Quantitative Aptitude Test

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Quantitative Aptitude Test - 14

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Question 1
1 / -0

Directions For Questions

Direction: What should come in place of the question mark (?) in the following number series?

...view full instructions

1, 5, 14, 46, 188, ?

A
944

B
756

C
940

D
752

E
None of these

Solution

The pattern of the number series is-

⇒ 1 × 1 + 4 = 5

⇒ 5 × 2 + 4 = 14

⇒ 14 × 3 + 4 = 46

⇒ 46 × 4 + 4 = 188

⇒ 188 × 5 + 4 = 944

∴ The value of ? is 944.
Question 2
1 / -0

Direction: In question below, find out the missing element in the series.

$\mathrm{B}_{2} \mathrm{CD}, \mathrm{C}_{4}, \mathrm{BCD}_{4}, \mathrm{~B}_{5} \mathrm{CD}, \mathrm{BC}_{6} \mathrm{D}$

A
$\mathrm{B}_{2} \mathrm{CD}$

B
$\mathrm{BCD}_{2}$

C
$\mathrm{BC}_{3} \mathrm{D}$

D
$\mathrm{BCD}_{7}$

E
None of the above

Solution

The pattern ‘BCD’ is written in every elements. Just the difference being that consecutive numbers have been written in subscript starting from 2. First, it’s written as a subscript of B, then C and then D. After that it’s written again as subscript of B and the similar cycle is continued as shown below:

Hence the missing term is BC₃D.
Question 3
1 / -0

A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.

CDDP DEER EFFT FGGV GHHX ?

A
HJJY

B
ZIIH

C
HIIZ

D
HIJZ

E
None of the above

Solution

Hence, “HIIZ” is missing term.
Question 4
1 / -0

Two cyclists started toward each other from two points A and B 144 km apart. First cyclist ride at speed of 8 km/hr but Second cyclist first covered 4 km in $1^{ {st }}$ hour, 5 km in $2^{\text {nd }}$ hour and 6 km in $3^{\text {rd }}$ hour and so on, So First cyclist will meet the second at what distance?

A
80 km

B
70 km

C
72 km

D
144 km

E

None of these

Solution

From the question, we get Arithmetic series of speed in each hour for B.

Arithmetic series of speed of B in every hour $=4,5,6, \ldots \ldots$

The first term of series $=\mathrm{a}=4$

Common difference $=\mathrm{d}=5-1=1$

Let, A and B meet after $\mathrm x$ hours.

Therefore, distance travelled by A in $\mathrm x$ hours is equal to the sum $\mathrm x$ terms of arithmetic series of B.

Total distance travelled by A in $\mathrm{x}$ hours $=$ Total distance travelled by B in $\mathrm{x}$ hours or sum of $\mathrm{x}$ terms of arithmetic series of B

$\Rightarrow 8 \times \mathrm x=\frac{\mathrm {x}}{2}(2 \times a+(\mathrm x-1) \mathrm d)$

$\Rightarrow 16=8+\mathrm x-1$

$\Rightarrow \mathrm x=9$

Therefore, A and B meet after $\mathrm 9$ hours.

So, at the distance, A and B meet together $=$ distance travelled by A in $\mathrm 9$ hours

At the distance, A and B meet together $=8 \times 9=72$ km
Question 5
1 / -0

The sum of 2 digits of a number is 15 and the difference between the digit is 3 then find the number.

A
58

B
69

C
72

D
36

E
63

Solution

Let the digits of the two-digit number be x and y.

So, the number can be xy.

From the question, we know

x+y = 15 …….(i)

and x-y = 3 …….(ii)

By adding equation (i) and equation (ii), we get

2x = 18

∴ x = $\frac{18}{2}$ = 9

x = 9

Putting the value of x in equation (i)

x+y = 15

9+y = 15

So, y = 15-9 = 6

y = 6

Therefore, x = 9 & y = 6

∴ The required number is 69.
Question 6
1 / -0

Sabu, Pinki and Billu got Rs. 1600 for completing a work in 20 days. Sabu worked only 4 days and received Rs. 320 and Pinki and Billu worked all 20 days. If only Pinki and Billu were assigned to the same work for the same money, then calculate the difference in Rs/day spent by the payer in both cases.

A
16

B
20

C
32

D
34

E
28

Solution

We know that the wages are distribute in the proportion of contribution to the total work done.

∴ Contribution of Sabu to total work = $\frac{\text{Wages of sabu}}{\text{Total wages distributed}}$ × Total work

⇒ Contribution of Sabu to total work $\frac{320}{1600}$ × 1 = $\frac{1}{5}$

∵ Sabu did $\frac{1}{5}$ part of total work in 4 days,

∴ Part of work done by Sabu in one day = $\frac{1}{5 \times 4} = \frac{1}{20}$

Let us assume Pinki and Billu can together complete the work in T days

Then part of work done by Pinki and Billu in one day = $\frac{1}{T}$

Also part of work done by Sabu, Pinki and Billu in one day = $\frac{1}{T} + \frac{1}{20}$

Total work done in 20 days = (part of work done by Sabu, Pinki and Billu in one day) × 4 + (part of work done by Pinki and Billu in one day) × 16

$\Rightarrow 1=\left(\frac{1}{T}+\frac{1}{20}\right) \times 4+\frac{1}{T} \times 16$

$\Rightarrow 1=\frac{4}{T}+\frac{4}{20}+\frac{16}{\mathrm{~T}}$

$\Rightarrow 1-\frac{4}{20}=\frac{4+16}{T}$

$\Rightarrow \frac{20-4}{20 \times 20}=\frac{1}{T}$

$\Rightarrow \frac{1}{T}=\frac{1}{25}$

So Billu and Pinki can complete the work in 25 days

Now calculating Rs/day spent by payer in both cases

Case I: when Sabu worked for 4 days out of 20 days.

