{x}-x+41\right)=\log _{10} 2^{x}\)
\(\Rightarrow>2^{x}-x+41=2^{x}\)
\(=>x=41\)
Hence, 41 is the correct answer.
Question 5
4 / -1

A sequence \(T_{1}, T_{2}, T 3, \ldots\) is defined as such that \(T_{n}=T_{n-1}-T_{n-2}+T_{n-3}\) for all \(n \geq 4\). If \(T_{1}=16, T_{2}=22, T_{3}=29,\) then find out what will be the value of
\(\left.T_{148}+T_{155}+T_{166}\right) ?\)

A
96

B
86

C
74

D
56

Solution

Given that \(T_{n}=T_{n-1}-T_{n-2}+T_{n-3}\)
For \(n=4\)
\(T_{4}=T_{3}-T_{2}+T_{1}\)
\(T_{4}=29-22+16\)
\(T_{1}=23\)
Similarly for \(n=5\)
\(T_{5}=T_{4}-T_{3}+T_{2}\)
\(T_{5}=23-29+22\)
\(T_{5}=16\)
for \(n=6\)
\(T_{6}=T_{5}-T_{4}+T_{3}\)
\(T_{6}=16-23+29\)
\(T_{6}=22\)
We can see that \(T_{1}=T_{5}=16\) and \(T_{2}=T_{6}=22\) and \(s o\) on \(\ldots\)
Hence we can say that after every 4 th term value repeat itself hence there is cyclicity
of
4. Therefore \(T_{k}=T_{k+4}\)
\(\Rightarrow T_{148}+T_{155}+T_{166}\)
\(\Rightarrow T_{4}+T_{3}+T_{2}\)
\(\Rightarrow 23+29+22\)
\(\Rightarrow 74\)
Hence, option \(\mathrm{C}\) is correct answer.
Question 6
4 / -1

A vessel contains milk and water in the ratio 7 : 3. 3 litres of this mixture is taken out and replaced with water. Again, 2 litres of the mixture is taken out and replaced with water. If the vessel contained 10 litres of milk-water solution initially, then what is the final ratio of milk and water in the vessel?

A
79 : 87

B
93 : 17

C
43 : 69

D
49 : 76

Solution

We know that Final quantity = Initial Quantity \(\left(1-\begin{array}{c}\text { quantity removed } \\ \text { total quantity }\end{array}\right)\) In the first case, 3 litres are removed and replaced with water and in the second case
2 litres are removed and replaced with water. Also, we know that the total volume of
the solution is 10 litres.
For milk, \(\Rightarrow\) Final proportion \(=\) Initial proportion \(\left(1-\frac{ \text { quantity removed in the }1 \text { st case })}{\text { total quantity }}\right)^{*}\) \(\left(1-\frac{\text { quantity removed in the } 2 \text { nd case }}{\text { total quantity }}\right)\) \(\Rightarrow\) Final proportion \(=\frac{7}{10} *\left(1-\frac{3}{10}\right) *\left(1-\frac{2}{10}\right)\)
\(\Rightarrow\) Final proportion \(=\frac{7}{10} * \frac{7}{10}+\frac{8}{10}=\frac{49}{125}\)
Therefore, ratio of milk and water \(=49: 76\)
Hence, option \(\mathrm{D}\) is the correct answer.
Question 7
4 / -1

In a triangle ABC, points P, Q and R lie on side AB, BC and CA respectively such that these points divide these sides in the ration of 1:2, 1:4 and 1:6 respectively. Find out the ratio of the area of triangle PQR to the area of triangle ABC?

A
6/15

B
7/15

C
8/15

D
11/15

Solution

Let's assume that length of side \(A B, B C\) and \(C A\) are \(3 x, 5 y\) and \(7 z\) respectively.

