Practice Model question set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Read the following information carefully to answer the questions that follow. The arithmetic mean, geometric mean and median of 6 positive numbers a, a, b, b, c, c, where a < b < c are $7/3$, 2, 2, respectively.What is the value of c?
(a)
(b)
(c)
(d)
a < b < c
Total numbers = 6
Increasing order a, a, b, b, c, c
∴ Median = ${\text"(6/2)th term + (6/2 + 1)th term"}/2$
= ${\text"3rd term + 4th term"}/2$
2 = ${b + b}/2$ = b
Arithmetic mean = ${a + a + b + b + c + c}/6$
⇒ $7/3 = {a + b + c}/3$
⇒ a + b + c = 7
⇒ a + c = 7 – 2 = 5 ... (i)
Geometric mean = $(a^2 × b^2 × c^2)^{1/6}$
⇒ 2 = $(abc)^{1/3}$
⇒ abc = 8
⇒ ac = $8/2$ = 4 ... (ii)
⇒ c = $4/a$
From equation (i),
a + $4/a$ = 5
⇒ ${a^2 + 4}/a$ = 5
⇒ $a^2$ – 5a + 4 = 0
⇒ $a^2$ – 4a – a + 4 = 0
⇒ a(a – 4) – 1(a – 4) = 0
⇒ (a – 4) (a – 1) = 0
if a = 1 then c = 4
a = 4 then c = 1
a = 1, c = 4 and b = 2
The value of c is 4.
Q-2) Consider the following data: x 1 2 3 4 5 f 3 5 9 - 2
If the arithmetic mean of the above distribution is 2.96, then what is the missing frequency?
| x | 1 | 2 | 3 | 4 | 5 |
| f | 3 | 5 | 9 | - | 2 |
(a)
(b)
(c)
(d)
| x | f | xf |
| 1 | 3 | 3 |
| 2 | 5 | 10 |
| 3 | 9 | 27 |
| 4 | $f_1$ | $4f_1$ |
| 5 | 2 | 10 |
| Total | 19 + $f_1$ | 50 + $4f_1$ |
∴ Mean = ${Σx_if_i}/{Σ_i}$
⇒ 2.96 = ${50 + 4f_1}/{19 + f_1}$ [given]
⇒ 56.24 + 2.96 $f_1 = 50 + 4f_1$
⇒ 6.24 = 1.04 $f_1$
⇒ $f_1$ = 6
Q-3) Read the following information carefully to answer the questions that follow. The arithmetic mean, geometric mean and median of 6 positive numbers a, a, b, b, c, c, where a < b < c are $7/3$, 2, 2, respectively.What is the mode?
(a)
(b)
(c)
(d)
a < b < c
Total numbers = 6
Increasing order a, a, b, b, c, c
∴ Median = ${\text"(6/2)th term + (6/2 + 1)th term"}/2$
= ${\text"3rd term + 4th term"}/2$
2 = ${b + b}/2$ = b
Arithmetic mean = ${a + a + b + b + c + c}/6$
⇒ $7/3 = {a + b + c}/3$
⇒ a + b + c = 7
⇒ a + c = 7 – 2 = 5 ... (i)
Geometric mean = $(a^2 × b^2 × c^2)^{1/6}$
⇒ 2 = $(abc)^{1/3}$
⇒ abc = 8
⇒ ac = $8/2$ = 4 ... (ii)
⇒ c = $4/a$
From equation (i),
a + $4/a$ = 5
⇒ ${a^2 + 4}/a$ = 5
⇒ $a^2$ – 5a + 4 = 0
⇒ $a^2$ – 4a – a + 4 = 0
⇒ a(a – 4) – 1(a – 4) = 0
⇒ (a – 4) (a – 1) = 0
if a = 1 then c = 4
a = 4 then c = 1
a = 1, c = 4 and b = 2
Mode = 3 (Median) – 2 (Mean)
= $3(2) - 2(7/3) = {18 - 14}/3 = 4/3$
Q-4) The following table gives the frequency distribution of life length in hours of 100 electric bulbs having median life 20 h.Life of bulbs(in hours) Number of bulbs 8-13 7 13-18 x 18-23 40 23-28 y 28-33 10 33-38 2
What is the missing frequency 'y'?