Rs/day spent by payer = $\frac{1600}{20}$ = Rs 80

Case II: when only Pinki and Billu worked for 25 days

Rs/day spent by payer = $\frac{1600}{25}$ = Rs 64

Difference in Rs/day spent by payer in both the cases = 80 - 64 = Rs. 16
Question 7
1 / -0

The ratio of the ages of Tina and Jatin is 23 : p. Tina is 10 years younger to Rahul. After 16 years, Rahul will be 72 years old. If the difference between the ages of Jatin and Tina is equal to the age of Rahul 4 years ago, find the value of p.

A
45

B
47

C
42

D
49

E
52

Solution

Given:

Let the ratio between the ages of Tina and Jatin be 23x : px, where x is a common variable.

Present age of Rahul = 72 - 16 = 56 years

Present age of Tina = 56 - 10 = 46 years

By equating the age of Tina with the ratio of the age of Tina and Jatin: 23x = 46 ⇒ x = 2

According to the question: 2p - 46 = 56 - 4

2p = 98

∴ p = 49
Question 8
1 / -0

Mayank threw two dice simultaneously. What is the probability of getting a total of 7?

A
$\frac{1}{6}$

B
$\frac{1}{4}$

C
$\frac{2}{3}$

D
$\frac{3}{4}$

E
None of these

Solution

On throwing two dice, n(S) = 6 × 6 = 36

Number of events of getting a total of 7, n(E) = 6, i.e., (1,6), (2,5) , (3,4), (4,3) , (5,2), (6,1)

∴ Probability of getting a total of 7 = $\frac{n (E)}{n (S)}$

= $(\frac{6}{36})$ = $(\frac{1}{6})$
Question 9
1 / -0

Directions For Questions

Direction: Read the following passage and answer the question given below.

There are two localities in Lucknow – Gomti Nagar and Indira Nagar. The total population of Indira Nagar is 400 more than the total population of Gomti Nagar. 30% of total population of Gomti Nagar is engineers and 60% of total population of Indira Nagar is engineers. The ratio of male to female engineers in Gomti Nagar and Indira Nagar is 2 : 3 and 7 : 5 respectively and the total population of Gomti Nagar is 800.

...view full instructions

What is the product of average of total male engineers from both localities and the average of female engineers from both localities?

A
57276

B
54256

C
54884

D
53678

E
54404

Solution

Average of male engineers = $\frac{(96 + 420)}{2}$ = 258

Average of female engineers = $\frac{(144 + 300)}{2}$ = 222

Required product = 258 × 222 = 57276

∴ The product of average of total male engineers from both localities and the average of female engineers from both localities is 57276.
Question 10
1 / -0

Directions For Questions

Direction: Read the following passage and answer the question given below.

There are two localities in Lucknow – Gomti Nagar and Indira Nagar. The total population of Indira Nagar is 400 more than the total population of Gomti Nagar. 30% of total population of Gomti Nagar is engineers and 60% of total population of Indira Nagar is engineers. The ratio of male to female engineers in Gomti Nagar and Indira Nagar is 2 : 3 and 7 : 5 respectively and the total population of Gomti Nagar is 800.

...view full instructions

50% of the number of male engineers in Gomti Nagar are what percent of the 150% of the number of female engineers in Indira Nagar?

A
13.25%

B
10.67%

C
8.33%

D
6.32%

E
12.56%

Solution

50% of the number of male engineers in Gomti Nagar = 96 × $\frac{50}{100}$ = 48

150% of the number of female engineers in Indira Nagar = 300 × $\frac{150}{100}$ = 450

Now required percentage = $\frac{48}{450}$ × 100 = 10.67%

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Quantitative Aptitude Test - 14

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Quantitative Aptitude Test - 14

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SHARING IS CARING

Question 1

944

756

940

752

None of these

Solution

Question 2

\(\mathrm{B}_{2} \mathrm{CD}\)

\(\mathrm{BCD}_{2}\)

\(\mathrm{BC}_{3} \mathrm{D}\)

\(\mathrm{BCD}_{7}\)

None of the above

Solution

Question 3

HJJY

ZIIH

HIIZ

HIJZ

None of the above

Solution

Question 4

80 km

70 km

72 km

144 km

None of these

Solution

Question 5

58

69

72

36

63

Solution

Question 6

16

20

32

34

28

Solution

Question 7

45

47

42

49

52

Solution

Question 8

16

14

23

34

None of these

Solution

Question 9

57276

54256

54884

53678

54404

Solution

Question 10

13.25%

10.67%

8.33%

6.32%

12.56%

Solution

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Question Analysis

$\frac{1}{6}$

$\frac{1}{4}$

$\frac{2}{3}$

$\frac{3}{4}$