Quantitative Ability Test

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Quantitative Ability Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

If the sum of n terms of an arithmetic pn+qn²progression in where p and q are constants then the common difference is :

A
3q

B
2q

C
q

D
4q

Solution

For the first term or the sum of one number put \(n=1\) \(S_{1}=T_{1}=p+q\)
for the sum of two numbers put \(n=2\) \(S_{2}=2 p+4 q\)
For 2 nd term \(=S_{2}-S_{1}\) \(T_{2}=2 p+4 q-p-q \Rightarrow p+3 q\)
Common difference(d) \(=T_{2}-T_{1}\) d \(=p+3 q-p-q \Rightarrow 2 q\)
Question 2
4 / -1

If \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) then \(\frac{x y}{y^{\prime} z^{\frac{y}{x}} z}=?\)

A
1 : 2 : 3

B
a²c : ab² : bc²

C
ab : bc : ca

D
a : b : c

Solution

\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(x=a k, y=b k, z=c k\)
\(\frac{x}{y}=\frac{a}{b}, \frac{y}{z}=\frac{b}{c}, \frac{z}{x}=\frac{c}{a}\)
\(\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}=\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \quad\) by multiply abc
\(\Rightarrow a^{2} c: a b^{2}: b c^{2}\)
Question 3
4 / -1

The ratio between the speed of a truck, car and train is 3 : 8 : 9. The car moved uniformly and covered a distance of 1040 km in 13 hours. What is the average speed of the truck and the train together?

A
75 km/hr

B
60 km/hr

C
48 km/hr

D
Cannot be determined

Solution

Speed of car =1040/13 =80kmph
Ratio of speed of truck, car and train =3 : 8 :9,
Now 8x =80
=> x =10
Hence, speed of truck =30 kmph.
speed of train =90 kmph.
Average speed of truck and train together
=30 +90/2 =120/2 =60 kmph.
Question 4
4 / -1

In , ΔABC ,G is centroid. AB = 15 cm, BC = 18 cm and AC = 25 cm, then find GD, while D is the mid-point of BC .

A
1/3√86

B
2/3√86

C
8/3√15

D
None

Solution

\(A B^{2}+A C^{2}=2\left(A D^{2}+B D^{2}\right)\)
\(225+625=2\left(A D^{2}+81\right)\)
\(A D=2 \sqrt{86}\)
we know that Centroid divides the median in 21 ratio from vertex \(\mathrm{AG} \cdot \mathrm{GD}=2: 1\)
\(G D=\frac{1}{3} A D\)
\(G D=\frac{2}{3} \sqrt{86}\)
Question 5
4 / -1

If \(\frac{x}{2 x+y+z}=\frac{y}{x+2 y+z}=\frac{z}{x+y+2 z}=a\) and \(x+y+z \neq 0\)
then a \(-?\)

A
1/3

B
1/4

C
1/2

D
1/8

Solution

\(x=a(2 x+y+z)\)
\(y=a(x+2 y+z)\)
\(z=a(x+y+2 z)\)
\((x+y+z)=a(4 x+4 y+4 z)\)
\((x+y+z)=4 a(x+y+z)\)
\(a=\frac{1}{4}\)
Question 6
4 / -1

Which one from the following is in ascending order.

A

\(\sqrt[4]{10}, \sqrt[3]{6}, \sqrt{3}\)

B
\(\sqrt{3}, \sqrt[4]{10}, \sqrt[3]{6}\)

C
\(\sqrt{3}, 6, \sqrt[1]{10}\)

D
\(\sqrt[4]{10}, \sqrt{3}, \sqrt[3]{6}\)

Solution

\(10^{\frac{1}{4}}, 6^{\frac{1}{3}}, 3^{\frac{1}{2}}\)
Surds are multiplied by \(\mathrm{LCM} 12\) \(10^{3}, 6^{4}, 3^{6}\)
1000,1296,729
So ascending order \(\sqrt{3}, \sqrt[4]{10}, \sqrt[3]{6}\)
Question 7
4 / -1

There are two containers of equal capacity. The ratio of milk to water in the first container is 3 :1, in the second container 5 : 2. If they are mixed up, the ratio of milk to water in the mixture will be

A
28 : 41

B
41 : 28

C
15 : 41

D
41 : 15

Solution

Let capacity of each container =x litre
In first container,
Milk =3x/4 ltres,
Water =x/4 litres
In second container,
Milk =5x/7 litres,
Water =2x/7 litres
On mixing both
Quantity of milk =3x/4 +5x/7 =41x/28
Quantity of water =7x+8x/28 litres =15x/28 litres
Required ratio =41x/28 :15x/28 =41:15.
Question 8
4 / -1

In a class there are 60 students, out of whom 15 percent are girls. Each girl's monthly fee is Rs. 250 and each boy's monthly fee is 34 percent more than a girl. What is the total monthly fees of girls and boys together?

A
Rs. 19,335

B
Rs. 18,435

C
Rs. 19,345

D
Rs. 19,435

Solution

Number of girls =60 x 15/100 =9
Total monthly fee of girl =250 x 9 =Rs 2,350
Number of boys =60 -9 =51
Monthly fee of one boy =250 x 134/100 =Rs 335
Total monthly fees of boys =51 x 335 =Rs 17,085
Sum =17085 + 2250 =Rs 19,335.
Question 9
4 / -1

If x is a real number then the minimum value of (x² - x +1) is :

A
3/4

B
0

C
1

D
1/4

Solution

Minimum value for \(x^{2}-x+1\) \(x^{2}-2\left(\frac{1}{2}\right) x+\frac{1}{4}-\frac{1}{4}+1\)
\(\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}\)
For minimum value of \(\left(x-\frac{1}{2}\right)^{2}\) should be zero because it cant be negative
so minimum value \(=\frac{3}{4}\)
Question 10
4 / -1

A cube with volume 1000 cm³. A new maximum cone is formed by this cube, then the volume of the cone will be (in cm³) :

A
280

B
2619

C
2691

D
2961

Solution

Volume of cube- (side) \(^{3}\) Side of cube(a)=10 cm
If the biggest cone is made from cube radius of cone \(r=\frac{1}{2} a=\frac{10}{2}=5\)
and height of cone \(h=a\) Volume of cone \(-\frac{1}{3} \pi r^{2} h\) \(=\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 10\)
\(\Rightarrow \frac{22 \times 260}{21}=261.91\)

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Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 7

Result

Quantitative Ability Test - 7

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SHARING IS CARING

Question 1

3q

2q

q

4q

Solution

Question 2

1 : 2 : 3

a2c : ab2 : bc2

ab : bc : ca

a : b : c

Solution

Question 3

75 km/hr

60 km/hr

48 km/hr

Cannot be determined

Solution

Question 4

1/3√86

2/3√86

8/3√15

None

Solution

Question 5

1/3

1/4

1/2

1/8

Solution

Question 6

\(\sqrt[4]{10}, \sqrt[3]{6}, \sqrt{3}\)

\(\sqrt{3}, \sqrt[4]{10}, \sqrt[3]{6}\)

\(\sqrt{3}, 6, \sqrt[1]{10}\)

\(\sqrt[4]{10}, \sqrt{3}, \sqrt[3]{6}\)

Solution

Question 7

28 : 41

41 : 28

15 : 41

41 : 15

Solution

Question 8

Rs. 19,335

Rs. 18,435

Rs. 19,345

Rs. 19,435

Solution

Question 9

3/4

0

1

1/4

Solution

Question 10

280

2619

2691

2961

Solution

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a²c : ab² : bc²