Then ratio of area of triangle APR to ABC:
\(\Rightarrow \frac{\Delta A P R}{\Delta A B C}=\frac{A P \times A R \times \sin A}{A B \times A C \times \sin A A}\)
\(\Rightarrow \frac{\Delta A P R}{\Delta A B C}=\frac{\pi}{3 x} \times \frac{6 z}{7 z}\)
\(\Rightarrow \frac{\Delta A P R}{\Delta A B C}=\frac{2}{7}\)
Similarly, in triangle BPQ to BAC \(\Rightarrow \frac{B P Q}{\Delta B A C}=\frac{B P}{B A} \times \frac{B Q}{B C}\)
\(\Rightarrow \frac{\Delta P Q}{\Delta B A C}=\frac{2}{15}\)
in triangle \(\mathrm{CRQ}\) to \(\mathrm{CAB}\)
\(\Rightarrow \frac{\triangle C R Q}{\Delta C A B}=\frac{C R}{C A} \times \frac{C Q}{C B}\)
\(\Rightarrow \frac{\triangle R R Q}{\Delta C A B}=\frac{4}{35}\)
Therefore the ratio of ratio of triangle PQR to triangle \(A B C\)
\(\Rightarrow 1-\left(\frac{\Delta A P R}{\Delta A B C}+\frac{\Delta B P Q}{\Delta B A C}+\frac{\Delta C R Q}{\Delta C A B}\right)\)
\(\Rightarrow 1-\left(\frac{2}{7}+\frac{2}{15}+\frac{4}{35}\right)\)
\(\Rightarrow \frac{7}{15}\)
Hence option B is correct answer.
Question 8
4 / -1

On selling 6 bats for Rs.600, a shopkeeper has a loss equal to the cost price of 2 bats. What should be the marked price of a bat if the shopkeeper has to gain a profit of 10% after giving a discount of 10%?

A
Rs. 165.75

B
Rs. 183.30

C
Rs. 190.25

D
Rs. 187.50

Solution

CP = SP + loss

So, CP of 6 bats = SP of 6 bats + CP of 2 bats

=> CP of 4 bats = SP of 6 bats

SP of 6 bats = Rs.600

=> CP of 4 bats = Rs.600

CP of 1 bat = Rs.150

Required gain = 10%

Required SP = Rs.150 + 10% of Rs.150 = Rs.165

According to the question,

MP - 10% of MP = Rs.165

Or, MP = Rs.1650/9= Rs.183.3

Hence, option B is the correct answer.
Question 9
4 / -1

\(\left(\log _{2} x\right)^{2}+3 \log _{2} x-4=0\)
What is the sum of all the possible values of \(\log _{x} 2 ?\)

A
3/4

B
1/2

C
1/3

D
1/4

Solution

Let \(\log _{2} x\) be \(y\)
Then, the equation reduces to \(y^{2}+3 y-4=0\)
On solving, we get \(y=1\) or \(y=-4\) \(\Rightarrow>\log _{2} x=1\) or -4
\(\log _{x} 2=-\frac{1}{4}\) or 1
\(\mathrm{Sum}=\frac{3}{4}\)
Hence, option \(A\) is the correct answer.
Question 10
4 / -1

Chintu distributed 75% of his birthday cake in the ratio 7 : 8 between his two friends and ate rest of the cake himself. Had he distributed the whole cake in the ratio 4 : 1: 3 among his friends and himself in that order, he would have received 187.5 g of cake extra. What was the total weight of the cake?

A
1.5 kg

B
1 kg

C
1.4 kg

D
1.2 kg

Solution

Let the weight of cake be 100x g

Cake received by Chintu = 25% of 100x g = 25x g

In the second case, the ratio of distribution is 4 : 1: 3

Share of Chintu = 3/8 * 100x g = 37.5x g

According to the question, 37.5x - 25x = 187.5g

=> x = 15

=> weight of cake = 1500g

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Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 15

Result

Quantitative Ability Test - 15

Score

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Rank

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SHARING IS CARING

Question 1

11

12

-11

10

Question 2

-3

2

1

0

Solution

Question 3

1

0

2

-1

Solution

Question 4

41

43

44

48

Solution

Question 5

96

86

74

56

Solution

Question 6

79 : 87

93 : 17

43 : 69

49 : 76

Solution

Question 7

6/15

7/15

8/15

11/15

Solution

Question 8

Rs. 165.75

Rs. 183.30

Rs. 190.25

Rs. 187.50

Solution

Question 9

3/4

1/2

1/3

1/4

Solution

Question 10

1.5 kg

1 kg

1.4 kg

1.2 kg

Solution

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