| Life of bulbs(in hours) | Number of bulbs |
| 8-13 | 7 |
| 13-18 | x |
| 18-23 | 40 |
| 23-28 | y |
| 28-33 | 10 |
| 33-38 | 2 |
(a)
(b)
(c)
(d)
Number of total bulbs = 100
∴ 7 + x + 40 + y + 10 + 2 = 100
⇒ x + y = 41 ... (i)
| Life of bulbs (in hours) | Number of bulbs | Cumulative Frequency |
| 8 - 13 | 7 | 7 |
| 13 - 18 | x | 7 + x |
| 18 - 23 | 40 | 47 + x |
| 23 - 28 | y | 47 + x + y |
| 28 - 33 | 10 | 57 + x + y |
| 33 - 38 | 2 | 59 + x + y |
| N = 100 |
The median life is 20 h, so median interval will be (18-23).
Here, l = 18, $N/2$ = 50
c = 7 + x, f = 40, h = 5
∴ Median = l + ${(N/2 - C)}/f$ × h
⇒ 20 = 18 + ${(50 – 7 – x)}/{40}$ × 5
⇒ 2 = ${50 – 7 – x}/8$
⇒ 16 = 50 – 7 – x
⇒ x = 43 – 16
⇒ x = 27
Missing frequency 'y' is 14.
Q-5) What is the difference between the number of boys studying Mathematics and the number of girls studying Physics?
(a)
(b)
(c)
(d)
Difference in the number of boys studying Mathematics and Physics = 180 – 150 = 30
Q-6) Read the following information and answer the four items that follow .Let the distribution of number of scooters of companies X and Y sold by 5 showrooms (A, B, C, D and E) in a certain year be denoted by S1 and the distribution of number of scooters of only company X sold by the five showrooms in the same year be denoted by S2Showroom A B C D E Total number of scooters sold S1 (in%) 19 21 15 33 12 6400 S2(in%) 24 18 20 30 8 3000
Number of scooters of company Y sold by showroom E is what per cent of the number of scooters of both companies sold by showroom C ?
| Showroom | A | B | C | D | E | Total number of scooters sold |
| S1 (in%) | 19 | 21 | 15 | 33 | 12 | 6400 |
| S2(in%) | 24 | 18 | 20 | 30 | 8 | 3000 |
(a)
(b)
(c)
(d)
No. of scooters of Y sold by E
= 12% × 6400 – 8% × 3600
= 528
No. of total scooters by showroom
c = 15% × 6400 = 960
Reqd. % = ${528}/{960}$ × 100 = 55%
Q-7) Consider the following grouped frequency distribution : x f 0-10 8 10-20 12 20-30 10 30-40 p 40-50 9
If the mean of the above data is 25. 2, then what is the value of p ?
| x | f |
| 0-10 | 8 |
| 10-20 | 12 |
| 20-30 | 10 |
| 30-40 | p |
| 40-50 | 9 |
(a)
(b)
(c)
(d)
Mean = (sum of ∞) / (sum of f) = (5*5 + 12*15 + 10*25 + p*35 + 9*45) / (8 + 12 + 10 + P + 9) = 25.2 (875 + 35P) / (39 + P) = 25.2 ⇒ P = 11
Q-8) What is the mode of the frequency distribution of SeriesII?
(a)
(b)
(c)
(d)
Mode of frequency distribution of series II is 46.
Q-9) What is the median of the data 3, 5, 9, 4, 6, 11, 18?
(a)
(b)
(c)
(d)
Arrange the number ascending order
3, 4, 5, 6, 9, 11, 18
Therefore $4^{th}$ term out of 7 is Median.
∴ Median = 6
Q-10) The arithmetic mean of 10 numbers was computed as 7.6. It was later discovered that a number 8 was wrongly read as 3 during the computation. What should be the correct mean?
(a)
(b)
(c)
(d)
Correct A.M. = ${n \ov{x} - \text"(Sum of wrong observations) + (Sum of correct observations)"}/n$
= ${10 × 7.6 - 3 + 8}/{10} = {81}/{10}$ = 8.1